Authors Thomas W. Judson
License GFDL-1.2-no-invariants-or-later
Abstract Algebra
Theory and Applications
Abstract Algebra
Theory and Applications
Thomas W. Judson
Stephen F. Austin State University
Sage Exercises for Abstract Algebra
Robert A. Beezer
University of Puget Sound
Traducción al español
Antonio Behn
Universidad de Chile
July 28, 2022
Edition: Annual Edition 2022
Website: abstract.pugetsound.edu
©1997–2022 Thomas W. Judson, Robert A. Beezer
Permission is granted to copy, distribute and/or modify this document under the terms
of the GNU Free Documentation License, Version 1.2 or any later version published by
the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no
Back-Cover Texts. A copy of the license is included in the appendix entitled “GNU Free
Documentation License.”
Acknowledgements
I would like to acknowledge the following reviewers for their helpful comments and sugges-
tions.
• David Anderson, University of Tennessee, Knoxville
• Robert Beezer, University of Puget Sound
• Myron Hood, California Polytechnic State University
• Herbert Kasube, Bradley University
• John Kurtzke, University of Portland
• Inessa Levi, University of Louisville
• Geoffrey Mason, University of California, Santa Cruz
• Bruce Mericle, Mankato State University
• Kimmo Rosenthal, Union College
• Mark Teply, University of Wisconsin
I would also like to thank Steve Quigley, Marnie Pommett, Cathie Griffin, Kelle Karshick,
and the rest of the staff at PWS Publishing for their guidance throughout this project. It
has been a pleasure to work with them.
Robert Beezer encouraged me to make Abstract Algebra: Theory and Applications avail-
able as an open source textbook, a decision that I have never regretted. With his assistance,
the book has been rewritten in PreTeXt (pretextbook.org1 ), making it possible to quickly
output print, web, pdf versions and more from the same source. The open source version
of this book has received support from the National Science Foundation (Awards #DUE-
1020957, #DUE–1625223, and #DUE–1821329).
1
pretextbook.org
iv
Preface
This text is intended for a one or two-semester undergraduate course in abstract algebra.
Traditionally, these courses have covered the theoretical aspects of groups, rings, and fields.
However, with the development of computing in the last several decades, applications that
involve abstract algebra and discrete mathematics have become increasingly important,
and many science, engineering, and computer science students are now electing to minor in
mathematics. Though theory still occupies a central role in the subject of abstract algebra
and no student should go through such a course without a good notion of what a proof is, the
importance of applications such as coding theory and cryptography has grown significantly.
Until recently most abstract algebra texts included few if any applications. However,
one of the major problems in teaching an abstract algebra course is that for many students it
is their first encounter with an environment that requires them to do rigorous proofs. Such
students often find it hard to see the use of learning to prove theorems and propositions;
applied examples help the instructor provide motivation.
This text contains more material than can possibly be covered in a single semester.
Certainly there is adequate material for a two-semester course, and perhaps more; however,
for a one-semester course it would be quite easy to omit selected chapters and still have a
useful text. The order of presentation of topics is standard: groups, then rings, and finally
fields. Emphasis can be placed either on theory or on applications. A typical one-semester
course might cover groups and rings while briefly touching on field theory, using Chapters 1
through 6, 9, 10, 11, 13 (the first part), 16, 17, 18 (the first part), 20, and 21. Parts of
these chapters could be deleted and applications substituted according to the interests of
the students and the instructor. A two-semester course emphasizing theory might cover
Chapters 1 through 6, 9, 10, 11, 13 through 18, 20, 21, 22 (the first part), and 23. On
the other hand, if applications are to be emphasized, the course might cover Chapters 1
through 14, and 16 through 22. In an applied course, some of the more theoretical results
could be assumed or omitted. A chapter dependency chart appears below. (A broken line
indicates a partial dependency.)
v
vi
Chapters 1–6
Chapter 8 Chapter 9 Chapter 7
Chapter 10
Chapter 11
Chapter 13 Chapter 16 Chapter 12 Chapter 14
Chapter 17 Chapter 15
Chapter 18 Chapter 20 Chapter 19
Chapter 21
Chapter 22
Chapter 23
Though there are no specific prerequisites for a course in abstract algebra, students
who have had other higher-level courses in mathematics will generally be more prepared
than those who have not, because they will possess a bit more mathematical sophistication.
Occasionally, we shall assume some basic linear algebra; that is, we shall take for granted an
elementary knowledge of matrices and determinants. This should present no great problem,
since most students taking a course in abstract algebra have been introduced to matrices
and determinants elsewhere in their career, if they have not already taken a sophomore or
junior-level course in linear algebra.
Exercise sections are the heart of any mathematics text. An exercise set appears at the
end of each chapter. The nature of the exercises ranges over several categories; computa-
tional, conceptual, and theoretical problems are included. A section presenting hints and
solutions to many of the exercises appears at the end of the text. Often in the solutions
a proof is only sketched, and it is up to the student to provide the details. The exercises
range in difficulty from very easy to very challenging. Many of the more substantial prob-
lems require careful thought, so the student should not be discouraged if the solution is not
forthcoming after a few minutes of work.
Ideally, students should read the relavent material before attending class. Reading ques-
tions have been added to each chapter before the exercises. To prepare for class, students
vii
should read the chapter before class and then answer the section’s reading questions to
prepare for the class.
There are additional exercises or computer projects at the ends of many of the chapters.
The computer projects usually require a knowledge of programming. All of these exercises
and projects are more substantial in nature and allow the exploration of new results and
theory.
Sage (sagemath.org2 ) is a free, open source, software system for advanced mathematics,
which is ideal for assisting with a study of abstract algebra. Sage can be used either on
your own computer, a local server, or on CoCalc (cocalc.com3 ). Robert Beezer has written
a comprehensive introduction to Sage and a selection of relevant exercises that appear at
the end of each chapter, including live Sage cells in the web version of the book. All of the
Sage code has been subject to automated tests of accuracy, using the most recent version
available at this time: SageMath Version 9.6 (released 2022-05-15).
Thomas W. Judson
Nacogdoches, Texas 2022
2
sagemath.org
3
cocalc.com
Contents
Acknowledgements iv
Preface v
1 Preliminaries 1
1.1 A Short Note on Proofs . . . . . . . . . . . . . . . . . . . . . 1
1.2 Sets and Equivalence Relations . . . . . . . . . . . . . . . . . . 3
1.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 13
1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5 References and Suggested Readings . . . . . . . . . . . . . . . . . 16
2 The Integers 17
2.1 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . 17
2.2 The Division Algorithm . . . . . . . . . . . . . . . . . . . . . 20
2.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 24
2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.5 Programming Exercises . . . . . . . . . . . . . . . . . . . . . 26
2.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 26
3 Groups 28
3.1 Integer Equivalence Classes and Symmetries . . . . . . . . . . . . . 28
3.2 Definitions and Examples. . . . . . . . . . . . . . . . . . . . . 33
3.3 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 40
3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.6 Additional Exercises: Detecting Errors . . . . . . . . . . . . . . . 43
3.7 References and Suggested Readings . . . . . . . . . . . . . . . . . 45
4 Cyclic Groups 46
4.1 Cyclic Subgroups . . . . . . . . . . . . . . . . . . . . . . . . 46
4.2 Multiplicative Group of Complex Numbers . . . . . . . . . . . . . . 49
4.3 The Method of Repeated Squares . . . . . . . . . . . . . . . . . 53
viii
CONTENTS ix
4.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 55
4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.6 Programming Exercises . . . . . . . . . . . . . . . . . . . . . 58
4.7 References and Suggested Readings . . . . . . . . . . . . . . . . . 58
5 Permutation Groups 59
5.1 Definitions and Notation . . . . . . . . . . . . . . . . . . . . . 59
5.2 Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . . 65
5.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 70
5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
6 Cosets and Lagrange’s Theorem 74
6.1 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
6.2 Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 76
6.3 Fermat’s and Euler’s Theorems . . . . . . . . . . . . . . . . . . 77
6.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 78
6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
7 Introduction to Cryptography 81
7.1 Private Key Cryptography . . . . . . . . . . . . . . . . . . . . 81
7.2 Public Key Cryptography . . . . . . . . . . . . . . . . . . . . 83
7.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 86
7.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
7.5 Additional Exercises: Primality and Factoring . . . . . . . . . . . . 88
7.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 89
8 Algebraic Coding Theory 91
8.1 Error-Detecting and Correcting Codes . . . . . . . . . . . . . . . 91
8.2 Linear Codes. . . . . . . . . . . . . . . . . . . . . . . . . . 98
8.3 Parity-Check and Generator Matrices . . . . . . . . . . . . . . . . 101
8.4 Efficient Decoding . . . . . . . . . . . . . . . . . . . . . . . 106
8.5 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 109
8.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
8.7 Programming Exercises . . . . . . . . . . . . . . . . . . . . . 113
8.8 References and Suggested Readings . . . . . . . . . . . . . . . . . 113
9 Isomorphisms 114
9.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . 114
9.2 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . 118
9.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 121
9.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
10 Normal Subgroups and Factor Groups 125
10.1 Factor Groups and Normal Subgroups. . . . . . . . . . . . . . . . 125
10.2 The Simplicity of the Alternating Group. . . . . . . . . . . . . . . 127
10.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 130
10.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
CONTENTS x
11 Homomorphisms 133
11.1 Group Homomorphisms . . . . . . . . . . . . . . . . . . . . . 133
11.2 The Isomorphism Theorems. . . . . . . . . . . . . . . . . . . . 135
11.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 138
11.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
11.5 Additional Exercises: Automorphisms . . . . . . . . . . . . . . . . 139
12 Matrix Groups and Symmetry 141
12.1 Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . . 141
12.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
12.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 154
12.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
12.5 References and Suggested Readings . . . . . . . . . . . . . . . . . 157
13 The Structure of Groups 158
13.1 Finite Abelian Groups . . . . . . . . . . . . . . . . . . . . . . 158
13.2 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . 162
13.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 165
13.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
13.5 Programming Exercises . . . . . . . . . . . . . . . . . . . . . 167
13.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 167
14 Group Actions 168
14.1 Groups Acting on Sets . . . . . . . . . . . . . . . . . . . . . . 168
14.2 The Class Equation . . . . . . . . . . . . . . . . . . . . . . . 170
14.3 Burnside’s Counting Theorem . . . . . . . . . . . . . . . . . . . 172
14.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 178
14.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
14.6 Programming Exercise . . . . . . . . . . . . . . . . . . . . . . 180
14.7 References and Suggested Reading . . . . . . . . . . . . . . . . . 180
15 The Sylow Theorems 181
15.1 The Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . 181
15.2 Examples and Applications . . . . . . . . . . . . . . . . . . . . 184
15.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 187
15.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
15.5 A Project . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
15.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 189
16 Rings 190
16.1 Rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
16.2 Integral Domains and Fields . . . . . . . . . . . . . . . . . . . 193
16.3 Ring Homomorphisms and Ideals. . . . . . . . . . . . . . . . . . 195
16.4 Maximal and Prime Ideals . . . . . . . . . . . . . . . . . . . . 198
16.5 An Application to Software Design . . . . . . . . . . . . . . . . . 200
16.6 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 203
16.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
CONTENTS xi
16.8 Programming Exercise . . . . . . . . . . . . . . . . . . . . . . 207
16.9 References and Suggested Readings . . . . . . . . . . . . . . . . . 207
17 Polynomials 209
17.1 Polynomial Rings . . . . . . . . . . . . . . . . . . . . . . . . 209
17.2 The Division Algorithm . . . . . . . . . . . . . . . . . . . . . 212
17.3 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . 215
17.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 220
17.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
17.6 Additional Exercises: Solving the Cubic and Quartic Equations . . . . . 222
18 Integral Domains 225
18.1 Fields of Fractions . . . . . . . . . . . . . . . . . . . . . . . 225
18.2 Factorization in Integral Domains . . . . . . . . . . . . . . . . . 228
18.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 235
18.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
18.5 References and Suggested Readings . . . . . . . . . . . . . . . . . 237
19 Lattices and Boolean Algebras 238
19.1 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
19.2 Boolean Algebras . . . . . . . . . . . . . . . . . . . . . . . . 241
19.3 The Algebra of Electrical Circuits . . . . . . . . . . . . . . . . . 246
19.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 248
19.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
19.6 Programming Exercises . . . . . . . . . . . . . . . . . . . . . 251
19.7 References and Suggested Readings . . . . . . . . . . . . . . . . . 251
20 Vector Spaces 252
20.1 Definitions and Examples. . . . . . . . . . . . . . . . . . . . . 252
20.2 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
20.3 Linear Independence . . . . . . . . . . . . . . . . . . . . . . 254
20.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 256
20.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
20.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 259
21 Fields 260
21.1 Extension Fields . . . . . . . . . . . . . . . . . . . . . . . . 260
21.2 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . 268
21.3 Geometric Constructions . . . . . . . . . . . . . . . . . . . . . 270
21.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 275
21.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
21.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 277
22 Finite Fields 278
22.1 Structure of a Finite Field . . . . . . . . . . . . . . . . . . . . 278
22.2 Polynomial Codes. . . . . . . . . . . . . . . . . . . . . . . . 282
22.3 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 289
CONTENTS xii
22.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
22.5 Additional Exercises: Error Correction for BCH Codes . . . . . . . . . 291
22.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 291
23 Galois Theory 293
23.1 Field Automorphisms . . . . . . . . . . . . . . . . . . . . . . 293
23.2 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . 297
23.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 303
23.4 Reading Questions . . . . . . . . . . . . . . . . . . . . . . . 307
23.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 308
23.6 References and Suggested Readings . . . . . . . . . . . . . . . . . 309
Appendices
A GNU Free Documentation License 311
B Hints and Answers to Selected Exercises 318
C Notation 336
Back Matter
Index 339
1
Preliminaries
A certain amount of mathematical maturity is necessary to find and study applications
of abstract algebra. A basic knowledge of set theory, mathematical induction, equivalence
relations, and matrices is a must. Even more important is the ability to read and understand
mathematical proofs. In this chapter we will outline the background needed for a course in
abstract algebra.
1.1 A Short Note on Proofs
Abstract mathematics is different from other sciences. In laboratory sciences such as chem-
istry and physics, scientists perform experiments to discover new principles and verify theo-
ries. Although mathematics is often motivated by physical experimentation or by computer
simulations, it is made rigorous through the use of logical arguments. In studying abstract
mathematics, we take what is called an axiomatic approach; that is, we take a collection
of objects S and assume some rules about their structure. These rules are called axioms.
Using the axioms for S, we wish to derive other information about S by using logical argu-
ments. We require that our axioms be consistent; that is, they should not contradict one
another. We also demand that there not be too many axioms. If a system of axioms is too
restrictive, there will be few examples of the mathematical structure.
A statement in logic or mathematics is an assertion that is either true or false. Consider
the following examples:
• 3 + 56 − 13 + 8/2.
• All cats are black.
• 2 + 3 = 5.
• 2x = 6 exactly when x = 4.
• If ax2 + bx + c = 0 and a 6= 0, then
√
−b ± b2 − 4ac
x= .
2a
• x3 − 4x2 + 5x − 6.
All but the first and last examples are statements, and must be either true or false.
A mathematical proof is nothing more than a convincing argument about the accuracy
of a statement. Such an argument should contain enough detail to convince the audience; for
1
CHAPTER 1. PRELIMINARIES 2
instance, we can see that the statement “2x = 6 exactly when x = 4” is false by evaluating
2 · 4 and noting that 6 6= 8, an argument that would satisfy anyone. Of course, audiences
may vary widely: proofs can be addressed to another student, to a professor, or to the
reader of a text. If more detail than needed is presented in the proof, then the explanation
will be either long-winded or poorly written. If too much detail is omitted, then the proof
may not be convincing. Again it is important to keep the audience in mind. High school
students require much more detail than do graduate students. A good rule of thumb for an
argument in an introductory abstract algebra course is that it should be written to convince
one’s peers, whether those peers be other students or other readers of the text.
Let us examine different types of statements. A statement could be as simple as “10/5 =
2;” however, mathematicians are usually interested in more complex statements such as “If
p, then q,” where p and q are both statements. If certain statements are known or assumed
to be true, we wish to know what we can say about other statements. Here p is called
the hypothesis and q is known as the conclusion. Consider the following statement: If
ax2 + bx + c = 0 and a 6= 0, then
√
−b ± b2 − 4ac
x= .
2a
The hypothesis is ax2 + bx + c = 0 and a 6= 0; the conclusion is
√
−b ± b2 − 4ac
x= .
2a
Notice that the statement says nothing about whether or not the hypothesis is true. How-
ever, if this entire statement is true and we can show that ax2 + bx + c = 0 with a 6= 0 is
true, then the conclusion must be true. A proof of this statement might simply be a series
of equations:
ax2 + bx + c = 0
b c
x2 + x = −
a a
2 2
b b b c
x2 + x + = −
a 2a 2a a
2
b b − 4ac
2
x+ =
2a 4a2
√
b ± b2 − 4ac
x+ =
2a 2a
√
−b ± b2 − 4ac
x= .
2a
If we can prove a statement true, then that statement is called a proposition. A propo-
sition of major importance is called a theorem. Sometimes instead of proving a theorem
or proposition all at once, we break the proof down into modules; that is, we prove several
supporting propositions, which are called lemmas, and use the results of these propositions
to prove the main result. If we can prove a proposition or a theorem, we will often, with
very little effort, be able to derive other related propositions called corollaries.
Some Cautions and Suggestions
There are several different strategies for proving propositions. In addition to using different
methods of proof, students often make some common mistakes when they are first learning
CHAPTER 1. PRELIMINARIES 3
how to prove theorems. To aid students who are studying abstract mathematics for the
first time, we list here some of the difficulties that they may encounter and some of the
strategies of proof available to them. It is a good idea to keep referring back to this list as
a reminder. (Other techniques of proof will become apparent throughout this chapter and
the remainder of the text.)
• A theorem cannot be proved by example; however, the standard way to show that a
statement is not a theorem is to provide a counterexample.
• Quantifiers are important. Words and phrases such as only, for all, for every, and for
some possess different meanings.
• Never assume any hypothesis that is not explicitly stated in the theorem. You cannot
take things for granted.
• Suppose you wish to show that an object exists and is unique. First show that there
actually is such an object. To show that it is unique, assume that there are two such
objects, say r and s, and then show that r = s.
• Sometimes it is easier to prove the contrapositive of a statement. Proving the state-
ment “If p, then q” is exactly the same as proving the statement “If not q, then not
p.”
• Although it is usually better to find a direct proof of a theorem, this task can some-
times be difficult. It may be easier to assume that the theorem that you are trying
to prove is false, and to hope that in the course of your argument you are forced to
make some statement that cannot possibly be true.
Remember that one of the main objectives of higher mathematics is proving theorems.
Theorems are tools that make new and productive applications of mathematics possible. We
use examples to give insight into existing theorems and to foster intuitions as to what new
theorems might be true. Applications, examples, and proofs are tightly interconnected—
much more so than they may seem at first appearance.
1.2 Sets and Equivalence Relations
Set Theory
A set is a well-defined collection of objects; that is, it is defined in such a manner that we
can determine for any given object x whether or not x belongs to the set. The objects that
belong to a set are called its elements or members. We will denote sets by capital letters,
such as A or X; if a is an element of the set A, we write a ∈ A.
A set is usually specified either by listing all of its elements inside a pair of braces or
by stating the property that determines whether or not an object x belongs to the set. We
might write
X = {x1 , x2 , . . . , xn }
for a set containing elements x1 , x2 , . . . , xn or
X = {x : x satisfies P}
if each x in X satisfies a certain property P. For example, if E is the set of even positive
integers, we can describe E by writing either
E = {2, 4, 6, . . .} or E = {x : x is an even integer and x > 0}.
CHAPTER 1. PRELIMINARIES 4
We write 2 ∈ E when we want to say that 2 is in the set E, and −3 ∈ / E to say that −3 is
not in the set E.
Some of the more important sets that we will consider are the following:
N = {n : n is a natural number} = {1, 2, 3, . . .};
Z = {n : n is an integer} = {. . . , −1, 0, 1, 2, . . .};
Q = {r : r is a rational number} = {p/q : p, q ∈ Z where q 6= 0};
R = {x : x is a real number};
C = {z : z is a complex number}.
We can find various relations between sets as well as perform operations on sets. A set
A is a subset of B, written A ⊂ B or B ⊃ A, if every element of A is also an element of B.
For example,
{4, 5, 8} ⊂ {2, 3, 4, 5, 6, 7, 8, 9}
and
N ⊂ Z ⊂ Q ⊂ R ⊂ C.
Trivially, every set is a subset of itself. A set B is a proper subset of a set A if B ⊂ A but
B 6= A. If A is not a subset of B, we write A 6⊂ B; for example, {4, 7, 9} 6⊂ {2, 4, 5, 8, 9}.
Two sets are equal, written A = B, if we can show that A ⊂ B and B ⊂ A.
It is convenient to have a set with no elements in it. This set is called the empty set
and is denoted by ∅. Note that the empty set is a subset of every set.
To construct new sets out of old sets, we can perform certain operations: the union
A ∪ B of two sets A and B is defined as
A ∪ B = {x : x ∈ A or x ∈ B};
the intersection of A and B is defined by
A ∩ B = {x : x ∈ A and x ∈ B}.
If A = {1, 3, 5} and B = {1, 2, 3, 9}, then
A ∪ B = {1, 2, 3, 5, 9} and A ∩ B = {1, 3}.
We can consider the union and the intersection of more than two sets. In this case we write
[n
Ai = A1 ∪ . . . ∪ An
i=1
and
\
n
Ai = A1 ∩ . . . ∩ An
i=1
for the union and intersection, respectively, of the sets A1 , . . . , An .
When two sets have no elements in common, they are said to be disjoint; for example,
if E is the set of even integers and O is the set of odd integers, then E and O are disjoint.
Two sets A and B are disjoint exactly when A ∩ B = ∅.
Sometimes we will work within one fixed set U , called the universal set. For any set
A ⊂ U , we define the complement of A, denoted by A′ , to be the set
A′ = {x : x ∈ U and x ∈
/ A}.
We define the difference of two sets A and B to be
A \ B = A ∩ B ′ = {x : x ∈ A and x ∈
/ B}.
CHAPTER 1. PRELIMINARIES 5
Example 1.1 Let R be the universal set and suppose that
A = {x ∈ R : 0 < x ≤ 3} and B = {x ∈ R : 2 ≤ x < 4}.
Then
A ∩ B = {x ∈ R : 2 ≤ x ≤ 3}
A ∪ B = {x ∈ R : 0 < x < 4}
A \ B = {x ∈ R : 0 < x < 2}
A′ = {x ∈ R : x ≤ 0 or x > 3}.
□
Proposition 1.2 Let A, B, and C be sets. Then
1. A ∪ A = A, A ∩ A = A, and A \ A = ∅;
2. A ∪ ∅ = A and A ∩ ∅ = ∅;
3. A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C;
4. A ∪ B = B ∪ A and A ∩ B = B ∩ A;
5. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C);
6. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
Proof. We will prove (1) and (3) and leave the remaining results to be proven in the
exercises.
(1) Observe that
A ∪ A = {x : x ∈ A or x ∈ A}
= {x : x ∈ A}
=A
and
A ∩ A = {x : x ∈ A and x ∈ A}
= {x : x ∈ A}
= A.
Also, A \ A = A ∩ A′ = ∅.
(3) For sets A, B, and C,
A ∪ (B ∪ C) = A ∪ {x : x ∈ B or x ∈ C}
= {x : x ∈ A or x ∈ B, or x ∈ C}
= {x : x ∈ A or x ∈ B} ∪ C
= (A ∪ B) ∪ C.
A similar argument proves that A ∩ (B ∩ C) = (A ∩ B) ∩ C. ■
Theorem 1.3 De Morgan’s Laws. Let A and B be sets. Then
1. (A ∪ B)′ = A′ ∩ B ′ ;
2. (A ∩ B)′ = A′ ∪ B ′ .
CHAPTER 1. PRELIMINARIES 6
Proof. (1) If A ∪ B = ∅, then the theorem follows immediately since both A and B are the
empty set. Otherwise, we must show that (A ∪ B)′ ⊂ A′ ∩ B ′ and (A ∪ B)′ ⊃ A′ ∩ B ′ . Let
x ∈ (A ∪ B)′ . Then x ∈ / A ∪ B. So x is neither in A nor in B, by the definition of the union
of sets. By the definition of the complement, x ∈ A′ and x ∈ B ′ . Therefore, x ∈ A′ ∩ B ′
and we have (A ∪ B)′ ⊂ A′ ∩ B ′ .
To show the reverse inclusion, suppose that x ∈ A′ ∩ B ′ . Then x ∈ A′ and x ∈ B ′ , and
so x ∈
/ A and x ∈/ B. Thus x ∈ / A ∪ B and so x ∈ (A ∪ B)′ . Hence, (A ∪ B)′ ⊃ A′ ∩ B ′ and
so (A ∪ B)′ = A′ ∩ B ′ .
The proof of (2) is left as an exercise. ■
Example 1.4 Other relations between sets often hold true. For example,
(A \ B) ∩ (B \ A) = ∅.
To see that this is true, observe that
(A \ B) ∩ (B \ A) = (A ∩ B ′ ) ∩ (B ∩ A′ )
= A ∩ A′ ∩ B ∩ B ′
= ∅.
□
Cartesian Products and Mappings
Given sets A and B, we can define a new set A × B, called the Cartesian product of A and
B, as a set of ordered pairs. That is,
A × B = {(a, b) : a ∈ A and b ∈ B}.
Example 1.5 If A = {x, y}, B = {1, 2, 3}, and C = ∅, then A × B is the set
{(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}
and
A × C = ∅.
□
We define the Cartesian product of n sets to be
A1 × · · · × An = {(a1 , . . . , an ) : ai ∈ Ai for i = 1, . . . , n}.
If A = A1 = A2 = · · · = An , we often write An for A × · · · × A (where A would be written
n times). For example, the set R3 consists of all of 3-tuples of real numbers.
Subsets of A × B are called relations. We will define a mapping or function f ⊂ A × B
from a set A to a set B to be the special type of relation where each element a ∈ A has
a unique element b ∈ B such that (a, b) ∈ f . Another way of saying this is that for every
f
element in A, f assigns a unique element in B. We usually write f : A → B or A → B.
Instead of writing down ordered pairs (a, b) ∈ A × B, we write f (a) = b or f : a 7→ b. The
set A is called the domain of f and
f (A) = {f (a) : a ∈ A} ⊂ B
is called the range or image of f . We can think of the elements in the function’s domain as
input values and the elements in the function’s range as output values.
CHAPTER 1. PRELIMINARIES 7
Example 1.6 Suppose A = {1, 2, 3} and B = {a, b, c}. In Figure 1.7 we define relations f
and g from A to B. The relation f is a mapping, but g is not because 1 ∈ A is not assigned
to a unique element in B; that is, g(1) = a and g(1) = b.
A B
f
1 a
2 b
3 c
A g B
1 a
2 b
3 c
Figure 1.7 Mappings and relations
□
Given a function f : A → B, it is often possible to write a list describing what the
function does to each specific element in the domain. However, not all functions can be
described in this manner. For example, the function f : R → R that sends each real number
to its cube is a mapping that must be described by writing f (x) = x3 or f : x 7→ x3 .
Consider the relation f : Q → Z given by f (p/q) = p. We know that 1/2 = 2/4, but
is f (1/2) = 1 or 2? This relation cannot be a mapping because it is not well-defined. A
relation is well-defined if each element in the domain is assigned to a unique element in the
range.
If f : A → B is a map and the image of f is B, i.e., f (A) = B, then f is said to
be onto or surjective. In other words, if there exists an a ∈ A for each b ∈ B such that
f (a) = b, then f is onto. A map is one-to-one or injective if a1 6= a2 implies f (a1 ) 6= f (a2 ).
Equivalently, a function is one-to-one if f (a1 ) = f (a2 ) implies a1 = a2 . A map that is both
one-to-one and onto is called bijective.
Example 1.8 Let f : Z → Q be defined by f (n) = n/1. Then f is one-to-one but not onto.
Define g : Q → Z by g(p/q) = p where p/q is a rational number expressed in its lowest
terms with a positive denominator. The function g is onto but not one-to-one. □
Given two functions, we can construct a new function by using the range of the first
function as the domain of the second function. Let f : A → B and g : B → C be mappings.
Define a new map, the composition of f and g from A to C, by (g ◦ f )(x) = g(f (x)).
CHAPTER 1. PRELIMINARIES 8
A B C
f g
1 a X
2 b Y
3 c Z
A C
g◦f
1 X
2 Y
3 Z
Figure 1.9 Composition of maps
Example 1.10 Consider the functions f : A → B and g : B → C that are defined in
Figure 1.9 (top). The composition of these functions, g ◦ f : A → C, is defined in Figure 1.9
(bottom). □
Example 1.11 Let f (x) = x2 and g(x) = 2x + 5. Then
(f ◦ g)(x) = f (g(x)) = (2x + 5)2 = 4x2 + 20x + 25
and
(g ◦ f )(x) = g(f (x)) = 2x2 + 5.
In general, order makes a difference; that is, in most cases f ◦ g 6= g ◦ f . □
√
Example 1.12 Sometimes it is the case that f ◦ g = g ◦ f . Let f (x) = x3 and g(x) = 3 x.
Then √ √
(f ◦ g)(x) = f (g(x)) = f ( 3 x ) = ( 3 x )3 = x
and √
3
(g ◦ f )(x) = g(f (x)) = g(x3 ) = x3 = x.
□
Example 1.13 Given a 2 × 2 matrix
a b
A= ,
c d
CHAPTER 1. PRELIMINARIES 9
we can define a map TA : R2 → R2 by
TA (x, y) = (ax + by, cx + dy)
for (x, y) in R2 . This is actually matrix multiplication; that is,
a b x ax + by
= .
c d y cx + dy
Maps from Rn to Rm given by matrices are called linear maps or linear transformations.
□
Example 1.14 Suppose that S = {1, 2, 3}. Define a map π : S → S by
π(1) = 2, π(2) = 1, π(3) = 3.
This is a bijective map. An alternative way to write π is
1 2 3 1 2 3
= .
π(1) π(2) π(3) 2 1 3
For any set S, a one-to-one and onto mapping π : S → S is called a permutation of S. □
Theorem 1.15 Let f : A → B, g : B → C, and h : C → D. Then
1. The composition of mappings is associative; that is, (h ◦ g) ◦ f = h ◦ (g ◦ f );
2. If f and g are both one-to-one, then the mapping g ◦ f is one-to-one;
3. If f and g are both onto, then the mapping g ◦ f is onto;
4. If f and g are bijective, then so is g ◦ f .
Proof. We will prove (1) and (3). Part (2) is left as an exercise. Part (4) follows directly
from (2) and (3).
(1) We must show that
h ◦ (g ◦ f ) = (h ◦ g) ◦ f .
For a ∈ A we have
(h ◦ (g ◦ f ))(a) = h((g ◦ f )(a))
= h(g(f (a)))
= (h ◦ g)(f (a))
= ((h ◦ g) ◦ f )(a).
(3) Assume that f and g are both onto functions. Given c ∈ C, we must show that
there exists an a ∈ A such that (g ◦ f )(a) = g(f (a)) = c. However, since g is onto, there
is an element b ∈ B such that g(b) = c. Similarly, there is an a ∈ A such that f (a) = b.
Accordingly,
(g ◦ f )(a) = g(f (a)) = g(b) = c.
■
If S is any set, we will use idS or id to denote the identity mapping from S to itself.
Define this map by id(s) = s for all s ∈ S. A map g : B → A is an inverse mapping of
f : A → B if g ◦ f = idA and f ◦ g = idB ; in other words, the inverse function of a function
simply “undoes” the function. A map is said to be invertible if it has an inverse. We usually
write f −1 for the inverse of f .
CHAPTER 1. PRELIMINARIES 10
√
Example 1.16 The function f (x) = x3 has inverse f −1 (x) = 3
x by Example 1.12. □
Example 1.17 The natural logarithm and the exponential functions, f (x) = ln x and
f −1 (x) = ex , are inverses of each other provided that we are careful about choosing domains.
Observe that
f (f −1 (x)) = f (ex ) = ln ex = x
and
f −1 (f (x)) = f −1 (ln x) = eln x = x
whenever composition makes sense. □
Example 1.18 Suppose that
3 1
A= .
5 2
Then A defines a map from R2 to R2 by
TA (x, y) = (3x + y, 5x + 2y).
We can find an inverse map of TA by simply inverting the matrix A; that is, TA−1 = TA−1 .
In this example,
−1 2 −1
A = ;
−5 3
hence, the inverse map is given by
TA−1 (x, y) = (2x − y, −5x + 3y).
It is easy to check that
TA−1 ◦ TA (x, y) = TA ◦ TA−1 (x, y) = (x, y).
Not every map has an inverse. If we consider the map
TB (x, y) = (3x, 0)
given by the matrix
3 0
B= ,
0 0
then an inverse map would have to be of the form
TB−1 (x, y) = (ax + by, cx + dy)
and
(x, y) = TB ◦ TB−1 (x, y) = (3ax + 3by, 0)
for all x and y. Clearly this is impossible because y might not be 0. □
Example 1.19 Given the permutation
1 2 3
π=
2 3 1
CHAPTER 1. PRELIMINARIES 11
on S = {1, 2, 3}, it is easy to see that the permutation defined by
−1 1 2 3
π =
3 1 2
is the inverse of π. In fact, any bijective mapping possesses an inverse, as we will see in the
next theorem. □
Theorem 1.20 A mapping is invertible if and only if it is both one-to-one and onto.
Proof. Suppose first that f : A → B is invertible with inverse g : B → A. Then
g ◦ f = idA is the identity map; that is, g(f (a)) = a. If a1 , a2 ∈ A with f (a1 ) = f (a2 ), then
a1 = g(f (a1 )) = g(f (a2 )) = a2 . Consequently, f is one-to-one. Now suppose that b ∈ B.
To show that f is onto, it is necessary to find an a ∈ A such that f (a) = b, but f (g(b)) = b
with g(b) ∈ A. Let a = g(b).
Conversely, let f be bijective and let b ∈ B. Since f is onto, there exists an a ∈ A such
that f (a) = b. Because f is one-to-one, a must be unique. Define g by letting g(b) = a. We
have now constructed the inverse of f . ■
Equivalence Relations and Partitions
A fundamental notion in mathematics is that of equality. We can generalize equality with
equivalence relations and equivalence classes. An equivalence relation on a set X is a
relation R ⊂ X × X such that
• (x, x) ∈ R for all x ∈ X (reflexive property);
• (x, y) ∈ R implies (y, x) ∈ R (symmetric property);
• (x, y) and (y, z) ∈ R imply (x, z) ∈ R (transitive property).
Given an equivalence relation R on a set X, we usually write x ∼ y instead of (x, y) ∈ R.
If the equivalence relation already has an associated notation such as =, ≡, or ∼
=, we will
use that notation.
Example 1.21 Let p, q, r, and s be integers, where q and s are nonzero. Define p/q ∼ r/s
if ps = qr. Clearly ∼ is reflexive and symmetric. To show that it is also transitive, suppose
that p/q ∼ r/s and r/s ∼ t/u, with q, s, and u all nonzero. Then ps = qr and ru = st.
Therefore,
psu = qru = qst.
Since s 6= 0, pu = qt. Consequently, p/q ∼ t/u. □
Example 1.22 Suppose that f and g are differentiable functions on R. We can define an
equivalence relation on such functions by letting f (x) ∼ g(x) if f ′ (x) = g ′ (x). It is clear that
∼ is both reflexive and symmetric. To demonstrate transitivity, suppose that f (x) ∼ g(x)
and g(x) ∼ h(x). From calculus we know that f (x) − g(x) = c1 and g(x) − h(x) = c2 , where
c1 and c2 are both constants. Hence,
f (x) − h(x) = (f (x) − g(x)) + (g(x) − h(x)) = c1 + c2
and f ′ (x) − h′ (x) = 0. Therefore, f (x) ∼ h(x). □
Example 1.23 For (x1 , y1 ) and (x2 , y2 ) in R2 , define (x1 , y1 ) ∼ (x2 , y2 ) if x21 + y12 = x22 + y22 .
Then ∼ is an equivalence relation on R2 . □
Example 1.24 Let A and B be 2 × 2 matrices with entries in the real numbers. We can
define an equivalence relation on the set of 2 × 2 matrices, by saying A ∼ B if there exists
CHAPTER 1. PRELIMINARIES 12
an invertible matrix P such that P AP −1 = B. For example, if
1 2 −18 33
A= and B = ,
−1 1 −11 20
then A ∼ B since P AP −1 = B for
2 5
P = .
1 3
Let I be the 2 × 2 identity matrix; that is,
1 0
I= .
0 1
Then IAI −1 = IAI = A; therefore, the relation is reflexive. To show symmetry, suppose
that A ∼ B. Then there exists an invertible matrix P such that P AP −1 = B. So
A = P −1 BP = P −1 B(P −1 )−1 .
Finally, suppose that A ∼ B and B ∼ C. Then there exist invertible matrices P and Q
such that P AP −1 = B and QBQ−1 = C. Since
C = QBQ−1 = QP AP −1 Q−1 = (QP )A(QP )−1 ,
the relation is transitive. Two matrices that are equivalent in this manner are said to be
similar. □
A partition
S P of a set X is a collection of nonempty sets X1 , X 2 , . . . such that Xi ∩Xj = ∅
for i 6= j and k Xk = X. Let ∼ be an equivalence relation on a set X and let x ∈ X. Then
[x] = {y ∈ X : y ∼ x} is called the equivalence class of x. We will see that an equivalence
relation gives rise to a partition via equivalence classes. Also, whenever a partition of a
set exists, there is some natural underlying equivalence relation, as the following theorem
demonstrates.
Theorem 1.25 Given an equivalence relation ∼ on a set X, the equivalence classes of X
form a partition of X. Conversely, if P = {Xi } is a partition of a set X, then there is an
equivalence relation on X with equivalence classes Xi .
Proof. Suppose there exists an equivalence relation ∼ on the set X. For any S x ∈ X, the
reflexive property shows that x ∈ [x] and so [x] is nonempty. Clearly X = x∈X [x]. Now
let x, y ∈ X. We need to show that either [x] = [y] or [x] ∩ [y] = ∅. Suppose that the
intersection of [x] and [y] is not empty and that z ∈ [x] ∩ [y]. Then z ∼ x and z ∼ y. By
symmetry and transitivity x ∼ y; hence, [x] ⊂ [y]. Similarly, [y] ⊂ [x] and so [x] = [y].
Therefore, any two equivalence classes are either disjoint or exactly the same.
Conversely, suppose that P = {Xi } is a partition of a set X. Let two elements be
equivalent if they are in the same partition. Clearly, the relation is reflexive. If x is in the
same partition as y, then y is in the same partition as x, so x ∼ y implies y ∼ x. Finally, if
x is in the same partition as y and y is in the same partition as z, then x must be in the
same partition as z, and transitivity holds. ■
Corollary 1.26 Two equivalence classes of an equivalence relation are either disjoint or
equal.
Let us examine some of the partitions given by the equivalence classes in the last set of
examples.
CHAPTER 1. PRELIMINARIES 13
Example 1.27 In the equivalence relation in Example 1.21, two pairs of integers, (p, q) and
(r, s), are in the same equivalence class when they reduce to the same fraction in its lowest
terms. □
Example 1.28 In the equivalence relation in Example 1.22, two functions f (x) and g(x)
are in the same partition when they differ by a constant. □
Example 1.29 We defined an equivalence class on R2 by (x1 , y1 ) ∼ (x2 , y2 ) if x21 + y12 =
x22 + y22 . Two pairs of real numbers are in the same partition when they lie on the same
circle about the origin. □
Example 1.30 Let r and s be two integers and suppose that n ∈ N. We say that r is
congruent to s modulo n, or r is congruent to s mod n, if r − s is evenly divisible by n; that
is, r − s = nk for some k ∈ Z. In this case we write r ≡ s (mod n). For example, 41 ≡ 17
(mod 8) since 41 − 17 = 24 is divisible by 8. We claim that congruence modulo n forms an
equivalence relation of Z. Certainly any integer r is equivalent to itself since r − r = 0 is
divisible by n. We will now show that the relation is symmetric. If r ≡ s (mod n), then
r − s = −(s − r) is divisible by n. So s − r is divisible by n and s ≡ r (mod n). Now
suppose that r ≡ s (mod n) and s ≡ t (mod n). Then there exist integers k and l such
that r − s = kn and s − t = ln. To show transitivity, it is necessary to prove that r − t is
divisible by n. However,
r − t = r − s + s − t = kn + ln = (k + l)n,
and so r − t is divisible by n.
If we consider the equivalence relation established by the integers modulo 3, then
[0] = {. . . , −3, 0, 3, 6, . . .},
[1] = {. . . , −2, 1, 4, 7, . . .},
[2] = {. . . , −1, 2, 5, 8, . . .}.
Notice that [0] ∪ [1] ∪ [2] = Z and also that the sets are disjoint. The sets [0], [1], and [2]
form a partition of the integers.
The integers modulo n are a very important example in the study of abstract algebra
and will become quite useful in our investigation of various algebraic structures such as
groups and rings. In our discussion of the integers modulo n we have actually assumed a
result known as the division algorithm, which will be stated and proved in Chapter 2. □
Sage. Sage is a powerful, open source, system for exact, numerical, and symbolic mathe-
matical computations. Electronic versions of this text contain comprehensive introductions
to the use of Sage to study abstract algebra, and include a set of exercises. These can be
found at the book’s website. Due to the format of this version of the text, at the end of
each chapter we have just included brief suggestions of how Sage might be employed.
1.3 Reading Questions
1. What do relations and mappings have in common?
2. What makes relations and mappings different?
3. State carefully the three defining properties of an equivalence relation. In other words,
do not just name the properties, give their definitions.
4. What is the big deal about equivalence relations? (Hint: Partitions.)
CHAPTER 1. PRELIMINARIES 14
5. Describe a general technique for proving that two sets are equal.
1.4 Exercises
1. Suppose that
A = {x : x ∈ N and x is even},
B = {x : x ∈ N and x is prime},
C = {x : x ∈ N and x is a multiple of 5}.
Describe each of the following sets.
(a) A ∩ B (c) A ∪ B
(b) B ∩ C (d) A ∩ (B ∪ C)
2. If A = {a, b, c}, B = {1, 2, 3}, C = {x}, and D = ∅, list all of the elements in each of
the following sets.
(a) A × B (c) A × B × C
(b) B × A (d) A × D
3. Find an example of two nonempty sets A and B for which A × B = B × A is true.
4. Prove A ∪ ∅ = A and A ∩ ∅ = ∅.
5. Prove A ∪ B = B ∪ A and A ∩ B = B ∩ A.
6. Prove A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
7. Prove A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
8. Prove A ⊂ B if and only if A ∩ B = A.
9. Prove (A ∩ B)′ = A′ ∪ B ′ .
10. Prove A ∪ B = (A ∩ B) ∪ (A \ B) ∪ (B \ A).
11. Prove (A ∪ B) × C = (A × C) ∪ (B × C).
12. Prove (A ∩ B) \ B = ∅.
13. Prove (A ∪ B) \ B = A \ B.
14. Prove A \ (B ∪ C) = (A \ B) ∩ (A \ C).
15. Prove A ∩ (B \ C) = (A ∩ B) \ (A ∩ C).
16. Prove (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B).
17. Which of the following relations f : Q → Q define a mapping? In each case, supply a
reason why f is or is not a mapping.
p+1 p+q
(a) f (p/q) = (c) f (p/q) =
p−2 q2
3p 3p2 p
(b) f (p/q) = (d) f (p/q) = 2 −
3q 7q q
18. Determine which of the following functions are one-to-one and which are onto. If the
function is not onto, determine its range.
(a) f : R → R defined by f (x) = ex
(b) f : Z → Z defined by f (n) = n2 + 3
(c) f : R → R defined by f (x) = sin x
(d) f : Z → Z defined by f (x) = x2
CHAPTER 1. PRELIMINARIES 15
19. Let f : A → B and g : B → C be invertible mappings; that is, mappings such that f −1
and g −1 exist. Show that (g ◦ f )−1 = f −1 ◦ g −1 .
20.
(a) Define a function f : N → N that is one-to-one but not onto.
(b) Define a function f : N → N that is onto but not one-to-one.
21. Prove the relation defined on R2 by (x1 , y1 ) ∼ (x2 , y2 ) if x21 + y12 = x22 + y22 is an
equivalence relation.
22. Let f : A → B and g : B → C be maps.
(a) If f and g are both one-to-one functions, show that g ◦ f is one-to-one.
(b) If g ◦ f is onto, show that g is onto.
(c) If g ◦ f is one-to-one, show that f is one-to-one.
(d) If g ◦ f is one-to-one and f is onto, show that g is one-to-one.
(e) If g ◦ f is onto and g is one-to-one, show that f is onto.
23. Define a function on the real numbers by
x+1
f (x) = .
x−1
What are the domain and range of f ? What is the inverse of f ? Compute f ◦ f −1 and
f −1 ◦ f .
24. Let f : X → Y be a map with A1 , A2 ⊂ X and B1 , B2 ⊂ Y .
(a) Prove f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ).
(b) Prove f (A1 ∩ A2 ) ⊂ f (A1 ) ∩ f (A2 ). Give an example in which equality fails.
(c) Prove f −1 (B1 ∪ B2 ) = f −1 (B1 ) ∪ f −1 (B2 ), where
f −1 (B) = {x ∈ X : f (x) ∈ B}.
(d) Prove f −1 (B1 ∩ B2 ) = f −1 (B1 ) ∩ f −1 (B2 ).
(e) Prove f −1 (Y \ B1 ) = X \ f −1 (B1 ).
25. Determine whether or not the following relations are equivalence relations on the given
set. If the relation is an equivalence relation, describe the partition given by it. If the
relation is not an equivalence relation, state why it fails to be one.
(a) x ∼ y in R if x ≥ y (c) x ∼ y in R if |x − y| ≤ 4
(b) m ∼ n in Z if mn > 0 (d) m ∼ n in Z if m ≡ n (mod 6)
26. Define a relation ∼ on R by stating that (a, b) ∼ (c, d) if and only if a2 + b2 ≤ c2 + d2 .
2
Show that ∼ is reflexive and transitive but not symmetric.
27. Show that an m × n matrix gives rise to a well-defined map from Rn to Rm .
28. Find the error in the following argument by providing a counterexample. “The reflexive
property is redundant in the axioms for an equivalence relation. If x ∼ y, then y ∼ x
by the symmetric property. Using the transitive property, we can deduce that x ∼ x.”
29. Projective Real Line. Define a relation on R2 \ {(0, 0)} by letting (x1 , y1 ) ∼ (x2 , y2 )
if there exists a nonzero real number λ such that (x1 , y1 ) = (λx2 , λy2 ). Prove that ∼
CHAPTER 1. PRELIMINARIES 16
defines an equivalence relation on R2 \ (0, 0). What are the corresponding equivalence
classes? This equivalence relation defines the projective line, denoted by P(R), which
is very important in geometry.
1.5 References and Suggested Readings
[1] Artin, M. Algebra (Classic Version). 2nd ed. Pearson, Upper Saddle River, NJ, 2018.
[2] Childs, L. A Concrete Introduction to Higher Algebra. 2nd ed. Springer-Verlag, New
York, 1995.
[3] Dummit, D. and Foote, R. Abstract Algebra. 3rd ed. Wiley, New York, 2003.
[4] Ehrlich, G. Fundamental Concepts of Algebra. PWS-KENT, Boston, 1991.
[5] Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper Saddle
River, NJ, 2003.
[6] Gallian, J. A. Contemporary Abstract Algebra. 7th ed. Brooks/Cole, Belmont, CA,
2009.
[7] Halmos, P. Naive Set Theory. Springer, New York, 1991. One of the best references
for set theory.
[8] Herstein, I. N. Abstract Algebra. 3rd ed. Wiley, New York, 1996.
[9] Hungerford, T. W. Algebra. Springer, New York, 1974. One of the standard graduate
algebra texts.
[10] Lang, S. Algebra. 3rd ed. Springer, New York, 2002. Another standard graduate text.
[11] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998.
[12] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[13] Nickelson, W. K. Introduction to Abstract Algebra. 3rd ed. Wiley, New York, 2006.
[14] Solow, D. How to Read and Do Proofs. 5th ed. Wiley, New York, 2009.
[15] van der Waerden, B. L. A History of Algebra. Springer-Verlag, New York, 1985. An
account of the historical development of algebra.
2
The Integers
The integers are the building blocks of mathematics. In this chapter we will investigate
the fundamental properties of the integers, including mathematical induction, the division
algorithm, and the Fundamental Theorem of Arithmetic.
2.1 Mathematical Induction
Suppose we wish to show that
n(n + 1)
1 + 2 + ··· + n =
2
for any natural number n. This formula is easily verified for small numbers such as n = 1,
2, 3, or 4, but it is impossible to verify for all natural numbers on a case-by-case basis. To
prove the formula true in general, a more generic method is required.
Suppose we have verified the equation for the first n cases. We will attempt to show
that we can generate the formula for the (n + 1)th case from this knowledge. The formula
is true for n = 1 since
1(1 + 1)
1= .
2
If we have verified the first n cases, then
n(n + 1)
1 + 2 + · · · + n + (n + 1) = +n+1
2
n2 + 3n + 2
=
2
(n + 1)[(n + 1) + 1]
= .
2
This is exactly the formula for the (n + 1)th case.
This method of proof is known as mathematical induction. Instead of attempting to
verify a statement about some subset S of the positive integers N on a case-by-case basis, an
impossible task if S is an infinite set, we give a specific proof for the smallest integer being
considered, followed by a generic argument showing that if the statement holds for a given
case, then it must also hold for the next case in the sequence. We summarize mathematical
induction in the following axiom.
Principle 2.1 First Principle of Mathematical Induction. Let S(n) be a statement
about integers for n ∈ N and suppose S(n0 ) is true for some integer n0 . If for all integers k
17
CHAPTER 2. THE INTEGERS 18
with k ≥ n0 , S(k) implies that S(k + 1) is true, then S(n) is true for all integers n greater
than or equal to n0 .
Example 2.2 For all integers n ≥ 3, 2n > n + 4. Since
8 = 23 > 3 + 4 = 7,
the statement is true for n0 = 3. Assume that 2k > k + 4 for k ≥ 3. Then 2k+1 = 2 · 2k >
2(k + 4). But
2(k + 4) = 2k + 8 > k + 5 = (k + 1) + 4
since k is positive. Hence, by induction, the statement holds for all integers n ≥ 3. □
Example 2.3 Every integer 10n+1 + 3 · 10n + 5 is divisible by 9 for n ∈ N. For n = 1,
101+1 + 3 · 10 + 5 = 135 = 9 · 15
is divisible by 9. Suppose that 10k+1 + 3 · 10k + 5 is divisible by 9 for k ≥ 1. Then
10(k+1)+1 + 3 · 10k+1 + 5 = 10k+2 + 3 · 10k+1 + 50 − 45
= 10(10k+1 + 3 · 10k + 5) − 45
is divisible by 9. □
Example 2.4 We will prove the binomial theorem using mathematical induction; that is,
n
X
n n
(a + b) = ak bn−k ,
k
k=0
where a and b are real numbers, n ∈ N, and
n n!
=
k k!(n − k)!
is the binomial coefficient. We first show that
n+1 n n
= + .
k k k−1
This result follows from
n n n! n!
+ = +
k k−1 k!(n − k)! (k − 1)!(n − k + 1)!
(n + 1)!
=
k!(n + 1 − k)!
n+1
= .
k
If n = 1, the binomial theorem is easy to verify. Now assume that the result is true for n
greater than or equal to 1. Then
(a + b)n+1 = (a + b)(a + b)n
n
!
X n k n−k
= (a + b) a b
k
k=0
CHAPTER 2. THE INTEGERS 19
n
X n
X
n n
= ak+1 bn−k + ak bn+1−k
k k
k=0 k=0
Xn Xn
n+1 n k n+1−k n k n+1−k
=a + a b + a b + bn+1
k−1 k
k=1 k=1
Xn
n n
= an+1 + + ak bn+1−k + bn+1
k−1 k
k=1
X n + 1
n+1
= ak bn+1−k .
k
k=0
□
We have an equivalent statement of the Principle of Mathematical Induction that is
often very useful.
Principle 2.5 Second Principle of Mathematical Induction. Let S(n) be a statement
about integers for n ∈ N and suppose S(n0 ) is true for some integer n0 . If S(n0 ), S(n0 +
1), . . . , S(k) imply that S(k + 1) for k ≥ n0 , then the statement S(n) is true for all integers
n ≥ n0 .
A nonempty subset S of Z is well-ordered if S contains a least element. Notice that the
set Z is not well-ordered since it does not contain a smallest element. However, the natural
numbers are well-ordered.
Principle 2.6 Principle of Well-Ordering. Every nonempty subset of the natural
numbers is well-ordered.
The Principle of Well-Ordering is equivalent to the Principle of Mathematical Induction.
Lemma 2.7 The Principle of Mathematical Induction implies that 1 is the least positive
natural number.
Proof. Let S = {n ∈ N : n ≥ 1}. Then 1 ∈ S. Assume that n ∈ S. Since 0 < 1, it must
be the case that n = n + 0 < n + 1. Therefore, 1 ≤ n < n + 1. Consequently, if n ∈ S,
then n + 1 must also be in S, and by the Principle of Mathematical Induction, and we have
S = N. ■
Theorem 2.8 The Principle of Mathematical Induction implies the Principle of Well-
Ordering. That is, every nonempty subset of N contains a least element.
Proof. We must show that if S is a nonempty subset of the natural numbers, then S
contains a least element. If S contains 1, then the theorem is true by Lemma 2.7. Assume
that if S contains an integer k such that 1 ≤ k ≤ n, then S contains a least element. We
will show that if a set S contains an integer less than or equal to n + 1, then S has a least
element. If S does not contain an integer less than n + 1, then n + 1 is the smallest integer
in S. Otherwise, since S is nonempty, S must contain an integer less than or equal to n. In
this case, by induction, S contains a least element. ■
Induction can also be very useful in formulating definitions. For instance, there are two
ways to define n!, the factorial of a positive integer n.
• The explicit definition: n! = 1 · 2 · 3 · · · (n − 1) · n.
• The inductive or recursive definition: 1! = 1 and n! = n(n − 1)! for n > 1.
Every good mathematician or computer scientist knows that looking at problems recursively,
as opposed to explicitly, often results in better understanding of complex issues.
CHAPTER 2. THE INTEGERS 20
2.2 The Division Algorithm
An application of the Principle of Well-Ordering that we will use often is the division
algorithm.
Theorem 2.9 Division Algorithm. Let a and b be integers, with b > 0. Then there
exist unique integers q and r such that
a = bq + r
where 0 ≤ r < b.
Proof. This is a perfect example of the existence-and-uniqueness type of proof. We must
first prove that the numbers q and r actually exist. Then we must show that if q ′ and r′
are two other such numbers, then q = q ′ and r = r′ .
Existence of q and r. Let
S = {a − bk : k ∈ Z and a − bk ≥ 0}.
If 0 ∈ S, then b divides a, and we can let q = a/b and r = 0. If 0 ∈ / S, we can use the
Well-Ordering Principle. We must first show that S is nonempty. If a > 0, then a − b · 0 ∈ S.
If a < 0, then a − b(2a) = a(1 − 2b) ∈ S. In either case S 6= ∅. By the Well-Ordering
Principle, S must have a smallest member, say r = a − bq. Therefore, a = bq + r, r ≥ 0.
We now show that r < b. Suppose that r > b. Then
a − b(q + 1) = a − bq − b = r − b > 0.
In this case we would have a − b(q + 1) in the set S. But then a − b(q + 1) < a − bq, which
would contradict the fact that r = a − bq is the smallest member of S. So r ≤ b. Since
0∈/ S, r 6= b and so r < b.
Uniqueness of q and r. Suppose there exist integers r, r′ , q, and q ′ such that
a = bq + r, 0 ≤ r < b and a = bq ′ + r′ , 0 ≤ r′ < b.
Then bq + r = bq ′ + r′ . Assume that r′ ≥ r. From the last equation we have b(q − q ′ ) = r′ − r;
therefore, b must divide r′ − r and 0 ≤ r′ − r ≤ r′ < b. This is possible only if r′ − r = 0.
Hence, r = r′ and q = q ′ . ■
Let a and b be integers. If b = ak for some integer k, we write a | b. An integer d
is called a common divisor of a and b if d | a and d | b. The greatest common divisor of
integers a and b is a positive integer d such that d is a common divisor of a and b and if d′
is any other common divisor of a and b, then d′ | d. We write d = gcd(a, b); for example,
gcd(24, 36) = 12 and gcd(120, 102) = 6. We say that two integers a and b are relatively
prime if gcd(a, b) = 1.
Theorem 2.10 Let a and b be nonzero integers. Then there exist integers r and s such that
gcd(a, b) = ar + bs.
Furthermore, the greatest common divisor of a and b is unique.
Proof. Let
S = {am + bn : m, n ∈ Z and am + bn > 0}.
Clearly, the set S is nonempty; hence, by the Well-Ordering Principle S must have a smallest
member, say d = ar + bs. We claim that d = gcd(a, b). Write a = dq + r′ where 0 ≤ r′ < d.
CHAPTER 2. THE INTEGERS 21
If r′ > 0, then
r′ = a − dq
= a − (ar + bs)q
= a − arq − bsq
= a(1 − rq) + b(−sq),
which is in S. But this would contradict the fact that d is the smallest member of S. Hence,
r′ = 0 and d divides a. A similar argument shows that d divides b. Therefore, d is a common
divisor of a and b.
Suppose that d′ is another common divisor of a and b, and we want to show that d′ | d.
If we let a = d′ h and b = d′ k, then
d = ar + bs = d′ hr + d′ ks = d′ (hr + ks).
So d′ must divide d. Hence, d must be the unique greatest common divisor of a and b. ■
Corollary 2.11 Let a and b be two integers that are relatively prime. Then there exist
integers r and s such that ar + bs = 1.
The Euclidean Algorithm
Among other things, Theorem 2.10 allows us to compute the greatest common divisor of
two integers.
Example 2.12 Let us compute the greatest common divisor of 945 and 2415. First observe
that
2415 = 945 · 2 + 525
945 = 525 · 1 + 420
525 = 420 · 1 + 105
420 = 105 · 4 + 0.
Reversing our steps, 105 divides 420, 105 divides 525, 105 divides 945, and 105 divides 2415.
Hence, 105 divides both 945 and 2415. If d were another common divisor of 945 and 2415,
then d would also have to divide 105. Therefore, gcd(945, 2415) = 105.
If we work backward through the above sequence of equations, we can also obtain num-
bers r and s such that 945r + 2415s = 105. Observe that
105 = 525 + (−1) · 420
= 525 + (−1) · [945 + (−1) · 525]
= 2 · 525 + (−1) · 945
= 2 · [2415 + (−2) · 945] + (−1) · 945
= 2 · 2415 + (−5) · 945.
So r = −5 and s = 2. Notice that r and s are not unique, since r = 41 and s = −16 would
also work. □
To compute gcd(a, b) = d, we are using repeated divisions to obtain a decreasing se-
quence of positive integers r1 > r2 > · · · > rn = d; that is,
b = aq1 + r1
CHAPTER 2. THE INTEGERS 22
a = r1 q2 + r2
r1 = r2 q3 + r3
..
.
rn−2 = rn−1 qn + rn
rn−1 = rn qn+1 .
To find r and s such that ar +bs = d, we begin with this last equation and substitute results
obtained from the previous equations:
d = rn
= rn−2 − rn−1 qn
= rn−2 − qn (rn−3 − qn−1 rn−2 )
= −qn rn−3 + (1 + qn qn−1 )rn−2
..
.
= ra + sb.
The algorithm that we have just used to find the greatest common divisor d of two integers
a and b and to write d as the linear combination of a and b is known as the Euclidean
algorithm.
Prime Numbers
Let p be an integer such that p > 1. We say that p is a prime number, or simply p is prime,
if the only positive numbers that divide p are 1 and p itself. An integer n > 1 that is not
prime is said to be composite.
Lemma 2.13 Euclid. Let a and b be integers and p be a prime number. If p | ab, then
either p | a or p | b.
Proof. Suppose that p does not divide a. We must show that p | b. Since gcd(a, p) = 1,
there exist integers r and s such that ar + ps = 1. So
b = b(ar + ps) = (ab)r + p(bs).
Since p divides both ab and itself, p must divide b = (ab)r + p(bs). ■
Theorem 2.14 Euclid. There exist an infinite number of primes.
Proof. We will prove this theorem by contradiction. Suppose that there are only a finite
number of primes, say p1 , p2 , . . . , pn . Let P = p1 p2 · · · pn + 1. Then P must be divisible
by some pi for 1 ≤ i ≤ n. In this case, pi must divide P − p1 p2 · · · pn = 1, which is a
contradiction. Hence, either P is prime or there exists an additional prime number p 6= pi
that divides P . ■
Theorem 2.15 Fundamental Theorem of Arithmetic. Let n be an integer such that
n > 1. Then
n = p1 p2 · · · pk ,
where p1 , . . . , pk are primes (not necessarily distinct). Furthermore, this factorization is
unique; that is, if
n = q1 q2 · · · ql ,
then k = l and the qi ’s are just the pi ’s rearranged.
CHAPTER 2. THE INTEGERS 23
Proof. Uniqueness. To show uniqueness we will use induction on n. The theorem is
certainly true for n = 2 since in this case n is prime. Now assume that the result holds for
all integers m such that 1 ≤ m < n, and
n = p 1 p 2 · · · p k = q1 q2 · · · ql ,
where p1 ≤ p2 ≤ · · · ≤ pk and q1 ≤ q2 ≤ · · · ≤ ql . By Lemma 2.13, p1 | qi for some
i = 1, . . . , l and q1 | pj for some j = 1, . . . , k. Since all of the pi ’s and qi ’s are prime, p1 = qi
and q1 = pj . Hence, p1 = q1 since p1 ≤ pj = q1 ≤ qi = p1 . By the induction hypothesis,
n′ = p2 · · · pk = q2 · · · ql
has a unique factorization. Hence, k = l and qi = pi for i = 1, . . . , k.
Existence. To show existence, suppose that there is some integer that cannot be written
as the product of primes. Let S be the set of all such numbers. By the Principle of Well-
Ordering, S has a smallest number, say a. If the only positive factors of a are a and 1, then
a is prime, which is a contradiction. Hence, a = a1 a2 where 1 < a1 < a and 1 < a2 < a.
Neither a1 ∈ S nor a2 ∈ S, since a is the smallest element in S. So
a1 = p1 · · · pr
a2 = q1 · · · qs .
Therefore,
a = a1 a2 = p1 · · · pr q1 · · · qs .
So a ∈
/ S, which is a contradiction. ■
Historical Note
Prime numbers were first studied by the ancient Greeks. Two important results from antiq-
uity are Euclid’s proof that an infinite number of primes exist and the Sieve of Eratosthenes,
a method of computing all of the prime numbers less than a fixed positive integer n. One
problem in number theory is to find a function f such that f (n) is prime for each integer n.
n
Pierre Fermat (1601?–1665) conjectured that 22 + 1 was prime for all n, but later it was
shown by Leonhard Euler (1707–1783) that
5
22 + 1 = 4,294,967,297
is a composite number. One of the many unproven conjectures about prime numbers is
Goldbach’s Conjecture. In a letter to Euler in 1742, Christian Goldbach stated the conjec-
ture that every even integer with the exception of 2 seemed to be the sum of two primes:
4 = 2 + 2, 6 = 3 + 3, 8 = 3 + 5, . . .. Although the conjecture has been verified for the
numbers up through 4 × 1018 , it has yet to be proven in general. Since prime numbers play
an important role in public key cryptography, there is currently a great deal of interest in
determining whether or not a large number is prime.
Sage. Sage’s original purpose was to support research in number theory, so it is perfect
for the types of computations with the integers that we have in this chapter.
CHAPTER 2. THE INTEGERS 24
2.3 Reading Questions
1. Use Sage to express 123456792 as a product of prime numbers.
2. Find the greatest common divisor of 84 and 52.
3. Find integers r and s so that r(84) + s(52) = gcd(84, 52).
4. Explain the use of the term “induction hypothesis.”
5. What is Goldbach’s Conjecture? And why is it called a “conjecture”?
2.4 Exercises
1. Prove that
n(n + 1)(2n + 1)
12 + 22 + · · · + n2 =
6
for n ∈ N.
2. Prove that
n2 (n + 1)2
13 + 23 + · · · + n3 =
4
for n ∈ N.
3. Prove that n! > 2n for n ≥ 4.
4. Prove that
n(3n − 1)x
x + 4x + 7x + · · · + (3n − 2)x =
2
for n ∈ N.
5. Prove that 10n+1 + 10n + 1 is divisible by 3 for n ∈ N.
6. Prove that 4 · 102n + 9 · 102n−1 + 5 is divisible by 99 for n ∈ N.
7. Show that
1X
n
√
n
a1 a2 · · · an ≤ ak .
n
k=1
8. Prove the Leibniz rule for f (n) (x), where f (n) is the nth derivative of f ; that is, show
that
n
X
(n) n (k)
(f g) (x) = f (x)g (n−k) (x).
k
k=0
9. Use induction to prove that 1 + 2 + 22 + · · · + 2n = 2n+1 − 1 for n ∈ N.
10. Prove that
1 1 1 n
+ + ··· + =
2 6 n(n + 1) n+1
for n ∈ N.
11. If x is a nonnegative real number, then show that (1 + x)n − 1 ≥ nx for n = 0, 1, 2, . . ..
12. Power Sets. Let X be a set. Define the power set of X, denoted P(X), to be the set
of all subsets of X. For example,
P({a, b}) = {∅, {a}, {b}, {a, b}}.
For every positive integer n, show that a set with exactly n elements has a power set
with exactly 2n elements.
CHAPTER 2. THE INTEGERS 25
13. Prove that the two principles of mathematical induction stated in Section 2.1 are
equivalent.
14. Show that the Principle of Well-Ordering for the natural numbers implies that 1 is the
smallest natural number. Use this result to show that the Principle of Well-Ordering
implies the Principle of Mathematical Induction; that is, show that if S ⊂ N such that
1 ∈ S and n + 1 ∈ S whenever n ∈ S, then S = N.
15. For each of the following pairs of numbers a and b, calculate gcd(a, b) and find integers
r and s such that gcd(a, b) = ra + sb.
(a) 14 and 39 (d) 471 and 562
(b) 234 and 165 (e) 23771 and 19945
(c) 1739 and 9923 (f) −4357 and 3754
16. Let a and b be nonzero integers. If there exist integers r and s such that ar + bs = 1,
show that a and b are relatively prime.
17. Fibonacci Numbers. The Fibonacci numbers are
1, 1, 2, 3, 5, 8, 13, 21, . . . .
We can define them inductively by f1 = 1, f2 = 1, and fn+2 = fn+1 + fn for n ∈ N.
(a) Prove that fn < 2n .
(b) Prove that fn+1 fn−1 = fn2 + (−1)n , n ≥ 2.
√ √ √
(c) Prove that fn = [(1 + 5 )n − (1 − 5 )n ]/2n 5.
√
(d) Show that ϕ = limn→∞ fn+1 /fn = ( 5 + 1)/2. The constant ϕ is known as the
golden ratio.
(e) Prove that fn and fn+1 are relatively prime.
18. Let a and b be integers such that gcd(a, b) = 1. Let r and s be integers such that
ar + bs = 1. Prove that
gcd(a, s) = gcd(r, b) = gcd(r, s) = 1.
19. Let x, y ∈ N be relatively prime. If xy is a perfect square, prove that x and y must
both be perfect squares.
20. Using the division algorithm, show that every perfect square is of the form 4k or 4k + 1
for some nonnegative integer k.
21. Suppose that a, b, r, s are pairwise relatively prime and that
a2 + b2 = r2
a2 − b2 = s2 .
Prove that a, r, and s are odd and b is even.
22. Let n ∈ N. Use the division algorithm to prove that every integer is congruent mod n
to precisely one of the integers 0, 1, . . . , n − 1. Conclude that if r is an integer, then
there is exactly one s in Z such that 0 ≤ s < n and [r] = [s]. Hence, the integers are
indeed partitioned by congruence mod n.
23. Define the least common multiple of two nonzero integers a and b, denoted by lcm(a, b),
to be the nonnegative integer m such that both a and b divide m, and if a and b divide
any other integer n, then m also divides n. Prove there exists a unique least common
CHAPTER 2. THE INTEGERS 26
multiple for any two integers a and b.
24. If d = gcd(a, b) and m = lcm(a, b), prove that dm = |ab|.
25. Show that lcm(a, b) = ab if and only if gcd(a, b) = 1.
26. Prove that gcd(a, c) = gcd(b, c) = 1 if and only if gcd(ab, c) = 1 for integers a, b, and
c.
27. Let a, b, c ∈ Z. Prove that if gcd(a, b) = 1 and a | bc, then a | c.
28. Let p ≥ 2. Prove that if 2p − 1 is prime, then p must also be prime.
29. Prove that there are an infinite number of primes of the form 6n + 5.
30. Prove that there are an infinite number of primes of the form 4n − 1.
31. Using the fact that 2 is prime, show that
√ there do not exist integers p and q such that
2 2
p = 2q . Demonstrate that therefore 2 cannot be a rational number.
2.5 Programming Exercises
1. The Sieve of Eratosthenes. One method of computing all of the prime numbers
less than a certain fixed positive integer N is to list all of the numbers n such that
1 < n < N . Begin by eliminating all of the multiples of 2. Next eliminate all of the
multiples of 3. Now eliminate all of the multiples of 5. Notice that 4 has already been
crossed out. Continue in this
√ manner, noticing that we do not have to go all the way
to N ; it suffices to stop at N . Using this method, compute all of the prime numbers
less than N = 250. We can also use this method to find all of the integers that are
relatively prime to an integer N . Simply eliminate the prime factors of N and all of
their multiples. Using this method, find all of the numbers that are relatively prime
to N = 120. Using the Sieve of Eratosthenes, write a program that will compute all
of the primes less than an integer N .
2. Let N0 = N ∪ {0}. Ackermann’s function is the function A : N0 × N0 → N0 defined by
the equations
A(0, y) = y + 1,
A(x + 1, 0) = A(x, 1),
A(x + 1, y + 1) = A(x, A(x + 1, y)).
Use this definition to compute A(3, 1). Write a program to evaluate Ackermann’s
function. Modify the program to count the number of statements executed in the
program when Ackermann’s function is evaluated. How many statements are executed
in the evaluation of A(4, 1)? What about A(5, 1)?
3. Write a computer program that will implement the Euclidean algorithm. The program
should accept two positive integers a and b as input and should output gcd(a, b) as
well as integers r and s such that
gcd(a, b) = ra + sb.
2.6 References and Suggested Readings
[1] Brookshear, J. G. Theory of Computation: Formal Languages, Automata, and Com-
plexity. Benjamin/Cummings, Redwood City, CA, 1989. Shows the relationships of
the theoretical aspects of computer science to set theory and the integers.
CHAPTER 2. THE INTEGERS 27
[2] Hardy, G. H. and Wright, E. M. An Introduction to the Theory of Numbers. 6th ed.
Oxford University Press, New York, 2008.
[3] Niven, I. and Zuckerman, H. S. An Introduction to the Theory of Numbers. 5th ed.
Wiley, New York, 1991.
[4] Vanden Eynden, C. Elementary Number Theory. 2nd ed. Waveland Press, Long
Grove IL, 2001.
3
Groups
We begin our study of algebraic structures by investigating sets associated with single
operations that satisfy certain reasonable axioms; that is, we want to define an operation
on a set in a way that will generalize such familiar structures as the integers Z together
with the single operation of addition, or invertible 2 × 2 matrices together with the single
operation of matrix multiplication. The integers and the 2 × 2 matrices, together with their
respective single operations, are examples of algebraic structures known as groups.
The theory of groups occupies a central position in mathematics. Modern group theory
arose from an attempt to find the roots of a polynomial in terms of its coefficients. Groups
now play a central role in such areas as coding theory, counting, and the study of symmetries;
many areas of biology, chemistry, and physics have benefited from group theory.
3.1 Integer Equivalence Classes and Symmetries
Let us now investigate some mathematical structures that can be viewed as sets with single
operations.
The Integers mod n
The integers mod n have become indispensable in the theory and applications of algebra.
In mathematics they are used in cryptography, coding theory, and the detection of errors
in identification codes.
We have already seen that two integers a and b are equivalent mod n if n divides a − b.
The integers mod n also partition Z into n different equivalence classes; we will denote the set
of these equivalence classes by Zn . Consider the integers modulo 12 and the corresponding
partition of the integers:
[0] = {. . . , −12, 0, 12, 24, . . .},
[1] = {. . . , −11, 1, 13, 25, . . .},
..
.
[11] = {. . . , −1, 11, 23, 35, . . .}.
When no confusion can arise, we will use 0, 1, . . . , 11 to indicate the equivalence classes
[0], [1], . . . , [11] respectively. We can do arithmetic on Zn . For two integers a and b, define
addition modulo n to be (a + b) (mod n); that is, the remainder when a + b is divided by
28
CHAPTER 3. GROUPS 29
n. Similarly, multiplication modulo n is defined as (ab) (mod n), the remainder when ab is
divided by n.
Example 3.1 The following examples illustrate integer arithmetic modulo n:
7 + 4 ≡ 1 (mod 5) 7 · 3 ≡ 1 (mod 5)
3 + 5 ≡ 0 (mod 8) 3 · 5 ≡ 7 (mod 8)
3 + 4 ≡ 7 (mod 12) 3 · 4 ≡ 0 (mod 12).
In particular, notice that it is possible that the product of two nonzero numbers modulo n
can be equivalent to 0 modulo n. □
Example 3.2 Most, but not all, of the usual laws of arithmetic hold for addition and
multiplication in Zn . For instance, it is not necessarily true that there is a multiplicative
inverse. Consider the multiplication table for Z8 in Figure 3.3. Notice that 2, 4, and 6 do
not have multiplicative inverses; that is, for n = 2, 4, or 6, there is no integer k such that
kn ≡ 1 (mod 8).
· 0 1 2 3 4 5 6 7
0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7
2 0 2 4 6 0 2 4 6
3 0 3 6 1 4 7 2 5
4 0 4 0 4 0 4 0 4
5 0 5 2 7 4 1 6 3
6 0 6 4 2 0 6 4 2
7 0 7 6 5 4 3 2 1
Figure 3.3 Multiplication table for Z8
□
Proposition 3.4 Let Zn be the set of equivalence classes of the integers mod n and a, b, c ∈
Zn .
1. Addition and multiplication are commutative:
a+b≡b+a (mod n)
ab ≡ ba (mod n).
2. Addition and multiplication are associative:
(a + b) + c ≡ a + (b + c) (mod n)
(ab)c ≡ a(bc) (mod n).
3. There are both additive and multiplicative identities:
a + 0 ≡ a (mod n)
a · 1 ≡ a (mod n).
4. Multiplication distributes over addition:
a(b + c) ≡ ab + ac (mod n).
CHAPTER 3. GROUPS 30
5. For every integer a there is an additive inverse −a:
a + (−a) ≡ 0 (mod n).
6. Let a be a nonzero integer. Then gcd(a, n) = 1 if and only if there exists a multiplicative
inverse b for a (mod n); that is, a nonzero integer b such that
ab ≡ 1 (mod n).
Proof. We will prove (1) and (6) and leave the remaining properties to be proven in the
exercises.
(1) Addition and multiplication are commutative modulo n since the remainder of a + b
divided by n is the same as the remainder of b + a divided by n.
(6) Suppose that gcd(a, n) = 1. Then there exist integers r and s such that ar + ns = 1.
Since ns = 1 − ar, it must be the case that ar ≡ 1 (mod n). Letting b be the equivalence
class of r, ab ≡ 1 (mod n).
Conversely, suppose that there exists an integer b such that ab ≡ 1 (mod n). Then n
divides ab − 1, so there is an integer k such that ab − nk = 1. Let d = gcd(a, n). Since d
divides ab − nk, d must also divide 1; hence, d = 1. ■
CHAPTER 3. GROUPS 31
Symmetries
A B A B
identity
D C D C
A B ◦
C D
180
rotation
D C B A
A B B A
reflection
vertical axis
D C C D
A B D C
reflection
horizontal axis
D C A B
Figure 3.5 Rigid motions of a rectangle
A symmetry of a geometric figure is a rearrangement of the figure preserving the arrange-
ment of its sides and vertices as well as its distances and angles. A map from the plane
to itself preserving the symmetry of an object is called a rigid motion. For example, if we
look at the rectangle in Figure 3.5, it is easy to see that a rotation of 180◦ or 360◦ returns
a rectangle in the plane with the same orientation as the original rectangle and the same
relationship among the vertices. A reflection of the rectangle across either the vertical axis
or the horizontal axis can also be seen to be a symmetry. However, a 90◦ rotation in either
direction cannot be a symmetry unless the rectangle is a square.
CHAPTER 3. GROUPS 32
B B
identity A B C
id =
A B C
A C A C
B A
rotation A B C
ρ1 =
B C A
A C C B
B C
rotation A B C
ρ2 =
C A B
A C B A
B C
reflection A B C
µ1 =
A C B
A C A B
B B
reflection A B C
µ2 =
C B A
A C C A
B A
reflection A B C
µ3 =
B A C
A C B C
Figure 3.6 Symmetries of a triangle
Let us find the symmetries of the equilateral triangle 4ABC. To find a symmetry of
CHAPTER 3. GROUPS 33
4ABC, we must first examine the permutations of the vertices A, B, and C and then ask
if a permutation extends to a symmetry of the triangle. Recall that a permutation of a set
S is a one-to-one and onto map π : S → S. The three vertices have 3! = 6 permutations,
so the triangle has at most six symmetries. To see that there are six permutations, observe
there are three different possibilities for the first vertex, and two for the second, and the
remaining vertex is determined by the placement of the first two. So we have 3·2·1 = 3! = 6
different arrangements. To denote the permutation of the vertices of an equilateral triangle
that sends A to B, B to C, and C to A, we write the array
A B C
.
B C A
Notice that this particular permutation corresponds to the rigid motion of rotating the
triangle by 120◦ in a clockwise direction. In fact, every permutation gives rise to a symmetry
of the triangle. All of these symmetries are shown in Figure 3.6.
A natural question to ask is what happens if one motion of the triangle 4ABC is
followed by another. Which symmetry is µ1 ρ1 ; that is, what happens when we do the
permutation ρ1 and then the permutation µ1 ? Remember that we are composing functions
here. Although we usually multiply left to right, we compose functions right to left. We have
(µ1 ρ1 )(A) = µ1 (ρ1 (A)) = µ1 (B) = C
(µ1 ρ1 )(B) = µ1 (ρ1 (B)) = µ1 (C) = B
(µ1 ρ1 )(C) = µ1 (ρ1 (C)) = µ1 (A) = A.
This is the same symmetry as µ2 . Suppose we do these motions in the opposite order,
ρ1 then µ1 . It is easy to determine that this is the same as the symmetry µ3 ; hence,
ρ1 µ1 6= µ1 ρ1 . A multiplication table for the symmetries of an equilateral triangle 4ABC is
given in Figure 3.7.
Notice that in the multiplication table for the symmetries of an equilateral triangle, for
every motion of the triangle α there is another motion β such that αβ = id; that is, for
every motion there is another motion that takes the triangle back to its original orientation.
◦ id ρ1 ρ2 µ1 µ2 µ3
id id ρ1 ρ2 µ1 µ2 µ3
ρ1 ρ1 ρ2 id µ3 µ1 µ2
ρ2 ρ2 id ρ1 µ2 µ3 µ1
µ1 µ1 µ2 µ 3 id ρ1 ρ2
µ2 µ2 µ3 µ 1 ρ2 id ρ1
µ3 µ3 µ1 µ 2 ρ1 ρ2 id
Figure 3.7 Symmetries of an equilateral triangle
3.2 Definitions and Examples
The integers mod n and the symmetries of a triangle or a rectangle are examples of groups.
A binary operation or law of composition on a set G is a function G × G → G that assigns
to each pair (a, b) ∈ G × G a unique element a ◦ b, or ab in G, called the composition of
a and b. A group (G, ◦) is a set G together with a law of composition (a, b) 7→ a ◦ b that
satisfies the following axioms.
CHAPTER 3. GROUPS 34
• The law of composition is associative. That is,
(a ◦ b) ◦ c = a ◦ (b ◦ c)
for a, b, c ∈ G.
• There exists an element e ∈ G, called the identity element, such that for any element
a∈G
e ◦ a = a ◦ e = a.
• For each element a ∈ G, there exists an inverse element in G, denoted by a−1 , such
that
a ◦ a−1 = a−1 ◦ a = e.
A group G with the property that a ◦ b = b ◦ a for all a, b ∈ G is called abelian or com-
mutative. Groups not satisfying this property are said to be nonabelian or noncommutative.
Example 3.8 The integers Z = {. . . , −1, 0, 1, 2, . . .} form a group under the operation of
addition. The binary operation on two integers m, n ∈ Z is just their sum. Since the integers
under addition already have a well-established notation, we will use the operator + instead
of ◦; that is, we shall write m + n instead of m ◦ n. The identity is 0, and the inverse of
n ∈ Z is written as −n instead of n−1 . Notice that the set of integers under addition have
the additional property that m + n = n + m and therefore form an abelian group. □
Most of the time we will write ab instead of a ◦ b; however, if the group already has a
natural operation such as addition in the integers, we will use that operation. That is, if
we are adding two integers, we still write m + n, −n for the inverse, and 0 for the identity
as usual. We also write m − n instead of m + (−n).
It is often convenient to describe a group in terms of an addition or multiplication table.
Such a table is called a Cayley table.
Example 3.9 The integers mod n form a group under addition modulo n. Consider Z5 ,
consisting of the equivalence classes of the integers 0, 1, 2, 3, and 4. We define the group
operation on Z5 by modular addition. We write the binary operation on the group additively;
that is, we write m + n. The element 0 is the identity of the group and each element in Z5
has an inverse. For instance, 2 + 3 = 3 + 2 = 0. Figure 3.10 is a Cayley table for Z5 . By
Proposition 3.4, Zn = {0, 1, . . . , n − 1} is a group under the binary operation of addition
mod n.
+ 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
Figure 3.10 Cayley table for (Z5 , +)
□
Example 3.11 Not every set with a binary operation is a group. For example, if we let
modular multiplication be the binary operation on Zn , then Zn fails to be a group. The
element 1 acts as a group identity since 1 · k = k · 1 = k for any k ∈ Zn ; however, a
multiplicative inverse for 0 does not exist since 0 · k = k · 0 = 0 for every k in Zn . Even if
we consider the set Zn \ {0}, we still may not have a group. For instance, let 2 ∈ Z6 . Then
CHAPTER 3. GROUPS 35
2 has no multiplicative inverse since
0·2=0 1·2=2
2·2=4 3·2=0
4·2=2 5 · 2 = 4.
By Proposition 3.4, every nonzero k does have an inverse in Zn if k is relatively prime to
n. Denote the set of all such nonzero elements in Zn by U (n). Then U (n) is a group called
the group of units of Zn . Figure 3.12 is a Cayley table for the group U (8).
· 1 3 5 7
1 1 3 5 7
3 3 1 7 5
5 5 7 1 3
7 7 5 3 1
Figure 3.12 Multiplication table for U (8)
□
Example 3.13 The symmetries of an equilateral triangle described in Section 3.1 form
a nonabelian group. As we observed, it is not necessarily true that αβ = βα for two
symmetries α and β. Using Figure 3.7, which is a Cayley table for this group, we can easily
check that the symmetries of an equilateral triangle are indeed a group. We will denote this
group by either S3 or D3 , for reasons that will be explained later. □
Example 3.14 We use M2 (R) to denote the set of all 2 × 2 matrices. Let GL2 (R) be the
subset of M2 (R) consisting of invertible matrices; that is, a matrix
a b
A=
c d
is in GL2 (R) if there exists a matrix A−1 such that AA−1 = A−1 A = I, where I is the 2 × 2
identity matrix. For A to have an inverse is equivalent to requiring that the determinant of
A be nonzero; that is, det A = ad − bc 6= 0. The set of invertible matrices forms a group
called the general linear group. The identity of the group is the identity matrix
1 0
I= .
0 1
The inverse of A ∈ GL2 (R) is
1 d −b
A−1 = .
ad − bc −c a
The product of two invertible matrices is again invertible. Matrix multiplication is associa-
tive, satisfying the other group axiom. For matrices it is not true in general that AB = BA;
hence, GL2 (R) is another example of a nonabelian group. □
Example 3.15 Let
1 0 0 1
1= I=
0 1 −1 0
CHAPTER 3. GROUPS 36
0 i i 0
J= K= ,
i 0 0 −i
where i2 = −1. Then the relations I 2 = J 2 = K 2 = −1, IJ = K, JK = I, KI = J,
JI = −K, KJ = −I, and IK = −J hold. The set Q8 = {±1, ±I, ±J, ±K} is a group
called the quaternion group. Notice that Q8 is noncommutative. □
Example 3.16 Let C∗ be the set of nonzero complex numbers. Under the operation of
multiplication C∗ forms a group. The identity is 1. If z = a + bi is a nonzero complex
number, then
a − bi
z −1 = 2
a + b2
is the inverse of z. It is easy to see that the remaining group axioms hold. □
A group is finite, or has finite order, if it contains a finite number of elements; otherwise,
the group is said to be infinite or to have infinite order. The order of a finite group is the
number of elements that it contains. If G is a group containing n elements, we write
|G| = n. The group Z5 is a finite group of order 5; the integers Z form an infinite group
under addition, and we sometimes write |Z| = ∞.
Basic Properties of Groups
Proposition 3.17 The identity element in a group G is unique; that is, there exists only
one element e ∈ G such that eg = ge = g for all g ∈ G.
Proof. Suppose that e and e′ are both identities in G. Then eg = ge = g and e′ g = ge′ = g
for all g ∈ G. We need to show that e = e′ . If we think of e as the identity, then ee′ = e′ ;
but if e′ is the identity, then ee′ = e. Combining these two equations, we have e = ee′ = e′ .
■
′ ′′
Inverses in a group are also unique. If g and g are both inverses of an element g
in a group G, then gg ′ = g ′ g = e and gg ′′ = g ′′ g = e. We want to show that g ′ = g ′′ ,
but g ′ = g ′ e = g ′ (gg ′′ ) = (g ′ g)g ′′ = eg ′′ = g ′′ . We summarize this fact in the following
proposition.
Proposition 3.18 If g is any element in a group G, then the inverse of g, denoted by g −1 ,
is unique.
Proposition 3.19 Let G be a group. If a, b ∈ G, then (ab)−1 = b−1 a−1 .
Proof. Let a, b ∈ G. Then abb−1 a−1 = aea−1 = aa−1 = e. Similarly, b−1 a−1 ab = e. But
by the previous proposition, inverses are unique; hence, (ab)−1 = b−1 a−1 . ■
Proposition 3.20 Let G be a group. For any a ∈ G, (a−1 )−1 = a.
Proof. Observe that a−1 (a−1 )−1 = e. Consequently, multiplying both sides of this equa-
tion by a, we have
(a−1 )−1 = e(a−1 )−1 = aa−1 (a−1 )−1 = ae = a.
■
It makes sense to write equations with group elements and group operations. If a and b
are two elements in a group G, does there exist an element x ∈ G such that ax = b? If such
an x does exist, is it unique? The following proposition answers both of these questions
positively.
Proposition 3.21 Let G be a group and a and b be any two elements in G. Then the
equations ax = b and xa = b have unique solutions in G.
CHAPTER 3. GROUPS 37
Proof. Suppose that ax = b. We must show that such an x exists. We can multiply both
sides of ax = b by a−1 to find x = ex = a−1 ax = a−1 b.
To show uniqueness, suppose that x1 and x2 are both solutions of ax = b; then ax1 =
b = ax2 . So x1 = a−1 ax1 = a−1 ax2 = x2 . The proof for the existence and uniqueness of
the solution of xa = b is similar. ■
Proposition 3.22 If G is a group and a, b, c ∈ G, then ba = ca implies b = c and ab = ac
implies b = c.
This proposition tells us that the right and left cancellation laws are true in groups. We
leave the proof as an exercise.
We can use exponential notation for groups just as we do in ordinary algebra. If G is a
group and g ∈ G, then we define g 0 = e. For n ∈ N, we define
gn = g · g · · · g
| {z }
n times
and
g −n = g −1 · g −1 · · · g −1 .
| {z }
n times
Theorem 3.23 In a group, the usual laws of exponents hold; that is, for all g, h ∈ G,
1. g m g n = g m+n for all m, n ∈ Z;
2. (g m )n = g mn for all m, n ∈ Z;
3. (gh)n = (h−1 g −1 )−n for all n ∈ Z. Furthermore, if G is abelian, then (gh)n = g n hn .
We will leave the proof of this theorem as an exercise. Notice that (gh)n 6= g n hn in
general, since the group may not be abelian. If the group is Z or Zn , we write the group
operation additively and the exponential operation multiplicatively; that is, we write ng
instead of g n . The laws of exponents now become
1. mg + ng = (m + n)g for all m, n ∈ Z;
2. m(ng) = (mn)g for all m, n ∈ Z;
3. m(g + h) = mg + mh for all n ∈ Z.
It is important to realize that the last statement can be made only because Z and Zn
are commutative groups.
Historical Note
Although the first clear axiomatic definition of a group was not given until the late 1800s,
group-theoretic methods had been employed before this time in the development of many
areas of mathematics, including geometry and the theory of algebraic equations.
Joseph-Louis Lagrange used group-theoretic methods in a 1770–1771 memoir to study meth-
ods of solving polynomial equations. Later, Évariste Galois (1811–1832) succeeded in devel-
oping the mathematics necessary to determine exactly which polynomial equations could be
solved in terms of the coefficients of the polynomial. Galois’ primary tool was group theory.
The study of geometry was revolutionized in 1872 when Felix Klein proposed that geo-
metric spaces should be studied by examining those properties that are invariant under
a transformation of the space. Sophus Lie, a contemporary of Klein, used group theory
to study solutions of partial differential equations. One of the first modern treatments of
group theory appeared in William Burnside’s The Theory of Groups of Finite Order [1],
first published in 1897.
CHAPTER 3. GROUPS 38
3.3 Subgroups
Definitions and Examples
Sometimes we wish to investigate smaller groups sitting inside a larger group. The set of
even integers 2Z = {. . . , −2, 0, 2, 4, . . .} is a group under the operation of addition. This
smaller group sits naturally inside of the group of integers under addition. We define a
subgroup H of a group G to be a subset H of G such that when the group operation of
G is restricted to H, H is a group in its own right. Observe that every group G with at
least two elements will always have at least two subgroups, the subgroup consisting of the
identity element alone and the entire group itself. The subgroup H = {e} of a group G
is called the trivial subgroup. A subgroup that is a proper subset of G is called a proper
subgroup. In many of the examples that we have investigated up to this point, there exist
other subgroups besides the trivial and improper subgroups.
Example 3.24 Consider the set of nonzero real numbers, R∗ , with the group operation of
multiplication. The identity of this group is 1 and the inverse of any element a ∈ R∗ is just
1/a. We will show that
Q∗ = {p/q : p and q are nonzero integers}
is a subgroup of R∗ . The identity of R∗ is 1; however, 1 = 1/1 is the quotient of two nonzero
integers. Hence, the identity of R∗ is in Q∗ . Given two elements in Q∗ , say p/q and r/s,
their product pr/qs is also in Q∗ . The inverse of any element p/q ∈ Q∗ is again in Q∗ since
(p/q)−1 = q/p. Since multiplication in R∗ is associative, multiplication in Q∗ is associative.
□
Example 3.25 Recall that C∗ is the multiplicative group of nonzero complex numbers. Let
H = {1, −1, i, −i}. Then H is a subgroup of C∗ . It is quite easy to verify that H is a group
under multiplication and that H ⊂ C∗ . □
Example 3.26 Let SL2 (R) be the subset of GL2 (R) consisting of matrices of determinant
one; that is, a matrix
a b
A=
c d
is in SL2 (R) exactly when ad − bc = 1. To show that SL2 (R) is a subgroup of the general
linear group, we must show that it is a group under matrix multiplication. The 2×2 identity
matrix is in SL2 (R), as is the inverse of the matrix A:
−1 d −b
A = .
−c a
It remains to show that multiplication is closed; that is, that the product of two matrices
of determinant one also has determinant one. We will leave this task as an exercise. The
group SL2 (R) is called the special linear group. □
Example 3.27 It is important to realize that a subset H of a group G can be a group
without being a subgroup of G. For H to be a subgroup of G, it must inherit the binary
operation of G. The set of all 2 × 2 matrices, M2 (R), forms a group under the operation of
addition. The 2 × 2 general linear group is a subset of M2 (R) and is a group under matrix
multiplication, but it is not a subgroup of M2 (R). If we add two invertible matrices, we do
CHAPTER 3. GROUPS 39
not necessarily obtain another invertible matrix. Observe that
1 0 −1 0 0 0
+ = ,
0 1 0 −1 0 0
but the zero matrix is not in GL2 (R). □
Example 3.28 One way of telling whether or not two groups are the same is by examining
their subgroups. Other than the trivial subgroup and the group itself, the group Z4 has
a single subgroup consisting of the elements 0 and 2. From the group Z2 , we can form
another group of four elements as follows. As a set this group is Z2 × Z2 . We perform the
group operation coordinatewise; that is, (a, b) + (c, d) = (a + c, b + d). Figure 3.29 is an
addition table for Z2 × Z2 . Since there are three nontrivial proper subgroups of Z2 × Z2 ,
H1 = {(0, 0), (0, 1)}, H2 = {(0, 0), (1, 0)}, and H3 = {(0, 0), (1, 1)}, Z4 and Z2 × Z2 must be
different groups.
+ (0, 0) (0, 1) (1, 0) (1, 1)
(0, 0) (0, 0) (0, 1) (1, 0) (1, 1)
(0, 1) (0, 1) (0, 0) (1, 1) (1, 0)
(1, 0) (1, 0) (1, 1) (0, 0) (0, 1)
(1, 1) (1, 1) (1, 0) (0, 1) (0, 0)
Figure 3.29 Addition table for Z2 × Z2
□
Some Subgroup Theorems
Let us examine some criteria for determining exactly when a subset of a group is a subgroup.
Proposition 3.30 A subset H of G is a subgroup if and only if it satisfies the following
conditions.
1. The identity e of G is in H.
2. If h1 , h2 ∈ H, then h1 h2 ∈ H.
3. If h ∈ H, then h−1 ∈ H.
Proof. First suppose that H is a subgroup of G. We must show that the three conditions
hold. Since H is a group, it must have an identity eH . We must show that eH = e, where e is
the identity of G. We know that eH eH = eH and that eeH = eH e = eH ; hence, eeH = eH eH .
By right-hand cancellation, e = eH . The second condition holds since a subgroup H is a
group. To prove the third condition, let h ∈ H. Since H is a group, there is an element
h′ ∈ H such that hh′ = h′ h = e. By the uniqueness of the inverse in G, h′ = h−1 .
Conversely, if the three conditions hold, we must show that H is a group under the same
operation as G; however, these conditions plus the associativity of the binary operation are
exactly the axioms stated in the definition of a group. ■
Proposition 3.31 Let H be a subset of a group G. Then H is a subgroup of G if and only
if H 6= ∅, and whenever g, h ∈ H then gh−1 is in H.
Proof. First assume that H is a subgroup of G. We wish to show that gh−1 ∈ H whenever
g and h are in H. Since h is in H, its inverse h−1 must also be in H. Because of the closure
of the group operation, gh−1 ∈ H.
Conversely, suppose that H ⊂ G such that H 6= ∅ and gh−1 ∈ H whenever g, h ∈ H. If
g ∈ H, then gg −1 = e is in H. If g ∈ H, then eg −1 = g −1 is also in H. Now let h1 , h2 ∈ H.
CHAPTER 3. GROUPS 40
We must show that their product is also in H. However, h1 (h−1
2 )
−1 = h h ∈ H. Hence, H
1 2
is a subgroup of G. ■
Sage. The first half of this text is about group theory. Sage includes Groups, Algorithms
and Programming (gap), a program designed primarly for just group theory, and in con-
tinuous development since 1986. Many of Sage’s computations for groups ultimately are
performed by GAP.
3.4 Reading Questions
1. In the group Z8 compute, (a) 6 + 7, and (b) 2−1 .
2. In the group U (16) compute, (a) 5 · 7, and (b) 3−1 .
3. State the definition of a group.
4. Explain a single method that will decide if a subset of a group is itself a subgroup.
5. Explain the origin of the term “abelian” for a commutative group.
6. Give an example of a group you have seen in your previous mathematical experience,
but that is not an example in this chapter.
3.5 Exercises
1. Find all x ∈ Z satisfying each of the following equations.
(a) 3x ≡ 2 (mod 7) (d) 9x ≡ 3 (mod 5)
(b) 5x + 1 ≡ 13 (mod 23) (e) 5x ≡ 1 (mod 6)
(c) 5x + 1 ≡ 13 (mod 26) (f) 3x ≡ 1 (mod 6)
2. Which of the following multiplication tables defined on the set G = {a, b, c, d} form a
group? Support your answer in each case.
(a) (c)
◦ a b c d ◦ a b c d
a a c d a a a b c d
b b b c d b b c d a
c c d a b c c d a b
d d a b c d d a b c
(b) (d)
◦ a b c d ◦ a b c d
a a b c d a a b c d
b b a d c b b a c d
c c d a b c c b a d
d d c b a d d d b c
3. Write out Cayley tables for groups formed by the symmetries of a rectangle and for
(Z4 , +). How many elements are in each group? Are the groups the same? Why or
why not?
4. Describe the symmetries of a rhombus and prove that the set of symmetries forms a
group. Give Cayley tables for both the symmetries of a rectangle and the symmetries
of a rhombus. Are the symmetries of a rectangle and those of a rhombus the same?
CHAPTER 3. GROUPS 41
5. Describe the symmetries of a square and prove that the set of symmetries is a group.
Give a Cayley table for the symmetries. How many ways can the vertices of a square be
permuted? Is each permutation necessarily a symmetry of the square? The symmetry
group of the square is denoted by D4 .
6. Give a multiplication table for the group U (12).
7. Let S = R \ {−1} and define a binary operation on S by a ∗ b = a + b + ab. Prove that
(S, ∗) is an abelian group.
8. Give an example of two elements A and B in GL2 (R) with AB 6= BA.
9. Prove that the product of two matrices in SL2 (R) has determinant one.
10. Prove that the set of matrices of the form
1 x y
0 1 z
0 0 1
is a group under matrix multiplication. This group, known as the Heisenberg group,
is important in quantum physics. Matrix multiplication in the Heisenberg group is
defined by
1 x y 1 x′ y ′ 1 x + x′ y + y ′ + xz ′
0 1 z 0 1 z ′ = 0 1 z + z′ .
0 0 1 0 0 1 0 0 1
11. Prove that det(AB) = det(A) det(B) in GL2 (R). Use this result to show that the
binary operation in the group GL2 (R) is closed; that is, if A and B are in GL2 (R),
then AB ∈ GL2 (R).
12. Let Zn2 = {(a1 , a2 , . . . , an ) : ai ∈ Z2 }. Define a binary operation on Zn2 by
(a1 , a2 , . . . , an ) + (b1 , b2 , . . . , bn ) = (a1 + b1 , a2 + b2 , . . . , an + bn ).
Prove that Zn2 is a group under this operation. This group is important in algebraic
coding theory.
13. Show that R∗ = R \ {0} is a group under the operation of multiplication.
14. Given the groups R∗ and Z, let G = R∗ × Z. Define a binary operation ◦ on G by
(a, m) ◦ (b, n) = (ab, m + n). Show that G is a group under this operation.
15. Prove or disprove that every group containing six elements is abelian.
16. Give a specific example of some group G and elements g, h ∈ G where (gh)n 6= g n hn .
17. Give an example of three different groups with eight elements. Why are the groups
different?
18. Show that there are n! permutations of a set containing n items.
19. Show that
0 + a ≡ a + 0 ≡ a (mod n)
for all a ∈ Zn .
20. Prove that there is a multiplicative identity for the integers modulo n:
a · 1 ≡ a (mod n).
21. For each a ∈ Zn find an element b ∈ Zn such that
a + b ≡ b + a ≡ 0 (mod n).
CHAPTER 3. GROUPS 42
22. Show that addition and multiplication mod n are well defined operations. That is,
show that the operations do not depend on the choice of the representative from the
equivalence classes mod n.
23. Show that addition and multiplication mod n are associative operations.
24. Show that multiplication distributes over addition modulo n:
a(b + c) ≡ ab + ac (mod n).
25. Let a and b be elements in a group G. Prove that abn a−1 = (aba−1 )n for n ∈ Z.
26. Let U (n) be the group of units in Zn . If n > 2, prove that there is an element k ∈ U (n)
such that k 2 = 1 and k 6= 1.
−1
27. Prove that the inverse of g1 g2 · · · gn is gn−1 gn−1 · · · g1−1 .
28. Prove the remainder of Proposition 3.21: if G is a group and a, b ∈ G, then the equation
xa = b has a unique solution in G.
29. Prove Theorem 3.23.
30. Prove the right and left cancellation laws for a group G; that is, show that in the group
G, ba = ca implies b = c and ab = ac implies b = c for elements a, b, c ∈ G.
31. Show that if a2 = e for all elements a in a group G, then G must be abelian.
32. Show that if G is a finite group of even order, then there is an a ∈ G such that a is
not the identity and a2 = e.
33. Let G be a group and suppose that (ab)2 = a2 b2 for all a and b in G. Prove that G is
an abelian group.
34. Find all the subgroups of Z3 × Z3 . Use this information to show that Z3 × Z3 is not the
same group as Z9 . (See Example 3.28 for a short description of the product of groups.)
35. Find all the subgroups of the symmetry group of an equilateral triangle.
36. Compute the subgroups of the symmetry group of a square.
37. Let H = {2k : k ∈ Z}. Show that H is a subgroup of Q∗ .
38. Let n = 0, 1, 2, . . . and nZ = {nk : k ∈ Z}. Prove that nZ is a subgroup of Z. Show
that these subgroups are the only subgroups of Z.
39. Let T = {z ∈ C∗ : |z| = 1}. Prove that T is a subgroup of C∗ .
40. Let G consist of the 2 × 2 matrices of the form
cos θ − sin θ
,
sin θ cos θ
where θ ∈ R. Prove that G is a subgroup of SL2 (R).
41. Prove that
√
G = {a + b 2 : a, b ∈ Q and a and b are not both zero}
is a subgroup of R∗ under the group operation of multiplication.
42. Let G be the group of 2 × 2 matrices under addition and
a b
H= :a+d=0 .
c d
Prove that H is a subgroup of G.
CHAPTER 3. GROUPS 43
43. Prove or disprove: SL2 (Z), the set of 2×2 matrices with integer entries and determinant
one, is a subgroup of SL2 (R).
44. List the subgroups of the quaternion group, Q8 .
45. Prove that the intersection of two subgroups of a group G is also a subgroup of G.
46. Prove or disprove: If H and K are subgroups of a group G, then H ∪ K is a subgroup
of G.
47. Prove or disprove: If H and K are subgroups of a group G, then HK = {hk : h ∈
H and k ∈ K} is a subgroup of G. What if G is abelian?
48. Let G be a group and g ∈ G. Show that
Z(G) = {x ∈ G : gx = xg for all g ∈ G}
is a subgroup of G. This subgroup is called the center of G.
49. Let a and b be elements of a group G. If a4 b = ba and a3 = e, prove that ab = ba.
50. Give an example of an infinite group in which every nontrivial subgroup is infinite.
51. If xy = x−1 y −1 for all x and y in G, prove that G must be abelian.
52. Prove or disprove: Every proper subgroup of a nonabelian group is nonabelian.
53. Let H be a subgroup of G and
C(H) = {g ∈ G : gh = hg for all h ∈ H}.
Prove C(H) is a subgroup of G. This subgroup is called the centralizer of H in G.
54. Let H be a subgroup of G. If g ∈ G, show that gHg −1 = {ghg −1 : h ∈ H} is also a
subgroup of G.
3.6 Additional Exercises: Detecting Errors
1. UPC Symbols. Universal Product Code (upc) symbols are found on most prod-
ucts in grocery and retail stores. The upc symbol is a 12-digit code identifying the
manufacturer of a product and the product itself (Figure 3.32). The first 11 digits
contain information about the product; the twelfth digit is used for error detection. If
d1 d2 · · · d12 is a valid upc number, then
3 · d1 + 1 · d2 + 3 · d3 + · · · + 3 · d11 + 1 · d12 ≡ 0 (mod 10).
(a) Show that the upc number 0-50000-30042-6, which appears in Figure 3.32, is a
valid upc number.
(b) Show that the number 0-50000-30043-6 is not a valid upc number.
(c) Write a formula to calculate the check digit, d12 , in the upc number.
(d) The upc error detection scheme can detect most transposition errors; that is, it
can determine if two digits have been interchanged. Show that the transposition
error 0-05000-30042-6 is not detected. Find a transposition error that is detected.
Can you find a general rule for the types of transposition errors that can be
detected?
(e) Write a program that will determine whether or not a upc number is valid.
CHAPTER 3. GROUPS 44
0 50000 30042 6
Figure 3.32 A upc code
2. It is often useful to use an inner product notation for this type of error detection
scheme; hence, we will use the notion
(d1 , d2 , . . . , dk ) · (w1 , w2 , . . . , wk ) ≡ 0 (mod n)
to mean
d1 w1 + d2 w2 + · · · + dk wk ≡ 0 (mod n).
Suppose that (d1 , d2 , . . . , dk ) · (w1 , w2 , . . . , wk ) ≡ 0 (mod n) is an error detection
scheme for the k-digit identification number d1 d2 · · · dk , where 0 ≤ di < n. Prove that
all single-digit errors are detected if and only if gcd(wi , n) = 1 for 1 ≤ i ≤ k.
3. Let (d1 , d2 , . . . , dk ) · (w1 , w2 , . . . , wk ) ≡ 0 (mod n) be an error detection scheme for the
k-digit identification number d1 d2 · · · dk , where 0 ≤ di < n. Prove that all transposition
errors of two digits di and dj are detected if and only if gcd(wi − wj , n) = 1 for i and
j between 1 and k.
4. ISBN Codes. Every book has an International Standard Book Number (isbn) code.
This is a 10-digit code indicating the book’s publisher and title. The tenth digit is a
check digit satisfying
(d1 , d2 , . . . , d10 ) · (10, 9, . . . , 1) ≡ 0 (mod 11).
One problem is that d10 might have to be a 10 to make the inner product zero; in this
case, 11 digits would be needed to make this scheme work. Therefore, the character X
is used for the eleventh digit. So isbn 3-540-96035-X is a valid isbn code.
(a) Is isbn 0-534-91500-0 a valid isbn code? What about isbn 0-534-91700-0 and
isbn 0-534-19500-0?
(b) Does this method detect all single-digit errors? What about all transposition
errors?
(c) How many different isbn codes are there?
(d) Write a computer program that will calculate the check digit for the first nine
digits of an isbn code.
(e) A publisher has houses in Germany and the United States. Its German prefix is 3-
540. If its United States prefix will be 0-abc, find abc such that the rest of the isbn
code will be the same for a book printed in Germany and in the United States.
Under the isbn coding method the first digit identifies the language; German is
3 and English is 0. The next group of numbers identifies the publisher, and the
last group identifies the specific book.
CHAPTER 3. GROUPS 45
3.7 References and Suggested Readings
[1] Burnside, W. Theory of Groups of Finite Order. 2nd ed. Cambridge University Press,
Cambridge, 1911; Dover, New York, 1953. A classic. Also available at books.google.com.
[2] Gallian, J. A. and Winters, S. “Modular Arithmetic in the Marketplace,” The Amer-
ican Mathematical Monthly 95 (1988): 548–51.
[3] Gallian, J. A. Contemporary Abstract Algebra. 7th ed. Brooks/Cole, Belmont, CA,
2009.
[4] Hall, M. Theory of Groups. 2nd ed. American Mathematical Society, Providence,
1959.
[5] Kurosh, A. E. The Theory of Groups, vols. I and II. American Mathematical Society,
Providence, 1979.
[6] Rotman, J. J. An Introduction to the Theory of Groups. 4th ed. Springer, New York,
1995.
4
Cyclic Groups
The groups Z and Zn , which are among the most familiar and easily understood groups, are
both examples of what are called cyclic groups. In this chapter we will study the properties
of cyclic groups and cyclic subgroups, which play a fundamental part in the classification
of all abelian groups.
4.1 Cyclic Subgroups
Often a subgroup will depend entirely on a single element of the group; that is, knowing
that particular element will allow us to compute any other element in the subgroup.
Example 4.1 Suppose that we consider 3 ∈ Z and look at all multiples (both positive and
negative) of 3. As a set, this is
3Z = {. . . , −3, 0, 3, 6, . . .}.
It is easy to see that 3Z is a subgroup of the integers. This subgroup is completely deter-
mined by the element 3 since we can obtain all of the other elements of the group by taking
multiples of 3. Every element in the subgroup is “generated” by 3. □
Example 4.2 If H = {2n : n ∈ Z}, then H is a subgroup of the multiplicative group of
nonzero rational numbers, Q∗ . If a = 2m and b = 2n are in H, then ab−1 = 2m 2−n = 2m−n
is also in H. By Proposition 3.31, H is a subgroup of Q∗ determined by the element 2. □
Theorem 4.3 Let G be a group and a be any element in G. Then the set
hai = {ak : k ∈ Z}
is a subgroup of G. Furthermore, hai is the smallest subgroup of G that contains a.
Proof. The identity is in hai since a0 = e. If g and h are any two elements in hai, then
by the definition of hai we can write g = am and h = an for some integers m and n. So
gh = am an = am+n is again in hai. Finally, if g = an in hai, then the inverse g −1 = a−n is
also in hai. Clearly, any subgroup H of G containing a must contain all the powers of a by
closure; hence, H contains hai. Therefore, hai is the smallest subgroup of G containing a.
■
Remark 4.4 If we are using the “+” notation, as in the case of the integers under addition,
we write hai = {na : n ∈ Z}.
For a ∈ G, we call hai the cyclic subgroup generated by a. If G contains some element
a such that G = hai, then G is a cyclic group. In this case a is a generator of G. If a is an
46
CHAPTER 4. CYCLIC GROUPS 47
element of a group G, we define the order of a to be the smallest positive integer n such
that an = e, and we write |a| = n. If there is no such integer n, we say that the order of a
is infinite and write |a| = ∞ to denote the order of a.
Example 4.5 Notice that a cyclic group can have more than a single generator. Both 1 and
5 generate Z6 ; hence, Z6 is a cyclic group. Not every element in a cyclic group is necessarily
a generator of the group. The order of 2 ∈ Z6 is 3. The cyclic subgroup generated by 2 is
h2i = {0, 2, 4}. □
The groups Z and Zn are cyclic groups. The elements 1 and −1 are generators for Z.
We can certainly generate Zn with 1 although there may be other generators of Zn , as in
the case of Z6 .
Example 4.6 The group of units, U (9), in Z9 is a cyclic group. As a set, U (9) is
{1, 2, 4, 5, 7, 8}. The element 2 is a generator for U (9) since
21 = 2 22 = 4
23 = 8 24 = 7
25 = 5 26 = 1.
□
Example 4.7 Not every group is a cyclic group. Consider the symmetry group of an
equilateral triangle S3 . The multiplication table for this group is Figure 3.7. The subgroups
of S3 are shown in Figure 4.8. Notice that every subgroup is cyclic; however, no single
element generates the entire group.
S3
{id, ρ1 , ρ2 } {id, µ1 } {id, µ2 } {id, µ3 }
{id}
Figure 4.8 Subgroups of S3
□
Theorem 4.9 Every cyclic group is abelian.
Proof. Let G be a cyclic group and a ∈ G be a generator for G. If g and h are in G, then
they can be written as powers of a, say g = ar and h = as . Since
gh = ar as = ar+s = as+r = as ar = hg,
G is abelian. ■
CHAPTER 4. CYCLIC GROUPS 48
Subgroups of Cyclic Groups
We can ask some interesting questions about cyclic subgroups of a group and subgroups of
a cyclic group. If G is a group, which subgroups of G are cyclic? If G is a cyclic group,
what type of subgroups does G possess?
Theorem 4.10 Every subgroup of a cyclic group is cyclic.
Proof. The main tools used in this proof are the division algorithm and the Principle of
Well-Ordering. Let G be a cyclic group generated by a and suppose that H is a subgroup
of G. If H = {e}, then trivially H is cyclic. Suppose that H contains some other element
g distinct from the identity. Then g can be written as an for some integer n. Since H is a
subgroup, g −1 = a−n must also be in H. Since either n or −n is positive, we can assume
that H contains positive powers of a and n > 0. Let m be the smallest natural number
such that am ∈ H. Such an m exists by the Principle of Well-Ordering.
We claim that h = am is a generator for H. We must show that every h′ ∈ H can be
written as a power of h. Since h′ ∈ H and H is a subgroup of G, h′ = ak for some integer
k. Using the division algorithm, we can find numbers q and r such that k = mq + r where
0 ≤ r < m; hence,
ak = amq+r = (am )q ar = hq ar .
So ar = ak h−q . Since ak and h−q are in H, ar must also be in H. However, m was the
smallest positive number such that am was in H; consequently, r = 0 and so k = mq.
Therefore,
h′ = ak = amq = hq
and H is generated by h. ■
Corollary 4.11 The subgroups of Z are exactly nZ for n = 0, 1, 2, . . ..
Proposition 4.12 Let G be a cyclic group of order n and suppose that a is a generator for
G. Then ak = e if and only if n divides k.
Proof. First suppose that ak = e. By the division algorithm, k = nq + r where 0 ≤ r < n;
hence,
e = ak = anq+r = anq ar = ear = ar .
Since the smallest positive integer m such that am = e is n, r = 0.
Conversely, if n divides k, then k = ns for some integer s. Consequently,
ak = ans = (an )s = es = e.
■
Theorem 4.13 Let G be a cyclic group of order n and suppose that a ∈ G is a generator
of the group. If b = ak , then the order of b is n/d, where d = gcd(k, n).
Proof. We wish to find the smallest integer m such that e = bm = akm . By Proposi-
tion 4.12, this is the smallest integer m such that n divides km or, equivalently, n/d divides
m(k/d). Since d is the greatest common divisor of n and k, n/d and k/d are relatively
prime. Hence, for n/d to divide m(k/d) it must divide m. The smallest such m is n/d. ■
Corollary 4.14 The generators of Zn are the integers r such that 1 ≤ r < n and gcd(r, n) =
1.
Example 4.15 Let us examine the group Z16 . The numbers 1, 3, 5, 7, 9, 11, 13, and 15
are the elements of Z16 that are relatively prime to 16. Each of these elements generates
Z16 . For example,
1·9=9 2·9=2 3 · 9 = 11
CHAPTER 4. CYCLIC GROUPS 49
4·9=4 5 · 9 = 13 6·9=6
7 · 9 = 15 8·9=8 9·9=1
10 · 9 = 10 11 · 9 = 3 12 · 9 = 12
13 · 9 = 5 14 · 9 = 14 15 · 9 = 7.
□
4.2 Multiplicative Group of Complex Numbers
The complex numbers are defined as
C = {a + bi : a, b ∈ R},
where i2 = −1. If z = a + bi, then a is the real part of z and b is the imaginary part of z.
To add two complex numbers z = a + bi and w = c + di, we just add the corresponding
real and imaginary parts:
z + w = (a + bi) + (c + di) = (a + c) + (b + d)i.
Remembering that i2 = −1, we multiply complex numbers just like polynomials. The
product of z and w is
(a + bi)(c + di) = ac + bdi2 + adi + bci = (ac − bd) + (ad + bc)i.
Every nonzero complex number z = a + bi has a multiplicative inverse; that is, there
exists a z −1 ∈ C∗ such that zz −1 = z −1 z = 1. If z = a + bi, then
a − bi
z −1 = .
a2 + b2
The complex conjugate of a complex number z √ = a + bi is defined to be z = a − bi. The
absolute value or modulus of z = a + bi is |z| = a2 + b2 .
Example 4.16 Let z = 2 + 3i and w = 1 − 2i. Then
z + w = (2 + 3i) + (1 − 2i) = 3 + i
and
zw = (2 + 3i)(1 − 2i) = 8 − i.
Also,
2 3
z −1 = − i
13 13
√
|z| = 13
z = 2 − 3i.
□
CHAPTER 4. CYCLIC GROUPS 50
y
z1 = 2 + 3i
z3 = −3 + 2i
0 x
z2 = 1 − 2i
Figure 4.17 Rectangular coordinates of a complex number
There are several ways of graphically representing complex numbers. We can represent
a complex number z = a + bi as an ordered pair on the xy plane where a is the x (or
real) coordinate and b is the y (or imaginary) coordinate. This is called the rectangular or
Cartesian representation. The rectangular representations of z1 = 2 + 3i, z2 = 1 − 2i, and
z3 = −3 + 2i are depicted in Figure 4.17.
y
a + bi
r
θ
0 x
Figure 4.18 Polar coordinates of a complex number
Nonzero complex numbers can also be represented using polar coordinates. To specify
CHAPTER 4. CYCLIC GROUPS 51
any nonzero point on the plane, it suffices to give an angle θ from the positive x axis in the
counterclockwise direction and a distance r from the origin, as in Figure 4.18. We can see
that
z = a + bi = r(cos θ + i sin θ).
Hence, p
r = |z| = a2 + b2
and
a = r cos θ
b = r sin θ.
We sometimes abbreviate r(cos θ + i sin θ) as r cis θ. To assure that the representation of z
is well-defined, we also require that 0◦ ≤ θ < 360◦ . If the measurement is in radians, then
0 ≤ θ < 2π.
Example 4.19 Suppose that z = 2 cis 60◦ . Then
a = 2 cos 60◦ = 1
and √
b = 2 sin 60◦ = 3.
√
Hence, the rectangular representation is z = 1 + 3 i.
Conversely, if we are given a rectangular representation √
of a complex
√ number, it is often
useful to know the number’s polar representation. If z = 3 2 − 3 2 i, then
p √
r = a2 + b2 = 36 = 6
and
b
θ = arctan = arctan(−1) = 315◦ ,
a
√ √
so 3 2 − 3 2 i = 6 cis 315◦ . □
The polar representation of a complex number makes it easy to find products and powers
of complex numbers. The proof of the following proposition is straightforward and is left
as an exercise.
Proposition 4.20 Let z = r cis θ and w = s cis ϕ be two nonzero complex numbers. Then
zw = rs cis(θ + ϕ).
Example 4.21 If z = 3 cis(π/3) and w = 2 cis(π/6), then zw = 6 cis(π/2) = 6i. □
Theorem 4.22 DeMoivre. Let z = r cis θ be a nonzero complex number. Then
[r cis θ]n = rn cis(nθ)
for n = 1, 2, . . ..
Proof. We will use induction on n. For n = 1 the theorem is trivial. Assume that the
theorem is true for all k such that 1 ≤ k ≤ n. Then
z n+1 = z n z
= rn (cos nθ + i sin nθ)r(cos θ + i sin θ)
= rn+1 [(cos nθ cos θ − sin nθ sin θ) + i(sin nθ cos θ + cos nθ sin θ)]
= rn+1 [cos(nθ + θ) + i sin(nθ + θ)]
CHAPTER 4. CYCLIC GROUPS 52
= rn+1 [cos(n + 1)θ + i sin(n + 1)θ].
■
Example 4.23 Suppose that z = 1+i and we wish to compute z 10 . Rather than computing
(1 + i)10 directly, it is much easier to switch to polar coordinates and calculate z 10 using
DeMoivre’s Theorem:
z 10 = (1 + i)10
√ π 10
= 2 cis
4
√ 10 5π
= ( 2 ) cis
2
π
= 32 cis
2
= 32i.
□
The Circle Group and the Roots of Unity
The multiplicative group of the complex numbers, C∗ , possesses some interesting subgroups.
Whereas Q∗ and R∗ have no interesting subgroups of finite order, C∗ has many. We first
consider the circle group,
T = {z ∈ C : |z| = 1}.
The following proposition is a direct result of Proposition 4.20.
Proposition 4.24 The circle group is a subgroup of C∗ .
Although the circle group has infinite order, it has many interesting finite subgroups.
Suppose that H = {1, −1, i, −i}. Then H is a subgroup of the circle group. Also, 1, −1, i,
and −i are exactly those complex numbers that satisfy the equation z 4 = 1. The complex
numbers satisfying the equation z n = 1 are called the nth roots of unity.
Theorem 4.25 If z n = 1, then the nth roots of unity are
2kπ
z = cis ,
n
where k = 0, 1, . . . , n − 1. Furthermore, the nth roots of unity form a cyclic subgroup of T
of order n
Proof. By DeMoivre’s Theorem,
n 2kπ
z = cis n = cis(2kπ) = 1.
n
The z’s are distinct since the numbers 2kπ/n are all distinct and are greater than or equal
to 0 but less than 2π. The fact that these are all of the roots of the equation z n = 1 follows
from from Corollary 17.9, which states that a polynomial of degree n can have at most n
roots. We will leave the proof that the nth roots of unity form a cyclic subgroup of T as an
exercise. ■
A generator for the group of the nth roots of unity is called a primitive nth root of
unity.
CHAPTER 4. CYCLIC GROUPS 53
Example 4.26 The 8th roots of unity can be represented as eight equally spaced points on
the unit circle (Figure 4.27). The primitive 8th roots of unity are
√ √
2 2
ω= + i
2√ 2√
2 2
ω3 = − + i
√2 √2
2 2
ω5 = − − i
√ 2 √ 2
2 2
ω7 = − i.
2 2
y
i
ω3 ω
−1 1
x
0
ω5 ω7
−i
Figure 4.27 8th roots of unity
□
4.3 The Method of Repeated Squares
Computing large powers can be very time-consuming. Just as anyone can compute 22 or
28 , everyone knows how to compute
1,000,000
22 .
However, such numbers are so large that we do not want to attempt the calculations;
moreover, past a certain point the computations would not be feasible even if we had
every computer in the world at our disposal. Even writing down the decimal representation
of a very large number may not be reasonable. It could be thousands or even millions of
CHAPTER 4. CYCLIC GROUPS 54
digits long. However, if we could compute something like
237,398,332 (mod 46,389),
we could very easily write the result down since it would be a number between 0 and 46,388.
If we want to compute powers modulo n quickly and efficiently, we will have to be clever.4
The first thing to notice is that any number a can be written as the sum of distinct
powers of 2; that is, we can write
a = 2k1 + 2k2 + · · · + 2kn ,
where k1 < k2 < · · · < kn . This is just the binary representation of a. For example, the
binary representation of 57 is 111001, since we can write 57 = 20 + 23 + 24 + 25 .
The laws of exponents still work in Zn ; that is, if b ≡ ax (mod n) and c ≡ ay (mod n),
k
then bc ≡ ax+y (mod n). We can compute a2 (mod n) in k multiplications by computing
0
a2 (mod n)
21
a (mod n)
..
.
k
a2 (mod n).
Each step involves squaring the answer obtained in the previous step, dividing by n, and
taking the remainder.
Example 4.28 We will compute 271321 (mod 481). Notice that
321 = 20 + 26 + 28 ;
hence, computing 271321 (mod 481) is the same as computing
0 +26 +28 0 6 8
2712 ≡ 2712 · 2712 · 2712 (mod 481).
i
So it will suffice to compute 2712 (mod 481) where i = 0, 6, 8. It is very easy to see that
1
2712 = 73,441 ≡ 329 (mod 481).
2
We can square this result to obtain a value for 2712 (mod 481):
2 1
2712 ≡ (2712 )2 (mod 481)
≡ (329) 2
(mod 481)
≡ 108,241 (mod 481)
≡ 16 (mod 481).
n n n+1
We are using the fact that (a2 )2 ≡ a2·2 ≡ a2 (mod n). Continuing, we can calculate
6
2712 ≡ 419 (mod 481)
and
8
2712 ≡ 16 (mod 481).
Therefore,
0 +26 +28
271321 ≡ 2712 (mod 481)
4
The results in this section are needed only in Chapter 7
CHAPTER 4. CYCLIC GROUPS 55
0 6 8
≡ 2712 · 2712 · 2712 (mod 481)
≡ 271 · 419 · 16 (mod 481)
≡ 1,816,784 (mod 481)
≡ 47 (mod 481).
□
The method of repeated squares will prove to be a very useful tool when we explore rsa
cryptography in Chapter 7. To encode and decode messages in a reasonable manner under
this scheme, it is necessary to be able to quickly compute large powers of integers mod n.
Sage. Sage support for cyclic groups is a little spotty — but we can still make effective
use of Sage and perhaps this situation could change soon.
4.4 Reading Questions
1. What is the order of the element 3 in U (20)?
2. What is the order of the element 5 in U (23)?
3. Find three generators of Z8 .
4. Find three generators of the 5th roots of unity.
5. Show how to compute 1540 (mod 23) efficiently by hand. Check your answer with
Sage.
4.5 Exercises
1. Prove or disprove each of the following statements.
(a) All of the generators of Z60 are prime.
(b) U (8) is cyclic.
(c) Q is cyclic.
(d) If every proper subgroup of a group G is cyclic, then G is a cyclic group.
(e) A group with a finite number of subgroups is finite.
2. Find the order of each of the following elements.
(a) 5 ∈ Z12 (d) −i ∈ C∗
√
(b) 3 ∈ R (e) 72 ∈ Z240
√
(c) 3 ∈ R∗ (f) 312 ∈ Z471
3. List all of the elements in each of the following subgroups.
(a) The subgroup of Z generated by 7
(b) The subgroup of Z24 generated by 15
(c) All subgroups of Z12
(d) All subgroups of Z60
(e) All subgroups of Z13
CHAPTER 4. CYCLIC GROUPS 56
(f) All subgroups of Z48
(g) The subgroup generated by 3 in U (20)
(h) The subgroup generated by 5 in U (18)
(i) The subgroup of R∗ generated by 7
(j) The subgroup of C∗ generated by i where i2 = −1
(k) The subgroup of C∗ generated by 2i
√
(l) The subgroup of C∗ generated by (1 + i)/ 2
√
(m) The subgroup of C∗ generated by (1 + 3 i)/2
4. Find the
subgroups
of GL2 (R) generated
by
each of the following matrices.
0 1 1 −1 1 −1
(a) (c) (e)
−1 0 1 0 −1 0
√
0 1/3 1 −1 3/2 1/2
(b) (d) (f) √
3 0 0 1 −1/2 3/2
5. Find the order of every element in Z18 .
6. Find the order of every element in the symmetry group of the square, D4 .
7. What are all of the cyclic subgroups of the quaternion group, Q8 ?
8. List all of the cyclic subgroups of U (30).
9. List every generator of each subgroup of order 8 in Z32 .
10. Find all elements of finite order in each of the following groups. Here the “∗” indicates
the set with zero removed.
(a) Z (b) Q∗ (c) R∗
11. If a24 = e in a group G, what are the possible orders of a?
12. Find a cyclic group with exactly one generator. Can you find cyclic groups with exactly
two generators? Four generators? How about n generators?
13. For n ≤ 20, which groups U (n) are cyclic? Make a conjecture as to what is true in
general. Can you prove your conjecture?
14. Let
0 1 0 −1
A= and B=
−1 0 1 −1
be elements in GL2 (R). Show that A and B have finite orders but AB does not.
15. Evaluate each of the following.
(a) (3 − 2i) + (5i − 6) (d) (9 − i)(9 − i)
(b) (4 − 5i) − (4i − 4) (e) i45
(c) (5 − 4i)(7 + 2i) (f) (1 + i) + (1 + i)
16. Convert the following complex numbers to the form a + bi.
(a) 2 cis(π/6) (c) 3 cis(π)
(b) 5 cis(9π/4) (d) cis(7π/4)/2
CHAPTER 4. CYCLIC GROUPS 57
17. Change the following complex numbers to polar representation.
(a) 1 − i (c) 2 + 2i (e) −3i
√ √
(b) −5 (d) 3 + i (f) 2i + 2 3
18. Calculate each of the following expressions.
(a) (1 + i)−1 (e) ((1 − i)/2)4
(b) (1 − i)6 √ √
√ (f) (− 2 − 2 i)12
(c) ( 3 + i)5
(d) (−i)10 (g) (−2 + 2i)−5
19. Prove each of the following statements.
(a) |z| = |z| (d) |z + w| ≤ |z| + |w|
(b) zz = |z|2 (e) |z − w| ≥ ||z| − |w||
(c) z −1 = z/|z|2 (f) |zw| = |z||w|
20. List and graph the 6th roots of unity. What are the generators of this group? What
are the primitive 6th roots of unity?
21. List and graph the 5th roots of unity. What are the generators of this group? What
are the primitive 5th roots of unity?
22. Calculate each of the following.
(a) 2923171 (mod 582) (c) 20719521 (mod 4724)
(b) 2557341 (mod 5681) (d) 971321 (mod 765)
23. Let a, b ∈ G. Prove the following statements.
(a) The order of a is the same as the order of a−1 .
(b) For all g ∈ G, |a| = |g −1 ag|.
(c) The order of ab is the same as the order of ba.
24. Let p and q be distinct primes. How many generators does Zpq have?
25. Let p be prime and r be a positive integer. How many generators does Zpr have?
26. Prove that Zp has no nontrivial subgroups if p is prime.
27. If g and h have orders 15 and 16 respectively in a group G, what is the order of hgi∩hhi?
28. Let a be an element in a group G. What is a generator for the subgroup ham i ∩ han i?
29. Prove that Zn has an even number of generators for n > 2.
30. Suppose that G is a group and let a, b ∈ G. Prove that if |a| = m and |b| = n with
gcd(m, n) = 1, then hai ∩ hbi = {e}.
31. Let G be an abelian group. Show that the elements of finite order in G form a subgroup.
This subgroup is called the torsion subgroup of G.
32. Let G be a finite cyclic group of order n generated by x. Show that if y = xk where
gcd(k, n) = 1, then y must be a generator of G.
33. If G is an abelian group that contains a pair of cyclic subgroups of order 2, show that
G must contain a subgroup of order 4. Does this subgroup have to be cyclic?
34. Let G be an abelian group of order pq where gcd(p, q) = 1. If G contains elements a
and b of order p and q respectively, then show that G is cyclic.
35. Prove that the subgroups of Z are exactly nZ for n = 0, 1, 2, . . ..
CHAPTER 4. CYCLIC GROUPS 58
36. Prove that the generators of Zn are the integers r such that 1 ≤ r < n and gcd(r, n) = 1.
37. Prove that if G has no proper nontrivial subgroups, then G is a cyclic group.
38. Prove that the order of an element in a cyclic group G must divide the order of the
group.
39. Prove that if G is a cyclic group of order m and d | m, then G must have a subgroup
of order d.
40. For what integers n is −1 an nth root of unity?
41. If z = r(cos θ + i sin θ) and w = s(cos ϕ + i sin ϕ) are two nonzero complex numbers,
show that
zw = rs[cos(θ + ϕ) + i sin(θ + ϕ)].
42. Prove that the circle group is a subgroup of C∗ .
43. Prove that the nth roots of unity form a cyclic subgroup of T of order n.
44. Let α ∈ T. Prove that αm = 1 and αn = 1 if and only if αd = 1 for d = gcd(m, n).
45. Let z ∈ C∗ . If |z| 6= 1, prove that the order of z is infinite.
46. Let z = cos θ + i sin θ be in T where θ ∈ Q. Prove that the order of z is infinite.
4.6 Programming Exercises
1. Write a computer program that will write any decimal number as the sum of distinct
powers of 2. What is the largest integer that your program will handle?
2. Write a computer program to calculate ax (mod n) by the method of repeated squares.
What are the largest values of n and x that your program will accept?
4.7 References and Suggested Readings
[1] Koblitz, N. A Course in Number Theory and Cryptography. 2nd ed. Springer, New
York, 1994.
[2] Pomerance, C. “Cryptology and Computational Number Theory—An Introduction,”
in Cryptology and Computational Number Theory, Pomerance, C., ed. Proceedings of
Symposia in Applied Mathematics, vol. 42, American Mathematical Society, Provi-
dence, RI, 1990. This book gives an excellent account of how the method of repeated
squares is used in cryptography.
5
Permutation Groups
Permutation groups are central to the study of geometric symmetries and to Galois the-
ory, the study of finding solutions of polynomial equations. They also provide abundant
examples of nonabelian groups.
Let us recall for a moment the symmetries of the equilateral triangle 4ABC from
Chapter 3. The symmetries actually consist of permutations of the three vertices, where a
permutation of the set S = {A, B, C} is a one-to-one and onto map π : S → S. The three
vertices have the following six permutations.
A B C A B C A B C
A B C C A B B C A
A B C A B C A B C
A C B C B A B A C
We have used the array
A B C
B C A
to denote the permutation that sends A to B, B to C, and C to A. That is,
A 7→ B
B 7→ C
C 7→ A.
The symmetries of a triangle form a group. In this chapter we will study groups of this
type.
5.1 Definitions and Notation
In general, the permutations of a set X form a group SX . If X is a finite set, we can assume
X = {1, 2, . . . , n}. In this case we write Sn instead of SX . The following theorem says that
Sn is a group. We call this group the symmetric group on n letters.
Theorem 5.1 The symmetric group on n letters, Sn , is a group with n! elements, where
the binary operation is the composition of maps.
Proof. The identity of Sn is just the identity map that sends 1 to 1, 2 to 2, . . ., n to n. If
f : Sn → Sn is a permutation, then f −1 exists, since f is one-to-one and onto; hence, every
permutation has an inverse. Composition of maps is associative, which makes the group
59
CHAPTER 5. PERMUTATION GROUPS 60
operation associative. We leave the proof that |Sn | = n! as an exercise. ■
A subgroup of Sn is called a permutation group.
Example 5.2 Consider the subgroup G of S5 consisting of the identity permutation id and
the permutations
1 2 3 4 5
σ=
1 2 3 5 4
1 2 3 4 5
τ=
3 2 1 4 5
1 2 3 4 5
µ= .
3 2 1 5 4
The following table tells us how to multiply elements in the permutation group G.
◦ id σ τ µ
id id σ τ µ
σ σ id µ τ
τ τ µ id σ
µ µ τ σ id
□
Remark 5.3 Though it is natural to multiply elements in a group from left to right,
functions are composed from right to left. Let σ and τ be permutations on a set X. To
compose σ and τ as functions, we calculate (σ ◦ τ )(x) = σ(τ (x)). That is, we do τ first, then
σ. There are several ways to approach this inconsistency. We will adopt the convention of
multiplying permutations right to left. To compute στ , do τ first and then σ. That is, by
στ (x) we mean σ(τ (x)). (Another way of solving this problem would be to write functions
on the right; that is, instead of writing σ(x), we could write (x)σ. We could also multiply
permutations left to right to agree with the usual way of multiplying elements in a group.
Certainly all of these methods have been used.
Example 5.4 Permutation multiplication is not usually commutative. Let
1 2 3 4
σ=
4 1 2 3
1 2 3 4
τ= .
2 1 4 3
Then
1 2 3 4
στ = ,
1 4 3 2
but
1 2 3 4
τσ = .
3 2 1 4
□
Cycle Notation
The notation that we have used to represent permutations up to this point is cumbersome,
to say the least. To work effectively with permutation groups, we need a more streamlined
CHAPTER 5. PERMUTATION GROUPS 61
method of writing down and manipulating permutations.
A permutation σ ∈ SX is a cycle of length k if there exist elements a1 , a2 , . . . , ak ∈ X
such that
σ(a1 ) = a2
σ(a2 ) = a3
..
.
σ(ak ) = a1
and σ(x) = x for all other elements x ∈ X. We will write (a1 , a2 , . . . , ak ) to denote the
cycle σ. Cycles are the building blocks of all permutations.
Example 5.5 The permutation
1 2 3 4 5 6 7
σ= = (1 6 2 3 5 4)
6 3 5 1 4 2 7
is a cycle of length 6, whereas
1 2 3 4 5 6
τ= = (2 4 3)
1 4 2 3 5 6
is a cycle of length 3.
Not every permutation is a cycle. Consider the permutation
1 2 3 4 5 6
= (1 2 4 3)(5 6).
2 4 1 3 6 5
This permutation actually contains a cycle of length 2 and a cycle of length 4. □
Example 5.6 It is very easy to compute products of cycles. Suppose that
σ = (1 3 5 2) and τ = (2 5 6).
If we think of σ as
1 7→ 3, 3 7→ 5, 5 7→ 2, 2 7→ 1,
and τ as
2 7→ 5, 5 7→ 6, 6 7→ 2,
then for στ remembering that we apply τ first and then σ, it must be the case that
1 7→ 3, 3 7→ 5, 5 7→ 6, 6 7→ 2 7→ 1,
or στ = (1 3 5 6). If µ = (1 6 3 4), then σµ = (1 6 5 2)(3 4). □
Two cycles in SX , σ = (a1 , a2 , . . . , ak ) and τ = (b1 , b2 , . . . , bl ), are disjoint if ai 6= bj for
all i and j.
Example 5.7 The cycles (1 3 5) and (2 7) are disjoint; however, the cycles (1 3 5) and (3 4 7)
are not. Calculating their products, we find that
(1 3 5)(2 7) = (1 3 5)(2 7)
(1 3 5)(3 4 7) = (1 3 4 7 5).
The product of two cycles that are not disjoint may reduce to something less complicated;
the product of disjoint cycles cannot be simplified. □
CHAPTER 5. PERMUTATION GROUPS 62
Proposition 5.8 Let σ and τ be two disjoint cycles in SX . Then στ = τ σ.
Proof. Let σ = (a1 , a2 , . . . , ak ) and τ = (b1 , b2 , . . . , bl ). We must show that στ (x) = τ σ(x)
for all x ∈ X. If x is neither in {a1 , a2 , . . . , ak } nor {b1 , b2 , . . . , bl }, then both σ and τ fix x.
That is, σ(x) = x and τ (x) = x. Hence,
στ (x) = σ(τ (x)) = σ(x) = x = τ (x) = τ (σ(x)) = τ σ(x).
Do not forget that we are multiplying permutations right to left, which is the opposite of the
order in which we usually multiply group elements. Now suppose that x ∈ {a1 , a2 , . . . , ak }.
Then σ(ai ) = a(i mod k)+1 ; that is,
a1 7→ a2
a2 →7 a3
..
.
ak−1 7→ ak
ak 7→ a1 .
However, τ (ai ) = ai since σ and τ are disjoint. Therefore,
στ (ai ) = σ(τ (ai ))
= σ(ai )
= a(i mod k)+1
= τ (a(i mod k)+1 )
= τ (σ(ai ))
= τ σ(ai ).
Similarly, if x ∈ {b1 , b2 , . . . , bl }, then σ and τ also commute. ■
Theorem 5.9 Every permutation in Sn can be written as the product of disjoint cycles.
Proof. We can assume that X = {1, 2, . . . , n}. If σ ∈ Sn and we define X1 to be
{σ(1), σ 2 (1), . . .}, then the set X1 is finite since X is finite. Now let i be the first integer in
X that is not in X1 and define X2 by {σ(i), σ 2 (i), . . .}. Again, X2 is a finite set. Continuing
in this manner, we can define finite disjoint sets X3 , X4 , . . .. Since X is a finite set, we are
guaranteed that this process will end and there will be only a finite number of these sets,
say r. If σi is the cycle defined by
(
σ(x) x ∈ Xi
σi (x) = ,
x x∈ / Xi
then σ = σ1 σ2 · · · σr . Since the sets X1 , X2 , . . . , Xr are disjoint, the cycles σ1 , σ2 , . . . , σr
must also be disjoint. ■
Example 5.10 Let
1 2 3 4 5 6
σ=
6 4 3 1 5 2
1 2 3 4 5 6
τ= .
3 2 1 5 6 4
CHAPTER 5. PERMUTATION GROUPS 63
Using cycle notation, we can write
σ = (1 6 2 4)
τ = (1 3)(4 5 6)
στ = (1 3 6)(2 4 5)
τ σ = (1 4 3)(2 5 6).
□
Remark 5.11 From this point forward we will find it convenient to use cycle notation to rep-
resent permutations. When using cycle notation, we often denote the identity permutation
by (1).
Transpositions
The simplest permutation is a cycle of length 2. Such cycles are called transpositions. Since
(a1 , a2 , . . . , an ) = (a1 , an )(a1 , an−1 ) · · · (a1 , a3 )(a1 , a2 ),
any cycle can be written as the product of transpositions, leading to the following proposi-
tion.
Proposition 5.12 Any permutation of a finite set containing at least two elements can be
written as the product of transpositions.
Example 5.13 Consider the permutation
(1 6)(2 5 3) = (1 6)(2 3)(2 5) = (1 6)(4 5)(2 3)(4 5)(2 5).
As we can see, there is no unique way to represent permutation as the product of transposi-
tions. For instance, we can write the identity permutation as (1 2)(1 2), as (1 3)(2 4)(1 3)(2 4),
and in many other ways. However, as it turns out, no permutation can be written as the
product of both an even number of transpositions and an odd number of transpositions.
For instance, we could represent the permutation (1 6) by
(2 3)(1 6)(2 3)
or by
(3 5)(1 6)(1 3)(1 6)(1 3)(3 5)(5 6),
but (1 6) will always be the product of an odd number of transpositions. □
Lemma 5.14 If the identity is written as the product of r transpositions,
id = τ1 τ2 · · · τr ,
then r is an even number.
Proof. We will employ induction on r. A transposition cannot be the identity; hence,
r > 1. If r = 2, then we are done. Suppose that r > 2. In this case the product of the last
two transpositions, τr−1 τr , must be one of the following cases:
(a, b)(a, b) = id
(b, c)(a, b) = (a, c)(b, c)
(c, d)(a, b) = (a, b)(c, d)
(a, c)(a, b) = (a, b)(b, c),
where a, b, c, and d are distinct.
CHAPTER 5. PERMUTATION GROUPS 64
The first equation simply says that a transposition is its own inverse. If this case occurs,
delete τr−1 τr from the product to obtain
id = τ1 τ2 · · · τr−3 τr−2 .
By induction r − 2 is even; hence, r must be even.
In each of the other three cases, we can replace τr−1 τr with the right-hand side of the
corresponding equation to obtain a new product of r transpositions for the identity. In this
new product the last occurrence of a will be in the next-to-the-last transposition. We can
continue this process with τr−2 τr−1 to obtain either a product of r − 2 transpositions or a
new product of r transpositions where the last occurrence of a is in τr−2 . If the identity is
the product of r − 2 transpositions, then again we are done, by our induction hypothesis;
otherwise, we will repeat the procedure with τr−3 τr−2 .
At some point either we will have two adjacent, identical transpositions canceling each
other out or a will be shuffled so that it will appear only in the first transposition. However,
the latter case cannot occur, because the identity would not fix a in this instance. Therefore,
the identity permutation must be the product of r − 2 transpositions and, again by our
induction hypothesis, we are done. ■
Theorem 5.15 If a permutation σ can be expressed as the product of an even number of
transpositions, then any other product of transpositions equaling σ must also contain an even
number of transpositions. Similarly, if σ can be expressed as the product of an odd number
of transpositions, then any other product of transpositions equaling σ must also contain an
odd number of transpositions.
Proof. Suppose that
σ = σ1 σ2 · · · σm = τ1 τ2 · · · τn ,
where m is even. We must show that n is also an even number. The inverse of σ is σm · · · σ1 .
Since
id = σσm · · · σ1 = τ1 · · · τn σm · · · σ1 ,
n must be even by Lemma 5.14. The proof for the case in which σ can be expressed as an
odd number of transpositions is left as an exercise. ■
In light of Theorem 5.15, we define a permutation to be even if it can be expressed
as an even number of transpositions and odd if it can be expressed as an odd number of
transpositions.
The Alternating Groups
One of the most important subgroups of Sn is the set of all even permutations, An . The
group An is called the alternating group on n letters.
Theorem 5.16 The set An is a subgroup of Sn .
Proof. Since the product of two even permutations must also be an even permutation,
An is closed. The identity is an even permutation and therefore is in An . If σ is an even
permutation, then
σ = σ1 σ2 · · · σr ,
where σi is a transposition and r is even. Since the inverse of any transposition is itself,
σ −1 = σr σr−1 · · · σ1
is also in An . ■
CHAPTER 5. PERMUTATION GROUPS 65
Proposition 5.17 The number of even permutations in Sn , n ≥ 2, is equal to the number
of odd permutations; hence, the order of An is n!/2.
Proof. Let An be the set of even permutations in Sn and Bn be the set of odd permutations.
If we can show that there is a bijection between these sets, they must contain the same
number of elements. Fix a transposition σ in Sn . Since n ≥ 2, such a σ exists. Define
λσ : An → Bn
by
λσ (τ ) = στ .
Suppose that λσ (τ ) = λσ (µ). Then στ = σµ and so
τ = σ −1 στ = σ −1 σµ = µ.
Therefore, λσ is one-to-one. We will leave the proof that λσ is surjective to the reader. ■
Example 5.18 The group A4 is the subgroup of S4 consisting of even permutations. There
are twelve elements in A4 :
(1) (1 2)(3 4) (1 3)(2 4) (1 4)(2 3)
(1 2 3) (1 3 2) (1 2 4) (1 4 2)
(1 3 4) (1 4 3) (2 3 4) (2 4 3).
One of the end-of-chapter exercises will be to write down all the subgroups of A4 . You will
find that there is no subgroup of order 6. Does this surprise you? □
Historical Note
Lagrange first thought of permutations as functions from a set to itself, but it was Cauchy
who developed the basic theorems and notation for permutations. He was the first to
use cycle notation. Augustin-Louis Cauchy (1789–1857) was born in Paris at the height
of the French Revolution. His family soon left Paris for the village of Arcueil to escape
the Reign of Terror. One of the family’s neighbors there was Pierre-Simon Laplace (1749–
1827), who encouraged him to seek a career in mathematics. Cauchy began his career as
a mathematician by solving a problem in geometry given to him by Lagrange. Cauchy
wrote over 800 papers on such diverse topics as differential equations, finite groups, applied
mathematics, and complex analysis. He was one of the mathematicians responsible for
making calculus rigorous. Perhaps more theorems and concepts in mathematics have the
name Cauchy attached to them than that of any other mathematician.
5.2 Dihedral Groups
Another special type of permutation group is the dihedral group. Recall the symmetry
group of an equilateral triangle in Chapter 3. Such groups consist of the rigid motions of a
regular n-sided polygon or n-gon. For n = 3, 4, . . ., we define the nth dihedral group to be
the group of rigid motions of a regular n-gon. We will denote this group by Dn . We can
number the vertices of a regular n-gon by 1, 2, . . . , n (Figure 5.19). Notice that there are
exactly n choices to replace the first vertex. If we replace the first vertex by k, then the
second vertex must be replaced either by vertex k + 1 or by vertex k − 1; hence, there are 2n
possible rigid motions of the n-gon. We summarize these results in the following theorem.
CHAPTER 5. PERMUTATION GROUPS 66
1
n 2
n−1 3
4
5
Figure 5.19 A regular n-gon
Theorem 5.20 The dihedral group, Dn , is a subgroup of Sn of order 2n.
Theorem 5.21 The group Dn , n ≥ 3, consists of all products of the two elements r and s,
where r has order n and s has order 2, and these two elements satisfy the relation srs = r−1 .
Proof. The possible motions of a regular n-gon are either reflections or rotations (Fig-
ure 5.22). There are exactly n possible rotations:
360◦ 360◦ 360◦
id, ,2 · , . . . , (n − 1) · .
n n n
We will denote the rotation 360◦ /n by r. The rotation r generates all of the other rotations.
That is,
360◦
rk = k · .
n
CHAPTER 5. PERMUTATION GROUPS 67
1 2
8 2 1
3
rotation
7 3 8
4
7
6 4 5
5 6
1 1
8 2 2 8
reflection
7 3 3 7
6 4 4 6
5 5
Figure 5.22 Rotations and reflections of a regular n-gon
Label the n reflections s1 , s2 , . . . , sn , where sk is the reflection that leaves vertex k fixed.
There are two cases of reflections, depending on whether n is even or odd. If there are an
even number of vertices, then two vertices are left fixed by a reflection, and s1 = sn/2+1 , s2 =
sn/2+2 , . . . , sn/2 = sn . If there are an odd number of vertices, then only a single vertex is left
fixed by a reflection and s1 , s2 , . . . , sn are distinct (Figure 5.23). In either case, the order of
each sk is two. Let s = s1 . Then s2 = 1 and rn = 1. Since any rigid motion t of the n-gon
replaces the first vertex by the vertex k, the second vertex must be replaced by either k + 1
or by k − 1. If the second vertex is replaced by k + 1, then t = rk . If the second vertex is
replaced by k − 1, then t = rk s.5 Hence, r and s generate Dn . That is, Dn consists of all
finite products of r and s,
Dn = {1, r, r2 , . . . , rn−1 , s, rs, r2 s, . . . , rn−1 s}.
We will leave the proof that srs = r−1 as an exercise.
CHAPTER 5. PERMUTATION GROUPS 68
1 1
6 2 2 6
5 3 3 5
4 4
1 1
2 6
5 2
4 3 3 4
Figure 5.23 Types of reflections of a regular n-gon
■
Example 5.24 The group of rigid motions of a square, D4 , consists of eight elements. With
the vertices numbered 1, 2, 3, 4 (Figure 5.25), the rotations are
r = (1 2 3 4)
r2 = (1 3)(2 4)
r3 = (1 4 3 2)
r4 = (1)
and the reflections are
s1 = (2 4)
s2 = (1 3).
The order of D4 is 8. The remaining two elements are
rs1 = (1 2)(3 4)
r3 s1 = (1 4)(2 3).
5
Since we are in an abstract group, we will adopt the convention that group elements are multiplied left
to right.
CHAPTER 5. PERMUTATION GROUPS 69
1 2
4 3
Figure 5.25 The group D4
□
The Motion Group of a Cube
We can investigate the groups of rigid motions of geometric objects other than a regular
n-sided polygon to obtain interesting examples of permutation groups. Let us consider the
group of rigid motions of a cube. By rigid motion, we mean a rotation with the axis of
rotation about opposing faces, edges, or vertices. One of the first questions that we can ask
about this group is “what is its order?” A cube has 6 sides. If a particular side is facing
upward, then there are four possible rotations of the cube that will preserve the upward-
facing side. Hence, the order of the group is 6 · 4 = 24. We have just proved the following
proposition.
Proposition 5.26 The group of rigid motions of a cube contains 24 elements.
Theorem 5.27 The group of rigid motions of a cube is S4 .
Proof. From Proposition 5.26, we already know that the motion group of the cube has 24
elements, the same number of elements as there are in S4 . There are exactly four diagonals
in the cube. If we label these diagonals 1, 2, 3, and 4, we must show that the motion group
of the cube will give us any permutation of the diagonals (Figure 5.28). If we can obtain
all of these permutations, then S4 and the group of rigid motions of the cube must be the
same. To obtain a transposition we can rotate the cube 180◦ about the axis joining the
midpoints of opposite edges (Figure 5.29). There are six such axes, giving all transpositions
in S4 . Since every element in S4 is the product of a finite number of transpositions, the
motion group of a cube must be S4 .
CHAPTER 5. PERMUTATION GROUPS 70
1 2
3 4
4 3
2 1
Figure 5.28 The motion group of a cube
1 2 2 1
4 3 4 3
3 4 3 4
2 1 1 2
Figure 5.29 Transpositions in the motion group of a cube
■
Sage. A permutation group is a very concrete representation of a group, and Sage support
for permutations groups is very good — making Sage a natural place for beginners to learn
about group theory.
5.3 Reading Questions
1. Express (1 3 4)(3 5 4) as a cycle, or a product of disjoint cycles. (Interpret the compo-
sition of functions in the order used by Sage, which is the reverse of the order used in
the book.)
2. What is a transposition?
3. What does it mean for a permutation to be even or odd?
4. Describe another group that is fundamentally the same as A3 .
CHAPTER 5. PERMUTATION GROUPS 71
5. Write the elements of the symmetry group of a pentagon using permutations in cycle
notation. Do this exercise by hand, and without the assistance of Sage.
5.4 Exercises
1. Write the following permutations in cycle notation.
(a) (c)
1 2 3 4 5 1 2 3 4 5
2 4 1 5 3 3 5 1 4 2
(b) (d)
1 2 3 4 5 1 2 3 4 5
4 2 5 1 3 1 4 3 2 5
2. Compute each of the following.
(a) (1 3 4 5)(2 3 4) (i) (1 2 3)(4 5)(1 2 5 4)−2
(b) (1 2)(1 2 5 3) (j) (1 2 5 4)100
(c) (1 4 3)(2 3)(2 4) (k) |(1 2 5 4)|
(d) (1 4 2 3)(3 4)(5 6)(1 3 2 4) (l) |(1 2 5 4)2 |
(e) (1 2 5 4)(1 3)(2 5) (m) (1 2)−1
(f) (1 2 5 4)(1 3)(2 5)2 (n) (1 2 5 3 7)−1
(g) (1 2 5 4)−1 (1 2 3)(4 5)(1 2 5 4) (o) [(1 2)(3 4)(1 2)(4 7)]−1
(h) (1 2 5 4)2 (1 2 3)(4 5) (p) [(1 2 3 5)(4 6 7)]−1
3. Express the following permutations as products of transpositions and identify them as
even or odd.
(a) (1 4 3 5 6) (d) (1 7 2 5 4)(1 4 2 3)(1 5 4 6 3 2)
(b) (1 5 6)(2 3 4)
(c) (1 4 2 6)(1 4 2) (e) (1 4 2 6 3 7)
4. Find (a1 , a2 , . . . , an )−1 .
5. List all of the subgroups of S4 . Find each of the following sets:
(a) {σ ∈ S4 : σ(1) = 3}
(b) {σ ∈ S4 : σ(2) = 2}
(c) {σ ∈ S4 : σ(1) = 3 and σ(2) = 2}.
Are any of these sets subgroups of S4 ?
6. Find all of the subgroups in A4 . What is the order of each subgroup?
7. Find all possible orders of elements in S7 and A7 .
8. Show that A10 contains an element of order 15.
9. Does A8 contain an element of order 26?
10. Find an element of largest order in Sn for n = 3, . . . , 10.
11. What are the possible cycle structures of elements of A5 ? What about A6 ?
CHAPTER 5. PERMUTATION GROUPS 72
12. Let σ ∈ Sn have order n. Show that for all integers i and j, σ i = σ j if and only if i ≡ j
(mod n).
13. Let σ = σ1 · · · σm ∈ Sn be the product of disjoint cycles. Prove that the order of σ is
the least common multiple of the lengths of the cycles σ1 , . . . , σm .
14. Using cycle notation, list the elements in D5 . What are r and s? Write every element
as a product of r and s.
15. If the diagonals of a cube are labeled as Figure 5.28, to which motion of the cube
does the permutation (12)(34) correspond? What about the other permutations of the
diagonals?
16. Find the group of rigid motions of a tetrahedron. Show that this is the same group as
A4 .
17. Prove that Sn is nonabelian for n ≥ 3.
18. Show that An is nonabelian for n ≥ 4.
19. Prove that Dn is nonabelian for n ≥ 3.
20. Let σ ∈ Sn be a cycle. Prove that σ can be written as the product of at most n − 1
transpositions.
21. Let σ ∈ Sn . If σ is not a cycle, prove that σ can be written as the product of at most
n − 2 transpositions.
22. If σ can be expressed as an odd number of transpositions, show that any other product
of transpositions equaling σ must also be odd.
23. If σ is a cycle of odd length, prove that σ 2 is also a cycle.
24. Show that a 3-cycle is an even permutation.
25. Prove that in An with n ≥ 3, any permutation is a product of cycles of length 3.
26. Prove that any element in Sn can be written as a finite product of the following
permutations.
(a) (1 2), (1 3), . . . , (1 n)
(b) (1 2), (2 3), . . . , (n − 1, n)
(c) (1 2), (1 2 . . . n)
27. Let G be a group and define a map λg : G → G by λg (a) = ga. Prove that λg is a
permutation of G.
28. Prove that there exist n! permutations of a set containing n elements.
29. Recall that the center of a group G is
Z(G) = {g ∈ G : gx = xg for all x ∈ G}.
Find the center of D8 . What about the center of D10 ? What is the center of Dn ?
30. Let τ = (a1 , a2 , . . . , ak ) be a cycle of length k.
(a) Prove that if σ is any permutation, then
στ σ −1 = (σ(a1 ), σ(a2 ), . . . , σ(ak ))
is a cycle of length k.
(b) Let µ be a cycle of length k. Prove that there is a permutation σ such that
στ σ −1 = µ.
CHAPTER 5. PERMUTATION GROUPS 73
31. For α and β in Sn , define α ∼ β if there exists an σ ∈ Sn such that σασ −1 = β. Show
that ∼ is an equivalence relation on Sn .
32. Let σ ∈ SX . If σ n (x) = y for some n ∈ Z, we will say that x ∼ y.
(a) Show that ∼ is an equivalence relation on X.
(b) Define the orbit of x ∈ X under σ ∈ SX to be the set
Ox,σ = {y : x ∼ y}.
Compute the orbits of each element in {1, 2, 3, 4, 5} under each of the following
elements in S5 :
α = (1 2 5 4)
β = (1 2 3)(4 5)
γ = (1 3)(2 5).
(c) If Ox,σ ∩ Oy,σ 6= ∅, prove that Ox,σ = Oy,σ . The orbits under a permutation σ
are the equivalence classes corresponding to the equivalence relation ∼.
(d) A subgroup H of SX is transitive if for every x, y ∈ X, there exists a σ ∈ H
such that σ(x) = y. Prove that hσi is transitive if and only if Ox,σ = X for some
x ∈ X.
33. Let α ∈ Sn for n ≥ 3. If αβ = βα for all β ∈ Sn , prove that α must be the identity
permutation; hence, the center of Sn is the trivial subgroup.
34. If α is even, prove that α−1 is also even. Does a corresponding result hold if α is odd?
35. If σ ∈ An and τ ∈ Sn , show that τ −1 στ ∈ An .
36. Show that α−1 β −1 αβ is even for α, β ∈ Sn .
37. Let r and s be the elements in Dn described in Theorem 5.21
(a) Show that srs = r−1 .
(b) Show that rk s = sr−k in Dn .
(c) Prove that the order of rk ∈ Dn is n/ gcd(k, n).
6
Cosets and Lagrange’s Theorem
Lagrange’s Theorem, one of the most important results in finite group theory, states that the
order of a subgroup must divide the order of the group. This theorem provides a powerful
tool for analyzing finite groups; it gives us an idea of exactly what type of subgroups we
might expect a finite group to possess. Central to understanding Lagranges’s Theorem is
the notion of a coset.
6.1 Cosets
Let G be a group and H a subgroup of G. Define a left coset of H with representative
g ∈ G to be the set
gH = {gh : h ∈ H}.
Right cosets can be defined similarly by
Hg = {hg : h ∈ H}.
If left and right cosets coincide or if it is clear from the context to which type of coset that
we are referring, we will use the word coset without specifying left or right.
Example 6.1 Let H be the subgroup of Z6 consisting of the elements 0 and 3. The cosets
are
0 + H = 3 + H = {0, 3}
1 + H = 4 + H = {1, 4}
2 + H = 5 + H = {2, 5}.
We will always write the cosets of subgroups of Z and Zn with the additive notation we have
used for cosets here. In a commutative group, left and right cosets are always identical. □
Example 6.2 Let H be the subgroup of S3 defined by the permutations {(1), (1 2 3), (1 3 2)}.
The left cosets of H are
(1)H = (1 2 3)H = (1 3 2)H = {(1), (1 2 3), (1 3 2)}
(1 2)H = (1 3)H = (2 3)H = {(1 2), (1 3), (2 3)}.
The right cosets of H are exactly the same as the left cosets:
H(1) = H(1 2 3) = H(1 3 2) = {(1), (1 2 3), (1 3 2)}
H(1 2) = H(1 3) = H(2 3) = {(1 2), (1 3), (2 3)}.
74
CHAPTER 6. COSETS AND LAGRANGE’S THEOREM 75
It is not always the case that a left coset is the same as a right coset. Let K be the
subgroup of S3 defined by the permutations {(1), (1 2)}. Then the left cosets of K are
(1)K = (1 2)K = {(1), (1 2)}
(1 3)K = (1 2 3)K = {(1 3), (1 2 3)}
(2 3)K = (1 3 2)K = {(2 3), (1 3 2)};
however, the right cosets of K are
K(1) = K(1 2) = {(1), (1 2)}
K(1 3) = K(1 3 2) = {(1 3), (1 3 2)}
K(2 3) = K(1 2 3) = {(2 3), (1 2 3)}.
□
The following lemma is quite useful when dealing with cosets. (We leave its proof as an
exercise.)
Lemma 6.3 Let H be a subgroup of a group G and suppose that g1 , g2 ∈ G. The following
conditions are equivalent.
1. g1 H = g2 H;
2. Hg1−1 = Hg2−1 ;
3. g1 H ⊂ g2 H;
4. g2 ∈ g1 H;
5. g1−1 g2 ∈ H.
In all of our examples the cosets of a subgroup H partition the larger group G. The
following theorem proclaims that this will always be the case.
Theorem 6.4 Let H be a subgroup of a group G. Then the left cosets of H in G partition
G. That is, the group G is the disjoint union of the left cosets of H in G.
Proof. Let g1 H and g2 H be two cosets of H in G. We must show that either g1 H ∩g2 H = ∅
or g1 H = g2 H. Suppose that g1 H ∩ g2 H 6= ∅ and a ∈ g1 H ∩ g2 H. Then by the definition
of a left coset, a = g1 h1 = g2 h2 for some elements h1 and h2 in H. Hence, g1 = g2 h2 h−1
1 or
g1 ∈ g2 H. By Lemma 6.3, g1 H = g2 H. ■
Remark 6.5 There is nothing special in this theorem about left cosets. Right cosets also
partition G; the proof of this fact is exactly the same as the proof for left cosets except that
all group multiplications are done on the opposite side of H.
Let G be a group and H be a subgroup of G. Define the index of H in G to be the
number of left cosets of H in G. We will denote the index by [G : H].
Example 6.6 Let G = Z6 and H = {0, 3}. Then [G : H] = 3. □
Example 6.7 Suppose that G = S3 , H = {(1), (1 2 3), (1 3 2)}, and K = {(1), (1 2)}. Then
[G : H] = 2 and [G : K] = 3. □
Theorem 6.8 Let H be a subgroup of a group G. The number of left cosets of H in G is
the same as the number of right cosets of H in G.
Proof. Let LH and RH denote the set of left and right cosets of H in G, respectively. If
we can define a bijective map ϕ : LH → RH , then the theorem will be proved. If gH ∈ LH ,
let ϕ(gH) = Hg −1 . By Lemma 6.3, the map ϕ is well-defined; that is, if g1 H = g2 H, then
CHAPTER 6. COSETS AND LAGRANGE’S THEOREM 76
Hg1−1 = Hg2−1 . To show that ϕ is one-to-one, suppose that
Hg1−1 = ϕ(g1 H) = ϕ(g2 H) = Hg2−1 .
Again by Lemma 6.3, g1 H = g2 H. The map ϕ is onto since ϕ(g −1 H) = Hg. ■
6.2 Lagrange’s Theorem
Proposition 6.9 Let H be a subgroup of G with g ∈ G and define a map ϕ : H → gH by
ϕ(h) = gh. The map ϕ is bijective; hence, the number of elements in H is the same as the
number of elements in gH.
Proof. We first show that the map ϕ is one-to-one. Suppose that ϕ(h1 ) = ϕ(h2 ) for
elements h1 , h2 ∈ H. We must show that h1 = h2 , but ϕ(h1 ) = gh1 and ϕ(h2 ) = gh2 . So
gh1 = gh2 , and by left cancellation h1 = h2 . To show that ϕ is onto is easy. By definition
every element of gH is of the form gh for some h ∈ H and ϕ(h) = gh. ■
Theorem 6.10 Lagrange. Let G be a finite group and let H be a subgroup of G. Then
|G|/|H| = [G : H] is the number of distinct left cosets of H in G. In particular, the number
of elements in H must divide the number of elements in G.
Proof. The group G is partitioned into [G : H] distinct left cosets. Each left coset has
|H| elements; therefore, |G| = [G : H]|H|. ■
Corollary 6.11 Suppose that G is a finite group and g ∈ G. Then the order of g must
divide the number of elements in G.
Corollary 6.12 Let |G| = p with p a prime number. Then G is cyclic and any g ∈ G such
that g 6= e is a generator.
Proof. Let g be in G such that g 6= e. Then by Corollary 6.11, the order of g must divide
the order of the group. Since |hgi| > 1, it must be p. Hence, g generates G. ■
Corollary 6.12 suggests that groups of prime order p must somehow look like Zp .
Corollary 6.13 Let H and K be subgroups of a finite group G such that G ⊃ H ⊃ K.
Then
[G : K] = [G : H][H : K].
Proof. Observe that
|G| |G| |H|
[G : K] = = · = [G : H][H : K].
|K| |H| |K|
■
Remark 6.14 The converse of Lagrange’s Theorem is false. The group A4 has order
12; however, it can be shown that it does not possess a subgroup of order 6. According to
Lagrange’s Theorem, subgroups of a group of order 12 can have orders of either 1, 2, 3, 4, or
6. However, we are not guaranteed that subgroups of every possible order exist. To prove
that A4 has no subgroup of order 6, we will assume that it does have such a subgroup H
and show that a contradiction must occur. Since A4 contains eight 3-cycles, we know that
H must contain a 3-cycle. We will show that if H contains one 3-cycle, then it must contain
more than 6 elements.
Proposition 6.15 The group A4 has no subgroup of order 6.
Proof. Since [A4 : H] = 2, there are only two cosets of H in A4 . Inasmuch as one of the
cosets is H itself, right and left cosets must coincide; therefore, gH = Hg or gHg −1 = H
for every g ∈ A4 . Since there are eight 3-cycles in A4 , at least one 3-cycle must be in H.
Without loss of generality, assume that (1 2 3) is in H. Then (1 2 3)−1 = (1 3 2) must also
CHAPTER 6. COSETS AND LAGRANGE’S THEOREM 77
be in H. Since ghg −1 ∈ H for all g ∈ A4 and all h ∈ H and
(1 2 4)(1 2 3)(1 2 4)−1 = (1 2 4)(1 2 3)(1 4 2) = (2 4 3)
(2 4 3)(1 2 3)(2 4 3)−1 = (2 4 3)(1 2 3)(2 3 4) = (1 4 2)
we can conclude that H must have at least seven elements
(1), (1 2 3), (1 3 2), (2 4 3), (2 4 3)−1 = (2 3 4), (1 4 2), (1 4 2)−1 = (1 2 4).
Therefore, A4 has no subgroup of order 6. ■
In fact, we can say more about when two cycles have the same length.
Theorem 6.16 Two cycles τ and µ in Sn have the same length if and only if there exists
a σ ∈ Sn such that µ = στ σ −1 .
Proof. Suppose that
τ = (a1 , a2 , . . . , ak )
µ = (b1 , b2 , . . . , bk ).
Define σ to be the permutation
σ(a1 ) = b1
σ(a2 ) = b2
..
.
σ(ak ) = bk .
Then µ = στ σ −1 .
Conversely, suppose that τ = (a1 , a2 , . . . , ak ) is a k-cycle and σ ∈ Sn . If σ(ai ) = b and
σ(a(i mod k)+1 ) = b′ , then µ(b) = b′ . Hence,
µ = (σ(a1 ), σ(a2 ), . . . , σ(ak )).
Since σ is one-to-one and onto, µ is a cycle of the same length as τ . ■
6.3 Fermat’s and Euler’s Theorems
The Euler ϕ-function is the map ϕ : N → N defined by ϕ(n) = 1 for n = 1, and, for n > 1,
ϕ(n) is the number of positive integers m with 1 ≤ m < n and gcd(m, n) = 1.
From Proposition 3.4, we know that the order of U (n), the group of units in Zn , is ϕ(n).
For example, |U (12)| = ϕ(12) = 4 since the numbers that are relatively prime to 12 are 1, 5,
7, and 11. For any prime p, ϕ(p) = p − 1. We state these results in the following theorem.
Theorem 6.17 Let U (n) be the group of units in Zn . Then |U (n)| = ϕ(n).
The following theorem is an important result in number theory, due to Leonhard Euler.
Theorem 6.18 Euler’s Theorem. Let a and n be integers such that n > 0 and gcd(a, n) =
1. Then aϕ(n) ≡ 1 (mod n).
Proof. By Theorem 6.17 the order of U (n) is ϕ(n). Consequently, aϕ(n) = 1 for all
a ∈ U (n); or aϕ(n) − 1 is divisible by n. Therefore, aϕ(n) ≡ 1 (mod n). ■
If we consider the special case of Euler’s Theorem in which n = p is prime and recall
that ϕ(p) = p − 1, we obtain the following result, due to Pierre de Fermat.
CHAPTER 6. COSETS AND LAGRANGE’S THEOREM 78
Theorem 6.19 Fermat’s Little Theorem. Let p be any prime number and suppose that
p ∤ a (p does not divide a). Then
ap−1 ≡ 1 (mod p).
Furthermore, for any integer b, bp ≡ b (mod p).
Sage. Sage can create all the subgroups of a group, so long as the group is not too large.
It can also create the cosets of a subgroup.
Historical Note
Joseph-Louis Lagrange (1736–1813), born in Turin, Italy, was of French and Italian descent.
His talent for mathematics became apparent at an early age. Leonhard Euler recognized
Lagrange’s abilities when Lagrange, who was only 19, communicated to Euler some work
that he had done in the calculus of variations. That year he was also named a professor at
the Royal Artillery School in Turin. At the age of 23 he joined the Berlin Academy. Frederick
the Great had written to Lagrange proclaiming that the “greatest king in Europe” should
have the “greatest mathematician in Europe” at his court. For 20 years Lagrange held the
position vacated by his mentor, Euler. His works include contributions to number theory,
group theory, physics and mechanics, the calculus of variations, the theory of equations, and
differential equations. Along with Laplace and Lavoisier, Lagrange was one of the people
responsible for designing the metric system. During his life Lagrange profoundly influenced
the development of mathematics, leaving much to the next generation of mathematicians
in the form of examples and new problems to be solved.
6.4 Reading Questions
1. State Lagrange’s Theorem in your own words.
2. Determine the left cosets of h3i in Z9 .
3. The set {(), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is a subgroup of S4 . What is its index in
S4 ?
4. Suppose G is a group of order 29. Describe G.
5. The number p = 137909 is prime. Explain how to compute 57137909 (mod 137909)
without a calculator.
6.5 Exercises
1. Suppose that G is a finite group with an element g of order 5 and an element h of order
7. Why must |G| ≥ 35?
2. Suppose that G is a finite group with 60 elements. What are the orders of possible
subgroups of G?
3. Prove or disprove: Every subgroup of the integers has finite index.
4. Prove or disprove: Every subgroup of the integers has finite order.
CHAPTER 6. COSETS AND LAGRANGE’S THEOREM 79
5. List the left and right cosets of the subgroups in each of the following.
(a) h8i in Z24 (e) An in Sn
(b) h3i in U (8) (f) D4 in S4
(c) 3Z in Z (g) T in C∗
(d) A4 in S4 (h) H = {(1), (1 2 3), (1 3 2)} in S4
6. Describe the left cosets of SL2 (R) in GL2 (R). What is the index of SL2 (R) in GL2 (R)?
7. Verify Euler’s Theorem for n = 15 and a = 4.
8. Use Fermat’s Little Theorem to show that if p = 4n + 3 is prime, there is no solution
to the equation x2 ≡ −1 (mod p).
9. Show that the integers have infinite index in the additive group of rational numbers.
10. Show that the additive group of real numbers has infinite index in the additive group
of the complex numbers.
11. Let H be a subgroup of a group G and suppose that g1 , g2 ∈ G. Prove that the
following conditions are equivalent.
(a) g1 H = g2 H
(b) Hg1−1 = Hg2−1
(c) g1 H ⊂ g2 H
(d) g2 ∈ g1 H
(e) g1−1 g2 ∈ H
12. If ghg −1 ∈ H for all g ∈ G and h ∈ H, show that right cosets are identical to left
cosets. That is, show that gH = Hg for all g ∈ G.
13. What fails in the proof of Theorem 6.8 if ϕ : LH → RH is defined by ϕ(gH) = Hg?
14. Suppose that g n = e. Show that the order of g divides n.
15. The cycle structure of a permutation σ is defined as the unordered list of the sizes of the
cycles in the cycle decomposition σ. For example, the permutation σ = (1 2)(3 4 5)(7 8)(9)
has cycle structure (2, 3, 2, 1) which can also be written as (1, 2, 2, 3).
Show that any two permutations α, β ∈ Sn have the same cycle structure if and
only if there exists a permutation γ such that β = γαγ −1 . If β = γαγ −1 for some
γ ∈ Sn , then α and β are conjugate.
16. If |G| = 2n, prove that the number of elements of order 2 is odd. Use this result to
show that G must contain a subgroup of order 2.
17. Suppose that [G : H] = 2. If a and b are not in H, show that ab ∈ H.
18. If [G : H] = 2, prove that gH = Hg.
19. Let H and K be subgroups of a group G. Prove that gH ∩ gK is a coset of H ∩ K in
G.
20. Let H and K be subgroups of a group G. Define a relation ∼ on G by a ∼ b if
there exists an h ∈ H and a k ∈ K such that hak = b. Show that this relation is an
equivalence relation. The corresponding equivalence classes are called double cosets.
Compute the double cosets of H = {(1), (1 2 3), (1 3 2)} in A4 .
21. Let G be a cyclic group of order n. Show that there are exactly ϕ(n) generators for G.
CHAPTER 6. COSETS AND LAGRANGE’S THEOREM 80
22. Let n = pe11 pe22 · · · pekk , where p1 , p2 , . . . , pk are distinct primes. Prove that
1 1 1
ϕ(n) = n 1 − 1− ··· 1 − .
p1 p2 pk
23. Show that X
n= ϕ(d)
d|n
for all positive integers n.
7
Introduction to Cryptography
Cryptography is the study of sending and receiving secret messages. The aim of cryptogra-
phy is to send messages across a channel so that only the intended recipient of the message
can read it. In addition, when a message is received, the recipient usually requires some
assurance that the message is authentic; that is, that it has not been sent by someone who
is trying to deceive the recipient. Modern cryptography is heavily dependent on abstract
algebra and number theory.
The message to be sent is called the plaintext message. The disguised message is called
the ciphertext. The plaintext and the ciphertext are both written in an alphabet, consisting
of letters or characters. Characters can include not only the familiar alphabetic characters
A, . . ., Z and a, . . ., z but also digits, punctuation marks, and blanks. A cryptosystem,
or cipher, has two parts: encryption, the process of transforming a plaintext message to
a ciphertext message, and decryption, the reverse transformation of changing a ciphertext
message into a plaintext message.
There are many different families of cryptosystems, each distinguished by a particular
encryption algorithm. Cryptosystems in a specified cryptographic family are distinguished
from one another by a parameter to the encryption function called a key. A classical
cryptosystem has a single key, which must be kept secret, known only to the sender and
the receiver of the message. If person A wishes to send secret messages to two different
people B and C, and does not wish to have B understand C’s messages or vice versa, A
must use two separate keys, so one cryptosystem is used for exchanging messages with B,
and another is used for exchanging messages with C.
Systems that use two separate keys, one for encoding and another for decoding, are
called public key cryptosystems. Since knowledge of the encoding key does not allow
anyone to guess at the decoding key, the encoding key can be made public. A public key
cryptosystem allows A and B to send messages to C using the same encoding key. Anyone
is capable of encoding a message to be sent to C, but only C knows how to decode such a
message.
7.1 Private Key Cryptography
In single or private key cryptosystems the same key is used for both encrypting and de-
crypting messages. To encrypt a plaintext message, we apply to the message some function
which is kept secret, say f . This function will yield an encrypted message. Given the en-
crypted form of the message, we can recover the original message by applying the inverse
transformation f −1 . The transformation f must be relatively easy to compute, as must f −1 ;
however, f must be extremely difficult to guess from available examples of coded messages.
81
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 82
Example 7.1 One of the first and most famous private key cryptosystems was the shift code
used by Julius Caesar. We first digitize the alphabet by letting A = 00, B = 01, . . . , Z = 25.
The encoding function will be
f (p) = p + 3 mod 26;
that is, A 7→ D, B 7→ E, . . . , Z 7→ C. The decoding function is then
f −1 (p) = p − 3 mod 26 = p + 23 mod 26.
Suppose we receive the encoded message DOJHEUD. To decode this message, we first
digitize it:
3, 14, 9, 7, 4, 20, 3.
Next we apply the inverse transformation to get
0, 11, 6, 4, 1, 17, 0,
or ALGEBRA. Notice here that there is nothing special about either of the numbers 3 or
26. We could have used a larger alphabet or a different shift. □
Cryptanalysis is concerned with deciphering a received or intercepted message. Methods
from probability and statistics are great aids in deciphering an intercepted message; for
example, the frequency analysis of the characters appearing in the intercepted message
often makes its decryption possible.
Example 7.2 Suppose we receive a message that we know was encrypted by using a shift
transformation on single letters of the 26-letter alphabet. To find out exactly what the shift
transformation was, we must compute b in the equation f (p) = p + b mod 26. We can do
this using frequency analysis. The letter E = 04 is the most commonly occurring letter
in the English language. Suppose that S = 18 is the most commonly occurring letter in
the ciphertext. Then we have good reason to suspect that 18 = 4 + b mod 26, or b = 14.
Therefore, the most likely encrypting function is
f (p) = p + 14 mod 26.
The corresponding decrypting function is
f −1 (p) = p + 12 mod 26.
It is now easy to determine whether or not our guess is correct. □
Simple shift codes are examples of monoalphabetic cryptosystems. In these ciphers a
character in the enciphered message represents exactly one character in the original message.
Such cryptosystems are not very sophisticated and are quite easy to break. In fact, in a
simple shift as described in Example 7.1, there are only 26 possible keys. It would be quite
easy to try them all rather than to use frequency analysis.
Let us investigate a slightly more sophisticated cryptosystem. Suppose that the encoding
function is given by
f (p) = ap + b mod 26.
We first need to find out when a decoding function f −1 exists. Such a decoding function
exists when we can solve the equation
c = ap + b mod 26
for p. By Proposition 3.4, this is possible exactly when a has an inverse or, equivalently,
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 83
when gcd(a, 26) = 1. In this case
f −1 (p) = a−1 p − a−1 b mod 26.
Such a cryptosystem is called an affine cryptosystem.
Example 7.3 Let us consider the affine cryptosystem f (p) = ap + b mod 26. For this
cryptosystem to work we must choose an a ∈ Z26 that is invertible. This is only possible if
gcd(a, 26) = 1. Recognizing this fact, we will let a = 5 since gcd(5, 26) = 1. It is easy to see
that a−1 = 21. Therefore, we can take our encryption function to be f (p) = 5p + 3 mod 26.
Thus, ALGEBRA is encoded as 3, 6, 7, 23, 8, 10, 3, or DGHXIKD. The decryption function
will be
f −1 (p) = 21p − 21 · 3 mod 26 = 21p + 15 mod 26.
□
A cryptosystem would be more secure if a ciphertext letter could represent more than
one plaintext letter. To give an example of this type of cryptosystem, called a polyalphabetic
cryptosystem, we will generalize affine codes by using matrices. The idea works roughly the
same as before; however, instead of encrypting one letter at a time we will encrypt pairs of
letters. We can store a pair of letters p1 and p2 in a vector
p1
p= .
p2
Let A be a 2 × 2 invertible matrix with entries in Z26 . We can define an encoding function
by
f (p) = Ap + b,
where b is a fixed column vector and matrix operations are performed in Z26 . The decoding
function must be
f −1 (p) = A−1 p − A−1 b.
Example 7.4 Suppose that we wish to encode the word HELP. The corresponding digit
string is 7, 4, 11, 15. If
3 5
A= ,
1 2
then
−1 2 21
A = .
25 3
If b = (2, 2)t , then our message is encrypted as RRGR. The encrypted letter R represents
more than one plaintext letter. □
Frequency analysis can still be performed on a polyalphabetic cryptosystem, because we
have a good understanding of how pairs of letters appear in the English language. The pair
th appears quite often; the pair qz never appears. To avoid decryption by a third party, we
must use a larger matrix than the one we used in Example 7.4.
7.2 Public Key Cryptography
If traditional cryptosystems are used, anyone who knows enough to encode a message will
also know enough to decode an intercepted message. In 1976, W. Diffie and M. Hellman
proposed public key cryptography, which is based on the observation that the encryption and
decryption procedures need not have the same key. This removes the requirement that the
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 84
encoding key be kept secret. The encoding function f must be relatively easy to compute,
but f −1 must be extremely difficult to compute without some additional information, so
that someone who knows only the encrypting key cannot find the decrypting key without
prohibitive computation. It is interesting to note that to date, no system has been proposed
that has been proven to be “one-way;” that is, for any existing public key cryptosystem,
it has never been shown to be computationally prohibitive to decode messages with only
knowledge of the encoding key.
The RSA Cryptosystem
The rsa cryptosystem introduced by R. Rivest, A. Shamir, and L. Adleman in 1978, is
based on the difficulty of factoring large numbers. Though it is not a difficult task to find
two large random primes and multiply them together, factoring a 150-digit number that is
the product of two large primes would take 100 million computers operating at 10 million
instructions per second about 50 million years under the fastest algorithms available in the
early 1990s. Although the algorithms have improved, factoring a number that is a product
of two large primes is still computationally prohibitive.
The rsa cryptosystem works as follows. Suppose that we choose two random 150-
digit prime numbers p and q. Next, we compute the product n = pq and also compute
ϕ(n) = m = (p − 1)(q − 1), where ϕ is the Euler ϕ-function. Now we start choosing random
integers E until we find one that is relatively prime to m; that is, we choose E such that
gcd(E, m) = 1. Using the Euclidean algorithm, we can find a number D such that DE ≡ 1
(mod m). The numbers n and E are now made public.
Suppose now that person B (Bob) wishes to send person A (Alice) a message over a
public line. Since E and n are known to everyone, anyone can encode messages. Bob
first digitizes the message according to some scheme, say A = 00, B = 02, . . . , Z = 25. If
necessary, he will break the message into pieces such that each piece is a positive integer
less than n. Suppose x is one of the pieces. Bob forms the number y = xE mod n and
sends y to Alice. For Alice to recover x, she need only compute x = y D mod n. Only Alice
knows D.
Example 7.5 Before exploring the theory behind the rsa cryptosystem or attempting to
use large integers, we will use some small integers just to see that the system does indeed
work. Suppose that we wish to send some message, which when digitized is 25. Let p = 23
and q = 29. Then
n = pq = 667
and
ϕ(n) = m = (p − 1)(q − 1) = 616.
We can let E = 487, since gcd(616, 487) = 1. The encoded message is computed to be
25487 mod 667 = 169.
This computation can be reasonably done by using the method of repeated squares as
described in Chapter 4. Using the Euclidean algorithm, we determine that 191E = 1+151m;
therefore, the decrypting key is (n, D) = (667, 191). We can recover the original message
by calculating
169191 mod 667 = 25.
□
Now let us examine why the rsa cryptosystem works. We know that DE ≡ 1 (mod m);
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 85
hence, there exists a k such that
DE = km + 1 = kϕ(n) + 1.
There are two cases to consider. In the first case assume that gcd(x, n) = 1. Then by
Theorem 6.18,
y D = (xE )D = xDE = xkm+1 = (xϕ(n) )k x = (1)k x = x mod n.
So we see that Alice recovers the original message x when she computes y D mod n.
For the other case, assume that gcd(x, n) 6= 1. Since n = pq and x < n, we know x is
a multiple of p or a multiple of q, but not both. We will describe the first possibility only,
since the second is entirely similar. There is then an integer r, with r < q and x = rp. Note
that we have gcd(x, q) = 1 and that m = ϕ(n) = (p − 1)(q − 1) = ϕ(p)ϕ(q). Then, using
Theorem 6.18, but now mod q,
xkm = xkϕ(p)ϕ(q) = (xϕ(q) )kϕ(p) = (1)kϕ(p) = 1 mod q.
So there is an integer t such that xkm = 1 + tq. Thus, Alice also recovers the message in
this case,
y D = xkm+1 = xkm x = (1 + tq)x = x + tq(rp) = x + trn = x mod n.
We can now ask how one would go about breaking the rsa cryptosystem. To find D
given n and E, we simply need to factor n and solve for D by using the Euclidean algorithm.
If we had known that 667 = 23 · 29 in Example 7.5, we could have recovered D.
Message Verification
There is a problem of message verification in public key cryptosystems. Since the encoding
key is public knowledge, anyone has the ability to send an encoded message. If Alice
receives a message from Bob, she would like to be able to verify that it was Bob who
actually sent the message. Suppose that Bob’s encrypting key is (n′ , E ′ ) and his decrypting
key is (n′ , D′ ). Also, suppose that Alice’s encrypting key is (n, E) and her decrypting key is
(n, D). Since encryption keys are public information, they can exchange coded messages at
their convenience. Bob wishes to assure Alice that the message he is sending is authentic.
Before Bob sends the message x to Alice, he decrypts x with his own key:
′
x′ = xD mod n′ .
Anyone can change x′ back to x just by encryption, but only Bob has the ability to form
x′ . Now Bob encrypts x′ with Alice’s encryption key to form
y ′ = x′ mod n,
E
a message that only Alice can decode. Alice decodes the message and then encodes the
result with Bob’s key to read the original message, a message that could have only been
sent by Bob.
Historical Note
Encrypting secret messages goes as far back as ancient Greece and Rome. As we know,
Julius Caesar used a simple shift code to send and receive messages. However, the formal
study of encoding and decoding messages probably began with the Arabs in the 1400s. In
the fifteenth and sixteenth centuries mathematicians such as Alberti and Viete discovered
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 86
that monoalphabetic cryptosystems offered no real security. In the 1800s, F. W. Kasiski
established methods for breaking ciphers in which a ciphertext letter can represent more
than one plaintext letter, if the same key was used several times. This discovery led to the
use of cryptosystems with keys that were used only a single time. Cryptography was placed
on firm mathematical foundations by such people as W. Friedman and L. Hill in the early
part of the twentieth century.
The period after World War I saw the development of special-purpose machines for encrypt-
ing and decrypting messages, and mathematicians were very active in cryptography during
World War II. Efforts to penetrate the cryptosystems of the Axis nations were organized
in England and in the United States by such notable mathematicians as Alan Turing and
A. A. Albert. The Allies gained a tremendous advantage in World War II by breaking the
ciphers produced by the German Enigma machine and the Japanese Purple ciphers.
By the 1970s, interest in commercial cryptography had begun to take hold. There was
a growing need to protect banking transactions, computer data, and electronic mail. In
the early 1970s, ibm developed and implemented luzifer, the forerunner of the National
Bureau of Standards’ Data Encryption Standard (DES).
The concept of a public key cryptosystem, due to Diffie and Hellman, is very recent (1976). It
was further developed by Rivest, Shamir, and Adleman with the rsa cryptosystem (1978).
It is not known how secure any of these systems are. The trapdoor knapsack cryptosys-
tem, developed by Merkle and Hellman, has been broken. It is still an open question
whether or not the rsa system can be broken. In 1991, rsa Laboratories published a list
of semiprimes (numbers with exactly two prime factors) with a cash prize for whoever was
able to provide a factorization (http://www.emc.com/emc-plus/rsa-labs/historical/the-rsa-
challenge-numbers.htma ). Although the challenge ended in 2007, many of these numbers
have not yet been factored.
There been a great deal of controversy about research in cryptography and cryptography
itself. In 1929, when Henry Stimson, Secretary of State under Herbert Hoover, dismissed the
Black Chamber (the State Department’s cryptography division) on the ethical grounds that
“gentlemen do not read each other’s mail.” During the last two decades of the twentieth
century, the National Security Agency wanted to keep information about cryptography
secret, whereas the academic community fought for the right to publish basic research.
Currently, research in mathematical cryptography and computational number theory is
very active, and mathematicians are free to publish their results in these areas.
a
www.emc.com/emc-plus/rsa-labs/historical/the-rsa-challenge-numbers.htm
Sage. Sage’s early development featured powerful routines for number theory, and later
included significant support for algebraic structures and other areas of discrete mathematics.
So it is a natural tool for the study of cryptology, including topics like RSA, elliptic curve
cryptography, and AES (Advanced Encryption Standard).
7.3 Reading Questions
1. Use the euler_phi()function in Sage to compute ϕ(893 456 123).
2. Use the power_mod()function in Sage to compute 7324 (mod 895).
3. Explain the mathematical basis for saying: encrypting a message using an rsa public
key is very simple computationally, while decrypting a communication without the
private key is very hard computationally.
4. Explain how in rsa message encoding differs from message verification.
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 87
5. Explain how one could be justified in saying that Diffie and Hellman’s proposal in 1976
was “revolutionary.”
7.4 Exercises
1. Encode IXLOVEXMATH using the cryptosystem in Example 7.1.
2. Decode ZLOOA WKLVA EHARQ WKHA ILQDO, which was encoded using the cryptosystem
in Example 7.1.
3. Assuming that monoalphabetic code was used to encode the following secret message,
what was the original message?
APHUO EGEHP PEXOV FKEUH CKVUE CHKVE APHUO
EGEHU EXOVL EXDKT VGEFT EHFKE UHCKF TZEXO
VEZDT TVKUE XOVKV ENOHK ZFTEH TEHKQ LEROF
PVEHP PEXOV ERYKP GERYT GVKEG XDRTE RGAGA
What is the significance of this message in the history of cryptography?
4. What is the total number of possible monoalphabetic cryptosystems? How secure are
such cryptosystems?
5. Prove that a 2×2 matrix A with entries in Z26 is invertible if and only if gcd(det(A), 26) =
1.
6. Given the matrix
3 4
A= ,
2 3
use the encryption function f (p) = Ap + b to encode the message CRYPTOLOGY, where
b = (2, 5)t . What is the decoding function?
7. Encrypt each of the following rsa messages x so that x is divided into blocks of integers
of length 2; that is, if x = 142528, encode 14, 25, and 28 separately.
(a) n = 3551, E = 629, x = 31
(b) n = 2257, E = 47, x = 23
(c) n = 120979, E = 13251, x = 142371
(d) n = 45629, E = 781, x = 231561
8. Compute the decoding key D for each of the encoding keys in Exercise 7.4.7.
9. Decrypt each of the following rsa messages y.
(a) n = 3551, D = 1997, y = 2791
(b) n = 5893, D = 81, y = 34
(c) n = 120979, D = 27331, y = 112135
(d) n = 79403, D = 671, y = 129381
10. For each of the following encryption keys (n, E) in the rsa cryptosystem, compute D.
(a) (n, E) = (451, 231)
(b) (n, E) = (3053, 1921)
(c) (n, E) = (37986733, 12371)
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 88
(d) (n, E) = (16394854313, 34578451)
11. Encrypted messages are often divided into blocks of n letters. A message such as THE
WORLD WONDERS WHY might be encrypted as JIW OCFRJ LPOEVYQ IOC but sent as JIW
OCF RJL POE VYQ IOC. What are the advantages of using blocks of n letters?
12. Find integers n, E, and X such that
XE ≡ X (mod n).
Is this a potential problem in the rsa cryptosystem?
13. Every person in the class should construct an rsa cryptosystem using primes that are
10 to 15 digits long. Hand in (n, E) and an encoded message. Keep D secret. See if
you can break one another’s codes.
7.5 Additional Exercises: Primality and Factoring
In the rsa cryptosystem it is important to be able to find large prime numbers easily. Also,
this cryptosystem is not secure if we can factor a composite number that is the product
of two large primes. The solutions to both of these problems are quite easy. To find out
if a number n is prime or to factor n, we can use trial division. We simply divide n by
√
d = 2, 3, . . . , n. Either a factorization will be obtained, or n is prime if no d divides n.
The problem is that such a computation is prohibitively time-consuming if n is very large.
1. A better algorithm for factoring odd positive integers is Fermat’s factorization algo-
rithm.
(a) Let n = ab be an odd composite number. Prove that n can be written as the
difference of two perfect squares:
n = x2 − y 2 = (x − y)(x + y).
Consequently, a positive odd integer can be factored exactly when we can find
integers x and y such that n = x2 − y 2 .
(b) Write a program to implement the following factorization algorithm based on the
observation in part (a). The expression ceiling(sqrt(n)) means the smallest
integer greater than or equal to the square root of n. Write another program to
do factorization using trial division and compare the speed of the two algorithms.
Which algorithm is faster and why?
x := ceiling ( sqrt ( n) )
y := 1
1 : while x ^2 - y ^2 > n do
y := y + 1
if x ^2 - y ^2 < n then
x := x + 1
y := 1
goto 1
else if x ^2 - y ^2 = 0 then
a := x - y
b := x + y
write n = a * b
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 89
2. Primality Testing. Recall Fermat’s Little Theorem from Chapter 6. Let p be prime
with gcd(a, p) = 1. Then ap−1 ≡ 1 (mod p). We can use Fermat’s Little Theorem as a
screening test for primes. For example, 15 cannot be prime since
215−1 ≡ 214 ≡ 4 (mod 15).
However, 17 is a potential prime since
217−1 ≡ 216 ≡ 1 (mod 17).
We say that an odd composite number n is a pseudoprime if
2n−1 ≡ 1 (mod n).
Which of the following numbers are primes and which are pseudoprimes?
(a) 342 (c) 601 (e) 771
(b) 811 (d) 561 (f) 631
3. Let n be an odd composite number and b be a positive integer such that gcd(b, n) = 1.
If bn−1 ≡ 1 (mod n), then n is a pseudoprime base b. Show that 341 is a pseudoprime
base 2 but not a pseudoprime base 3.
4. Write a program to determine all primes less than 2000 using trial division. Write a
second program that will determine all numbers less than 2000 that are either primes
or pseudoprimes. Compare the speed of the two programs. How many pseudoprimes
are there below 2000?
There exist composite numbers that are pseudoprimes for all bases to which they are
relatively prime. These numbers are called Carmichael numbers. The first Carmichael
number is 561 = 3 · 11 · 17. In 1992, Alford, Granville, and Pomerance proved that
there are an infinite number of Carmichael numbers [4]. However, Carmichael numbers
are very rare. There are only 2163 Carmichael numbers less than 25 × 109 . For more
sophisticated primality tests, see [1], [6], or [7].
7.6 References and Suggested Readings
[1] Bressoud, D. M. Factorization and Primality Testing. Springer-Verlag, New York,
1989.
[2] Diffie, W. and Hellman, M. E. “New Directions in Cryptography,” IEEE Trans. In-
form. Theory 22 (1976), 644–54.
[3] Gardner, M. “Mathematical games: A new kind of cipher that would take millions of
years to break,” Scientific American 237 (1977), 120–24.
[4] Granville, A. “Primality Testing and Carmichael Numbers,” Notices of the American
Mathematical Society 39 (1992), 696–700.
[5] Hellman, M. E. “The Mathematics of Public Key Cryptography,” Scientific American
241 (1979), 130–39.
[6] Koblitz, N. A Course in Number Theory and Cryptography. 2nd ed. Springer, New
York, 1994.
[7] Pomerance, C., ed. “Cryptology and Computational Number Theory”, Proceedings
of Symposia in Applied Mathematics 42 (1990) American Mathematical Society, Prov-
idence, RI.
CHAPTER 7. INTRODUCTION TO CRYPTOGRAPHY 90
[8] Rivest, R. L., Shamir, A., and Adleman, L., “A Method for Obtaining Signatures and
Public-key Cryptosystems,” Comm. ACM 21 (1978), 120–26.
8
Algebraic Coding Theory
Coding theory is an application of algebra that has become increasingly important over the
last several decades. When we transmit data, we are concerned about sending a message over
a channel that could be affected by “noise.” We wish to be able to encode and decode the
information in a manner that will allow the detection, and possibly the correction, of errors
caused by noise. This situation arises in many areas of communications, including radio,
telephone, television, computer communications, and digital media technology. Probability,
combinatorics, group theory, linear algebra, and polynomial rings over finite fields all play
important roles in coding theory.
8.1 Error-Detecting and Correcting Codes
Let us examine a simple model of a communications system for transmitting and receiving
coded messages (Figure 8.1).
91
CHAPTER 8. ALGEBRAIC CODING THEORY 92
m-digit message
Encoder
n-digit code word
Transmitter
Noise
Receiver
n-digit received word
Decoder
m-digit received message or error
Figure 8.1 Encoding and decoding messages
Uncoded messages may be composed of letters or characters, but typically they consist
of binary m-tuples. These messages are encoded into codewords, consisting of binary n-
tuples, by a device called an encoder. The message is transmitted and then decoded. We
will consider the occurrence of errors during transmission. An error occurs if there is a
change in one or more bits in the codeword. A decoding scheme is a method that either
converts an arbitrarily received n-tuple into a meaningful decoded message or gives an error
message for that n-tuple. If the received message is a codeword (one of the special n-tuples
allowed to be transmitted), then the decoded message must be the unique message that
was encoded into the codeword. For received non-codewords, the decoding scheme will give
an error indication, or, if we are more clever, will actually try to correct the error and
reconstruct the original message. Our goal is to transmit error-free messages as cheaply
and quickly as possible.
Example 8.2 One possible coding scheme would be to send a message several times and to
compare the received copies with one another. Suppose that the message to be encoded is
a binary n-tuple (x1 , x2 , . . . , xn ). The message is encoded into a binary 3n-tuple by simply
repeating the message three times:
(x1 , x2 , . . . , xn ) 7→ (x1 , x2 , . . . , xn , x1 , x2 , . . . , xn , x1 , x2 , . . . , xn ).
To decode the message, we choose as the ith digit the one that appears in the ith place
in at least two of the three transmissions. For example, if the original message is (0110),
CHAPTER 8. ALGEBRAIC CODING THEORY 93
then the transmitted message will be (0110 0110 0110). If there is a transmission error in
the fifth digit, then the received codeword will be (0110 1110 0110), which will be correctly
decoded as (0110).6 This triple-repetition method will automatically detect and correct all
single errors, but it is slow and inefficient: to send a message consisting of n bits, 2n extra
bits are required, and we can only detect and correct single errors. We will see that it is
possible to find an encoding scheme that will encode a message of n bits into m bits with
m much smaller than 3n. □
Example 8.3 Even parity, a commonly used coding scheme, is much more efficient than the
simple repetition scheme. The ascii (American Standard Code for Information Interchange)
coding system uses binary 8-tuples, yielding 28 = 256 possible 8-tuples. However, only seven
bits are needed since there are only 27 = 128 ascii characters. What can or should be done
with the extra bit? Using the full eight bits, we can detect single transmission errors. For
example, the ascii codes for A, B, and C are
A = 6510 = 010000012 ,
B = 6610 = 010000102 ,
C = 6710 = 010000112 .
Notice that the leftmost bit is always set to 0; that is, the 128 ascii characters have codes
000000002 = 010 ,
..
.
011111112 = 12710 .
The bit can be used for error checking on the other seven bits. It is set to either 0 or 1
so that the total number of 1 bits in the representation of a character is even. Using even
parity, the codes for A, B, and C now become
A = 010000012 ,
B = 010000102 ,
C = 110000112 .
Suppose an A is sent and a transmission error in the sixth bit is caused by noise over the
communication channel so that (0100 0101) is received. We know an error has occurred since
the received word has an odd number of 1s, and we can now request that the codeword be
transmitted again. When used for error checking, the leftmost bit is called a parity check
bit.
By far the most common error-detecting codes used in computers are based on the
addition of a parity bit. Typically, a computer stores information in m-tuples called words.
Common word lengths are 8, 16, and 32 bits. One bit in the word is set aside as the parity
check bit, and is not used to store information. This bit is set to either 0 or 1, depending
on the number of 1s in the word.
Adding a parity check bit allows the detection of all single errors because changing a
single bit either increases or decreases the number of 1s by one, and in either case the parity
has been changed from even to odd, so the new word is not a codeword. (We could also
construct an error detection scheme based on odd parity; that is, we could set the parity
check bit so that a codeword always has an odd number of 1s.) □
The even parity system is easy to implement, but has two drawbacks. First, multiple
7
We will adopt the convention that bits are numbered left to right in binary n-tuples.
CHAPTER 8. ALGEBRAIC CODING THEORY 94
errors are not detectable. Suppose an A is sent and the first and seventh bits are changed
from 0 to 1. The received word is a codeword, but will be decoded into a C instead of an A.
Second, we do not have the ability to correct errors. If the 8-tuple (1001 1000) is received,
we know that an error has occurred, but we have no idea which bit has been changed. We
will now investigate a coding scheme that will not only allow us to detect transmission errors
but will actually correct the errors.
Example 8.4 Suppose that our original message is either a 0 or a 1, and that 0 encodes
to (000) and 1 encodes to (111). If only a single error occurs during transmission, we can
detect and correct the error. For example, if a (101) is received, then the second bit must
have been changed from a 1 to a 0. The originally transmitted codeword must have been
(111). This method will detect and correct all single errors.
Table 8.5 A repetition code
Transmitted Received Word
Codeword 000 001 010 011 100 101 110 111
000 0 1 1 2 1 2 2 3
111 3 2 2 1 2 1 1 0
In Table 8.5, we present all possible words that might be received for the transmitted
codewords (000) and (111). Table 8.5 also shows the number of bits by which each received
3-tuple differs from each original codeword. □
Maximum-Likelihood Decoding
The coding scheme presented in Example 8.4 is not a complete solution to the problem
because it does not account for the possibility of multiple errors. For example, either a
(000) or a (111) could be sent and a (001) received. We have no means of deciding from the
received word whether there was a single error in the third bit or two errors, one in the first
bit and one in the second. No matter what coding scheme is used, an incorrect message
could be received. We could transmit a (000), have errors in all three bits, and receive
the codeword (111). It is important to make explicit assumptions about the likelihood and
distribution of transmission errors so that, in a particular application, it will be known
whether a given error detection scheme is appropriate. We will assume that transmission
errors are rare, and, that when they do occur, they occur independently in each bit; that is,
if p is the probability of an error in one bit and q is the probability of an error in a different
bit, then the probability of errors occurring in both of these bits at the same time is pq.
We will also assume that a received n-tuple is decoded into a codeword that is closest to it;
that is, we assume that the receiver uses maximum-likelihood decoding. 7
7
This section requires a knowledge of probability, but can be skipped without loss of continuity.
CHAPTER 8. ALGEBRAIC CODING THEORY 95
p
0 0
q
q
1 p 1
Figure 8.6 Binary symmetric channel
A binary symmetric channel is a model that consists of a transmitter capable of sending
a binary signal, either a 0 or a 1, together with a receiver. Let p be the probability that
the signal is correctly received. Then q = 1 − p is the probability of an incorrect reception.
If a 1 is sent, then the probability that a 1 is received is p and the probability that a 0 is
received is q (Figure 8.6). The probability that no errors occur during the transmission of
a binary codeword of length n is pn . For example, if p = 0.999 and a message consisting of
10,000 bits is sent, then the probability of a perfect transmission is
(0.999)10,000 ≈ 0.00005.
Theorem 8.7 If a binary n-tuple (x1 , . . . , xn ) is transmitted across a binary symmetric
channel with probability p that no error will occur in each coordinate, then the probability
that there are errors in exactly k coordinates is
n k n−k
q p .
k
Proof. Fix k different coordinates. We first compute the probability that an error has
occurred in this fixed set of coordinates. The probability of an error occurring in a particular
one of these k coordinates is q; the probability that an error will not occur in any of the
remaining n − k coordinates is p. The probability of each of these n independent events is
q k pn−k . The number of possible error patterns with exactly k errors occurring is equal to
n n!
= ,
k k!(n − k)!
the number of combinations of n things taken k at a time. Each of these error patterns has
probability q k pn−k of occurring; hence, the probability of all of these error patterns is
n k n−k
q p .
k
■
Example 8.8 Suppose that p = 0.995 and a 500-bit message is sent. The probability that
the message was sent error-free is
pn = (0.995)500 ≈ 0.082.
The probability of exactly one error occurring is
n
qpn−1 = 500(0.005)(0.995)499 ≈ 0.204.
1
CHAPTER 8. ALGEBRAIC CODING THEORY 96
The probability of exactly two errors is
n 2 n−2 500 · 499
q p = (0.005)2 (0.995)498 ≈ 0.257.
2 2
The probability of more than two errors is approximately
1 − 0.082 − 0.204 − 0.257 = 0.457.
□
Block Codes
If we are to develop efficient error-detecting and error-correcting codes, we will need more
sophisticated mathematical tools. Group theory will allow faster methods of encoding and
decoding messages. A code is an (n, m)-block code if the information that is to be coded
can be divided into blocks of m binary digits, each of which can be encoded into n binary
digits. More specifically, an (n, m)-block code consists of an encoding function
E : Zm
2 → Z2
n
and a decoding function
D : Zn2 → Zm
2 .
A codeword is any element in the image of E. We also require that E be one-to-one so that
two information blocks will not be encoded into the same codeword. If our code is to be
error-correcting, then D must be onto.
Example 8.9 The even-parity coding system developed to detect single errors in ascii
characters is an (8, 7)-block code. The encoding function is
E(x7 , x6 , . . . , x1 ) = (x8 , x7 , . . . , x1 ),
where x8 = x7 + x6 + · · · + x1 with addition in Z2 . □
Let x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) be binary n-tuples. The Hamming distance or
distance, d(x, y), between x and y is the number of bits in which x and y differ. The distance
between two codewords is the minimum number of transmission errors required to change
one codeword into the other. The minimum distance for a code, dmin , is the minimum of
all distances d(x, y), where x and y are distinct codewords. The weight, w(x), of a binary
codeword x is the number of 1s in x. Clearly, w(x) = d(x, 0), where 0 = (00 · · · 0).
Example 8.10 Let x = (10101), y = (11010), and z = (00011) be all of the codewords in
some code C. Then we have the following Hamming distances:
d(x, y) = 4, d(x, z) = 3, d(y, z) = 3.
The minimum distance for this code is 3. We also have the following weights:
w(x) = 3, w(y) = 3, w(z) = 2.
□
The following proposition lists some basic properties about the weight of a codeword
and the distance between two codewords. The proof is left as an exercise.
Proposition 8.11 Let x, y, and z be binary n-tuples. Then
1. w(x) = d(x, 0);
CHAPTER 8. ALGEBRAIC CODING THEORY 97
2. d(x, y) ≥ 0;
3. d(x, y) = 0 exactly when x = y;
4. d(x, y) = d(y, x);
5. d(x, y) ≤ d(x, z) + d(z, y).
The weights in a particular code are usually much easier to compute than the Hamming
distances between all codewords in the code. If a code is set up carefully, we can use this
fact to our advantage.
Suppose that x = (1101) and y = (1100) are codewords in some code. If we transmit
(1101) and an error occurs in the rightmost bit, then (1100) will be received. Since (1100) is
a codeword, the decoder will decode (1100) as the transmitted message. This code is clearly
not very appropriate for error detection. The problem is that d(x, y) = 1. If x = (1100) and
y = (1010) are codewords, then d(x, y) = 2. If x is transmitted and a single error occurs,
then y can never be received. Table 8.12 gives the distances between all 4-bit codewords
in which the first three bits carry information and the fourth is an even parity check bit.
We can see that the minimum distance here is 2; hence, the code is suitable as a single
error-detecting code.
Table 8.12 Distances between 4-bit codewords
0000 0011 0101 0110 1001 1010 1100 1111
0000 0 2 2 2 2 2 2 4
0011 2 0 2 2 2 2 4 2
0101 2 2 0 2 2 4 2 2
0110 2 2 2 0 4 2 2 2
1001 2 2 2 4 0 2 2 2
1010 2 2 4 2 2 0 2 2
1100 2 4 2 2 2 2 0 2
1111 4 2 2 2 2 2 2 0
To determine exactly what the error-detecting and error-correcting capabilities for a
code are, we need to analyze the minimum distance for the code. Let x and y be codewords.
If d(x, y) = 1 and an error occurs where x and y differ, then x is changed to y. The received
codeword is y and no error message is given. Now suppose d(x, y) = 2. Then a single error
cannot change x to y. Therefore, if dmin = 2, we have the ability to detect single errors.
However, suppose that d(x, y) = 2, y is sent, and a noncodeword z is received such that
d(x, z) = d(y, z) = 1.
Then the decoder cannot decide between x and y. Even though we are aware that an error
has occurred, we do not know what the error is.
Suppose dmin ≥ 3. Then the maximum-likelihood decoding scheme corrects all single
errors. Starting with a codeword x, an error in the transmission of a single bit gives y with
d(x, y) = 1, but d(z, y) ≥ 2 for any other codeword z 6= x. If we do not require the correction
of errors, then we can detect multiple errors when a code has a minimum distance that is
greater than or equal to 3.
Theorem 8.13 Let C be a code with dmin = 2n + 1. Then C can correct any n or fewer
errors. Furthermore, any 2n or fewer errors can be detected in C.
Proof. Suppose that a codeword x is sent and the word y is received with at most n errors.
Then d(x, y) ≤ n. If z is any codeword other than x, then
2n + 1 ≤ d(x, z) ≤ d(x, y) + d(y, z) ≤ n + d(y, z).
CHAPTER 8. ALGEBRAIC CODING THEORY 98
Hence, d(y, z) ≥ n + 1 and y will be correctly decoded as x. Now suppose that x is
transmitted and y is received and that at least one error has occurred, but not more than
2n errors. Then 1 ≤ d(x, y) ≤ 2n. Since the minimum distance between codewords is 2n + 1,
y cannot be a codeword. Consequently, the code can detect between 1 and 2n errors. ■
Example 8.14 In Table 8.15, the codewords c1 = (00000), c2 = (00111), c3 = (11100), and
c4 = (11011) determine a single error-correcting code.
Table 8.15 Hamming distances for an error-correcting code
00000 00111 11100 11011
00000 0 3 3 4
00111 3 0 4 3
11100 3 4 0 3
11011 4 3 3 0
□
Historical Note
Modern coding theory began in 1948 with C. Shannon’s paper, “A Mathematical Theory
of Information” [7]. This paper offered an example of an algebraic code, and Shannon’s
Theorem proclaimed exactly how good codes could be expected to be. Richard Hamming
began working with linear codes at Bell Labs in the late 1940s and early 1950s after becoming
frustrated because the programs that he was running could not recover from simple errors
generated by noise. Coding theory has grown tremendously in the past several decades.
The Theory of Error-Correcting Codes, by MacWilliams and Sloane [5], published in 1977,
already contained over 1500 references. Linear codes (Reed-Muller (32, 6)-block codes) were
used on NASA’s Mariner space probes. More recent space probes such as Voyager have used
what are called convolution codes. Currently, very active research is being done with Goppa
codes, which are heavily dependent on algebraic geometry.
8.2 Linear Codes
To gain more knowledge of a particular code and develop more efficient techniques of encod-
ing, decoding, and error detection, we need to add additional structure to our codes. One
way to accomplish this is to require that the code also be a group. A group code is a code
that is also a subgroup of Zn2 .
To check that a code is a group code, we need only verify one thing. If we add any
two elements in the code, the result must be an n-tuple that is again in the code. It is not
necessary to check that the inverse of the n-tuple is in the code, since every codeword is its
own inverse, nor is it necessary to check that 0 is a codeword. For instance,
(11000101) + (11000101) = (00000000).
Example 8.16 Suppose that we have a code that consists of the following 7-tuples:
(0000000) (0001111) (0010101) (0011010)
(0100110) (0101001) (0110011) (0111100)
(1000011) (1001100) (1010110) (1011001)
(1100101) (1101010) (1110000) (1111111).
It is a straightforward though tedious task to verify that this code is also a subgroup of Z72
and, therefore, a group code. This code is a single error-detecting and single error-correcting
CHAPTER 8. ALGEBRAIC CODING THEORY 99
code, but it is a long and tedious process to compute all of the distances between pairs of
codewords to determine that dmin = 3. It is much easier to see that the minimum weight
of all the nonzero codewords is 3. As we will soon see, this is no coincidence. However, the
relationship between weights and distances in a particular code is heavily dependent on the
fact that the code is a group. □
Lemma 8.17 Let x and y be binary n-tuples. Then w(x + y) = d(x, y).
Proof. Suppose that x and y are binary n-tuples. Then the distance between x and y
is exactly the number of places in which x and y differ. But x and y differ in a particular
coordinate exactly when the sum in the coordinate is 1, since
1+1=0
0+0=0
1+0=1
0 + 1 = 1.
Consequently, the weight of the sum must be the distance between the two codewords. ■
Theorem 8.18 Let dmin be the minimum distance for a group code C. Then dmin is the
minimum weight of all the nonzero codewords in C. That is,
dmin = min{w(x) : x 6= 0}.
Proof. Observe that
dmin = min{d(x, y) : x 6= y}
= min{d(x, y) : x + y 6= 0}
= min{w(x + y) : x + y 6= 0}
= min{w(z) : z 6= 0}.
■
Linear Codes
From Example 8.16, it is now easy to check that the minimum nonzero weight is 3; hence,
the code does indeed detect and correct all single errors. We have now reduced the problem
of finding “good” codes to that of generating group codes. One easy way to generate group
codes is to employ a bit of matrix theory.
Define the inner product of two binary n-tuples to be
x · y = x1 y1 + · · · + xn yn ,
where x = (x1 , x2 , . . . , xn )t and y = (y1 , y2 , . . . , yn )t are column vectors.8 For example, if
x = (011001)t and y = (110101)t , then x · y = 0. We can also look at an inner product as
the product of a row matrix with a column matrix; that is,
x · y = xt y
y1
y2
= x1 x2 · · · xn .
..
yn
8
Since we will be working with matrices, we will write binary n-tuples as column vectors for the remainder
of this chapter.
CHAPTER 8. ALGEBRAIC CODING THEORY 100
= x1 y1 + x2 y2 + · · · + xn yn .
Example 8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that
our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to
obtain an even number of 1s. Notice that an arbitrary n-tuple x = (x1 , x2 , . . . , xn )t has an
even number of 1s exactly when x1 + x2 + · · · + xn = 0; hence, a 4-tuple x = (x1 , x2 , x3 , x4 )t
has an even number of 1s if x1 + x2 + x3 + x4 = 0, or
1
1
x · 1 = xt 1 = x1 x2 x3 x4 = 0.
1
1
This example leads us to hope that there is a connection between matrices and coding
theory. □
Let Mm×n (Z2 ) denote the set of all m × n matrices with entries in Z2 . We do matrix
operations as usual except that all our addition and multiplication operations occur in Z2 .
Define the null space of a matrix H ∈ Mm×n (Z2 ) to be the set of all binary n-tuples x such
that Hx = 0. We denote the null space of a matrix H by Null(H).
Example 8.20 Suppose that
0 1 0 1 0
H = 1 1 1 1 0 .
0 0 1 1 1
For a 5-tuple x = (x1 , x2 , x3 , x4 , x5 )t to be in the null space of H, Hx = 0. Equivalently,
the following system of equations must be satisfied:
x2 + x4 = 0
x1 + x2 + x3 + x4 = 0
x3 + x4 + x5 = 0.
The set of binary 5-tuples satisfying these equations is
(00000) (11110) (10101) (01011).
This code is easily determined to be a group code. □
Theorem 8.21 Let H be in Mm×n (Z2 ). Then the null space of H is a group code.
Proof. Since each element of Zn2 is its own inverse, the only thing that really needs to be
checked here is closure. Let x, y ∈ Null(H) for some matrix H in Mm×n (Z2 ). Then Hx = 0
and Hy = 0. So
H(x + y) = Hx + Hy = 0 + 0 = 0.
Hence, x + y is in the null space of H and therefore must be a codeword. ■
A code is a linear code if it is determined by the null space of some matrix H ∈
Mm×n (Z2 ).
Example 8.22 Let C be the code given by the matrix
0 0 0 1 1 1
H = 0 1 1 0 1 1 .
1 0 1 0 0 1
CHAPTER 8. ALGEBRAIC CODING THEORY 101
Suppose that the 6-tuple x = (010011)t is received. It is a simple matter of matrix multi-
plication to determine whether or not x is a codeword. Since
0
Hx = 1 ,
1
the received word is not a codeword. We must either attempt to correct the word or request
that it be transmitted again. □
8.3 Parity-Check and Generator Matrices
We need to find a systematic way of generating linear codes as well as fast methods of
decoding. By examining the properties of a matrix H and by carefully choosing H, it is
possible to develop very efficient methods of encoding and decoding messages. To this end,
we will introduce standard generator and canonical parity-check matrices.
Suppose that H is an m × n matrix with entries in Z2 and n > m. If the last m
columns of the matrix form the m × m identity matrix, Im , then the matrix is a canonical
parity-check matrix. More specifically, H = (A | Im ), where A is the m × (n − m) matrix
a11 a12 ··· a1,n−m
a21 a22 ··· a2,n−m
. .. .. ..
.. . . .
am1 am2 · · · am,n−m
and Im is the m × m identity matrix
1 0 ··· 0
0 1 · · · 0
. . . .. .
. .
. . . . .
0 0 ··· 1
With each canonical parity-check matrix we can associate an n×(n−m) standard generator
matrix
In−m
G= .
A
Our goal will be to show that an x satisfying Gx = y exists if and only if Hy = 0. Given a
message block x to be encoded, the matrix G will allow us to quickly encode it into a linear
codeword y.
Example 8.23 Suppose that we have the following eight words to be encoded:
(000), (001), (010), . . . , (111).
For
0 1 1
A = 1 1 0 ,
1 0 1
CHAPTER 8. ALGEBRAIC CODING THEORY 102
the associated standard generator and canonical parity-check matrices are
1 0 0
0 1 0
0 0 1
G=
0 1 1
1 1 0
1 0 1
and
0 1 1 1 0 0
H = 1 1 0 0 1 0 ,
1 0 1 0 0 1
respectively.
Observe that the rows in H represent the parity checks on certain bit positions in a
6-tuple. The 1s in the identity matrix serve as parity checks for the 1s in the same row. If
x = (x1 , x2 , x3 , x4 , x5 , x6 ), then
x2 + x3 + x4
0 = Hx = x1 + x2 + x5 ,
x1 + x3 + x6
which yields a system of equations:
x2 + x3 + x4 = 0
x1 + x2 + x5 = 0
x1 + x3 + x6 = 0.
Here x4 serves as a check bit for x2 and x3 ; x5 is a check bit for x1 and x2 ; and x6 is a check
bit for x1 and x3 . The identity matrix keeps x4 , x5 , and x6 from having to check on each
other. Hence, x1 , x2 , and x3 can be arbitrary but x4 , x5 , and x6 must be chosen to ensure
parity. The null space of H is easily computed to be
(000000) (001101) (010110) (011011)
(100011) (101110) (110101) (111000).
An even easier way to compute the null space is with the generator matrix G (Table 8.24).
□
Table 8.24 A matrix-generated code
Message Word x Codeword Gx
000 000000
001 001101
010 010110
011 011011
100 100011
101 101110
110 110101
111 111000
CHAPTER 8. ALGEBRAIC CODING THEORY 103
Theorem 8.25 If H ∈ Mm×n (Z2 ) is a canonical parity-check matrix, then Null(H) consists
of all x ∈ Zn2 whose first n − m bits are arbitrary but whose last m bits are determined by
Hx = 0. Each of the last m bits serves as an even parity check bit for some of the first
n − m bits. Hence, H gives rise to an (n, n − m)-block code.
We leave the proof of this theorem as an exercise. In light of the theorem, the first
n − m bits in x are called information bits and the last m bits are called check bits. In
Example 8.23, the first three bits are the information bits and the last three are the check
bits.
Theorem
8.26 Suppose that G is an n × k standard generator matrix. Then C =
y : Gx = y for x ∈ Zk2 is an (n, k)-block code. More specifically, C is a group code.
Proof. Let Gx1 = y1 and Gx2 = y2 be two codewords. Then y1 + y2 is in C since
G(x1 + x2 ) = Gx1 + Gx2 = y1 + y2 .
We must also show that two message blocks cannot be encoded into the same codeword.
That is, we must show that if Gx = Gy, then x = y. Suppose that Gx = Gy. Then
Gx − Gy = G(x − y) = 0.
However, the first k coordinates in G(x − y) are exactly x1 − y1 , . . . , xk − yk , since they are
determined by the identity matrix, Ik , part of G. Hence, G(x − y) = 0 exactly when x = y.
■
Before we can prove the relationship between canonical parity-check matrices and stan-
dard generating matrices, we need to prove a lemma.
Lemma 8.27 Let H = (A | Im ) be an m×n canonical parity-check matrix and G = In−m A
be the corresponding n × (n − m) standard generator matrix. Then HG = 0.
Proof. Let C = HG. The ijth entry in C is
X
n
cij = hik gkj
k=1
X
n−m X
n
= hik gkj + hik gkj
k=1 k=n−m+1
X
n−m Xn
= aik δkj + δi−(m−n),k akj
k=1 k=n−m+1
= aij + aij
= 0,
where (
1 i=j
δij =
0 i 6= j
is the Kronecker delta. ■
Theorem
8.28 Let H = (A | Im ) be an m × n canonical parity-check matrix and let
In−m
G= A be the n × (n − m) standard generator matrix associated with H. Let C be the
code generated by G. Then y is in C if and only if Hy = 0. In particular, C is a linear
code with canonical parity-check matrix H.
CHAPTER 8. ALGEBRAIC CODING THEORY 104
Proof. First suppose that y ∈ C. Then Gx = y for some x ∈ Zm 2 . By Lemma 8.27,
Hy = HGx = 0.
Conversely, suppose that y = (y1 , . . . , yn )t is in the null space of H. We need to find
an x in Zn−m
2 such that Gxt = y. Since Hy = 0, the following set of equations must be
satisfied:
a11 y1 + a12 y2 + · · · + a1,n−m yn−m + yn−m+1 = 0
a21 y1 + a22 y2 + · · · + a2,n−m yn−m + yn−m+2 = 0
..
.
am1 y1 + am2 y2 + · · · + am,n−m yn−m + yn−m+m = 0.
Equivalently, yn−m+1 , . . . , yn are determined by y1 , . . . , yn−m :
yn−m+1 = a11 y1 + a12 y2 + · · · + a1,n−m yn−m
yn−m+2 = a21 y1 + a22 y2 + · · · + a2,n−m yn−m
..
.
yn = am1 y1 + am2 y2 + · · · + am,n−m yn−m .
Consequently, we can let xi = yi for i = 1, . . . , n − m. ■
It would be helpful if we could compute the minimum distance of a linear code directly
from its matrix H in order to determine the error-detecting and error-correcting capabilities
of the code. Suppose that
e1 = (100 · · · 00)t
e2 = (010 · · · 00)t
..
.
en = (000 · · · 01)t
are the n-tuples in Zn2 of weight 1. For an m × n binary matrix H, Hei is exactly the ith
column of the matrix H.
Example 8.29 Observe that
0
1 1 1 0 0 1 1
1 0 0 1 0
0 = 0 .
1 1 0 0 1 0 1
0
□
We state this result in the following proposition and leave the proof as an exercise.
Proposition 8.30 Let ei be the binary n-tuple with a 1 in the ith coordinate and 0’s
elsewhere and suppose that H ∈ Mm×n (Z2 ). Then Hei is the ith column of the matrix H.
Theorem 8.31 Let H be an m × n binary matrix. Then the null space of H is a single
error-detecting code if and only if no column of H consists entirely of zeros.
Proof. Suppose that Null(H) is a single error-detecting code. Then the minimum distance
of the code must be at least 2. Since the null space is a group code, it is sufficient to require
that the code contain no codewords of less than weight 2 other than the zero codeword.
That is, ei must not be a codeword for i = 1, . . . , n. Since Hei is the ith column of H, the
CHAPTER 8. ALGEBRAIC CODING THEORY 105
only way in which ei could be in the null space of H would be if the ith column were all
zeros, which is impossible; hence, the code must have the capability to detect at least single
errors.
Conversely, suppose that no column of H is the zero column. By Proposition 8.30,
Hei 6= 0. ■
Example 8.32 If we consider the matrices
1 1 1 0 0
H1 = 1 0 0 1 0
1 1 0 0 1
and
1 1 1 0 0
H2 = 1 0 0 0 0 ,
1 1 0 0 1
then the null space of H1 is a single error-detecting code and the null space of H2 is not.
□
We can even do better than Theorem 8.31. This theorem gives us conditions on a matrix
H that tell us when the minimum weight of the code formed by the null space of H is 2.
We can also determine when the minimum distance of a linear code is 3 by examining the
corresponding matrix.
Example 8.33 If we let
1 1 1 0
H = 1 0 0 1
1 1 0 0
and want to determine whether or not H is the canonical parity-check matrix for an error-
correcting code, it is necessary to make certain that Null(H) does not contain any 4-tuples
of weight 2. That is, (1100), (1010), (1001), (0110), (0101), and (0011) must not be in
Null(H). The next theorem states that we can indeed determine that the code generated
by H is error-correcting by examining the columns of H. Notice in this example that not
only does H have no zero columns, but also that no two columns are the same. □
Theorem 8.34 Let H be a binary matrix. The null space of H is a single error-correcting
code if and only if H does not contain any zero columns and no two columns of H are
identical.
Proof. The n-tuple ei + ej has 1s in the ith and jth entries and 0s elsewhere, and
w(ei + ej ) = 2 for i 6= j. Since
0 = H(ei + ej ) = Hei + Hej
can only occur if the ith and jth columns are identical, the null space of H is a single
error-correcting code. ■
Suppose now that we have a canonical parity-check matrix H with three rows. Then
we might ask how many more columns we can add to the matrix and still have a null space
that is a single error-detecting and single error-correcting code. Since each column has three
entries, there are 23 = 8 possible distinct columns. We cannot add the columns
0 1 0 0
0 , 0 , 1 , 0 .
0 0 0 1
CHAPTER 8. ALGEBRAIC CODING THEORY 106
So we can add as many as four columns and still maintain a minimum distance of 3.
In general, if H is an m × n canonical parity-check matrix, then there are n − m informa-
tion positions in each codeword. Each column has m bits, so there are 2m possible distinct
columns. It is necessary that the columns 0, e1 , . . . , em be excluded, leaving 2m − (1 + m)
remaining columns for information if we are still to maintain the ability not only to detect
but also to correct single errors.
8.4 Efficient Decoding
We are now at the stage where we are able to generate linear codes that detect and correct
errors fairly easily, but it is still a time-consuming process to decode a received n-tuple and
determine which is the closest codeword, because the received n-tuple must be compared to
each possible codeword to determine the proper decoding. This can be a serious impediment
if the code is very large.
Example 8.35 Given the binary matrix
1 1 1 0 0
H = 0 1 0 1 0
1 0 0 0 1
and the 5-tuples x = (11011)t and y = (01011)t , we can compute
0 1
Hx = 0 and Hy = 0 .
0 1
Hence, x is a codeword and y is not, since x is in the null space and y is not. Notice that
Hy is identical to the first column of H. In fact, this is where the error occurred. If we flip
the first bit in y from 0 to 1, then we obtain x. □
If H is an m × n matrix and x ∈ Z2 , then we say that the syndrome of x is Hx. The
n
following proposition allows the quick detection and correction of errors.
Proposition 8.36 Let the m × n binary matrix H determine a linear code and let x be the
received n-tuple. Write x as x = c + e, where c is the transmitted codeword and e is the
transmission error. Then the syndrome Hx of the received codeword x is also the syndrome
of the error e.
Proof. The proof follows from the fact that
Hx = H(c + e) = Hc + He = 0 + He = He.
■
This proposition tells us that the syndrome of a received word depends solely on the
error and not on the transmitted codeword. The proof of the following theorem follows
immediately from Proposition 8.36 and from the fact that He is the ith column of the
matrix H.
Theorem 8.37 Let H ∈ Mm×n (Z2 ) and suppose that the linear code corresponding to H
is single error-correcting. Let r be a received n-tuple that was transmitted with at most one
error. If the syndrome of r is 0, then no error has occurred; otherwise, if the syndrome of r
is equal to some column of H, say the ith column, then the error has occurred in the ith bit.
CHAPTER 8. ALGEBRAIC CODING THEORY 107
Example 8.38 Consider the matrix
1 0 1 1 0 0
H = 0 1 1 0 1 0
1 1 1 0 0 1
and suppose that the 6-tuples x = (111110)t , y = (111111)t , and z = (010111)t have been
received. Then
1 1 1
Hx = 1 , Hy = 1 , Hz = 0 .
1 0 0
Hence, x has an error in the third bit and z has an error in the fourth bit. The transmitted
codewords for x and z must have been (110110) and (010011), respectively. The syndrome
of y does not occur in any of the columns of the matrix H, so multiple errors must have
occurred to produce y. □
Coset Decoding
We can use group theory to obtain another way of decoding messages. A linear code C is
a subgroup of Zn2 . Coset or standard decoding uses the cosets of C in Zn2 to implement
maximum-likelihood decoding. Suppose that C is an (n, m)-linear code. A coset of C in Zn2
is written in the form x + C, where x ∈ Zn2 . By Lagrange’s Theorem (Theorem 6.10), there
are 2n−m distinct cosets of C in Zn2 .
Example 8.39 Let C be the (5, 3)-linear code given by the parity-check matrix
0 1 1 0 0
H = 1 0 0 1 0 .
1 1 0 0 1
The code consists of the codewords
(00000) (01101) (10011) (11110).
There are 25−2 = 23 cosets of C in Z52 , each with order 22 = 4. These cosets are listed in
Table 8.40. □
Table 8.40 Cosets of C
Coset Coset
Representative
C (00000)(01101)(10011)(11110)
(10000) + C (10000)(11101)(00011)(01110)
(01000) + C (01000)(00101)(11011)(10110)
(00100) + C (00100)(01001)(10111)(11010)
(00010) + C (00010)(01111)(10001)(11100)
(00001) + C (00001)(01100)(10010)(11111)
(10100) + C (00111)(01010)(10100)(11001)
(00110) + C (00110)(01011)(10101)(11000)
Our task is to find out how knowing the cosets might help us to decode a message.
Suppose that x was the original codeword sent and that r is the n-tuple received. If e is
CHAPTER 8. ALGEBRAIC CODING THEORY 108
the transmission error, then r = e + x or, equivalently, x = e + r. However, this is exactly
the statement that r is an element in the coset e + C. In maximum-likelihood decoding
we expect the error e to be as small as possible; that is, e will have the least weight. An
n-tuple of least weight in a coset is called a coset leader. Once we have determined a coset
leader for each coset, the decoding process becomes a task of calculating r + e to obtain x.
Example 8.41 In Table 8.40, notice that we have chosen a representative of the least
possible weight for each coset. These representatives are coset leaders. Now suppose that
r = (01111) is the received word. To decode r, we find that it is in the coset (00010) + C;
hence, the originally transmitted codeword must have been (01101) = (01111) + (00010).
□
A potential problem with this method of decoding is that we might have to examine every
coset for the received codeword. The following proposition gives a method of implementing
coset decoding. It states that we can associate a syndrome with each coset; hence, we can
make a table that designates a coset leader corresponding to each syndrome. Such a list is
called a decoding table.
Table 8.42 Syndromes for each coset
Syndrome Coset Leader
(000) (00000)
(001) (00001)
(010) (00010)
(011) (10000)
(100) (00100)
(101) (01000)
(110) (00110)
(111) (10100)
Proposition 8.43 Let C be an (n, k)-linear code given by the matrix H and suppose that
x and y are in Zn2 . Then x and y are in the same coset of C if and only if Hx = Hy. That
is, two n-tuples are in the same coset if and only if their syndromes are the same.
Proof. Two n-tuples x and y are in the same coset of C exactly when x − y ∈ C; however,
this is equivalent to H(x − y) = 0 or Hx = Hy. ■
Example 8.44 Table 8.42 is a decoding table for the code C given in Example 8.39. If
x = (01111) is received, then its syndrome can be computed to be
0
Hx = 1 .
0
Examining the decoding table, we determine that the coset leader is (00010). It is now easy
to decode the received codeword. □
Given an (n, k)-block code, the question arises of whether or not coset decoding is a
manageable scheme. A decoding table requires a list of cosets and syndromes, one for each
of the 2n−k cosets of C. Suppose that we have a (32, 24)-block code. We have a huge
number of codewords, 224 , yet there are only 232−24 = 28 = 256 cosets.
Sage. Sage has a substantial repertoire of commands for coding theory, including the
ability to build many different families of codes.
CHAPTER 8. ALGEBRAIC CODING THEORY 109
8.5 Reading Questions
1. Suppose a binary code has minimum distance d = 6. How many errors can be detected?
How many errors can be corrected?
2. Explain why it is impossible for the 8-bit string with decimal value 5610 to be an
ascii code for a character. Assume the leftmost bit of the string is being used as a
parity-check bit.
3. Suppose we receive the 8-bit string with decimal value 5610 when we are expecting
ascii characters with a parity-check bit in the first bit (leftmost). We know an error
has occurred in transmission. Give one of the probable guesses for the character which
was actually sent (other than ‘8’), under the assumption that any individual bit is
rarely sent in error. Explain the logic of your answer. (You may need to consult a
table of ascii values online.)
4. Suppose a linear code C is created as the null space of the parity-check matrix
0 1 0 1 0
H = 1 1 1 1 0
0 0 1 1 1
Then x = 11100 is not a codeword. Describe a computation, and give the result of
that computation, which verifies that x is not a codeword of the code C.
5. For H and x as in the previous question, suppose that x is received as a message. Give
a maximum likelihood decoding of the received message.
8.6 Exercises
1. Why is the following encoding scheme not acceptable?
Information 0 1 2 3 4 5 6 7 8
Codeword 000 001 010 011 101 110 111 000 001
2. Without doing any addition, explain why the following set of 4-tuples in Z42 cannot be
a group code.
(0110) (1001) (1010) (1100)
3. Compute the Hamming distances between the following pairs of n-tuples.
(a) (011010), (011100) (c) (00110), (01111)
(b) (11110101), (01010100) (d) (1001), (0111)
4. Compute the weights of the following n-tuples.
(a) (011010) (c) (01111)
(b) (11110101) (d) (1011)
5. Suppose that a linear code C has a minimum weight of 7. What are the error-detection
and error-correction capabilities of C?
6. In each of the following codes, what is the minimum distance for the code? What
is the best situation we might hope for in connection with error detection and error
correction?
(a) (011010) (011100) (110111) (110000)
(b) (011100) (011011) (111011) (100011)
(000000) (010101) (110100) (110011)
CHAPTER 8. ALGEBRAIC CODING THEORY 110
(c) (000000) (011100) (110101) (110001)
(d) (0110110) (0111100) (1110000) (1111111)
(1001001) (1000011) (0001111) (0000000)
7. Compute the null space of each of the following matrices. What type of (n, k)-block
codes are the null spaces? Can you find a matrix (not necessarily a standard generator
matrix) that generates each code? Are your generator matrices unique?
(a) (c)
0 1 0 0 0 1 0 0 1 1
1 0 1 0 1
0 1 0 1 1
1 0 0 1 0
(b) (d)
1 0 1 0 0 0 0 0 0 1 1 1 1
1 1 0 1 0 0 0 1 1 0 0 1 1
0 1 0 0 1 0 1 0 1 0 1 0 1
1 1 0 0 0 1 0 1 1 0 0 1 1
8. Construct a (5, 2)-block code. Discuss both the error-detection and error-correction
capabilities of your code.
9. Let C be the code obtained from the null space of the matrix
0 1 0 0 1
H = 1 0 1 0 1 .
0 0 1 1 1
Decode the message
01111 10101 01110 00011
if possible.
10. Suppose that a 1000-bit binary message is transmitted. Assume that the probability of
a single error is p and that the errors occurring in different bits are independent of one
another. If p = 0.01, what is the probability of more than one error occurring? What
is the probability of exactly two errors occurring? Repeat this problem for p = 0.0001.
11. Which matrices are canonical parity-check matrices? For those matrices that are canon-
ical parity-check matrices, what are the corresponding standard generator matrices?
What are the error-detection and error-correction capabilities of the code generated by
each of these matrices?
(a) (c)
1 1 0 0 0
1 1 1 0
0 0 1 0 0
1 0 0 1
0 0 0 1 0
1 0 0 0 1
(d)
(b)
0 1 1 0 0 0 0 0 0 1 0 0 0
1 1 0 1 0 0 0 1 1 0 1 0 0
0 1 0 0 1 0 1 0 1 0 0 1 0
1 1 0 0 0 1 0 1 1 0 0 0 1
12. List all possible syndromes for the codes generated by each of the matrices in Exer-
cise 8.6.11.
CHAPTER 8. ALGEBRAIC CODING THEORY 111
13. Let
0 1 1 1 1
H = 0 0 0 1 1 .
1 0 1 0 1
Compute the syndrome caused by each of the following transmission errors.
(a) An error in the first bit.
(b) An error in the third bit.
(c) An error in the last bit.
(d) Errors in the third and fourth bits.
14. Let C be the group code in Z32 defined by the codewords (000) and (111). Compute
the cosets of C in Z32 . Why was there no need to specify right or left cosets? Give the
single transmission error, if any, to which each coset corresponds.
15. For each of the following matrices, find the cosets of the corresponding code C. Give
a decoding table for each code if possible.
(a) (c)
0 1 0 0 0 1 0 0 1 1
1 0 1 0 1
0 1 0 1 1
1 0 0 1 0
(b) (d)
0 0 1 0 0 1 0 0 1 1 1 1
1 1 0 1 0 1 1 1 0 0 1 1
0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 1 1 0 0 1 0
16. Let x, y, and z be binary n-tuples. Prove each of the following statements.
(a) w(x) = d(x, 0)
(b) d(x, y) = d(x + z, y + z)
(c) d(x, y) = w(x − y)
17. A metric on a set X is a map d : X × X → R satisfying the following conditions.
(a) d(x, y) ≥ 0 for all x, y ∈ X;
(b) d(x, y) = 0 exactly when x = y;
(c) d(x, y) = d(y, x);
(d) d(x, y) ≤ d(x, z) + d(z, y).
In other words, a metric is simply a generalization of the notion of distance. Prove
that Hamming distance is a metric on Zn2 . Decoding a message actually reduces to
deciding which is the closest codeword in terms of distance.
18. Let C be a linear code. Show that either the ith coordinates in the codewords of C
are all zeros or exactly half of them are zeros.
19. Let C be a linear code. Show that either every codeword has even weight or exactly
half of the codewords have even weight.
20. Show that the codewords of even weight in a linear code C are also a linear code.
CHAPTER 8. ALGEBRAIC CODING THEORY 112
21. If we are to use an error-correcting linear code to transmit the 128 ascii characters,
what size matrix must be used? What size matrix must be used to transmit the
extended ascii character set of 256 characters? What if we require only error detection
in both cases?
22. Find the canonical parity-check matrix that gives the even parity check bit code with
three information positions. What is the matrix for seven information positions? What
are the corresponding standard generator matrices?
23. How many check positions are needed for a single error-correcting code with 20 infor-
mation positions? With 32 information positions?
24. Let ei be the binary n-tuple with a 1 in the ith coordinate and 0’s elsewhere and
suppose that H ∈ Mm×n (Z2 ). Show that Hei is the ith column of the matrix H.
25. Let C be an (n, k)-linear code. Define the dual or orthogonal code of C to be
C ⊥ = {x ∈ Zn2 : x · y = 0 for all y ∈ C}.
(a) Find the dual code of the linear code C where C is given by the matrix
1 1 1 0 0
0 0 1 0 1 .
1 0 0 1 0
(b) Show that C ⊥ is an (n, n − k)-linear code.
(c) Find the standard generator and parity-check matrices of C and C ⊥ . What
happens in general? Prove your conjecture.
26. Let H be an m × n matrix over Z2 , where the ith column is the number i written in
binary with m bits. The null space of such a matrix is called a Hamming code.
(a) Show that the matrix
0 0 0 1 1 1
H = 0 1 1 0 0 1
1 0 1 0 1 0
generates a Hamming code. What are the error-correcting properties of a Ham-
ming code?
(b) The column corresponding to the syndrome also marks the bit that was in error;
that is, the ith column of the matrix is i written as a binary number, and the
syndrome immediately tells us which bit is in error. If the received word is
(101011), compute the syndrome. In which bit did the error occur in this case,
and what codeword was originally transmitted?
(c) Give a binary matrix H for the Hamming code with six information positions and
four check positions. What are the check positions and what are the information
positions? Encode the messages (101101) and (001001). Decode the received
words (0010000101) and (0000101100). What are the possible syndromes for this
code?
(d) What is the number of check bits and the number of information bits in an (m, n)-
block Hamming code? Give both an upper and a lower bound on the number of
information bits in terms of the number of check bits. Hamming codes having
the maximum possible number of information bits with k check bits are called
CHAPTER 8. ALGEBRAIC CODING THEORY 113
perfect. Every possible syndrome except 0 occurs as a column. If the number of
information bits is less than the maximum, then the code is called shortened. In
this case, give an example showing that some syndromes can represent multiple
errors.
8.7 Programming Exercises
1. Write a program to implement a (16, 12)-linear code. Your program should be able to
encode and decode messages using coset decoding. Once your program is written, write
a program to simulate a binary symmetric channel with transmission noise. Compare
the results of your simulation with the theoretically predicted error probability.
8.8 References and Suggested Readings
[1] Blake, I. F. “Codes and Designs,” Mathematics Magazine 52 (1979), 81–95.
[2] Hill, R. A First Course in Coding Theory. Oxford University Press, Oxford, 1990.
[3] Levinson, N. “Coding Theory: A Counterexample to G. H. Hardy’s Conception of
Applied Mathematics,” American Mathematical Monthly 77 (1970), 249–58.
[4] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998.
[5] MacWilliams, F. J. and Sloane, N. J. A. The Theory of Error-Correcting Codes. North-
Holland Mathematical Library, 16, Elsevier, Amsterdam, 1983.
[6] Roman, S. Coding and Information Theory. Springer-Verlag, New York, 1992.
[7] Shannon, C. E. “A Mathematical Theory of Communication,” Bell System Technical
Journal 27 (1948), 379–423, 623–56.
[8] Thompson, T. M. From Error-Correcting Codes through Sphere Packing to Simple
Groups. Carus Monograph Series, No. 21. Mathematical Association of America,
Washington, DC, 1983.
[9] van Lint, J. H. Introduction to Coding Theory. Springer, New York, 1999.
9
Isomorphisms
Many groups may appear to be different at first glance, but can be shown to be the same
by a simple renaming of the group elements. For example, Z4 and the subgroup of the
circle group T generated by i can be shown to be the same by demonstrating a one-to-one
correspondence between the elements of the two groups and between the group operations.
In such a case we say that the groups are isomorphic.
9.1 Definition and Examples
Two groups (G, ·) and (H, ◦) are isomorphic if there exists a one-to-one and onto map
ϕ : G → H such that the group operation is preserved; that is,
ϕ(a · b) = ϕ(a) ◦ ϕ(b)
for all a and b in G. If G is isomorphic to H, we write G ∼
= H. The map ϕ is called an
isomorphism.
Example 9.1 To show that Z4 ∼ = hii, define a map ϕ : Z4 → hii by ϕ(n) = in . We must
show that ϕ is bijective and preserves the group operation. The map ϕ is one-to-one and
onto because
ϕ(0) = 1
ϕ(1) = i
ϕ(2) = −1
ϕ(3) = −i.
Since
ϕ(m + n) = im+n = im in = ϕ(m)ϕ(n),
the group operation is preserved. □
Example 9.2 We can define an isomorphism ϕ from the additive group of real numbers
(R, +) to the multiplicative group of positive real numbers (R+ , ·) with the exponential map;
that is,
ϕ(x + y) = ex+y = ex ey = ϕ(x)ϕ(y).
Of course, we must still show that ϕ is one-to-one and onto, but this can be determined
using calculus. □
114
CHAPTER 9. ISOMORPHISMS 115
Example 9.3 The integers are isomorphic to the subgroup of Q∗ consisting of elements of
the form 2n . Define a map ϕ : Z → Q∗ by ϕ(n) = 2n . Then
ϕ(m + n) = 2m+n = 2m 2n = ϕ(m)ϕ(n).
By definition the map ϕ is onto the subset {2n : n ∈ Z} of Q∗ . To show that the map is
injective, assume that m 6= n. If we can show that ϕ(m) 6= ϕ(n), then we are done. Suppose
that m > n and assume that ϕ(m) = ϕ(n). Then 2m = 2n or 2m−n = 1, which is impossible
since m − n > 0. □
Example 9.4 The groups Z8 and Z12 cannot be isomorphic since they have different orders;
however, it is true that U (8) ∼
= U (12). We know that
U (8) = {1, 3, 5, 7}
U (12) = {1, 5, 7, 11}.
An isomorphism ϕ : U (8) → U (12) is then given by
1 7→ 1
3 7→ 5
5 7→ 7
7 7→ 11.
The map ϕ is not the only possible isomorphism between these two groups. We could define
another isomorphism ψ by ψ(1) = 1, ψ(3) = 11, ψ(5) = 5, ψ(7) = 7. In fact, both of these
groups are isomorphic to Z2 × Z2 (see Example 3.28 in Chapter 3). □
Example 9.5 Even though S3 and Z6 possess the same number of elements, we would
suspect that they are not isomorphic, because Z6 is abelian and S3 is nonabelian. To
demonstrate that this is indeed the case, suppose that ϕ : Z6 → S3 is an isomorphism. Let
a, b ∈ S3 be two elements such that ab 6= ba. Since ϕ is an isomorphism, there exist elements
m and n in Z6 such that
ϕ(m) = a and ϕ(n) = b.
However,
ab = ϕ(m)ϕ(n) = ϕ(m + n) = ϕ(n + m) = ϕ(n)ϕ(m) = ba,
which contradicts the fact that a and b do not commute. □
Theorem 9.6 Let ϕ : G → H be an isomorphism of two groups. Then the following
statements are true.
1. ϕ−1 : H → G is an isomorphism.
2. |G| = |H|.
3. If G is abelian, then H is abelian.
4. If G is cyclic, then H is cyclic.
5. If G has a subgroup of order n, then H has a subgroup of order n.
Proof. Assertions (1) and (2) follow from the fact that ϕ is a bijection. We will prove (3)
here and leave the remainder of the theorem to be proved in the exercises.
(3) Suppose that h1 and h2 are elements of H. Since ϕ is onto, there exist elements
CHAPTER 9. ISOMORPHISMS 116
g1 , g2 ∈ G such that ϕ(g1 ) = h1 and ϕ(g2 ) = h2 . Therefore,
h1 h2 = ϕ(g1 )ϕ(g2 ) = ϕ(g1 g2 ) = ϕ(g2 g1 ) = ϕ(g2 )ϕ(g1 ) = h2 h1 .
■
We are now in a position to characterize all cyclic groups.
Theorem 9.7 All cyclic groups of infinite order are isomorphic to Z.
Proof. Let G be a cyclic group with infinite order and suppose that a is a generator of G.
Define a map ϕ : Z → G by ϕ : n 7→ an . Then
ϕ(m + n) = am+n = am an = ϕ(m)ϕ(n).
To show that ϕ is injective, suppose that m and n are two elements in Z, where m 6= n. We
can assume that m > n. We must show that am 6= an . Let us suppose the contrary; that
is, am = an . In this case am−n = e, where m − n > 0, which contradicts the fact that a
has infinite order. Our map is onto since any element in G can be written as an for some
integer n and ϕ(n) = an . ■
Theorem 9.8 If G is a cyclic group of order n, then G is isomorphic to Zn .
Proof. Let G be a cyclic group of order n generated by a and define a map ϕ : Zn → G
by ϕ : k 7→ ak , where 0 ≤ k < n. The proof that ϕ is an isomorphism is one of the
end-of-chapter exercises. ■
Corollary 9.9 If G is a group of order p, where p is a prime number, then G is isomorphic
to Zp .
Proof. The proof is a direct result of Corollary 6.12. ■
The main goal in group theory is to classify all groups; however, it makes sense to
consider two groups to be the same if they are isomorphic. We state this result in the
following theorem, whose proof is left as an exercise.
Theorem 9.10 The isomorphism of groups determines an equivalence relation on the class
of all groups.
Hence, we can modify our goal of classifying all groups to classifying all groups up to
isomorphism; that is, we will consider two groups to be the same if they are isomorphic.
Cayley’s Theorem
Cayley proved that if G is a group, it is isomorphic to a group of permutations on some set;
hence, every group is a permutation group. Cayley’s Theorem is what we call a represen-
tation theorem. The aim of representation theory is to find an isomorphism of some group
G that we wish to study into a group that we know a great deal about, such as a group of
permutations or matrices.
Example 9.11 Consider the group Z3 . The Cayley table for Z3 is as follows.
+ 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1
The addition table of Z3 suggests that it is the same as the permutation group G =
CHAPTER 9. ISOMORPHISMS 117
{(0), (012), (021)}. The isomorphism here is
0 1 2
0 7→ = (0)
0 1 2
0 1 2
1 7→ = (0 1 2)
1 2 0
0 1 2
2 7→ = (0 2 1).
2 0 1
□
Theorem 9.12 Cayley. Every group is isomorphic to a group of permutations.
Proof. Let G be a group. We must find a group of permutations G that is isomorphic
to G. For any g ∈ G, define a function λg : G → G by λg (a) = ga. We claim that λg is a
permutation of G. To show that λg is one-to-one, suppose that λg (a) = λg (b). Then
ga = λg (a) = λg (b) = gb.
Hence, a = b. To show that λg is onto, we must prove that for each a ∈ G, there is a b such
that λg (b) = a. Let b = g −1 a.
Now we are ready to define our group G. Let
G = {λg : g ∈ G}.
We must show that G is a group under composition of functions and find an isomorphism
between G and G. We have closure under composition of functions since
(λg ◦ λh )(a) = λg (ha) = gha = λgh (a).
Also,
λe (a) = ea = a
and
(λg−1 ◦ λg )(a) = λg−1 (ga) = g −1 ga = a = λe (a).
We can define an isomorphism from G to G by ϕ : g 7→ λg . The group operation is
preserved since
ϕ(gh) = λgh = λg λh = ϕ(g)ϕ(h).
It is also one-to-one, because if ϕ(g)(a) = ϕ(h)(a), then
ga = λg a = λh a = ha.
Hence, g = h. That ϕ is onto follows from the fact that ϕ(g) = λg for any λg ∈ G. ■
The isomorphism g 7→ λg is known as the left regular representation of G.
Historical Note
Arthur Cayley was born in England in 1821, though he spent much of the first part of his
life in Russia, where his father was a merchant. Cayley was educated at Cambridge, where
he took the first Smith’s Prize in mathematics. A lawyer for much of his adult life, he
wrote several papers in his early twenties before entering the legal profession at the age of
25. While practicing law he continued his mathematical research, writing more than 300
papers during this period of his life. These included some of his best work. In 1863 he left
CHAPTER 9. ISOMORPHISMS 118
law to become a professor at Cambridge. Cayley wrote more than 900 papers in fields such
as group theory, geometry, and linear algebra. His legal knowledge was very valuable to
Cambridge; he participated in the writing of many of the university’s statutes. Cayley was
also one of the people responsible for the admission of women to Cambridge.
9.2 Direct Products
Given two groups G and H, it is possible to construct a new group from the Cartesian
product of G and H, G × H. Conversely, given a large group, it is sometimes possible to
decompose the group; that is, a group is sometimes isomorphic to the direct product of
two smaller groups. Rather than studying a large group G, it is often easier to study the
component groups of G.
External Direct Products
If (G, ·) and (H, ◦) are groups, then we can make the Cartesian product of G and H into a
new group. As a set, our group is just the ordered pairs (g, h) ∈ G × H where g ∈ G and
h ∈ H. We can define a binary operation on G × H by
(g1 , h1 )(g2 , h2 ) = (g1 · g2 , h1 ◦ h2 );
that is, we just multiply elements in the first coordinate as we do in G and elements in the
second coordinate as we do in H. We have specified the particular operations · and ◦ in
each group here for the sake of clarity; we usually just write (g1 , h1 )(g2 , h2 ) = (g1 g2 , h1 h2 ).
Proposition 9.13 Let G and H be groups. The set G × H is a group under the operation
(g1 , h1 )(g2 , h2 ) = (g1 g2 , h1 h2 ) where g1 , g2 ∈ G and h1 , h2 ∈ H.
Proof. Clearly the binary operation defined above is closed. If eG and eH are the identities
of the groups G and H respectively, then (eG , eH ) is the identity of G × H. The inverse of
(g, h) ∈ G × H is (g −1 , h−1 ). The fact that the operation is associative follows directly from
the associativity of G and H. ■
Example 9.14 Let R be the group of real numbers under addition. The Cartesian product
of R with itself, R × R = R2 , is also a group, in which the group operation is just addition in
each coordinate; that is, (a, b) + (c, d) = (a + c, b + d). The identity is (0, 0) and the inverse
of (a, b) is (−a, −b). □
Example 9.15 Consider
Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)}.
Although Z2 × Z2 and Z4 both contain four elements, they are not isomorphic. Every
element (a, b) in Z2 × Z2 other than the identity has order 2, since (a, b) + (a, b) = (0, 0);
however, Z4 is cyclic. □
The group G × H is called the external direct product of G and H. Notice that there
is nothing special about the fact that we have used only two groups to build a new group.
The direct product
Yn
Gi = G1 × G2 × · · · × Gn
i=1
of the groups G1 , G2 , . . . , Gn is defined in exactly the same manner. If G = G1 = G2 =
· · · = Gn , we often write Gn instead of G1 × G2 × · · · × Gn .
CHAPTER 9. ISOMORPHISMS 119
Example 9.16 The group Zn2 , considered as a set, is just the set of all binary n-tuples.
The group operation is the “exclusive or” of two binary n-tuples. For example,
(01011101) + (01001011) = (00010110).
This group is important in coding theory, in cryptography, and in many areas of computer
science. □
Theorem 9.17 Let (g, h) ∈ G × H. If g and h have finite orders r and s respectively, then
the order of (g, h) in G × H is the least common multiple of r and s.
Proof. Suppose that m is the least common multiple of r and s and let n = |(g, h)|. Then
(g, h)m = (g m , hm ) = (eG , eH )
(g n , hn ) = (g, h)n = (eG , eH ).
Hence, n must divide m, and n ≤ m. However, by the second equation, both r and s must
divide n; therefore, n is a common multiple of r and s. Since m is the least common multiple
of r and s, m ≤ n. Consequently, m must be equal to n. ■
Q
Corollary 9.18 Q Let (g1 , . . . , gn ) ∈ Gi . If gi has finite order ri in Gi , then the order of
(g1 , . . . , gn ) in Gi is the least common multiple of r1 , . . . , rn .
Example 9.19 Let (8, 56) ∈ Z12 × Z60 . Since gcd(8, 12) = 4, the order of 8 is 12/4 = 3 in
Z12 . Similarly, the order of 56 in Z60 is 15. The least common multiple of 3 and 15 is 15;
hence, (8, 56) has order 15 in Z12 × Z60 . □
Example 9.20 The group Z2 × Z3 consists of the pairs
(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2).
In this case, unlike that of Z2 × Z2 and Z4 , it is true that Z2 × Z3 ∼
= Z6 . We need only show
that Z2 × Z3 is cyclic. It is easy to see that (1, 1) is a generator for Z2 × Z3 . □
The next theorem tells us exactly when the direct product of two cyclic groups is cyclic.
Theorem 9.21 The group Zm × Zn is isomorphic to Zmn if and only if gcd(m, n) = 1.
Proof. We will first show that if Zm × Zn ∼ = Zmn , then gcd(m, n) = 1. We will prove the
contrapositive; that is, we will show that if gcd(m, n) = d > 1, then Zm ×Zn cannot be cyclic.
Notice that mn/d is divisible by both m and n; hence, for any element (a, b) ∈ Zm × Zn ,
(a, b) + (a, b) + · · · + (a, b) = (0, 0).
| {z }
mn/d times
Therefore, no (a, b) can generate all of Zm × Zn .
The converse follows directly from Theorem 9.17 since lcm(m, n) = mn if and only if
gcd(m, n) = 1. ■
Corollary 9.22 Let n1 , . . . , nk be positive integers. Then
Y
k
Zni ∼
= Zn1 ···nk
i=1
if and only if gcd(ni , nj ) = 1 for i 6= j.
Corollary 9.23 If
m = pe11 · · · pekk ,
CHAPTER 9. ISOMORPHISMS 120
where the pi s are distinct primes, then
Zm ∼
= Zpe11 × · · · × Zpek .
k
e
Proof. Since the greatest common divisor of pei i and pj j is 1 for i 6= j, the proof follows
from Corollary 9.22. ■
In Chapter 13, we will prove that all finite abelian groups are isomorphic to direct
products of the form
Zpe1 × · · · × Zpek
1 k
where p1 , . . . , pk are (not necessarily distinct) primes.
Internal Direct Products
The external direct product of two groups builds a large group out of two smaller groups.
We would like to be able to reverse this process and conveniently break down a group into
its direct product components; that is, we would like to be able to say when a group is
isomorphic to the direct product of two of its subgroups.
Let G be a group with subgroups H and K satisfying the following conditions.
• G = HK = {hk : h ∈ H, k ∈ K};
• H ∩ K = {e};
• hk = kh for all k ∈ K and h ∈ H.
Then G is the internal direct product of H and K.
Example 9.24 The group U (8) is the internal direct product of
H = {1, 3} and K = {1, 5}.
□
Example 9.25 The dihedral group D6 is an internal direct product of its two subgroups
H = {id, r3 } and K = {id, r2 , r4 , s, r2 s, r4 s}.
It can easily be shown that K ∼
= S3 ; consequently, D6 ∼
= Z2 × S3 . □
Example 9.26 Not every group can be written as the internal direct product of two of its
proper subgroups. If the group S3 were an internal direct product of its proper subgroups
H and K, then one of the subgroups, say H, would have to have order 3. In this case H is
the subgroup {(1), (123), (132)}. The subgroup K must have order 2, but no matter which
subgroup we choose for K, the condition that hk = kh will never be satisfied for h ∈ H and
k ∈ K. □
Theorem 9.27 Let G be the internal direct product of subgroups H and K. Then G is
isomorphic to H × K.
Proof. Since G is an internal direct product, we can write any element g ∈ G as g = hk
for some h ∈ H and some k ∈ K. Define a map ϕ : G → H × K by ϕ(g) = (h, k).
The first problem that we must face is to show that ϕ is a well-defined map; that is, we
must show that h and k are uniquely determined by g. Suppose that g = hk = h′ k ′ . Then
h−1 h′ = k(k ′ )−1 is in both H and K, so it must be the identity. Therefore, h = h′ and
k = k ′ , which proves that ϕ is, indeed, well-defined.
To show that ϕ preserves the group operation, let g1 = h1 k1 and g2 = h2 k2 and observe
CHAPTER 9. ISOMORPHISMS 121
that
ϕ(g1 g2 ) = ϕ(h1 k1 h2 k2 )
= ϕ(h1 h2 k1 k2 )
= (h1 h2 , k1 k2 )
= (h1 , k1 )(h2 , k2 )
= ϕ(g1 )ϕ(g2 ).
We will leave the proof that ϕ is one-to-one and onto as an exercise. ■
Example 9.28 The group Z6 is an internal direct product isomorphic to {0, 2, 4} × {0, 3}.
□
We can extend the definition of an internal direct product of G to a collection of sub-
groups H1 , H2 , . . . , Hn of G, by requiring that
• G = H1 H2 · · · Hn = {h1 h2 · · · hn : hi ∈ Hi };
• Hi ∩ h∪j̸=i Hj i = {e};
• hi hj = hj hi for all hi ∈ Hi and hj ∈ Hj .
We will leave the proof of the following theorem as an exercise.
Theorem 9.29 Let G beQthe internal direct product of subgroups Hi , where i = 1, 2, . . . , n.
Then G is isomorphic to i Hi .
Sage. Sage can quickly determine if two permutation groups are isomorphic, even though
this should, in theory, be a very difficult computation.
9.3 Reading Questions
1. Determine the order of (1, 2) in Z4 × Z8 .
2. List three properties of a group that are preserved by an isomorphism.
3. Find a group isomorphic to Z15 that is an external direct product of two non-trivial
groups.
4. Explain why we can now say “the infinite cyclic group”?
5. Compare and contrast external direct products and internal direct products.
9.4 Exercises
1. Prove that Z ∼
= nZ for n 6= 0.
2. Prove that C∗ is isomorphic to the subgroup of GL2 (R) consisting of matrices of the
form
a b
.
−b a
3. Prove or disprove: U (8) ∼
= Z4 .
4. Prove that U (8) is isomorphic to the group of matrices
1 0 1 0 −1 0 −1 0
, , , .
0 1 0 −1 0 1 0 −1
CHAPTER 9. ISOMORPHISMS 122
5. Show that U (5) is isomorphic to U (10), but U (12) is not.
6. Show that the nth roots of unity are isomorphic to Zn .
7. Show that any cyclic group of order n is isomorphic to Zn .
8. Prove that Q is not isomorphic to Z.
9. Let G = R \ {−1} and define a binary operation on G by
a ∗ b = a + b + ab.
Prove that G is a group under this operation. Show that (G, ∗) is isomorphic to the
multiplicative group of nonzero real numbers.
10. Show that the matrices
1 0 0 1 0 0 0 1 0
0 1 0 0 0 1 1 0 0
0 0 1 0 1 0 0 0 1
0 0 1 0 0 1 0 1 0
1 0 0 0 1 0 0 0 1
0 1 0 1 0 0 1 0 0
form a group. Find an isomorphism of G with a more familiar group of order 6.
11. Find five non-isomorphic groups of order 8.
12. Prove S4 is not isomorphic to D12 .
13. Let ω = cis(2π/n) be a primitive nth root of unity. Prove that the matrices
ω 0 0 1
A= and B=
0 ω −1 1 0
generate a multiplicative group isomorphic to Dn .
14. Show that the set of all matrices of the form
±1 k
,
0 1
is a group isomorphic to Dn , where all entries in the matrix are in Zn .
15. List all of the elements of Z4 × Z2 .
16. Find the order of each of the following elements.
(a) (3, 4) in Z4 × Z6
(b) (6, 15, 4) in Z30 × Z45 × Z24
(c) (5, 10, 15) in Z25 × Z25 × Z25
(d) (8, 8, 8) in Z10 × Z24 × Z80
17. Prove that D4 cannot be the internal direct product of two of its proper subgroups.
18. Prove that the subgroup of Q∗ consisting of elements of the form 2m 3n for m, n ∈ Z is
an internal direct product isomorphic to Z × Z.
19. Prove that S3 × Z2 is isomorphic to D6 . Can you make a conjecture about D2n ? Prove
your conjecture.
CHAPTER 9. ISOMORPHISMS 123
20. Prove or disprove: Every abelian group of order divisible by 3 contains a subgroup of
order 3.
21. Prove or disprove: Every nonabelian group of order divisible by 6 contains a subgroup
of order 6.
22. Let G be a group of order 20. If G has subgroups H and K of orders 4 and 5 respectively
such that hk = kh for all h ∈ H and k ∈ K, prove that G is the internal direct product
of H and K.
23. Prove or disprove the following assertion. Let G, H, and K be groups. If G × K ∼ =
H × K, then G ∼ = H.
24. Prove or disprove: There is a noncyclic abelian group of order 51.
25. Prove or disprove: There is a noncyclic abelian group of order 52.
26. Let ϕ : G → H be a group isomorphism. Show that ϕ(x) = eH if and only if x = eG ,
where eG and eH are the identities of G and H, respectively.
∼ H. Show that if G is cyclic, then so is H.
27. Let G =
28. Prove that any group G of order p, p prime, must be isomorphic to Zp .
29. Show that Sn is isomorphic to a subgroup of An+2 .
30. Prove that Dn is isomorphic to a subgroup of Sn .
31. Let ϕ : G1 → G2 and ψ : G2 → G3 be isomorphisms. Show that ϕ−1 and ψ ◦ ϕ are both
isomorphisms. Using these results, show that the isomorphism of groups determines
an equivalence relation on the class of all groups.
32. Prove U (5) ∼
= Z4 . Can you generalize this result for U (p), where p is prime?
33. Write out the permutations associated with each element of S3 in the proof of Cayley’s
Theorem.
34. An automorphism of a group G is an isomorphism with itself. Prove that complex
conjugation is an automorphism of the additive group of complex numbers; that is,
show that the map ϕ(a + bi) = a − bi is an isomorphism from C to C.
35. Prove that a + ib 7→ a − ib is an automorphism of C∗ .
36. Prove that A 7→ B −1 AB is an automorphism of SL2 (R) for all B in GL2 (R).
37. We will denote the set of all automorphisms of G by Aut(G). Prove that Aut(G) is a
subgroup of SG , the group of permutations of G.
38. Find Aut(Z6 ).
39. Find Aut(Z).
40. Find two nonisomorphic groups G and H such that Aut(G) ∼
= Aut(H).
41. Let G be a group and g ∈ G. Define a map ig : G → G by ig (x) = gxg −1 . Prove that ig
defines an automorphism of G. Such an automorphism is called an inner automorphism.
The set of all inner automorphisms is denoted by Inn(G).
42. Prove that Inn(G) is a subgroup of Aut(G).
43. What are the inner automorphisms of the quaternion group Q8 ? Is Inn(G) = Aut(G)
in this case?
44. Let G be a group and g ∈ G. Define maps λg : G → G and ρg : G → G by λg (x) = gx
and ρg (x) = xg −1 . Show that ig = ρg ◦ λg is an automorphism of G. The isomorphism
g 7→ ρg is called the right regular representation of G.
45. Let G be the internal direct product of subgroups H and K. Show that the map
ϕ : G → H × K defined by ϕ(g) = (h, k) for g = hk, where h ∈ H and k ∈ K, is
one-to-one and onto.
CHAPTER 9. ISOMORPHISMS 124
46. Let G and H be isomorphic groups. If G has a subgroup of order n, prove that H must
also have a subgroup of order n.
47. If G ∼
= G and H ∼ = H, show that G × H ∼= G × H.
48. Prove that G × H is isomorphic to H × G.
49. Let n1 , . . . , nk be positive integers. Show that
Y
k
Zni ∼
= Zn1 ···nk
i=1
if and only if gcd(ni , nj ) = 1 for i 6= j.
50. Prove that A × B is abelian if and only if A and B are abelian.
51. If
Q G is the internal direct product of H1 , H2 , . . . , Hn , prove that G is isomorphic to
i Hi .
52. Let H1 and H2 be subgroups of G1 and G2 , respectively. Prove that H1 × H2 is a
subgroup of G1 × G2 .
53. Let m, n ∈ Z. Prove that hm, ni = hdi if and only if d = gcd(m, n).
54. Let m, n ∈ Z. Prove that hmi ∩ hni = hli if and only if l = lcm(m, n).
55. Groups of order 2p. In this series of exercises we will classify all groups of order 2p,
where p is an odd prime.
(a) Assume G is a group of order 2p, where p is an odd prime. If a ∈ G, show that
a must have order 1, 2, p, or 2p.
(b) Suppose that G has an element of order 2p. Prove that G is isomorphic to Z2p .
Hence, G is cyclic.
(c) Suppose that G does not contain an element of order 2p. Show that G must
contain an element of order p. Hint: Assume that G does not contain an element
of order p.
(d) Suppose that G does not contain an element of order 2p. Show that G must
contain an element of order 2.
(e) Let P be a subgroup of G with order p and y ∈ G have order 2. Show that
yP = P y.
(f) Suppose that G does not contain an element of order 2p and P = hzi is a subgroup
of order p generated by z. If y is an element of order 2, then yz = z k y for some
2 ≤ k < p.
(g) Suppose that G does not contain an element of order 2p. Prove that G is not
abelian.
(h) Suppose that G does not contain an element of order 2p and P = hzi is a subgroup
of order p generated by z and y is an element of order 2. Show that we can list
the elements of G as {z i y j | 0 ≤ i < p, 0 ≤ j < 2}.
(i) Suppose that G does not contain an element of order 2p and P = hzi is a
subgroup of order p generated by z and y is an element of order 2. Prove that
the product (z i y j )(z r y s ) can be expressed as a uniquely as z m y n for some non
negative integers m, n. Thus, conclude that there is only one possibility for a
non-abelian group of order 2p, it must therefore be the one we have seen already,
the dihedral group.
10
Normal Subgroups and Factor Groups
If H is a subgroup of a group G, then right cosets are not always the same as left cosets;
that is, it is not always the case that gH = Hg for all g ∈ G. The subgroups for which this
property holds play a critical role in group theory—they allow for the construction of a new
class of groups, called factor or quotient groups. Factor groups may be studied directly or
by using homomorphisms, a generalization of isomorphisms. We will study homomorphisms
in Chapter 11.
10.1 Factor Groups and Normal Subgroups
Normal Subgroups
A subgroup H of a group G is normal in G if gH = Hg for all g ∈ G. That is, a normal
subgroup of a group G is one in which the right and left cosets are precisely the same.
Example 10.1 Let G be an abelian group. Every subgroup H of G is a normal subgroup.
Since gh = hg for all g ∈ G and h ∈ H, it will always be the case that gH = Hg. □
Example 10.2 Let H be the subgroup of S3 consisting of elements (1) and (12). Since
(123)H = {(1 2 3), (1 3)} and H(1 2 3) = {(1 2 3), (2 3)},
H cannot be a normal subgroup of S3 . However, the subgroup N , consisting of the permu-
tations (1), (1 2 3), and (1 3 2), is normal since the cosets of N are
N = {(1), (1 2 3), (1 3 2)}
(1 2)N = N (1 2) = {(1 2), (1 3), (2 3)}.
□
The following theorem is fundamental to our understanding of normal subgroups.
Theorem 10.3 Let G be a group and N be a subgroup of G. Then the following statements
are equivalent.
1. The subgroup N is normal in G.
2. For all g ∈ G, gN g −1 ⊂ N .
3. For all g ∈ G, gN g −1 = N .
125
CHAPTER 10. NORMAL SUBGROUPS AND FACTOR GROUPS 126
Proof. (1) ⇒ (2). Since N is normal in G, gN = N g for all g ∈ G. Hence, for a given
g ∈ G and n ∈ N , there exists an n′ in N such that gn = n′ g. Therefore, gng −1 = n′ ∈ N
or gN g −1 ⊂ N .
(2) ⇒ (3). Let g ∈ G. Since gN g −1 ⊂ N , we need only show N ⊂ gN g −1 . For n ∈ N ,
g −1 ng = g −1 n(g −1 )−1 ∈ N . Hence, g −1 ng = n′ for some n′ ∈ N . Therefore, n = gn′ g −1 is
in gN g −1 .
(3) ⇒ (1). Suppose that gN g −1 = N for all g ∈ G. Then for any n ∈ N there exists an
n′ ∈ N such that gng −1 = n′ . Consequently, gn = n′ g or gN ⊂ N g. Similarly, N g ⊂ gN .
■
Factor Groups
If N is a normal subgroup of a group G, then the cosets of N in G form a group G/N under
the operation (aN )(bN ) = abN . This group is called the factor or quotient group of G and
N . Our first task is to prove that G/N is indeed a group.
Theorem 10.4 Let N be a normal subgroup of a group G. The cosets of N in G form a
group G/N of order [G : N ].
Proof. The group operation on G/N is (aN )(bN ) = abN . This operation must be shown
to be well-defined; that is, group multiplication must be independent of the choice of coset
representative. Let aN = bN and cN = dN . We must show that
(aN )(cN ) = acN = bdN = (bN )(dN ).
Then a = bn1 and c = dn2 for some n1 and n2 in N . Hence,
acN = bn1 dn2 N
= bn1 dN
= bn1 N d
= bN d
= bdN .
The remainder of the theorem is easy: eN = N is the identity and g −1 N is the inverse of
gN . The order of G/N is, of course, the number of cosets of N in G. ■
It is very important to remember that the elements in a factor group are sets of elements
in the original group.
Example 10.5 Consider the normal subgroup of S3 , N = {(1), (1 2 3), (1 3 2)}. The cosets
of N in S3 are N and (12)N . The factor group S3 /N has the following multiplication table.
N (1 2)N
N N (1 2)N
(1 2)N (1 2)N N
This group is isomorphic to Z2 . At first, multiplying cosets seems both complicated and
strange; however, notice that S3 /N is a smaller group. The factor group displays a certain
amount of information about S3 . Actually, N = A3 , the group of even permutations, and
(1 2)N = {(1 2), (1 3), (2 3)} is the set of odd permutations. The information captured in
G/N is parity; that is, multiplying two even or two odd permutations results in an even
permutation, whereas multiplying an odd permutation by an even permutation yields an
odd permutation. □
CHAPTER 10. NORMAL SUBGROUPS AND FACTOR GROUPS 127
Example 10.6 Consider the normal subgroup 3Z of Z. The cosets of 3Z in Z are
0 + 3Z = {. . . , −3, 0, 3, 6, . . .}
1 + 3Z = {. . . , −2, 1, 4, 7, . . .}
2 + 3Z = {. . . , −1, 2, 5, 8, . . .}.
The group Z/3Z is given by the Cayley table below.
+ 0 + 3Z 1 + 3Z 2 + 3Z
0 + 3Z 0 + 3Z 1 + 3Z 2 + 3Z
1 + 3Z 1 + 3Z 2 + 3Z 0 + 3Z
2 + 3Z 2 + 3Z 0 + 3Z 1 + 3Z
In general, the subgroup nZ of Z is normal. The cosets of Z/nZ are
nZ
1 + nZ
2 + nZ
..
.
(n − 1) + nZ.
The sum of the cosets k + nZ and l + nZ is k + l + nZ. Notice that we have written our
cosets additively, because the group operation is integer addition. □
Example 10.7 Consider the dihedral group Dn , generated by the two elements r and s,
satisfying the relations
rn = id
s2 = id
srs = r−1 .
The element r actually generates the cyclic subgroup of rotations, Rn , of Dn . Since srs−1 =
srs = r−1 ∈ Rn , the group of rotations is a normal subgroup of Dn ; therefore, Dn /Rn is a
group. Since there are exactly two elements in this group, it must be isomorphic to Z2 . □
10.2 The Simplicity of the Alternating Group
Of special interest are groups with no nontrivial normal subgroups. Such groups are called
simple groups. Of course, we already have a whole class of examples of simple groups, Zp ,
where p is prime. These groups are trivially simple since they have no proper subgroups
other than the subgroup consisting solely of the identity. Other examples of simple groups
are not so easily found. We can, however, show that the alternating group, An , is simple
for n ≥ 5. The proof of this result requires several lemmas.
Lemma 10.8 The alternating group An is generated by 3-cycles for n ≥ 3.
Proof. To show that the 3-cycles generate An , we need only show that any pair of
transpositions can be written as the product of 3-cycles. Since (a, b) = (b, a), every pair of
transpositions must be one of the following:
(a, b)(a, b) = id
(a, b)(c, d) = (a, c, b)(a, c, d)
CHAPTER 10. NORMAL SUBGROUPS AND FACTOR GROUPS 128
(a, b)(a, c) = (a, c, b).
■
Lemma 10.9 Let N be a normal subgroup of An , where n ≥ 3. If N contains a 3-cycle,
then N = An .
Proof. We will first show that An is generated by 3-cycles of the specific form (i, j, k),
where i and j are fixed in {1, 2, . . . , n} and we let k vary. Every 3-cycle is the product of
3-cycles of this form, since
(i, a, j) = (i, j, a)2
(i, a, b) = (i, j, b)(i, j, a)2
(j, a, b) = (i, j, b)2 (i, j, a)
(a, b, c) = (i, j, a)2 (i, j, c)(i, j, b)2 (i, j, a).
Now suppose that N is a nontrivial normal subgroup of An for n ≥ 3 such that N contains
a 3-cycle of the form (i, j, a). Using the normality of N , we see that
[(i, j)(a, k)](i, j, a)2 [(i, j)(a, k)]−1 = (i, j, k)
is in N . Hence, N must contain all of the 3-cycles (i, j, k) for 1 ≤ k ≤ n. By Lemma 10.8,
these 3-cycles generate An ; hence, N = An . ■
Lemma 10.10 For n ≥ 5, every nontrivial normal subgroup N of An contains a 3-cycle.
Proof. Let σ be an arbitrary element in a normal subgroup N . There are several possible
cycle structures for σ.
• σ is a 3-cycle.
• σ is the product of disjoint cycles, σ = τ (a1 , a2 , . . . , ar ) ∈ N , where r > 3.
• σ is the product of disjoint cycles, σ = τ (a1 , a2 , a3 )(a4 , a5 , a6 ).
• σ = τ (a1 , a2 , a3 ), where τ is the product of disjoint 2-cycles.
• σ = τ (a1 , a2 )(a3 , a4 ), where τ is the product of an even number of disjoint 2-cycles.
If σ is a 3-cycle, then we are done. If N contains a product of disjoint cycles, σ, and at least
one of these cycles has length greater than 3, say σ = τ (a1 , a2 , . . . , ar ), then
(a1 , a2 , a3 )σ(a1 , a2 , a3 )−1
is in N since N is normal; hence,
σ −1 (a1 , a2 , a3 )σ(a1 , a2 , a3 )−1
is also in N . Since
σ −1 (a1 , a2 , a3 )σ(a1 , a2 , a3 )−1 = σ −1 (a1 , a2 , a3 )σ(a1 , a3 , a2 )
= (a1 , a2 , . . . , ar )−1 τ −1 (a1 , a2 , a3 )τ (a1 , a2 , . . . , ar )(a1 , a3 , a2 )
= (a1 , ar , ar−1 , . . . , a2 )(a1 , a2 , a3 )(a1 , a2 , . . . , ar )(a1 , a3 , a2 )
= (a1 , a3 , ar ),
N must contain a 3-cycle; hence, N = An .
CHAPTER 10. NORMAL SUBGROUPS AND FACTOR GROUPS 129
Now suppose that N contains a disjoint product of the form
σ = τ (a1 , a2 , a3 )(a4 , a5 , a6 ).
Then
σ −1 (a1 , a2 , a4 )σ(a1 , a2 , a4 )−1 ∈ N
since
(a1 , a2 , a4 )σ(a1 , a2 , a4 )−1 ∈ N .
So
σ −1 (a1 , a2 , a4 )σ(a1 , a2 , a4 )−1 = [τ (a1 , a2 , a3 )(a4 , a5 , a6 )]−1 (a1 , a2 , a4 )τ (a1 , a2 , a3 )(a4 , a5 , a6 )(a1 , a2 , a4 )−1
= (a4 , a6 , a5 )(a1 , a3 , a2 )τ −1 (a1 , a2 , a4 )τ (a1 , a2 , a3 )(a4 , a5 , a6 )(a1 , a4 , a2 )
= (a4 , a6 , a5 )(a1 , a3 , a2 )(a1 , a2 , a4 )(a1 , a2 , a3 )(a4 , a5 , a6 )(a1 , a4 , a2 )
= (a1 , a4 , a2 , a6 , a3 ).
So N contains a disjoint cycle of length greater than 3, and we can apply the previous case.
Suppose N contains a disjoint product of the form σ = τ (a1 , a2 , a3 ), where τ is the
product of disjoint 2-cycles. Since σ ∈ N , σ 2 ∈ N , and
σ 2 = τ (a1 , a2 , a3 )τ (a1 , a2 , a3 )
= (a1 , a3 , a2 ).
So N contains a 3-cycle.
The only remaining possible case is a disjoint product of the form
σ = τ (a1 , a2 )(a3 , a4 ),
where τ is the product of an even number of disjoint 2-cycles. But
σ −1 (a1 , a2 , a3 )σ(a1 , a2 , a3 )−1
is in N since (a1 , a2 , a3 )σ(a1 , a2 , a3 )−1 is in N ; and so
σ −1 (a1 , a2 , a3 )σ(a1 , a2 , a3 )−1 = τ −1 (a1 , a2 )(a3 , a4 )(a1 , a2 , a3 )τ (a1 , a2 )(a3 , a4 )(a1 , a2 , a3 )−1
= (a1 , a3 )(a2 , a4 ).
Since n ≥ 5, we can find b ∈ {1, 2, . . . , n} such that b 6= a1 , a2 , a3 , a4 . Let µ = (a1 , a3 , b).
Then
µ−1 (a1 , a3 )(a2 , a4 )µ(a1 , a3 )(a2 , a4 ) ∈ N
and
µ−1 (a1 , a3 )(a2 , a4 )µ(a1 , a3 )(a2 , a4 ) = (a1 , ba3 )(a1 , a3 )(a2 , a4 )(a1 , a3 , b)(a1 , a3 )(a2 , a4 )
= (a1 a3 b).
Therefore, N contains a 3-cycle. This completes the proof of the lemma. ■
Theorem 10.11 The alternating group, An , is simple for n ≥ 5.
Proof. Let N be a normal subgroup of An . By Lemma 10.10, N contains a 3-cycle. By
Lemma 10.9, N = An ; therefore, An contains no proper nontrivial normal subgroups for
n ≥ 5. ■
CHAPTER 10. NORMAL SUBGROUPS AND FACTOR GROUPS 130
Sage. Sage can easily determine if a subgroup is normal or not. If so, it can create the
quotient group. However, the construction creates a new permuation group, isomorphic to
the quotient group, so its utility is limited.
Historical Note
One of the foremost problems of group theory has been to classify all simple finite groups.
This problem is over a century old and has been solved only in the last few decades of
the twentieth century. In a sense, finite simple groups are the building blocks of all finite
groups. The first nonabelian simple groups to be discovered were the alternating groups.
Galois was the first to prove that A5 was simple. Later, mathematicians such as C. Jordan
and L. E. Dickson found several infinite families of matrix groups that were simple. Other
families of simple groups were discovered in the 1950s. At the turn of the century, William
Burnside conjectured that all nonabelian simple groups must have even order. In 1963, W.
Feit and J. Thompson proved Burnside’s conjecture and published their results in the paper
“Solvability of Groups of Odd Order,” which appeared in the Pacific Journal of Mathematics.
Their proof, running over 250 pages, gave impetus to a program in the 1960s and 1970s
to classify all finite simple groups. Daniel Gorenstein was the organizer of this remarkable
effort. One of the last simple groups was the “Monster,” discovered by R. Greiss. The
Monster, a 196,833 × 196,833 matrix group, is one of the 26 sporadic, or special, simple
groups. These sporadic simple groups are groups that fit into no infinite family of simple
groups. Some of the sporadic groups play an important role in physics.
10.3 Reading Questions
1. Let G be the group of symmetries of an equilateral triangle, expressed as permutations
of the vertices numbered 1, 2, 3. Let H be the subgroup H = h(1 2)i. Build the left
and right cosets of H in G.
2. Based on your answer to the previous question, is H normal in G? Explain why or
why not.
3. The subgroup 8Z is normal in Z. In the factor group Z/8Z perform the computation
(3 + 8Z) + (7 + 8Z).
4. List two statements about a group G and a subgroup H that are equivalent to “H is
normal in G.”
5. In your own words, what is a factor group?
10.4 Exercises
1. For each of the following groups G, determine whether H is a normal subgroup of G.
If H is a normal subgroup, write out a Cayley table for the factor group G/H.
(a) G = S4 and H = A4
(b) G = A5 and H = {(1), (1 2 3), (1 3 2)}
(c) G = S4 and H = D4
(d) G = Q8 and H = {1, −1, I, −I}
(e) G = Z and H = 5Z
CHAPTER 10. NORMAL SUBGROUPS AND FACTOR GROUPS 131
2. Find all the subgroups of D4 . Which subgroups are normal? What are all the factor
groups of D4 up to isomorphism?
3. Find all the subgroups of the quaternion group, Q8 . Which subgroups are normal?
What are all the factor groups of Q8 up to isomorphism?
4. Let T be the group of nonsingular upper triangular 2 × 2 matrices with entries in R;
that is, matrices of the form
a b
,
0 c
where a, b, c ∈ R and ac 6= 0. Let U consist of matrices of the form
1 x
,
0 1
where x ∈ R.
(a) Show that U is a subgroup of T .
(b) Prove that U is abelian.
(c) Prove that U is normal in T .
(d) Show that T /U is abelian.
(e) Is T normal in GL2 (R)?
5. Show that the intersection of two normal subgroups is a normal subgroup.
6. If G is abelian, prove that G/H must also be abelian.
7. Prove or disprove: If H is a normal subgroup of G such that H and G/H are abelian,
then G is abelian.
8. If G is cyclic, prove that G/H must also be cyclic.
9. Prove or disprove: If H and G/H are cyclic, then G is cyclic.
10. Let H be a subgroup of index 2 of a group G. Prove that H must be a normal subgroup
of G. Conclude that Sn is not simple for n ≥ 3.
11. If a group G has exactly one subgroup H of order k, prove that H is normal in G.
12. Define the centralizer of an element g in a group G to be the set
C(g) = {x ∈ G : xg = gx}.
Show that C(g) is a subgroup of G. If g generates a normal subgroup of G, prove that
C(g) is normal in G.
13. Recall that the center of a group G is the set
Z(G) = {x ∈ G : xg = gx for all g ∈ G}.
(a) Calculate the center of S3 .
(b) Calculate the center of GL2 (R).
(c) Show that the center of any group G is a normal subgroup of G.
(d) If G/Z(G) is cyclic, show that G is abelian.
CHAPTER 10. NORMAL SUBGROUPS AND FACTOR GROUPS 132
14. Let G be a group and let G′ = haba−1 b−1 i; that is, G′ is the subgroup of all finite prod-
ucts of elements in G of the form aba−1 b−1 . The subgroup G′ is called the commutator
subgroup of G.
(a) Show that G′ is a normal subgroup of G.
(b) Let N be a normal subgroup of G. Prove that G/N is abelian if and only if N
contains the commutator subgroup of G.
11
Homomorphisms
One of the basic ideas of algebra is the concept of a homomorphism, a natural generalization
of an isomorphism. If we relax the requirement that an isomorphism of groups be bijective,
we have a homomorphism.
11.1 Group Homomorphisms
A homomorphism between groups (G, ·) and (H, ◦) is a map ϕ : G → H such that
ϕ(g1 · g2 ) = ϕ(g1 ) ◦ ϕ(g2 )
for g1 , g2 ∈ G. The range of ϕ in H is called the homomorphic image of ϕ.
Two groups are related in the strongest possible way if they are isomorphic; however, a
weaker relationship may exist between two groups. For example, the symmetric group Sn
and the group Z2 are related by the fact that Sn can be divided into even and odd permu-
tations that exhibit a group structure like that Z2 , as shown in the following multiplication
table.
even odd
even even odd
odd odd even
We use homomorphisms to study relationships such as the one we have just described.
Example 11.1 Let G be a group and g ∈ G. Define a map ϕ : Z → G by ϕ(n) = g n . Then
ϕ is a group homomorphism, since
ϕ(m + n) = g m+n = g m g n = ϕ(m)ϕ(n).
This homomorphism maps Z onto the cyclic subgroup of G generated by g. □
Example 11.2 Let G = GL2 (R). If
a b
A=
c d
is in G, then the determinant is nonzero; that is, det(A) = ad − bc 6= 0. Also, for any two
elements A and B in G, det(AB) = det(A) det(B). Using the determinant, we can define a
homomorphism ϕ : GL2 (R) → R∗ by A 7→ det(A). □
133
CHAPTER 11. HOMOMORPHISMS 134
Example 11.3 Recall that the circle group T consists of all complex numbers z such that
|z| = 1. We can define a homomorphism ϕ from the additive group of real numbers R to T
by ϕ : θ 7→ cos θ + i sin θ. Indeed,
ϕ(α + β) = cos(α + β) + i sin(α + β)
= (cos α cos β − sin α sin β) + i(sin α cos β + cos α sin β)
= (cos α + i sin α)(cos β + i sin β)
= ϕ(α)ϕ(β).
Geometrically, we are simply wrapping the real line around the circle in a group-theoretic
fashion. □
The following proposition lists some basic properties of group homomorphisms.
Proposition 11.4 Let ϕ : G1 → G2 be a homomorphism of groups. Then
1. If e is the identity of G1 , then ϕ(e) is the identity of G2 ;
2. For any element g ∈ G1 , ϕ(g −1 ) = [ϕ(g)]−1 ;
3. If H1 is a subgroup of G1 , then ϕ(H1 ) is a subgroup of G2 ;
4. If H2 is a subgroup of G2 , then ϕ−1 (H2 ) = {g ∈ G1 : ϕ(g) ∈ H2 } is a subgroup of G1 .
Furthermore, if H2 is normal in G2 , then ϕ−1 (H2 ) is normal in G1 .
Proof. (1) Suppose that e and e′ are the identities of G1 and G2 , respectively; then
e′ ϕ(e) = ϕ(e) = ϕ(ee) = ϕ(e)ϕ(e).
By cancellation, ϕ(e) = e′ .
(2) This statement follows from the fact that
ϕ(g −1 )ϕ(g) = ϕ(g −1 g) = ϕ(e) = e′ .
(3) The set ϕ(H1 ) is nonempty since the identity of G2 is in ϕ(H1 ). Suppose that H1
is a subgroup of G1 and let x and y be in ϕ(H1 ). There exist elements a, b ∈ H1 such that
ϕ(a) = x and ϕ(b) = y. Since
xy −1 = ϕ(a)[ϕ(b)]−1 = ϕ(ab−1 ) ∈ ϕ(H1 ),
ϕ(H1 ) is a subgroup of G2 by Proposition 3.31.
(4) Let H2 be a subgroup of G2 and define H1 to be ϕ−1 (H2 ); that is, H1 is the set of
all g ∈ G1 such that ϕ(g) ∈ H2 . The identity is in H1 since ϕ(e) = e′ . If a and b are in H1 ,
then ϕ(ab−1 ) = ϕ(a)[ϕ(b)]−1 is in H2 since H2 is a subgroup of G2 . Therefore, ab−1 ∈ H1
and H1 is a subgroup of G1 . If H2 is normal in G2 , we must show that g −1 hg ∈ H1 for
h ∈ H1 and g ∈ G1 . But
ϕ(g −1 hg) = [ϕ(g)]−1 ϕ(h)ϕ(g) ∈ H2 ,
since H2 is a normal subgroup of G2 . Therefore, g −1 hg ∈ H1 . ■
Let ϕ : G → H be a group homomorphism and suppose that e is the identity of H. By
Proposition 11.4, ϕ−1 ({e}) is a subgroup of G. This subgroup is called the kernel of ϕ and
will be denoted by ker ϕ. In fact, this subgroup is a normal subgroup of G since the trivial
subgroup is normal in H. We state this result in the following theorem, which says that
with every homomorphism of groups we can naturally associate a normal subgroup.
CHAPTER 11. HOMOMORPHISMS 135
Theorem 11.5 Let ϕ : G → H be a group homomorphism. Then the kernel of ϕ is a
normal subgroup of G.
Example 11.6 Let us examine the homomorphism ϕ : GL2 (R) → R∗ defined by A 7→
det(A). Since 1 is the identity of R∗ , the kernel of this homomorphism is all 2 × 2 matrices
having determinant one. That is, ker ϕ = SL2 (R). □
Example 11.7 The kernel of the group homomorphism ϕ : R → C∗ defined by ϕ(θ) =
cos θ + i sin θ is {2πn : n ∈ Z}. Notice that ker ϕ ∼
= Z. □
Example 11.8 Suppose that we wish to determine all possible homomorphisms ϕ from Z7
to Z12 . Since the kernel of ϕ must be a subgroup of Z7 , there are only two possible kernels,
{0} and all of Z7 . The image of a subgroup of Z7 must be a subgroup of Z12 . Hence, there
is no injective homomorphism; otherwise, Z12 would have a subgroup of order 7, which
is impossible. Consequently, the only possible homomorphism from Z7 to Z12 is the one
mapping all elements to zero. □
Example 11.9 Let G be a group. Suppose that g ∈ G and ϕ is the homomorphism from Z
to G given by ϕ(n) = g n . If the order of g is infinite, then the kernel of this homomorphism
is {0} since ϕ maps Z onto the cyclic subgroup of G generated by g. However, if the order
of g is finite, say n, then the kernel of ϕ is nZ. □
11.2 The Isomorphism Theorems
Although it is not evident at first, factor groups correspond exactly to homomorphic images,
and we can use factor groups to study homomorphisms. We already know that with every
group homomorphism ϕ : G → H we can associate a normal subgroup of G, ker ϕ. The con-
verse is also true; that is, every normal subgroup of a group G gives rise to homomorphism
of groups.
Let H be a normal subgroup of G. Define the natural or canonical homomorphism
ϕ : G → G/H
by
ϕ(g) = gH.
This is indeed a homomorphism, since
ϕ(g1 g2 ) = g1 g2 H = g1 Hg2 H = ϕ(g1 )ϕ(g2 ).
The kernel of this homomorphism is H. The following theorems describe the relationships
between group homomorphisms, normal subgroups, and factor groups.
Theorem 11.10 First Isomorphism Theorem. If ψ : G → H is a group homomorphism
with K = ker ψ, then K is normal in G. Let ϕ : G → G/K be the canonical homomorphism.
Then there exists a unique isomorphism η : G/K → ψ(G) such that ψ = ηϕ.
Proof. We already know that K is normal in G. Define η : G/K → ψ(G) by η(gK) = ψ(g).
We first show that η is a well-defined map. If g1 K = g2 K, then for some k ∈ K, g1 k = g2 ;
consequently,
η(g1 K) = ψ(g1 ) = ψ(g1 )ψ(k) = ψ(g1 k) = ψ(g2 ) = η(g2 K).
Thus, η does not depend on the choice of coset representatives and the map η : G/K → ψ(G)
is uniquely defined since ψ = ηϕ. We must also show that η is a homomorphism. Indeed,
η(g1 Kg2 K) = η(g1 g2 K)
CHAPTER 11. HOMOMORPHISMS 136
= ψ(g1 g2 )
= ψ(g1 )ψ(g2 )
= η(g1 K)η(g2 K).
Clearly, η is onto ψ(G). To show that η is one-to-one, suppose that η(g1 K) = η(g2 K).
Then ψ(g1 ) = ψ(g2 ). This implies that ψ(g1−1 g2 ) = e, or g1−1 g2 is in the kernel of ψ; hence,
g1−1 g2 K = K; that is, g1 K = g2 K. ■
Mathematicians often use diagrams called commutative diagrams to describe such the-
orems. The following diagram “commutes” since ψ = ηϕ.
ψ
G H
ϕ η
G/K
Example 11.11 Let G be a cyclic group with generator g. Define a map ϕ : Z → G by
n 7→ g n . This map is a surjective homomorphism since
ϕ(m + n) = g m+n = g m g n = ϕ(m)ϕ(n).
Clearly ϕ is onto. If |g| = m, then g m = e. Hence, ker ϕ = mZ and Z/ ker ϕ = Z/mZ ∼ = G.
On the other hand, if the order of g is infinite, then ker ϕ = 0 and ϕ is an isomorphism of
G and Z. Hence, two cyclic groups are isomorphic exactly when they have the same order.
Up to isomorphism, the only cyclic groups are Z and Zn . □
Theorem 11.12 Second Isomorphism Theorem. Let H be a subgroup of a group G
(not necessarily normal in G) and N a normal subgroup of G. Then HN is a subgroup of
G, H ∩ N is a normal subgroup of H, and
H/H ∩ N ∼ = HN /N .
Proof. We will first show that HN = {hn : h ∈ H, n ∈ N } is a subgroup of G. Suppose
that h1 n1 , h2 n2 ∈ HN . Since N is normal, (h2 )−1 n1 h2 ∈ N . So
(h1 n1 )(h2 n2 ) = h1 h2 ((h2 )−1 n1 h2 )n2
is in HN . The inverse of hn ∈ HN is in HN since
(hn)−1 = n−1 h−1 = h−1 (hn−1 h−1 ).
Next, we prove that H ∩ N is normal in H. Let h ∈ H and n ∈ H ∩ N . Then
h−1 nh ∈ H since each element is in H. Also, h−1 nh ∈ N since N is normal in G; therefore,
h−1 nh ∈ H ∩ N .
Now define a map ϕ from H to HN /N by h 7→ hN . The map ϕ is onto, since any coset
hnN = hN is the image of h in H. We also know that ϕ is a homomorphism because
ϕ(hh′ ) = hh′ N = hN h′ N = ϕ(h)ϕ(h′ ).
By the First Isomorphism Theorem, the image of ϕ is isomorphic to H/ ker ϕ; that is,
HN /N = ϕ(H) ∼
= H/ ker ϕ.
CHAPTER 11. HOMOMORPHISMS 137
Since
ker ϕ = {h ∈ H : h ∈ N } = H ∩ N ,
HN /N = ϕ(H) ∼
= H/H ∩ N . ■
Theorem 11.13 Correspondence Theorem. Let N be a normal subgroup of a group
G. Then H 7→ H/N is a one-to-one correspondence between the set of subgroups H of G
containing N and the set of subgroups of G/N . Furthermore, the normal subgroups of G
containing N correspond to normal subgroups of G/N .
Proof. Let H be a subgroup of G containing N . Since N is normal in H, H/N is a factor
group. Let aN and bN be elements of H/N . Then (aN )(b−1 N ) = ab−1 N ∈ H/N ; hence,
H/N is a subgroup of G/N .
Let S be a subgroup of G/N . This subgroup is a set of cosets of N . If H = {g ∈ G :
gN ∈ S}, then for h1 , h2 ∈ H, we have that (h1 N )(h2 N ) = h1 h2 N ∈ S and h−1
1 N ∈ S.
Therefore, H must be a subgroup of G. Clearly, H contains N . Therefore, S = H/N .
Consequently, the map H 7→ H/N is onto.
Suppose that H1 and H2 are subgroups of G containing N such that H1 /N = H2 /N .
If h1 ∈ H1 , then h1 N ∈ H1 /N . Hence, h1 N = h2 N ⊂ H2 for some h2 in H2 . However,
since N is contained in H2 , we know that h1 ∈ H2 or H1 ⊂ H2 . Similarly, H2 ⊂ H1 . Since
H1 = H2 , the map H 7→ H/N is one-to-one.
Suppose that H is normal in G and N is a subgroup of H. Then it is easy to verify
that the map G/N → G/H defined by gN 7→ gH is a homomorphism. The kernel of this
homomorphism is H/N , which proves that H/N is normal in G/N .
Conversely, suppose that H/N is normal in G/N . The homomorphism given by
G/N
G → G/N →
H/N
has kernel H. Hence, H must be normal in G. ■
Notice that in the course of the proof of Theorem 11.13, we have also proved the following
theorem.
Theorem 11.14 Third Isomorphism Theorem. Let G be a group and N and H be
normal subgroups of G with N ⊂ H. Then
G/N
G/H ∼
= .
H/N
Example 11.15 By the Third Isomorphism Theorem,
Z/mZ ∼
= (Z/mnZ)/(mZ/mnZ).
Since |Z/mnZ| = mn and |Z/mZ| = m, we have |mZ/mnZ| = n. □
Sage. Sage can create homomorphisms between groups, which can be used directly as
functions, and then queried for their kernels and images. So there is great potential for
exploring the many fundamental relationships between groups, normal subgroups, quotient
groups and properties of homomorphisms.
CHAPTER 11. HOMOMORPHISMS 138
11.3 Reading Questions
1. Consider the function ϕ : Z10 → Z10 defined by ϕ(x) = x + x. Prove that ϕ is a group
homomorphism.
2. For ϕ defined in the previous question, explain why ϕ is not a group isomorphism.
3. Compare and contrast isomorphisms and homomorphisms.
4. Paraphrase the First Isomorphism Theorem using only words. No symbols allowed at
all.
5. “For every normal subgroup there is a homomorphism, and for every homomorphism
there is a normal subgroup.” Explain the (precise) basis for this (vague) statement.
11.4 Exercises
1. Prove that det(AB) = det(A) det(B) for A, B ∈ GL2 (R). This shows that the deter-
minant is a homomorphism from GL2 (R) to R∗ .
2. Which of the following maps are homomorphisms? If the map is a homomorphism,
what is the kernel?
(a) ϕ : R∗ → GL2 (R) defined by
1 0
ϕ(a) =
0 a
(b) ϕ : R → GL2 (R) defined by
1 0
ϕ(a) =
a 1
(c) ϕ : GL2 (R) → R defined by
a b
ϕ =a+d
c d
(d) ϕ : GL2 (R) → R∗ defined by
a b
ϕ = ad − bc
c d
(e) ϕ : M2 (R) → R defined by
a b
ϕ = b,
c d
where M2 (R) is the additive group of 2 × 2 matrices with entries in R.
3. Let A be an m × n matrix. Show that matrix multiplication, x 7→ Ax, defines a
homomorphism ϕ : Rn → Rm .
4. Let ϕ : Z → Z be given by ϕ(n) = 7n. Prove that ϕ is a group homomorphism. Find
the kernel and the image of ϕ.
5. Describe all of the homomorphisms from Z24 to Z18 .
6. Describe all of the homomorphisms from Z to Z12 .
CHAPTER 11. HOMOMORPHISMS 139
7. In the group Z24 , let H = h4i and N = h6i.
(a) List the elements in HN (we usually write H + N for these additive groups) and
H ∩ N.
(b) List the cosets in HN /N , showing the elements in each coset.
(c) List the cosets in H/(H ∩ N ), showing the elements in each coset.
(d) Give the correspondence between HN /N and H/(H ∩ N ) described in the proof
of the Second Isomorphism Theorem.
8. If G is an abelian group and n ∈ N, show that ϕ : G → G defined by g 7→ g n is a group
homomorphism.
9. If ϕ : G → H is a group homomorphism and G is abelian, prove that ϕ(G) is also
abelian.
10. If ϕ : G → H is a group homomorphism and G is cyclic, prove that ϕ(G) is also cyclic.
11. Show that a homomorphism defined on a cyclic group is completely determined by its
action on the generator of the group.
12. If a group G has exactly one subgroup H of order k, prove that H is normal in G.
13. Prove or disprove: Q/Z ∼= Q.
14. Let G be a finite group and N a normal subgroup of G. If H is a subgroup of G/N ,
prove that ϕ−1 (H) is a subgroup in G of order |H| · |N |, where ϕ : G → G/N is the
canonical homomorphism.
15. Let G1 and G2 be groups, and let H1 and H2 be normal subgroups of G1 and G2 respec-
tively. Let ϕ : G1 → G2 be a homomorphism. Show that ϕ induces a homomorphism
ϕ : (G1 /H1 ) → (G2 /H2 ) if ϕ(H1 ) ⊂ H2 .
16. If H and K are normal subgroups of G and H ∩ K = {e}, prove that G is isomorphic
to a subgroup of G/H × G/K.
17. Let ϕ : G1 → G2 be a surjective group homomorphism. Let H1 be a normal subgroup
of G1 and suppose that ϕ(H1 ) = H2 . Prove or disprove that G1 /H1 ∼
= G2 /H2 .
18. Let ϕ : G → H be a group homomorphism. Show that ϕ is one-to-one if and only if
ϕ−1 (e) = {e}.
19. Given a homomorphism ϕ : G → H define a relation ∼ on G by a ∼ b if ϕ(a) = ϕ(b)
for a, b ∈ G. Show this relation is an equivalence relation and describe the equivalence
classes.
11.5 Additional Exercises: Automorphisms
1. Let Aut(G) be the set of all automorphisms of G; that is, isomorphisms from G to
itself. Prove this set forms a group and is a subgroup of the group of permutations of
G; that is, Aut(G) ≤ SG .
2. An inner automorphism of G,
ig : G → G,
is defined by the map
ig (x) = gxg −1 ,
for g ∈ G. Show that ig ∈ Aut(G).
CHAPTER 11. HOMOMORPHISMS 140
3. The set of all inner automorphisms is denoted by Inn(G). Show that Inn(G) is a
subgroup of Aut(G).
4. Find an automorphism of a group G that is not an inner automorphism.
5. Let G be a group and ig be an inner automorphism of G, and define a map
G → Aut(G)
by
g 7→ ig .
Prove that this map is a homomorphism with image Inn(G) and kernel Z(G). Use this
result to conclude that
G/Z(G) ∼= Inn(G).
6. Compute Aut(S3 ) and Inn(S3 ). Do the same thing for D4 .
7. Find all of the homomorphisms ϕ : Z → Z. What is Aut(Z)?
8. Find all of the automorphisms of Z8 . Prove that Aut(Z8 ) ∼
= U (8).
9. For k ∈ Zn , define a map ϕk : Zn → Zn by a 7→ ka. Prove that ϕk is a homomorphism.
10. Prove that ϕk is an isomorphism if and only if k is a generator of Zn .
11. Show that every automorphism of Zn is of the form ϕk , where k is a generator of Zn .
12. Prove that ψ : U (n) → Aut(Zn ) is an isomorphism, where ψ : k 7→ ϕk .
12
Matrix Groups and Symmetry
When Felix Klein (1849–1925) accepted a chair at the University of Erlangen, he outlined in
his inaugural address a program to classify different geometries. Central to Klein’s program
was the theory of groups: he considered geometry to be the study of properties that are
left invariant under transformation groups. Groups, especially matrix groups, have now
become important in the study of symmetry and have found applications in such disciplines
as chemistry and physics. In the first part of this chapter, we will examine some of the
classical matrix groups, such as the general linear group, the special linear group, and the
orthogonal group. We will then use these matrix groups to investigate some of the ideas
behind geometric symmetry.
12.1 Matrix Groups
Some Facts from Linear Algebra
Before we study matrix groups, we must recall some basic facts from linear algebra. One
of the most fundamental ideas of linear algebra is that of a linear transformation. A linear
transformation or linear map T : Rn → Rm is a map that preserves vector addition and
scalar multiplication; that is, for vectors x and y in Rn and a scalar α ∈ R,
T (x + y) = T (x) + T (y)
T (αy) = αT (y).
An m × n matrix with entries in R represents a linear transformation from Rn to Rm . If
we write vectors x = (x1 , . . . , xn )t and y = (y1 , . . . , yn )t in Rn as column matrices, then an
m × n matrix
a11 a12 · · · a1n
a21 a22 · · · a2n
A= . .. .. ..
.. . . .
am1 am2 · · · amn
maps the vectors to Rm linearly by matrix multiplication. Observe that if α is a real number,
A(x + y) = Ax + Ay and αAx = A(αx),
141
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 142
where
x1
x2
x = . .
..
xn
We will often abbreviate the matrix A by writing (aij ).
Conversely, if T : Rn → Rm is a linear map, we can associate a matrix A with T by
considering what T does to the vectors
e1 = (1, 0, . . . , 0)t
e2 = (0, 1, . . . , 0)t
..
.
en = (0, 0, . . . , 1)t .
We can write any vector x = (x1 , . . . , xn )t as
x1 e1 + x2 e2 + · · · + xn en .
Consequently, if
T (e1 ) = (a11 , a21 , . . . , am1 )t ,
T (e2 ) = (a12 , a22 , . . . , am2 )t ,
..
.
T (en ) = (a1n , a2n , . . . , amn )t ,
then
T (x) = T (x1 e1 + x2 e2 + · · · + xn en )
= x1 T (e1 ) + x2 T (e2 ) + · · · + xn T (en )
!t
Xn X
n
= a1k xk , . . . , amk xk
k=1 k=1
= Ax.
Example 12.1 If we let T : R2 → R2 be the map given by
T (x1 , x2 ) = (2x1 + 5x2 , −4x1 + 3x2 ),
the axioms that T must satisfy to be a linear transformation are easily verified. The column
vectors T e1 = (2, −4)t and T e2 = (5, 3)t tell us that T is given by the matrix
2 5
A= .
−4 3
□
Since we are interested in groups of matrices, we need to know which matrices have
multiplicative inverses. Recall that an n × n matrix A is invertible exactly when there
exists another matrix A−1 such that AA−1 = A−1 A = I, where
1 0 ··· 0
0 1 · · · 0
I = . . . .
.. .. . . ..
0 0 ··· 1
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 143
is the n × n identity matrix. From linear algebra we know that A is invertible if and only if
the determinant of A is nonzero. Sometimes an invertible matrix is said to be nonsingular.
Example 12.2 If A is the matrix
2 1
,
5 3
then the inverse of A is
−1 3 −1
A = .
−5 2
We are guaranteed that A−1 exists, since det(A) = 2 · 3 − 5 · 1 = 1 is nonzero. □
Some other facts about determinants will also prove useful in the course of this chapter.
Let A and B be n × n matrices. From linear algebra we have the following properties of
determinants.
• The determinant is a homomorphism into the multiplicative group of real numbers;
that is, det(AB) = (det A)(det B).
• If A is an invertible matrix, then det(A−1 ) = 1/ det A.
• If we define the transpose of a matrix A = (aij ) to be At = (aji ), then det(At ) = det A.
• Let T be the linear transformation associated with an n × n matrix A. Then T
multiplies volumes by a factor of | det A|. In the case of R2 , this means that T
multiplies areas by | det A|.
Linear maps, matrices, and determinants are covered in any elementary linear algebra
text; however, if you have not had a course in linear algebra, it is a straightforward process
to verify these properties directly for 2 × 2 matrices, the case with which we are most
concerned.
The General and Special Linear Groups
The set of all n × n invertible matrices forms a group called the general linear group.
We will denote this group by GLn (R). The general linear group has several important
subgroups. The multiplicative properties of the determinant imply that the set of matrices
with determinant one is a subgroup of the general linear group. Stated another way, suppose
that det(A) = 1 and det(B) = 1. Then det(AB) = det(A) det(B) = 1 and det(A−1 ) =
1/ det A = 1. This subgroup is called the special linear group and is denoted by SLn (R).
Example 12.3 Given a 2 × 2 matrix
a b
A= ,
c d
the determinant of A is ad − bc. The group GL2 (R) consists of those matrices in which
ad − bc 6= 0. The inverse of A is
−1 1 d −b
A = .
ad − bc −c a
If A is in SL2 (R), then
−1 d −b
A = .
−c a
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 144
Geometrically, SL2 (R) is the group that preserves the areas of parallelograms. Let
1 1
A=
0 1
be in SL2 (R). In Figure 12.4, the unit square corresponding to the vectors x = (1, 0)t
and y = (0, 1)t is taken by A to the parallelogram with sides (1, 0)t and (1, 1)t ; that is,
Ax = (1, 0)t and Ay = (1, 1)t . Notice that these two parallelograms have the same area. □
y y
(1, 1)
(0, 1)
(1, 0) x (1, 0) x
Figure 12.4 SL2 (R) acting on the unit square
The Orthogonal Group O(n)
Another subgroup of GLn (R) is the orthogonal group. A matrix A is orthogonal if A−1 = At .
The orthogonal group consists of the set of all orthogonal matrices. We write O(n) for
the n × n orthogonal group. We leave as an exercise the proof that O(n) is a subgroup of
GLn (R).
Example 12.5 The following matrices are orthogonal:
√ √
√ −1/ 2 0 1/ 2
3/5 −4/5 1/2 − 3/2 √ √ √
, √ , 1/ 6 −2/ 6 1/ 6 .
4/5 3/5 3/2 1/2 √ √ √
1/ 3 1/ 3 1/ 3
□
There is a more geometric way of viewing the group O(n). The orthogonal matrices are
exactly those matrices that preserve the length of vectors. We can define the length of a
vector using the Euclidean inner product, or dot product, of two vectors. The Euclidean
inner product of two vectors x = (x1 , . . . , xn )t and y = (y1 , . . . , yn )t is
y1
y2
hx, yi = xt y = (x1 , x2 , . . . , xn ) . = x1 y1 + · · · + xn yn .
..
yn
We define the length of a vector x = (x1 , . . . , xn )t to be
p q
kxk = hx, xi = x21 + · · · + x2n .
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 145
Associated with the notion of the length of a vector is the idea of the distance between two
vectors. We define the distance between two vectors x and y to be kx − yk. We leave as
an exercise the proof of the following proposition about the properties of Euclidean inner
products.
Proposition 12.6 Let x, y, and w be vectors in Rn and α ∈ R. Then
1. hx, yi = hy, xi.
2. hx, y + wi = hx, yi + hx, wi.
3. hαx, yi = hx, αyi = αhx, yi.
4. hx, xi ≥ 0 with equality exactly when x = 0.
5. If hx, yi = 0 for all x in Rn , then y = 0.
√
Example 12.7 The vector x = (3, 4)t has length 32 + 42 = 5. We can also see that the
orthogonal matrix
3/5 −4/5
A=
4/5 3/5
preserves the length of this vector. The vector Ax = (−7/5, 24/5)t also has length 5. □
t t
Since det(AA ) = det(I) = 1 and det(A) = det(A ), the determinant of any orthogonal
matrix is either 1 or −1. Consider the column vectors
a1j
a2j
aj = .
..
anj
of the orthogonal matrix A = (aij ). Since AAt = I, har , as i = δrs , where
(
1 r=s
δrs =
0 r 6= s
is the Kronecker delta. Accordingly, column vectors of an orthogonal matrix all have length
1; and the Euclidean inner product of distinct column vectors is zero. Any set of vectors
satisfying these properties is called an orthonormal set. Conversely, given an n × n matrix
A whose columns form an orthonormal set, it follows that A−1 = At .
We say that a matrix A is distance-preserving, length-preserving, or inner product-
preserving when kAx − Ayk = kx − yk, kAxk = kxk, or hAx, Ayi = hx, yi, respectively. The
following theorem, which characterizes the orthogonal group, says that these notions are
the same.
Theorem 12.8 Let A be an n × n matrix. The following statements are equivalent.
1. The columns of the matrix A form an orthonormal set.
2. A−1 = At .
3. For vectors x and y, hAx, Ayi = hx, yi.
4. For vectors x and y, kAx − Ayk = kx − yk.
5. For any vector x, kAxk = kxk.
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 146
Proof. We have already shown (1) and (2) to be equivalent.
(2) ⇒ (3).
hAx, Ayi = (Ax)t Ay
= xt At Ay
= xt y
= hx, yi.
(3) ⇒ (2). Since
hx, xi = hAx, Axi
= xt At Ax
= hx, At Axi,
we know that hx, (At A − I)xi = 0 for all x. Therefore, At A − I = 0 or A−1 = At .
(3) ⇒ (4). If A is inner product-preserving, then A is distance-preserving, since
kAx − Ayk2 = kA(x − y)k2
= hA(x − y), A(x − y)i
= hx − y, x − yi
= kx − yk2 .
(4) ⇒ (5). If A is distance-preserving, then A is length-preserving. Letting y = 0, we
have
kAxk = kAx − Ayk = kx − yk = kxk.
(5) ⇒ (3). We use the following identity to show that length-preserving implies inner
product-preserving:
1
hx, yi = kx + yk2 − kxk2 − kyk2 .
2
Observe that
1
hAx, Ayi = kAx + Ayk2 − kAxk2 − kAyk2
2
1
= kA(x + y)k2 − kAxk2 − kAyk2
2
1
= kx + yk2 − kxk2 − kyk2
2
= hx, yi.
■
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 147
y y
(sin θ, − cos θ)
(a, b) (cos θ, sin θ)
θ
x x
(a, −b)
Figure 12.9 O(2) acting on R2
Example 12.10 Let us examine the orthogonal group on R2 a bit more closely. An element
A ∈ O(2) is determined by its action on e1 = (1, 0)t and e2 = (0, 1)t . If Ae1 = (a, b)t , then
a2 + b2 = 1, since the length of a vector must be preserved when it is multiplied by A. Since
multiplication of an element of O(2) preserves length and orthogonality, Ae2 = ±(−b, a)t .
If we choose Ae2 = (−b, a)t , then
a −b cos θ − sin θ
A= = ,
b a sin θ cos θ
where 0 ≤ θ < 2π. The matrix A rotates a vector in R2 counterclockwise about the origin
by an angle of θ (Figure 12.9).
If we choose Ae2 = (b, −a)t , then we obtain the matrix
a b cos θ sin θ
B= = .
b −a sin θ − cos θ
Here, det B = −1 and
1 0
B2 = .
0 1
A reflection about the horizontal axis is given by the matrix
1 0
C= ,
0 −1
and B = AC (see Figure 12.9). Thus, a reflection about a line ℓ is simply a reflection about
the horizontal axis followed by a rotation. □
Two of the other matrix or matrix-related groups that we will consider are the special
orthogonal group and the group of Euclidean motions. The special orthogonal group, SO(n),
is just the intersection of O(n) and SLn (R); that is, those elements in O(n) with determinant
one. The Euclidean group, E(n), can be written as ordered pairs (A, x), where A is in
O(n) and x is in Rn . We define multiplication by
(A, x)(B, y) = (AB, Ay + x).
The identity of the group is (I, 0); the inverse of (A, x) is (A−1 , −A−1 x). In Exercise 12.4.6,
you are asked to check that E(n) is indeed a group under this operation.
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 148
y y
x+y
x
x x
Figure 12.11 Translations in R2
12.2 Symmetry
An isometry or rigid motion in Rn is a distance-preserving function f from Rn to Rn . This
means that f must satisfy
kf (x) − f (y)k = kx − yk
for all x, y ∈ Rn . It is not difficult to show that f must be a one-to-one map. By Theo-
rem 12.8, any element in O(n) is an isometry on Rn ; however, O(n) does not include all
possible isometries on Rn . Translation by a vector x, Ty (x) = x + y is also an isometry
(Figure 12.11); however, T cannot be in O(n) since it is not a linear map.
We are mostly interested in isometries in R2 . In fact, the only isometries in R2 are
rotations and reflections about the origin, translations, and combinations of the two. For
example, a glide reflection is a translation followed by a reflection (Figure 12.12). In Rn all
isometries are given in the same manner. The proof is very easy to generalize.
y y
x
x x
T (x)
Figure 12.12 Glide reflections
Lemma 12.13 An isometry f that fixes the origin in R2 is a linear transformation. In
particular, f is given by an element in O(2).
Proof. Let f be an isometry in R2 fixing the origin. We will first show that f preserves
inner products. Since f (0) = 0, kf (x)k = kxk; therefore,
kxk2 − 2hf (x), f (y)i + kyk2 = kf (x)k2 − 2hf (x), f (y)i + kf (y)k2
= hf (x) − f (y), f (x) − f (y)i
= kf (x) − f (y)k2
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 149
= kx − yk2
= hx − y, x − yi
= kxk2 − 2hx, yi + kyk2 .
Consequently,
hf (x), f (y)i = hx, yi.
Now let e1 and e2 be (1, 0)t and (0, 1)t , respectively. If
x = (x1 , x2 ) = x1 e1 + x2 e2 ,
then
f (x) = hf (x), f (e1 )if (e1 ) + hf (x), f (e2 )if (e2 ) = x1 f (e1 ) + x2 f (e2 ).
The linearity of f easily follows. ■
For any arbitrary isometry, f , Tx f will fix the origin for some vector x in R ; hence,
2
Tx f (y) = Ay for some matrix A ∈ O(2). Consequently, f (y) = Ay + x. Given the isometries
f (y) = Ay + x1
g(y) = By + x2 ,
their composition is
f (g(y)) = f (By + x2 ) = ABy + Ax2 + x1 .
This last computation allows us to identify the group of isometries on R2 with E(2).
Theorem 12.14 The group of isometries on R2 is the Euclidean group, E(2).
A symmetry group in Rn is a subgroup of the group of isometries on Rn that fixes a
set of points X ⊂ Rn . It is important to realize that the symmetry group of X depends
both on Rn and on X. For example, the symmetry group of the origin in R1 is Z2 , but the
symmetry group of the origin in R2 is O(2).
Theorem 12.15 The only finite symmetry groups in R2 are Zn and Dn .
Proof. We simply need to find all of the finite subgroups G of E(2). Any finite symmetry
group G in R2 must fix the origin and must be a finite subgroup of O(2), since translations
and glide reflections have infinite order. By Example 12.10, elements in O(2) are either
rotations of the form
cos θ − sin θ
Rθ =
sin θ cos θ
or reflections of the form
cos ϕ − sin ϕ 1 0 cos ϕ sin ϕ
Tϕ = = .
sin ϕ cos ϕ 0 −1 sin ϕ − cos ϕ
Notice that det(Rθ ) = 1, det(Tϕ ) = −1, and Tϕ2 = I. We can divide the proof up into two
cases. In the first case, all of the elements in G have determinant one. In the second case,
there exists at least one element in G with determinant −1.
Case 1. The determinant of every element in G is one. In this case every element in G must
be a rotation. Since G is finite, there is a smallest angle, say θ0 , such that the corresponding
element Rθ0 is the smallest rotation in the positive direction. We claim that Rθ0 generates
G. If not, then for some positive integer n there is an angle θ1 between nθ0 and (n + 1)θ0 .
If so, then (n + 1)θ0 − θ1 corresponds to a rotation smaller than θ0 , which contradicts the
minimality of θ0 .
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 150
Case 2. The group G contains a reflection T . The kernel of the homomorphism ϕ :
G → {−1, 1} given by A 7→ det(A) consists of elements whose determinant is 1. Therefore,
|G/ ker ϕ| = 2. We know that the kernel is cyclic by the first case and is a subgroup of G
of, say, order n. Hence, |G| = 2n. The elements of G are
Rθ , . . . , Rθn−1 , T Rθ , . . . , T Rθn−1 .
These elements satisfy the relation
T Rθ T = Rθ−1 .
Consequently, G must be isomorphic to Dn in this case. ■
The Wallpaper Groups
Suppose that we wish to study wallpaper patterns in the plane or crystals in three dimen-
sions. Wallpaper patterns are simply repeating patterns in the plane (Figure 12.16). The
analogs of wallpaper patterns in R3 are crystals, which we can think of as repeating pat-
terns of molecules in three dimensions (Figure 12.17). The mathematical equivalent of a
wallpaper or crystal pattern is called a lattice.
Figure 12.16 A wallpaper pattern in R2
Figure 12.17 A crystal structure in R3
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 151
Let us examine wallpaper patterns in the plane a little more closely. Suppose that x
and y are linearly independent vectors in R2 ; that is, one vector cannot be a scalar multiple
of the other. A lattice of x and y is the set of all linear combinations mx + ny, where m
and n are integers. The vectors x and y are said to be a basis for the lattice.
Notice that a lattice can have several bases. For example, the vectors (1, 1)t and (2, 0)t
have the same lattice as the vectors (−1, 1)t and (−1, −1)t (Figure 12.18). However, any
lattice is completely determined by a basis. Given two bases for the same lattice, say {x1 , x2 }
and {y1 , y2 }, we can write
y1 = α1 x1 + α2 x2
y2 = β1 x1 + β2 x2 ,
where α1 , α2 , β1 , and β2 are integers. The matrix corresponding to this transformation is
α1 α2
U= .
β1 β2
If we wish to give x1 and x2 in terms of y1 and y2 , we need only calculate U −1 ; that is,
−1 y1 x1
U = .
y2 x2
Since U has integer entries, U −1 must also have integer entries; hence the determinants of
both U and U −1 must be integers. Because U U −1 = I,
det(U U −1 ) = det(U ) det(U −1 ) = 1;
consequently, det(U ) = ±1. A matrix with determinant ±1 and integer entries is called
unimodular. For example, the matrix
3 1
5 2
is unimodular. It should be clear that there is a minimum length for vectors in a lattice.
(−1, 1) (1, 1)
(2, 0)
(−1, −1)
Figure 12.18 A lattice in R2
We can classify lattices by studying their symmetry groups. The symmetry group of
a lattice is the subgroup of E(2) that maps the lattice to itself. We consider two lattices
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 152
in R2 to be equivalent if they have the same symmetry group. Similarly, classification of
crystals in R3 is accomplished by associating a symmetry group, called a space group, with
each type of crystal. Two lattices are considered different if their space groups are not the
same. The natural question that now arises is how many space groups exist.
A space group is composed of two parts: a translation subgroup and a point. The
translation subgroup is an infinite abelian subgroup of the space group made up of the
translational symmetries of the crystal; the point group is a finite group consisting of ro-
tations and reflections of the crystal about a point. More specifically, a space group is a
subgroup of G ⊂ E(2) whose translations are a set of the form {(I, t) : t ∈ L}, where L is a
lattice. Space groups are, of course, infinite. Using geometric arguments, we can prove the
following theorem (see [5] or [6]).
Theorem 12.19 Every translation group in R2 is isomorphic to Z × Z.
The point group of G is G0 = {A : (A, b) ∈ G for some b}. In particular, G0 must be a
subgroup of O(2). Suppose that x is a vector in a lattice L with space group G, translation
group H, and point group G0 . For any element (A, y) in G,
(A, y)(I, x)(A, y)−1 = (A, Ax + y)(A−1 , −A−1 y)
= (AA−1 , −AA−1 y + Ax + y)
= (I, Ax);
hence, (I, Ax) is in the translation group of G. More specifically, Ax must be in the lattice
L. It is important to note that G0 is not usually a subgroup of the space group G; however,
if T is the translation subgroup of G, then G/T ∼ = G0 . The proof of the following theorem
can be found in [2], [5], or [6].
Theorem 12.20 The point group in the wallpaper groups is isomorphic to Zn or Dn , where
n = 1, 2, 3, 4, 6.
To answer the question of how the point groups and the translation groups can be
combined, we must look at the different types of lattices. Lattices can be classified by the
structure of a single lattice cell. The possible cell shapes are parallelogram, rectangular,
square, rhombic, and hexagonal (Figure 12.21). The wallpaper groups can now be classified
according to the types of reflections that occur in each group: these are ordinarily reflections,
glide reflections, both, or none.
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 153
Rectangular
Square Rhombic
Parallelogram
Hexagonal
Figure 12.21 Types of lattices in R2
Table 12.22 The 17 wallpaper groups
Notation and Reflections or
Space Groups Point Group Lattice Type Glide Reflections?
p1 Z1 parallelogram none
p2 Z2 parallelogram none
p3 Z3 hexagonal none
p4 Z4 square none
p6 Z6 hexagonal none
pm D1 rectangular reflections
pg D1 rectangular glide reflections
cm D1 rhombic both
pmm D2 rectangular reflections
pmg D2 rectangular glide reflections
pgg D2 rectangular both
c2mm D2 rhombic both
p3m1, p31m D3 hexagonal both
p4m, p4g D4 square both
p6m D6 hexagonal both
Theorem 12.23 There are exactly 17 wallpaper groups.
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 154
p4m p4g
Figure 12.24 The wallpaper groups p4m and p4g
The 17 wallpaper groups are listed in Table 12.22. The groups p3m1 and p31m can
be distinguished by whether or not all of their threefold centers lie on the reflection axes:
those of p3m1 must, whereas those of p31m may not. Similarly, the fourfold centers of p4m
must lie on the reflection axes whereas those of p4g need not (Figure 12.24). The complete
proof of this theorem can be found in several of the references at the end of this chapter,
including [5], [6], [10], and [11].
Sage. We have not seen how to use Sage profitably with the material in this chapter.
Historical Note
Symmetry groups have intrigued mathematicians for a long time. Leonardo da Vinci was
probably the first person to know all of the point groups. At the International Congress of
Mathematicians in 1900, David Hilbert gave a now-famous address outlining 23 problems
to guide mathematics in the twentieth century. Hilbert’s eighteenth problem asked whether
or not crystallographic groups in n dimensions were always finite. In 1910, L. Bieberbach
proved that crystallographic groups are finite in every dimension. Finding out how many
of these groups there are in each dimension is another matter. In R3 there are 230 different
space groups; in R4 there are 4783. No one has been able to compute the number of
space groups for R5 and beyond. It is interesting to note that the crystallographic groups
were found mathematically for R3 before the 230 different types of crystals were actually
discovered in nature.
12.3 Reading Questions
1. What is a nonsingular matrix? Give an example of a 2 × 2 nonsingular matrix. How
do you know your example is nonsingular?
2. What is an isometry in Rn ? Can you give an example of an isometry in R2 ?
3. What is an orthonormal set of vectors?
4. What is the difference between the orthogonal group and the special orthogonal group?
5. What is a lattice?
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 155
12.4 Exercises
1. Prove the identity
1
hx, yi = kx + yk2 − kxk2 − kyk2 .
2
2. Show that O(n) is a group.
3. Prove that the following matrices are orthogonal. Are any of these matrices in SO(n)?
(a) (c)
√ √ 4/5 0 3/5
1/ 2 −1/ 2 −3/5
√ √ 0 4/5
1/ 2 1/ 2
0 −1 0
(d)
(b)
1/3 2/3 −2/3
√ √
1/ 5 2/ 5 −2/3 2/3 1/3
√ √
−2/ 5 1/ 5 2/3 1/3 2/3
4. Determine the symmetry group of each of the figures below.
(a)
(c)
(b)
5. Let x, y, and w be vectors in Rn and α ∈ R. Prove each of the following properties of
inner products.
(a) hx, yi = hy, xi.
(b) hx, y + wi = hx, yi + hx, wi.
(c) hαx, yi = hx, αyi = αhx, yi.
(d) hx, xi ≥ 0 with equality exactly when x = 0.
(e) If hx, yi = 0 for all x in Rn , then y = 0.
6. Verify that
E(n) = {(A, x) : A ∈ O(n) and x ∈ Rn }
is a group.
7. Prove that {(2, 1), (1, 1)} and {(12, 5), (7, 3)} are bases for the same lattice.
8. Let G be a subgroup of E(2) and suppose that T is the translation subgroup of G.
Prove that the point group of G is isomorphic to G/T .
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 156
9. Let A ∈ SL2 (R) and suppose that the vectors x and y form two sides of a parallelo-
gram in R2 . Prove that the area of this parallelogram is the same as the area of the
parallelogram with sides Ax and Ay.
10. Prove that SO(n) is a normal subgroup of O(n).
11. Show that any isometry f in Rn is a one-to-one map.
12. Prove or disprove: an element in E(2) of the form (A, x), where x 6= 0, has infinite
order.
13. Prove or disprove: There exists an infinite abelian subgroup of O(n).
14. Let x = (x1 , x2 ) be a point on the unit circle in R2 ; that is, x21 + x22 = 1. If A ∈ O(2),
show that Ax is also a point on the unit circle.
15. Let G be a group with a subgroup H (not necessarily normal) and a normal subgroup
N . Then G is a semidirect product of N by H if
• H ∩ N = {id};
• HN = G.
Show that each of the following is true.
(a) S3 is the semidirect product of A3 by H = {(1), (1 2)}.
(b) The quaternion group, Q8 , cannot be written as a semidirect product.
(c) E(2) is the semidirect product of O(2) by H, where H consists of all translations
in R2 .
16. Determine which of the 17 wallpaper groups preserves the symmetry of the pattern in
Figure 12.16.
17. Determine which of the 17 wallpaper groups preserves the symmetry of the pattern in
Figure 12.25.
Figure 12.25
18. Find the rotation group of a dodecahedron.
19. For each of the 17 wallpaper groups, draw a wallpaper pattern having that group as a
symmetry group.
CHAPTER 12. MATRIX GROUPS AND SYMMETRY 157
12.5 References and Suggested Readings
[1] Coxeter, H. M. and Moser, W. O. J. Generators and Relations for Discrete Groups,
3rd ed. Springer-Verlag, New York, 1972.
[2] Grove, L. C. and Benson, C. T. Finite Reflection Groups. 2nd ed. Springer-Verlag,
New York, 1985.
[3] Hiller, H. “Crystallography and Cohomology of Groups,” American Mathematical
Monthly 93 (1986), 765–79.
[4] Lockwood, E. H. and Macmillan, R. H. Geometric Symmetry. Cambridge University
Press, Cambridge, 1978.
[5] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[6] Martin, G. Transformation Groups: An Introduction to Symmetry. Springer-Verlag,
New York, 1982.
[7] Milnor, J. “Hilbert’s Problem 18: On Crystallographic Groups, Fundamental Domains,
and Sphere Packing,” t Proceedings of Symposia in Pure Mathematics 18, American
Mathematical Society, 1976.
[8] Phillips, F. C. An Introduction to Crystallography. 4th ed. Wiley, New York, 1971.
[9] Rose, B. I. and Stafford, R. D. “An Elementary Course in Mathematical Symmetry,”
American Mathematical Monthly 88 (1980), 54–64.
[10] Schattschneider, D. “The Plane Symmetry Groups: Their Recognition and Their
Notation,” American Mathematical Monthly 85 (1978), 439–50.
[11] Schwarzenberger, R. L. “The 17 Plane Symmetry Groups,” Mathematical Gazette 58
(1974), 123–31.
[12] Weyl, H. Symmetry. Princeton University Press, Princeton, NJ, 1952.
13
The Structure of Groups
The ultimate goal of group theory is to classify all groups up to isomorphism; that is, given a
particular group, we should be able to match it up with a known group via an isomorphism.
For example, we have already proved that any finite cyclic group of order n is isomorphic
to Zn ; hence, we “know” all finite cyclic groups. It is probably not reasonable to expect
that we will ever know all groups; however, we can often classify certain types of groups or
distinguish between groups in special cases.
In this chapter we will characterize all finite abelian groups. We shall also investigate
groups with sequences of subgroups. If a group has a sequence of subgroups, say
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e},
where each subgroup Hi is normal in Hi+1 and each of the factor groups Hi+1 /Hi is abelian,
then G is a solvable group. In addition to allowing us to distinguish between certain classes
of groups, solvable groups turn out to be central to the study of solutions to polynomial
equations.
13.1 Finite Abelian Groups
In our investigation of cyclic groups we found that every group of prime order was isomorphic
to Zp , where p was a prime number. We also determined that Zmn ∼ = Zm × Zn when
gcd(m, n) = 1. In fact, much more is true. Every finite abelian group is isomorphic to a
direct product of cyclic groups of prime power order; that is, every finite abelian group is
isomorphic to a group of the type
Zpα1 × · · · × Zpαnn ,
1
where each pk is prime (not necessarily distinct).
First, let us examine a slight generalization of finite abelian groups. Suppose that G
is a group and let {gi } be a set of elements in G, where i is in some index set I (not
necessarily finite). The smallest subgroup of G containing all of the gi ’s is the subgroup of
G generated by the gi ’s. If this subgroup of G is in fact all of G, then G is generated by
the set {gi : i ∈ I}. In this case the gi ’s are said to be the generators of G. If there is a
finite set {gi : i ∈ I} that generates G, then G is finitely generated.
Example 13.1 Obviously, all finite groups are finitely generated. For example, the group
S3 is generated by the permutations (1 2) and (1 2 3). The group Z × Zn is an infinite group
but is finitely generated by {(1, 0), (0, 1)}. □
158
CHAPTER 13. THE STRUCTURE OF GROUPS 159
Example 13.2 Not all groups are finitely generated. Consider the rational numbers Q
under the operation of addition. Suppose that Q is finitely generated with generators
p1 /q1 , . . . , pn /qn , where each pi /qi is a fraction expressed in its lowest terms. Let p be some
prime that does not divide any of the denominators q1 , . . . , qn . We claim that 1/p cannot
be in the subgroup of Q that is generated by p1 /q1 , . . . , pn /qn , since p does not divide the
denominator of any element in this subgroup. This fact is easy to see since the sum of any
two generators is
pi /qi + pj /qj = (pi qj + pj qi )/(qi qj ).
□
Proposition 13.3 Let H be the subgroup of a group G that is generated by {gi ∈ G : i ∈ I}.
Then h ∈ H exactly when it is a product of the form
h = giα11 · · · giαnn ,
where the gik s are not necessarily distinct.
Proof. Let K be the set of all products of the form giα11 · · · giαnn , where the gik s are not
necessarily distinct. Certainly K is a subset of H. We need only show that K is a subgroup
of G. If this is the case, then K = H, since H is the smallest subgroup containing all the
gi s.
Clearly, the set K is closed under the group operation. Since gi0 = 1, the identity is in
K. It remains to show that the inverse of an element g = gik11 · · · giknn in K must also be in
K. However,
g −1 = (gik11 · · · giknn )−1 = (gi−k
n
n
· · · gi−k
1
1
).
■
The reason that powers of a fixed gi may occur several times in the product is that we
may have a nonabelian group. However, if the group is abelian, then the gi s need occur
only once. For example, a product such as a−3 b5 a7 in an abelian group could always be
simplified (in this case, to a4 b5 ).
Now let us restrict our attention to finite abelian groups. We can express any finite
abelian group as a finite direct product of cyclic groups. More specifically, letting p be
prime, we define a group G to be a p-group if every element in G has as its order a power
of p. For example, both Z2 × Z2 and Z4 are 2-groups, whereas Z27 is a 3-group. We shall
prove the Fundamental Theorem of Finite Abelian Groups which tells us that every finite
abelian group is isomorphic to a direct product of cyclic p-groups.
Theorem 13.4 Fundamental Theorem of Finite Abelian Groups. Every finite
abelian group G is isomorphic to a direct product of cyclic groups of the form
Zpα1 × Zpα2 × · · · × Zpαnn
1 2
here the pi ’s are primes (not necessarily distinct).
Example 13.5 Suppose that we wish to classify all abelian groups of order 540 = 22 · 33 · 5.
The Fundamental Theorem of Finite Abelian Groups tells us that we have the following six
possibilities.
• Z2 × Z2 × Z3 × Z3 × Z3 × Z5 ;
• Z2 × Z2 × Z3 × Z9 × Z5 ;
• Z2 × Z2 × Z27 × Z5 ;
• Z4 × Z3 × Z3 × Z3 × Z5 ;
CHAPTER 13. THE STRUCTURE OF GROUPS 160
• Z4 × Z3 × Z9 × Z5 ;
• Z4 × Z27 × Z5 .
□
The proof of the Fundamental Theorem of Finite Abelian Groups depends on several
lemmas.
Lemma 13.6 Let G be a finite abelian group of order n. If p is a prime that divides n,
then G contains an element of order p.
Proof. We will prove this lemma by induction. If n = 1, then there is nothing to show.
Now suppose that the lemma is true for all groups of order k, where k < n. Furthermore,
let p be a prime that divides n.
If G has no proper nontrivial subgroups, then G = hai, where a is any element other
than the identity. By Exercise 4.5.39, the order of G must be prime. Since p divides n, we
know that p = n, and G contains p − 1 elements of order p.
Now suppose that G contains a nontrivial proper subgroup H. Then 1 < |H| < n. If
p | |H|, then H contains an element of order p by induction and the lemma is true. Suppose
that p does not divide the order of H. Since G is abelian, it must be the case that H is a
normal subgroup of G, and |G| = |H| · |G/H|. Consequently, p must divide |G/H|. Since
|G/H| < |G| = n, we know that G/H contains an element aH of order p by the induction
hypothesis. Thus,
H = (aH)p = ap H,
and ap ∈ H but a ∈ / H. If |H| = r, then p and r are relatively prime, and there exist
integers s and t such that sp + tr = 1. Furthermore, the order of ap must divide r, and
(ap )r = (ar )p = 1.
We claim that ar has order p. We must show that ar 6= 1. Suppose ar = 1. Then
a = asp+tr
= asp atr
= (ap )s (ar )t
= (ap )s 1
= (ap )s .
Since ap ∈ H, it must be the case that a = (ap )s ∈ H, which is a contradiction. Therefore,
ar 6= 1 is an element of order p in G. ■
Lemma 13.6 is a special case of Cauchy’s Theorem (Theorem 15.1), which states that
if G is a finite group and p a prime such that p divides the order of G, then G contains a
subgroup of order p. We will prove Cauchy’s Theorem in Chapter 15.
Lemma 13.7 A finite abelian group is a p-group if and only if its order is a power of p.
Proof. If |G| = pn then by Lagrange’s theorem, then the order of any g ∈ G must divide
pn , and therefore must be a power of p. Conversely, if |G| is not a power of p, then it has
some other prime divisor q, so by Lemma 13.6, G has an element of order q and thus is not
a p-group. ■
Lemma 13.8 Let G be a finite abelian group of order n = pα1 1 · · · pαk k , where where p1 , . . . , pk
are distinct primes and α1 , α2 , . . . , αk are positive integers. Then G is the internal direct
product of subgroups G1 , G2 , . . . , Gk , where Gi is the subgroup of G consisting of all elements
of order pri for some integer r.
Proof. Since G is an abelian group, we are guaranteed that Gi is a subgroup of G for
i = 1, . . . , k. Since the identity has order p0i = 1, we know that 1 ∈ Gi . If g ∈ Gi has order
CHAPTER 13. THE STRUCTURE OF GROUPS 161
pri , then g −1 must also have order pri . Finally, if h ∈ Gi has order psi , then
t t t
(gh)pi = g pi hpi = 1 · 1 = 1,
where t is the maximum of r and s.
We must show that
G = G 1 G2 · · · Gk
and Gi ∩ Gj = {1} for i 6= j. Suppose that g1 ∈ G1 is in the subgroup generated by
G2 , G3 , . . . , Gk . Then g1 = g2 g3 · · · gk for gi ∈ Gi . Since gi has order pαi , we know that
α2 αk
αi p ···p
gip = 1 for i = 2, 3, . . . , k, and g1 2 k
= 1. Since the order of g1 is a power of p1
α2 αk
and gcd(p1 , p2 · · · pk ) = 1, it must be the case that g1 = 1 and the intersection of G1
with any of the subgroups G2 , G3 , . . . , Gk is the identity. A similar argument shows that
Gi ∩ Gj = {1} for i 6= j.
Next, we must show that it possible to write every g ∈ G as a product g1 · · · gk , where
gi ∈ Gi . Since the order of g divides the order of G, we know that
|g| = pβ1 1 pβ2 2 · · · pβk k
for some integers β1 , . . . , βk . Letting ai = |g|/pβi i , the ai ’s are relatively prime; hence, there
exist integers b1 , . . . , bk such that a1 b1 + · · · + ak bk = 1. Consequently,
g = g a1 b1 +···+ak bk = g a1 b1 · · · g ak bk .
Since βi
g (ai bi )pi = g bi |g| = e,
it follows that g ai bi must be in Gi . Let gi = g ai bi . Then g = g1 · · · gk ∈ G1 G2 · · · Gk .
Therefore, G = G1 G2 · · · Gk is an internal direct product of subgroups. ■
If remains for us to determine the possible structure of each pi -group Gi in Lemma 13.8.
Lemma 13.9 Let G be a finite abelian p-group and suppose that g ∈ G has maximal order.
Then G is isomorphic to hgi × H for some subgroup H of G.
Proof. By Lemma 13.7, we may assume that the order of G is pn . We shall induct on n.
If n = 1, then G is cyclic of order p and must be generated by g. Suppose now that the
statement of the lemma holds for all integers k with 1 ≤ k < n and let g be of maximal
m
order in G, say |g| = pm . Then ap = e for all a ∈ G. Now choose h in G such that h ∈ / hgi,
where h has the smallest possible order. Certainly such an h exists; otherwise, G = hgi and
we are done. Let H = hhi.
We claim that hgi ∩ H = {e}. It suffices to show that |H| = p. Since |hp | = |h|/p, the
order of hp is smaller than the order of h and must be in hgi by the minimality of h; that
is, hp = g r for some number r. Hence,
m−1 m−1 m
(g r )p = (hp )p = hp = e,
and the order of g r must be less than or equal to pm−1 . Therefore, g r cannot generate hgi.
Notice that p must occur as a factor of r, say r = ps, and hp = g r = g ps . Define a to be
g −s h. Then a cannot be in hgi; otherwise, h would also have to be in hgi. Also,
ap = g −sp hp = g −r hp = h−p hp = e.
We have now formed an element a with order p such that a ∈ / hgi. Since h was chosen to
have the smallest order of all of the elements that are not in hgi, |H| = p.
CHAPTER 13. THE STRUCTURE OF GROUPS 162
Now we will show that the order of gH in the factor group G/H must be the same as
the order of g in G. If |gH| < |g| = pm , then
m−1 m−1
H = (gH)p = gp H;
m−1
hence, g p must be in hgi ∩ H = {e}, which contradicts the fact that the order of g is pm .
Therefore, gH must have maximal order in G/H. By the Correspondence Theorem and our
induction hypothesis,
G/H ∼= hgHi × K/H
for some subgroup K of G containing H. We claim that hgi ∩ K = {e}. If b ∈ hgi ∩ K, then
bH ∈ hgHi ∩ K/H = {H} and b ∈ hgi ∩ H = {e}. It follows that G = hgiK implies that
G∼ = hgi × K. ■
The proof of the Fundamental Theorem of Finite Abelian Groups follows very quickly
from Lemma 13.9. Suppose that G is a finite abelian group and let g be an element of
maximal order in G. If hgi = G, then we are done; otherwise, G ∼ = Z|g| × H for some
subgroup H contained in G by the lemma. Since |H| < |G|, we can apply mathematical
induction.
We now state the more general theorem for all finitely generated abelian groups. The
proof of this theorem can be found in any of the references at the end of this chapter.
Theorem 13.10 The Fundamental Theorem of Finitely Generated Abelian
Groups. Every finitely generated abelian group G is isomorphic to a direct product of
cyclic groups of the form
Zpα1 × Zpα2 × · · · × Zpαnn × Z × · · · × Z,
1 2
where the pi ’s are primes (not necessarily distinct).
13.2 Solvable Groups
A subnormal series of a group G is a finite sequence of subgroups
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e},
where Hi is a normal subgroup of Hi+1 . If each subgroup Hi is normal in G, then the
series is called a normal series. The length of a subnormal or normal series is the number
of proper inclusions.
Example 13.11 Any series of subgroups of an abelian group is a normal series. Consider
the following series of groups:
Z ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0},
Z24 ⊃ h2i ⊃ h6i ⊃ h12i ⊃ {0}.
□
Example 13.12 A subnormal series need not be a normal series. Consider the following
subnormal series of the group D4 :
D4 ⊃ {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ⊃ {(1), (1 2)(3 4)} ⊃ {(1)}.
The subgroup {(1), (1 2)(3 4)} is not normal in D4 ; consequently, this series is not a normal
series. □
CHAPTER 13. THE STRUCTURE OF GROUPS 163
A subnormal (normal) series {Kj } is a refinement of a subnormal (normal) series {Hi }
if {Hi } ⊂ {Kj }. That is, each Hi is one of the Kj .
Example 13.13 The series
Z ⊃ 3Z ⊃ 9Z ⊃ 45Z ⊃ 90Z ⊃ 180Z ⊃ {0}
is a refinement of the series
Z ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0}.
□
The best way to study a subnormal or normal series of subgroups, {Hi } of G, is actually
to study the factor groups Hi+1 /Hi . We say that two subnormal (normal) series {Hi }
and {Kj } of a group G are isomorphic if there is a one-to-one correspondence between the
collections of factor groups {Hi+1 /Hi } and {Kj+1 /Kj }.
Example 13.14 The two normal series
Z60 ⊃ h3i ⊃ h15i ⊃ {0}
Z60 ⊃ h4i ⊃ h20i ⊃ {0}
of the group Z60 are isomorphic since
Z60 /h3i ∼
= h20i/{0} ∼
= Z3
h3i/h15i = h4i/h20i ∼
∼ = Z5
h15i/{0} ∼
= Z60 /h4i ∼
= Z4 .
□
A subnormal series {Hi } of a group G is a composition series if all the factor groups
are simple; that is, if none of the factor groups of the series contains a normal subgroup. A
normal series {Hi } of G is a principal series if all the factor groups are simple.
Example 13.15 The group Z60 has a composition series
Z60 ⊃ h3i ⊃ h15i ⊃ h30i ⊃ {0}
with factor groups
Z60 /h3i ∼
= Z3
h3i/h15i ∼= Z5
h15i/h30i ∼
= Z2
h30i/{0} ∼
= Z2 .
Since Z60 is an abelian group, this series is automatically a principal series. Notice that a
composition series need not be unique. The series
Z60 ⊃ h2i ⊃ h4i ⊃ h20i ⊃ {0}
is also a composition series. □
Example 13.16 For n ≥ 5, the series
Sn ⊃ An ⊃ {(1)}
CHAPTER 13. THE STRUCTURE OF GROUPS 164
is a composition series for Sn since Sn /An ∼
= Z2 and An is simple. □
Example 13.17 Not every group has a composition series or a principal series. Suppose
that
{0} = H0 ⊂ H1 ⊂ · · · ⊂ Hn−1 ⊂ Hn = Z
is a subnormal series for the integers under addition. Then H1 must be of the form kZ
for some k ∈ N. In this case H1 /H0 ∼ = kZ is an infinite cyclic group with many nontrivial
proper normal subgroups. □
Although composition series need not be unique as in the case of Z60 , it turns out that
any two composition series are related. The factor groups of the two composition series
for Z60 are Z2 , Z2 , Z3 , and Z5 ; that is, the two composition series are isomorphic. The
Jordan-Hölder Theorem says that this is always the case.
Theorem 13.18 Jordan-Hölder. Any two composition series of G are isomorphic.
Proof. We shall employ mathematical induction on the length of the composition series.
If the length of a composition series is 1, then G must be a simple group. In this case any
two composition series are isomorphic.
Suppose now that the theorem is true for all groups having a composition series of length
k, where 1 ≤ k < n. Let
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}
G = Km ⊃ Km−1 ⊃ · · · ⊃ K1 ⊃ K0 = {e}
be two composition series for G. We can form two new subnormal series for G since Hi ∩
Km−1 is normal in Hi+1 ∩ Km−1 and Kj ∩ Hn−1 is normal in Kj+1 ∩ Hn−1 :
G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e}
G = Km ⊃ Km−1 ⊃ Km−1 ∩ Hn−1 ⊃ · · · ⊃ K0 ∩ Hn−1 = {e}.
Since Hi ∩ Km−1 is normal in Hi+1 ∩ Km−1 , the Second Isomorphism Theorem (Theo-
rem 11.12) implies that
(Hi+1 ∩ Km−1 )/(Hi ∩ Km−1 ) = (Hi+1 ∩ Km−1 )/(Hi ∩ (Hi+1 ∩ Km−1 ))
∼
= Hi (Hi+1 ∩ Km−1 )/Hi ,
where Hi is normal in Hi (Hi+1 ∩ Km−1 ). Since {Hi } is a composition series, Hi+1 /Hi
must be simple; consequently, Hi (Hi+1 ∩ Km−1 )/Hi is either Hi+1 /Hi or Hi /Hi . That is,
Hi (Hi+1 ∩ Km−1 ) must be either Hi or Hi+1 . Removing any nonproper inclusions from the
series
Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e},
we have a composition series for Hn−1 . Our induction hypothesis says that this series must
be equivalent to the composition series
Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}.
Hence, the composition series
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}
and
G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e}
are equivalent. If Hn−1 = Km−1 , then the composition series {Hi } and {Kj } are equivalent
and we are done; otherwise, Hn−1 Km−1 is a normal subgroup of G properly containing
CHAPTER 13. THE STRUCTURE OF GROUPS 165
Hn−1 . In this case Hn−1 Km−1 = G and we can apply the Second Isomorphism Theorem
once again; that is,
Km−1 /(Km−1 ∩ Hn−1 ) ∼
= (Hn−1 Km−1 )/Hn−1 = G/Hn−1 .
Therefore,
G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e}
and
G = Km ⊃ Km−1 ⊃ Km−1 ∩ Hn−1 ⊃ · · · ⊃ K0 ∩ Hn−1 = {e}
are equivalent and the proof of the theorem is complete. ■
A group G is solvable if it has a subnormal series {Hi } such that all of the factor groups
Hi+1 /Hi are abelian. Solvable groups will play a fundamental role when we study Galois
theory and the solution of polynomial equations.
Example 13.19 The group S4 is solvable since
S4 ⊃ A4 ⊃ {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ⊃ {(1)}
has abelian factor groups; however, for n ≥ 5 the series
Sn ⊃ An ⊃ {(1)}
is a composition series for Sn with a nonabelian factor group. Therefore, Sn is not a solvable
group for n ≥ 5. □
Sage. Sage is able to create direct products of cyclic groups, though they are realized as
permutation groups. This is a situation that should improve. However, with a classification
of finite abelian groups, we can describe how to construct in Sage every group of order less
than 16.
13.3 Reading Questions
1. How many abelian groups are there of order 200 = 23 52 ?
2. How many abelian groups are there of order 729 = 36 ?
3. Find a subgroup of order 6 in Z8 × Z3 × Z3 .
4. It can be shown that an abelian group of order 72 contains a subgroup of order 8.
What are the possibilities for this subgroup?
5. What is a principal series of the group G? Your answer should not use new terms
defined in this chapter.
13.4 Exercises
1. Find all of the abelian groups of order less than or equal to 40 up to isomorphism.
2. Find all of the abelian groups of order 200 up to isomorphism.
3. Find all of the abelian groups of order 720 up to isomorphism.
CHAPTER 13. THE STRUCTURE OF GROUPS 166
4. Find all of the composition series for each of the following groups.
(a) Z12 (e) S3 × Z4
(b) Z48 (f) S4
(c) The quaternions, Q8 (g) Sn , n ≥ 5
(d) D4 (h) Q
5. Show that the infinite direct product G = Z2 × Z2 × · · · is not finitely generated.
6. Let G be an abelian group of order m. If n divides m, prove that G has a subgroup of
order n.
7. A group G is a torsion group if every element of G has finite order. Prove that a
finitely generated abelian torsion group must be finite.
8. Let G, H, and K be finitely generated abelian groups. Show that if G × H ∼ = G × K,
∼
then H = K. Give a counterexample to show that this cannot be true in general.
9. Let G and H be solvable groups. Show that G × H is also solvable.
10. If G has a composition (principal) series and if N is a proper normal subgroup of G,
show there exists a composition (principal) series containing N .
11. Prove or disprove: Let N be a normal subgroup of G. If N and G/N have composition
series, then G must also have a composition series.
12. Let N be a normal subgroup of G. If N and G/N are solvable groups, show that G is
also a solvable group.
13. Prove that G is a solvable group if and only if G has a series of subgroups
G = Pn ⊃ Pn−1 ⊃ · · · ⊃ P1 ⊃ P0 = {e}
where Pi is normal in Pi+1 and the order of Pi+1 /Pi is prime.
14. Let G be a solvable group. Prove that any subgroup of G is also solvable.
15. Let G be a solvable group and N a normal subgroup of G. Prove that G/N is solvable.
16. Prove that Dn is solvable for all integers n.
17. Suppose that G has a composition series. If N is a normal subgroup of G, show that
N and G/N also have composition series.
18. Let G be a cyclic p-group with subgroups H and K. Prove that either H is contained
in K or K is contained in H.
19. Suppose that G is a solvable group with order n ≥ 2. Show that G contains a normal
nontrivial abelian subgroup.
20. Recall that the commutator subgroup G′ of a group G is defined as the subgroup of
G generated by elements of the form a−1 b−1 ab for a, b ∈ G. We can define a series of
subgroups of G by G(0) = G, G(1) = G′ , and G(i+1) = (G(i) )′ .
(a) Prove that G(i+1) is normal in (G(i) )′ . The series of subgroups
G(0) = G ⊃ G(1) ⊃ G(2) ⊃ · · ·
is called the derived series of G.
(b) Show that G is solvable if and only if G(n) = {e} for some integer n.
21. Suppose that G is a solvable group with order n ≥ 2. Show that G contains a normal
nontrivial abelian factor group.
CHAPTER 13. THE STRUCTURE OF GROUPS 167
22. Zassenhaus Lemma. Let H and K be subgroups of a group G. Suppose also that
H ∗ and K ∗ are normal subgroups of H and K respectively. Then
(a) H ∗ (H ∩ K ∗ ) is a normal subgroup of H ∗ (H ∩ K).
(b) K ∗ (H ∗ ∩ K) is a normal subgroup of K ∗ (H ∩ K).
(c) H ∗ (H ∩K)/H ∗ (H ∩K ∗ ) ∼
= K ∗ (H ∩K)/K ∗ (H ∗ ∩K) ∼
= (H ∩K)/(H ∗ ∩K)(H ∩K ∗ ).
23. Schreier’s Theorem. Use the Zassenhaus Lemma to prove that two subnormal
(normal) series of a group G have isomorphic refinements.
24. Use Schreier’s Theorem to prove the Jordan-Hölder Theorem.
13.5 Programming Exercises
1. Write a program that will compute all possible abelian groups of order n. What is the
largest n for which your program will work?
13.6 References and Suggested Readings
[1] Hungerford, T. W. Algebra. Springer, New York, 1974.
[2] Lang, S. Algebra. 3rd ed. Springer, New York, 2002.
[3] Rotman, J. J. An Introduction to the Theory of Groups. 4th ed. Springer, New York,
1995.
14
Group Actions
Group actions generalize group multiplication. If G is a group and X is an arbitrary set, a
group action of an element g ∈ G and x ∈ X is a product, gx, living in X. Many problems
in algebra are best be attacked via group actions. For example, the proofs of the Sylow
theorems and of Burnside’s Counting Theorem are most easily understood when they are
formulated in terms of group actions.
14.1 Groups Acting on Sets
Let X be a set and G be a group. A (left) action of G on X is a map G × X → X given
by (g, x) 7→ gx, where
1. ex = x for all x ∈ X;
2. (g1 g2 )x = g1 (g2 x) for all x ∈ X and all g1 , g2 ∈ G.
Under these considerations X is called a G-set. Notice that we are not requiring X to be
related to G in any way. It is true that every group G acts on every set X by the trivial
action (g, x) 7→ x; however, group actions are more interesting if the set X is somehow
related to the group G.
Example 14.1 Let G = GL2 (R) and X = R2 . Then G acts on X by left multiplication. If
v ∈ R2 and I is the identity matrix, then Iv = v. If A and B are 2 × 2 invertible matrices,
then (AB)v = A(Bv) since matrix multiplication is associative. □
Example 14.2 Let G = D4 be the symmetry group of a square. If X = {1, 2, 3, 4} is the set
of vertices of the square, then we can consider D4 to consist of the following permutations:
{(1), (1 3), (2 4), (1 4 3 2), (1 2 3 4), (1 2)(3 4), (1 4)(2 3), (1 3)(2 4)}.
The elements of D4 act on X as functions. The permutation (1 3)(2 4) acts on vertex 1 by
sending it to vertex 3, on vertex 2 by sending it to vertex 4, and so on. It is easy to see that
the axioms of a group action are satisfied. □
In general, if X is any set and G is a subgroup of SX , the group of all permutations
acting on X, then X is a G-set under the group action
(σ, x) 7→ σ(x)
for σ ∈ G and x ∈ X.
168
CHAPTER 14. GROUP ACTIONS 169
Example 14.3 If we let X = G, then every group G acts on itself by the left regular
representation; that is, (g, x) 7→ λg (x) = gx, where λg is left multiplication:
e · x = λe x = ex = x
(gh) · x = λgh x = λg λh x = λg (hx) = g · (h · x).
If H is a subgroup of G, then G is an H-set under left multiplication by elements of H. □
Example 14.4 Let G be a group and suppose that X = G. If H is a subgroup of G, then
G is an H-set under conjugation; that is, we can define an action of H on G,
H × G → G,
via
(h, g) 7→ hgh−1
for h ∈ H and g ∈ G. Clearly, the first axiom for a group action holds. Observing that
(h1 h2 , g) = h1 h2 g(h1 h2 )−1
= h1 (h2 gh−1 −1
2 )h1
= (h1 , (h2 , g)),
we see that the second condition is also satisfied. □
Example 14.5 Let H be a subgroup of G and LH the set of left cosets of H. The set LH
is a G-set under the action
(g, xH) 7→ gxH.
Again, it is easy to see that the first axiom is true. Since (gg ′ )xH = g(g ′ xH), the second
axiom is also true. □
If G acts on a set X and x, y ∈ X, then x is said to be G-equivalent to y if there exists
a g ∈ G such that gx = y. We write x ∼G y or x ∼ y if two elements are G-equivalent.
Proposition 14.6 Let X be a G-set. Then G-equivalence is an equivalence relation on X.
Proof. The relation ∼ is reflexive since ex = x. Suppose that x ∼ y for x, y ∈ X. Then
there exists a g such that gx = y. In this case g −1 y = x; hence, y ∼ x. To show that the
relation is transitive, suppose that x ∼ y and y ∼ z. Then there must exist group elements
g and h such that gx = y and hy = z. So z = hy = (hg)x, and x is equivalent to z. ■
If X is a G-set, then each partition of X associated with G-equivalence is called an orbit
of X under G. We will denote the orbit that contains an element x of X by Ox .
Example 14.7 Let G be the permutation group defined by
G = {(1), (1 2 3), (1 3 2), (4 5), (1 2 3)(4 5), (1 3 2)(4 5)}
and X = {1, 2, 3, 4, 5}. Then X is a G-set. The orbits are O1 = O2 = O3 = {1, 2, 3} and
O4 = O5 = {4, 5}. □
Now suppose that G is a group acting on a set X and let g be an element of G. The
fixed point set of g in X, denoted by Xg , is the set of all x ∈ X such that gx = x. We can
also study the group elements g that fix a given x ∈ X. This set is more than a subset of
G, it is a subgroup. This subgroup is called the stabilizer subgroup or isotropy subgroup
of x. We will denote the stabilizer subgroup of x by Gx .
Remark 14.8 It is important to remember that Xg ⊂ X and Gx ⊂ G.
CHAPTER 14. GROUP ACTIONS 170
Example 14.9 Let X = {1, 2, 3, 4, 5, 6} and suppose that G is the permutation group given
by the permutations
{(1), (1 2)(3 4 5 6), (3 5)(4 6), (1 2)(3 6 5 4)}.
Then the fixed point sets of X under the action of G are
X(1) = X,
X(3 5)(4 6) = {1, 2},
X(1 2)(3 4 5 6) = X(1 2)(3 6 5 4) = ∅,
and the stabilizer subgroups are
G1 = G2 = {(1), (3 5)(4 6)},
G3 = G4 = G5 = G6 = {(1)}.
It is easily seen that Gx is a subgroup of G for each x ∈ X. □
Proposition 14.10 Let G be a group acting on a set X and x ∈ X. The stabilizer group
of x, Gx , is a subgroup of G.
Proof. Clearly, e ∈ Gx since the identity fixes every element in the set X. Let g, h ∈ Gx .
Then gx = x and hx = x. So (gh)x = g(hx) = gx = x; hence, the product of two elements
in Gx is also in Gx . Finally, if g ∈ Gx , then x = ex = (g −1 g)x = (g −1 )gx = g −1 x. So g −1
is in Gx . ■
We will denote the number of elements in the fixed point set of an element g ∈ G by
|Xg | and denote the number of elements in the orbit of x ∈ X by |Ox |. The next theorem
demonstrates the relationship between orbits of an element x ∈ X and the left cosets of Gx
in G.
Theorem 14.11 Let G be a finite group and X a finite G-set. If x ∈ X, then |Ox | = [G :
Gx ].
Proof. We know that |G|/|Gx | is the number of left cosets of Gx in G by Lagrange’s
Theorem (Theorem 6.10). We will define a bijective map ϕ between the orbit Ox of X and
the set of left cosets LGx of Gx in G. Let y ∈ Ox . Then there exists a g in G such that
gx = y. Define ϕ by ϕ(y) = gGx . To show that ϕ is one-to-one, assume that ϕ(y1 ) = ϕ(y2 ).
Then
ϕ(y1 ) = g1 Gx = g2 Gx = ϕ(y2 ),
where g1 x = y1 and g2 x = y2 . Since g1 Gx = g2 Gx , there exists a g ∈ Gx such that g2 = g1 g,
y2 = g2 x = g1 gx = g1 x = y1 ;
consequently, the map ϕ is one-to-one. Finally, we must show that the map ϕ is onto. Let
gGx be a left coset. If gx = y, then ϕ(y) = gGx . ■
14.2 The Class Equation
Let X be a finite G-set and XG be the set of fixed points in X; that is,
XG = {x ∈ X : gx = x for all g ∈ G}.
CHAPTER 14. GROUP ACTIONS 171
Since the orbits of the action partition X,
X
n
|X| = |XG | + |Oxi |,
i=k
where xk , . . . , xn are representatives from the distinct nontrivial orbits of X.
Now consider the special case in which G acts on itself by conjugation, (g, x) 7→ gxg −1 .
The center of G,
Z(G) = {x : xg = gx for all g ∈ G},
is the set of points that are fixed by conjugation. The nontrivial orbits of the action are
called the conjugacy classes of G. If x1 , . . . , xk are representatives from each of the nontrivial
conjugacy classes of G and |Ox1 | = n1 , . . . , |Oxk | = nk , then
|G| = |Z(G)| + n1 + · · · + nk .
The stabilizer subgroups of each of the xi ’s, C(xi ) = {g ∈ G : gxi = xi g}, are called the
centralizer subgroups of the xi ’s. From Theorem 14.11, we obtain the class equation:
|G| = |Z(G)| + [G : C(x1 )] + · · · + [G : C(xk )].
One of the consequences of the class equation is that the order of each conjugacy class must
divide the order of G.
Example 14.12 It is easy to check that the conjugacy classes in S3 are the following:
{(1)}, {(1 2 3), (1 3 2)}, {(1 2), (1 3), (2 3)}.
The class equation is 6 = 1 + 2 + 3. □
Example 14.13 The center of D4 is {(1), (1 3)(2 4)}, and the conjugacy classes are
{(1 3), (2 4)}, {(1 4 3 2), (1 2 3 4)}, {(1 2)(3 4), (1 4)(2 3)}.
Thus, the class equation for D4 is 8 = 2 + 2 + 2 + 2. □
Example 14.14 For Sn it takes a bit of work to find the conjugacy classes. We begin with
cycles. Suppose that σ = (a1 , . . . , ak ) is a cycle and let τ ∈ Sn . By Theorem 6.16,
τ στ −1 = (τ (a1 ), . . . , τ (ak )).
Consequently, any two cycles of the same length are conjugate. Now let σ = σ1 σ2 · · · σr be
a cycle decomposition, where the length of each cycle σi is ri . Then σ is conjugate to every
other τ ∈ Sn whose cycle decomposition has the same lengths.
The number of conjugate classes in Sn is the number of ways in which n can be parti-
tioned into sums of positive integers. In the case of S3 for example, we can partition the
integer 3 into the following three sums:
3=1+1+1
3=1+2
3 = 3;
therefore, there are three conjugacy classes. There are variations to problem of finding
the number of such partitions for any positive integer n that are what computer scientists
call NP-complete. This effectively means that the problem cannot be solved for a large n
because the computations would be too time-consuming for even the largest computer. □
CHAPTER 14. GROUP ACTIONS 172
Theorem 14.15 Let G be a group of order pn where p is prime. Then G has a nontrivial
center.
Proof. We apply the class equation
|G| = |Z(G)| + n1 + · · · + nk .
Since each ni > 1 and ni | |G|, it follows that p must divide each ni . Also, p | |G|; hence, p
must divide |Z(G)|. Since the identity is always in the center of G, |Z(G)| ≥ 1. Therefore,
|Z(G)| ≥ p, and there exists some g ∈ Z(G) such that g 6= 1. ■
Corollary 14.16 Let G be a group of order p2 where p is prime. Then G is abelian.
Proof. By Theorem 14.15, |Z(G)| = p or p2 . Suppose that |Z(G)| = p. Then Z(G) and
G/Z(G) both have order p and must both be cyclic groups. Choosing a generator aZ(G)
for G/Z(G), we can write any element gZ(G) in the quotient group as am Z(G) for some
integer m; hence, g = am x for some x in the center of G. Similarly, if hZ(G) ∈ G/Z(G),
there exists a y in Z(G) such that h = an y for some integer n. Since x and y are in the
center of G, they commute with all other elements of G; therefore,
gh = am xan y = am+n xy = an yam x = hg,
and G must be abelian. Hence, |Z(G)| = p2 . ■
14.3 Burnside’s Counting Theorem
Suppose that we wish to color the vertices of a square with two different colors, say black and
white. We might suspect that there would be 24 = 16 different colorings. However, some of
these colorings are equivalent. If we color the first vertex black and the remaining vertices
white, it is the same as coloring the second vertex black and the remaining ones white since
we could obtain the second coloring simply by rotating the square 90◦ (Figure 14.17).
B W W B
W W W W
W W W W
B W W B
Figure 14.17 Equivalent colorings of square
CHAPTER 14. GROUP ACTIONS 173
Burnside’s Counting Theorem offers a method of computing the number of distinguish-
able ways in which something can be done. In addition to its geometric applications, the
theorem has interesting applications to areas in switching theory and chemistry. The proof
of Burnside’s Counting Theorem depends on the following lemma.
Lemma 14.18 Let X be a G-set and suppose that x ∼ y. Then Gx is isomorphic to Gy .
In particular, |Gx | = |Gy |.
Proof. Let G act on X by (g, x) 7→ g · x. Since x ∼ y, there exists a g ∈ G such that
g · x = y. Let a ∈ Gx . Since
gag −1 · y = ga · g −1 y = ga · x = g · x = y,
we can define a map ϕ : Gx → Gy by ϕ(a) = gag −1 . The map ϕ is a homomorphism since
ϕ(ab) = gabg −1 = gag −1 gbg −1 = ϕ(a)ϕ(b).
Suppose that ϕ(a) = ϕ(b). Then gag −1 = gbg −1 or a = b; hence, the map is injective. To
show that ϕ is onto, let b be in Gy ; then g −1 bg is in Gx since
g −1 bg · x = g −1 b · gx = g −1 b · y = g −1 · y = x;
and ϕ(g −1 bg) = b. ■
Theorem 14.19 Burnside. Let G be a finite group acting on a set X and let k denote
the number of orbits of X. Then
1 X
k= |Xg |.
|G|
g∈G
Proof. We look at all the fixed points x of all the elements in g ∈ G; that is, we look at
all g’s and all x’s such that gx = x. If viewed in terms of fixed point sets, the number of all
g’s fixing x’s is X
|Xg |.
g∈G
However, if viewed in terms of the stabilizer subgroups, this number is
X
|Gx |;
x∈X
P P
hence, g∈G |Xg | = x∈X |Gx |. By Lemma 14.18,
X
|Gy | = |Ox | · |Gx |.
y∈Ox
By Theorem 14.11 and Lagrange’s Theorem, this expression is equal to |G|. Summing over
all of the k distinct orbits, we conclude that
X X
|Xg | = |Gx | = k · |G|.
g∈G x∈X
■
Example 14.20 Let X = {1, 2, 3, 4, 5} and suppose that G is the permutation group
G = {(1), (1 3), (1 3)(2 5), (2 5)}. The orbits of X are {1, 3}, {2, 5}, and {4}. The fixed point
CHAPTER 14. GROUP ACTIONS 174
sets are
X(1) = X
X(1 3) = {2, 4, 5}
X(1 3)(2 5) = {4}
X(2 5) = {1, 3, 4}.
Burnside’s Theorem says that
1 X 1
k= |Xg | = (5 + 3 + 1 + 3) = 3.
|G| 4
g∈G
□
A Geometric Example
Before we apply Burnside’s Theorem to switching-theory problems, let us examine the
number of ways in which the vertices of a square can be colored black or white. Notice
that we can sometimes obtain equivalent colorings by simply applying a rigid motion to the
square. For instance, as we have pointed out, if we color one of the vertices black and the
remaining three white, it does not matter which vertex was colored black since a rotation
will give an equivalent coloring.
The symmetry group of a square, D4 , is given by the following permutations:
(1) (1 3) (2 4) (1 4 3 2)
(1 2 3 4) (1 2)(3 4) (1 4)(2 3) (1 3)(2 4)
The group G acts on the set of vertices {1, 2, 3, 4} in the usual manner. We can describe
the different colorings by mappings from X into Y = {B, W } where B and W represent
the colors black and white, respectively. Each map f : X → Y describes a way to color
the corners of the square. Every σ ∈ D4 induces a permutation σ e of the possible colorings
given by σe(f ) = f ◦ σ for f : X → Y . For example, suppose that f is defined by
f (1) = B
f (2) = W
f (3) = W
f (4) = W
and σ = (12)(34). Then σ e(f ) = f ◦ σ sends vertex 2 to B and the remaining vertices to
e is a permutation group G
W . The set of all such σ e on the set of possible colorings. Let Xe
e is the set of all possible maps from X
denote the set of all possible colorings; that is, X to
e
Y . Now we must compute the number of G-equivalence classes.
e(1) = X
1. X e since the identity fixes every possible coloring. |X|
e = 24 = 16.
2. Xe(1 2 3 4) consists of all f ∈ X
e such that f is unchanged by the permutation (1 2 3 4).
In this case f (1) = f (2) = f (3) = f (4), so that all values of f must be the same; that
is, either f (x) = B or f (x) = W for every vertex x of the square. So |X e(1 2 3 4) | = 2.
e(1 4 3 2) | = 2.
3. |X
e(1 3)(2 4) , f (1) = f (3) and f (2) = f (4). Thus, |X
4. For X e(1 3)(2 4) | = 22 = 4.
CHAPTER 14. GROUP ACTIONS 175
e(1 2)(3 4) | = 4.
5. |X
e(1 4)(2 3) | = 4.
6. |X
7. For Xe(1 3) , f (1) = f (3) and the other corners can be of any color; hence, |X
e(1 3) | =
23 = 8.
e(2 4) | = 8.
8. |X
By Burnside’s Theorem, we can conclude that there are exactly
1 4
(2 + 21 + 22 + 21 + 22 + 22 + 23 + 23 ) = 6
8
ways to color the vertices of the square.
Proposition 14.21 Let G be a permutation group of X and X e the set of functions from X
e
to Y . Then G induces a group G that permutes the elements of X, e where σ e∈G e is defined
e
e(f ) = f ◦ σ for σ ∈ G and f ∈ X. Furthermore, if n is the number of cycles in the
by σ
cycle decomposition of σ, then |Xeσ | = |Y |n .
Proof. Let σ ∈ G and f ∈ X. e Since σ permutes the elements of X, f ◦ σ must also be in
e
X. Suppose that g is another function from X to Y such that σ e(f ) = σ
e(g). Then for each
x ∈ X,
f (σ(x)) = σe(f )(x) = σ
e(g)(x) = g(σ(x)).
Since σ is a permutation of X, every element x′ in X is the image of some x in X under σ;
hence, f and g agree on all elements of X. Therefore, f = g and σ e is injective. The map
σ 7→ σ
e is onto, since the two sets are the same size.
Suppose that σ is a permutation of X with cycle decomposition σ = σ1 σ2 · · · σn . Any
f in Xeσ must have the same value on each cycle of σ. Since there are n cycles and |Y |
eσ | = |Y |n .
possible values for each cycle, |X ■
Example 14.22 Let X = {1, 2, . . . , 7} and suppose that Y = {A, B, C}. If g is the
eg must
permutation of X given by (1 3)(2 4 5) = (1 3)(2 4 5)(6)(7), then n = 4. Any f ∈ X
have the same value on each cycle in g. There are |Y | = 3 such choices for any value, so
eg | = 34 = 81.
|X □
Example 14.23 Suppose that we wish to color the vertices of a square using four different
colors. By Proposition 14.21, we can immediately decide that there are
1 4
(4 + 41 + 42 + 41 + 42 + 42 + 43 + 43 ) = 55
8
possible ways. □
Switching Functions
In switching theory we are concerned with the design of electronic circuits with binary
inputs and outputs. The simplest of these circuits is a switching function that has n inputs
and a single output (Figure 14.24). Large electronic circuits can often be constructed by
combining smaller modules of this kind. The inherent problem here is that even for a simple
circuit a large number of different switching functions can be constructed. With only four
inputs and a single output, we can construct 65,536 different switching functions. However,
we can often replace one switching function with another merely by permuting the input
leads to the circuit (Figure 14.25).
CHAPTER 14. GROUP ACTIONS 176
x1
x2
.. f f (x1 , x2 , . . . , xn )
.
xn
Figure 14.24 A switching function of n variables
We define a switching or Boolean function of n variables to be a function from Zn2
to Z2 . Since any switching function can have two possible values for each binary n-tuple
n
and there are 2n binary n-tuples, 22 switching functions are possible for n variables. In
general, allowing permutations of the inputs greatly reduces the number of different kinds
of modules that are needed to build a large circuit.
a a
f f (a, b) f f (b, a) = g(a, b)
b b
Figure 14.25 Switching functions in two variables
The possible switching functions with two input variables a and b are listed in Ta-
ble 14.26. Two switching functions f and g are equivalent if g can be obtained from f
by a permutation of the input variables. For example, g(a, b, c) = f (b, c, a). In this case
g ∼ f via the permutation (a, c, b). In the case of switching functions of two variables, the
permutation (a, b) reduces 16 possible switching functions to 12 equivalent functions since
f2 ∼ f4
f3 ∼ f5
f10 ∼ f12
f11 ∼ f13 .
Table 14.26 Switching functions in two variables
Inputs Outputs
f0 f1 f2 f3 f4 f5 f6 f7
0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1
Inputs Outputs
f8 f9 f10 f11 f12 f13 f14 f15
0 0 1 1 1 1 1 1 1 1
0 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1
3
For three input variables there are 22 = 256 possible switching functions; in the case
4
of four variables there are 22 = 65,536. The number of equivalence classes is too large to
reasonably calculate directly. It is necessary to employ Burnside’s Theorem.
CHAPTER 14. GROUP ACTIONS 177
Consider a switching function with three possible inputs, a, b, and c. As we have
mentioned, two switching functions f and g are equivalent if a permutation of the input
variables of f gives g. It is important to notice that a permutation of the switching functions
is not simply a permutation of the input values {a, b, c}. A switching function is a set of
output values for the inputs a, b, and c, so when we consider equivalent switching functions,
we are permuting 23 possible outputs, not just three input values. For example, each binary
triple (a, b, c) has a specific output associated with it. The permutation (acb) changes
outputs as follows:
(0, 0, 0) 7→ (0, 0, 0)
(0, 0, 1) 7→ (0, 1, 0)
(0, 1, 0) 7→ (1, 0, 0)
..
.
(1, 1, 0) 7→ (1, 0, 1)
(1, 1, 1) 7→ (1, 1, 1).
Let X be the set of output values for a switching function in n variables. Then |X| = 2n .
We can enumerate these values as follows:
(0, . . . , 0, 1) 7→ 0
(0, . . . , 1, 0) 7→ 1
(0, . . . , 1, 1) 7→ 2
..
.
(1, . . . , 1, 1) 7→ 2n − 1.
Now let us consider a circuit with four input variables and a single output. Suppose that
we can permute the leads of any circuit according to the following permutation group:
(a), (a, c), (b, d), (a, d, c, b),
(a, b, c, d), (a, b)(c, d), (a, d)(b, c), (a, c)(b, d).
The permutations of the four possible input variables induce the permutations of the output
values in Table 14.27.
Hence, there are
1 16
(2 + 2 · 212 + 2 · 26 + 3 · 210 ) = 9616
8
possible switching functions of four variables under this group of permutations. This number
will be even smaller if we consider the full symmetric group on four letters.
Table 14.27 Permutations of switching functions in four variables
Group Number
Permutation Switching Function Permutation of Cycles
(a) (0) 16
(a, c) (2, 8)(3, 9)(6, 12)(7, 13) 12
(b, d) (1, 4)(3, 6)(9, 12)(11, 14) 12
(a, d, c, b) (1, 2, 4, 8)(3, 6.12, 9)(5, 10)(7, 14, 13, 11) 6
(a, b, c, d) (1, 8, 4, 2)(3, 9, 12, 6)(5, 10)(7, 11, 13, 14) 6
(a, b)(c, d) (1, 2)(4, 8)(5, 10)(6, 9)(7, 11)(13, 14) 10
(a, d)(b, c) (1, 8)(2, 4)(3, 12)(5, 10)(7, 14)(11, 13) 10
(a, c)(b, d) (1, 4)(2, 8)(3, 12)(6, 9)(7, 13)(11, 14) 10
CHAPTER 14. GROUP ACTIONS 178
Sage. Sage has many commands related to conjugacy, which is a group action. It also has
commands for orbits and stabilizers of permutation groups. In the supplement, we illustrate
the automorphism group of a (combinatorial) graph as another example of a group action
on the vertex set of the graph.
Historical Note
William Burnside was born in London in 1852. He attended Cambridge University from
1871 to 1875 and won the Smith’s Prize in his last year. After his graduation he lectured
at Cambridge. He was made a member of the Royal Society in 1893. Burnside wrote
approximately 150 papers on topics in applied mathematics, differential geometry, and
probability, but his most famous contributions were in group theory. Several of Burnside’s
conjectures have stimulated research to this day. One such conjecture was that every group
of odd order is solvable; that is, for a group G of odd order, there exists a sequence of
subgroups
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}
such that Hi is normal in Hi+1 and Hi+1 /Hi is abelian. This conjecture was finally proven
by W. Feit and J. Thompson in 1963. Burnside’s The Theory of Groups of Finite Order,
published in 1897, was one of the first books to treat groups in a modern context as opposed
to permutation groups. The second edition, published in 1911, is still a classic.
14.4 Reading Questions
1. Give an informal description of a group action.
2. Describe the class equation.
3. What are the groups of order 49?
4. How many switching functions are there with 5 inputs? (Give both a simple expression
and the total number as a single integer.)
5. The “Historical Note” mentions the proof of Burnside’s Conjecture. How long was the
proof?
14.5 Exercises
1. Examples 14.1–14.5 in the first section each describe an action of a group G on a set
X, which will give rise to the equivalence relation defined by G-equivalence. For each
example, compute the equivalence classes of the equivalence relation, the G-equivalence
classes.
2. Compute all Xg and all Gx for each of the following permutation groups.
(a) X = {1, 2, 3}, G = S3 = {(1), (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}
(b) X = {1, 2, 3, 4, 5, 6}, G = {(1), (1 2), (3 4 5), (3 5 4), (1 2)(3 4 5), (1 2)(3 5 4)}
3. Compute the G-equivalence classes of X for each of the G-sets in Exercise 14.5.2. For
each x ∈ X verify that |G| = |Ox | · |Gx |.
4. Let G be the additive group of real numbers. Let the action of θ ∈ G on the real plane
R2 be given by rotating the plane counterclockwise about the origin through θ radians.
Let P be a point on the plane other than the origin.
(a) Show that R2 is a G-set.
CHAPTER 14. GROUP ACTIONS 179
(b) Describe geometrically the orbit containing P .
(c) Find the group GP .
5. Let G = A4 and suppose that G acts on itself by conjugation; that is, (g, h) 7→ ghg −1 .
(a) Determine the conjugacy classes (orbits) of each element of G.
(b) Determine all of the isotropy subgroups for each element of G.
6. Find the conjugacy classes and the class equation for each of the following groups.
(a) S4 (b) D5 (c) Z9 (d) Q8
7. Write the class equation for S5 and for A5 .
8. If a square remains fixed in the plane, how many different ways can the corners of the
square be colored if three colors are used?
9. How many ways can the vertices of an equilateral triangle be colored using three
different colors?
10. Find the number of ways a six-sided die can be constructed if each side is marked
differently with 1, . . . , 6 dots.
11. Up to a rotation, how many ways can the faces of a cube be colored with three different
colors?
12. Consider 12 straight wires of equal lengths with their ends soldered together to form
the edges of a cube. Either silver or copper wire can be used for each edge. How many
different ways can the cube be constructed?
13. Suppose that we color each of the eight corners of a cube. Using three different colors,
how many ways can the corners be colored up to a rotation of the cube?
14. Each of the faces of a regular tetrahedron can be painted either red or white. Up to a
rotation, how many different ways can the tetrahedron be painted?
15. Suppose that the vertices of a regular hexagon are to be colored either red or white.
How many ways can this be done up to a symmetry of the hexagon?
16. A molecule of benzene is made up of six carbon atoms and six hydrogen atoms, linked
together in a hexagonal shape as in Figure 14.28.
(a) How many different compounds can be formed by replacing one or more of the
hydrogen atoms with a chlorine atom?
(b) Find the number of different chemical compounds that can be formed by replacing
three of the six hydrogen atoms in a benzene ring with a CH3 radical.
H
H H
H H
H
Figure 14.28 A benzene ring
CHAPTER 14. GROUP ACTIONS 180
17. How many equivalence classes of switching functions are there if the input variables x1 ,
x2 , and x3 can be permuted by any permutation in S3 ? What if the input variables
x1 , x2 , x3 , and x4 can be permuted by any permutation in S4 ?
18. How many equivalence classes of switching functions are there if the input variables x1 ,
x2 , x3 , and x4 can be permuted by any permutation in the subgroup of S4 generated
by the permutation (x1 , x2 , x3 , x4 )?
19. A striped necktie has 12 bands of color. Each band can be colored by one of four
possible colors. How many possible different-colored neckties are there?
20. A group acts faithfully on a G-set X if the identity is the only element of G that leaves
every element of X fixed. Show that G acts faithfully on X if and only if no two
distinct elements of G have the same action on each element of X.
21. Let p be prime. Show that the number of different abelian groups of order pn (up to
isomorphism) is the same as the number of conjugacy classes in Sn .
22. Let a ∈ G. Show that for any g ∈ G, gC(a)g −1 = C(gag −1 ).
23. Let |G| = pn be a nonabelian group for p prime. Prove that |Z(G)| < pn−1 .
24. Let G be a group with order pn where p is prime and X a finite G-set. If XG = {x ∈
X : gx = x for all g ∈ G} is the set of elements in X fixed by the group action, then
prove that |X| ≡ |XG | (mod p).
25. If G is a group of order pn , where p is prime and n ≥ 2, show that G must have a
proper subgroup of order p. If n ≥ 3, is it true that G will have a proper subgroup of
order p2 ?
14.6 Programming Exercise
1. Write a program to compute the number of conjugacy classes in Sn . What is the largest
n for which your program will work?
14.7 References and Suggested Reading
[1] De Bruijin, N. G. “Pólya’s Theory of Counting,” in Applied Combinatorial Mathemat-
ics, Beckenbach, E. F., ed. Wiley, New York, 1964.
[2] Eidswick, J. A. “Cubelike Puzzles—What Are They and How Do You Solve Them?”
American Mathematical Monthly 93 (1986), 157–76.
[3] Harary, F., Palmer, E. M., and Robinson, R. W. “Pólya’s Contributions to Chem-
ical Enumeration,” in Chemical Applications of Graph Theory, Balaban, A. T., ed.
Academic Press, London, 1976.
[4] Gårding, L. and Tambour, T. Algebra for Computer Science. Springer-Verlag, New
York, 1988.
[5] Laufer, H. B. Discrete Mathematics and Applied Modern Algebra. PWS-Kent, Boston,
1984.
[6] Pólya, G. and Read, R. C. Combinatorial Enumeration of Groups, Graphs, and Chem-
ical Compounds. Springer-Verlag, New York, 1985.
[7] Shapiro, L. W. “Finite Groups Acting on Sets with Applications,” Mathematics Mag-
azine, May–June 1973, 136–47.
15
The Sylow Theorems
We already know that the converse of Lagrange’s Theorem is false. If G is a group of
order m and n divides m, then G does not necessarily possess a subgroup of order n. For
example, A4 has order 12 but does not possess a subgroup of order 6. However, the Sylow
Theorems do provide a partial converse for Lagrange’s Theorem—in certain cases they
guarantee us subgroups of specific orders. These theorems yield a powerful set of tools for
the classification of all finite nonabelian groups.
15.1 The Sylow Theorems
We will use what we have learned about group actions to prove the Sylow Theorems. Recall
for a moment what it means for G to act on itself by conjugation and how conjugacy
classes are distributed in the group according to the class equation, discussed in Chapter 14.
A group G acts on itself by conjugation via the map (g, x) 7→ gxg −1 . Let x1 , . . . , xk be
representatives from each of the distinct conjugacy classes of G that consist of more than
one element. Then the class equation can be written as
|G| = |Z(G)| + [G : C(x1 )] + · · · + [G : C(xk )],
where Z(G) = {g ∈ G : gx = xg for all x ∈ G} is the center of G and C(xi ) = {g ∈ G :
gxi = xi g} is the centralizer subgroup of xi .
We begin our investigation of the Sylow Theorems by examining subgroups of order p,
where p is prime. A group G is a p-group if every element in G has as its order a power of
p, where p is a prime number. A subgroup of a group G is a p-subgroup if it is a p-group.
Theorem 15.1 Cauchy. Let G be a finite group and p a prime such that p divides the
order of G. Then G contains a subgroup of order p.
Proof. We will use induction on the order of G. If |G| = p, then clearly G itself is the
required subgroup. We now assume that every group of order k, where p ≤ k < n and p
divides k, has an element of order p. Assume that |G| = n and p | n and consider the class
equation of G:
|G| = |Z(G)| + [G : C(x1 )] + · · · + [G : C(xk )].
We have two cases.
Case 1. Suppose the order of one of the centralizer subgroups, C(xi ), is divisible by p for
some i, i = 1, . . . , k. In this case, by our induction hypothesis, we are done. Since C(xi )
is a proper subgroup of G and p divides |C(xi )|, C(xi ) must contain an element of order p.
Hence, G must contain an element of order p.
181
CHAPTER 15. THE SYLOW THEOREMS 182
Case 2. Suppose the order of no centralizer subgroup is divisible by p. Then p divides
[G : C(xi )], the order of each conjugacy class in the class equation; hence, p must divide
the center of G, Z(G). Since Z(G) is abelian, it must have a subgroup of order p by the
Fundamental Theorem of Finite Abelian Groups. Therefore, the center of G contains an
element of order p. ■
Corollary 15.2 Let G be a finite group. Then G is a p-group if and only if |G| = pn .
Example 15.3 Let us consider the group A5 . We know that |A5 | = 60 = 22 · 3 · 5. By
Cauchy’s Theorem, we are guaranteed that A5 has subgroups of orders 2, 3 and 5. The
Sylow Theorems will give us even more information about the possible subgroups of A5 . □
We are now ready to state and prove the first of the Sylow Theorems. The proof is very
similar to the proof of Cauchy’s Theorem.
Theorem 15.4 First Sylow Theorem. Let G be a finite group and p a prime such that
pr divides |G|. Then G contains a subgroup of order pr .
Proof. We induct on the order of G once again. If |G| = p, then we are done. Now
suppose that the order of G is n with n > p and that the theorem is true for all groups of
order less than n, where p divides n. We shall apply the class equation once again:
|G| = |Z(G)| + [G : C(x1 )] + · · · + [G : C(xk )].
First suppose that p does not divide [G : C(xi )] for some i. Then pr | |C(xi )|, since pr
divides |G| = |C(xi )| · [G : C(xi )]. Now we can apply the induction hypothesis to C(xi ).
Hence, we may assume that p divides [G : C(xi )] for all i. Since p divides |G|, the class
equation says that p must divide |Z(G)|; hence, by Cauchy’s Theorem, Z(G) has an element
of order p, say g. Let N be the group generated by g. Clearly, N is a normal subgroup
of Z(G) since Z(G) is abelian; therefore, N is normal in G since every element in Z(G)
commutes with every element in G. Now consider the factor group G/N of order |G|/p. By
the induction hypothesis, G/N contains a subgroup H of order pr−1 . The inverse image of
H under the canonical homomorphism ϕ : G → G/N is a subgroup of order pr in G. ■
A Sylow p-subgroup P of a group G is a maximal p-subgroup of G. To prove the other
two Sylow Theorems, we need to consider conjugate subgroups as opposed to conjugate
elements in a group. For a group G, let S be the collection of all subgroups of G. For any
subgroup H, S is a H-set, where H acts on S by conjugation. That is, we have an action
H ×S →S
defined by
h · K 7→ hKh−1
for K in S.
The set
N (H) = {g ∈ G : gHg −1 = H}
is a subgroup of G called the the normalizer of H in G. Notice that H is a normal subgroup
of N (H). In fact, N (H) is the largest subgroup of G in which H is normal.
Lemma 15.5 Let P be a Sylow p-subgroup of a finite group G and let x have as its order
a power of p. If x−1 P x = P , then x ∈ P .
Proof. Certainly x ∈ N (P ), and the cyclic subgroup, hxP i ⊂ N (P )/P , has as its order a
power of p. By the Correspondence Theorem there exists a subgroup H of N (P ) containing
P such that H/P = hxP i. Since |H| = |P | · |hxP i|, the order of H must be a power of p.
However, P is a Sylow p-subgroup contained in H. Since the order of P is the largest power
of p dividing |G|, H = P . Therefore, H/P is the trivial subgroup and xP = P , or x ∈ P .
CHAPTER 15. THE SYLOW THEOREMS 183
■
Lemma 15.6 Let H and K be subgroups of G. The number of distinct H-conjugates of K
is [H : N (K) ∩ H].
Proof. We define a bijection between the conjugacy classes of K and the right cosets of
N (K) ∩ H by h−1 Kh 7→ (N (K) ∩ H)h. To show that this map is a bijection, let h1 , h2 ∈ H
and suppose that (N (K) ∩ H)h1 = (N (K) ∩ H)h2 . Then h2 h−1 1 ∈ N (K). Therefore,
K = h2 h−1 1 Kh 1 h−1
2 or h −1
1 Kh 1 = h −1
2 Kh 2 , and the map is an injection. It is easy to
see that this map is surjective; hence, we have a one-to-one and onto map between the
H-conjugates of K and the right cosets of N (K) ∩ H in H. ■
Theorem 15.7 Second Sylow Theorem. Let G be a finite group and p a prime dividing
|G|. Then all Sylow p-subgroups of G are conjugate. That is, if P1 and P2 are two Sylow
p-subgroups, there exists a g ∈ G such that gP1 g −1 = P2 .
Proof. Let P be a Sylow p-subgroup of G and suppose that |G| = pr m with |P | = pr . Let
S = {P = P1 , P2 , . . . , Pk }
consist of the distinct conjugates of P in G. By Lemma 15.6, k = [G : N (P )]. Notice that
|G| = pr m = |N (P )| · [G : N (P )] = |N (P )| · k.
Since pr divides |N (P )|, p cannot divide k.
Given any other Sylow p-subgroup Q, we must show that Q ∈ S. Consider the Q-
conjugacy classes of each Pi . Clearly, these conjugacy classes partition S. The size of the
partition containing Pi is [Q : N (Pi ) ∩ Q] by Lemma 15.6, and Lagrange’s Theorem tells us
that |Q| = [Q : N (Pi ) ∩ Q]|N (Pi ) ∩ Q|. Thus, [Q : N (Pi ) ∩ Q] must be a divisor of |Q| = pr .
Hence, the number of conjugates in every equivalence class of the partition is a power of
p. However, since p does not divide k, one of these equivalence classes must contain only a
single Sylow p-subgroup, say Pj . In this case, x−1 Pj x = Pj for all x ∈ Q. By Lemma 15.5,
Pj = Q. ■
Theorem 15.8 Third Sylow Theorem. Let G be a finite group and let p be a prime
dividing the order of G. Then the number of Sylow p-subgroups is congruent to 1 (mod p)
and divides |G|.
Proof. Let P be a Sylow p-subgroup acting on the set of Sylow p-subgroups,
S = {P = P1 , P2 , . . . , Pk },
by conjugation. From the proof of the Second Sylow Theorem, the only P -conjugate of P
is itself and the order of the other P -conjugacy classes is a power of p. Each P -conjugacy
class contributes a positive power of p toward |S| except the equivalence class {P }. Since
|S| is the sum of positive powers of p and 1, |S| ≡ 1 (mod p).
Now suppose that G acts on S by conjugation. Since all Sylow p-subgroups are conjugate,
there can be only one orbit under this action. For P ∈ S,
|S| = |orbit of P | = [G : N (P )]
by Lemma 15.6. But [G : N (P )] is a divisor of |G|; consequently, the number of Sylow
p-subgroups of a finite group must divide the order of the group. ■
CHAPTER 15. THE SYLOW THEOREMS 184
Historical Note
Peter Ludvig Mejdell Sylow was born in 1832 in Christiania, Norway (now Oslo). After
attending Christiania University, Sylow taught high school. In 1862 he obtained a temporary
appointment at Christiania University. Even though his appointment was relatively brief,
he influenced students such as Sophus Lie (1842–1899). Sylow had a chance at a permanent
chair in 1869, but failed to obtain the appointment. In 1872, he published a 10-page paper
presenting the theorems that now bear his name. Later Lie and Sylow collaborated on a
new edition of Abel’s works. In 1898, a chair at Christiania University was finally created
for Sylow through the efforts of his student and colleague Lie. Sylow died in 1918.
15.2 Examples and Applications
Example 15.9 Using the Sylow Theorems, we can determine that A5 has subgroups of
orders 2, 3, 4, and 5. The Sylow p-subgroups of A5 have orders 3, 4, and 5. The Third
Sylow Theorem tells us exactly how many Sylow p-subgroups A5 has. Since the number of
Sylow 5-subgroups must divide 60 and also be congruent to 1 (mod 5), there are either one
or six Sylow 5-subgroups in A5 . All Sylow 5-subgroups are conjugate. If there were only
a single Sylow 5-subgroup, it would be conjugate to itself; that is, it would be a normal
subgroup of A5 . Since A5 has no normal subgroups, this is impossible; hence, we have
determined that there are exactly six distinct Sylow 5-subgroups of A5 . □
The Sylow Theorems allow us to prove many useful results about finite groups. By
using them, we can often conclude a great deal about groups of a particular order if certain
hypotheses are satisfied.
Theorem 15.10 If p and q are distinct primes with p < q, then every group G of order
pq has a single subgroup of order q and this subgroup is normal in G. Hence, G cannot be
simple. Furthermore, if q 6≡ 1 (mod p), then G is cyclic.
Proof. We know that G contains a subgroup H of order q. The number of conjugates of
H divides pq and is equal to 1 + kq for k = 0, 1, . . .. However, 1 + q is already too large to
divide the order of the group; hence, H can only be conjugate to itself. That is, H must be
normal in G.
The group G also has a Sylow p-subgroup, say K. The number of conjugates of K must
divide q and be equal to 1 + kp for k = 0, 1, . . .. Since q is prime, either 1 + kp = q or
1 + kp = 1. If 1 + kp = 1, then K is normal in G. In this case, we can easily show that G
satisfies the criteria, given in Chapter 9, for the internal direct product of H and K. Since
H is isomorphic to Zq and K is isomorphic to Zp , G ∼ = Zp × Zq ∼= Zpq by Theorem 9.21. ■
Example 15.11 Every group of order 15 is cyclic. This is true because 15 = 5 · 3 and 5 6≡ 1
(mod 3). □
Example 15.12 Let us classify all of the groups of order 99 = 32 · 11 up to isomorphism.
First we will show that every group G of order 99 is abelian. By the Third Sylow Theorem,
there are 1 + 3k Sylow 3-subgroups, each of order 9, for some k = 0, 1, 2, . . .. Also, 1 + 3k
must divide 11; hence, there can only be a single normal Sylow 3-subgroup H in G. Similarly,
there are 1 + 11k Sylow 11-subgroups and 1 + 11k must divide 9. Consequently, there is only
one Sylow 11-subgroup K in G. By Corollary 14.16, any group of order p2 is abelian for p
prime; hence, H is isomorphic either to Z3 × Z3 or to Z9 . Since K has order 11, it must
be isomorphic to Z11 . Therefore, the only possible groups of order 99 are Z3 × Z3 × Z11 or
Z9 × Z11 up to isomorphism. □
To determine all of the groups of order 5 · 7 · 47 = 1645, we need the following theorem.
CHAPTER 15. THE SYLOW THEOREMS 185
Theorem 15.13 Let G′ = haba−1 b−1 : a, b ∈ Gi be the subgroup consisting of all finite
products of elements of the form aba−1 b−1 in a group G. Then G′ is a normal subgroup of
G and G/G′ is abelian.
The subgroup G′ of G is called the commutator subgroup of G. We leave the proof of
this theorem as an exercise (Exercise 10.4.14 in Chapter 10).
Example 15.14 We will now show that every group of order 5 · 7 · 47 = 1645 is abelian, and
cyclic by Theorem 9.21. By the Third Sylow Theorem, G has only one subgroup H1 of order
47. So G/H1 has order 35 and must be abelian by Theorem 15.10. Hence, the commutator
subgroup of G is contained in H which tells us that |G′ | is either 1 or 47. If |G′ | = 1, we
are done. Suppose that |G′ | = 47. The Third Sylow Theorem tells us that G has only one
subgroup of order 5 and one subgroup of order 7. So there exist normal subgroups H2 and
H3 in G, where |H2 | = 5 and |H3 | = 7. In either case the quotient group is abelian; hence,
G′ must be a subgroup of Hi , i = 1, 2. Therefore, the order of G′ is 1, 5, or 7. However, we
already have determined that |G′ | = 1 or 47. So the commutator subgroup of G is trivial,
and consequently G is abelian. □
Finite Simple Groups
Given a finite group, one can ask whether or not that group has any normal subgroups.
Recall that a simple group is one with no proper nontrivial normal subgroups. As in the
case of A5 , proving a group to be simple can be a very difficult task; however, the Sylow
Theorems are useful tools for proving that a group is not simple. Usually, some sort of
counting argument is involved.
Example 15.15 Let us show that no group G of order 20 can be simple. By the Third
Sylow Theorem, G contains one or more Sylow 5-subgroups. The number of such subgroups
is congruent to 1 (mod 5) and must also divide 20. The only possible such number is 1.
Since there is only a single Sylow 5-subgroup and all Sylow 5-subgroups are conjugate, this
subgroup must be normal. □
Example 15.16 Let G be a finite group of order pn , n > 1 and p prime. By Theorem 14.15,
G has a nontrivial center. Since the center of any group G is a normal subgroup, G cannot
be a simple group. Therefore, groups of orders 4, 8, 9, 16, 25, 27, 32, 49, 64, and 81 are not
simple. In fact, the groups of order 4, 9, 25, and 49 are abelian by Corollary 14.16. □
Example 15.17 No group of order 56 = 23 · 7 is simple. We have seen that if we can show
that there is only one Sylow p-subgroup for some prime p dividing 56, then this must be a
normal subgroup and we are done. By the Third Sylow Theorem, there are either one or
eight Sylow 7-subgroups. If there is only a single Sylow 7-subgroup, then it must be normal.
On the other hand, suppose that there are eight Sylow 7-subgroups. Then each of these
subgroups must be cyclic; hence, the intersection of any two of these subgroups contains
only the identity of the group. This leaves 8 · 6 = 48 distinct elements in the group, each
of order 7. Now let us count Sylow 2-subgroups. There are either one or seven Sylow
2-subgroups. Any element of a Sylow 2-subgroup other than the identity must have as its
order a power of 2; and therefore cannot be one of the 48 elements of order 7 in the Sylow
7-subgroups. Since a Sylow 2-subgroup has order 8, there is only enough room for a single
Sylow 2-subgroup in a group of order 56. If there is only one Sylow 2-subgroup, it must be
normal. □
For other groups G, it is more difficult to prove that G is not simple. Suppose G has
order 48. In this case the technique that we employed in the last example will not work.
We need the following lemma to prove that no group of order 48 is simple.
CHAPTER 15. THE SYLOW THEOREMS 186
Lemma 15.18 Let H and K be finite subgroups of a group G. Then
|H| · |K|
|HK| = .
|H ∩ K|
Proof. Recall that
HK = {hk : h ∈ H, k ∈ K}.
Certainly, |HK| ≤ |H| · |K| since some element in HK could be written as the product of
different elements in H and K. It is quite possible that h1 k1 = h2 k2 for h1 , h2 ∈ H and
k1 , k2 ∈ K. If this is the case, let
a = (h1 )−1 h2 = k1 (k2 )−1 .
Notice that a ∈ H ∩ K, since (h1 )−1 h2 is in H and k2 (k1 )−1 is in K; consequently,
h2 = h1 a−1
k2 = ak1 .
Conversely, let h = h1 b−1 and k = bk1 for b ∈ H ∩ K. Then hk = h1 k1 , where h ∈ H
and k ∈ K. Hence, any element hk ∈ HK can be written in the form hi ki for hi ∈ H and
ki ∈ K, as many times as there are elements in H ∩ K; that is, |H ∩ K| times. Therefore,
|HK| = (|H| · |K|)/|H ∩ K|. ■
Example 15.19 To demonstrate that a group G of order 48 is not simple, we will show
that G contains either a normal subgroup of order 8 or a normal subgroup of order 16. By
the Third Sylow Theorem, G has either one or three Sylow 2-subgroups of order 16. If there
is only one subgroup, then it must be a normal subgroup.
Suppose that the other case is true, and two of the three Sylow 2-subgroups are H and
K. We claim that |H ∩ K| = 8. If |H ∩ K| ≤ 4, then by Lemma 15.18,
16 · 16
|HK| ≥ = 64,
4
which is impossible. Notice that H ∩ K has index two in both of H and K, so is normal in
both, and thus H and K are each in the normalizer of H ∩ K. Because H is a subgroup of
N (H ∩ K) and because N (H ∩ K) has strictly more than 16 elements, |N (H ∩ K)| must
be a multiple of 16 greater than 1, as well as dividing 48. The only possibility is that
|N (H ∩ K)| = 48. Hence, N (H ∩ K) = G. □
The following famous conjecture of Burnside was proved in a long and difficult paper
by Feit and Thompson [2].
Theorem 15.20 Odd Order Theorem. Every finite simple group of nonprime order
must be of even order.
The proof of this theorem laid the groundwork for a program in the 1960s and 1970s
that classified all finite simple groups. The success of this program is one of the outstanding
achievements of modern mathematics.
Sage. Sage will compute a single Sylow p-subgroup for each prime divisor p of the order
of the group. Then, with conjugacy, all of the Sylow p-subgroups can be enumerated. It is
also possible to compute the normalizer of a subgroup.
CHAPTER 15. THE SYLOW THEOREMS 187
15.3 Reading Questions
1. State Sylow’s First Theorem.
2. How many groups are there of order 69? Why?
3. Give two descriptions, fundamentally different in character, of the normalizer of a
subgroup.
4. Suppose that G is an abelian group. What is the commutator subgroup of G, and how
do you know?
5. What’s all the fuss about Sylow’s Theorems?
15.4 Exercises
1. What are the orders of all Sylow p-subgroups where G has order 18, 24, 54, 72, and
80?
2. Find all the Sylow 3-subgroups of S4 and show that they are all conjugate.
3. Show that every group of order 45 has a normal subgroup of order 9.
4. Let H be a Sylow p-subgroup of G. Prove that H is the only Sylow p-subgroup of G
contained in N (H).
5. Prove that no group of order 96 is simple.
6. Prove that no group of order 160 is simple.
7. If H is a normal subgroup of a finite group G and |H| = pk for some prime p, show
that H is contained in every Sylow p-subgroup of G.
8. Let G be a group of order p2 q 2 , where p and q are distinct primes such that q ∤ p2 − 1
and p ∤ q 2 − 1. Prove that G must be abelian. Find a pair of primes for which this is
true.
9. Show that a group of order 33 has only one Sylow 3-subgroup.
10. Let H be a subgroup of a group G. Prove or disprove that the normalizer of H is
normal in G.
11. Let G be a finite group whose order is divisible by a prime p. Prove that if there is
only one Sylow p-subgroup in G, it must be a normal subgroup of G.
12. Let G be a group of order pr , p prime. Prove that G contains a normal subgroup of
order pr−1 .
13. Suppose that G is a finite group of order pn k, where k < p. Show that G must contain
a normal subgroup.
14. Let H be a subgroup of a finite group G. Prove that gN (H)g −1 = N (gHg −1 ) for any
g ∈ G.
15. Prove that a group of order 108 must have a normal subgroup.
16. Classify all the groups of order 175 up to isomorphism.
17. Show that every group of order 255 is cyclic.
18. Let G have order pe11 · · · penn and suppose that G has n Sylow p-subgroups P1 , . . . , Pn
where |Pi | = pei i . Prove that G is isomorphic to P1 × · · · × Pn .
19. Let P be a normal Sylow p-subgroup of G. Prove that every inner automorphism of G
fixes P .
20. What is the smallest possible order of a group G such that G is nonabelian and |G| is
odd? Can you find such a group?
CHAPTER 15. THE SYLOW THEOREMS 188
21. The Frattini Lemma. If H is a normal subgroup of a finite group G and P is
a Sylow p-subgroup of H, for each g ∈ G show that there is an h in H such that
gP g −1 = hP h−1 . Also, show that if N is the normalizer of P , then G = HN .
22. Show that if the order of G is pn q, where p and q are primes and p > q, then G contains
a normal subgroup.
23. Prove that the number of distinct conjugates of a subgroup H of a finite group G is
[G : N (H)].
24. Prove that a Sylow 2-subgroup of S5 is isomorphic to D4 .
25. Another Proof of the Sylow Theorems.
(a) Suppose p is prime and p does not divide m. Show that
k
p m
p∤ .
pk
(b) Let S denote the set of all pk element subsets of G. Show that p does not divide
|S|.
(c) Define an action of G on S by left multiplication, aT = {at : t ∈ T } for a ∈ G
and T ∈ S. Prove that this is a group action.
(d) Prove p ∤ |OT | for some T ∈ S.
(e) Let {T1 , . . . , Tu } be an orbit such that p ∤ u and H = {g ∈ G : gT1 = T1 }. Prove
that H is a subgroup of G and show that |G| = u|H|.
(f) Show that pk divides |H| and pk ≤ |H|.
(g) Show that |H| = |OT | ≤ pk ; conclude that therefore pk = |H|.
26. Let G be a group. Prove that G′ = haba−1 b−1 : a, b ∈ Gi is a normal subgroup of
G and G/G′ is abelian. Find an example to show that {aba−1 b−1 : a, b ∈ G} is not
necessarily a group.
15.5 A Project
The main objective of finite group theory is to classify all possible finite groups up to
isomorphism. This problem is very difficult even if we try to classify the groups of order
less than or equal to 60. However, we can break the problem down into several intermediate
problems. This is a challenging project that requires a working knowledge of the group
theory you have learned up to this point. Even if you do not complete it, it will teach you
a great deal about finite groups. You can use Table 15.21 as a guide.
CHAPTER 15. THE SYLOW THEOREMS 189
Table 15.21 Numbers of distinct groups G, |G| ≤ 60
Order Number Order Number Order Number Order Number
1 ? 16 14 31 1 46 2
2 ? 17 1 32 51 47 1
3 ? 18 ? 33 1 48 52
4 ? 19 ? 34 ? 49 ?
5 ? 20 5 35 1 50 5
6 ? 21 ? 36 14 51 ?
7 ? 22 2 37 1 52 ?
8 ? 23 1 38 ? 53 ?
9 ? 24 ? 39 2 54 15
10 ? 25 2 40 14 55 2
11 ? 26 2 41 1 56 ?
12 5 27 5 42 ? 57 2
13 ? 28 ? 43 1 58 ?
14 ? 29 1 44 4 59 1
15 1 30 4 45 ? 60 13
1. Find all simple groups G ( |G| ≤ 60). Do not use the Odd Order Theorem unless you
are prepared to prove it.
2. Find the number of distinct groups G, where the order of G is n for n = 1, . . . , 60.
3. Find the actual groups (up to isomorphism) for each n.
15.6 References and Suggested Readings
[1] Edwards, H. “A Short History of the Fields Medal,” Mathematical Intelligencer 1
(1978), 127–29.
[2] Feit, W. and Thompson, J. G. “Solvability of Groups of Odd Order,” Pacific Journal
of Mathematics 13 (1963), 775–1029.
[3] Gallian, J. A. “The Search for Finite Simple Groups,” Mathematics Magazine 49
(1976), 163–79.
[4] Gorenstein, D. “Classifying the Finite Simple Groups,” Bulletin of the American
Mathematical Society 14 (1986), 1–98.
[5] Gorenstein, D. Finite Groups. AMS Chelsea Publishing, Providence RI, 1968.
[6] Gorenstein, D., Lyons, R., and Solomon, R. The Classification of Finite Simple Groups.
American Mathematical Society, Providence RI, 1994.
16
Rings
Up to this point we have studied sets with a single binary operation satisfying certain axioms,
but we are often more interested in working with sets that have two binary operations. For
example, one of the most natural algebraic structures to study is the integers with the
operations of addition and multiplication. These operations are related to one another by
the distributive property. If we consider a set with two such related binary operations
satisfying certain axioms, we have an algebraic structure called a ring. In a ring we add
and multiply elements such as real numbers, complex numbers, matrices, and functions.
16.1 Rings
A nonempty set R is a ring if it has two closed binary operations, addition and multiplication,
satisfying the following conditions.
1. a + b = b + a for a, b ∈ R.
2. (a + b) + c = a + (b + c) for a, b, c ∈ R.
3. There is an element 0 in R such that a + 0 = a for all a ∈ R.
4. For every element a ∈ R, there exists an element −a in R such that a + (−a) = 0.
5. (ab)c = a(bc) for a, b, c ∈ R.
6. For a, b, c ∈ R,
a(b + c) = ab + ac
(a + b)c = ac + bc.
This last condition, the distributive axiom, relates the binary operations of addition and
multiplication. Notice that the first four axioms simply require that a ring be an abelian
group under addition, so we could also have defined a ring to be an abelian group (R, +)
together with a second binary operation satisfying the fifth and sixth conditions given above.
If there is an element 1 ∈ R such that 1 6= 0 and 1a = a1 = a for each element a ∈ R,
we say that R is a ring with unity or identity. A ring R for which ab = ba for all a, b in
R is called a commutative ring. A commutative ring R with identity is called an integral
domain if, for every a, b ∈ R such that ab = 0, either a = 0 or b = 0. A division ring
is a ring R, with an identity, in which every nonzero element in R is a unit; that is, for
each a ∈ R with a 6= 0, there exists a unique element a−1 such that a−1 a = aa−1 = 1. A
190
CHAPTER 16. RINGS 191
commutative division ring is called a field. The relationship among rings, integral domains,
division rings, and fields is shown in Figure 16.1.
Rings
Commutative Rings with
Rings Identity
Integral Division
Domains Rings
Fields
Figure 16.1 Types of rings
Example 16.2 As we have mentioned previously, the integers form a ring. In fact, Z is
an integral domain. Certainly if ab = 0 for two integers a and b, either a = 0 or b = 0.
However, Z is not a field. There is no integer that is the multiplicative inverse of 2, since
1/2 is not an integer. The only integers with multiplicative inverses are 1 and −1. □
Example 16.3 Under the ordinary operations of addition and multiplication, all of the
familiar number systems are rings: the rationals, Q; the real numbers, R; and the complex
numbers, C. Each of these rings is a field. □
Example 16.4 We can define the product of two elements a and b in Zn by ab (mod n).
For instance, in Z12 , 5 · 7 ≡ 11 (mod 12). This product makes the abelian group Zn into
a ring. Certainly Zn is a commutative ring; however, it may fail to be an integral domain.
If we consider 3 · 4 ≡ 0 (mod 12) in Z12 , it is easy to see that a product of two nonzero
elements in the ring can be equal to zero. □
A nonzero element a in a commutative ring R is called a zero divisor if there is a nonzero
element b in R such that ab = 0. In the previous example, 3 and 4 are zero divisors in Z12 .
Example 16.5 In calculus the continuous real-valued functions on an interval [a, b] form a
commutative ring. We add or multiply two functions by adding or multiplying the values
of the functions. If f (x) = x2 and g(x) = cos x, then (f + g)(x) = f (x) + g(x) = x2 + cos x
and (f g)(x) = f (x)g(x) = x2 cos x. □
Example 16.6 The 2 × 2 matrices with entries in R form a ring under the usual operations
of matrix addition and multiplication. This ring is noncommutative, since it is usually the
case that AB 6= BA. Also, notice that we can have AB = 0 when neither A nor B is zero.
□
Example 16.7 For an example of a noncommutative division ring, let
1 0 0 1 0 i i 0
1= , i= , j= , k= ,
0 1 −1 0 i 0 0 −i
where i2 = −1. These elements satisfy the following relations:
i2 = j2 = k2 = −1
CHAPTER 16. RINGS 192
ij = k
jk = i
ki = j
ji = −k
kj = −i
ik = −j.
Let H consist of elements of the form a + bi + cj + dk, where a, b, c, d are real numbers.
Equivalently, H can be considered to be the set of all 2 × 2 matrices of the form
α β
,
−β α
where α = a + di and β = b + ci are complex numbers. We can define addition and
multiplication on H either by the usual matrix operations or in terms of the generators 1, i,
j, and k:
(a1 + b1 i + c1 j + d1 k) + (a2 + b2 i + c2 j + d2 k)
= (a1 + a2 ) + (b1 + b2 )i + (c1 + c2 )j + (d1 + d2 )k
and
(a1 + b1 i + c1 j + d1 k)(a2 + b2 i + c2 j + d2 k) = α + βi + γj + δk,
where
α = a1 a2 − b1 b2 − c1 c2 − d1 d2
β = a1 b2 + a2 b1 + c1 d2 − d1 c2
γ = a1 c2 − b1 d2 + c1 a2 + d1 b2
δ = a1 d2 + b1 c2 − c1 b2 + d1 a2 .
Though multiplication looks complicated, it is actually a straightforward computation if
we remember that we just add and multiply elements in H like polynomials and keep in
mind the relationships between the generators i, j, and k. The ring H is called the ring of
quaternions.
To show that the quaternions are a division ring, we must be able to find an inverse for
each nonzero element. Notice that
(a + bi + cj + dk)(a − bi − cj − dk) = a2 + b2 + c2 + d2 .
This element can be zero only if a, b, c, and d are all zero. So if a + bi + cj + dk 6= 0,
a − bi − cj − dk
(a + bi + cj + dk) = 1.
a2 + b2 + c2 + d2
□
Proposition 16.8 Let R be a ring with a, b ∈ R. Then
1. a0 = 0a = 0;
2. a(−b) = (−a)b = −ab;
3. (−a)(−b) = ab.
CHAPTER 16. RINGS 193
Proof. To prove (1), observe that
a0 = a(0 + 0) = a0 + a0;
hence, a0 = 0. Similarly, 0a = 0. For (2), we have ab + a(−b) = a(b − b) = a0 = 0;
consequently, −ab = a(−b). Similarly, −ab = (−a)b. Part (3) follows directly from (2) since
(−a)(−b) = −(a(−b)) = −(−ab) = ab. ■
Just as we have subgroups of groups, we have an analogous class of substructures for
rings. A subring S of a ring R is a subset S of R such that S is also a ring under the
inherited operations from R.
Example 16.9 The ring nZ is a subring of Z. Notice that even though the original ring
may have an identity, we do not require that its subring have an identity. We have the
following chain of subrings:
Z ⊂ Q ⊂ R ⊂ C.
□
The following proposition gives us some easy criteria for determining whether or not
a subset of a ring is indeed a subring. (We will leave the proof of this proposition as an
exercise.)
Proposition 16.10 Let R be a ring and S a subset of R. Then S is a subring of R if and
only if the following conditions are satisfied.
1. S 6= ∅.
2. rs ∈ S for all r, s ∈ S.
3. r − s ∈ S for all r, s ∈ S.
Example 16.11 Let R = M2 (R) be the ring of 2 × 2 matrices with entries in R. If T is the
set of upper triangular matrices in R; i.e.,
a b
T = : a, b, c ∈ R ,
0 c
then T is a subring of R. If
′ ′
a b a b
A= and B=
0 c 0 c′
are in T , then clearly A − B is also in T . Also,
′
aa ab′ + bc′
AB =
0 cc′
is in T . □
16.2 Integral Domains and Fields
Let us briefly recall some definitions. If R is a commutative ring and r is a nonzero element
in R, then r is said to be a zero divisor if there is some nonzero element s ∈ R such that
rs = 0. A commutative ring with identity is said to be an integral domain if it has no zero
divisors. If an element a in a ring R with identity has a multiplicative inverse, we say that
a is a unit. If every nonzero element in a ring R is a unit, then R is called a division ring.
CHAPTER 16. RINGS 194
A commutative division ring is called a field.
Example 16.12 If i2 = −1, then the set Z[i] = {m + ni : m, n ∈ Z} forms a ring known
as the Gaussian integers. It is easily seen that the Gaussian integers are a subring of the
complex numbers since they are closed under addition and multiplication. Let α = a + bi
be a unit in Z[i]. Then α = a − bi is also a unit since if αβ = 1, then αβ = 1. If β = c + di,
then
1 = αβαβ = (a2 + b2 )(c2 + d2 ).
Therefore, a2 + b2 must either be 1 or −1; or, equivalently, a + bi = ±1 or a + bi = ±i.
Therefore, units of this ring are ±1 and ±i; hence, the Gaussian integers are not a field. We
will leave it as an exercise to prove that the Gaussian integers are an integral domain. □
Example 16.13 The set of matrices
1 0 1 1 0 1 0 0
F = , , ,
0 1 1 0 1 1 0 0
with entries in Z2 forms a field. □
√ √
Example
√ √ The set Q( 2 ) = {a + b 2 : a, b ∈ Q} is a field. The inverse of an element
16.14
a + b 2 in Q( 2 ) is
a −b √
+ 2 2.
a − 2b
2 2 a − 2b2
□
We have the following alternative characterization of integral domains.
Proposition 16.15 Cancellation Law. Let D be a commutative ring with identity. Then
D is an integral domain if and only if for all nonzero elements a ∈ D with ab = ac, we have
b = c.
Proof. Let D be an integral domain. Then D has no zero divisors. Let ab = ac with
a 6= 0. Then a(b − c) = 0. Hence, b − c = 0 and b = c.
Conversely, let us suppose that cancellation is possible in D. That is, suppose that
ab = ac implies b = c. Let ab = 0. If a 6= 0, then ab = a0 or b = 0. Therefore, a cannot be
a zero divisor. ■
The following surprising theorem is due to Wedderburn.
Theorem 16.16 Every finite integral domain is a field.
Proof. Let D be a finite integral domain and D∗ be the set of nonzero elements of D. We
must show that every element in D∗ has an inverse. For each a ∈ D∗ we can define a map
λa : D∗ → D∗ by λa (d) = ad. This map makes sense, because if a 6= 0 and d 6= 0, then
ad 6= 0. The map λa is one-to-one, since for d1 , d2 ∈ D∗ ,
ad1 = λa (d1 ) = λa (d2 ) = ad2
implies d1 = d2 by left cancellation. Since D∗ is a finite set, the map λa must also be
onto; hence, for some d ∈ D∗ , λa (d) = ad = 1. Therefore, a has a left inverse. Since D is
commutative, d must also be a right inverse for a. Consequently, D is a field. ■
For any nonnegative integer n and any element r in a ring R we write r + · · · + r (n
times) as nr. We define the characteristic of a ring R to be the least positive integer n such
that nr = 0 for all r ∈ R. If no such integer exists, then the characteristic of R is defined
to be 0. We will denote the characteristic of R by char R.
Example 16.17 For every prime p, Zp is a field of characteristic p. By Proposition 3.4,
every nonzero element in Zp has an inverse; hence, Zp is a field. If a is any nonzero element
CHAPTER 16. RINGS 195
in the field, then pa = 0, since the order of any nonzero element in the abelian group Zp is
p. □
Lemma 16.18 Let R be a ring with identity. If 1 has order n, then the characteristic of R
is n.
Proof. If 1 has order n, then n is the least positive integer such that n1 = 0. Thus, for
all r ∈ R,
nr = n(1r) = (n1)r = 0r = 0.
On the other hand, if no positive n exists such that n1 = 0, then the characteristic of R is
zero. ■
Theorem 16.19 The characteristic of an integral domain is either prime or zero.
Proof. Let D be an integral domain and suppose that the characteristic of D is n with
n 6= 0. If n is not prime, then n = ab, where 1 < a < n and 1 < b < n. By Lemma 16.18,
we need only consider the case n1 = 0. Since 0 = n1 = (ab)1 = (a1)(b1) and there are no
zero divisors in D, either a1 = 0 or b1 = 0. Hence, the characteristic of D must be less than
n, which is a contradiction. Therefore, n must be prime. ■
16.3 Ring Homomorphisms and Ideals
In the study of groups, a homomorphism is a map that preserves the operation of the
group. Similarly, a homomorphism between rings preserves the operations of addition and
multiplication in the ring. More specifically, if R and S are rings, then a ring homomorphism
is a map ϕ : R → S satisfying
ϕ(a + b) = ϕ(a) + ϕ(b)
ϕ(ab) = ϕ(a)ϕ(b)
for all a, b ∈ R. If ϕ : R → S is a one-to-one and onto homomorphism, then ϕ is called an
isomorphism of rings.
The set of elements that a ring homomorphism maps to 0 plays a fundamental role in
the theory of rings. For any ring homomorphism ϕ : R → S, we define the kernel of a ring
homomorphism to be the set
ker ϕ = {r ∈ R : ϕ(r) = 0}.
Example 16.20 For any integer n we can define a ring homomorphism ϕ : Z → Zn by
a 7→ a (mod n). This is indeed a ring homomorphism, since
ϕ(a + b) = (a + b) (mod n)
= a (mod n) + b (mod n)
= ϕ(a) + ϕ(b)
and
ϕ(ab) = ab (mod n)
=a (mod n) · b (mod n)
= ϕ(a)ϕ(b).
The kernel of the homomorphism ϕ is nZ. □
Example 16.21 Let C[a, b] be the ring of continuous real-valued functions on an interval
[a, b] as in Example 16.5. For a fixed α ∈ [a, b], we can define a ring homomorphism
CHAPTER 16. RINGS 196
ϕα : C[a, b] → R by ϕα (f ) = f (α). This is a ring homomorphism since
ϕα (f + g) = (f + g)(α) = f (α) + g(α) = ϕα (f ) + ϕα (g)
ϕα (f g) = (f g)(α) = f (α)g(α) = ϕα (f )ϕα (g).
Ring homomorphisms of the type ϕα are called evaluation homomorphisms. □
In the next proposition we will examine some fundamental properties of ring homomor-
phisms. The proof of the proposition is left as an exercise.
Proposition 16.22 Let ϕ : R → S be a ring homomorphism.
1. If R is a commutative ring, then ϕ(R) is a commutative ring.
2. ϕ(0) = 0.
3. Let 1R and 1S be the identities for R and S, respectively. If ϕ is onto, then ϕ(1R ) = 1S .
4. If R is a field and ϕ(R) 6= {0}, then ϕ(R) is a field.
In group theory we found that normal subgroups play a special role. These subgroups
have nice characteristics that make them more interesting to study than arbitrary subgroups.
In ring theory the objects corresponding to normal subgroups are a special class of subrings
called ideals. An ideal in a ring R is a subring I of R such that if a is in I and r is in R,
then both ar and ra are in I; that is, rI ⊂ I and Ir ⊂ I for all r ∈ R.
Example 16.23 Every ring R has at least two ideals, {0} and R. These ideals are called
the trivial ideals. □
Let R be a ring with identity and suppose that I is an ideal in R such that 1 is in I.
Since for any r ∈ R, r1 = r ∈ I by the definition of an ideal, I = R.
Example 16.24 If a is any element in a commutative ring R with identity, then the set
hai = {ar : r ∈ R}
is an ideal in R. Certainly, hai is nonempty since both 0 = a0 and a = a1 are in hai. The
sum of two elements in hai is again in hai since ar + ar′ = a(r + r′ ). The inverse of ar is
−ar = a(−r) ∈ hai. Finally, if we multiply an element ar ∈ hai by an arbitrary element
s ∈ R, we have s(ar) = a(sr). Therefore, hai satisfies the definition of an ideal. □
If R is a commutative ring with identity, then an ideal of the form hai = {ar : r ∈ R}
is called a principal ideal.
Theorem 16.25 Every ideal in the ring of integers Z is a principal ideal.
Proof. The zero ideal {0} is a principal ideal since h0i = {0}. If I is any nonzero ideal
in Z, then I must contain some positive integer m. There exists a least positive integer n
in I by the Principle of Well-Ordering. Now let a be any element in I. Using the division
algorithm, we know that there exist integers q and r such that
a = nq + r
where 0 ≤ r < n. This equation tells us that r = a − nq ∈ I, but r must be 0 since n is the
least positive element in I. Therefore, a = nq and I = hni. ■
Example 16.26 The set nZ is ideal in the ring of integers. If na is in nZ and b is in Z,
then nab is in nZ as required. In fact, by Theorem 16.25, these are the only ideals of Z. □
Proposition 16.27 The kernel of any ring homomorphism ϕ : R → S is an ideal in R.
CHAPTER 16. RINGS 197
Proof. We know from group theory that ker ϕ is an additive subgroup of R. Suppose that
r ∈ R and a ∈ ker ϕ. Then we must show that ar and ra are in ker ϕ. However,
ϕ(ar) = ϕ(a)ϕ(r) = 0ϕ(r) = 0
and
ϕ(ra) = ϕ(r)ϕ(a) = ϕ(r)0 = 0.
■
Remark 16.28 In our definition of an ideal we have required that rI ⊂ I and Ir ⊂ I for
all r ∈ R. Such ideals are sometimes referred to as two-sided ideals. We can also consider
one-sided ideals; that is, we may require only that either rI ⊂ I or Ir ⊂ I for r ∈ R hold
but not both. Such ideals are called left ideals and right ideals, respectively. Of course, in a
commutative ring any ideal must be two-sided. In this text we will concentrate on two-sided
ideals.
Theorem 16.29 Let I be an ideal of R. The factor group R/I is a ring with multiplication
defined by
(r + I)(s + I) = rs + I.
Proof. We already know that R/I is an abelian group under addition. Let r + I and
s + I be in R/I. We must show that the product (r + I)(s + I) = rs + I is independent of
the choice of coset; that is, if r′ ∈ r + I and s′ ∈ s + I, then r′ s′ must be in rs + I. Since
r′ ∈ r + I, there exists an element a in I such that r′ = r + a. Similarly, there exists a b ∈ I
such that s′ = s + b. Notice that
r′ s′ = (r + a)(s + b) = rs + as + rb + ab
and as + rb + ab ∈ I since I is an ideal; consequently, r′ s′ ∈ rs + I. We will leave as an
exercise the verification of the associative law for multiplication and the distributive laws.
■
The ring R/I in Theorem 16.29 is called the factor or quotient ring. Just as with
group homomorphisms and normal subgroups, there is a relationship between ring homo-
morphisms and ideals.
Theorem 16.30 Let I be an ideal of R. The map ϕ : R → R/I defined by ϕ(r) = r + I is
a ring homomorphism of R onto R/I with kernel I.
Proof. Certainly ϕ : R → R/I is a surjective abelian group homomorphism. It remains
to show that ϕ works correctly under ring multiplication. Let r and s be in R. Then
ϕ(r)ϕ(s) = (r + I)(s + I) = rs + I = ϕ(rs),
which completes the proof of the theorem. ■
The map ϕ : R → R/I is often called the natural or canonical homomorphism. In ring
theory we have isomorphism theorems relating ideals and ring homomorphisms similar to
the isomorphism theorems for groups that relate normal subgroups and homomorphisms in
Chapter 11. We will prove only the First Isomorphism Theorem for rings in this chapter
and leave the proofs of the other two theorems as exercises. All of the proofs are similar to
the proofs of the isomorphism theorems for groups.
Theorem 16.31 First Isomorphism Theorem. Let ψ : R → S be a ring homomorphism.
Then ker ψ is an ideal of R. If ϕ : R → R/ ker ψ is the canonical homomorphism, then there
exists a unique isomorphism η : R/ ker ψ → ψ(R) such that ψ = ηϕ.
CHAPTER 16. RINGS 198
Proof. Let K = ker ψ. By the First Isomorphism Theorem for groups, there exists a
well-defined group homomorphism η : R/K → ψ(R) defined by η(r + K) = ψ(r) for the
additive abelian groups R and R/K. To show that this is a ring homomorphism, we need
only show that η((r + K)(s + K)) = η(r + K)η(s + K); but
η((r + K)(s + K)) = η(rs + K)
= ψ(rs)
= ψ(r)ψ(s)
= η(r + K)η(s + K).
■
Theorem 16.32 Second Isomorphism Theorem. Let I be a subring of a ring R and
J an ideal of R. Then I ∩ J is an ideal of I and
I/I ∩ J ∼
= (I + J)/J.
Theorem 16.33 Third Isomorphism Theorem. Let R be a ring and I and J be ideals
of R where J ⊂ I. Then
R/J
R/I ∼
= .
I/J
Theorem 16.34 Correspondence Theorem. Let I be an ideal of a ring R. Then
S 7→ S/I is a one-to-one correspondence between the set of subrings S containing I and the
set of subrings of R/I. Furthermore, the ideals of R containing I correspond to ideals of
R/I.
16.4 Maximal and Prime Ideals
In this particular section we are especially interested in certain ideals of commutative rings.
These ideals give us special types of factor rings. More specifically, we would like to char-
acterize those ideals I of a commutative ring R such that R/I is an integral domain or a
field.
A proper ideal M of a ring R is a maximal ideal of R if the ideal M is not a proper
subset of any ideal of R except R itself. That is, M is a maximal ideal if for any ideal I
properly containing M , I = R. The following theorem completely characterizes maximal
ideals for commutative rings with identity in terms of their corresponding factor rings.
Theorem 16.35 Let R be a commutative ring with identity and M an ideal in R. Then
M is a maximal ideal of R if and only if R/M is a field.
Proof. Let M be a maximal ideal in R. If R is a commutative ring, then R/M must also
be a commutative ring. Clearly, 1 + M acts as an identity for R/M . We must also show
that every nonzero element in R/M has an inverse. If a + M is a nonzero element in R/M ,
then a ∈
/ M . Define I to be the set {ra + m : r ∈ R and m ∈ M }. We will show that I is
an ideal in R. The set I is nonempty since 0a + 0 = 0 is in I. If r1 a + m1 and r2 a + m2 are
two elements in I, then
(r1 a + m1 ) − (r2 a + m2 ) = (r1 − r2 )a + (m1 − m2 )
is in I. Also, for any r ∈ R it is true that rI ⊂ I; hence, I is closed under multiplication
and satisfies the necessary conditions to be an ideal. Therefore, by Proposition 16.10 and
the definition of an ideal, I is an ideal properly containing M . Since M is a maximal ideal,
I = R; consequently, by the definition of I there must be an m in M and an element b in
CHAPTER 16. RINGS 199
R such that 1 = ab + m. Therefore,
1 + M = ab + M = ba + M = (a + M )(b + M ).
Conversely, suppose that M is an ideal and R/M is a field. Since R/M is a field, it
must contain at least two elements: 0 + M = M and 1 + M . Hence, M is a proper ideal
of R. Let I be any ideal properly containing M . We need to show that I = R. Choose
a in I but not in M . Since a + M is a nonzero element in a field, there exists an element
b + M in R/M such that (a + M )(b + M ) = ab + M = 1 + M . Consequently, there exists
an element m ∈ M such that ab + m = 1 and 1 is in I. Therefore, r1 = r ∈ I for all r ∈ R.
Consequently, I = R. ■
Example 16.36 Let pZ be an ideal in Z, where p is prime. Then pZ is a maximal ideal
since Z/pZ ∼
= Zp is a field. □
A proper ideal P in a commutative ring R is called a prime ideal if whenever ab ∈ P ,
then either a ∈ P or b ∈ P .9
Example 16.37 It is easy to check that the set P = {0, 2, 4, 6, 8, 10} is an ideal in Z12 .
This ideal is prime. In fact, it is a maximal ideal. □
Proposition 16.38 Let R be a commutative ring with identity 1, where 1 6= 0. Then P is
a prime ideal in R if and only if R/P is an integral domain.
Proof. First let us assume that P is an ideal in R and R/P is an integral domain. Suppose
that ab ∈ P . If a+P and b+P are two elements of R/P such that (a+P )(b+P ) = 0+P = P ,
then either a + P = P or b + P = P . This means that either a is in P or b is in P , which
shows that P must be prime.
Conversely, suppose that P is prime and
(a + P )(b + P ) = ab + P = 0 + P = P .
Then ab ∈ P . If a ∈
/ P , then b must be in P by the definition of a prime ideal; hence,
b + P = 0 + P and R/P is an integral domain. ■
Example 16.39 Every ideal in Z is of the form nZ. The factor ring Z/nZ ∼ = Zn is an
integral domain only when n is prime. It is actually a field. Hence, the nonzero prime ideals
in Z are the ideals pZ, where p is prime. This example really justifies the use of the word
“prime” in our definition of prime ideals. □
Since every field is an integral domain, we have the following corollary.
Corollary 16.40 Every maximal ideal in a commutative ring with identity is also a prime
ideal.
Historical Note
Amalie Emmy Noether, one of the outstanding mathematicians of the twentieth century,
was born in Erlangen, Germany in 1882. She was the daughter of Max Noether (1844–
1921), a distinguished mathematician at the University of Erlangen. Together with Paul
Gordon (1837–1912), Emmy Noether’s father strongly influenced her early education. She
entered the University of Erlangen at the age of 18. Although women had been admitted
to universities in England, France, and Italy for decades, there was great resistance to
their presence at universities in Germany. Noether was one of only two women among
the university’s 986 students. After completing her doctorate under Gordon in 1907, she
continued to do research at Erlangen, occasionally lecturing when her father was ill.
9
It is possible to define prime ideals in a noncommutative ring. See [1] or [3].
CHAPTER 16. RINGS 200
Noether went to Göttingen to study in 1916. David Hilbert and Felix Klein tried unsuccess-
fully to secure her an appointment at Göttingen. Some of the faculty objected to women
lecturers, saying, “What will our soldiers think when they return to the university and are
expected to learn at the feet of a woman?” Hilbert, annoyed at the question, responded,
“Meine Herren, I do not see that the sex of a candidate is an argument against her admission
as a Privatdozent. After all, the Senate is not a bathhouse.” At the end of World War I,
attitudes changed and conditions greatly improved for women. After Noether passed her
habilitation examination in 1919, she was given a title and was paid a small sum for her
lectures.
In 1922, Noether became a Privatdozent at Göttingen. Over the next 11 years she used
axiomatic methods to develop an abstract theory of rings and ideals. Though she was not
good at lecturing, Noether was an inspiring teacher. One of her many students was B. L.
van der Waerden, author of the first text treating abstract algebra from a modern point of
view. Some of the other mathematicians Noether influenced or closely worked with were
Alexandroff, Artin, Brauer, Courant, Hasse, Hopf, Pontryagin, von Neumann, and Weyl.
One of the high points of her career was an invitation to address the International Congress
of Mathematicians in Zurich in 1932. In spite of all the recognition she received from her
colleagues, Noether’s abilities were never recognized as they should have been during her
lifetime. She was never promoted to full professor by the Prussian academic bureaucracy.
In 1933, Noether, who was Jewish, was banned from participation in all academic activities
in Germany. She emigrated to the United States, took a position at Bryn Mawr College,
and became a member of the Institute for Advanced Study at Princeton. Noether died
suddenly on April 14, 1935. After her death she was eulogized by such notable scientists as
Albert Einstein.
16.5 An Application to Software Design
The Chinese Remainder Theorem is a result from elementary number theory about the
solution of systems of simultaneous congruences. The Chinese mathematician Sun-tsï wrote
about the theorem in the first century A.D. This theorem has some interesting consequences
in the design of software for parallel processors.
Lemma 16.41 Let m and n be positive integers such that gcd(m, n) = 1. Then for a, b ∈ Z
the system
x ≡ a (mod m)
x ≡ b (mod n)
has a solution. If x1 and x2 are two solutions of the system, then x1 ≡ x2 (mod mn).
Proof. The equation x ≡ a (mod m) has a solution since a + km satisfies the equation for
all k ∈ Z. We must show that there exists an integer k1 such that
a + k1 m ≡ b (mod n).
This is equivalent to showing that
k1 m ≡ (b − a) (mod n)
has a solution for k1 . Since m and n are relatively prime, there exist integers s and t such
that ms + nt = 1. Consequently,
(b − a)ms = (b − a) − (b − a)nt,
CHAPTER 16. RINGS 201
or
[(b − a)s]m ≡ (b − a) (mod n).
Now let k1 = (b − a)s.
To show that any two solutions are congruent modulo mn, let c1 and c2 be two solutions
of the system. That is,
ci ≡ a (mod m)
ci ≡ b (mod n)
for i = 1, 2. Then
c2 ≡ c1 (mod m)
c2 ≡ c1 (mod n).
Therefore, both m and n divide c1 − c2 . Consequently, c2 ≡ c1 (mod mn). ■
Example 16.42 Let us solve the system
x≡3 (mod 4)
x≡4 (mod 5).
Using the Euclidean algorithm, we can find integers s and t such that 4s + 5t = 1. Two
such integers are s = 4 and t = −3. Consequently,
x = a + k1 m = 3 + 4k1 = 3 + 4[(5 − 4)4] = 19.
□
Theorem 16.43 Chinese Remainder Theorem. Let n1 , n2 , . . . , nk be positive integers
such that gcd(ni , nj ) = 1 for i 6= j. Then for any integers a1 , . . . , ak , the system
x ≡ a1 (mod n1 )
x ≡ a2 (mod n2 )
..
.
x ≡ ak (mod nk )
has a solution. Furthermore, any two solutions of the system are congruent modulo n1 n2 · · · nk .
Proof. We will use mathematical induction on the number of equations in the system. If
there are k = 2 equations, then the theorem is true by Lemma 16.41. Now suppose that
the result is true for a system of k equations or less and that we wish to find a solution of
x ≡ a1 (mod n1 )
x ≡ a2 (mod n2 )
..
.
x ≡ ak+1 (mod nk+1 ).
Considering the first k equations, there exists a solution that is unique modulo n1 · · · nk ,
say a. Since n1 · · · nk and nk+1 are relatively prime, the system
x≡a (mod n1 · · · nk )
x ≡ ak+1 (mod nk+1 )
has a solution that is unique modulo n1 . . . nk+1 by the lemma. ■
CHAPTER 16. RINGS 202
Example 16.44 Let us solve the system
x≡3 (mod 4)
x≡4 (mod 5)
x≡1 (mod 9)
x≡5 (mod 7).
From Example 16.42 we know that 19 is a solution of the first two congruences and any
other solution of the system is congruent to 19 (mod 20). Hence, we can reduce the system
to a system of three congruences:
x ≡ 19 (mod 20)
x≡1 (mod 9)
x≡5 (mod 7).
Solving the next two equations, we can reduce the system to
x ≡ 19 (mod 180)
x ≡ 5 (mod 7).
Solving this last system, we find that 19 is a solution for the system that is unique up to
modulo 1260. □
One interesting application of the Chinese Remainder Theorem in the design of computer
software is that the theorem allows us to break up a calculation involving large integers into
several less formidable calculations. A computer will handle integer calculations only up to
a certain size due to the size of its processor chip, which is usually a 32 or 64-bit processor
chip. For example, the largest integer available on a computer with a 64-bit processor chip
is
263 − 1 = 9,223,372,036,854,775,807.
Larger processors such as 128 or 256-bit have been proposed or are under development.
There is even talk of a 512-bit processor chip. The largest integer that such a chip could
store with be 2511 − 1, which would be a 154 digit number. However, we would need to deal
with much larger numbers to break sophisticated encryption schemes.
Special software is required for calculations involving larger integers which cannot be
added directly by the machine. By using the Chinese Remainder Theorem we can break
down large integer additions and multiplications into calculations that the computer can
handle directly. This is especially useful on parallel processing computers which have the
ability to run several programs concurrently.
Most computers have a single central processing unit (CPU) containing one processor
chip and can only add two numbers at a time. To add a list of ten numbers, the CPU must
do nine additions in sequence. However, a parallel processing computer has more than one
CPU. A computer with 10 CPUs, for example, can perform 10 different additions at the
same time. If we can take a large integer and break it down into parts, sending each part
to a different CPU, then by performing several additions or multiplications simultaneously
on those parts, we can work with an integer that the computer would not be able to handle
as a whole.
Example 16.45 Suppose that we wish to multiply 2134 by 1531. We will use the integers
95, 97, 98, and 99 because they are relatively prime. We can break down each integer into
CHAPTER 16. RINGS 203
four parts:
2134 ≡ 44 (mod 95)
2134 ≡ 0 (mod 97)
2134 ≡ 76 (mod 98)
2134 ≡ 55 (mod 99)
and
1531 ≡ 11 (mod 95)
1531 ≡ 76 (mod 97)
1531 ≡ 61 (mod 98)
1531 ≡ 46 (mod 99).
Multiplying the corresponding equations, we obtain
2134 · 1531 ≡ 44 · 11 ≡ 9 (mod 95)
2134 · 1531 ≡ 0 · 76 ≡ 0 (mod 97)
2134 · 1531 ≡ 76 · 61 ≡ 30 (mod 98)
2134 · 1531 ≡ 55 · 46 ≡ 55 (mod 99).
Each of these four computations can be sent to a different processor if our computer has
several CPUs. By the above calculation, we know that 2134·1531 is a solution of the system
x ≡ 9 (mod 95)
x ≡ 0 (mod 97)
x ≡ 30 (mod 98)
x ≡ 55 (mod 99).
The Chinese Remainder Theorem tells us that solutions are unique up to modulo 95 · 97 ·
98 · 99 = 89,403,930. Solving this system of congruences for x tells us that 2134 · 1531 =
3,267,154.
The conversion of the computation into the four subcomputations will take some com-
puting time. In addition, solving the system of congruences can also take considerable time.
However, if we have many computations to be performed on a particular set of numbers, it
makes sense to transform the problem as we have done above and to perform the necessary
calculations simultaneously. □
Sage. Rings are at the heart of Sage’s design, so you will find a wide range of possibil-
ities for computing with rings and fields. Ideals, quotients, and homomorphisms are all
available.
16.6 Reading Questions
1. What is the fundamental difference between groups and rings?
2. Give two characterizations of an integral domain.
3. Provide two examples of fields, one infinite, one finite.
4. Who was Emmy Noether?
CHAPTER 16. RINGS 204
5. Speculate on a computer program that might use the Chinese Remainder Theorem to
speed up computations with large integers.
16.7 Exercises
1. Which of the following sets are rings with respect to the usual operations of addition
and multiplication? If the set is a ring, is it also a field?
(a) 7Z
(b) Z18
√ √
(c) Q( 2 ) = {a + b 2 : a, b ∈ Q}
√ √ √ √ √
(d) Q( 2, 3 ) = {a + b 2 + c 3 + d 6 : a, b, c, d ∈ Q}
√ √
(e) Z[ 3 ] = {a + b 3 : a, b ∈ Z}
√
(f) R = {a + b 3 : a, b ∈ Q}
3
(g) Z[i] = {a + bi : a, b ∈ Z and i2 = −1}
√ √ √
(h) Q( 3 ) = {a + b 3 + c 9 : a, b, c ∈ Q}
3 3 3
2. Let R be the ring of 2 × 2 matrices of the form
a b
,
0 0
where a, b ∈ R. Show that although R is a ring that has no identity, we can find a
subring S of R with an identity.
3. List or characterize all of the units in each of the following rings.
(a) Z10
(b) Z12
(c) Z7
(d) M2 (Z), the 2 × 2 matrices with entries in Z
(e) M2 (Z2 ), the 2 × 2 matrices with entries in Z2
4. Find all of the ideals in each of the following rings. Which of these ideals are maximal
and which are prime?
(a) Z18
(b) Z25
(c) M2 (R), the 2 × 2 matrices with entries in R
(d) M2 (Z), the 2 × 2 matrices with entries in Z
(e) Q
5. For each of the following rings R with ideal I, give an addition table and a multiplication
table for R/I.
(a) R = Z and I = 6Z
CHAPTER 16. RINGS 205
(b) R = Z12 and I = {0, 3, 6, 9}
6. Find all homomorphisms ϕ : Z/6Z → Z/15Z.
7. Prove that R is not isomorphic to C.
√ √
8. Prove
√ or disprove:√ The ring Q( 2 ) = {a + b 2 : a, b ∈ Q} is isomorphic to the ring
Q( 3 ) = {a + b 3 : a, b ∈ Q}.
9. What is the characteristic of the field formed by the set of matrices
1 0 1 1 0 1 0 0
F = , , ,
0 1 1 0 1 1 0 0
with entries in Z2 ?
10. Define a map ϕ : C → M2 (R) by
a b
ϕ(a + bi) = .
−b a
Show that ϕ is an isomorphism of C with its image in M2 (R).
11. Prove that the Gaussian integers, Z[i], are an integral domain.
√ √
12. Prove that Z[ 3 i] = {a + b 3 i : a, b ∈ Z} is an integral domain.
13. Solve each of the following systems of congruences.
(a) x ≡ 4 (mod 7)
x ≡ 2 (mod 5) x ≡ 7 (mod 9)
x≡6 (mod 11) x ≡ 5 (mod 11)
(b)
x ≡ 3 (mod 7) (d)
x≡0 (mod 8)
x≡5 (mod 15) x ≡ 3 (mod 5)
x ≡ 0 (mod 8)
(c) x ≡ 1 (mod 11)
x ≡ 2 (mod 4) x ≡ 5 (mod 13)
14. Use the method of parallel computation outlined in the text to calculate 2234 + 4121
by dividing the calculation into four separate additions modulo 95, 97, 98, and 99.
15. Explain why the method of parallel computation outlined in the text fails for 2134·1531
if we attempt to break the calculation down into two smaller calculations modulo 98
and 99.
16. If R is a field, show that the only two ideals of R are {0} and R itself.
17. Let a be any element in a ring R with identity. Show that (−1)a = −a.
18. Let ϕ : R → S be a ring homomorphism. Prove each of the following statements.
(a) If R is a commutative ring, then ϕ(R) is a commutative ring.
(b) ϕ(0) = 0.
(c) Let 1R and 1S be the identities for R and S, respectively. If ϕ is onto, then
ϕ(1R ) = 1S .
(d) If R is a field and ϕ(R) 6= 0, then ϕ(R) is a field.
CHAPTER 16. RINGS 206
19. Prove that the associative law for multiplication and the distributive laws hold in R/I.
20. Prove the Second Isomorphism Theorem for rings: Let I be a subring of a ring R and
J an ideal in R. Then I ∩ J is an ideal in I and
I/I ∩ J ∼
= I + J/J.
21. Prove the Third Isomorphism Theorem for rings: Let R be a ring and I and J be ideals
of R, where J ⊂ I. Then
R/J
R/I ∼
= .
I/J
22. Prove the Correspondence Theorem: Let I be an ideal of a ring R. Then S → S/I is
a one-to-one correspondence between the set of subrings S containing I and the set of
subrings of R/I. Furthermore, the ideals of R correspond to ideals of R/I.
23. Let R be a ring and S a subset of R. Show that S is a subring of R if and only if each
of the following conditions is satisfied.
(a) S 6= ∅.
(b) rs ∈ S for all r, s ∈ S.
(c) r − s ∈ S for all r, s ∈ S.
T
24. Let R be a ring with a collection of subrings {Rα }. Prove that Rα is a subring of R.
Give an example to show that the union of two subrings is not necessarily a subring.
T
25. Let {Iα }α∈A be a collection of ideals in a ring R. Prove that α∈A Iα is also an ideal
in R. Give an example to show that if I1 and I2 are ideals in R, then I1 ∪ I2 may not
be an ideal.
26. Let R be an integral domain. Show that if the only ideals in R are {0} and R itself, R
must be a field.
27. Let R be a commutative ring. An element a in R is nilpotent if an = 0 for some
positive integer n. Show that the set of all nilpotent elements forms an ideal in R.
28. A ring R is a Boolean ring if for every a ∈ R, a2 = a. Show that every Boolean ring
is a commutative ring.
29. Let R be a ring, where a3 = a for all a ∈ R. Prove that R must be a commutative
ring.
30. Let R be a ring with identity 1R and S a subring of R with identity 1S . Prove or
disprove that 1R = 1S .
31. If we do not require the identity of a ring to be distinct from 0, we will not have a
very interesting mathematical structure. Let R be a ring such that 1 = 0. Prove that
R = {0}.
32. Let R be a ring. Define the center of R to be
Z(R) = {a ∈ R : ar = ra for all r ∈ R}.
Prove that Z(R) is a commutative subring of R.
33. Let p be prime. Prove that
Z(p) = {a/b : a, b ∈ Z and gcd(b, p) = 1}
is a ring. The ring Z(p) is called the ring of integers localized at p.
34. Prove or disprove: Every finite integral domain is isomorphic to Zp .
CHAPTER 16. RINGS 207
35. Let R be a ring with identity.
(a) Let u be a unit in R. Define a map iu : R → R by r 7→ uru−1 . Prove that iu is an
automorphism of R. Such an automorphism of R is called an inner automorphism
of R. Denote the set of all inner automorphisms of R by Inn(R).
(b) Denote the set of all automorphisms of R by Aut(R). Prove that Inn(R) is a
normal subgroup of Aut(R).
(c) Let U (R) be the group of units in R. Prove that the map
ϕ : U (R) → Inn(R)
defined by u 7→ iu is a homomorphism. Determine the kernel of ϕ.
(d) Compute Aut(Z), Inn(Z), and U (Z).
36. Let R and S be arbitrary rings. Show that their Cartesian product is a ring if we
define addition and multiplication in R × S by
(a) (r, s) + (r′ , s′ ) = (r + r′ , s + s′ )
(b) (r, s)(r′ , s′ ) = (rr′ , ss′ )
37. An element x in a ring is called an idempotent if x2 = x. Prove that the only idempo-
tents in an integral domain are 0 and 1. Find a ring with a idempotent x not equal to
0 or 1.
38. Let gcd(a, n) = d and gcd(b, d) 6= 1. Prove that ax ≡ b (mod n) does not have a
solution.
39. The Chinese Remainder Theorem for Rings. Let R be a ring and I and J be
ideals in R such that I + J = R.
(a) Show that for any r and s in R, the system of equations
x≡r (mod I)
x ≡ s (mod J)
has a solution.
(b) In addition, prove that any two solutions of the system are congruent modulo
I ∩ J.
(c) Let I and J be ideals in a ring R such that I + J = R. Show that there exists a
ring isomorphism
R/(I ∩ J) ∼
= R/I × R/J.
16.8 Programming Exercise
1. Write a computer program implementing fast addition and multiplication using the
Chinese Remainder Theorem and the method outlined in the text.
16.9 References and Suggested Readings
[1] Anderson, F. W. and Fuller, K. R. Rings and Categories of Modules. 2nd ed. Springer,
New York, 1992.
CHAPTER 16. RINGS 208
[2] Atiyah, M. F. and MacDonald, I. G. Introduction to Commutative Algebra. Westview
Press, Boulder, CO, 1994.
[3] Herstein, I. N. Noncommutative Rings. Mathematical Association of America, Wash-
ington, DC, 1994.
[4] Kaplansky, I. Commutative Rings. Revised edition. University of Chicago Press,
Chicago, 1974.
[5] Knuth, D. E. The Art of Computer Programming: Semi-Numerical Algorithms, vol.
2. 3rd ed. Addison-Wesley Professional, Boston, 1997.
[6] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998. A
good source for applications.
[7] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[8] McCoy, N. H. Rings and Ideals. Carus Monograph Series, No. 8. Mathematical
Association of America, Washington, DC, 1968.
[9] McCoy, N. H. The Theory of Rings. Chelsea, New York, 1972.
[10] Zariski, O. and Samuel, P. Commutative Algebra, vols. I and II. Springer, New York,
1975, 1960.
17
Polynomials
Most people are fairly familiar with polynomials by the time they begin to study abstract
algebra. When we examine polynomial expressions such as
p(x) = x3 − 3x + 2
q(x) = 3x2 − 6x + 5,
we have a pretty good idea of what p(x) + q(x) and p(x)q(x) mean. We just add and
multiply polynomials as functions; that is,
(p + q)(x) = p(x) + q(x)
= (x3 − 3x + 2) + (3x2 − 6x + 5)
= x3 + 3x2 − 9x + 7
and
(pq)(x) = p(x)q(x)
= (x3 − 3x + 2)(3x2 − 6x + 5)
= 3x5 − 6x4 − 4x3 + 24x2 − 27x + 10.
It is probably no surprise that polynomials form a ring. In this chapter we shall emphasize
the algebraic structure of polynomials by studying polynomial rings. We can prove many
results for polynomial rings that are similar to the theorems we proved for the integers.
Analogs of prime numbers, the division algorithm, and the Euclidean algorithm exist for
polynomials.
17.1 Polynomial Rings
Throughout this chapter we shall assume that R is a commutative ring with identity. Any
expression of the form
X
n
f (x) = ai xi = a0 + a1 x + a2 x2 + · · · + an xn ,
i=0
where ai ∈ R and an 6= 0, is called a polynomial over R with indeterminate x. The
elements a0 , a1 , . . . , an are called the coefficients of f . The coefficient an is called the
leading coefficient. A polynomial is called monic if the leading coefficient is 1. If n is the
209
CHAPTER 17. POLYNOMIALS 210
largest nonnegative number for which an 6= 0, we say that the degree of f is n and write
deg f (x) = n. If no such n exists—that is, if f = 0 is the zero polynomial—then the degree
of f is defined to be −∞. We will denote the set of all polynomials with coefficients in a
ring R by R[x]. Two polynomials are equal exactly when their corresponding coefficients
are equal; that is, if we let
p(x) = a0 + a1 x + · · · + an xn
q(x) = b0 + b1 x + · · · + bm xm ,
then p(x) = q(x) if and only if ai = bi for all i ≥ 0.
To show that the set of all polynomials forms a ring, we must first define addition and
multiplication. We define the sum of two polynomials as follows. Let
p(x) = a0 + a1 x + · · · + an xn
q(x) = b0 + b1 x + · · · + bm xm .
Then the sum of p(x) and q(x) is
p(x) + q(x) = c0 + c1 x + · · · + ck xk ,
where ci = ai + bi for each i. We define the product of p(x) and q(x) to be
p(x)q(x) = c0 + c1 x + · · · + cm+n xm+n ,
where
X
i
ci = ak bi−k = a0 bi + a1 bi−1 + · · · + ai−1 b1 + ai b0
k=0
for each i. Notice that in each case some of the coefficients may be zero.
Example 17.1 Suppose that
p(x) = 3 + 0x + 0x2 + 2x3 + 0x4
and
q(x) = 2 + 0x − x2 + 0x3 + 4x4
are polynomials in Z[x]. If the coefficient of some term in a polynomial is zero, then we
usually just omit that term. In this case we would write p(x) = 3+2x3 and q(x) = 2−x2 +4x4 .
The sum of these two polynomials is
p(x) + q(x) = 5 − x2 + 2x3 + 4x4 .
The product,
p(x)q(x) = (3 + 2x3 )(2 − x2 + 4x4 ) = 6 − 3x2 + 4x3 + 12x4 − 2x5 + 8x7 ,
can be calculated either by determining the ci s in the definition or by simply multiplying
polynomials in the same way as we have always done. □
Example 17.2 Let
p(x) = 3 + 3x3 and q(x) = 4 + 4x2 + 4x4
be polynomials in Z12 [x]. The sum of p(x) and q(x) is 7 + 4x2 + 3x3 + 4x4 . The product of
the two polynomials is the zero polynomial. This example tells us that we can not expect
CHAPTER 17. POLYNOMIALS 211
R[x] to be an integral domain if R is not an integral domain. □
Theorem 17.3 Let R be a commutative ring with identity. Then R[x] is a commutative
ring with identity.
Proof. Our first task is to show that R[x] is an abelian group under polynomial addi-
tion.
Pn Thei zero polynomial, f (x) = 0, is the additive identity. P Given a polynomial Pnp(x) =i
n
a
i=0 i x , the inverse of p(x) is easily verified to be −p(x) = i=0 (−ai )x i = −
i=0 ai x .
Commutativity and associativity follow immediately from the definition of polynomial ad-
dition and from the fact that addition in R is both commutative and associative.
To show that polynomial multiplication is associative, let
X
m
p(x) = ai xi ,
i=0
X
n
q(x) = bi xi ,
i=0
X
p
r(x) = ci xi .
i=0
Then
" ! !# !
X
m X
n X
p
[p(x)q(x)]r(x) = ai xi bi xi ci xi
i=0 i=0 i=0
!
X X
m+n i X
p
= aj bi−j x
i
ci x i
i=0 j=0 i=0
!
X X
m+n+p i X
j
= ak bj−k ci−j xi
i=0 j=0 k=0
X
m+n+p X
= aj bk cl xi
i=0 j+k+l=i
!
X X
m+n+p i X
i−j
= aj bk ci−j−k xi
i=0 j=0 k=0
! n+p i
X
m X X
= ai xi bj ci−j xi
i=0 i=0 j=0
!" ! !#
X
m X
n X
p
i i i
= ai x bi x ci x
i=0 i=0 i=0
= p(x)[q(x)r(x)]
The commutativity and distribution properties of polynomial multiplication are proved in
a similar manner. We shall leave the proofs of these properties as an exercise. ■
Proposition 17.4 Let p(x) and q(x) be polynomials in R[x], where R is an integral domain.
Then deg p(x) + deg q(x) = deg(p(x)q(x)). Furthermore, R[x] is an integral domain.
Proof. Suppose that we have two nonzero polynomials
p(x) = am xm + · · · + a1 x + a0
CHAPTER 17. POLYNOMIALS 212
and
q(x) = bn xn + · · · + b1 x + b0
with am 6= 0 and bn 6= 0. The degrees of p(x) and q(x) are m and n, respectively. The
leading term of p(x)q(x) is am bn xm+n , which cannot be zero since R is an integral domain;
hence, the degree of p(x)q(x) is m + n, and p(x)q(x) 6= 0. Since p(x) 6= 0 and q(x) 6= 0
imply that p(x)q(x) 6= 0, we know that R[x] must also be an integral domain. ■
We also want to consider polynomials in two or more variables, such as x − 3xy + 2y 3 .
2
Let R be a ring and suppose that we are given two indeterminates x and y. Certainly
we can form the ring (R[x])[y]. It is straightforward but perhaps tedious to show that
(R[x])[y] ∼
= R([y])[x]. We shall identify these two rings by this isomorphism and simply
write R[x, y]. The ring R[x, y] is called the ring of polynomials in two indeterminates x and
y with coefficients in R. We can define the ring of polynomials in n indeterminates with
coefficients in R similarly. We shall denote this ring by R[x1 , x2 , . . . , xn ].
Theorem 17.5 Let R be a commutative ring with identity and α ∈ R. Then we have a
ring homomorphism ϕα : R[x] → R defined by
ϕα (p(x)) = p(α) = an αn + · · · + a1 α + a0 ,
where p(x) = an xn + · · · + a1 x + a0 .
P P
Proof. Let p(x) = ni=0 ai xi and q(x) = m i
i=0 bi x . It is easy to show that ϕα (p(x) +
q(x)) = ϕα (p(x)) + ϕα (q(x)). To show that multiplication is preserved under the map ϕα ,
observe that
ϕα (p(x))ϕα (q(x)) = p(α)q(α)
! !
X
n X
m
= ai α i bi αi
i=0 i=0
!
X
m+n X
i
= ak bi−k αi
i=0 k=0
= ϕα (p(x)q(x)).
■
The map ϕα : R[x] → R is called the evaluation homomorphism at α.
17.2 The Division Algorithm
Recall that the division algorithm for integers (Theorem 2.9) says that if a and b are integers
with b > 0, then there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.
The algorithm by which q and r are found is just long division. A similar theorem exists for
polynomials. The division algorithm for polynomials has several important consequences.
Since its proof is very similar to the corresponding proof for integers, it is worthwhile to
review Theorem 2.9 at this point.
Theorem 17.6 Division Algorithm. Let f (x) and g(x) be polynomials in F [x], where F
is a field and g(x) is a nonzero polynomial. Then there exist unique polynomials q(x), r(x) ∈
F [x] such that
f (x) = g(x)q(x) + r(x),
where either deg r(x) < deg g(x) or r(x) is the zero polynomial.
CHAPTER 17. POLYNOMIALS 213
Proof. We will first consider the existence of q(x) and r(x). If f (x) is the zero polynomial,
then
0 = 0 · g(x) + 0;
hence, both q and r must also be the zero polynomial. Now suppose that f (x) is not the zero
polynomial and that deg f (x) = n and deg g(x) = m. If m > n, then we can let q(x) = 0
and r(x) = f (x). Hence, we may assume that m ≤ n and proceed by induction on n. If
f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0
g(x) = bm xm + bm−1 xm−1 + · · · + b1 x + b0
the polynomial
an n−m
f ′ (x) = f (x) − x g(x)
bm
has degree less than n or is the zero polynomial. By induction, there exist polynomials q ′ (x)
and r(x) such that
f ′ (x) = q ′ (x)g(x) + r(x),
where r(x) = 0 or the degree of r(x) is less than the degree of g(x). Now let
an n−m
q(x) = q ′ (x) + x .
bm
Then
f (x) = g(x)q(x) + r(x),
with r(x) the zero polynomial or deg r(x) < deg g(x).
To show that q(x) and r(x) are unique, suppose that there exist two other polynomials
q1 (x) and r1 (x) such that f (x) = g(x)q1 (x) + r1 (x) with deg r1 (x) < deg g(x) or r1 (x) = 0,
so that
f (x) = g(x)q(x) + r(x) = g(x)q1 (x) + r1 (x),
and
g(x)[q(x) − q1 (x)] = r1 (x) − r(x).
If q(x) − q1 (x) is not the zero polynomial, then
deg(g(x)[q(x) − q1 (x)]) = deg(r1 (x) − r(x)) ≥ deg g(x).
However, the degrees of both r(x) and r1 (x) are strictly less than the degree of g(x); there-
fore, r(x) = r1 (x) and q(x) = q1 (x). ■
Example 17.7 The division algorithm merely formalizes long division of polynomials, a
task we have been familiar with since high school. For example, suppose that we divide
x3 − x2 + 2x − 3 by x − 2.
x2 + x + 4
x − 2 x 3 − x 2 + 2x − 3
x3 − 2x2
x2 + 2x − 3
x2 − 2x
4x − 3
4x − 8
5
Hence, x3 − x2 + 2x − 3 = (x − 2)(x2 + x + 4) + 5. □
CHAPTER 17. POLYNOMIALS 214
Let p(x) be a polynomial in F [x] and α ∈ F . We say that α is a zero or root of p(x) if
p(x) is in the kernel of the evaluation homomorphism ϕα . All we are really saying here is
that α is a zero of p(x) if p(α) = 0.
Corollary 17.8 Let F be a field. An element α ∈ F is a zero of p(x) ∈ F [x] if and only if
x − α is a factor of p(x) in F [x].
Proof. Suppose that α ∈ F and p(α) = 0. By the division algorithm, there exist polyno-
mials q(x) and r(x) such that
p(x) = (x − α)q(x) + r(x)
and the degree of r(x) must be less than the degree of x − α. Since the degree of r(x) is
less than 1, r(x) = a for a ∈ F ; therefore,
p(x) = (x − α)q(x) + a.
But
0 = p(α) = 0 · q(α) + a = a;
consequently, p(x) = (x − α)q(x), and x − α is a factor of p(x).
Conversely, suppose that x − α is a factor of p(x); say p(x) = (x − α)q(x). Then
p(α) = 0 · q(α) = 0. ■
Corollary 17.9 Let F be a field. A nonzero polynomial p(x) of degree n in F [x] can have
at most n distinct zeros in F .
Proof. We will use induction on the degree of p(x). If deg p(x) = 0, then p(x) is a constant
polynomial and has no zeros. Let deg p(x) = 1. Then p(x) = ax + b for some a and b in F .
If α1 and α2 are zeros of p(x), then aα1 + b = aα2 + b or α1 = α2 .
Now assume that deg p(x) > 1. If p(x) does not have a zero in F , then we are done.
On the other hand, if α is a zero of p(x), then p(x) = (x − α)q(x) for some q(x) ∈ F [x] by
Corollary 17.8. The degree of q(x) is n − 1 by Proposition 17.4. Let β be some other zero
of p(x) that is distinct from α. Then p(β) = (β − α)q(β) = 0. Since α 6= β and F is a field,
q(β) = 0. By our induction hypothesis, q(x) can have at most n − 1 zeros in F that are
distinct from α. Therefore, p(x) has at most n distinct zeros in F . ■
Let F be a field. A monic polynomial d(x) is a greatest common divisor of polynomials
p(x), q(x) ∈ F [x] if d(x) evenly divides both p(x) and q(x); and, if for any other polynomial
d′ (x) dividing both p(x) and q(x), d′ (x) | d(x). We write d(x) = gcd(p(x), q(x)). Two
polynomials p(x) and q(x) are relatively prime if gcd(p(x), q(x)) = 1.
Proposition 17.10 Let F be a field and suppose that d(x) is a greatest common divisor
of two polynomials p(x) and q(x) in F [x]. Then there exist polynomials r(x) and s(x) such
that
d(x) = r(x)p(x) + s(x)q(x).
Furthermore, the greatest common divisor of two polynomials is unique.
Proof. Let d(x) be the monic polynomial of smallest degree in the set
S = {f (x)p(x) + g(x)q(x) : f (x), g(x) ∈ F [x]}.
We can write d(x) = r(x)p(x) + s(x)q(x) for two polynomials r(x) and s(x) in F [x]. We
need to show that d(x) divides both p(x) and q(x). We shall first show that d(x) divides
p(x). By the division algorithm, there exist polynomials a(x) and b(x) such that p(x) =
a(x)d(x) + b(x), where b(x) is either the zero polynomial or deg b(x) < deg d(x). Therefore,
b(x) = p(x) − a(x)d(x)
CHAPTER 17. POLYNOMIALS 215
= p(x) − a(x)(r(x)p(x) + s(x)q(x))
= p(x) − a(x)r(x)p(x) − a(x)s(x)q(x)
= p(x)(1 − a(x)r(x)) + q(x)(−a(x)s(x))
is a linear combination of p(x) and q(x) and therefore must be in S. However, b(x) must
be the zero polynomial since d(x) was chosen to be of smallest degree; consequently, d(x)
divides p(x). A symmetric argument shows that d(x) must also divide q(x); hence, d(x) is
a common divisor of p(x) and q(x).
To show that d(x) is a greatest common divisor of p(x) and q(x), suppose that d′ (x)
is another common divisor of p(x) and q(x). We will show that d′ (x) | d(x). Since d′ (x)
is a common divisor of p(x) and q(x), there exist polynomials u(x) and v(x) such that
p(x) = u(x)d′ (x) and q(x) = v(x)d′ (x). Therefore,
d(x) = r(x)p(x) + s(x)q(x)
= r(x)u(x)d′ (x) + s(x)v(x)d′ (x)
= d′ (x)[r(x)u(x) + s(x)v(x)].
Since d′ (x) | d(x), d(x) is a greatest common divisor of p(x) and q(x).
Finally, we must show that the greatest common divisor of p(x) and q(x) is unique.
Suppose that d′ (x) is another greatest common divisor of p(x) and q(x). We have just
shown that there exist polynomials u(x) and v(x) in F [x] such that d(x) = d′ (x)[r(x)u(x) +
s(x)v(x)]. Since
deg d(x) = deg d′ (x) + deg[r(x)u(x) + s(x)v(x)]
and d(x) and d′ (x) are both greatest common divisors, deg d(x) = deg d′ (x). Since d(x) and
d′ (x) are both monic polynomials of the same degree, it must be the case that d(x) = d′ (x).
■
Notice the similarity between the proof of Proposition 17.10 and the proof of Theo-
rem 2.10.
17.3 Irreducible Polynomials
A nonconstant polynomial f (x) ∈ F [x] is irreducible over a field F if f (x) cannot be
expressed as a product of two polynomials g(x) and h(x) in F [x], where the degrees of g(x)
and h(x) are both smaller than the degree of f (x). Irreducible polynomials function as the
“prime numbers” of polynomial rings.
Example 17.11 The polynomial x2 − 2 ∈ Q[x] is irreducible since it cannot be factored
any further over the rational numbers. Similarly, x2 + 1 is irreducible over the real numbers.
□
Example 17.12 The polynomial p(x) = x3 + x2 + 2 is irreducible over Z3 [x]. Suppose that
this polynomial was reducible over Z3 [x]. By the division algorithm there would have to be
a factor of the form x − a, where a is some element in Z3 [x]. Hence, it would have to be
true that p(a) = 0. However,
p(0) = 2
p(1) = 1
p(2) = 2.
Therefore, p(x) has no zeros in Z3 and must be irreducible. □
CHAPTER 17. POLYNOMIALS 216
Lemma 17.13 Let p(x) ∈ Q[x]. Then
r
p(x) = (a0 + a1 x + · · · + an xn ),
s
where r, s, a0 , . . . , an are integers, the ai ’s are relatively prime, and r and s are relatively
prime.
Proof. Suppose that
b0 b1 bn
p(x) = + x + · · · + xn ,
c0 c1 cn
where the bi ’s and the ci ’s are integers. We can rewrite p(x) as
1
p(x) = (d0 + d1 x + · · · + dn xn ),
c0 · · · cn
where d0 , . . . , dn are integers. Let d be the greatest common divisor of d0 , . . . , dn . Then
d
p(x) = (a0 + a1 x + · · · + an xn ),
c0 · · · cn
where di = dai and the ai ’s are relatively prime. Reducing d/(c0 · · · cn ) to its lowest terms,
we can write
r
p(x) = (a0 + a1 x + · · · + an xn ),
s
where gcd(r, s) = 1. ■
Theorem 17.14 Gauss’s Lemma. Let p(x) ∈ Z[x] be a monic polynomial such that p(x)
factors into a product of two polynomials α(x) and β(x) in Q[x], where the degrees of both
α(x) and β(x) are less than the degree of p(x). Then p(x) = a(x)b(x), where a(x) and b(x)
are monic polynomials in Z[x] with deg α(x) = deg a(x) and deg β(x) = deg b(x).
Proof. By Lemma 17.13, we can assume that
c1 c1
α(x) = (a0 + a1 x + · · · + am xm ) = α1 (x)
d1 d1
c2 c2
β(x) = (b0 + b1 x + · · · + bn x ) = β1 (x),
n
d2 d2
where the ai ’s are relatively prime and the bi ’s are relatively prime. Consequently,
c1 c2 c
p(x) = α(x)β(x) = α1 (x)β1 (x) = α1 (x)β1 (x),
d1 d2 d
where c/d is the product of c1 /d1 and c2 /d2 expressed in lowest terms. Hence, dp(x) =
cα1 (x)β1 (x).
If d = 1, then cam bn = 1 since p(x) is a monic polynomial. Hence, either c = 1
or c = −1. If c = 1, then either am = bn = 1 or am = bn = −1. In the first case
p(x) = α1 (x)β1 (x), where α1 (x) and β1 (x) are monic polynomials with deg α(x) = deg α1 (x)
and deg β(x) = deg β1 (x). In the second case a(x) = −α1 (x) and b(x) = −β1 (x) are the
correct monic polynomials since p(x) = (−α1 (x))(−β1 (x)) = a(x)b(x). The case in which
c = −1 can be handled similarly.
Now suppose that d 6= 1. Since gcd(c, d) = 1, there exists a prime p such that p | d and
p ∤ c. Also, since the coefficients of α1 (x) are relatively prime, there exists a coefficient ai
such that p ∤ ai . Similarly, there exists a coefficient bj of β1 (x) such that p ∤ bj . Let α1′ (x)
and β1′ (x) be the polynomials in Zp [x] obtained by reducing the coefficients of α1 (x) and
β1 (x) modulo p. Since p | d, α1′ (x)β1′ (x) = 0 in Zp [x]. However, this is impossible since
neither α1′ (x) nor β1′ (x) is the zero polynomial and Zp [x] is an integral domain. Therefore,
d = 1 and the theorem is proven. ■
CHAPTER 17. POLYNOMIALS 217
Corollary 17.15 Let p(x) = xn + an−1 xn−1 + · · · + a0 be a polynomial with coefficients in
Z and a0 6= 0. If p(x) has a zero in Q, then p(x) also has a zero α in Z. Furthermore, α
divides a0 .
Proof. Let p(x) have a zero a ∈ Q. Then p(x) must have a linear factor x − a. By Gauss’s
Lemma, p(x) has a factorization with a linear factor in Z[x]. Hence, for some α ∈ Z
p(x) = (x − α)(xn−1 + · · · − a0 /α).
Thus a0 /α ∈ Z and so α | a0 . ■
Example 17.16 Let p(x) = x4 −2x3 +x+1. We shall show that p(x) is irreducible over Q[x].
Assume that p(x) is reducible. Then either p(x) has a linear factor, say p(x) = (x − α)q(x),
where q(x) is a polynomial of degree three, or p(x) has two quadratic factors.
If p(x) has a linear factor in Q[x], then it has a zero in Z. By Corollary 17.15, any zero
must divide 1 and therefore must be ±1; however, p(1) = 1 and p(−1) = 3. Consequently,
we have eliminated the possibility that p(x) has any linear factors.
Therefore, if p(x) is reducible it must factor into two quadratic polynomials, say
p(x) = (x2 + ax + b)(x2 + cx + d)
= x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd,
where each factor is in Z[x] by Gauss’s Lemma. Hence,
a + c = −2
ac + b + d = 0
ad + bc = 1
bd = 1.
Since bd = 1, either b = d = 1 or b = d = −1. In either case b = d and so
ad + bc = b(a + c) = 1.
Since a + c = −2, we know that −2b = 1. This is impossible since b is an integer. Therefore,
p(x) must be irreducible over Q. □
Theorem 17.17 Eisenstein’s Criterion. Let p be a prime and suppose that
f (x) = an xn + · · · + a0 ∈ Z[x].
If p | ai for i = 0, 1, . . . , n − 1, but p ∤ an and p2 ∤ a0 , then f (x) is irreducible over Q.
Proof. By Gauss’s Lemma, we need only show that f (x) does not factor into polynomials
of lower degree in Z[x]. Let
f (x) = (br xr + · · · + b0 )(cs xs + · · · + c0 )
be a factorization in Z[x], with br and cs not equal to zero and r, s < n. Since p2 does not
divide a0 = b0 c0 , either b0 or c0 is not divisible by p. Suppose that p ∤ b0 and p | c0 . Since
p ∤ an and an = br cs , neither br nor cs is divisible by p. Let m be the smallest value of k
such that p ∤ ck . Then
am = b0 cm + b1 cm−1 + · · · + bm c0
is not divisible by p, since each term on the right-hand side of the equation is divisible by p
except for b0 cm . Therefore, m = n since ai is divisible by p for m < n. Hence, f (x) cannot
be factored into polynomials of lower degree and therefore must be irreducible. ■
CHAPTER 17. POLYNOMIALS 218
Example 17.18 The polynomial
f (x) = 16x5 − 9x4 + 3x2 + 6x − 21
is easily seen to be irreducible over Q by Eisenstein’s Criterion if we let p = 3. □
Eisenstein’s Criterion is more useful in constructing irreducible polynomials of a certain
degree over Q than in determining the irreducibility of an arbitrary polynomial in Q[x]:
given an arbitrary polynomial, it is not very likely that we can apply Eisenstein’s Crite-
rion. The real value of Theorem 17.17 is that we now have an easy method of generating
irreducible polynomials of any degree.
Ideals in F [x]
Let F be a field. Recall that a principal ideal in F [x] is an ideal hp(x)i generated by some
polynomial p(x); that is,
hp(x)i = {p(x)q(x) : q(x) ∈ F [x]}.
Example 17.19 The polynomial x2 in F [x] generates the ideal hx2 i consisting of all poly-
nomials with no constant term or term of degree 1. □
Theorem 17.20 If F is a field, then every ideal in F [x] is a principal ideal.
Proof. Let I be an ideal of F [x]. If I is the zero ideal, the theorem is easily true. Suppose
that I is a nontrivial ideal in F [x], and let p(x) ∈ I be a nonzero element of minimal degree.
If deg p(x) = 0, then p(x) is a nonzero constant and 1 must be in I. Since 1 generates all of
F [x], h1i = I = F [x] and I is again a principal ideal.
Now assume that deg p(x) ≥ 1 and let f (x) be any element in I. By the division
algorithm there exist q(x) and r(x) in F [x] such that f (x) = p(x)q(x) + r(x) and deg r(x) <
deg p(x). Since f (x), p(x) ∈ I and I is an ideal, r(x) = f (x)−p(x)q(x) is also in I. However,
since we chose p(x) to be of minimal degree, r(x) must be the zero polynomial. Since we
can write any element f (x) in I as p(x)q(x) for some q(x) ∈ F [x], it must be the case that
I = hp(x)i. ■
Example 17.21 It is not the case that every ideal in the ring F [x, y] is a principal ideal.
Consider the ideal of F [x, y] generated by the polynomials x and y. This is the ideal of
F [x, y] consisting of all polynomials with no constant term. Since both x and y are in the
ideal, no single polynomial can generate the entire ideal. □
Theorem 17.22 Let F be a field and suppose that p(x) ∈ F [x]. Then the ideal generated
by p(x) is maximal if and only if p(x) is irreducible.
Proof. Suppose that p(x) generates a maximal ideal of F [x]. Then hp(x)i is also a prime
ideal of F [x]. Since a maximal ideal must be properly contained inside F [x], p(x) cannot
be a constant polynomial. Let us assume that p(x) factors into two polynomials of lesser
degree, say p(x) = f (x)g(x). Since hp(x)i is a prime ideal one of these factors, say f (x), is
in hp(x)i and therefore be a multiple of p(x). But this would imply that hp(x)i ⊂ hf (x)i,
which is impossible since hp(x)i is maximal.
Conversely, suppose that p(x) is irreducible over F [x]. Let I be an ideal in F [x] contain-
ing hp(x)i. By Theorem 17.20, I is a principal ideal; hence, I = hf (x)i for some f (x) ∈ F [x].
Since p(x) ∈ I, it must be the case that p(x) = f (x)g(x) for some g(x) ∈ F [x]. However,
p(x) is irreducible; hence, either f (x) or g(x) is a constant polynomial. If f (x) is constant,
then I = F [x] and we are done. If g(x) is constant, then f (x) is a constant multiple of I
and I = hp(x)i. Thus, there are no proper ideals of F [x] that properly contain hp(x)i. ■
CHAPTER 17. POLYNOMIALS 219
Sage. Polynomial rings are very important for computational approaches to algebra, and
so Sage makes it very easy to compute with polynomials, over rings, or over fields. And it
is trivial to check if a polynomial is irreducible.
Historical Note
Throughout history, the solution of polynomial equations has been a challenging problem.
The Babylonians knew how to solve the equation ax2 + bx + c = 0. Omar Khayyam (1048–
1131) devised methods of solving cubic equations through the use of geometric constructions
and conic sections. The algebraic solution of the general cubic equation ax3 +bx2 +cx+d = 0
was not discovered until the sixteenth century. An Italian mathematician, Luca Pacioli (ca.
1445–1509), wrote in Summa de Arithmetica that the solution of the cubic was impossible.
This was taken as a challenge by the rest of the mathematical community.
Scipione del Ferro (1465–1526), of the University of Bologna, solved the “depressed cubic,”
ax3 + cx + d = 0.
He kept his solution an absolute secret. This may seem surprising today, when mathe-
maticians are usually very eager to publish their results, but in the days of the Italian
Renaissance secrecy was customary. Academic appointments were not easy to secure and
depended on the ability to prevail in public contests. Such challenges could be issued at
any time. Consequently, any major new discovery was a valuable weapon in such a contest.
If an opponent presented a list of problems to be solved, del Ferro could in turn present a
list of depressed cubics. He kept the secret of his discovery throughout his life, passing it
on only on his deathbed to his student Antonio Fior (ca. 1506–?).
Although Fior was not the equal of his teacher, he immediately issued a challenge to Niccolo
Fontana (1499–1557). Fontana was known as Tartaglia (the Stammerer). As a youth he
had suffered a blow from the sword of a French soldier during an attack on his village. He
survived the savage wound, but his speech was permanently impaired. Tartaglia sent Fior
a list of 30 various mathematical problems; Fior countered by sending Tartaglia a list of
30 depressed cubics. Tartaglia would either solve all 30 of the problems or absolutely fail.
After much effort Tartaglia finally succeeded in solving the depressed cubic and defeated
Fior, who faded into obscurity.
At this point another mathematician, Gerolamo Cardano (1501–1576), entered the story.
Cardano wrote to Tartaglia, begging him for the solution to the depressed cubic. Tartaglia
refused several of his requests, then finally revealed the solution to Cardano after the latter
swore an oath not to publish the secret or to pass it on to anyone else. Using the knowledge
that he had obtained from Tartaglia, Cardano eventually solved the general cubic
ax3 + bx2 + cx + d = 0.
Cardano shared the secret with his student, Ludovico Ferrari (1522–1565), who solved the
general quartic equation,
ax4 + bx3 + cx2 + dx + e = 0.
In 1543, Cardano and Ferrari examined del Ferro’s papers and discovered that he had also
solved the depressed cubic. Cardano felt that this relieved him of his obligation to Tartaglia,
so he proceeded to publish the solutions in Ars Magna (1545), in which he gave credit to
del Ferro for solving the special case of the cubic. This resulted in a bitter dispute between
Cardano and Tartaglia, who published the story of the oath a year later.
CHAPTER 17. POLYNOMIALS 220
17.4 Reading Questions
1. Suppose p(x) is a polynomial of degree n with coefficients from any field. How many
roots can p(x) have? How does this generalize your high school algebra experience?
2. What is the definition of an irreducible polynomial?
3. Find the remainder upon division of 8 x5 − 18 x4 + 20 x3 − 25 x2 + 20 by 4 x2 − x − 2.
4. A single theorem in this chapter connects many of the ideas of this chapter to many
of the ideas of the previous chapter. State a paraphrased version of this theorem.
5. Early in this chapter, we say, “We can prove many results for polynomial rings that
are similar to the theorems we proved for the integers.” Write a short essay (or a very
long paragraph) justifying this assertion.
17.5 Exercises
1. List all of the polynomials of degree 3 or less in Z2 [x].
2. Compute each of the following.
(a) (5x2 + 3x − 4) + (4x2 − x + 9) in Z12 [x]
(b) (5x2 + 3x − 4)(4x2 − x + 9) in Z12 [x]
(c) (7x3 + 3x2 − x) + (6x2 − 8x + 4) in Z9 [x]
(d) (3x2 + 2x − 4) + (4x2 + 2) in Z5 [x]
(e) (3x2 + 2x − 4)(4x2 + 2) in Z5 [x]
(f) (5x2 + 3x − 2)2 in Z12 [x]
3. Use the division algorithm to find q(x) and r(x) such that a(x) = q(x)b(x) + r(x) with
deg r(x) < deg b(x) for each of the following pairs of polynomials.
(a) a(x) = 5x3 + 6x2 − 3x + 4 and b(x) = x − 2 in Z7 [x]
(b) a(x) = 6x4 − 2x3 + x2 − 3x + 1 and b(x) = x2 + x − 2 in Z7 [x]
(c) a(x) = 4x5 − x3 + x2 + 4 and b(x) = x3 − 2 in Z5 [x]
(d) a(x) = x5 + x3 − x2 − x and b(x) = x3 + x in Z2 [x]
4. Find the greatest common divisor of each of the following pairs p(x) and q(x) of
polynomials. If d(x) = gcd(p(x), q(x)), find two polynomials a(x) and b(x) such that
a(x)p(x) + b(x)q(x) = d(x).
(a) p(x) = x3 −6x2 +14x−15 and q(x) = x3 −8x2 +21x−18, where p(x), q(x) ∈ Q[x]
(b) p(x) = x3 + x2 − x + 1 and q(x) = x3 + x − 1, where p(x), q(x) ∈ Z2 [x]
(c) p(x) = x3 + x2 − 4x + 4 and q(x) = x3 + 3x − 2, where p(x), q(x) ∈ Z5 [x]
(d) p(x) = x3 − 2x + 4 and q(x) = 4x3 + x + 3, where p(x), q(x) ∈ Q[x]
5. Find all of the zeros for each of the following polynomials.
(a) 5x3 + 4x2 − x + 9 in Z12 [x] (c) 5x4 + 2x2 − 3 in Z7 [x]
(b) 3x3 − 4x2 − x + 4 in Z5 [x] (d) x3 + x + 1 in Z2 [x]
CHAPTER 17. POLYNOMIALS 221
6. Find all of the units in Z[x].
7. Find a unit p(x) in Z4 [x] such that deg p(x) > 1.
8. Which of the following polynomials are irreducible over Q[x]?
(a) x4 − 2x3 + 2x2 + x + 4 (c) 3x5 − 4x3 − 6x2 + 6
(b) x4 − 5x3 + 3x − 2 (d) 5x5 − 6x4 − 3x2 + 9x − 15
9. Find all of the irreducible polynomials of degrees 2 and 3 in Z2 [x].
10. Give two different factorizations of x2 + x + 8 in Z10 [x].
11. Prove or disprove: There exists a polynomial p(x) in Z6 [x] of degree n with more than
n distinct zeros.
12. If F is a field, show that F [x1 , . . . , xn ] is an integral domain.
13. Show that the division algorithm does not hold for Z[x]. Why does it fail?
14. Prove or disprove: xp + a is irreducible for any a ∈ Zp , where p is prime.
15. Let f (x) be irreducible in F [x], where F is a field. If f (x) | p(x)q(x), prove that either
f (x) | p(x) or f (x) | q(x).
16. Suppose that R and S are isomorphic rings. Prove that R[x] ∼ = S[x].
17. Let F be a field and a ∈ F . If p(x) ∈ F [x], show that p(a) is the remainder obtained
when p(x) is divided by x − a.
18. The Rational Root Theorem. Let
p(x) = an xn + an−1 xn−1 + · · · + a0 ∈ Z[x],
where an 6= 0. Prove that if p(r/s) = 0, where gcd(r, s) = 1, then r | a0 and s | an .
19. Let Q∗ be the multiplicative group of positive rational numbers. Prove that Q∗ is
isomorphic to (Z[x], +).
20. Cyclotomic Polynomials. The polynomial
xn − 1
Φn (x) = = xn−1 + xn−2 + · · · + x + 1
x−1
is called the cyclotomic polynomial. Show that Φp (x) is irreducible over Q for any
prime p.
21. If F is a field, show that there are infinitely many irreducible polynomials in F [x].
22. Let R be a commutative ring with identity. Prove that multiplication is commutative
in R[x].
23. Let R be a commutative ring with identity. Prove that multiplication is distributive
in R[x].
24. Show that xp − x has p distinct zeros in Zp , for any prime p. Conclude that
xp − x = x(x − 1)(x − 2) · · · (x − (p − 1)).
25. Let F be a field and f (x) = a0 + a1 x + · · · + an xn be in F [x]. Define f ′ (x) = a1 +
2a2 x + · · · + nan xn−1 to be the derivative of f (x).
(a) Prove that
(f + g)′ (x) = f ′ (x) + g ′ (x).
Conclude that we can define a homomorphism of abelian groups D : F [x] → F [x]
by D(f (x)) = f ′ (x).
CHAPTER 17. POLYNOMIALS 222
(b) Calculate the kernel of D if char F = 0.
(c) Calculate the kernel of D if char F = p.
(d) Prove that
(f g)′ (x) = f ′ (x)g(x) + f (x)g ′ (x).
(e) Suppose that we can factor a polynomial f (x) ∈ F [x] into linear factors, say
f (x) = a(x − a1 )(x − a2 ) · · · (x − an ).
Prove that f (x) has no repeated factors if and only if f (x) and f ′ (x) are relatively
prime.
26. Let F be a field. Show that F [x] is never a field.
27. Let R be an integral domain. Prove that R[x1 , . . . , xn ] is an integral domain.
28. Let R be a commutative ring with identity. Show that R[x] has a subring R′ isomorphic
to R.
29. Let p(x) and q(x) be polynomials in R[x], where R is a commutative ring with identity.
Prove that deg(p(x) + q(x)) ≤ max(deg p(x), deg q(x)).
17.6 Additional Exercises: Solving the Cubic and Quartic Equa-
tions
1. Complete the square to solve the general quadratic equation
ax2 + bx + c = 0
to obtain √
−b ± b2 − 4ac
x= .
2a
The discriminant of the quadratic equation ∆ = b2 − 4ac determines the nature of the
solutions of the equation. If ∆ > 0, the equation has two distinct real solutions. If
∆ = 0, the equation has a single repeated real root. If ∆ < 0, there are two distinct
imaginary solutions.
2. Show that any cubic equation of the form
x3 + bx2 + cx + d = 0
can be reduced to the form y 3 + py + q = 0 by making the substitution x = y − b/3.
3. Prove that the cube roots of 1 are given by
√
−1 + i 3
ω=
2 √
−1 − i 3
ω2 =
2
3
ω = 1.
4. Make the substitution
p
y=z−
3z
for y in the equation y 3 + py + q = 0 and obtain two solutions A and B for z 3 .
CHAPTER 17. POLYNOMIALS 223
√
5. Show that the product of the solutions obtained in (4) is −p3 /27, deducing that 3
AB =
−p/3.
6. Prove that the possible solutions for z in (4) are given by
√
3
√3
√3
√3
√
3
√
3
A, ω A, ω 2 A, B, ω B, ω2 B
and use this result to show that the three possible solutions for y are
s r s r
3 q p3 q 2 3 q p3 q2
ωi − + + + ω 2i − − + ,
2 27 4 2 27 4
where i = 0, 1, 2.
7. The discriminant of the cubic equation is
p3 q2
∆= + .
27 4
Show that y 3 + py + q = 0
(a) has three real roots, at least two of which are equal, if ∆ = 0.
(b) has one real root and two conjugate imaginary roots if ∆ > 0.
(c) has three distinct real roots if ∆ < 0.
8. Solve the following cubic equations.
(a) x3 − 4x2 + 11x + 30 = 0
(b) x3 − 3x + 5 = 0
(c) x3 − 3x + 2 = 0
(d) x3 + x + 3 = 0
9. Show that the general quartic equation
x4 + ax3 + bx2 + cx + d = 0
can be reduced to
y 4 + py 2 + qy + r = 0
by using the substitution x = y − a/4.
10. Show that
1 2 1 2
y + z = (z − p)y − qy +
2 2
z −r .
2 4
11. Show that the right-hand side of Exercise 17.6.10 can be put in the form (my + k)2 if
and only if
1 2
q − 4(z − p)
2
z − r = 0.
4
12. From Exercise 17.6.11 obtain the resolvent cubic equation
z 3 − pz 2 − 4rz + (4pr − q 2 ) = 0.
Solving the resolvent cubic equation, put the equation found in Exercise 17.6.10 in the
CHAPTER 17. POLYNOMIALS 224
form 2
2 1
y + z = (my + k)2
2
to obtain the solution of the quartic equation.
13. Use this method to solve the following quartic equations.
(a) x4 − x2 − 3x + 2 = 0
(b) x4 + x3 − 7x2 − x + 6 = 0
(c) x4 − 2x2 + 4x − 3 = 0
(d) x4 − 4x3 + 3x2 − 5x + 2 = 0
18
Integral Domains
One of the most important rings we study is the ring of integers. It was our first example of
an algebraic structure: the first polynomial ring that we examined was Z[x]. We also know
that the integers sit naturally inside the field of rational numbers, Q. The ring of integers
is the model for all integral domains. In this chapter we will examine integral domains in
general, answering questions about the ideal structure of integral domains, polynomial rings
over integral domains, and whether or not an integral domain can be embedded in a field.
18.1 Fields of Fractions
Every field is also an integral domain; however, there are many integral domains that are
not fields. For example, the integers Z form an integral domain but not a field. A question
that naturally arises is how we might associate an integral domain with a field. There is a
natural way to construct the rationals Q from the integers: the rationals can be represented
as formal quotients of two integers. The rational numbers are certainly a field. In fact, it
can be shown that the rationals are the smallest field that contains the integers. Given an
integral domain D, our question now becomes how to construct a smallest field F containing
D. We will do this in the same way as we constructed the rationals from the integers.
An element p/q ∈ Q is the quotient of two integers p and q; however, different pairs of
integers can represent the same rational number. For instance, 1/2 = 2/4 = 3/6. We know
that
a c
=
b d
if and only if ad = bc. A more formal way of considering this problem is to examine fractions
in terms of equivalence relations. We can think of elements in Q as ordered pairs in Z × Z.
A quotient p/q can be written as (p, q). For instance, (3, 7) would represent the fraction
3/7. However, there are problems if we consider all possible pairs in Z × Z. There is no
fraction 5/0 corresponding to the pair (5, 0). Also, the pairs (3, 6) and (2, 4) both represent
the fraction 1/2. The first problem is easily solved if we require the second coordinate to
be nonzero. The second problem is solved by considering two pairs (a, b) and (c, d) to be
equivalent if ad = bc.
If we use the approach of ordered pairs instead of fractions, then we can study integral
domains in general. Let D be any integral domain and let
S = {(a, b) : a, b ∈ D and b 6= 0}.
Define a relation on S by (a, b) ∼ (c, d) if ad = bc.
Lemma 18.1 The relation ∼ between elements of S is an equivalence relation.
225
CHAPTER 18. INTEGRAL DOMAINS 226
Proof. Since D is commutative, ab = ba; hence, ∼ is reflexive on D. Now suppose that
(a, b) ∼ (c, d). Then ad = bc or cb = da. Therefore, (c, d) ∼ (a, b) and the relation is
symmetric. Finally, to show that the relation is transitive, let (a, b) ∼ (c, d) and (c, d) ∼
(e, f ). In this case ad = bc and cf = de. Multiplying both sides of ad = bc by f yields
af d = adf = bcf = bde = bed.
Since D is an integral domain, we can deduce that af = be or (a, b) ∼ (e, f ). ■
We will denote the set of equivalence classes on S by FD . We now need to define
the operations of addition and multiplication on FD . Recall how fractions are added and
multiplied in Q:
a c ad + bc
+ = ;
b d bd
a c ac
· = .
b d bd
It seems reasonable to define the operations of addition and multiplication on FD in a
similar manner. If we denote the equivalence class of (a, b) ∈ S by [a, b], then we are led to
define the operations of addition and multiplication on FD by
[a, b] + [c, d] = [ad + bc, bd]
and
[a, b] · [c, d] = [ac, bd],
respectively. The next lemma demonstrates that these operations are independent of the
choice of representatives from each equivalence class.
Lemma 18.2 The operations of addition and multiplication on FD are well-defined.
Proof. We will prove that the operation of addition is well-defined. The proof that
multiplication is well-defined is left as an exercise. Let [a1 , b1 ] = [a2 , b2 ] and [c1 , d1 ] = [c2 , d2 ].
We must show that
[a1 d1 + b1 c1 , b1 d1 ] = [a2 d2 + b2 c2 , b2 d2 ]
or, equivalently, that
(a1 d1 + b1 c1 )(b2 d2 ) = (b1 d1 )(a2 d2 + b2 c2 ).
Since [a1 , b1 ] = [a2 , b2 ] and [c1 , d1 ] = [c2 , d2 ], we know that a1 b2 = b1 a2 and c1 d2 = d1 c2 .
Therefore,
(a1 d1 + b1 c1 )(b2 d2 ) = a1 d1 b2 d2 + b1 c1 b2 d2
= a1 b2 d1 d2 + b1 b2 c1 d2
= b1 a2 d1 d2 + b1 b2 d1 c2
= (b1 d1 )(a2 d2 + b2 c2 ).
■
Lemma 18.3 The set of equivalence classes of S, FD , under the equivalence relation ∼,
together with the operations of addition and multiplication defined by
[a, b] + [c, d] = [ad + bc, bd]
[a, b] · [c, d] = [ac, bd],
is a field.
CHAPTER 18. INTEGRAL DOMAINS 227
Proof. The additive and multiplicative identities are [0, 1] and [1, 1], respectively. To show
that [0, 1] is the additive identity, observe that
[a, b] + [0, 1] = [a1 + b0, b1] = [a, b].
It is easy to show that [1, 1] is the multiplicative identity. Let [a, b] ∈ FD such that a 6= 0.
Then [b, a] is also in FD and [a, b] · [b, a] = [1, 1]; hence, [b, a] is the multiplicative inverse for
[a, b]. Similarly, [−a, b] is the additive inverse of [a, b]. We leave as exercises the verification
of the associative and commutative properties of multiplication in FD . We also leave it to
the reader to show that FD is an abelian group under addition.
It remains to show that the distributive property holds in FD ; however,
[a, b][e, f ] + [c, d][e, f ] = [ae, bf ] + [ce, df ]
= [aedf + bf ce, bdf 2 ]
= [aed + bce, bdf ]
= [ade + bce, bdf ]
= ([a, b] + [c, d])[e, f ]
and the lemma is proved. ■
The field FD in Lemma 18.3 is called the field of fractions or field of quotients of the
integral domain D.
Theorem 18.4 Let D be an integral domain. Then D can be embedded in a field of
fractions FD , where any element in FD can be expressed as the quotient of two elements
in D. Furthermore, the field of fractions FD is unique in the sense that if E is any field
containing D, then there exists a map ψ : FD → E giving an isomorphism with a subfield
of E such that ψ(a) = a for all elements a ∈ D, where we identify a with its image in FD .
Proof. We will first demonstrate that D can be embedded in the field FD . Define a map
ϕ : D → FD by ϕ(a) = [a, 1]. Then for a and b in D,
ϕ(a + b) = [a + b, 1] = [a, 1] + [b, 1] = ϕ(a) + ϕ(b)
and
ϕ(ab) = [ab, 1] = [a, 1][b, 1] = ϕ(a)ϕ(b);
hence, ϕ is a homomorphism. To show that ϕ is one-to-one, suppose that ϕ(a) = ϕ(b).
Then [a, 1] = [b, 1], or a = a1 = 1b = b. Finally, any element of FD can be expressed as the
quotient of two elements in D, since
ϕ(a)[ϕ(b)]−1 = [a, 1][b, 1]−1 = [a, 1] · [1, b] = [a, b].
Now let E be a field containing D and define a map ψ : FD → E by ψ([a, b]) = ab−1 .
To show that ψ is well-defined, let [a1 , b1 ] = [a2 , b2 ]. Then a1 b2 = b1 a2 . Therefore, a1 b−1
1 =
a2 b−1
2 and ψ([a1 , b 1 ]) = ψ([a 2 , b 2 ]).
If [a, b] and [c, d] are in FD , then
ψ([a, b] + [c, d]) = ψ([ad + bc, bd])
= (ad + bc)(bd)−1
= ab−1 + cd−1
= ψ([a, b]) + ψ([c, d])
and
ψ([a, b] · [c, d]) = ψ([ac, bd])
CHAPTER 18. INTEGRAL DOMAINS 228
= (ac)(bd)−1
= ab−1 cd−1
= ψ([a, b])ψ([c, d]).
Therefore, ψ is a homomorphism.
To complete the proof of the theorem, we need to show that ψ is one-to-one. Suppose
that ψ([a, b]) = ab−1 = 0. Then a = 0b = 0 and [a, b] = [0, b]. Therefore, the kernel of ψ is
the zero element [0, b] in FD , and ψ is injective. ■
Example 18.5 Since Q is a field, Q[x] is an integral domain. The field of fractions of Q[x]
is the set of all rational expressions p(x)/q(x), where p(x) and q(x) are polynomials over
the rationals and q(x) is not the zero polynomial. We will denote this field by Q(x). □
We will leave the proofs of the following corollaries of Theorem 18.4 as exercises.
Corollary 18.6 Let F be a field of characteristic zero. Then F contains a subfield isomor-
phic to Q.
Corollary 18.7 Let F be a field of characteristic p. Then F contains a subfield isomorphic
to Zp .
18.2 Factorization in Integral Domains
The building blocks of the integers are the prime numbers. If F is a field, then irreducible
polynomials in F [x] play a role that is very similar to that of the prime numbers in the
ring of integers. Given an arbitrary integral domain, we are led to the following series of
definitions.
Let R be a commutative ring with identity, and let a and b be elements in R. We say
that a divides b, and write a | b, if there exists an element c ∈ R such that b = ac. A unit
in R is an element that has a multiplicative inverse. Two elements a and b in R are said to
be associates if there exists a unit u in R such that a = ub.
Let D be an integral domain. A nonzero element p ∈ D that is not a unit is said to be
irreducible provided that whenever p = ab, either a or b is a unit. Furthermore, p is prime
if whenever p | ab either p | a or p | b.
Example 18.8 It is important to notice that prime and irreducible elements do not always
coincide. Let R be the subring (with identity) of Q[x, y] generated by x2 , y 2 , and xy. Each
of these elements is irreducible in R; however, xy is not prime, since xy divides x2 y 2 but
does not divide either x2 or y 2 . □
The Fundamental Theorem of Arithmetic states that every positive integer n > 1 can
be factored into a product of prime numbers p1 · · · pk , where the pi ’s are not necessarily
distinct. We also know that such factorizations are unique up to the order of the pi ’s. We
can easily extend this result to the integers. The question arises of whether or not such
factorizations are possible in other rings. Generalizing this definition, we say an integral
domain D is a unique factorization domain, or ufd, if D satisfies the following criteria.
1. Let a ∈ D such that a 6= 0 and a is not a unit. Then a can be written as the product
of irreducible elements in D.
2. Let a = p1 · · · pr = q1 · · · qs , where the pi ’s and the qi ’s are irreducible. Then r = s
and there is a π ∈ Sr such that pi and qπ(j) are associates for j = 1, . . . , r.
Example 18.9 The integers are a unique factorization domain by the Fundamental Theo-
rem of Arithmetic. □
CHAPTER 18. INTEGRAL DOMAINS 229
Example
√ 18.10 √
Not every integral domain is a unique factorization domain. The subring
Z[ 3 i] = {a + b 3 i} of the √ complex numbers√is an integral domain (Exercise 16.7.12,
Chapter 16). Let z = a + b 3 i and define ν : Z[ 3 i] → N ∪ {0} by ν(z) = |z| = a2 + 3b2 .
2
It is clear that ν(z) ≥ 0 with equality when z = 0. Also, from our knowledge of complex
numbers we know that ν(zw) = ν(z)ν(w).
√ It is easy to show that if ν(z) = 1, then z is a
unit, and that the only units of Z[ 3 i] are 1 and −1.
We claim that 4 has two distinct factorizations into irreducible elements:
√ √
4 = 2 · 2 = (1 − 3 i)(1 + 3 i).
√
We must show that each of these factors is an irreducible
√ element in Z[ 3 i]. If 2 is not
irreducible, then 2 = zw for elements z, w√in Z[ 3 i] where ν(z) = ν(w) = 2. However,
there does not exist an element in z in Z[ 3 i] such that ν(z) = 2 because the equation
a2 + 3b2 = 2 has no integer
√ solutions.
√ Therefore, 2 must be irreducible. A similar argument
shows√ that both√1 − 3 i and 1 + 3 i are irreducible. Since 2 is not a unit multiple of either
1 − 3 i or 1 + 3 i, 4 has at least two distinct factorizations into irreducible elements. □
Principal Ideal Domains
Let R be a commutative ring with identity. Recall that a principal ideal generated by
a ∈ R is an ideal of the form hai = {ra : r ∈ R}. An integral domain in which every ideal
is principal is called a principal ideal domain, or pid.
Lemma 18.11 Let D be an integral domain and let a, b ∈ D. Then
1. a | b if and only if hbi ⊂ hai.
2. a and b are associates if and only if hbi = hai.
3. a is a unit in D if and only if hai = D.
Proof. (1) Suppose that a | b. Then b = ax for some x ∈ D. Hence, for every r in
D, br = (ax)r = a(xr) and hbi ⊂ hai. Conversely, suppose that hbi ⊂ hai. Then b ∈ hai.
Consequently, b = ax for some x ∈ D. Thus, a | b.
(2) Since a and b are associates, there exists a unit u such that a = ub. Therefore, b | a
and hai ⊂ hbi. Similarly, hbi ⊂ hai. It follows that hai = hbi. Conversely, suppose that
hai = hbi. By part (1), a | b and b | a. Then a = bx and b = ay for some x, y ∈ D. Therefore,
a = bx = ayx. Since D is an integral domain, xy = 1; that is, x and y are units and a and
b are associates.
(3) An element a ∈ D is a unit if and only if a is an associate of 1. However, a is an
associate of 1 if and only if hai = h1i = D. ■
Theorem 18.12 Let D be a pid and hpi be a nonzero ideal in D. Then hpi is a maximal
ideal if and only if p is irreducible.
Proof. Suppose that hpi is a maximal ideal. If some element a in D divides p, then
hpi ⊂ hai. Since hpi is maximal, either D = hai or hpi = hai. Consequently, either a and p
are associates or a is a unit. Therefore, p is irreducible.
Conversely, let p be irreducible. If hai is an ideal in D such that hpi ⊂ hai ⊂ D, then
a | p. Since p is irreducible, either a must be a unit or a and p are associates. Therefore,
either D = hai or hpi = hai. Thus, hpi is a maximal ideal. ■
Corollary 18.13 Let D be a pid. If p is irreducible, then p is prime.
Proof. Let p be irreducible and suppose that p | ab. Then habi ⊂ hpi. By Corollary 16.40,
since hpi is a maximal ideal, hpi must also be a prime ideal. Thus, either a ∈ hpi or b ∈ hpi.
Hence, either p | a or p | b. ■
CHAPTER 18. INTEGRAL DOMAINS 230
Lemma 18.14 Let D be a pid. Let I1 , I2 , . . . be a set of ideals such that I1 ⊂ I2 ⊂ · · ·.
N such that In = IN for all n ≥ N .
Then there exists an integer S
Proof. We claim that I = ∞ i=1 Ii is an ideal of D. Certainly I is not empty, since I1 ⊂ I
and 0 ∈ I. If a, b ∈ I, then a ∈ Ii and b ∈ Ij for some i and j in N. Without loss of
generality we can assume that i ≤ j. Hence, a and b are both in Ij and so a − b is also in Ij .
Now let r ∈ D and a ∈ I. Again, we note that a ∈ Ii for some positive integer i. Since Ii is
an ideal, ra ∈ Ii and hence must be in I. Therefore, we have shown that I is an ideal in D.
Since D is a principal ideal domain, there exists an element a ∈ D that generates I.
Since a is in IN for some N ∈ N, we know that IN = I = hai. Consequently, In = IN for
n ≥ N. ■
Any commutative ring satisfying the condition in Lemma 18.14 is said to satisfy the
ascending chain condition, or ACC. Such rings are called Noetherian rings, after Emmy
Noether.
Theorem 18.15 Every pid is a ufd.
Proof. Existence of a factorization. Let D be a pid and a be a nonzero element in D that
is not a unit. If a is irreducible, then we are done. If not, then there exists a factorization
a = a1 b1 , where neither a1 nor b1 is a unit. Hence, hai ⊂ ha1 i. By Lemma 18.11, we know
that hai 6= ha1 i; otherwise, a and a1 would be associates and b1 would be a unit, which
would contradict our assumption. Now suppose that a1 = a2 b2 , where neither a2 nor b2 is a
unit. By the same argument as before, ha1 i ⊂ ha2 i. We can continue with this construction
to obtain an ascending chain of ideals
hai ⊂ ha1 i ⊂ ha2 i ⊂ · · · .
By Lemma 18.14, there exists a positive integer N such that han i = haN i for all n ≥ N .
Consequently, aN must be irreducible. We have now shown that a is the product of two
elements, one of which must be irreducible.
Now suppose that a = c1 p1 , where p1 is irreducible. If c1 is not a unit, we can repeat
the preceding argument to conclude that hai ⊂ hc1 i. Either c1 is irreducible or c1 = c2 p2 ,
where p2 is irreducible and c2 is not a unit. Continuing in this manner, we obtain another
chain of ideals
hai ⊂ hc1 i ⊂ hc2 i ⊂ · · · .
This chain must satisfy the ascending chain condition; therefore,
a = p1 p2 · · · pr
for irreducible elements p1 , . . . , pr .
Uniqueness of the factorization. To show uniqueness, let
a = p1 p2 · · · pr = q1 q2 · · · qs ,
where each pi and each qi is irreducible. Without loss of generality, we can assume that
r < s. Since p1 divides q1 q2 · · · qs , by Corollary 18.13 it must divide some qi . By rearranging
the qi ’s, we can assume that p1 | q1 ; hence, q1 = u1 p1 for some unit u1 in D. Therefore,
a = p1 p2 · · · pr = u1 p1 q2 · · · qs
or
p2 · · · pr = u1 q2 · · · qs .
Continuing in this manner, we can arrange the qi ’s such that p2 = q2 , p3 = q3 , . . . , pr = qr ,
to obtain
u1 u2 · · · ur qr+1 · · · qs = 1.
In this case qr+1 · · · qs is a unit, which contradicts the fact that qr+1 , . . . , qs are irreducibles.
Therefore, r = s and the factorization of a is unique. ■
CHAPTER 18. INTEGRAL DOMAINS 231
Corollary 18.16 Let F be a field. Then F [x] is a ufd.
Example 18.17 Every pid is a ufd, but it is not the case that every ufd is a pid. In
Corollary 18.31, we will prove that Z[x] is a ufd. However, Z[x] is not a pid. Let I =
{5f (x) + xg(x) : f (x), g(x) ∈ Z[x]}. We can easily show that I is an ideal of Z[x]. Suppose
that I = hp(x)i. Since 5 ∈ I, 5 = f (x)p(x). In this case p(x) = p must be a constant. Since
x ∈ I, x = pg(x); consequently, p = ±1. However, it follows from this fact that hp(x)i = Z[x].
But this would mean that 3 is in I. Therefore, we can write 3 = 5f (x) + xg(x) for some f (x)
and g(x) in Z[x]. Examining the constant term of this polynomial, we see that 3 = 5f (x),
which is impossible. □
Euclidean Domains
We have repeatedly used the division algorithm when proving results about either Z or F [x],
where F is a field. We should now ask when a division algorithm is available for an integral
domain.
Let D be an integral domain such that there is a function ν : D \ {0} → N satisfying
the following conditions.
1. If a and b are nonzero elements in D, then ν(a) ≤ ν(ab).
2. Let a, b ∈ D and suppose that b 6= 0. Then there exist elements q, r ∈ D such that
a = bq + r and either r = 0 or ν(r) < ν(b).
Then D is called a Euclidean domain and ν is called a Euclidean valuation.
Example 18.18 Absolute value on Z is a Euclidean valuation. □
Example 18.19 Let F be a field. Then the degree of a polynomial in F [x] is a Euclidean
valuation. □
Example 18.20 Recall that the Gaussian integers in Example 16.12 of Chapter 16 are
defined by
Z[i] = {a + bi : a, b ∈ Z}.
√ usually measure√the size of a complex number a + bi by its absolute value,. |a + bi| =
We
a2 + b2 ; however, a2 + b2 may not be an integer. For our valuation we will let ν(a+bi) =
a + b2 to ensure that we have an integer.
2
We claim that ν(a + bi) = a2 + b2 is a Euclidean valuation on Z[i]. Let z, w ∈ Z[i].
Then ν(zw) = |zw|2 = |z|2 |w|2 = ν(z)ν(w). Since ν(z) ≥ 1 for every nonzero z ∈ Z[i],
ν(z) ≤ ν(z)ν(w).
Next, we must show that for any z = a + bi and w = c + di in Z[i] with w 6= 0, there
exist elements q and r in Z[i] such that z = qw + r with either r = 0 or ν(r) < ν(w). We
can view z and w as elements in Q(i) = {p + qi : p, q ∈ Q}, the field of fractions of Z[i].
Observe that
c − di
zw−1 = (a + bi)
c2 + d2
ac + bd bc − ad
= 2 + 2 i
c + d2
c +d2
n1 n2
= m1 + 2 + m2 + 2 i
c + d2 c + d2
n1 n2
= (m1 + m2 i) + 2 2
+ 2 i
c +d c + d2
= (m1 + m2 i) + (s + ti)
CHAPTER 18. INTEGRAL DOMAINS 232
in Q(i). In the last steps we are writing the real and imaginary parts as an integer plus a
proper fraction. That is, we take the closest integer mi such that the fractional part satisfies
|ni /(a2 + b2 )| ≤ 1/2. For example, we write
9 1
=1+
8 8
15 1
=2− .
8 8
Thus, s and t are the “fractional parts” of zw−1 = (m1 + m2 i) + (s + ti). We also know that
s2 + t2 ≤ 1/4 + 1/4 = 1/2. Multiplying by w, we have
z = zw−1 w = w(m1 + m2 i) + w(s + ti) = qw + r,
where q = m1 + m2 i and r = w(s + ti). Since z and qw are in Z[i], r must be in Z[i]. Finally,
we need to show that either r = 0 or ν(r) < ν(w). However,
1
ν(r) = ν(w)ν(s + ti) ≤ ν(w) < ν(w).
2
□
Theorem 18.21 Every Euclidean domain is a principal ideal domain.
Proof. Let D be a Euclidean domain and let ν be a Euclidean valuation on D. Suppose
I is a nontrivial ideal in D and choose a nonzero element b ∈ I such that ν(b) is minimal
for all a ∈ I. Since D is a Euclidean domain, there exist elements q and r in D such that
a = bq + r and either r = 0 or ν(r) < ν(b). But r = a − bq is in I since I is an ideal;
therefore, r = 0 by the minimality of b. It follows that a = bq and I = hbi. ■
Corollary 18.22 Every Euclidean domain is a unique factorization domain.
Factorization in D[x]
One of the most important polynomial rings is Z[x]. One of the first questions that come
to mind about Z[x] is whether or not it is a ufd. We will prove a more general statement
here. Our first task is to obtain a more general version of Gauss’s Lemma (Theorem 17.14).
Let D be a unique factorization domain and suppose that
p(x) = an xn + · · · + a1 x + a0
in D[x]. Then the content of p(x) is the greatest common divisor of a0 , . . . , an . We say that
p(x) is primitive if gcd(a0 , . . . , an ) = 1.
Example 18.23 In Z[x] the polynomial p(x) = 5x4 − 3x3 + x − 4 is a primitive polynomial
since the greatest common divisor of the coefficients is 1; however, the polynomial q(x) =
4x2 − 6x + 8 is not primitive since the content of q(x) is 2. □
Theorem 18.24 Gauss’s Lemma. Let D be a ufd and let f (x) and g(x) be primitive
polynomials in D[x]. Then f (x)g(x) is primitive.
P Pn
Proof. Let f (x) = m i
i=0 ai x and g(x) =
i
i=0 bi x . Suppose that p is a prime dividing
the coefficients of f (x)g(x). Let r be the smallest integer such that p ∤ ar and s be the
smallest integer such that p ∤ bs . The coefficient of xr+s in f (x)g(x) is
cr+s = a0 br+s + a1 br+s−1 + · · · + ar+s−1 b1 + ar+s b0 .
Since p divides a0 , . . . , ar−1 and b0 , . . . , bs−1 , p divides every term of cr+s except for the
term ar bs . However, since p | cr+s , either p divides ar or p divides bs . But this is impossible.
■
CHAPTER 18. INTEGRAL DOMAINS 233
Lemma 18.25 Let D be a ufd, and let p(x) and q(x) be in D[x]. Then the content of
p(x)q(x) is equal to the product of the contents of p(x) and q(x).
Proof. Let p(x) = cp1 (x) and q(x) = dq1 (x), where c and d are the contents of p(x)
and q(x), respectively. Then p1 (x) and q1 (x) are primitive. We can now write p(x)q(x) =
cdp1 (x)q1 (x). Since p1 (x)q1 (x) is primitive, the content of p(x)q(x) must be cd. ■
Lemma 18.26 Let D be a ufd and F its field of fractions. Suppose that p(x) ∈ D[x] and
p(x) = f (x)g(x), where f (x) and g(x) are in F [x]. Then p(x) = f1 (x)g1 (x), where f1 (x)
and g1 (x) are in D[x]. Furthermore, deg f (x) = deg f1 (x) and deg g(x) = deg g1 (x).
Proof. Let a and b be nonzero elements of D such that af (x), bg(x) are in D[x]. We
can find a1 , b1 ∈ D such that af (x) = a1 f1 (x) and bg(x) = b1 g1 (x), where f1 (x) and g1 (x)
are primitive polynomials in D[x]. Therefore, abp(x) = (a1 f1 (x))(b1 g1 (x)). Since f1 (x)
and g1 (x) are primitive polynomials, it must be the case that ab | a1 b1 by Gauss’s Lemma.
Thus there exists a c ∈ D such that p(x) = cf1 (x)g1 (x). Clearly, deg f (x) = deg f1 (x) and
deg g(x) = deg g1 (x). ■
The following corollaries are direct consequences of Lemma 18.26.
Corollary 18.27 Let D be a ufd and F its field of fractions. A primitive polynomial p(x)
in D[x] is irreducible in F [x] if and only if it is irreducible in D[x].
Corollary 18.28 Let D be a ufd and F its field of fractions. If p(x) is a monic polynomial
in D[x] with p(x) = f (x)g(x) in F [x], then p(x) = f1 (x)g1 (x), where f1 (x) and g1 (x) are in
D[x]. Furthermore, deg f (x) = deg f1 (x) and deg g(x) = deg g1 (x).
Theorem 18.29 If D is a ufd, then D[x] is a ufd.
Proof. Let p(x) be a nonzero polynomial in D[x]. If p(x) is a constant polynomial,
then it must have a unique factorization since D is a ufd. Now suppose that p(x) is
a polynomial of positive degree in D[x]. Let F be the field of fractions of D, and let
p(x) = f1 (x)f2 (x) · · · fn (x) by a factorization of p(x), where each fi (x) is irreducible. Choose
ai ∈ D such that ai fi (x) is in D[x]. There exist b1 , . . . , bn ∈ D such that ai fi (x) = bi gi (x),
where gi (x) is a primitive polynomial in D[x]. By Corollary 18.27, each gi (x) is irreducible
in D[x]. Consequently, we can write
a1 · · · an p(x) = b1 · · · bn g1 (x) · · · gn (x).
Let b = b1 · · · bn . Since g1 (x) · · · gn (x) is primitive, a1 · · · an divides b. Therefore, p(x) =
ag1 (x) · · · gn (x), where a ∈ D. Since D is a ufd, we can factor a as uc1 · · · ck , where u is a
unit and each of the ci ’s is irreducible in D.
We will now show the uniqueness of this factorization. Let
p(x) = a1 · · · am f1 (x) · · · fn (x) = b1 · · · br g1 (x) · · · gs (x)
be two factorizations of p(x), where all of the factors are irreducible in D[x]. By Corol-
lary 18.27, each of the fi ’s and gi ’s is irreducible in F [x]. The ai ’s and the bi ’s are units
in F . Since F [x] is a pid, it is a ufd; therefore, n = s. Now rearrange the gi (x)’s so that
fi (x) and gi (x) are associates for i = 1, . . . , n. Then there exist c1 , . . . , cn and d1 , . . . , dn in
D such that (ci /di )fi (x) = gi (x) or ci fi (x) = di gi (x). The polynomials fi (x) and gi (x) are
primitive; hence, ci and di are associates in D. Thus, a1 · · · am = ub1 · · · br in D, where u is
a unit in D. Since D is a unique factorization domain, m = s. Finally, we can reorder the
bi ’s so that ai and bi are associates for each i. This completes the uniqueness part of the
proof. ■
The theorem that we have just proven has several obvious but important corollaries.
CHAPTER 18. INTEGRAL DOMAINS 234
Corollary 18.30 Let F be a field. Then F [x] is a ufd.
Corollary 18.31 The ring of polynomials over the integers, Z[x], is a ufd.
Corollary 18.32 Let D be a ufd. Then D[x1 , . . . , xn ] is a ufd.
Remark 18.33 It is important to notice that every Euclidean domain is a pid and every
pid is a ufd. However, as demonstrated by our examples, the converse of each of these
statements fails. There are principal ideal domains that are not Euclidean domains, and
there are unique factorization domains that are not principal ideal domains (Z[x]).
Sage. Sage supports distinctions between “plain” rings, domains, principal ideal domains
and fields. Support is often very good for constructions and computations with PID’s,
but sometimes problems get significantly harder (computationally) when a ring has less
structure that that of a PID. So be aware when using Sage that some questions may go
unanswered for rings with less structure.
Historical Note
Karl Friedrich Gauss, born in Brunswick, Germany on April 30, 1777, is considered to be
one of the greatest mathematicians who ever lived. Gauss was truly a child prodigy. At
the age of three he was able to detect errors in the books of his father’s business. Gauss
entered college at the age of 15. Before the age of 20, Gauss was able to construct a regular
17-sided polygon with a ruler and compass. This was the first new construction of a regular
n-sided polygon since the time of the ancient Greeks. Gauss succeeded in showing that if
n
N = 22 + 1 was prime, then it was possible to construct a regular N -sided polygon.
Gauss obtained his Ph.D. in 1799 under the direction of Pfaff at the University of Helmstedt.
In his dissertation he gave the first complete proof of the Fundamental Theorem of Algebra,
which states that every polynomial with real coefficients can be factored into linear factors
over the complex numbers. The acceptance of complex numbers √ was brought about by
Gauss, who was the first person to use the notation of i for −1.
Gauss then turned his attention toward number theory; in 1801, he published his famous
book on number theory, Disquisitiones Arithmeticae. Throughout his life Gauss was in-
trigued with this branch of mathematics. He once wrote, “Mathematics is the queen of the
sciences, and the theory of numbers is the queen of mathematics.”
In 1807, Gauss was appointed director of the Observatory at the University of Göttingen, a
position he held until his death. This position required him to study applications of math-
ematics to the sciences. He succeeded in making contributions to fields such as astronomy,
mechanics, optics, geodesy, and magnetism. Along with Wilhelm Weber, he coinvented the
first practical electric telegraph some years before a better version was invented by Samuel
F. B. Morse.
Gauss was clearly the most prominent mathematician in the world in the early nineteenth
century. His status naturally made his discoveries subject to intense scrutiny. Gauss’s cold
and distant personality many times led him to ignore the work of his contemporaries, making
him many enemies. He did not enjoy teaching very much, and young mathematicians who
sought him out for encouragement were often rebuffed. Nevertheless, he had many outstand-
ing students, including Eisenstein, Riemann, Kummer, Dirichlet, and Dedekind. Gauss also
offered a great deal of encouragement to Sophie Germain (1776–1831), who overcame the
many obstacles facing women in her day to become a very prominent mathematician. Gauss
died at the age of 78 in Göttingen on February 23, 1855.
CHAPTER 18. INTEGRAL DOMAINS 235
18.3 Reading Questions
1. Integral domains are an abstraction of which two fundamental rings that we have
already studied?
2. What are the various types of integral domains defined in this section?
3. The field of fractions of a ring abstracts what idea from basic mathematics?
4. In the previous chapter we had a theorem about irreducible polynomials generating
maximal ideals. Which theorem in this chapter generalizes this previous result?
5. Describe an example which is a ufd, but not a pid.
18.4 Exercises
√ √
1. Let z = a + b 3 i be in √Z[ 3 i]. If a2 + 3b2 = 1, show that z must be a unit. Show
that the only units of Z[ 3 i] are 1 and −1.
2. The Gaussian integers, Z[i], are a ufd. Factor each of the following elements in Z[i]
into a product of irreducibles.
(a) 5 (c) 6 + 8i
(b) 1 + 3i (d) 2
3. Let D be an integral domain.
(a) Prove that FD is an abelian group under the operation of addition.
(b) Show that the operation of multiplication is well-defined in the field of fractions,
FD .
(c) Verify the associative and commutative properties for multiplication in FD .
4. Prove or disprove: Any subring of a field F containing 1 is an integral domain.
5. Prove or disprove: If D is an integral domain, then every prime element in D is also
irreducible in D.
6. Let F be a field of characteristic zero. Prove that F contains a subfield isomorphic to
Q.
7. Let F be a field.
(a) Prove that the field of fractions of F [x], denoted by F (x), is isomorphic to the
set all rational expressions p(x)/q(x), where q(x) is not the zero polynomial.
(b) Let p(x1 , . . . , xn ) and q(x1 , . . . , xn ) be polynomials in F [x1 , . . . , xn ]. Show that
the set of all rational expressions p(x1 , . . . , xn )/q(x1 , . . . , xn ) is isomorphic to the
field of fractions of F [x1 , . . . , xn ]. We denote the field of fractions of F [x1 , . . . , xn ]
by F (x1 , . . . , xn ).
8. Let p be prime and denote the field of fractions of Zp [x] by Zp (x). Prove that Zp (x) is
an infinite field of characteristic p.
9. Prove that the field of fractions of the Gaussian integers, Z[i], is
Q(i) = {p + qi : p, q ∈ Q}.
10. A field F is called a prime field if it has no proper subfields. If E is a subfield of F
and E is a prime field, then E is a prime subfield of F .
(a) Prove that every field contains a unique prime subfield.
CHAPTER 18. INTEGRAL DOMAINS 236
(b) If F is a field of characteristic 0, prove that the prime subfield of F is isomorphic
to the field of rational numbers, Q.
(c) If F is a field of characteristic p, prove that the prime subfield of F is isomorphic
to Zp .
√ √
11. Let Z[ 2 ] = {a + b 2 : a, b ∈ Z}.
√
(a) Prove that Z[ 2 ] is an integral domain.
√
(b) Find all of the units in Z[ 2 ].
√
(c) Determine the field of fractions of Z[ 2 ].
√
(d) Prove
√ that Z[ 2i] is a Euclidean domain under the Euclidean valuation ν(a +
b 2 i) = a2 + 2b2 .
12. Let D be a ufd. An element d ∈ D is a greatest common divisor of a and b in D if
d | a and d | b and d is divisible by any other element dividing both a and b.
(a) If D is a pid and a and b are both nonzero elements of D, prove there exists a
unique greatest common divisor of a and b up to associates. That is, if d and d′
are both greatest common divisors of a and b, then d and d′ are associates. We
write gcd(a, b) for the greatest common divisor of a and b.
(b) Let D be a pid and a and b be nonzero elements of D. Prove that there exist
elements s and t in D such that gcd(a, b) = as + bt.
13. Let D be an integral domain. Define a relation on D by a ∼ b if a and b are associates
in D. Prove that ∼ is an equivalence relation on D.
14. Let D be a Euclidean domain with Euclidean valuation ν. If u is a unit in D, show
that ν(u) = ν(1).
15. Let D be a Euclidean domain with Euclidean valuation ν. If a and b are associates in
D, prove that ν(a) = ν(b).
√
16. Show that Z[ 5 i] is not a unique factorization domain.
17. Prove or disprove: Every subdomain of a ufd is also a ufd.
18. An ideal of a commutative ring R is said to be finitely generated if there exist elements
a1 , . . . , an in R such that every element r in the ideal can be written as a1 r1 +· · ·+an rn
for some r1 , . . . , rn in R. Prove that R satisfies the ascending chain condition if and
only if every ideal of R is finitely generated.
19. Let D be an integral domain with a descending chain of ideals I1 ⊃ I2 ⊃ I3 ⊃ · · ·.
Suppose that there exists an N such that Ik = IN for all k ≥ N . A ring satisfying this
condition is said to satisfy the descending chain condition, or DCC. Rings satisfying
the DCC are called Artinian rings, after Emil Artin. Show that if D satisfies the
descending chain condition, it must satisfy the ascending chain condition.
20. Let R be a commutative ring with identity. We define a multiplicative subset of R to
be a subset S such that 1 ∈ S and ab ∈ S if a, b ∈ S.
(a) Define a relation ∼ on R × S by (a, s) ∼ (a′ , s′ ) if there exists an s∗ ∈ S such
that s∗ (s′ a − sa′ ) = 0. Show that ∼ is an equivalence relation on R × S.
(b) Let a/s denote the equivalence class of (a, s) ∈ R × S and let S −1 R be the set of
all equivalence classes with respect to ∼. Define the operations of addition and
CHAPTER 18. INTEGRAL DOMAINS 237
multiplication on S −1 R by
a b at + bs
+ =
s t st
ab ab
= ,
st st
respectively. Prove that these operations are well-defined on S −1 R and that
S −1 R is a ring with identity under these operations. The ring S −1 R is called the
ring of quotients of R with respect to S.
(c) Show that the map ψ : R → S −1 R defined by ψ(a) = a/1 is a ring homomor-
phism.
(d) If R has no zero divisors and 0 ∈
/ S, show that ψ is one-to-one.
(e) Prove that P is a prime ideal of R if and only if S = R \ P is a multiplicative
subset of R.
(f) If P is a prime ideal of R and S = R \ P , show that the ring of quotients S −1 R
has a unique maximal ideal. Any ring that has a unique maximal ideal is called
a local ring.
18.5 References and Suggested Readings
[1] Atiyah, M. F. and MacDonald, I. G. Introduction to Commutative Algebra. Westview
Press, Boulder, CO, 1994.
[2] Zariski, O. and Samuel, P. Commutative Algebra, vols. I and II. Springer, New York,
1975, 1960.
19
Lattices and Boolean Algebras
The axioms of a ring give structure to the operations of addition and multiplication on a
set. However, we can construct algebraic structures, known as lattices and Boolean algebras,
that generalize other types of operations. For example, the important operations on sets
are inclusion, union, and intersection. Lattices are generalizations of order relations on
algebraic spaces, such as set inclusion in set theory and inequality in the familiar number
systems N, Z, Q, and R. Boolean algebras generalize the operations of intersection and
union. Lattices and Boolean algebras have found applications in logic, circuit theory, and
probability.
19.1 Lattices
Partially Ordered Sets
We begin the study of lattices and Boolean algebras by generalizing the idea of inequality.
Recall that a relation on a set X is a subset of X × X. A relation P on X is called a partial
order of X if it satisfies the following axioms.
1. The relation is reflexive: (a, a) ∈ P for all a ∈ X.
2. The relation is antisymmetric: if (a, b) ∈ P and (b, a) ∈ P , then a = b.
3. The relation is transitive: if (a, b) ∈ P and (b, c) ∈ P , then (a, c) ∈ P .
We will usually write a b to mean (a, b) ∈ P unless some symbol is naturally associated
with a particular partial order, such as a ≤ b with integers a and b, or A ⊂ B with sets A
and B. A set X together with a partial order is called a partially ordered set, or poset.
Example 19.1 The set of integers (or rationals or reals) is a poset where a ≤ b has the
usual meaning for two integers a and b in Z. □
Example 19.2 Let X be any set. We will define the power set of X to be the set of all
subsets of X. We denote the power set of X by P(X). For example, let X = {a, b, c}. Then
P(X) is the set of all subsets of the set {a, b, c}:
∅ {a} {b} {c}
{a, b} {a, c} {b, c} {a, b, c}.
On any power set of a set X, set inclusion, ⊂, is a partial order. We can represent the order
on {a, b, c} schematically by a diagram such as the one in Figure 19.3. □
238
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 239
{a, b, c}
{a, b} {a, c} {b, c}
{a} {b} {c}
∅
Figure 19.3 Partial order on P({a, b, c})
Example 19.4 Let G be a group. The set of subgroups of G is a poset, where the partial
order is set inclusion. □
Example 19.5 There can be more than one partial order on a particular set. We can form
a partial order on N by a b if a | b. The relation is certainly reflexive since a | a for all
a ∈ N. If m | n and n | m, then m = n; hence, the relation is also antisymmetric. The
relation is transitive, because if m | n and n | p, then m | p. □
Example 19.6 Let X = {1, 2, 3, 4, 6, 8, 12, 24} be the set of divisors of 24 with the partial
order defined in Example 19.5. Figure 19.7 shows the partial order on X. □
24
8 12
4 6
2 3
1
Figure 19.7 A partial order on the divisors of 24
Let Y be a subset of a poset X. An element u in X is an upper bound of Y if a u for
every element a ∈ Y . If u is an upper bound of Y such that u v for every other upper
bound v of Y , then u is called a least upper bound or supremum of Y . An element l in
X is said to be a lower bound of Y if l a for all a ∈ Y . If l is a lower bound of Y such
that k l for every other lower bound k of Y , then l is called a greatest lower bound or
infimum of Y .
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 240
Example 19.8 Let Y = {2, 3, 4, 6} be contained in the set X of Example 19.6. Then Y
has upper bounds 12 and 24, with 12 as a least upper bound. The only lower bound is 1;
hence, it must be a greatest lower bound. □
As it turns out, least upper bounds and greatest lower bounds are unique if they exist.
Theorem 19.9 Let Y be a nonempty subset of a poset X. If Y has a least upper bound,
then Y has a unique least upper bound. If Y has a greatest lower bound, then Y has a unique
greatest lower bound.
Proof. Let u1 and u2 be least upper bounds for Y . By the definition of the least upper
bound, u1 u for all upper bounds u of Y . In particular, u1 u2 . Similarly, u2 u1 .
Therefore, u1 = u2 by antisymmetry. A similar argument show that the greatest lower
bound is unique. ■
On many posets it is possible to define binary operations by using the greatest lower
bound and the least upper bound of two elements. A lattice is a poset L such that every
pair of elements in L has a least upper bound and a greatest lower bound. The least upper
bound of a, b ∈ L is called the join of a and b and is denoted by a ∨ b. The greatest lower
bound of a, b ∈ L is called the meet of a and b and is denoted by a ∧ b.
Example 19.10 Let X be a set. Then the power set of X, P(X), is a lattice. For two sets
A and B in P(X), the least upper bound of A and B is A ∪ B. Certainly A ∪ B is an upper
bound of A and B, since A ⊂ A ∪ B and B ⊂ A ∪ B. If C is some other set containing both
A and B, then C must contain A ∪ B; hence, A ∪ B is the least upper bound of A and B.
Similarly, the greatest lower bound of A and B is A ∩ B. □
Example 19.11 Let G be a group and suppose that X is the set of subgroups of G. Then
X is a poset ordered by set-theoretic inclusion, ⊂. The set of subgroups of G is also a lattice.
If H and K are subgroups of G, the greatest lower bound of H and K is H ∩ K. The set
H ∪ K may not be a subgroup of G. We leave it as an exercise to show that the least upper
bound of H and K is the subgroup generated by H ∪ K. □
In set theory we have certain duality conditions. For example, by De Morgan’s laws,
any statement about sets that is true about (A ∪ B)′ must also be true about A′ ∩ B ′ . We
also have a duality principle for lattices.
Axiom 19.12 Principle of Duality. Any statement that is true for all lattices remains
true when is replaced by and ∨ and ∧ are interchanged throughout the statement.
The following theorem tells us that a lattice is an algebraic structure with two binary
operations that satisfy certain axioms.
Theorem 19.13 If L is a lattice, then the binary operations ∨ and ∧ satisfy the following
properties for a, b, c ∈ L.
1. Commutative laws: a ∨ b = b ∨ a and a ∧ b = b ∧ a.
2. Associative laws: a ∨ (b ∨ c) = (a ∨ b) ∨ c and a ∧ (b ∧ c) = (a ∧ b) ∧ c.
3. Idempotent laws: a ∨ a = a and a ∧ a = a.
4. Absorption laws: a ∨ (a ∧ b) = a and a ∧ (a ∨ b) = a.
Proof. By the Principle of Duality, we need only prove the first statement in each part.
(1) By definition a ∨ b is the least upper bound of {a, b}, and b ∨ a is the least upper
bound of {b, a}; however, {a, b} = {b, a}.
(2) We will show that a ∨ (b ∨ c) and (a ∨ b) ∨ c are both least upper bounds of {a, b, c}.
Let d = a ∨ b. Then c d ∨ c = (a ∨ b) ∨ c. We also know that
a a ∨ b = d d ∨ c = (a ∨ b) ∨ c.
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 241
A similar argument demonstrates that b (a ∨ b) ∨ c. Therefore, (a ∨ b) ∨ c is an upper
bound of {a, b, c}. We now need to show that (a ∨ b) ∨ c is the least upper bound of {a, b, c}.
Let u be some other upper bound of {a, b, c}. Then a u and b u; hence, d = a ∨ b u.
Since c u, it follows that (a ∨ b) ∨ c = d ∨ c u. Therefore, (a ∨ b) ∨ c must be the least
upper bound of {a, b, c}. The argument that shows a ∨ (b ∨ c) is the least upper bound of
{a, b, c} is the same. Consequently, a ∨ (b ∨ c) = (a ∨ b) ∨ c.
(3) The join of a and a is the least upper bound of {a}; hence, a ∨ a = a.
(4) Let d = a ∧ b. Then a a ∨ d. On the other hand, d = a ∧ b a, and so a ∨ d a.
Therefore, a ∨ (a ∧ b) = a. ■
Given any arbitrary set L with operations ∨ and ∧, satisfying the conditions of the
previous theorem, it is natural to ask whether or not this set comes from some lattice. The
following theorem says that this is always the case.
Theorem 19.14 Let L be a nonempty set with two binary operations ∨ and ∧ satisfying the
commutative, associative, idempotent, and absorption laws. We can define a partial order
on L by a b if a ∨ b = b. Furthermore, L is a lattice with respect to if for all a, b ∈ L,
we define the least upper bound and greatest lower bound of a and b by a ∨ b and a ∧ b,
respectively.
Proof. We first show that L is a poset under . Since a ∨ a = a, a a and is reflexive.
To show that is antisymmetric, let a b and b a. Then a ∨ b = b and b ∨ a = a. By
the commutative law, b = a ∨ b = b ∨ a = a. Finally, we must show that is transitive. Let
a b and b c. Then a ∨ b = b and b ∨ c = c. Thus,
a ∨ c = a ∨ (b ∨ c) = (a ∨ b) ∨ c = b ∨ c = c,
or a c.
To show that L is a lattice, we must prove that a ∨ b and a ∧ b are, respectively, the least
upper and greatest lower bounds of a and b. Since a = (a ∨ b) ∧ a = a ∧ (a ∨ b), it follows
that a a ∨ b. Similarly, b a ∨ b. Therefore, a ∨ b is an upper bound for a and b. Let u
be any other upper bound of both a and b. Then a u and b u. But a ∨ b u since
(a ∨ b) ∨ u = a ∨ (b ∨ u) = a ∨ u = u.
The proof that a ∧ b is the greatest lower bound of a and b is left as an exercise. ■
19.2 Boolean Algebras
Let us investigate the example of the power set, P(X), of a set X more closely. The power
set is a lattice that is ordered by inclusion. By the definition of the power set, the largest
element in P(X) is X itself and the smallest element is ∅, the empty set. For any set A in
P(X), we know that A ∩ X = A and A ∪ ∅ = A. This suggests the following definition for
lattices. An element I in a poset X is a largest element if a I for all a ∈ X. An element
O is a smallest element of X if O a for all a ∈ X.
Let A be in P(X). Recall that the complement of A is
A′ = X \ A = {x : x ∈ X and x ∈
/ A}.
We know that A ∪ A′ = X and A ∩ A′ = ∅. We can generalize this example for lattices.
A lattice L with a largest element I and a smallest element O is complemented if for each
a ∈ L, there exists an a′ such that a ∨ a′ = I and a ∧ a′ = O.
In a lattice L, the binary operations ∨ and ∧ satisfy commutative and associative laws;
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 242
however, they need not satisfy the distributive law
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c);
however, in P(X) the distributive law is satisfied since
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
for A, B, C ∈ P(X). We will say that a lattice L is distributive if the following distributive
law holds:
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
for all a, b, c ∈ L.
Theorem 19.15 A lattice L is distributive if and only if
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c)
for all a, b, c ∈ L.
Proof. Let us assume that L is a distributive lattice.
a ∨ (b ∧ c) = [a ∨ (a ∧ c)] ∨ (b ∧ c)
= a ∨ [(a ∧ c) ∨ (b ∧ c)]
= a ∨ [(c ∧ a) ∨ (c ∧ b)]
= a ∨ [c ∧ (a ∨ b)]
= a ∨ [(a ∨ b) ∧ c]
= [(a ∨ b) ∧ a] ∨ [(a ∨ b) ∧ c]
= (a ∨ b) ∧ (a ∨ c).
The converse follows directly from the Duality Principle. ■
A Boolean algebra is a lattice B with a greatest element I and a smallest element O
such that B is both distributive and complemented. The power set of X, P(X), is our
prototype for a Boolean algebra. As it turns out, it is also one of the most important
Boolean algebras. The following theorem allows us to characterize Boolean algebras in
terms of the binary relations ∨ and ∧ without mention of the fact that a Boolean algebra
is a poset.
Theorem 19.16 A set B is a Boolean algebra if and only if there exist binary operations
∨ and ∧ on B satisfying the following axioms.
1. a ∨ b = b ∨ a and a ∧ b = b ∧ a for a, b ∈ B.
2. a ∨ (b ∨ c) = (a ∨ b) ∨ c and a ∧ (b ∧ c) = (a ∧ b) ∧ c for a, b, c ∈ B.
3. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) and a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) for a, b, c ∈ B.
4. There exist elements I and O such that a ∨ O = a and a ∧ I = a for all a ∈ B.
5. For every a ∈ B there exists an a′ ∈ B such that a ∨ a′ = I and a ∧ a′ = O.
Proof. Let B be a set satisfying (1)–(5) in the theorem. One of the idempotent laws is
satisfied since
a=a∨O
= a ∨ (a ∧ a′ )
= (a ∨ a) ∧ (a ∨ a′ )
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 243
= (a ∨ a) ∧ I
= a ∨ a.
Observe that
I ∨ b = (b ∨ b′ ) ∨ b = (b′ ∨ b) ∨ b = b′ ∨ (b ∨ b) = b′ ∨ b = I.
Consequently, the first of the two absorption laws holds, since
a ∨ (a ∧ b) = (a ∧ I) ∨ (a ∧ b)
= a ∧ (I ∨ b)
=a∧I
= a.
The other idempotent and absorption laws are proven similarly. Since B also satisfies (1)–
(3), the conditions of Theorem 19.14 are met; therefore, B must be a lattice. Condition (4)
tells us that B is a distributive lattice.
For a ∈ B, O ∨ a = a; hence, O a and O is the smallest element in B. To show that I
is the largest element in B, we will first show that a ∨ b = b is equivalent to a ∧ b = a. Since
a ∨ I = a for all a ∈ B, using the absorption laws we can determine that
a ∨ I = (a ∧ I) ∨ I = I ∨ (I ∧ a) = I
or a I for all a in B. Finally, since we know that B is complemented by (5), B must be
a Boolean algebra.
Conversely, suppose that B is a Boolean algebra. Let I and O be the greatest and least
elements in B, respectively. If we define a ∨ b and a ∧ b as least upper and greatest lower
bounds of {a, b}, then B is a Boolean algebra by Theorem 19.14, Theorem 19.15, and our
hypothesis. ■
Many other identities hold in Boolean algebras. Some of these identities are listed in
the following theorem.
Theorem 19.17 Let B be a Boolean algebra. Then
1. a ∨ I = I and a ∧ O = O for all a ∈ B.
2. If a ∨ b = a ∨ c and a ∧ b = a ∧ c for a, b, c ∈ B, then b = c.
3. If a ∨ b = I and a ∧ b = O, then b = a′ .
4. (a′ )′ = a for all a ∈ B.
5. I ′ = O and O′ = I.
6. (a ∨ b)′ = a′ ∧ b′ and (a ∧ b)′ = a′ ∨ b′ (De Morgan’s Laws).
Proof. We will prove only (2). The rest of the identities are left as exercises. For
a ∨ b = a ∨ c and a ∧ b = a ∧ c, we have
b = b ∨ (b ∧ a)
= b ∨ (a ∧ b)
= b ∨ (a ∧ c)
= (b ∨ a) ∧ (b ∨ c)
= (a ∨ b) ∧ (b ∨ c)
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 244
= (a ∨ c) ∧ (b ∨ c)
= (c ∨ a) ∧ (c ∨ b)
= c ∨ (a ∧ b)
= c ∨ (a ∧ c)
= c ∨ (c ∧ a)
= c.
■
Finite Boolean Algebras
A Boolean algebra is a finite Boolean algebra if it contains a finite number of elements
as a set. Finite Boolean algebras are particularly nice since we can classify them up to
isomorphism.
Let B and C be Boolean algebras. A bijective map ϕ : B → C is an isomorphism of
Boolean algebras if
ϕ(a ∨ b) = ϕ(a) ∨ ϕ(b)
ϕ(a ∧ b) = ϕ(a) ∧ ϕ(b)
for all a and b in B.
We will show that any finite Boolean algebra is isomorphic to the Boolean algebra
obtained by taking the power set of some finite set X. We will need a few lemmas and
definitions before we prove this result. Let B be a finite Boolean algebra. An element
a ∈ B is an atom of B if a 6= O and a ∧ b = a for all b ∈ B with b 6= O. Equivalently, a is
an atom of B if there is no b ∈ B with b 6= O distinct from a such that O b a.
Lemma 19.18 Let B be a finite Boolean algebra. If b is a element of B with b 6= O, then
there is an atom a in B such that a b.
Proof. If b is an atom, let a = b. Otherwise, choose an element b1 , not equal to O or b,
such that b1 b. We are guaranteed that this is possible since b is not an atom. If b1 is an
atom, then we are done. If not, choose b2 , not equal to O or b1 , such that b2 b1 . Again,
if b2 is an atom, let a = b2 . Continuing this process, we can obtain a chain
O · · · b3 b2 b1 b.
Since B is a finite Boolean algebra, this chain must be finite. That is, for some k, bk is an
atom. Let a = bk . ■
Lemma 19.19 Let a and b be atoms in a finite Boolean algebra B such that a 6= b. Then
a ∧ b = O.
Proof. Since a ∧ b is the greatest lower bound of a and b, we know that a ∧ b a. Hence,
either a ∧ b = a or a ∧ b = O. However, if a ∧ b = a, then either a b or a = O. In either
case we have a contradiction because a and b are both atoms; therefore, a ∧ b = O. ■
Lemma 19.20 Let B be a Boolean algebra and a, b ∈ B. The following statements are
equivalent.
1. a b.
2. a ∧ b′ = O.
3. a′ ∨ b = I.
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 245
Proof. (1) ⇒ (2). If a b, then a ∨ b = b. Therefore,
a ∧ b′ = a ∧ (a ∨ b)′
= a ∧ (a′ ∧ b′ )
= (a ∧ a′ ) ∧ b′
= O ∧ b′
= O.
(2) ⇒ (3). If a ∧ b′ = O, then a′ ∨ b = (a ∧ b′ )′ = O′ = I.
(3) ⇒ (1). If a′ ∨ b = I, then
a = a ∧ (a′ ∨ b)
= (a ∧ a′ ) ∨ (a ∧ b)
= O ∨ (a ∧ b)
= a ∧ b.
Thus, a b. ■
Lemma 19.21 Let B be a Boolean algebra and b and c be elements in B such that b 6 c.
Then there exists an atom a ∈ B such that a b and a 6 c.
Proof. By Lemma 19.20, b ∧ c′ 6= O. Hence, there exists an atom a such that a b ∧ c′ .
Consequently, a b and a 6 c. ■
Lemma 19.22 Let b ∈ B and a1 , . . . , an be the atoms of B such that ai b. Then
b = a1 ∨ · · · ∨ an . Furthermore, if a, a1 , . . . , an are atoms of B such that a b, ai b, and
b = a ∨ a1 ∨ · · · ∨ an , then a = ai for some i = 1, . . . , n.
Proof. Let b1 = a1 ∨ · · · ∨ an . Since ai b for each i, we know that b1 b. If we can show
that b b1 , then the lemma is true by antisymmetry. Assume b 6 b1 . Then there exists
an atom a such that a b and a 6 b1 . Since a is an atom and a b, we can deduce that
a = ai for some ai . However, this is impossible since a b1 . Therefore, b b1 .
Now suppose that b = a1 ∨ · · · ∨ an . If a is an atom less than b,
a = a ∧ b = a ∧ (a1 ∨ · · · ∨ an ) = (a ∧ a1 ) ∨ · · · ∨ (a ∧ an ).
But each term is O or a with a ∧ ai occurring for only one ai . Hence, by Lemma 19.19,
a = ai for some i. ■
Theorem 19.23 Let B be a finite Boolean algebra. Then there exists a set X such that B
is isomorphic to P(X).
Proof. We will show that B is isomorphic to P(X), where X is the set of atoms of B. Let
a ∈ B. By Lemma 19.22, we can write a uniquely as a = a1 ∨ · · · ∨ an for a1 , . . . , an ∈ X.
Consequently, we can define a map ϕ : B → P(X) by
ϕ(a) = ϕ(a1 ∨ · · · ∨ an ) = {a1 , . . . , an }.
Clearly, ϕ is onto.
Now let a = a1 ∨ · · · ∨ an and b = b1 ∨ · · · ∨ bm be elements in B, where each ai and each
bi is an atom. If ϕ(a) = ϕ(b), then {a1 , . . . , an } = {b1 , . . . , bm } and a = b. Consequently, ϕ
is injective.
The join of a and b is preserved by ϕ since
ϕ(a ∨ b) = ϕ(a1 ∨ · · · ∨ an ∨ b1 ∨ · · · ∨ bm )
= {a1 , . . . , an , b1 , . . . , bm }
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 246
= {a1 , . . . , an } ∪ {b1 , . . . , bm }
= ϕ(a1 ∨ · · · ∨ an ) ∪ ϕ(b1 ∧ · · · ∨ bm )
= ϕ(a) ∪ ϕ(b).
Similarly, ϕ(a ∧ b) = ϕ(a) ∩ ϕ(b). ■
We leave the proof of the following corollary as an exercise.
Corollary 19.24 The order of any finite Boolean algebra must be 2n for some positive
integer n.
19.3 The Algebra of Electrical Circuits
The usefulness of Boolean algebras has become increasingly apparent over the past several
decades with the development of the modern computer. The circuit design of computer
chips can be expressed in terms of Boolean algebras. In this section we will develop the
Boolean algebra of electrical circuits and switches; however, these results can easily be
generalized to the design of integrated computer circuitry.
A switch is a device, located at some point in an electrical circuit, that controls the flow
of current through the circuit. Each switch has two possible states: it can be open, and
not allow the passage of current through the circuit, or a it can be closed, and allow the
passage of current. These states are mutually exclusive. We require that every switch be
in one state or the other—a switch cannot be open and closed at the same time. Also, if
one switch is always in the same state as another, we will denote both by the same letter;
that is, two switches that are both labeled with the same letter a will always be open at
the same time and closed at the same time.
Given two switches, we can construct two fundamental types of circuits. Two switches
a and b are in series if they make up a circuit of the type that is illustrated in Figure 19.25.
Current can pass between the terminals A and B in a series circuit only if both of the
switches a and b are closed. We will denote this combination of switches by a ∧ b. Two
switches a and b are in parallel if they form a circuit of the type that appears in Figure 19.26.
In the case of a parallel circuit, current can pass between A and B if either one of the switches
is closed. We denote a parallel combination of circuits a and b by a ∨ b.
A a b B
Figure 19.25 a ∧ b
a
A B
b
Figure 19.26 a ∨ b
We can build more complicated electrical circuits out of series and parallel circuits by
replacing any switch in the circuit with one of these two fundamental types of circuits.
Circuits constructed in this manner are called series-parallel circuits.
We will consider two circuits equivalent if they act the same. That is, if we set the
switches in equivalent circuits exactly the same we will obtain the same result. For example,
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 247
in a series circuit a∧b is exactly the same as b∧a. Notice that this is exactly the commutative
law for Boolean algebras. In fact, the set of all series-parallel circuits forms a Boolean algebra
under the operations of ∨ and ∧. We can use diagrams to verify the different axioms of
a Boolean algebra. The distributive law, a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c), is illustrated in
Figure 19.27. If a is a switch, then a′ is the switch that is always open when a is closed and
always closed when a is open. A circuit that is always closed is I in our algebra; a circuit
that is always open is O. The laws for a ∧ a′ = O and a ∨ a′ = I are shown in Figure 19.28
and Figure 19.29, respectively.
b a b
a
c a c
Figure 19.27 a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
a
a a′
Figure 19.28 a ∧ a′ = O
a′
Figure 19.29 a ∨ a′ = I
Example 19.30 Every Boolean expression represents a switching circuit. For example,
given the expression (a ∨ b) ∧ (a ∨ b′ ) ∧ (a ∨ b), we can construct the circuit in Figure 19.33.
□
Theorem 19.31 The set of all circuits is a Boolean algebra.
We leave as an exercise the proof of this theorem for the Boolean algebra axioms not
yet verified. We can now apply the techniques of Boolean algebras to switching theory.
Example 19.32 Given a complex circuit, we can now apply the techniques of Boolean
algebra to reduce it to a simpler one. Consider the circuit in Figure 19.33. Since
(a ∨ b) ∧ (a ∨ b′ ) ∧ (a ∨ b) = (a ∨ b) ∧ (a ∨ b) ∧ (a ∨ b′ )
= (a ∨ b) ∧ (a ∨ b′ )
= a ∨ (b ∧ b′ )
=a∨O
= a,
we can replace the more complicated circuit with a circuit containing the single switch a
and achieve the same function. □
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 248
a a a
b b′ b
Figure 19.33 (a ∨ b) ∧ (a ∨ b′ ) ∧ (a ∨ b)
Sage. Sage has a full suite of functionality for both posets and lattices, all as part of its ex-
cellent support for combinatorics. There is little in this chapter that cannot be investigated
with Sage.
Historical Note
George Boole (1815–1864) was the first person to study lattices. In 1847, he published The
Investigation of the Laws of Thought, a book in which he used lattices to formalize logic
and the calculus of propositions. Boole believed that mathematics was the study of form
rather than of content; that is, he was not so much concerned with what he was calculating
as with how he was calculating it. Boole’s work was carried on by his friend Augustus De
Morgan (1806–1871). De Morgan observed that the principle of duality often held in set
theory, as is illustrated by De Morgan’s laws for set theory. He believed, as did Boole, that
mathematics was the study of symbols and abstract operations.
Set theory and logic were further advanced by such mathematicians as Alfred North White-
head (1861–1947), Bertrand Russell (1872–1970), and David Hilbert (1862–1943). In Prin-
cipia Mathematica, Whitehead and Russell attempted to show the connection between math-
ematics and logic by the deduction of the natural number system from the rules of formal
logic. If the natural numbers could be determined from logic itself, then so could much
of the rest of existing mathematics. Hilbert attempted to build up mathematics by using
symbolic logic in a way that would prove the consistency of mathematics. His approach
was dealt a mortal blow by Kurt Gödel (1906–1978), who proved that there will always
be “undecidable” problems in any sufficiently rich axiomatic system; that is, that in any
mathematical system of any consequence, there will always be statements that can never
be proven either true or false.
As often occurs, this basic research in pure mathematics later became indispensable in a
wide variety of applications. Boolean algebras and logic have become essential in the design
of the large-scale integrated circuitry found on today’s computer chips. Sociologists have
used lattices and Boolean algebras to model social hierarchies; biologists have used them to
describe biosystems.
19.4 Reading Questions
1. Describe succinctly what a poset is. Do not just list the defining properties, but
give a description that another student of algebra who has never seen a poset might
understand. For example, part of your answer might include what type of common
algebraic topics a poset generalizes, and your answer should be short on symbols.
2. How does a lattice differ from a poset? Answer this in the spirit of the previous
question.
3. How does a Boolean algebra differ from a lattice? Again, answer this in the spirit of
the previous two questions.
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 249
4. Give two (perhaps related) reasons why any discussion of finite Boolean algebras might
center on the example of the power set of a finite set.
5. Describe a major innovation of the middle twentieth century made possible by Boolean
algebra.
19.5 Exercises
1. Draw the lattice diagram for the power set of X = {a, b, c, d} with the set inclusion
relation, ⊂.
2. Draw the diagram for the set of positive integers that are divisors of 30. Is this poset
a Boolean algebra?
3. Draw a diagram of the lattice of subgroups of Z12 .
4. Let B be the set of positive integers that are divisors of 210. Define an order on B
by a b if a | b. Prove that B is a Boolean algebra. Find a set X such that B is
isomorphic to P(X).
5. Prove or disprove: Z is a poset under the relation a b if a | b.
6. Draw the switching circuit for each of the following Boolean expressions.
(a) (a ∨ b ∨ a′ ) ∧ a (c) a ∨ (a ∧ b)
(b) (a ∨ b)′ ∧ (a ∨ b) (d) (c ∨ a ∨ b) ∧ c′ ∧ (a ∨ b)′
7. Draw a circuit that will be closed exactly when only one of three switches a, b, and c
are closed.
8. Prove or disprove that the two circuits shown are equivalent.
a b c a b
a′ b
a c′ a c′
9. Let X be a finite set containing n elements. Prove that |P(X)| = 2n . Conclude that
the order of any finite Boolean algebra must be 2n for some n ∈ N.
10. For each of the following circuits, write a Boolean expression. If the circuit can be
replaced by one with fewer switches, give the Boolean expression and draw a diagram
for the new circuit.
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 250
a b′
a′
b
a a b
a′
b a′ b
a b c
a′ b′ c
a b′ c′
11. Prove or disprove: The set of all nonzero integers is a lattice, where a b is defined
by a | b.
12. Let L be a nonempty set with two binary operations ∨ and ∧ satisfying the commuta-
tive, associative, idempotent, and absorption laws. We can define a partial order on L,
as in Theorem 19.14, by a b if a ∨ b = b. Prove that the greatest lower bound of a
and b is a ∧ b.
13. Let G be a group and X be the set of subgroups of G ordered by set-theoretic inclusion.
If H and K are subgroups of G, show that the least upper bound of H and K is the
subgroup generated by H ∪ K.
14. Let R be a ring and suppose that X is the set of ideals of R. Show that X is a poset
ordered by set-theoretic inclusion, ⊂. Define the meet of two ideals I and J in X by
I ∩ J and the join of I and J by I + J. Prove that the set of ideals of R is a lattice
under these operations.
15. Let B be a Boolean algebra. Prove each of the following identities.
(a) a ∨ I = I and a ∧ O = O for all a ∈ B.
(b) If a ∨ b = I and a ∧ b = O, then b = a′ .
(c) (a′ )′ = a for all a ∈ B.
(d) I ′ = O and O′ = I.
(e) (a ∨ b)′ = a′ ∧ b′ and (a ∧ b)′ = a′ ∨ b′ (De Morgan’s laws).
16. By drawing the appropriate diagrams, complete the proof of Theorem 19.31 to show
that the switching functions form a Boolean algebra.
17. Let B be a Boolean algebra. Define binary operations + and · on B by
a + b = (a ∧ b′ ) ∨ (a′ ∧ b)
a · b = a ∧ b.
Prove that B is a commutative ring under these operations satisfying a2 = a for all
CHAPTER 19. LATTICES AND BOOLEAN ALGEBRAS 251
a ∈ B.
18. Let X be a poset such that for every a and b in X, either a b or b a. Then X is
said to be a totally ordered set.
(a) Is a | b a total order on N?
(b) Prove that N, Z, Q, and R are totally ordered sets under the usual ordering ≤.
19. Let X and Y be posets. A map ϕ : X → Y is order-preserving if a b implies that
ϕ(a) ϕ(b). Let L and M be lattices. A map ψ : L → M is a lattice homomorphism
if ψ(a ∨ b) = ψ(a) ∨ ψ(b) and ψ(a ∧ b) = ψ(a) ∧ ψ(b). Show that every lattice homo-
morphism is order-preserving, but that it is not the case that every order-preserving
homomorphism is a lattice homomorphism.
20. Let B be a Boolean algebra. Prove that a = b if and only if (a ∧ b′ ) ∨ (a′ ∧ b) = O for
a, b ∈ B.
21. Let B be a Boolean algebra. Prove that a = O if and only if (a ∧ b′ ) ∨ (a′ ∧ b) = b for
all b ∈ B.
22. Let L and M be lattices. Define an order relation on L × M by (a, b) (c, d) if a c
and b d. Show that L × M is a lattice under this partial order.
19.6 Programming Exercises
1. A Boolean or switching function on n variables is a map f : {O, I}n → {0, I}. A
Boolean polynomial is a special type of Boolean function: it is any type of Boolean
expression formed from a finite combination of variables x1 , . . . , xn together with O
and I, using the operations ∨, ∧, and ′ . The values of the functions are defined in
Table 19.34. Write a program to evaluate Boolean polynomials.
Table 19.34 Boolean polynomials
x y x′ x∨y x∧y
0 0 1 0 0
0 1 1 1 0
1 0 0 1 0
1 1 0 1 1
19.7 References and Suggested Readings
[1] Donnellan, T. Lattice Theory . Pergamon Press, Oxford, 1968.
[2] Halmos, P. R. “The Basic Concepts of Algebraic Logic,” American Mathematical
Monthly 53 (1956), 363–87.
[3] Hohn, F. “Some Mathematical Aspects of Switching,” American Mathematical Monthly
62 (1955), 75–90.
[4] Hohn, F. Applied Boolean Algebra. 2nd ed. Macmillan, New York, 1966.
[5] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998.
[6] Whitesitt, J. Boolean Algebra and Its Applications. Dover, Mineola, NY, 2010.
20
Vector Spaces
In a physical system a quantity can often be described with a single number. For example,
we need to know only a single number to describe temperature, mass, or volume. However,
for some quantities, such as location, we need several numbers. To give the location of
a point in space, we need x, y, and z coordinates. Temperature distribution over a solid
object requires four numbers: three to identify each point within the object and a fourth
to describe the temperature at that point. Often n-tuples of numbers, or vectors, also have
certain algebraic properties, such as addition or scalar multiplication.
In this chapter we will examine mathematical structures called vector spaces. As with
groups and rings, it is desirable to give a simple list of axioms that must be satisfied to
make a set of vectors a structure worth studying.
20.1 Definitions and Examples
A vector space V over a field F is an abelian group with a scalar product α · v or αv defined
for all α ∈ F and all v ∈ V satisfying the following axioms.
• α(βv) = (αβ)v;
• (α + β)v = αv + βv;
• α(u + v) = αu + αv;
• 1v = v;
where α, β ∈ F and u, v ∈ V .
The elements of V are called vectors; the elements of F are called scalars. It is important
to notice that in most cases two vectors cannot be multiplied. In general, it is only possible
to multiply a vector with a scalar. To differentiate between the scalar zero and the vector
zero, we will write them as 0 and 0, respectively.
Let us examine several examples of vector spaces. Some of them will be quite familiar;
others will seem less so.
Example 20.1 The n-tuples of real numbers, denoted by Rn , form a vector space over R.
Given vectors u = (u1 , . . . , un ) and v = (v1 , . . . , vn ) in Rn and α in R, we can define vector
addition by
u + v = (u1 , . . . , un ) + (v1 , . . . , vn ) = (u1 + v1 , . . . , un + vn )
and scalar multiplication by
αu = α(u1 , . . . , un ) = (αu1 , . . . , αun ).
252
CHAPTER 20. VECTOR SPACES 253
□
Example 20.2 If F is a field, then F [x] is a vector space over F . The vectors in F [x]
are simply polynomials, and vector addition is just polynomial addition. If α ∈ F and
p(x) ∈ F [x], then scalar multiplication is defined by αp(x). □
Example 20.3 The set of all continuous real-valued functions on a closed interval [a, b] is a
vector space over R. If f (x) and g(x) are continuous on [a, b], then (f + g)(x) is defined to
be f (x) + g(x). Scalar multiplication is defined by (αf )(x) = αf (x) for α ∈ R. For example,
if f (x) = sin x and g(x) = x2 , then (2f + 5g)(x) = 2 sin x + 5x2 . □
√ √
Example 20.4 √ Let V = Q( 2√) = {a + b 2 : a, b ∈ Q}. Then V is √ a vector space over Q.
If u = a + b 2 and v = c + d 2, then u + v = (a + c) + (b + d) 2 is again in V . Also,
for α ∈ Q, αv is in V . We will leave it as an exercise to verify that all of the vector space
axioms hold for V . □
Proposition 20.5 Let V be a vector space over F . Then each of the following statements
is true.
1. 0v = 0 for all v ∈ V .
2. α0 = 0 for all α ∈ F .
3. If αv = 0, then either α = 0 or v = 0.
4. (−1)v = −v for all v ∈ V .
5. −(αv) = (−α)v = α(−v) for all α ∈ F and all v ∈ V .
Proof. To prove (1), observe that
0v = (0 + 0)v = 0v + 0v;
consequently, 0 + 0v = 0v + 0v. Since V is an abelian group, 0 = 0v.
The proof of (2) is almost identical to the proof of (1). For (3), we are done if α = 0.
Suppose that α 6= 0. Multiplying both sides of αv = 0 by 1/α, we have v = 0.
To show (4), observe that
v + (−1)v = 1v + (−1)v = (1 − 1)v = 0v = 0,
and so −v = (−1)v. We will leave the proof of (5) as an exercise. ■
20.2 Subspaces
Just as groups have subgroups and rings have subrings, vector spaces also have substructures.
Let V be a vector space over a field F , and W a subset of V . Then W is a subspace of V if
it is closed under vector addition and scalar multiplication; that is, if u, v ∈ W and α ∈ F ,
it will always be the case that u + v and αv are also in W .
Example 20.6 Let W be the subspace of R3 defined by W = {(x1 , 2x1 + x2 , x1 − x2 ) :
x1 , x2 ∈ R}. We claim that W is a subspace of R3 . Since
α(x1 , 2x1 + x2 , x1 − x2 ) = (αx1 , α(2x1 + x2 ), α(x1 − x2 ))
= (αx1 , 2(αx1 ) + αx2 , αx1 − αx2 ),
W is closed under scalar multiplication. To show that W is closed under vector addition,
CHAPTER 20. VECTOR SPACES 254
let u = (x1 , 2x1 + x2 , x1 − x2 ) and v = (y1 , 2y1 + y2 , y1 − y2 ) be vectors in W . Then
u + v = (x1 + y1 , 2(x1 + y1 ) + (x2 + y2 ), (x1 + y1 ) − (x2 + y2 )).
□
Example 20.7 Let W be the subset of polynomials of F [x] with no odd-power terms. If
p(x) and q(x) have no odd-power terms, then neither will p(x) + q(x). Also, αp(x) ∈ W for
α ∈ F and p(x) ∈ W . □
Let V be any vector space over a field F and suppose that v1 , v2 , . . . , vn are vectors in
V and α1 , α2 , . . . , αn are scalars in F . Any vector w in V of the form
X
n
w= αi vi = α1 v1 + α2 v2 + · · · + αn vn
i=1
is called a linear combination of the vectors v1 , v2 , . . . , vn . The spanning set of vectors
v1 , v2 , . . . , vn is the set of vectors obtained from all possible linear combinations of v1 , v2 , . . . , vn .
If W is the spanning set of v1 , v2 , . . . , vn , then we say that W is spanned by v1 , v2 , . . . , vn .
Proposition 20.8 Let S = {v1 , v2 , . . . , vn } be vectors in a vector space V . Then the span
of S is a subspace of V .
Proof. Let u and v be in S. We can write both of these vectors as linear combinations of
the vi ’s:
u = α1 v1 + α2 v2 + · · · + αn vn
v = β1 v1 + β2 v2 + · · · + βn vn .
Then
u + v = (α1 + β1 )v1 + (α2 + β2 )v2 + · · · + (αn + βn )vn
is a linear combination of the vi ’s. For α ∈ F ,
αu = (αα1 )v1 + (αα2 )v2 + · · · + (ααn )vn
is in the span of S. ■
20.3 Linear Independence
Let S = {v1 , v2 , . . . , vn } be a set of vectors in a vector space V . If there exist scalars
α1 , α2 . . . αn ∈ F such that not all of the αi ’s are zero and
α1 v1 + α2 v2 + · · · + αn vn = 0,
then S is said to be linearly dependent. If the set S is not linearly dependent, then it is
said to be linearly independent. More specifically, S is a linearly independent set if
α1 v1 + α2 v2 + · · · + αn vn = 0
implies that
α1 = α2 = · · · = αn = 0
for any set of scalars {α1 , α2 . . . αn }.
Proposition 20.9 Let {v1 , v2 , . . . , vn } be a set of linearly independent vectors in a vector
CHAPTER 20. VECTOR SPACES 255
space. Suppose that
v = α1 v1 + α2 v2 + · · · + αn vn = β1 v1 + β2 v2 + · · · + βn vn .
Then α1 = β1 , α2 = β2 , . . . , αn = βn .
Proof. If
v = α1 v1 + α2 v2 + · · · + αn vn = β1 v1 + β2 v2 + · · · + βn vn ,
then
(α1 − β1 )v1 + (α2 − β2 )v2 + · · · + (αn − βn )vn = 0.
Since v1 , . . . , vn are linearly independent, αi − βi = 0 for i = 1, . . . , n. ■
The definition of linear dependence makes more sense if we consider the following propo-
sition.
Proposition 20.10 A set {v1 , v2 , . . . , vn } of vectors in a vector space V is linearly dependent
if and only if one of the vi ’s is a linear combination of the rest.
Proof. Suppose that {v1 , v2 , . . . , vn } is a set of linearly dependent vectors. Then there
exist scalars α1 , . . . , αn such that
α1 v1 + α2 v2 + · · · + αn vn = 0,
with at least one of the αi ’s not equal to zero. Suppose that αk 6= 0. Then
α1 αk−1 αk+1 αn
vk = − v1 − · · · − vk−1 − vk+1 − · · · − vn .
αk αk αk αk
Conversely, suppose that
vk = β1 v1 + · · · + βk−1 vk−1 + βk+1 vk+1 + · · · + βn vn .
Then
β1 v1 + · · · + βk−1 vk−1 − vk + βk+1 vk+1 + · · · + βn vn = 0.
■
The following proposition is a consequence of the fact that any system of homogeneous
linear equations with more unknowns than equations will have a nontrivial solution. We
leave the details of the proof for the end-of-chapter exercises.
Proposition 20.11 Suppose that a vector space V is spanned by n vectors. If m > n, then
any set of m vectors in V must be linearly dependent.
A set {e1 , e2 , . . . , en } of vectors in a vector space V is called a basis for V if {e1 , e2 , . . . , en }
is a linearly independent set that spans V .
Example 20.12 The vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1) form a basis
for R3 . The set certainly spans R3 , since any arbitrary vector (x1 , x2 , x3 ) in R3 can be
written as x1 e1 + x2 e2 + x3 e3 . Also, none of the vectors e1 , e2 , e3 can be written as a linear
combination of the other two; hence, they are linearly independent. The vectors e1 , e2 , e3
are not the only basis of R3 : the set {(3, 2, 1), (3, 2, 0), (1, 1, 1)} is also a basis for R3 . □
√ √ √ √ √
Example 20.13 Let√Q( 2 ) = {a+b 2 : a, b ∈ Q}. The sets {1, 2 } and {1 + 2, 1− 2 }
are both bases of Q( 2 ). □
From the last two examples it should be clear that a given vector space has several bases.
In fact, there are an infinite number of bases for both of these examples. In general, there
is no unique basis for a vector space. However, every basis of R3 consists of exactly three
CHAPTER 20. VECTOR SPACES 256
√
vectors, and every basis of Q( 2 ) consists of exactly two vectors. This is a consequence of
the next proposition.
Proposition 20.14 Let {e1 , e2 , . . . , em } and {f1 , f2 , . . . , fn } be two bases for a vector space
V . Then m = n.
Proof. Since {e1 , e2 , . . . , em } is a basis, it is a linearly independent set. By Proposi-
tion 20.11, n ≤ m. Similarly, {f1 , f2 , . . . , fn } is a linearly independent set, and the last
proposition implies that m ≤ n. Consequently, m = n. ■
If {e1 , e2 , . . . , en } is a basis for a vector space V , then we say that the dimension of V is
n and we write dim V = n. We will leave the proof of the following theorem as an exercise.
Theorem 20.15 Let V be a vector space of dimension n.
1. If S = {v1 , . . . , vn } is a set of linearly independent vectors for V , then S is a basis for
V.
2. If S = {v1 , . . . , vn } spans V , then S is a basis for V .
3. If S = {v1 , . . . , vk } is a set of linearly independent vectors for V with k < n, then
there exist vectors vk+1 , . . . , vn such that
{v1 , . . . , vk , vk+1 , . . . , vn }
is a basis for V .
Sage. Many of Sage’s computations, in a wide variety of algebraic settings, come from
solving problems in linear algebra. So you will find a wealth of linear algebra functionality.
Further, you can use structures such as finite fields to find vector spaces in new settings.
20.4 Reading Questions
1. Why do the axioms of a vector space appear to only have four conditions, rather than
the ten you may have seen the first time you saw an axiomatic definition?
√ √
2. The set V = Q( 11) = {a + b 11 | a, b ∈ Q} is a vector space. Carefully define the
operations on this set that√will make this possible. Describe the subspace spanned by
S = {u}, where u = 3 + 72 11 ∈ V .
3. Write a long paragraph, or a short essay, on the importance of linear independence in
linear algebra.
4. Write a long paragraph, or a short essay, on the importance of spanning sets in linear
algebra.
5. “Linear algebra is all about linear combinations.” Explain why you might say this.
20.5 Exercises
1. If F is a field, show that F [x] is a vector space over F , where the vectors in F [x]
are polynomials. Vector addition is polynomial addition, and scalar multiplication is
defined by αp(x) for α ∈ F .
√
2. Prove that Q( 2 ) is a vector space.
√ √ √ √ √
3. Let Q( 2, 3 ) be the field generated by elements
√ √ of the form a + b 2 + c 3 + d 6,
where a, b, c, d are in Q. Prove that Q( 2, 3 ) is a vector space of dimension 4 over
CHAPTER 20. VECTOR SPACES 257
√ √
Q. Find a basis for Q( 2, 3 ).
4. Prove that the complex numbers are a vector space of dimension 2 over R.
5. Prove that the set Pn of all polynomials of degree less than n form a subspace of the
vector space F [x]. Find a basis for Pn and compute the dimension of Pn .
6. Let F be a field and denote the set of n-tuples of F by F n . Given vectors u =
(u1 , . . . , un ) and v = (v1 , . . . , vn ) in F n and α in F , define vector addition by
u + v = (u1 , . . . , un ) + (v1 , . . . , vn ) = (u1 + v1 , . . . , un + vn )
and scalar multiplication by
αu = α(u1 , . . . , un ) = (αu1 , . . . , αun ).
Prove that F n is a vector space of dimension n under these operations.
7. Which of the following sets are subspaces of R3 ? If the set is indeed a subspace, find
a basis for the subspace and compute its dimension.
(a) {(x1 , x2 , x3 ) : 3x1 − 2x2 + x3 = 0}
(b) {(x1 , x2 , x3 ) : 3x1 + 4x3 = 0, 2x1 − x2 + x3 = 0}
(c) {(x1 , x2 , x3 ) : x1 − 2x2 + 2x3 = 2}
(d) {(x1 , x2 , x3 ) : 3x1 − 2x22 = 0}
8. Show that the set of all possible solutions (x, y, z) ∈ R3 of the equations
Ax + By + Cz = 0
Dx + Ey + Cz = 0
form a subspace of R3 .
9. Let W be the subset of continuous functions on [0, 1] such that f (0) = 0. Prove that
W is a subspace of C[0, 1].
10. Let V be a vector space over F . Prove that −(αv) = (−α)v = α(−v) for all α ∈ F
and all v ∈ V .
11. Let V be a vector space of dimension n. Prove each of the following statements.
(a) If S = {v1 , . . . , vn } is a set of linearly independent vectors for V , then S is a basis
for V .
(b) If S = {v1 , . . . , vn } spans V , then S is a basis for V .
(c) If S = {v1 , . . . , vk } is a set of linearly independent vectors for V with k < n, then
there exist vectors vk+1 , . . . , vn such that
{v1 , . . . , vk , vk+1 , . . . , vn }
is a basis for V .
12. Prove that any set of vectors containing 0 is linearly dependent.
13. Let V be a vector space. Show that {0} is a subspace of V of dimension zero.
14. If a vector space V is spanned by n vectors, show that any set of m vectors in V must
be linearly dependent for m > n.
CHAPTER 20. VECTOR SPACES 258
15. Linear Transformations. Let V and W be vector spaces over a field F , of dimensions
m and n, respectively. If T : V → W is a map satisfying
T (u + v) = T (u) + T (v)
T (αv) = αT (v)
for all α ∈ F and all u, v ∈ V , then T is called a linear transformation from V into W .
(a) Prove that the kernel of T , ker(T ) = {v ∈ V : T (v) = 0}, is a subspace of V .
The kernel of T is sometimes called the null space of T .
(b) Prove that the range or range space of T , R(V ) = {w ∈ W : T (v) = w for some v ∈
V }, is a subspace of W .
(c) Show that T : V → W is injective if and only if ker(T ) = {0}.
(d) Let {v1 , . . . , vk } be a basis for the null space of T . We can extend this basis to be
a basis {v1 , . . . , vk , vk+1 , . . . , vm } of V . Why? Prove that {T (vk+1 ), . . . , T (vm )}
is a basis for the range of T . Conclude that the range of T has dimension m − k.
(e) Let dim V = dim W . Show that a linear transformation T : V → W is injective
if and only if it is surjective.
16. Let V and W be finite dimensional vector spaces of dimension n over a field F . Suppose
that T : V → W is a vector space isomorphism. If {v1 , . . . , vn } is a basis of V , show
that {T (v1 ), . . . , T (vn )} is a basis of W . Conclude that any vector space over a field F
of dimension n is isomorphic to F n .
17. Direct Sums. Let U and V be subspaces of a vector space W . The sum of U and V ,
denoted U + V , is defined to be the set of all vectors of the form u + v, where u ∈ U
and v ∈ V .
(a) Prove that U + V and U ∩ V are subspaces of W .
(b) If U + V = W and U ∩ V = 0, then W is said to be the direct sum. In this case,
we write W = U ⊕ V . Show that every element w ∈ W can be written uniquely
as w = u + v, where u ∈ U and v ∈ V .
(c) Let U be a subspace of dimension k of a vector space W of dimension n. Prove
that there exists a subspace V of dimension n − k such that W = U ⊕ V . Is the
subspace V unique?
(d) If U and V are arbitrary subspaces of a vector space W , show that
dim(U + V ) = dim U + dim V − dim(U ∩ V ).
18. Dual Spaces. Let V and W be finite dimensional vector spaces over a field F .
(a) Show that the set of all linear transformations from V into W , denoted by
Hom(V, W ), is a vector space over F , where we define vector addition as fol-
lows:
(S + T )(v) = S(v) + T (v)
(αS)(v) = αS(v),
where S, T ∈ Hom(V, W ), α ∈ F , and v ∈ V .
(b) Let V be an F -vector space. Define the dual space of V to be V ∗ = Hom(V, F ).
Elements in the dual space of V are called linear functionals. Let v1 , . . . , vn be
CHAPTER 20. VECTOR SPACES 259
an ordered basis for V . If v = α1 v1 + · · · + αn vn is any vector in V , define a
linear functional ϕi : V → F by ϕi (v) = αi . Show that the ϕi ’s form a basis for
V ∗ . This basis is called the dual basis of v1 , . . . , vn (or simply the dual basis if
the context makes the meaning clear).
(c) Consider the basis {(3, 1), (2, −2)} for R2 . What is the dual basis for (R2 )∗ ?
(d) Let V be a vector space of dimension n over a field F and let V ∗∗ be the dual
space of V ∗ . Show that each element v ∈ V gives rise to an element λv in V ∗∗
and that the map v 7→ λv is an isomorphism of V with V ∗∗ .
20.6 References and Suggested Readings
[1] Beezer, R. A First Course in Linear Algebra . Available online at linear.ups.edu.
2004–2014.
[2] Bretscher, O. Linear Algebra with Applications. 4th ed. Pearson, Upper Saddle River,
NJ, 2009.
[3] Curtis, C. W. Linear Algebra: An Introductory Approach. 4th ed. Springer, New
York, 1984.
[4] Hoffman, K. and Kunze, R. Linear Algebra. 2nd ed. Prentice-Hall, Englewood Cliffs,
NJ, 1971.
[5] Johnson, L. W., Riess, R. D., and Arnold, J. T. Introduction to Linear Algebra. 6th
ed. Pearson, Upper Saddle River, NJ, 2011.
[6] Leon, S. J. Linear Algebra with Applications. 8th ed. Pearson, Upper Saddle River,
NJ, 2010.
21
Fields
It is natural to ask whether or not some field F is contained in a larger field. We think of
the rational numbers, which reside inside the real numbers, while in turn, the real numbers
live inside the complex numbers. We can also study the fields between Q and R and inquire
as to the nature of these fields.
More specifically if we are given a field F and a polynomial p(x) ∈ F [x], we can ask
whether or not we can find a field E containing F such that p(x) factors into linear factors
over E[x]. For example, if we consider the polynomial
p(x) = x4 − 5x2 + 6
in Q[x], then p(x) factors as (x2 − 2)(x2 − 3). However, both of these factors are irreducible
in Q[x]. If we wish to find a zero of p(x), we must go to a larger field. Certainly the field
of real numbers will work, since
√ √ √ √
p(x) = (x − 2)(x + 2)(x − 3)(x + 3).
It is possible to find a smaller field in which p(x) has a zero, namely
√ √
Q( 2) = {a + b 2 : a, b ∈ Q}.
We wish to be able to compute and study such fields for arbitrary polynomials over a field
F.
21.1 Extension Fields
A field E is an extension field of a field F if F is a subfield of E. The field F is called the
base field. We write F ⊂ E.
Example 21.1 For example, let
√ √
F = Q( 2 ) = {a + b 2 : a, b ∈ Q}
√ √ √ √
and let E = Q( 2 + 3 ) be the smallest field containing both Q and 2 + 3. Both E and
F are extension fields of the rational
√ numbers. We claim
√ √ that E is an extension
√ √ field of√ F. √
To
see this, we need only show that 2 is in E. Since √2+ √3 is in E,
√ 1/( √2+ 3 ) = 3− √2
must√also be in E. Taking linear combinations of 2 + 3 and 3 − 2, we find that 2
and 3 must both be in E. □
260
CHAPTER 21. FIELDS 261
Example 21.2 Let p(x) = x2 + x + 1 ∈ Z2 [x]. Since neither 0 nor 1 is a root of this
polynomial, we know that p(x) is irreducible over Z2 . We will construct a field extension
of Z2 containing an element α such that p(α) = 0. By Theorem 17.22, the ideal hp(x)i
generated by p(x) is maximal; hence, Z2 [x]/hp(x)i is a field. Let f (x) + hp(x)i be an
arbitrary element of Z2 [x]/hp(x)i. By the division algorithm,
f (x) = (x2 + x + 1)q(x) + r(x),
where the degree of r(x) is less than the degree of x2 + x + 1. Therefore,
f (x) + hx2 + x + 1i = r(x) + hx2 + x + 1i.
The only possibilities for r(x) are then 0, 1, x, and 1+x. Consequently, E = Z2 [x]/hx2 +x+1i
is a field with four elements and must be a field extension of Z2 , containing a zero α of p(x).
The field Z2 (α) consists of elements
0 + 0α = 0
1 + 0α = 1
0 + 1α = α
1 + 1α = 1 + α.
Notice that α2 + α + 1 = 0; hence, if we compute (1 + α)2 ,
(1 + α)(1 + α) = 1 + α + α + (α)2 = α.
Other calculations are accomplished in a similar manner. We summarize these computations
in the following tables, which tell us how to add and multiply elements in E. □
+ 0 1 α 1+α
0 0 1 α 1+α
1 1 0 1+α α
α α 1+α 0 1
1+α 1+α α 1 0
Figure 21.3 Addition Table for Z2 (α)
· 0 1 α 1+α
0 0 0 0 0
1 0 1 α 1+α
α 0 α 1+α 1
1+α 0 1+α 1 α
Figure 21.4 Multiplication Table for Z2 (α)
The following theorem, due to Kronecker, is so important and so basic to our under-
standing of fields that it is often known as the Fundamental Theorem of Field Theory.
Theorem 21.5 Let F be a field and let p(x) be a nonconstant polynomial in F [x]. Then
there exists an extension field E of F and an element α ∈ E such that p(α) = 0.
Proof. To prove this theorem, we will employ the method that we used to construct
Example 21.2. Clearly, we can assume that p(x) is an irreducible polynomial. We wish to
find an extension field E of F containing an element α such that p(α) = 0. The ideal hp(x)i
generated by p(x) is a maximal ideal in F [x] by Theorem 17.22; hence, F [x]/hp(x)i is a field.
CHAPTER 21. FIELDS 262
We claim that E = F [x]/hp(x)i is the desired field.
We first show that E is a field extension of F . We can define a homomorphism of
commutative rings by the map ψ : F → F [x]/hp(x)i, where ψ(a) = a + hp(x)i for a ∈ F . It
is easy to check that ψ is indeed a ring homomorphism. Observe that
ψ(a) + ψ(b) = (a + hp(x)i) + (b + hp(x)i) = (a + b) + hp(x)i = ψ(a + b)
and
ψ(a)ψ(b) = (a + hp(x)i)(b + hp(x)i) = ab + hp(x)i = ψ(ab).
To prove that ψ is one-to-one, assume that
a + hp(x)i = ψ(a) = ψ(b) = b + hp(x)i.
Then a − b is a multiple of p(x), since it lives in the ideal hp(x)i. Since p(x) is a nonconstant
polynomial, the only possibility is that a − b = 0. Consequently, a = b and ψ is injective.
Since ψ is one-to-one, we can identify F with the subfield {a + hp(x)i : a ∈ F } of E and
view E as an extension field of F .
It remains for us to prove that p(x) has a zero α ∈ E. Set α = x + hp(x)i. Then α is in
E. If p(x) = a0 + a1 x + · · · + an xn , then
p(α) = a0 + a1 (x + hp(x)i) + · · · + an (x + hp(x)i)n
= a0 + (a1 x + hp(x)i) + · · · + (an xn + hp(x)i)
= a0 + a1 x + · · · + an xn + hp(x)i
= 0 + hp(x)i.
Therefore, we have found an element α ∈ E = F [x]/hp(x)i such that α is a zero of p(x). ■
Example 21.6 Let p(x) = x5 + x4 + 1 ∈ Z2 [x]. Then p(x) has irreducible factors x2 + x + 1
and x3 + x + 1. For a field extension E of Z2 such that p(x) has a root in E, we can let E
be either Z2 [x]/hx2 + x + 1i or Z2 [x]/hx3 + x + 1i. We will leave it as an exercise to show
that Z2 [x]/hx3 + x + 1i is a field with 23 = 8 elements. □
Algebraic Elements
An element α in an extension field E over F is algebraic over F if f (α) = 0 for some nonzero
polynomial f (x) ∈ F [x]. An element in E that is not algebraic over F is transcendental
over F . An extension field E of a field F is an algebraic extension of F if every element in
E is algebraic over F . If E is a field extension of F and α1 , . . . , αn are contained in E, we
denote the smallest field containing F and α1 , . . . , αn by F (α1 , . . . , αn ). If E = F (α) for
some α ∈ E, then E is a simple extension of F .
√
Example 21.7 Both 2 and i are algebraic over Q since they are zeros of the polynomials
x2 − 2 and x2 +1, respectively. Clearly π and e are algebraic over the real numbers; however,
it is a nontrivial fact that they are transcendental over Q. Numbers in R that are algebraic
over Q are in fact quite rare. Almost all real numbers are transcendental over Q.10 (In many
cases we do not know whether or not a particular number is transcendental; for example, it
is still not known whether π + e is transcendental or algebraic.) □
A complex number that is algebraic over Q is an algebraic number. A transcendental
number is an element of C that is transcendental over Q.
11
The probability that a real number chosen at random from the interval [0, 1] will be transcendental over
the rational numbers is one.
CHAPTER 21. FIELDS 263
p √ p √
Example√21.8 We will show that √ 2 + 3 is algebraic over Q. If α = 2 + 3, then
α2 = 2 + 3. Hence, α2 − 2 = 3 and (α2 − 2)2 = 3. Since α4 − 4α2 + 1 = 0, it must be
true that α is a zero of the polynomial x4 − 4x2 + 1 ∈ Q[x]. □
It is very easy to give an example of an extension field E over a field F , where E contains
an element transcendental over F . The following theorem characterizes transcendental
extensions.
Theorem 21.9 Let E be an extension field of F and α ∈ E. Then α is transcendental over
F if and only if F (α) is isomorphic to F (x), the field of fractions of F [x].
Proof. Let ϕα : F [x] → E be the evaluation homomorphism for α. Then α is transcenden-
tal over F if and only if ϕα (p(x)) = p(α) 6= 0 for all nonconstant polynomials p(x) ∈ F [x].
This is true if and only if ker ϕα = {0}; that is, it is true exactly when ϕα is one-to-one.
Hence, E must contain a copy of F [x]. The smallest field containing F [x] is the field of
fractions F (x). By Theorem 18.4, E must contain a copy of this field. ■
We have a more interesting situation in the case of algebraic extensions.
Theorem 21.10 Let E be an extension field of a field F and α ∈ E with α algebraic over
F . Then there is a unique irreducible monic polynomial p(x) ∈ F [x] of smallest degree such
that p(α) = 0. If f (x) is another polynomial in F [x] such that f (α) = 0, then p(x) divides
f (x).
Proof. Let ϕα : F [x] → E be the evaluation homomorphism. The kernel of ϕα is a
principal ideal generated by some p(x) ∈ F [x] with deg p(x) ≥ 1. We know that such a
polynomial exists, since F [x] is a principal ideal domain and α is algebraic. The ideal hp(x)i
consists exactly of those elements of F [x] having α as a zero. If f (α) = 0 and f (x) is not
the zero polynomial, then f (x) ∈ hp(x)i and p(x) divides f (x). So p(x) is a polynomial of
minimal degree having α as a zero. Any other polynomial of the same degree having α as
a zero must have the form βp(x) for some β ∈ F .
Suppose now that p(x) = r(x)s(x) is a factorization of p(x) into polynomials of lower
degree. Since p(α) = 0, r(α)s(α) = 0; consequently, either r(α) = 0 or s(α) = 0, which
contradicts the fact that p is of minimal degree. Therefore, p(x) must be irreducible. ■
Let E be an extension field of F and α ∈ E be algebraic over F . The unique monic
polynomial p(x) of the last theorem is called the minimal polynomial for α over F . The
degree of p(x) is the degree of α over F .
− 2 and g(x) = x4 − 4x2 + 1. These polynomials are the
Example 21.11 Let f (x) = x2 p
√ √
minimal polynomials of 2 and 2 + 3, respectively. □
Proposition 21.12 Let E be a field extension of F and α ∈ E be algebraic over F . Then
F (α) ∼
= F [x]/hp(x)i, where p(x) is the minimal polynomial of α over F .
Proof. Let ϕα : F [x] → E be the evaluation homomorphism. The kernel of this map is
hp(x)i, where p(x) is the minimal polynomial of α. By the First Isomorphism Theorem for
rings, the image of ϕα in E is isomorphic to F (α) since it contains both F and α. ■
Theorem 21.13 Let E = F (α) be a simple extension of F , where α ∈ E is algebraic over
F . Suppose that the degree of α over F is n. Then every element β ∈ E can be expressed
uniquely in the form
β = b0 + b1 α + · · · + bn−1 αn−1
for bi ∈ F .
Proof. Since ϕα (F [x]) ∼= F (α), every element in E = F (α) must be of the form ϕα (f (x)) =
f (α), where f (α) is a polynomial in α with coefficients in F . Let
p(x) = xn + an−1 xn−1 + · · · + a0
CHAPTER 21. FIELDS 264
be the minimal polynomial of α. Then p(α) = 0; hence,
αn = −an−1 αn−1 − · · · − a0 .
Similarly,
αn+1 = ααn
= −an−1 αn − an−2 αn−1 − · · · − a0 α
= −an−1 (−an−1 αn−1 − · · · − a0 ) − an−2 αn−1 − · · · − a0 α.
Continuing in this manner, we can express every monomial αm , m ≥ n, as a linear combi-
nation of powers of α that are less than n. Hence, any β ∈ F (α) can be written as
β = b0 + b1 α + · · · + bn−1 αn−1 .
To show uniqueness, suppose that
β = b0 + b1 α + · · · + bn−1 αn−1 = c0 + c1 α + · · · + cn−1 αn−1
for bi and ci in F . Then
g(x) = (b0 − c0 ) + (b1 − c1 )x + · · · + (bn−1 − cn−1 )xn−1
is in F [x] and g(α) = 0. Since the degree of g(x) is less than the degree of p(x), the
irreducible polynomial of α, g(x) must be the zero polynomial. Consequently,
b0 − c0 = b1 − c1 = · · · = bn−1 − cn−1 = 0,
or bi = ci for i = 0, 1, . . . , n − 1. Therefore, we have shown uniqueness. ■
Example 21.14 Since x2 + 1 is irreducible over R, hx2 + 1i is a maximal ideal in R[x]. So
E = R[x]/hx2 +1i is a field extension of R that contains a root of x2 +1. Let α = x+hx2 +1i.
We can identify E with the complex numbers. By Proposition 21.12, E is isomorphic to
R(α) = {a + bα : a, b ∈ R}. We know that α2 = −1 in E, since
α2 + 1 = (x + hx2 + 1i)2 + (1 + hx2 + 1i)
= (x2 + 1) + hx2 + 1i
= 0.
Hence, we have an isomorphism of R(α) with C defined by the map that takes a + bα to
a + bi. □
Let E be a field extension of a field F . If we regard E as a vector space over F , then we
can bring the machinery of linear algebra to bear on the problems that we will encounter in
our study of fields. The elements in the field E are vectors; the elements in the field F are
scalars. We can think of addition in E as adding vectors. When we multiply an element in
E by an element of F , we are multiplying a vector by a scalar. This view of field extensions
is especially fruitful if a field extension E of F is a finite dimensional vector space over F ,
and Theorem 21.13 states that E = F (α) is finite dimensional vector space over F with
basis {1, α, α2 , . . . , αn−1 }.
If an extension field E of a field F is a finite dimensional vector space over F of dimension
n, then we say that E is a finite extension of degree n over F . We write
[E : F ] = n.
CHAPTER 21. FIELDS 265
to indicate the dimension of E over F .
Theorem 21.15 Every finite extension field E of a field F is an algebraic extension.
Proof. Let α ∈ E. Since [E : F ] = n, the elements
1, α, . . . , αn
cannot be linearly independent. Hence, there exist ai ∈ F , not all zero, such that
an αn + an−1 αn−1 + · · · + a1 α + a0 = 0.
Therefore,
p(x) = an xn + · · · + a0 ∈ F [x]
is a nonzero polynomial with p(α) = 0. ■
Remark 21.16 Theorem 21.15 says that every finite extension of a field F is an algebraic
extension. The converse is false, however. We will leave it as an exercise to show that the
set of all elements in R that are algebraic over Q forms an infinite field extension of Q.
The next theorem is a counting theorem, similar to Lagrange’s Theorem in group theory.
Theorem 21.17 will prove to be an extremely useful tool in our investigation of finite field
extensions.
Theorem 21.17 If E is a finite extension of F and K is a finite extension of E, then K
is a finite extension of F and
[K : F ] = [K : E][E : F ].
Proof. Let {α1 , . . . , αn } be a basis for E as a vector space over F and {β1 , . . . , βm } be a
basis for K as a vector space over E. We claim that {αi βj } is Pa basis for K over P F . We will
first show that these vectors span K. Let u ∈ K. Then u = m b
j=1 j jβ and bj = n
i=1 aij αi ,
where bj ∈ E and aij ∈ F . Then
!
Xm Xn X
u= aij αi βj = aij (αi βj ).
j=1 i=1 i,j
So the mn vectors αi βj must span K over F .
We must show that {αi βj } are linearly independent. Recall that a set of vectors
v1 , v2 , . . . , vn in a vector space V are linearly independent if
c1 v1 + c2 v2 + · · · + cn vn = 0
implies that
c1 = c2 = · · · = cn = 0.
Let X
u= cij (αi βj ) = 0
i,j
for cij ∈ F . We need to prove that all of the cij ’s are zero. We can rewrite u as
!
Xm Xn
cij αi βj = 0,
j=1 i=1
P
where i cij αi ∈ E. Since the βj ’s are linearly independent over E, it must be the case
that
X
n
cij αi = 0
i=1
CHAPTER 21. FIELDS 266
for all j. However, the αj are also linearly independent over F . Therefore, cij = 0 for all i
and j, which completes the proof. ■
The following corollary is easily proved using mathematical induction.
Corollary 21.18 If Fi is a field for i = 1, . . . , k and Fi+1 is a finite extension of Fi , then
Fk is a finite extension of F1 and
[Fk : F1 ] = [Fk : Fk−1 ] · · · [F2 : F1 ].
Corollary 21.19 Let E be an extension field of F . If α ∈ E is algebraic over F with
minimal polynomial p(x) and β ∈ F (α) with minimal polynomial q(x), then deg q(x) divides
deg p(x).
Proof. We know that deg p(x) = [F (α) : F ] and deg q(x) = [F (β) : F ]. Since F ⊂ F (β) ⊂
F (α),
[F (α) : F ] = [F (α) : F (β)][F (β) : F ].
■
√ √
Example 21.20 Let us determine an extension √ field√ of Q containing 3 + 5. It is easy
to determine that the minimal polynomial of 3 + 5 is x4 − 16x2 + 4. It follows that
√ √
[Q( 3 + 5 ) : Q] = 4.
√ √ √ √
We√know that {1, 3 } is√ a basis for Q( 3 )√over Q. Hence, 3 + 5√ cannot be in
Q( 3 ).√ It√ follows that √ 5√ cannot be √ in Q( 3 ) either.
√ √ Therefore,
√ √ √{1, 5 } is a basis
for√Q( √3, 5 ) = √ (Q( 3√))( 5 ) over Q( 3 ) and {1, 3, 5, 3 5 = 15 } is a basis for
Q( 3, 5 ) = Q( 3 + 5 ) over Q. This example shows that it is possible that some
extension F (α1 , . . . , αn ) is actually a simple extension of F even though n > 1. □
√ √ √
Example 21.21 √ Let us compute a basis for Q( 3 5, 5 i), where √ is the positive square
5√
root of 5 and 5 is the real cube root of 5. We know that 5 i ∈
3
/ Q( 5 ), so
3
√ √ √
[Q( 5, 5 i) : Q( 5 )] = 2.
3 3
√ √ √ √
It is easy√to determine
√ that {1, 5i }√ is a basis for Q( 3 5, 5 i) over Q( √3
5 ).√We also know
that {1, 5, ( 5 ) } is a basis for Q( 5 ) over Q. Hence, a basis for Q( 3 5, 5 i) over Q is
3 3 2 3
√ √ √ √ √ √ √
{1, 5 i, 5, ( 5 )2 , ( 5 )5 i, ( 5 )7 i = 5 5 i or 5 i}.
3 3 6 6 6 6
√
Notice that 6 5 i is a zero of x6 + 5. We can show that this polynomial is irreducible over
Q using Eisenstein’s Criterion, where we let p = 5. Consequently,
√ √ √
Q ⊂ Q( 5 i) ⊂ Q( 5, 5 i).
6 3
√ √ √
But it must be the case that Q( 6 5 i) = Q( 3 5, 5 i), since the degree of both of these
extensions is 6. □
Theorem 21.22 Let E be a field extension of F . Then the following statements are
equivalent.
1. E is a finite extension of F .
2. There exists a finite number of algebraic elements α1 , . . . , αn ∈ E such that E =
F (α1 , . . . , αn ).
3. There exists a sequence of fields
E = F (α1 , . . . , αn ) ⊃ F (α1 , . . . , αn−1 ) ⊃ · · · ⊃ F (α1 ) ⊃ F ,
CHAPTER 21. FIELDS 267
where each field F (α1 , . . . , αi ) is algebraic over F (α1 , . . . , αi−1 ).
Proof. (1) ⇒ (2). Let E be a finite algebraic extension of F . Then E is a finite dimensional
vector space over F and there exists a basis consisting of elements α1 , . . . , αn in E such that
E = F (α1 , . . . , αn ). Each αi is algebraic over F by Theorem 21.15.
(2) ⇒ (3). Suppose that E = F (α1 , . . . , αn ), where every αi is algebraic over F . Then
E = F (α1 , . . . , αn ) ⊃ F (α1 , . . . , αn−1 ) ⊃ · · · ⊃ F (α1 ) ⊃ F ,
where each field F (α1 , . . . , αi ) is algebraic over F (α1 , . . . , αi−1 ).
(3) ⇒ (1). Let
E = F (α1 , . . . , αn ) ⊃ F (α1 , . . . , αn−1 ) ⊃ · · · ⊃ F (α1 ) ⊃ F ,
where each field F (α1 , . . . , αi ) is algebraic over F (α1 , . . . , αi−1 ). Since
F (α1 , . . . , αi ) = F (α1 , . . . , αi−1 )(αi )
is simple extension and αi is algebraic over F (α1 , . . . , αi−1 ), it follows that
[F (α1 , . . . , αi ) : F (α1 , . . . , αi−1 )]
is finite for each i. Therefore, [E : F ] is finite. ■
Algebraic Closure
Given a field F , the question arises as to whether or not we can find a field E such that
every polynomial p(x) has a root in E. This leads us to the following theorem.
Theorem 21.23 Let E be an extension field of F . The set of elements in E that are
algebraic over F form a field.
Proof. Let α, β ∈ E be algebraic over F . Then F (α, β) is a finite extension of F . Since
every element of F (α, β) is algebraic over F , α ± β, αβ, and α/β (β 6= 0) are all algebraic
over F . Consequently, the set of elements in E that are algebraic over F form a field. ■
Corollary 21.24 The set of all algebraic numbers forms a field; that is, the set of all
complex numbers that are algebraic over Q makes up a field.
Let E be a field extension of a field F . We define the algebraic closure of a field F in
E to be the field consisting of all elements in E that are algebraic over F . A field F is
algebraically closed if every nonconstant polynomial in F [x] has a root in F .
Theorem 21.25 A field F is algebraically closed if and only if every nonconstant polynomial
in F [x] factors into linear factors over F [x].
Proof. Let F be an algebraically closed field. If p(x) ∈ F [x] is a nonconstant polynomial,
then p(x) has a zero in F , say α. Therefore, x − α must be a factor of p(x) and so
p(x) = (x − α)q1 (x), where deg q1 (x) = deg p(x) − 1. Continue this process with q1 (x) to
find a factorization
p(x) = (x − α)(x − β)q2 (x),
where deg q2 (x) = deg p(x) − 2. The process must eventually stop since the degree of p(x)
is finite.
Conversely, suppose that every nonconstant polynomial p(x) in F [x] factors into linear
factors. Let ax − b be such a factor. Then p(b/a) = 0. Consequently, F is algebraically
closed. ■
Corollary 21.26 An algebraically closed field F has no proper algebraic extension E.
CHAPTER 21. FIELDS 268
Proof. Let E be an algebraic extension of F ; then F ⊂ E. For α ∈ E, the minimal
polynomial of α is x − α. Therefore, α ∈ F and F = E. ■
Theorem 21.27 Every field F has a unique algebraic closure.
It is a nontrivial fact that every field has a unique algebraic closure. The proof is not
extremely difficult, but requires some rather sophisticated set theory. We refer the reader
to [3], [4], or [8] for a proof of this result.
We now state the Fundamental Theorem of Algebra, first proven by Gauss at the age
of 22 in his doctoral thesis. This theorem states that every polynomial with coefficients in
the complex numbers has a root in the complex numbers. The proof of this theorem will
be given in Chapter 23.
Theorem 21.28 Fundamental Theorem of Algebra. The field of complex numbers is
algebraically closed.
21.2 Splitting Fields
Let F be a field and p(x) be a nonconstant polynomial in F [x]. We already know that we
can find a field extension of F that contains a root of p(x). However, we would like to know
whether an extension E of F containing all of the roots of p(x) exists. In other words, can
we find a field extension of F such that p(x) factors into a product of linear polynomials?
What is the “smallest” extension containing all the roots of p(x)?
Let F be a field and p(x) = a0 + a1 x + · · · + an xn be a nonconstant polynomial in F [x].
An extension field E of F is a splitting field of p(x) if there exist elements α1 , . . . , αn in E
such that E = F (α1 , . . . , αn ) and
p(x) = (x − α1 )(x − α2 ) · · · (x − αn ).
A polynomial p(x) ∈ F [x] splits in E if it is the product of linear factors in E[x].
Example 21.29 Let p(x) = x4 + 2x2 − √ 8 be in Q[x]. Then p(x) has irreducible factors
x − 2 and x + 4. Therefore, the field Q( 2, i) is a splitting field for p(x).
2 2 □
√
Example 21.30 Let p(x) = x3 − 3 be in Q[x]. Then p(x) has a root in the field Q( 3 3 ).
However, this field is not a splitting field for p(x) since the complex cube roots of 3,
√ √
− 3 3 ± ( 6 3 )5 i
,
2
√
are not in Q( 3 3 ). □
Theorem 21.31 Let p(x) ∈ F [x] be a nonconstant polynomial. Then there exists a splitting
field E for p(x).
Proof. We will use mathematical induction on the degree of p(x). If deg p(x) = 1, then
p(x) is a linear polynomial and E = F . Assume that the theorem is true for all polynomials
of degree k with 1 ≤ k < n and let deg p(x) = n. We can assume that p(x) is irreducible;
otherwise, by our induction hypothesis, we are done. By Theorem 21.5, there exists a field
K such that p(x) has a zero α1 in K. Hence, p(x) = (x − α1 )q(x), where q(x) ∈ K[x].
Since deg q(x) = n − 1, there exists a splitting field E ⊃ K of q(x) that contains the zeros
α2 , . . . , αn of p(x) by our induction hypothesis. Consequently,
E = K(α2 , . . . , αn ) = F (α1 , . . . , αn )
is a splitting field of p(x). ■
CHAPTER 21. FIELDS 269
The question of uniqueness now arises for splitting fields. This question is answered in
the affirmative. Given two splitting fields K and L of a polynomial p(x) ∈ F [x], there exists
a field isomorphism ϕ : K → L that preserves F . In order to prove this result, we must first
prove a lemma.
Lemma 21.32 Let ϕ : E → F be an isomorphism of fields. Let K be an extension field
of E and α ∈ K be algebraic over E with minimal polynomial p(x). Suppose that L is
an extension field of F such that β is root of the polynomial in F [x] obtained from p(x)
under the image of ϕ. Then ϕ extends to a unique isomorphism ϕ : E(α) → F (β) such that
ϕ(α) = β and ϕ agrees with ϕ on E.
Proof. If p(x) has degree n, then by Theorem 21.13 we can write any element in E(α)
as a linear combination of 1, α, . . . , αn−1 . Therefore, the isomorphism that we are seeking
must be
ϕ(a0 + a1 α + · · · + an−1 αn−1 ) = ϕ(a0 ) + ϕ(a1 )β + · · · + ϕ(an−1 )β n−1 ,
where
a0 + a1 α + · · · + an−1 αn−1
is an element in E(α). The fact that ϕ is an isomorphism could be checked by direct
computation; however, it is easier to observe that ϕ is a composition of maps that we
already know to be isomorphisms.
We can extend ϕ to be an isomorphism from E[x] to F [x], which we will also denote by
ϕ, by letting
ϕ(a0 + a1 x + · · · + an xn ) = ϕ(a0 ) + ϕ(a1 )x + · · · + ϕ(an )xn .
This extension agrees with the original isomorphism ϕ : E → F , since constant polynomials
get mapped to constant polynomials. By assumption, ϕ(p(x)) = q(x); hence, ϕ maps hp(x)i
onto hq(x)i. Consequently, we have an isomorphism ψ : E[x]/hp(x)i → F [x]/hq(x)i. By
Proposition 21.12, we have isomorphisms σ : E[x]/hp(x)i → E(α) and τ : F [x]/hq(x)i →
F (β), defined by evaluation at α and β, respectively. Therefore, ϕ = τ ψσ −1 is the required
isomorphism (see Figure 21.33).
ψ
E[x]/hp(x)i −−−−→ F [x]/hq(x)i
σ τ
y y
ϕ
E(α) −−−−→ F (β)
y y
ϕ
E −−−−→ F
Figure 21.33
We leave the proof of uniqueness as a exercise. ■
Theorem 21.34 Let ϕ : E → F be an isomorphism of fields and let p(x) be a nonconstant
polynomial in E[x] and q(x) the corresponding polynomial in F [x] under the isomorphism.
If K is a splitting field of p(x) and L is a splitting field of q(x), then ϕ extends to an
isomorphism ψ : K → L.
Proof. We will use mathematical induction on the degree of p(x). We can assume that
p(x) is irreducible over E. Therefore, q(x) is also irreducible over F . If deg p(x) = 1, then
by the definition of a splitting field, K = E and L = F and there is nothing to prove.
CHAPTER 21. FIELDS 270
Assume that the theorem holds for all polynomials of degree less than n. Since K is a
splitting field of p(x), all of the roots of p(x) are in K. Choose one of these roots, say α, such
that E ⊂ E(α) ⊂ K. Similarly, we can find a root β of q(x) in L such that F ⊂ F (β) ⊂ L.
By Lemma 21.32, there exists an isomorphism ϕ : E(α) → F (β) such that ϕ(α) = β and ϕ
agrees with ϕ on E (see Figure 21.35).
ψ
K −−−−→ L
σ τ
y y
ϕ
E(α) −−−−→ F (β)
y y
ϕ
E −−−−→ F
Figure 21.35
Now write p(x) = (x − α)f (x) and q(x) = (x − β)g(x), where the degrees of f (x) and
g(x) are less than the degrees of p(x) and q(x), respectively. The field extension K is a
splitting field for f (x) over E(α), and L is a splitting field for g(x) over F (β). By our
induction hypothesis there exists an isomorphism ψ : K → L such that ψ agrees with ϕ on
E(α). Hence, there exists an isomorphism ψ : K → L such that ψ agrees with ϕ on E. ■
Corollary 21.36 Let p(x) be a polynomial in F [x]. Then there exists a splitting field K of
p(x) that is unique up to isomorphism.
21.3 Geometric Constructions
In ancient Greece, three classic problems were posed. These problems are geometric in
nature and involve straightedge-and-compass constructions from what is now high school
geometry; that is, we are allowed to use only a straightedge and compass to solve them.
The problems can be stated as follows.
1. Given an arbitrary angle, can one trisect the angle into three equal subangles using
only a straightedge and compass?
2. Given an arbitrary circle, can one construct a square with the same area using only a
straightedge and compass?
3. Given a cube, can one construct the edge of another cube having twice the volume of
the original? Again, we are only allowed to use a straightedge and compass to do the
construction.
After puzzling mathematicians for over two thousand years, each of these constructions was
finally shown to be impossible. We will use the theory of fields to provide a proof that the
solutions do not exist. It is quite remarkable that the long-sought solution to each of these
three geometric problems came from abstract algebra.
First, let us determine more specifically what we mean by a straightedge and compass,
and also examine the nature of these problems in a bit more depth. To begin with, a
straightedge is not a ruler. We cannot measure arbitrary lengths with a straightedge. It is
merely a tool for drawing a line through two points. The statement that the trisection of
an arbitrary angle is impossible means that there is at least one angle that is impossible to
trisect with a straightedge-and-compass construction. Certainly it is possible to trisect an
angle in special cases. We can construct a 30◦ angle; hence, it is possible to trisect a 90◦
CHAPTER 21. FIELDS 271
angle. However, we will show that it is impossible to construct a 20◦ angle. Therefore, we
cannot trisect a 60◦ angle.
Constructible Numbers
A real number α is constructible if we can construct a line segment of length |α| in a finite
number of steps from a segment of unit length by using a straightedge and compass.
Theorem 21.37 The set of all constructible real numbers forms a subfield F of the field of
real numbers.
Proof. Let α and β be constructible numbers. We must show that α + β, α − β, αβ, and
α/β (β 6= 0) are also constructible numbers. We can assume that both α and β are positive
with α > β. It is quite obvious how to construct α + β and α − β. To find a line segment
with length αβ, we assume that β > 1 and construct the triangle in Figure 21.38 such that
triangles 4ABC and 4ADE are similar. Since α/1 = x/β, the line segment x has length
αβ. A similar construction can be made if β < 1. We will leave it as an exercise to show
that the same triangle can be used to construct α/β for β 6= 0. ■
D
β B
1
α C
A E
x
Figure 21.38 Construction of products
√
Lemma 21.39 If α is a constructible number, then α is a constructible number.
Proof. In Figure 21.40 the triangles 4ABD, 4BCD, and 4ABC are similar; hence,
1/x = x/α, or x2 = α. ■
CHAPTER 21. FIELDS 272
B
x
1 α
A D C
Figure 21.40 Construction of roots
By Theorem 21.37, we can locate in the plane any point P = (p, q) that has rational
coordinates p and q. We need to know what other points can be constructed with a compass
and straightedge from points with rational coordinates.
Lemma 21.41 Let F be a subfield of R.
1. If a line contains two points in F , then it has the equation ax + by + c = 0, where a,
b, and c are in F .
2. If a circle has a center at a point with coordinates in F and a radius that is also in F ,
then it has the equation x2 + y 2 + dx + ey + f = 0, where d, e, and f are in F .
Proof. Let (x1 , y1 ) and (x2 , y2 ) be points on a line whose coordinates are in F . If x1 = x2 ,
then the equation of the line through the two points is x − x1 = 0, which has the form
ax + by + c = 0. If x1 6= x2 , then the equation of the line through the two points is given by
y2 − y1
y − y1 = (x − x1 ),
x2 − x1
which can also be put into the proper form.
To prove the second part of the lemma, suppose that (x1 , y1 ) is the center of a circle of
radius r. Then the circle has the equation
(x − x1 )2 + (y − y1 )2 − r2 = 0.
This equation can easily be put into the appropriate form. ■
Starting with a field of constructible numbers F , we have three possible ways of con-
structing additional points in R with a compass and straightedge.
1. To find possible new points in R, we can take the intersection of two lines, each of
which passes through two known points with coordinates in F .
2. The intersection of a line that passes through two points that have coordinates in F
and a circle whose center has coordinates in F with radius of a length in F will give
new points in R.
3. We can obtain new points in R by intersecting two circles whose centers have coordi-
nates in F and whose radii are of lengths in F .
CHAPTER 21. FIELDS 273
The first case gives no new points in R, since the solution of two equations of the form
ax + by + c = 0 having coefficients in F will always be in F . The third case can be reduced
to the second case. Let
x2 + y 2 + d1 x + e1 y + f1 = 0
x2 + y 2 + d2 x + e2 y + f2 = 0
be the equations of two circles, where di , ei , and fi are in F for i = 1, 2. These circles have
the same intersection as the circle
x2 + y 2 + d1 x + e1 x + f1 = 0
and the line
(d1 − d2 )x + b(e2 − e1 )y + (f2 − f1 ) = 0.
The last equation is that of the chord passing through the intersection points of the two
circles. Hence, the intersection of two circles can be reduced to the case of an intersection
of a line with a circle.
Considering the case of the intersection of a line and a circle, we must determine the
nature of the solutions of the equations
ax + by + c = 0
2 2
x + y + dx + ey + f = 0.
If we eliminate y from these equations, we obtain an equation of the form Ax2 +Bx+C = 0,
where A, B, and C are in F . The x coordinate of the intersection points is given by
√
−B ± B 2 − 4AC
x=
2A
√
and is in F ( α ), where α = B 2 − 4AC > 0. We have proven the following lemma.
Lemma 21.42 Let F be a field of constructible numbers. Then the points determined by
√
the intersections of lines and circles in F lie in the field F ( α ) for some α in F .
Theorem 21.43 A real number α is a constructible number if and only if there exists a
sequence of fields
Q = F0 ⊂ F1 ⊂ · · · ⊂ Fk
√
such that Fi = Fi−1 ( αi ) with αi ∈ Fi and α ∈ Fk . In particular, there exists an integer
k > 0 such that [Q(α) : Q] = 2k .
Proof. The existence of the Fi ’s and the αi ’s is a direct consequence of Lemma 21.42 and
of the fact that
[Fk : Q] = [Fk : Fk−1 ][Fk−1 : Fk−2 ] · · · [F1 : Q] = 2k .
■
Corollary 21.44 The field of all constructible numbers is an algebraic extension of Q.
As we can see by the field of constructible numbers, not every algebraic extension of a
field is a finite extension.
Doubling the Cube and Squaring the Circle
We are now ready to investigate the classical problems of doubling the cube and squaring
the circle. We can use the field of constructible numbers to show exactly when a particular
geometric construction can be accomplished.
CHAPTER 21. FIELDS 274
Doubling the cube is impossible. Given the edge of the cube, it is impossible to
construct with a straightedge and compass the edge of the cube that has twice the volume
of the original cube. Let the original cube have an edge of length 1 and, therefore, a volume
of 1. If we could construct
√ a cube
√ having a volume of 2, then this new cube would have
an edge of length 3 2. However, 3 2 is a zero of the irreducible polynomial x3 − 2 over Q;
hence, √
[Q( 2 ) : Q] = 3
3
This is impossible, since 3 is not a power of 2.
Squaring the circle. Suppose that we have a circle of radius 1. The area of the circle
√
is π; therefore, we must be able to construct a square with side π. This is impossible
√
since π and consequently π are both transcendental. Therefore, using a straightedge and
compass, it is not possible to construct a square with the same area as the circle.
Trisecting an Angle
Trisecting an arbitrary angle is impossible. We will show that it is impossible to construct
a 20◦ angle. Consequently, a 60◦ angle cannot be trisected. We first need to calculate the
triple-angle formula for the cosine:
cos 3θ = cos(2θ + θ)
= cos 2θ cos θ − sin 2θ sin θ
= (2 cos2 θ − 1) cos θ − 2 sin2 θ cos θ
= (2 cos2 θ − 1) cos θ − 2(1 − cos2 θ) cos θ
= 4 cos3 θ − 3 cos θ.
The angle θ can be constructed if and only if α = cos θ is constructible. Let θ = 20◦ . Then
cos 3θ = cos 60◦ = 1/2. By the triple-angle formula for the cosine,
1
4α3 − 3α = .
2
Therefore, α is a zero of 8x3 − 6x − 1. This polynomial has no factors in Z[x], and hence
is irreducible over Q[x]. Thus, [Q(α) : Q] = 3. Consequently, α cannot be a constructible
number.
Sage. Extensions of the field of rational numbers are a central object of study in number
theory, so with Sage’s roots in this discipline, it is no surprise that there is extensive support
for fields and for extensions of the rationals. Sage also contains an implementation of the
entire field of algebraic numbers, with exact representations.
Historical Note
Algebraic number theory uses the tools of algebra to solve problems in number theory.
Modern algebraic number theory began with Pierre de Fermat (1601–1665). Certainly we
can find many positive integers that satisfy the equation x2 + y 2 = z 2 ; Fermat conjectured
that the equation xn + y n = z n has no positive integer solutions for n ≥ 3. He stated in the
margin of his copy of the Latin translation of Diophantus’ Arithmetica that he had found a
marvelous proof of this theorem, but that the margin of the book was too narrow to contain
it. Building on work of other mathematicians, it was Andrew Wiles who finally succeeded
in proving Fermat’s Last Theorem in the 1990s. Wiles’s achievement was reported on the
front page of the New York Times.
CHAPTER 21. FIELDS 275
Attempts to prove Fermat’s Last Theorem have led to important contributions to algebraic
number theory by such notable mathematicians as Leonhard Euler (1707–1783). Significant
advances in the understanding of Fermat’s Last Theorem were made by Ernst Kummer
(1810–1893). Kummer’s student, Leopold Kronecker (1823–1891), became one of the leading
algebraists of the nineteenth century. Kronecker’s theory of ideals and his study of algebraic
number theory added much to the understanding of fields.
David Hilbert (1862–1943) and Hermann Minkowski (1864–1909) were among the mathe-
maticians who led the way in this subject at the beginning of the twentieth century. Hilbert
and Minkowski were both mathematicians at Göttingen University in Germany. Göttingen
was truly one the most important centers of mathematical research during the last two cen-
turies. The large number of exceptional mathematicians who studied there included Gauss,
Dirichlet, Riemann, Dedekind, Noether, and Weyl.
André Weil answered questions in number theory using algebraic geometry, a field of math-
ematics that studies geometry by studying commutative rings. From about 1955 to 1970,
Alexander Grothendieck dominated the field of algebraic geometry. Pierre Deligne, a stu-
dent of Grothendieck, solved several of Weil’s number-theoretic conjectures. One of the
most recent contributions to algebra and number theory is Gerd Falting’s proof of the
Mordell-Weil conjecture. This conjecture of Mordell and Weil essentially says that certain
polynomials p(x, y) in Z[x, y] have only a finite number of integral solutions.
21.4 Reading Questions
1. What does it mean for an extension field E of a field F to be a simple extension of F ?
2. What is the definition of a minimal polynomial of an element α ∈ E, where E is an
extension of F , and α is algebraic over F ?
3. Describe how linear algebra enters into this chapter. What critical result relies on a
proof that is almost entirely linear algebra?
4. What is the definition of an algebraically closed field?
5. What is a splitting field of a polynomial p(x) ∈ F [x]?
21.5 Exercises
1. Show that each of the following numbers is algebraic over Q by finding the minimal
polynomial of the number over Q.
q √
(a) 1/3 + 7
√ √
3
(b) 3 + 5
√ √
(c) 3 + 2 i
(d) cos θ + i sin θ for θ = 2π/n with n ∈ N
q
√
2−i
3
(e)
2. Find a basis for each of the following field extensions. What is the degree of each
extension?
√ √
(a) Q( 3, 6 ) over Q
√ √
(b) Q( 3 2, 3 3 ) over Q
CHAPTER 21. FIELDS 276
√
(c) Q( 2, i) over Q
√ √ √
(d) Q( 3, 5, 7 ) over Q
√ √
(e) Q( 2, 3 2 ) over Q
√ √
(f) Q( 8 ) over Q( 2 )
√ √
(g) Q(i, 2 + i, 3 + i) over Q
√ √ √
(h) Q( 2 + 5 ) over Q( 5 )
√ √ √ √ √
(i) Q( 2, 6 + 10 ) over Q( 3 + 5 )
3. Find the splitting field for each of the following polynomials.
(a) x4 − 10x2 + 21 over Q (c) x3 + 2x + 2 over Z3
(b) x4 + 1 over Q (d) x3 − 3 over Q
√
4. Consider the field extension Q( 4 3, i) over Q.
√ √
(a) Find a basis for the field extension Q( 4 3, i) over Q. Conclude that [Q( 4 3, i) :
Q] = 8.
√
(b) Find all subfields F of Q( 4 3, i) such that [F : Q] = 2.
√
(c) Find all subfields F of Q( 4 3, i) such that [F : Q] = 4.
5. Show that Z2 [x]/hx3 + x + 1i is a field with eight elements. Construct a multiplication
table for the multiplicative group of the field.
6. Show that the regular 9-gon is not constructible with a straightedge and compass, but
that the regular 20-gon is constructible.
7. Prove that the cosine of one degree (cos 1◦ ) is algebraic over Q but not constructible.
8. Can a cube be constructed with three times the volume of a given cube?
√ √ √
9. Prove that Q( 3, 4 3, 8 3, . . .) is an algebraic extension of Q but not a finite extension.
10. Prove or disprove: π is algebraic over Q(π 3 ).
11. Let p(x) be a nonconstant polynomial of degree n in F [x]. Prove that there exists a
splitting field E for p(x) such that [E : F ] ≤ n!.
√ √
12. Prove or disprove: Q( 2 ) ∼ = Q( 3 ).
√ √
13. Prove that the fields Q( 4 3 ) and Q( 4 3 i) are isomorphic but not equal.
14. Let K be an algebraic extension of E, and E an algebraic extension of F . Prove that
K is algebraic over F . [Caution: Do not assume that the extensions are finite.]
15. Prove or disprove: Z[x]/hx3 − 2i is a field.
16. Let F be a field of characteristic p. Prove that p(x) = xp − a either is irreducible over
F or splits in F .
17. Let E be the algebraic closure of a field F . Prove that every polynomial p(x) in F [x]
splits in E.
18. If every irreducible polynomial p(x) in F [x] is linear, show that F is an algebraically
closed field.
19. Prove that if α and β are constructible numbers such that β = 6 0, then so is α/β.
20. Show that the set of all elements in R that are algebraic over Q form a field extension
of Q that is not finite.
CHAPTER 21. FIELDS 277
21. Let E be an algebraic extension of a field F , and let σ be an automorphism of E leaving
F fixed. Let α ∈ E. Show that σ induces a permutation of the set of all zeros of the
minimal polynomial of α that are in E.
√ √ √ √ √ √
22. Show that
√ Q( 3, 7 ) = Q( 3 + 7 ). Extend your proof to show that Q( a, b ) =
√
Q( a + b ), where a 6= b and neither a nor b is a perfect square.
23. Let E be a finite extension of a field F . If [E : F ] = 2, show that E is a splitting field
of F for some polynomial f (x) ∈ F [x].
24. Prove or disprove: Given a polynomial p(x) in Z6 [x], it is possible to construct a ring
R such that p(x) has a root in R.
25. Let E be a field extension of F and α ∈ E. Determine [F (α) : F (α3 )].
26. Let α, β be transcendental over Q. Prove that either αβ or α + β is also transcendental.
27. Let E be an extension field of F and α ∈ E be transcendental over F . Prove that
every element in F (α) that is not in F is also transcendental over F .
28. Let α be a root of an irreducible monic polynomial p(x) ∈ F [x], with deg p = n. Prove
that [F (α) : F ] = n.
21.6 References and Suggested Readings
[1] Dean, R. A. Elements of Abstract Algebra . Wiley, New York, 1966.
[2] Dudley, U. A Budget of Trisections. Springer-Verlag, New York, 1987. An interesting
and entertaining account of how not to trisect an angle.
[3] Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper Saddle
River, NJ, 2003.
[4] Kaplansky, I. Fields and Rings, 2nd ed. University of Chicago Press, Chicago, 1972.
[5] Klein, F. Famous Problems of Elementary Geometry. Chelsea, New York, 1955.
[6] Martin, G. Geometric Constructions. Springer, New York, 1998.
[7] H. Pollard and H. G. Diamond. Theory of Algebraic Numbers, Dover, Mineola, NY,
2010.
[8] Walker, E. A. Introduction to Abstract Algebra. Random House, New York, 1987.
This work contains a proof showing that every field has an algebraic closure.
22
Finite Fields
Finite fields appear in many applications of algebra, including coding theory and cryptog-
raphy. We already know one finite field, Zp , where p is prime. In this chapter we will show
that a unique finite field of order pn exists for every prime p, where n is a positive integer.
Finite fields are also called Galois fields in honor of Évariste Galois, who was one of the first
mathematicians to investigate them.
22.1 Structure of a Finite Field
Recall that a field F has characteristic p if p is the smallest positive integer such that for
every nonzero element α in F , we have pα = 0. If no such integer exists, then F has
characteristic 0. From Theorem 16.19 we know that p must be prime. Suppose that F is a
finite field with n elements. Then nα = 0 for all α in F . Consequently, the characteristic of
F must be p, where p is a prime dividing n. This discussion is summarized in the following
proposition.
Proposition 22.1 If F is a finite field, then the characteristic of F is p, where p is prime.
Throughout this chapter we will assume that p is a prime number unless otherwise
stated.
Proposition 22.2 If F is a finite field of characteristic p, then the order of F is pn for
some n ∈ N.
Proof. Let ϕ : Z → F be the ring homomorphism defined by ϕ(n) = n · 1. Since the
characteristic of F is p, the kernel of ϕ must be pZ and the image of ϕ must be a subfield
of F isomorphic to Zp . We will denote this subfield by K. Since F is a finite field, it
must be a finite extension of K and, therefore, an algebraic extension of K. Suppose that
[F : K] = n is the dimension of F , where F is a K vector space. There must exist elements
α1 , . . . , αn ∈ F such that any element α in F can be written uniquely in the form
α = a1 α1 + · · · + an αn ,
where the ai ’s are in K. Since there are p elements in K, there are pn possible linear
combinations of the αi ’s. Therefore, the order of F must be pn . ■
Lemma 22.3 Freshman’s Dream. Let p be prime and D be an integral domain of
characteristic p. Then
n n n
ap + bp = (a + b)p
for all positive integers n.
278
CHAPTER 22. FINITE FIELDS 279
Proof. We will prove this lemma using mathematical induction on n. We can use the
binomial formula (see Chapter 2, Example 2.4) to verify the case for n = 1; that is,
p
X
p p k p−k
(a + b) = a b .
k
k=0
If 0 < k < p, then
p p!
=
k k!(p − k)!
must be divisible by p, since p cannot divide k!(p − k)!. Note that D is an integral domain
of characteristic p, so all but the first and last terms in the sum must be zero. Therefore,
(a + b)p = ap + bp .
Now suppose that the result holds for all k, where 1 ≤ k ≤ n. By the induction
hypothesis,
n+1 n n n n n+1 n+1
(a + b)p = ((a + b)p )p = (ap + bp )p = (ap )p + (bp )p = ap + bp .
Therefore, the lemma is true for n + 1 and the proof is complete. ■
Let F be a field. A polynomial f (x) ∈ F [x] of degree n is separable if it has n distinct
roots in the splitting field of f (x); that is, f (x) is separable when it factors into distinct
linear factors over the splitting field of f . An extension E of F is a separable extension of
F if every element in E is the root of a separable polynomial in F [x].
√ √
2 −2 is separable over Q since it factors as (x− 2 )(x+ 2 ).
Example 22.4√ The polynomial x √ √
In fact, Q( 2 ) is a separable extension of Q. Let α = a + b 2 be any element in Q( 2 ).
If b = 0, then α is a root of x − a. If b 6= 0, then α is the root of the separable polynomial
√ √
x2 − 2ax + a2 − 2b2 = (x − (a + b 2 ))(x − (a − b 2 )).
□
Fortunately, we have an easy test to determine the separability of any polynomial. Let
f (x) = a0 + a1 x + · · · + an xn
be any polynomial in F [x]. Define the derivative of f (x) to be
f ′ (x) = a1 + 2a2 x + · · · + nan xn−1 .
Lemma 22.5 Let F be a field and f (x) ∈ F [x]. Then f (x) is separable if and only if f (x)
and f ′ (x) are relatively prime.
Proof. Let f (x) be separable. Then f (x) factors over some extension field of F as
f (x) = (x − α1 )(x − α2 ) · · · (x − αn ), where αi 6= αj for i 6= j. Taking the derivative of f (x),
we see that
f ′ (x) = (x − α2 ) · · · (x − αn )
+ (x − α1 )(x − α3 ) · · · (x − αn )
+ · · · + (x − α1 ) · · · (x − αn−1 ).
Hence, f (x) and f ′ (x) can have no common factors.
To prove the converse, we will show that the contrapositive of the statement is true.
Suppose that f (x) = (x − α)k g(x), where k > 1. Differentiating, we have
f ′ (x) = k(x − α)k−1 g(x) + (x − α)k g ′ (x).
Therefore, f (x) and f ′ (x) have a common factor. ■
CHAPTER 22. FINITE FIELDS 280
Theorem 22.6 For every prime p and every positive integer n, there exists a finite field F
with pn elements. Furthermore, any field of order pn is isomorphic to the splitting field of
n
xp − x over Zp .
n
Proof. Let f (x) = xp − x and let F be the splitting field of f (x). Then by Lemma 22.5,
f (x) has pn distinct zeros in F , since f ′ (x) = pn xp −1 − 1 = −1 is relatively prime to
n
f (x). We claim that the roots of f (x) form a subfield of F . Certainly 0 and 1 are zeros
of f (x). If α and β are zeros of f (x), then α + β and αβ are also zeros of f (x), since
n n n n n n
αp + β p = (α + β)p and αp β p = (αβ)p . We also need to show that the additive inverse
and the multiplicative inverse of each root of f (x) are roots of f (x). For any zero α of f (x),
we know that −α is also a zero of f (x), since
n n n
f (−α) = (−α)p − (−α) = −αp + α = −(αp − α) = 0,
provided p is odd. If p = 2, then
n
f (−α) = (−α)2 − (−α) = α + α = 0.
If α 6= 0, then (α−1 )p = (αp )−1 = α−1 . Since the zeros of f (x) form a subfield of F and
n n
f (x) splits in this subfield, the subfield must be all of F .
Let E be any other field of order pn . To show that E is isomorphic to F , we must show
that every element in E is a root of f (x). Certainly 0 is a root of f (x). Let α be a nonzero
element of E. The order of the multiplicative group of nonzero elements of E is pn − 1;
hence, αp −1 = 1 or αp − α = 0. Since E contains pn elements, E must be a splitting field
n n
of f (x); however, by Corollary 21.36, the splitting field of any polynomial is unique up to
isomorphism. ■
n n
The unique finite field with p elements is called the Galois field of order p . We will
denote this field by GF(pn ).
Theorem 22.7 Every subfield of the Galois field GF(pn ) has pm elements, where m divides
n. Conversely, if m | n for m > 0, then there exists a unique subfield of GF(pn ) isomorphic
to GF(pm ).
Proof. Let F be a subfield of E = GF(pn ). Then F must be a field extension of K that
contains pm elements, where K is isomorphic to Zp . Then m | n, since [E : K] = [E : F ][F :
K].
To prove the converse, suppose that m | n for some m > 0. Then pm − 1 divides pn − 1.
Consequently, xp −1 − 1 divides xp −1 − 1. Therefore, xp − x must divide xp − x, and
m n m n
m n
every zero of xp − x is also a zero of xp − x. Thus, GF(pn ) contains, as a subfield, a
m
splitting field of xp − x, which must be isomorphic to GF(pm ). ■
Example 22.8 The lattice of subfields of GF(p24 ) is given in Figure 22.9. □
CHAPTER 22. FINITE FIELDS 281
GF(p24 )
GF(p8 ) GF(p12 )
GF(p4 ) GF(p6 )
GF(p2 ) GF(p3 )
GF(p)
Figure 22.9 Subfields of GF(p24 )
With each field F we have a multiplicative group of nonzero elements of F which we
will denote by F ∗ . The multiplicative group of any finite field is cyclic. This result follows
from the more general result that we will prove in the next theorem.
Theorem 22.10 If G is a finite subgroup of F ∗ , the multiplicative group of nonzero elements
of a field F , then G is cyclic.
Proof. Let G be a finite subgroup of F ∗ of order n. By the Fundamental Theorem of
Finite Abelian Groups (Theorem 13.4),
G∼
= Zpe1 × · · · × Zpek ,
1 k
where n = pe11 · · · pekk and the p1 , . . . , pk are (not necessarily distinct) primes. Let m be the
least common multiple of pe11 , . . . , pekk . Then G contains an element of order m. Since every
α in G satisfies xr − 1 for some r dividing m, α must also be a root of xm − 1. Since xm − 1
has at most m roots in F , n ≤ m. On the other hand, we know that m ≤ |G|; therefore,
m = n. Thus, G contains an element of order n and must be cyclic. ■
Corollary 22.11 The multiplicative group of all nonzero elements of a finite field is cyclic.
Corollary 22.12 Every finite extension E of a finite field F is a simple extension of F .
Proof. Let α be a generator for the cyclic group E ∗ of nonzero elements of E. Then
E = F (α). ■
Example 22.13 The finite field GF(24 ) is isomorphic to the field Z2 /h1+x+x4 i. Therefore,
the elements of GF(24 ) can be taken to be
{a0 + a1 α + a2 α2 + a3 α3 : ai ∈ Z2 and 1 + α + α4 = 0}.
Remembering that 1 + α + α4 = 0, we add and multiply elements of GF(24 ) exactly as we
add and multiply polynomials. The multiplicative group of GF(24 ) is isomorphic to Z15
with generator α:
α1 = α α6 = α2 + α3 α11 = α + α2 + α3
α2 = α2 α7 = 1 + α + α3 α12 = 1 + α + α2 + α3
α3 = α3 α8 = 1 + α2 α13 = 1 + α2 + α3
CHAPTER 22. FINITE FIELDS 282
α4 = 1 + α α9 = α + α3 α14 = 1 + α3
α5 = α + α2 α10 = 1 + α + α2 α15 = 1.
□
22.2 Polynomial Codes
With knowledge of polynomial rings and finite fields, it is now possible to derive more
sophisticated codes than those of Chapter 8. First let us recall that an (n, k)-block code
consists of a one-to-one encoding function E : Zk2 → Zn2 and a decoding function D : Zn2 →
Zk2 . The code is error-correcting if D is onto. A code is a linear code if it is the null space
of a matrix H ∈ Mk×n (Z2 ).
We are interested in a class of codes known as cyclic codes. Let ϕ : Zk2 → Zn2 be a binary
(n, k)-block code. Then ϕ is a cyclic code if for every codeword (a1 , a2 , . . . , an ), the cyclically
shifted n-tuple (an , a1 , a2 , . . . , an−1 ) is also a codeword. Cyclic codes are particularly easy
to implement on a computer using shift registers [2, 3].
Example 22.14 Consider the (6, 3)-linear codes generated by the two matrices
1 0 0 1 0 0
0 1 0 1 1 0
0 0 1 1 1 1
G1 = and G2 = .
1 0 0 1 1 1
0 1 0 0 1 1
0 0 1 0 0 1
Messages in the first code are encoded as follows:
(000) 7 → (000000) (100) 7→ (100100)
(001) 7→ (001001) (101) 7 → (101101)
(010) 7 → (010010) (110) 7 → (110110)
(011) 7 → (011011) (111) 7 → (111111).
It is easy to see that the codewords form a cyclic code. In the second code, 3-tuples are
encoded in the following manner:
(000) 7 → (000000) (100) 7→ (111100)
(001) 7→ (001111) (101) 7 → (110011)
(010) 7 → (011110) (110) 7 → (100010)
(011) 7 → (010001) (111) 7 → (101101).
This code cannot be cyclic, since (101101) is a codeword but (011011) is not a codeword.
□
Polynomial Codes
We would like to find an easy method of obtaining cyclic linear codes. To accomplish this, we
can use our knowledge of finite fields and polynomial rings over Z2 . Any binary n-tuple can
be interpreted as a polynomial in Z2 [x]. Stated another way, the n-tuple (a0 , a1 , . . . , an−1 )
corresponds to the polynomial
f (x) = a0 + a1 x + · · · + an−1 xn−1 ,
CHAPTER 22. FINITE FIELDS 283
where the degree of f (x) is at most n − 1. For example, the polynomial corresponding to
the 5-tuple (10011) is
1 + 0x + 0x2 + 1x3 + 1x4 = 1 + x3 + x4 .
Conversely, with any polynomial f (x) ∈ Z2 [x] with deg f (x) < n we can associate a binary
n-tuple. The polynomial x + x2 + x4 corresponds to the 5-tuple (01101).
Let us fix a nonconstant polynomial g(x) in Z2 [x] of degree n − k. We can define an
(n, k)-code C in the following manner. If (a0 , . . . , ak−1 ) is a k-tuple to be encoded, then
f (x) = a0 + a1 x + · · · + ak−1 xk−1 is the corresponding polynomial in Z2 [x]. To encode f (x),
we multiply by g(x). The codewords in C are all those polynomials in Z2 [x] of degree less
than n that are divisible by g(x). Codes obtained in this manner are called polynomial
codes.
Example 22.15 If we let g(x) = 1 + x3 , we can define a (6, 3)-code C as follows. To encode
a 3-tuple (a0 , a1 , a2 ), we multiply the corresponding polynomial f (x) = a0 + a1 x + a2 x2 by
1 + x3 . We are defining a map ϕ : Z32 → Z62 by ϕ : f (x) 7→ g(x)f (x). It is easy to check that
this map is a group homomorphism. In fact, if we regard Zn2 as a vector space over Z2 , ϕ is
a linear transformation of vector spaces (see Exercise 20.5.15, Chapter 20). Let us compute
the kernel of ϕ. Observe that ϕ(a0 , a1 , a2 ) = (000000) exactly when
0 + 0x + 0x2 + 0x3 + 0x4 + 0x5 = (1 + x3 )(a0 + a1 x + a2 x2 )
= a0 + a1 x + a2 x2 + a0 x3 + a1 x4 + a2 x5 .
Since the polynomials over a field form an integral domain, a0 + a1 x + a2 x2 must be the
zero polynomial. Therefore, ker ϕ = {(000)} and ϕ is one-to-one.
To calculate a generator matrix for C, we merely need to examine the way the polyno-
mials 1, x, and x2 are encoded:
(1 + x3 ) · 1 = 1 + x3
(1 + x3 )x = x + x4
(1 + x3 )x2 = x2 + x5 .
We obtain the code corresponding to the generator matrix G1 in Example 22.14. The
parity-check matrix for this code is
1 0 0 1 0 0
H = 0 1 0 0 1 0 .
0 0 1 0 0 1
Since the smallest weight of any nonzero codeword is 2, this code has the ability to detect
all single errors. □
Rings of polynomials have a great deal of structure; therefore, our immediate goal
is to establish a link between polynomial codes and ring theory. Recall that xn − 1 =
(x − 1)(xn−1 + · · · + x + 1). The factor ring
Rn = Z2 [x]/hxn − 1i
can be considered to be the ring of polynomials of the form
f (t) = a0 + a1 t + · · · + an−1 tn−1
CHAPTER 22. FINITE FIELDS 284
that satisfy the condition tn = 1. It is an easy exercise to show that Zn2 and Rn are isomor-
phic as vector spaces. We will often identify elements in Zn2 with elements in Z[x]/hxn − 1i.
In this manner we can interpret a linear code as a subset of Z[x]/hxn − 1i.
The additional ring structure on polynomial codes is very powerful in describing cyclic
codes. A cyclic shift of an n-tuple can be described by polynomial multiplication. If
f (t) = a0 + a1 t + · · · + an−1 tn−1 is a code polynomial in Rn , then
tf (t) = an−1 + a0 t + · · · + an−2 tn−1
is the cyclically shifted word obtained from multiplying f (t) by t. The following theorem
gives a beautiful classification of cyclic codes in terms of the ideals of Rn .
Theorem 22.16 A linear code C in Zn2 is cyclic if and only if it is an ideal in Rn =
Z[x]/hxn − 1i.
Proof. Let C be a linear cyclic code and suppose that f (t) is in C. Then tf (t) must also
be in C. Consequently, tk f (t) is in C for all k ∈ N. Since C is a linear code, any linear
combination of the codewords f (t), tf (t), t2 f (t), . . . , tn−1 f (t) is also a codeword; therefore,
for every polynomial p(t), p(t)f (t) is in C. Hence, C is an ideal.
Conversely, let C be an ideal in Z2 [x]/hxn + 1i. Suppose that f (t) = a0 + a1 t + · · · +
an−1 tn−1 is a codeword in C. Then tf (t) is a codeword in C; that is, (a1 , . . . , an−1 , a0 ) is
in C. ■
Theorem 22.16 tells us that knowing the ideals of Rn is equivalent to knowing the linear
cyclic codes in Zn2 . Fortunately, the ideals in Rn are easy to describe. The natural ring
homomorphism ϕ : Z2 [x] → Rn defined by ϕ[f (x)] = f (t) is a surjective homomorphism.
The kernel of ϕ is the ideal generated by xn − 1. By Theorem 16.34, every ideal C in Rn is
of the form ϕ(I), where I is an ideal in Z2 [x] that contains hxn − 1i. By Theorem 17.20, we
know that every ideal I in Z2 [x] is a principal ideal, since Z2 is a field. Therefore, I = hg(x)i
for some unique monic polynomial in Z2 [x]. Since hxn − 1i is contained in I, it must be the
case that g(x) divides xn − 1. Consequently, every ideal C in Rn is of the form
C = hg(t)i = {f (t)g(t) : f (t) ∈ Rn and g(x) | (xn − 1) in Z2 [x]}.
The unique monic polynomial of the smallest degree that generates C is called the minimal
generator polynomial of C.
Example 22.17 If we factor x7 − 1 into irreducible components, we have
x7 − 1 = (1 + x)(1 + x + x3 )(1 + x2 + x3 ).
We see that g(t) = (1 + t + t3 ) generates an ideal C in R7 . This code is a (7, 4)-block code.
As in Example 22.15, it is easy to calculate a generator matrix by examining what g(t) does
to the polynomials 1, t, t2 , and t3 . A generator matrix for C is
1 0 0 0
1 1 0 0
0 1 1 0
G = 1 0 1 1 .
0 1 0 1
0 0 1 0
0 0 0 1
□
CHAPTER 22. FINITE FIELDS 285
In general, we can determine a generator matrix for an (n, k)-code C by the manner in
which the elements tk are encoded. Let xn − 1 = g(x)h(x) in Z2 [x]. If g(x) = g0 + g1 x +
· · · + gn−k xn−k and h(x) = h0 + h1 x + · · · + hk xk , then the n × k matrix
g0 0 ··· 0
g1 g0 ··· 0
. .. .. ..
.. . .
.
G = gn−k gn−k−1 · · · g0
0 gn−k · · · g1
.. .. .. ..
. . . .
0 0 · · · gn−k
is a generator matrix for the code C with generator polynomial g(t). The parity-check
matrix for C is the (n − k) × n matrix
0 ··· 0 0 hk · · · h0
0 · · · 0 hk · · · h0 0
H= .
· · · · · · · · · · · · · · · · · · · · ·
hk · · · h0 0 0 ··· 0
We will leave the details of the proof of the following proposition as an exercise.
Proposition 22.18 Let C = hg(t)i be a cyclic code in Rn and suppose that xn −1 = g(x)h(x).
Then G and H are generator and parity-check matrices for C, respectively. Furthermore,
HG = 0.
Example 22.19 In Example 22.17,
x7 − 1 = g(x)h(x) = (1 + x + x3 )(1 + x + x2 + x4 ).
Therefore, a parity-check matrix for this code is
0 0 1 0 1 1 1
H = 0 1 0 1 1 1 0 .
1 0 1 1 1 0 0
□
To determine the error-detecting and error-correcting capabilities of a cyclic code, we
need to know something about determinants. If α1 , . . . , αn are elements in a field F , then
the n × n matrix
1 1 ··· 1
α1 α2 · · · αn
2
α1 α22 · · · αn2
.. .. .. ..
. . . .
α1n−1 α2n−1 · · · αnn−1
is called the Vandermonde matrix. The determinant of this matrix is called the Van-
dermonde determinant. We will need the following lemma in our investigation of cyclic
codes.
CHAPTER 22. FINITE FIELDS 286
Lemma 22.20 Let α1 , . . . , αn be elements in a field F with n ≥ 2. Then
1 1 ··· 1
α1 α2 · · · αn
2 Y
det
α1 α22 · · · αn2 =
(αi − αj ).
.. .. .. .. 1≤j<i≤n
. . . .
α1n−1 n−1
α2 · · · αn
n−1
In particular, if the αi ’s are distinct, then the determinant is nonzero.
Proof. We will induct on n. If n = 2, then the determinant is α2 − α1 . Let us assume the
result for n − 1 and consider the polynomial p(x) defined by
1 1 ··· 1 1
α1 α2 · · · αn−1 x
2
p(x) = det α22 · · · αn−1 x2 .
α1 2
.. .. .. .. ..
. . . . .
α1n−1 α2n−1 · · · αn−1
n−1
xn−1
Expanding this determinant by cofactors on the last column, we see that p(x) is a polynomial
of at most degree n − 1. Moreover, the roots of p(x) are α1 , . . . , αn−1 , since the substitution
of any one of these elements in the last column will produce a column identical to the last
column in the matrix. Remember that the determinant of a matrix is zero if it has two
identical columns. Therefore,
p(x) = (x − α1 )(x − α2 ) · · · (x − αn−1 )β,
where
1 1 ··· 1
α1 α2 · · · αn−1
β = (−1)n+n det α12 α22 · · · αn−1 .
2
.. .. .. ..
. . . .
α1n−2 α2n−2 · · · αn−1
n−2
By our induction hypothesis,
Y
β = (−1)n+n (αi − αj ).
1≤j<i≤n−1
If we let x = αn , the result now follows immediately. ■
The following theorem gives us an estimate on the error detection and correction capa-
bilities for a particular generator polynomial.
Theorem 22.21 Let C = hg(t)i be a cyclic code in Rn and suppose that ω is a primitive
nth root of unity over Z2 . If s consecutive powers of ω are roots of g(x), then the minimum
distance of C is at least s + 1.
Proof. Suppose that
g(ω r ) = g(ω r+1 ) = · · · = g(ω r+s−1 ) = 0.
Let f (x) be some polynomial in C with s or fewer nonzero coefficients. We can assume that
f (x) = ai0 xi0 + ai1 xi1 + · · · + ais−1 xis−1
CHAPTER 22. FINITE FIELDS 287
be some polynomial in C. It will suffice to show that all of the ai ’s must be 0. Since
g(ω r ) = g(ω r+1 ) = · · · = g(ω r+s−1 ) = 0
and g(x) divides f (x),
f (ω r ) = f (ω r+1 ) = · · · = f (ω r+s−1 ) = 0.
Equivalently, we have the following system of equations:
ai0 (ω r )i0 + ai1 (ω r )i1 + · · · + ais−1 (ω r )is−1 = 0
ai0 (ω r+1 )i0 + ai1 (ω r+1 )i2 + · · · + ais−1 (ω r+1 )is−1 = 0
..
.
ai0 (ω r+s−1 )i0 + ai1 (ω r+s−1 )i1 + · · · + ais−1 (ω r+s−1 )is−1 = 0.
Therefore, (ai0 , ai1 , . . . , ais−1 ) is a solution to the homogeneous system of linear equations
(ω i0 )r x0 + (ω i1 )r x1 + · · · + (ω is−1 )r xn−1 = 0
(ω i0 )r+1 x0 + (ω i1 )r+1 x1 + · · · + (ω is−1 )r+1 xn−1 = 0
..
.
(ω i0 )r+s−1 x0 + (ω i1 )r+s−1 x1 + · · · + (ω is−1 )r+s−1 xn−1 = 0.
However, this system has a unique solution, since the determinant of the matrix
(ω i0 )r (ω i1 )r ··· (ω is−1 )r
(ω i0 )r+1 (ω i1 )r+1 ··· (ω is−1 )r+1
.. .. .. ..
. . . .
(ω i0 )r+s−1 (ω i1 )r+s−1 · · · (ω is−1 )r+s−1
can be shown to be nonzero using Lemma 22.20 and the basic properties of determinants
(Exercise). Therefore, this solution must be ai0 = ai1 = · · · = ais−1 = 0. ■
BCH Codes
Some of the most important codes, discovered independently by A. Hocquenghem in 1959
and by R. C. Bose and D. V. Ray-Chaudhuri in 1960, are bch codes. The European
and transatlantic communication systems both use bch codes. Information words to be
encoded are of length 231, and a polynomial of degree 24 is used to generate the code.
Since 231 + 24 = 255 = 28 − 1, we are dealing with a (255, 231)-block code. This bch code
will detect six errors and has a failure rate of 1 in 16 million. One advantage of bch codes
is that efficient error correction algorithms exist for them.
The idea behind bch codes is to choose a generator polynomial of smallest degree that
has the largest error detection and error correction capabilities. Let d = 2r + 1 for some
r ≥ 0. Suppose that ω is a primitive nth root of unity over Z2 , and let mi (x) be the minimal
polynomial over Z2 of ω i . If
g(x) = lcm[m1 (x), m2 (x), . . . , m2r (x)],
then the cyclic code hg(t)i in Rn is called the bch code of length n and distance d. By
Theorem 22.21, the minimum distance of C is at least d.
CHAPTER 22. FINITE FIELDS 288
Theorem 22.22 Let C = hg(t)i be a cyclic code in Rn . The following statements are
equivalent.
1. The code C is a bch code whose minimum distance is at least d.
2. A code polynomial f (t) is in C if and only if f (ω i ) = 0 for 1 ≤ i < d.
3. The matrix
1 ω ω2 ··· ω n−1
1 ω2 ω4 ··· ω (n−1)(2)
H=
1 ω3 ω6 ··· ω (n−1)(3)
.. .. .. .. ..
. . . . .
1 ω 2r ω 4r ··· ω (n−1)(2r)
is a parity-check matrix for C.
Proof. (1) ⇒ (2). If f (t) is in C, then g(x) | f (x) in Z2 [x]. Hence, for i = 1, . . . , 2r,
f (ω i ) = 0 since g(ω i ) = 0. Conversely, suppose that f (ω i ) = 0 for 1 ≤ i ≤ d. Then f (x) is
divisible by each mi (x), since mi (x) is the minimal polynomial of ω i . Therefore, g(x) | f (x)
by the definition of g(x). Consequently, f (x) is a codeword.
(2) ⇒ (3). Let f (t) = a0 + a1 t + · · · + an−1 vtn−1 be in Rn . The corresponding n-tuple
in Zn2 is x = (a0 a1 · · · an−1 )t . By (2),
a0 + a1 ω + · · · + an−1 ω n−1 f (ω)
a0 + a1 ω 2 + · · · + an−1 (ω 2 )n−1 f (ω 2 )
Hx = .. = . =0
. ..
a0 + a1 ω 2r + · · · + an−1 (ω 2r )n−1 f (ω 2r )
exactly when f (t) is in C. Thus, H is a parity-check matrix for C.
(3) ⇒ (1). By (3), a code polynomial f (t) = a0 +a1 t+· · ·+an−1 tn−1 is in C exactly when
f (ω i ) = 0 for i = 1, . . . , 2r. The smallest such polynomial is g(t) = lcm[m1 (t), . . . , m2r (t)].
Therefore, C = hg(t)i. ■
Example 22.23 It is easy to verify that x15 − 1 ∈ Z2 [x] has a factorization
x15 − 1 = (x + 1)(x2 + x + 1)(x4 + x + 1)(x4 + x3 + 1)(x4 + x3 + x2 + x + 1),
where each of the factors is an irreducible polynomial. Let ω be a root of 1 + x + x4 . The
Galois field GF(24 ) is
{a0 + a1 ω + a2 ω 2 + a3 ω 3 : ai ∈ Z2 and 1 + ω + ω 4 = 0}.
By Example 22.8, ω is a primitive 15th root of unity. The minimal polynomial of ω is
m1 (x) = 1 + x + x4 . It is easy to see that ω 2 and ω 4 are also roots of m1 (x). The minimal
polynomial of ω 3 is m2 (x) = 1 + x + x2 + x3 + x4 . Therefore,
g(x) = m1 (x)m2 (x) = 1 + x4 + x6 + x7 + x8
has roots ω, ω 2 , ω 3 , ω 4 . Since both m1 (x) and m2 (x) divide x15 − 1, the bch code is a
(15, 7)-code. If x15 − 1 = g(x)h(x), then h(x) = 1 + x4 + x6 + x7 ; therefore, a parity-check
CHAPTER 22. FINITE FIELDS 289
matrix for this code is
0 0 0 0 0 0 0 1 1 0 1 0 0 0 1
0 0 0 0 0 0 1 1 0 1 0 0 0 1 0
0
0 0 0 0 1 1 0 1 0 0 0 1 0 0
0 0 0 0 1 1 0 1 0 0 0 1 0 0 0
.
0 0 0 1 1 0 1 0 0 0 1 0 0 0 0
0 0 1 1 0 1 0 0 0 1 0 0 0 0 0
0 1 1 0 1 0 0 0 1 0 0 0 0 0 0
1 1 0 1 0 0 0 1 0 0 0 0 0 0 0
□
Sage. Finite fields are important in a variety of applied disciplines, such as cryptography
and coding theory (see introductions to these topics in other chapters). Sage has excellent
support for finite fields allowing for a wide variety of computations.
22.3 Reading Questions
1. When is a field extension separable?
2. What are the possible orders for subfields of a finite field?
3. What is the structure of the non-zero elements of a finite field?
4. Provide a characterization of finite fields using the concept of a splitting field.
5. Why is a theorem in this chapter titled “The Freshman’s Dream?”
22.4 Exercises
1. Calculate each of the following.
(a) [GF(36 ) : GF(33 )] (c) [GF(625) : GF(25)]
(b) [GF(128) : GF(16)] (d) [GF(p12 ) : GF(p2 )]
2. Calculate [GF(pm ) : GF(pn )], where n | m.
3. What is the lattice of subfields for GF(p30 )?
4. Let α be a zero of x3 + x2 + 1 over Z2 . Construct a finite field of order 8. Show that
x3 + x2 + 1 splits in Z2 (α).
5. Construct a finite field of order 27.
6. Prove or disprove: Q∗ is cyclic.
7. Factor each of the following polynomials in Z2 [x].
(a) x5 − 1 (c) x9 − 1
(b) x6 + x5 + x4 + x3 + x2 + x + 1 (d) x4 + x3 + x2 + x + 1
8. Prove or disprove: Z2 [x]/hx3 + x + 1i ∼
= Z2 [x]/hx3 + x2 + 1i.
9. Determine the number of cyclic codes of length n for n = 6, 7, 8, 10.
10. Prove that the ideal ht + 1i in Rn is the code in Zn2 consisting of all words of even
parity.
11. Construct all bch codes of
(a) length 7. (b) length 15.
CHAPTER 22. FINITE FIELDS 290
12. Prove or disprove: There exists a finite field that is algebraically closed.
13. Let p be prime. Prove that the field of rational functions Zp (x) is an infinite field of
characteristic p.
n n n
14. Let D be an integral domain of characteristic p. Prove that (a − b)p = ap − bp for
all a, b ∈ D.
15. Show that every element in a finite field can be written as the sum of two squares.
16. Let E and F be subfields of a finite field K. If E is isomorphic to F , show that E = F .
17. Let F ⊂ E ⊂ K be fields. If K is a separable extension of F , show that K is also
separable extension of E.
18. Let E be an extension of a finite field F , where F has q elements. Let α ∈ E be
algebraic over F of degree n. Prove that F (α) has q n elements.
19. Show that every finite extension of a finite field F is simple; that is, if E is a finite
extension of a finite field F , prove that there exists an α ∈ E such that E = F (α).
20. Show that for every n there exists an irreducible polynomial of degree n in Zp [x].
21. Prove that the Frobenius map Φ : GF(pn ) → GF(pn ) given by Φ : α 7→ αp is an
automorphism of order n.
22. Show that every element in GF(pn ) can be written in the form ap for some unique
a ∈ GF(pn ).
23. Let E and F be subfields of GF(pn ). If |E| = pr and |F | = ps , what is the order of
E ∩ F?
24. Wilson’s Theorem. Let p be prime. Prove that (p − 1)! ≡ −1 (mod p).
25. If g(t) is the minimal generator polynomial for a cyclic code C in Rn , prove that the
constant term of g(x) is 1.
26. Often it is conceivable that a burst of errors might occur during transmission, as in
the case of a power surge. Such a momentary burst of interference might alter several
consecutive bits in a codeword. Cyclic codes permit the detection of such error bursts.
Let C be an (n, k)-cyclic code. Prove that any error burst up to n − k digits can be
detected.
27. Prove that the rings Rn and Zn2 are isomorphic as vector spaces.
28. Let C be a code in Rn that is generated by g(t). If hf (t)i is another code in Rn , show
that hg(t)i ⊂ hf (t)i if and only if f (x) divides g(x) in Z2 [x].
29. Let C = hg(t)i be a cyclic code in Rn and suppose that xn − 1 = g(x)h(x), where
g(x) = g0 + g1 x + · · · + gn−k xn−k and h(x) = h0 + h1 x + · · · + hk xk . Define G to be
the n × k matrix
g0 0 ··· 0
g1 g0 ··· 0
. .. .. ..
.. . .
.
G = gn−k gn−k−1 · · · g0
0 g · · · g
n−k 1
.. .. .. ..
. . . .
0 0 · · · gn−k
CHAPTER 22. FINITE FIELDS 291
and H to be the (n − k) × n matrix
0 ··· 0 0 hk · · · h0
0 · · · 0 hk · · · h0 0
H= .
· · · · · · · · · · · · · · · · · · · · ·
hk · · · h0 0 0 ··· 0
(a) Prove that G is a generator matrix for C.
(b) Prove that H is a parity-check matrix for C.
(c) Show that HG = 0.
22.5 Additional Exercises: Error Correction for BCH Codes
bch codes have very attractive error correction algorithms. Let C be a bch code in Rn ,
and suppose that a code polynomial c(t) = c0 + c1 t + · · · + cn−1 tn−1 is transmitted. Let
w(t) = w0 + w1 t + · · · wn−1 tn−1 be the polynomial in Rn that is received. If errors have
occurred in bits a1 , . . . , ak , then w(t) = c(t) + e(t), where e(t) = ta1 + ta2 + · · · + tak is the
error polynomial. The decoder must determine the integers ai and then recover c(t) from
w(t) by flipping the ai th bit. From w(t) we can compute w(ω i ) = si for i = 1, . . . , 2r, where
ω is a primitive nth root of unity over Z2 . We say the syndrome of w(t) is s1 , . . . , s2r .
1. Show that w(t) is a code polynomial if and only if si = 0 for all i.
2. Show that
si = w(ω i ) = e(ω i ) = ω ia1 + ω ia2 + · · · + ω iak
for i = 1, . . . , 2r. The error-locator polynomial is defined to be
s(x) = (x + ω a1 )(x + ω a2 ) · · · (x + ω ak ).
3. Recall the (15, 7)-block bch code in Example 22.19. By Theorem 8.13, this code is
capable of correcting two errors. Suppose that these errors occur in bits a1 and a2 .
The error-locator polynomial is s(x) = (x + ω a1 )(x + ω a2 ). Show that
2 2 s3
s(x) = x + s1 x + s1 + .
s1
4. Let w(t) = 1 + t2 + t4 + t5 + t7 + t12 + t13 . Determine what the originally transmitted
code polynomial was.
22.6 References and Suggested Readings
[1] Childs, L. A Concrete Introduction to Higher Algebra. 2nd ed. Springer-Verlag, New
York, 1995.
[2] Gåding, L. and Tambour, T. Algebra for Computer Science. Springer-Verlag, New
York, 1988.
[3] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998.
An excellent presentation of finite fields and their applications.
[4] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[5] Roman, S. Coding and Information Theory. Springer-Verlag, New York, 1992.
CHAPTER 22. FINITE FIELDS 292
[6] van Lint, J. H. Introduction to Coding Theory. Springer, New York, 1999.
23
Galois Theory
A classic problem of algebra is to find the solutions of a polynomial equation. The solution
to the quadratic equation was known in antiquity. Italian mathematicians found general
solutions to the general cubic and quartic equations in the sixteenth century; however,
attempts to solve the general fifth-degree, or quintic, polynomial were repulsed for the next
three hundred years. Certainly, equations such as x5 − 1 = 0 or x6 − x3 − 6 = 0 could be
solved, but no solution like the quadratic formula was found for the general quintic,
ax5 + bx4 + cx3 + dx2 + ex + f = 0.
Finally, at the beginning of the nineteenth century, Ruffini and Abel both found quintics
that could not be solved with any formula. It was Galois, however, who provided the full
explanation by showing which polynomials could and could not be solved by formulas. He
discovered the connection between groups and field extensions. Galois theory demonstrates
the strong interdependence of group and field theory, and has had far-reaching implications
beyond its original purpose.
In this chapter we will prove the Fundamental Theorem of Galois Theory. This result
will be used to establish the insolvability of the quintic and to prove the Fundamental
Theorem of Algebra.
23.1 Field Automorphisms
Our first task is to establish a link between group theory and field theory by examining
automorphisms of fields.
Proposition 23.1 The set of all automorphisms of a field F is a group under composition
of functions.
Proof. If σ and τ are automorphisms of F , then so are στ and σ −1 . The identity is
certainly an automorphism; hence, the set of all automorphisms of a field F is indeed a
group. ■
Proposition 23.2 Let E be a field extension of F . Then the set of all automorphisms of
E that fix F elementwise is a group; that is, the set of all automorphisms σ : E → E such
that σ(α) = α for all α ∈ F is a group.
Proof. We need only show that the set of automorphisms of E that fix F elementwise is
a subgroup of the group of all automorphisms of E. Let σ and τ be two automorphisms
of E such that σ(α) = α and τ (α) = α for all α ∈ F . Then στ (α) = σ(α) = α and
σ −1 (α) = α. Since the identity fixes every element of E, the set of automorphisms of E
that leave elements of F fixed is a subgroup of the entire group of automorphisms of E. ■
293
CHAPTER 23. GALOIS THEORY 294
Let E be a field extension of F . We will denote the full group of automorphisms of E
by Aut(E). We define the Galois group of E over F to be the group of automorphisms of
E that fix F elementwise; that is,
G(E/F ) = {σ ∈ Aut(E) : σ(α) = α for all α ∈ F }.
If f (x) is a polynomial in F [x] and E is the splitting field of f (x) over F , then we define
the Galois group of f (x) to be G(E/F ).
Example 23.3 Complex conjugation, defined by σ : a + bi 7→ a − bi, is an automorphism
of the complex numbers. Since
σ(a) = σ(a + 0i) = a − 0i = a,
the automorphism defined by complex conjugation must be in G(C/R). □
√ √ √ √
Example 23.4 Consider the fields Q ⊂ Q( 5 ) ⊂ Q( 3, 5 ). Then for a, b ∈ Q( 5 ),
√ √
σ(a + b 3 ) = a − b 3
√ √ √
is an automorphism of Q( 3, 5 ) leaving Q( 5 ) fixed. Similarly,
√ √
τ (a + b 5 ) = a − b 5
√ √ √
is an automorphism
√ √ of Q( 3, 5 ) leaving Q( 3 ) fixed. The automorphism µ = στ√ moves
√
both 3 and 5. It will soon be clear that {id, σ, τ, µ} is the Galois group of Q( 3, 5 )
over Q. The following table shows that this group is isomorphic to Z2 × Z2 .
id σ τ µ
id id σ τ µ
σ σ id µ τ
τ τ µ id σ
µ µ τ σ id
√ √ √ √ √
We may also regard the field Q( √3, √5 ) as a vector √ √over Q that has basis {1, 3, 5, 15 }.
space
It is no coincidence that |G(Q( 3, 5 )/Q)| = [Q( 3, 5 ) : Q)] = 4. □
Proposition 23.5 Let E be a field extension of F and f (x) be a polynomial in F [x]. Then
any automorphism in G(E/F ) defines a permutation of the roots of f (x) that lie in E.
Proof. Let
f (x) = a0 + a1 x + a2 x2 + · · · + an xn
and suppose that α ∈ E is a zero of f (x). Then for σ ∈ G(E/F ),
0 = σ(0)
= σ(f (α))
= σ(a0 + a1 α + a2 α2 + · · · + an αn )
= a0 + a1 σ(α) + a2 [σ(α)]2 + · · · + an [σ(α)]n ;
therefore, σ(α) is also a zero of f (x). ■
Let E be an algebraic extension of a field F . Two elements α, β ∈ E are √ conjugate over
F
√ if they have
√ the same minimal polynomial. For example, in the field Q( 2 ) the elements
2 and − 2 are conjugate over Q since they are both roots of the irreducible polynomial
x2 − 2.
CHAPTER 23. GALOIS THEORY 295
A converse of the last proposition exists. The proof follows directly from Lemma 21.32.
Proposition 23.6 If α and β are conjugate over F , there exists an isomorphism σ : F (α) →
F (β) such that σ is the identity when restricted to F .
Theorem 23.7 Let f (x) be a polynomial in F [x] and suppose that E is the splitting field
for f (x) over F . If f (x) has no repeated roots, then
|G(E/F )| = [E : F ].
Proof. We will use mathematical induction on [E : F ]. If [E : F ] = 1, then E = F and
there is nothing to show. If [E : F ] > 1, let f (x) = p(x)q(x), where p(x) is irreducible of
degree d. We may assume that d > 1; otherwise, f (x) splits over F and [E : F ] = 1. Let α
be a root of p(x). If ϕ : F (α) → E is any injective homomorphism, then ϕ(α) = β is a root
of p(x), and ϕ : F (α) → F (β) is a field automorphism. Since f (x) has no repeated roots,
p(x) has exactly d roots β ∈ E. By Proposition 23.5, there are exactly d isomorphisms
ϕ : F (α) → F (βi ) that fix F , one for each root β1 , . . . , βd of p(x) (see Figure 23.8).
ψ
E −−−−→ E
y y
ϕ
F (α) −−−−→ F (β)
y y
identity
F −−−−−→ F
Figure 23.8
Since E is a splitting field of f (x) over F , it is also a splitting field over F (α). Similarly,
E is a splitting field of f (x) over F (β). Since [E : F (α)] = [E : F ]/d, induction shows that
each of the d isomorphisms ϕ has exactly [E : F ]/d extensions, ψ : E → E, and we have
constructed [E : F ] isomorphisms that fix F . Finally, suppose that σ is any automorphism
fixing F . Then σ restricted to F (α) is ϕ for some ϕ : F (α) → F (β). ■
Corollary 23.9 Let F be a finite field with a finite extension E such that [E : F ] = k.
Then G(E/F ) is cyclic of order k.
Proof. Let p be the characteristic of E and F and assume that the orders of E and F are
pm and pn , respectively. Then nk = m. We can also assume that E is the splitting field of
m m
xp − x over a subfield of order p. Therefore, E must also be the splitting field of xp − x
over F . Applying Theorem 23.7, we find that |G(E/F )| = k.
To prove that G(E/F ) is cyclic, we must find a generator for G(E/F ). Let σ : E → E
n
be defined by σ(α) = αp . We claim that σ is the element in G(E/F ) that we are seeking.
We first need to show that σ is in Aut(E). If α and β are in E,
n n n
σ(α + β) = (α + β)p = αp + β p = σ(α) + σ(β)
by Lemma 22.3. Also, it is easy to show that σ(αβ) = σ(α)σ(β). Since σ is a nonzero
homomorphism of fields, it must be injective. It must also be onto, since E is a finite field.
n
We know that σ must be in G(E/F ), since F is the splitting field of xp − x over the base
field of order p. This means that σ leaves every element in F fixed. Finally, we must show
that the order of σ is k. By Theorem 23.7, we know that
nk m
σ k (α) = αp = αp = α
is the identity of G(E/F ). However, σ r cannot be the identity for 1 ≤ r < k; otherwise,
nr
xp − x would have pm roots, which is impossible. ■
CHAPTER 23. GALOIS THEORY 296
√ √
Example 23.10 We can now confirm that the Galois group of Q( 3, 5 ) over Q in Exam-
ple 23.4 √
is indeed
√ isomorphic to Z2 × Z2 . Certainly the√ √ H = {id, σ, τ, µ} is a subgroup
group
of G(Q( 3, 5 )/Q); however, H must be all of G(Q( 3, 5 )/Q), since
√ √ √ √
|H| = [Q( 3, 5 ) : Q] = |G(Q( 3, 5 )/Q)| = 4.
□
Example 23.11 Let us compute the Galois group of
f (x) = x4 + x3 + x2 + x + 1
over Q. We know that f (x) is irreducible by Exercise 17.5.20 in Chapter 17. Furthermore,
since (x − 1)f (x) = x5 − 1, we can use DeMoivre’s Theorem to determine that the roots of
f (x) are ω i , where i = 1, . . . , 4 and
ω = cos(2π/5) + i sin(2π/5).
Hence, the splitting field of f (x) must be Q(ω). We can define automorphisms σi of Q(ω) by
σi (ω) = ω i for i = 1, . . . , 4. It is easy to check that these are indeed distinct automorphisms
in G(Q(ω)/Q). Since
[Q(ω) : Q] = |G(Q(ω)/Q)| = 4,
the σi ’s must be all of G(Q(ω)/Q). Therefore, G(Q(ω)/Q) ∼
= Z4 since ω is a generator for
the Galois group. □
Separable Extensions
Many of the results that we have just proven depend on the fact that a polynomial f (x) in
F [x] has no repeated roots in its splitting field. It is evident that we need to know exactly
when a polynomial factors into distinct linear factors in its splitting field. Let E be the
splitting field of a polynomial f (x) in F [x]. Suppose that f (x) factors over E as
Y
r
f (x) = (x − α1 )n1 (x − α2 )n2 · · · (x − αr )nr = (x − αi )ni .
i=1
We define the multiplicity of a root αi of f (x) to be ni . A root with multiplicity 1 is called
a simple root. Recall that a polynomial f (x) ∈ F [x] of degree n is separable if it has n
distinct roots in its splitting field E. Equivalently, f (x) is separable if it factors into distinct
linear factors over E[x]. An extension E of F is a separable extension of F if every element
in E is the root of a separable polynomial in F [x]. Also recall that f (x) is separable if and
only if gcd(f (x), f ′ (x)) = 1 (Lemma 22.5).
Proposition 23.12 Let f (x) be an irreducible polynomial over F . If the characteristic of
F is 0, then f (x) is separable. If the characteristic of F is p and f (x) 6= g(xp ) for some
g(x) in F [x], then f (x) is also separable.
Proof. First assume that char F = 0. Since deg f ′ (x) < deg f (x) and f (x) is irreducible,
the only way gcd(f (x), f ′ (x)) 6= 1 is if f ′ (x) is the zero polynomial; however, this is impos-
sible in a field of characteristic zero. If char F = p, then f ′ (x) can be the zero polynomial if
every coefficient of f ′ (x) is a multiple of p. This can happen only if we have a polynomial
of the form f (x) = a0 + a1 xp + a2 x2p + · · · + an xnp . ■
Certainly extensions of a field F of the form F (α) are some of the easiest to study and
understand. Given a field extension E of F , the obvious question to ask is when it is possible
CHAPTER 23. GALOIS THEORY 297
to find an element α ∈ E such that E = F (α). In this case, α is called a primitive element.
We already know that primitive elements exist for certain extensions. For example,
√ √ √ √
Q( 3, 5 ) = Q( 3 + 5 )
and √ √ √
Q( 5, 5 i) = Q( 5 i).
3 6
Corollary 22.12 tells us that there exists a primitive element for any finite extension of a
finite field. The next theorem tells us that we can often find a primitive element.
Theorem 23.13 Primitive Element Theorem. Let E be a finite separable extension
of a field F . Then there exists an α ∈ E such that E = F (α).
Proof. We already know that there is no problem if F is a finite field. Suppose that E
is a finite extension of an infinite field. We will prove the result for F (α, β). The general
case easily follows when we use mathematical induction. Let f (x) and g(x) be the minimal
polynomials of α and β, respectively. Let K be the field in which both f (x) and g(x) split.
Suppose that f (x) has zeros α = α1 , . . . , αn in K and g(x) has zeros β = β1 , . . . , βm in K.
All of these zeros have multiplicity 1, since E is separable over F . Since F is infinite, we
can find an a in F such that
αi − α
a 6=
β − βj
for all i and j with j 6= 1. Therefore, a(β − βj ) 6= αi − α. Let γ = α + aβ. Then
γ = α + aβ 6= αi + aβj ;
hence, γ − aβj 6= αi for all i, j with j 6= 1. Define h(x) ∈ F (γ)[x] by h(x) = f (γ − ax).
Then h(β) = f (α) = 0. However, h(βj ) 6= 0 for j 6= 1. Hence, h(x) and g(x) have a single
common factor in F (γ)[x]; that is, the minimal polynomial of β over F (γ) must be linear,
since β is the only zero common to both g(x) and h(x). So β ∈ F (γ) and α = γ − aβ is in
F (γ). Hence, F (α, β) = F (γ). ■
23.2 The Fundamental Theorem
The goal of this section is to prove the Fundamental Theorem of Galois Theory. This
theorem explains the connection between the subgroups of G(E/F ) and the intermediate
fields between E and F .
Proposition 23.14 Let {σi : i ∈ I} be a collection of automorphisms of a field F . Then
F{σi } = {a ∈ F : σi (a) = a for all σi }
is a subfield of F .
Proof. Let σi (a) = a and σi (b) = b. Then
σi (a ± b) = σi (a) ± σi (b) = a ± b
and
σi (ab) = σi (a)σi (b) = ab.
If a 6= 0, then σi (a−1 ) = [σi (a)]−1 = a−1 . Finally, σi (0) = 0 and σi (1) = 1 since σi is an
automorphism. ■
Corollary 23.15 Let F be a field and let G be a subgroup of Aut(F ). Then
FG = {α ∈ F : σ(α) = α for all σ ∈ G}
CHAPTER 23. GALOIS THEORY 298
is a subfield of F .
The subfield F{σi } of F is called the fixed field of {σi }. The field fixed by a subgroup G
of Aut(F ) will be denoted by FG .
√ √ √ √ √
Example
√ 23.16√ Let σ : Q( 3, 5 ) → Q(
√ √ 3, 5 ) be the automorphism that maps 3 to
− 3. Then Q( 5 ) is the subfield of Q( 3, 5 ) left fixed by σ. □
Proposition 23.17 Let E be a splitting field over F of a separable polynomial. Then
EG(E/F ) = F .
Proof. Let G = G(E/F ). Clearly, F ⊂ EG ⊂ E. Also, E must be a splitting field of EG
and G(E/F ) = G(E/EG ). By Theorem 23.7,
|G| = [E : EG ] = [E : F ].
Therefore, [EG : F ] = 1. Consequently, EG = F . ■
A large number of mathematicians first learned Galois theory from Emil Artin’s mono-
graph on the subject [1]. The very clever proof of the following lemma is due to Artin.
Lemma 23.18 Let G be a finite group of automorphisms of E and let F = EG . Then
[E : F ] ≤ |G|.
Proof. Let |G| = n. We must show that any set of n + 1 elements α1 , . . . , αn+1 in E is
linearly dependent over F ; that is, we need to find elements ai ∈ F , not all zero, such that
a1 α1 + a2 α2 + · · · + an+1 αn+1 = 0.
Suppose that σ1 = id, σ2 , . . . , σn are the automorphisms in G. The homogeneous system of
linear equations
σ1 (α1 )x1 + σ1 (α2 )x2 + · · · + σ1 (αn+1 )xn+1 = 0
σ2 (α1 )x1 + σ2 (α2 )x2 + · · · + σ2 (αn+1 )xn+1 = 0
..
.
σn (α1 )x1 + σn (α2 )x2 + · · · + σn (αn+1 )xn+1 = 0
has more unknowns than equations. From linear algebra we know that this system has a
nontrivial solution, say xi = ai for i = 1, 2, . . . , n + 1. Since σ1 is the identity, the first
equation translates to
a1 α1 + a2 α2 + · · · + an+1 αn+1 = 0.
The problem is that some of the ai ’s may be in E but not in F . We must show that this is
impossible.
Suppose that at least one of the ai ’s is in E but not in F . By rearranging the αi ’s we
may assume that a1 is nonzero. Since any nonzero multiple of a solution is also a solution,
we can also assume that a1 = 1. Of all possible solutions fitting this description, we choose
the one with the smallest number of nonzero terms. Again, by rearranging α2 , . . . , αn+1 if
necessary, we can assume that a2 is in E but not in F . Since F is the subfield of E that
is fixed elementwise by G, there exists a σi in G such that σi (a2 ) 6= a2 . Applying σi to
each equation in the system, we end up with the same homogeneous system, since G is a
group. Therefore, x1 = σi (a1 ) = 1, x2 = σi (a2 ), . . ., xn+1 = σi (an+1 ) is also a solution of
the original system. We know that a linear combination of two solutions of a homogeneous
system is also a solution; consequently,
x1 = 1 − 1 = 0
x2 = a2 − σi (a2 )
CHAPTER 23. GALOIS THEORY 299
..
.
xn+1 = an+1 − σi (an+1 )
must be another solution of the system. This is a nontrivial solution because σi (a2 ) 6= a2 ,
and has fewer nonzero entries than our original solution. This is a contradiction, since the
number of nonzero solutions to our original solution was assumed to be minimal. We can
therefore conclude that a1 , . . . , an+1 ∈ F . ■
Let E be an algebraic extension of F . If every irreducible polynomial in F [x] with a
root in E has all of its roots in E, then E is called a normal extension of F ; that is, every
irreducible polynomial in F [x] containing a root in E is the product of linear factors in
E[x].
Theorem 23.19 Let E be a field extension of F . Then the following statements are
equivalent.
1. E is a finite, normal, separable extension of F .
2. E is a splitting field over F of a separable polynomial.
3. F = EG for some finite group G of automorphisms of E.
Proof. (1) ⇒ (2). Let E be a finite, normal, separable extension of F . By the Primitive
Element Theorem, we can find an α in E such that E = F (α). Let f (x) be the minimal
polynomial of α over F . The field E must contain all of the roots of f (x) since it is a normal
extension F ; hence, E is a splitting field for f (x).
(2) ⇒ (3). Let E be the splitting field over F of a separable polynomial. By Proposi-
tion 23.17, EG(E/F ) = F . Since |G(E/F )| = [E : F ], this is a finite group.
(3) ⇒ (1). Let F = EG for some finite group of automorphisms G of E. Since [E : F ] ≤
|G|, E is a finite extension of F . To show that E is a finite, normal extension of F , let
f (x) ∈ F [x] be an irreducible monic polynomial that has a root α in E. We must show that
f (x) is the product of distinct linear factors in E[x]. By Proposition 23.5, automorphisms in
G permute the roots of f (x) lying in E. Hence, Q if we let G act on α, we can obtain distinct
roots α1 = α, α2 , . . . , αn in E. Let g(x) = ni=1 (x − αi ). Then g(x) is separable over F
and g(α) = 0. Any automorphism σ in G permutes the factors of g(x) since it permutes
these roots; hence, when σ acts on g(x), it must fix the coefficients of g(x). Therefore,
the coefficients of g(x) must be in F . Since deg g(x) ≤ deg f (x) and f (x) is the minimal
polynomial of α, f (x) = g(x). ■
Corollary 23.20 Let K be a field extension of F such that F = KG for some finite group
of automorphisms G of K. Then G = G(K/F ).
Proof. Since F = KG , G is a subgroup of G(K/F ). Hence,
[K : F ] ≤ |G| ≤ |G(K/F )| = [K : F ].
It follows that G = G(K/F ), since they must have the same order. ■
Before we determine the exact correspondence between field extensions and automor-
phisms of fields, let us return to a familiar example.
√ √
Example 23.21 In Example 23.4 we examined the automorphisms of Q( 3, 5 ) fixing Q.
√ compares the lattice of field extensions of Q with the lattice of subgroups of
Figure√23.22
G(Q( 3, 5 )/Q). The Fundamental Theorem of Galois Theory tells us what the relation-
ship is between the two lattices. □
CHAPTER 23. GALOIS THEORY 300
√ √
{id, σ, τ, µ} Q( 3, 5 )
√ √ √
{id, σ} {id, τ } {id, µ} Q( 3 ) Q( 5 ) Q( 15 )
{id} Q
√ √
Figure 23.22 G(Q( 3, 5 )/Q)
We are now ready to state and prove the Fundamental Theorem of Galois Theory.
Theorem 23.23 Fundamental Theorem of Galois Theory. Let F be a finite field
or a field of characteristic zero. If E is a finite normal extension of F with Galois group
G(E/F ), then the following statements are true.
1. The map K 7→ G(E/K) is a bijection of subfields K of E containing F with the
subgroups of G(E/F ).
2. If F ⊂ K ⊂ E, then
[E : K] = |G(E/K)| and [K : F ] = [G(E/F ) : G(E/K)].
3. F ⊂ K ⊂ L ⊂ E if and only if {id} ⊂ G(E/L) ⊂ G(E/K) ⊂ G(E/F ).
4. K is a normal extension of F if and only if G(E/K) is a normal subgroup of G(E/F ).
In this case
G(K/F ) ∼= G(E/F )/G(E/K).
Proof. (1) Suppose that G(E/K) = G(E/L) = G. Both K and L are fixed fields of
G; hence, K = L and the map defined by K 7→ G(E/K) is one-to-one. To show that
the map is onto, let G be a subgroup of G(E/F ) and K be the field fixed by G. Then
F ⊂ K ⊂ E; consequently, E is a normal extension of K. Thus, G(E/K) = G and the map
K 7→ G(E/K) is a bijection.
(2) By Theorem Theorem 23.7, |G(E/K)| = [E : K]; therefore,
|G(E/F )| = [G(E/F ) : G(E/K)] · |G(E/K)| = [E : F ] = [E : K][K : F ].
Thus, [K : F ] = [G(E/F ) : G(E/K)].
Statement (3) is illustrated in Figure 23.24. We leave the proof of this property as an
exercise.
(4) This part takes a little more work. Let K be a normal extension of F . If σ is in
G(E/F ) and τ is in G(E/K), we need to show that σ −1 τ σ is in G(E/K); that is, we need
to show that σ −1 τ σ(α) = α for all α ∈ K. Suppose that f (x) is the minimal polynomial of
α over F . Then σ(α) is also a root of f (x) lying in K, since K is a normal extension of F .
Hence, τ (σ(α)) = σ(α) or σ −1 τ σ(α) = α.
Conversely, let G(E/K) be a normal subgroup of G(E/F ). We need to show that
F = KG(K/F ) . Let τ ∈ G(E/K). For all σ ∈ G(E/F ) there exists a τ ∈ G(E/K) such that
τ σ = στ . Consequently, for all α ∈ K
τ (σ(α)) = σ(τ (α)) = σ(α);
CHAPTER 23. GALOIS THEORY 301
hence, σ(α) must be in the fixed field of G(E/K). Let σ be the restriction of σ to K. Then
σ is an automorphism of K fixing F , since σ(α) ∈ K for all α ∈ K; hence, σ ∈ G(K/F ).
Next, we will show that the fixed field of G(K/F ) is F . Let β be an element in K that
is fixed by all automorphisms in G(K/F ). In particular, σ(β) = β for all σ ∈ G(E/F ).
Therefore, β belongs to the fixed field F of G(E/F ).
Finally, we must show that when K is a normal extension of F ,
G(K/F ) ∼
= G(E/F )/G(E/K).
For σ ∈ G(E/F ), let σK be the automorphism of K obtained by restricting σ to K. Since K
is a normal extension, the argument in the preceding paragraph shows that σK ∈ G(K/F ).
Consequently, we have a map ϕ : G(E/F ) → G(K/F ) defined by σ 7→ σK . This map is a
group homomorphism since
ϕ(στ ) = (στ )K = σK τK = ϕ(σ)ϕ(τ ).
The kernel of ϕ is G(E/K). By (2),
|G(E/F )|/|G(E/K)| = [K : F ] = |G(K/F )|.
Hence, the image of ϕ is G(K/F ) and ϕ is onto. Applying the First Isomorphism Theorem,
we have
G(K/F ) ∼= G(E/F )/G(E/K).
■
E −−−−→ {id}
x
y
L −−−−→ G(E/L)
x
y
K −−−−→ G(E/K)
x
y
F −−−−→ G(E/F )
Figure 23.24 Subgroups of G(E/F ) and subfields of E
Example 23.25 In this example we will illustrate the Fundamental Theorem of Galois
Theory by determining the lattice of subgroups of the Galois group of f (x) = x4 − 2. We
will compare this lattice to the lattice of field extensions √ of Q that are contained in the
splitting field of√x4 − 2. The
√ splitting field of f (x) is Q( 4
2, i).
√ To see this,
√ notice that f (x)
factors as√(x + 2 )(x − 2 ); hence, the roots of f (x) are ±√ 2 and ± 2 i. We first adjoin
2 2 4 4
the root 4 √ 2 to Q and then
√ adjoin the root i of x2 + 1 to Q( 4 2 ). The splitting field of f (x)
is then Q( 2√4
)(i) = Q( 2, i).
4
√ √
Since [Q( 2 ) : Q] =√ 4 and i is not in Q( 4 2 ), it must be the case that [Q( 4 2, i) :
√
4
Q( 4 2 )] = 2. Hence, [Q( 4 2, i) : Q] = 8. The set
√ √ √ √ √ √
{1, 2, ( 2 )2 , ( 2 )3 , i, i 2, i( 2 )2 , i( 2 )3 }
4 4 4 4 4 4
√ √
is a basis of Q( 4 2, i) over Q. The lattice of field extensions of Q contained in Q( 4 2, i) is
illustrated in Figure 23.26(a).
The Galois group G of f (x) must be of order 8. Let σ be the automorphism defined by
CHAPTER 23. GALOIS THEORY 302
√ √
σ( 4 2 ) = i 4 2 and σ(i) = i, and τ be the automorphism defined by complex conjugation;
that is, τ (i) = −i. Then G has an element of order 4 and an element of order 2. It is easy
to verify by direct computation that the elements of G are {id, σ, σ 2 , σ 3 , τ, στ, σ 2 τ, σ 3 τ }
and that the relations τ 2 = id, σ 4 = id, and τ στ = σ −1 are satisfied; hence, G must be
isomorphic to D4 . The lattice of subgroups of G is illustrated in Figure 23.26(b). □
√
Q( 4 2, i)
√ √ √ √ √
Q( 4 2 ) Q( 4 2 i) Q( 2, i) Q((1 + i) 4 2 ) Q((1 − i) 4 2 )
√ √
Q( 2 ) Q(i) Q( 2 i)
Q (a)
D4
{id, σ 2 , τ, σ 2 τ } {id, σ, σ 2 , σ 3 }{id, σ 2 , στ, σ 3 τ }
{id, τ } {id, σ 2 τ } {id, σ 2 } {id, στ } {id, σ 3 τ }
{id} (b)
Figure 23.26 Galois group of x4 − 2
Historical Note
Solutions for the cubic and quartic equations were discovered in the 1500s. Attempts to
find solutions for the quintic equations puzzled some of history’s best mathematicians. In
1798, P. Ruffini submitted a paper that claimed no such solution could be found; however,
the paper was not well received. In 1826, Niels Henrik Abel (1802–1829) finally offered the
first correct proof that quintics are not always solvable by radicals.
Abel inspired the work of Évariste Galois. Born in 1811, Galois began to display extraor-
dinary mathematical talent at the age of 14. He applied for entrance to the École Poly-
technique several times; however, he had great difficulty meeting the formal entrance re-
CHAPTER 23. GALOIS THEORY 303
quirements, and the examiners failed to recognize his mathematical genius. He was finally
accepted at the École Normale in 1829.
Galois worked to develop a theory of solvability for polynomials. In 1829, at the age of
17, Galois presented two papers on the solution of algebraic equations to the Académie des
Sciences de Paris. These papers were sent to Cauchy, who subsequently lost them. A third
paper was submitted to Fourier, who died before he could read the paper. Another paper
was presented, but was not published until 1846.
Galois’ democratic sympathies led him into the Revolution of 1830. He was expelled from
school and sent to prison for his part in the turmoil. After his release in 1832, he was drawn
into a duel possibly over a love affair. Certain that he would be killed, he spent the evening
before his death outlining his work and his basic ideas for research in a long letter to his
friend Chevalier. He was indeed dead the next day, at the age of 20.
23.3 Applications
Solvability by Radicals
Throughout this section we shall assume that all fields have characteristic zero to ensure
that irreducible polynomials do not have multiple roots. The immediate goal of this section
is to determine when the roots of a polynomial f (x) can be computed with a finite number of
operations on the coefficients of f (x). The allowable operations are addition, subtraction,
multiplication, division, and the extraction of nth roots. Certainly the solution to the
quadratic equation, ax2 + bx + c = 0, illustrates this process:
√
−b ± b2 − 4ac
x= .
2a
The only one of these operations that might demand a larger field is the taking of nth roots.
We are led to the following definition.
An extension field E of a field F is an extension by radicals if there exists a chain of
subfields
F = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fr = E
such for i = 1, 2, . . . , r, we have Fi = Fi−1 (αi ) and αini ∈ Fi−1 for some positive integer ni .
A polynomial f (x) is solvable by radicals over F if the splitting field K of f (x) over F is
contained in an extension of F by radicals. Our goal is to arrive at criteria that will tell
us whether or not a polynomial f (x) is solvable by radicals by examining the Galois group
f (x).
The easiest polynomial to solve by radicals is one of the form xn − a. As we discussed
in Chapter 4, the roots of xn − 1 are called the nth roots of unity. These roots are a finite
subgroup of the splitting field of xn − 1. By Corollary 22.11, the nth roots of unity form a
cyclic group. Any generator of this group is called a primitive nth root of unity.
Example 23.27 The polynomial xn − 1 is solvable by radicals over Q. The roots of this
polynomial are 1, ω, ω 2 , . . . , ω n−1 , where
2π 2π
ω = cos + i sin .
n n
The splitting field of xn − 1 over Q is Q(ω). □
We shall prove that a polynomial is solvable by radicals if its Galois group is solvable.
Recall that a subnormal series of a group G is a finite sequence of subgroups
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e},
CHAPTER 23. GALOIS THEORY 304
where Hi is normal in Hi+1 . A group G is solvable if it has a subnormal series {Hi } such
that all of the factor groups Hi+1 /Hi are abelian. For example, if we examine the series
{id} ⊂ A3 ⊂ S3 , we see that S3 is solvable. On the other hand, S5 is not solvable, by
Theorem 10.11.
Lemma 23.28 Let F be a field of characteristic zero and E be the splitting field of xn − a
over F with a ∈ F . Then G(E/F ) is a solvable group.
√ √ √
Proof. The roots of xn − a are n a, ω n a, . . . , ω n−1 n a, where ω is a primitive nth root
of unity. Suppose that F contains all of its nth roots of unity. If ζ is one of the roots of
xn − a, then distinct roots of xn − a are ζ, ωζ, . . . , ω n−1 ζ, and E = F (ζ). Since G(E/F )
permutes the roots xn − a, the elements in G(E/F ) must be determined by their action on
these roots. Let σ and τ be in G(E/F ) and suppose that σ(ζ) = ω i ζ and τ (ζ) = ω j ζ. If F
contains the roots of unity, then
στ (ζ) = σ(ω j ζ) = ω j σ(ζ) = ω i+j ζ = ω i τ (ζ) = τ (ω i ζ) = τ σ(ζ).
Therefore, στ = τ σ and G(E/F ) is abelian, and G(E/F ) must be solvable.
Now suppose that F does not contain a primitive nth root of unity. Let ω be a generator
of the cyclic group of the nth roots of unity. Let α be a zero of xn − a. Since α and ωα
are both in the splitting field of xn − a, ω = (ωα)/α is also in E. Let K = F (ω). Then
F ⊂ K ⊂ E. Since K is the splitting field of xn − 1, K is a normal extension of F .
Therefore, any automorphism σ in G(F (ω)/F ) is determined by σ(ω). It must be the case
that σ(ω) = ω i for some integer i since all of the zeros of xn −1 are powers of ω. If τ (ω) = ω j
is in G(F (ω)/F ), then
στ (ω) = σ(ω j ) = [σ(ω)]j = ω ij = [τ (ω)]i = τ (ω i ) = τ σ(ω).
Therefore, G(F (ω)/F ) is abelian. By the Fundamental Theorem of Galois Theory the series
{id} ⊂ G(E/F (ω)) ⊂ G(E/F )
is a normal series. By our previous argument, G(E/F (ω)) is abelian. Since
G(E/F )/G(E/F (ω)) ∼
= G(F (ω)/F )
is also abelian, G(E/F ) is solvable. ■
Lemma 23.29 Let F be a field of characteristic zero and let
F = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fr = E
a radical extension of F . Then there exists a normal radical extension
F = K0 ⊂ K1 ⊂ K2 ⊂ · · · ⊂ Kr = K
such that K that contains E and Ki is a normal extension of Ki−1 .
Proof. Since E is a radical extension of F , there exists a chain of subfields
F = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fr = E
such for i = 1, 2, . . . , r, we have Fi = Fi−1 (αi ) and αini ∈ Fi−1 for some positive integer ni .
We will build a normal radical extension of F ,
F = K0 ⊂ K1 ⊂ K2 ⊂ · · · ⊂ Kr = K
CHAPTER 23. GALOIS THEORY 305
such that K ⊇ E. Define K1 for be the splitting field of xn1 − α1n1 . The roots of this
polynomial are α1 , α1 ω, α1 ω 2 , . . . , α1 ω n1 −1 , where ω is a primitive n1 th root of unity. If F
contains all of its n1 roots of unity, then K1 = F (α1 ). On the other hand, suppose that
F does not contain a primitive n1 th root of unity. If β is a root of xn1 − α1n1 , then all of
the roots of xn1 − α1n1 must be β, ωβ, . . . , ω n1 −1 , where ω is a primitive n1 th root of unity.
In this case, K1 = F (ωβ). Thus, K1 is a normal radical extension of F containing F1 .
Continuing in this manner, we obtain
F = K0 ⊂ K1 ⊂ K2 ⊂ · · · ⊂ Kr = K
such that Ki is a normal extension of Ki−1 and Ki ⊇ Fi for i = 1, 2, . . . , r. ■
We will now prove the main theorem about solvability by radicals.
Theorem 23.30 Let f (x) be in F [x], where char F = 0. If f (x) is solvable by radicals,
then the Galois group of f (x) over F is solvable.
Proof. Since f (x) is solvable by radicals there exists an extension E of F by radicals
F = F0 ⊂ F1 ⊂ · · · ⊂ Fn = E. By Lemma 23.29, we can assume that E is a splitting field
f (x) and Fi is normal over Fi−1 . By the Fundamental Theorem of Galois Theory, G(E/Fi )
is a normal subgroup of G(E/Fi−1 ). Therefore, we have a subnormal series of subgroups of
G(E/F ):
{id} ⊂ G(E/Fn−1 ) ⊂ · · · ⊂ G(E/F1 ) ⊂ G(E/F ).
Again by the Fundamental Theorem of Galois Theory, we know that
G(E/Fi−1 )/G(E/Fi ) ∼
= G(Fi /Fi−1 ).
By Lemma 23.28, G(Fi /Fi−1 ) is solvable; hence, G(E/F ) is also solvable. ■
The converse of Theorem 23.30 is also true. For a proof, see any of the references at the
end of this chapter.
Insolvability of the Quintic
We are now in a position to find a fifth-degree polynomial that is not solvable by radicals.
We merely need to find a polynomial whose Galois group is S5 . We begin by proving a
lemma.
Lemma 23.31 If p is prime, then any subgroup of Sp that contains a transposition and a
cycle of length p must be all of Sp .
Proof. Let G be a subgroup of Sp that contains a transposition σ and τ a cycle of length
p. We may assume that σ = (1 2). The order of τ is p and τ n must be a cycle of length p for
1 ≤ n < p. Therefore, we may assume that µ = τ n = (1, 2, i3 , . . . , ip ) for some n, where 1 ≤
n < p (see Exercise 5.4.13 in Chapter 5). Noting that (1 2)(1, 2, i3 , . . . , ip ) = (2, i3 , . . . , ip )
and (2, i3 , . . . , ip )k (1 2)(2, i3 , . . . , ip )−k = (1 ik ), we can obtain all the transpositions of the
form (1n) for 1 ≤ n < p. However, these transpositions generate all transpositions in Sp ,
since (1 j)(1 i)(1 j) = (i j). The transpositions generate Sp . ■
CHAPTER 23. GALOIS THEORY 306
y
60
f(x) = x5 −6x3 −27x−3
40
20
x
-3 -2 -1 1 2 3
-20
-40
-60
Figure 23.32 The graph of f (x) = x5 − 6x3 − 27x − 3
Example 23.33 We will show that f (x) = x5 −6x3 −27x−3 ∈ Q[x] is not solvable. We claim
that the Galois group of f (x) over Q is S5 . By Eisenstein’s Criterion, f (x) is irreducible
and, therefore, must be separable. The derivative of f (x) is f ′ (x) = 5x4 − 18x2 − 27; hence,
setting f ′ (x) = 0 and solving, we find that the only real roots of f ′ (x) are
s √
6 6+9
x=± .
5
Therefore, f (x) can have at most one maximum and one minimum. It is easy to show that
f (x) changes sign between −3 and −2, between −2 and 0, and once again between 0 and
4 (Figure 23.32). Therefore, f (x) has exactly three distinct real roots. The remaining two
roots of f (x) must be complex conjugates. Let K be the splitting field of f (x). Since f (x)
has five distinct roots in K and every automorphism of K fixing Q is determined by the
way it permutes the roots of f (x), we know that G(K/Q) is a subgroup of S5 . Since f is
irreducible, there is an element in σ ∈ G(K/Q) such that σ(a) = b for two roots a and b
of f (x). The automorphism of C that takes a + bi 7→ a − bi leaves the real roots fixed and
interchanges the complex roots; consequently, G(K/Q) contains a transposition. If α is one
of the real roots of f (x), then [Q(α) : Q] = 5 by Exercise 21.5.28. Since Q(α) is a subfield
of K, it must be the case that [K : Q] is divisible by 5. Since [K : Q] = |G(K/Q)| and
G(K/Q) ⊂ S5 , we know that G(K/Q) contains a cycle of length 5. By Lemma 23.31, S5 is
generated by a transposition and an element of order 5; therefore, G(K/Q) must be all of
CHAPTER 23. GALOIS THEORY 307
S5 . By Theorem 10.11, S5 is not solvable. Consequently, f (x) cannot be solved by radicals.
□
The Fundamental Theorem of Algebra
It seems fitting that the last theorem that we will state and prove is the Fundamental
Theorem of Algebra. This theorem was first proven by Gauss in his doctoral thesis. Prior
to Gauss’s proof, mathematicians suspected that there might exist polynomials over the real
and complex numbers having no solutions. The Fundamental Theorem of Algebra states
that every polynomial over the complex numbers factors into distinct linear factors.
Theorem 23.34 Fundamental Theorem of Algebra. The field of complex numbers is
algebraically closed; that is, every polynomial in C[x] has a root in C.
Proof. Suppose that E is a proper finite field extension of the complex numbers. Since
any finite extension of a field of characteristic zero is a simple extension, there exists an
α ∈ E such that E = C(α) with α the root of an irreducible polynomial f (x) in C[x]. The
splitting field L of f (x) is a finite normal separable extension of C that contains E. We
must show that it is impossible for L to be a proper extension of C.
Suppose that L is a proper extension of C. Since L is the splitting field of f (x) = (x2 +1)
over R, L is a finite normal separable extension of R. Let K be the fixed field of a Sylow
2-subgroup G of G(L/R). Then L ⊃ K ⊃ R and |G(L/K)| = [L : K]. Since [L : R] = [L :
K][K : R], we know that [K : R] must be odd. Consequently, K = R(β) with β having a
minimal polynomial f (x) of odd degree. Therefore, K = R.
We now know that G(L/R) must be a 2-group. It follows that G(L/C) is a 2-group. We
have assumed that L 6= C; therefore, |G(L/C)| ≥ 2. By the first Sylow Theorem and the
Fundamental Theorem of Galois Theory, there exists a subgroup G of G(L/C) of index 2
and a field E fixed elementwise by G. Then [E : C] = 2 and there exists an element √ γ∈E
with minimal polynomial x2 + bx + c in C[x]. This polynomial has roots (−b ± b2 − 4c )/2
that are in C, since b2 − 4c is in C. This is impossible; hence, L = C. ■
Although our proof was strictly algebraic, we were forced to rely on results from cal-
culus. It is necessary to assume the completeness axiom from analysis to show that every
polynomial of odd degree has a real root and that every positive real number has a square
root. It seems that there is no possible way to avoid this difficulty and formulate a purely
algebraic argument. It is somewhat amazing that there are several elegant proofs of the
Fundamental Theorem of Algebra that use complex analysis. It is also interesting to note
that we can obtain a proof of such an important theorem from two very different fields of
mathematics.
Sage. Fields, field extensions, roots of polynomials, and group theory — Sage has it all,
and so it is possible to carefully study very complicated examples from Galois Theory with
Sage.
23.4 Reading Questions
1. What is the Galois group of a field extension?
2. When are two elements of a field extension conjugate? (In other words, what is the
definition?)
3. Summarize the nature and importance of the Fundamental Theorem of Galois Theory.
Capture the essence of the result without getting bogged down in too many details.
CHAPTER 23. GALOIS THEORY 308
4. Why are “solvable” groups so named? Paraphrasing the relevant theorem would be a
good answer.
5. Argue the following statement, both pro and con. Which side wins the debate?
Everything we have done in this entire course has been in preparation for
this chapter.
23.5 Exercises
1. Compute each of the following Galois groups. Which of these field extensions are
normal field extensions? If the extension is not normal, find a normal extension of Q
in which the
√ extension field is contained. √ √ 3
(a) G(Q( 30 )/Q) (d) G(Q( 2, 2, i)/Q)
√4
(b) G(Q( 5 )/Q)
√ √ √ √
(c) G(Q( 2, 3, 5 )/Q) (e) G(Q( 6, i)/Q)
2. Determine the separability of each of the following polynomials.
(a) x3 + 2x2 − x − 2 over Q (c) x4 + x2 + 1 over Z3
(b) x4 + 2x2 + 1 over Q (d) x3 + x2 + 1 over Z2
3. Give the order and describe a generator of the Galois group of GF(729) over GF(9).
4. Determine the Galois groups of each of the following polynomials in Q[x]; hence, de-
termine the solvability by radicals of each of the polynomials.
(a) x5 − 12x2 + 2 (f) (x2 − 2)(x2 + 2)
(b) x5 − 4x4 + 2x + 2
(g) x8 − 1
(c) x3 − 5
(h) x8 + 1
(d) x4 − x2 − 6
(e) x5 + 1 (i) x4 − 3x2 − 10
5. Find a primitive element in the splitting field of each of the following polynomials in
Q[x].
(a) x4 − 1 (c) x4 − 2x2 − 15
(b) x4 − 8x2 + 15 (d) x3 − 2
6. Prove that the Galois group of an irreducible quadratic polynomial is isomorphic to
Z2 .
7. Prove that the Galois group of an irreducible cubic polynomial is isomorphic to S3 or
Z3 .
8. Let F ⊂ K ⊂ E be fields. If E is a normal extension of F , show that E must also be
a normal extension of K.
9. Let G be the Galois group of a polynomial of degree n. Prove that |G| divides n!.
10. Let F ⊂ E. If f (x) is solvable over F , show that f (x) is also solvable over E.
11. Construct a polynomial f (x) in Q[x] of degree 7 that is not solvable by radicals.
12. Let p be prime. Prove that there exists a polynomial f (x) ∈ Q[x] of degree p with
Galois group isomorphic to Sp . Conclude that for each prime p with p ≥ 5 there exists
a polynomial of degree p that is not solvable by radicals.
13. Let p be a prime and Zp (t) be the field of rational functions over Zp . Prove that
f (x) = xp − t is an irreducible polynomial in Zp (t)[x]. Show that f (x) is not separable.
CHAPTER 23. GALOIS THEORY 309
14. Let E be an extension field of F . Suppose that K and L are two intermediate fields.
If there exists an element σ ∈ G(E/F ) such that σ(K) = L, then K and L are said
to be conjugate fields. Prove that K and L are conjugate if and only if G(E/K) and
G(E/L) are conjugate subgroups of G(E/F ).
15. Let σ ∈ Aut(R). If a is a positive real number, show that σ(a) > 0.
16. Let K be the splitting field of x3 + x2 + 1 ∈ Z2 [x]. Prove or disprove that K is an
extension by radicals.
17. Let F be a field such that char(F ) 6= 2. Prove that the splitting field of f (x) =
√
ax2 + bx + c is F ( α ), where α = b2 − 4ac.
18. Prove or disprove: Two different subgroups of a Galois group will have different fixed
fields.
19. Let K be the splitting field of a polynomial over F . If E is a field extension of F
contained in K and [E : F ] = 2, then E is the splitting field of some polynomial in
F [x].
20. We know that the cyclotomic polynomial
xp − 1
Φp (x) = = xp−1 + xp−2 + · · · + x + 1
x−1
is irreducible over Q for every prime p. Let ω be a zero of Φp (x), and consider the field
Q(ω).
(a) Show that ω, ω 2 , . . . , ω p−1 are distinct zeros of Φp (x), and conclude that they are
all the zeros of Φp (x).
(b) Show that G(Q(ω)/Q) is abelian of order p − 1.
(c) Show that the fixed field of G(Q(ω)/Q) is Q.
21. Let F be a finite field or a field of characteristic zero. Let E be a finite normal
extension of F with Galois group G(E/F ). Prove that F ⊂ K ⊂ L ⊂ E if and only if
{id} ⊂ G(E/L) ⊂ G(E/K) ⊂ G(E/F ).
22. Let F be a field of characteristic zero and let f (x) ∈ F [x] be a separable polynomial
of degree Q
n. If E is the splitting field of f (x), let α1 , . . . , αn be the roots of f (x) in E.
Let ∆ = i<j (αi − αj ). We define the discriminant of f (x) to be ∆2 .
(a) If f (x) = x2 + bx + c, show that ∆2 = b2 − 4c.
(b) If f (x) = x3 + px + q, show that ∆2 = −4p3 − 27q 2 .
(c) Prove that ∆2 is in F .
(d) If σ ∈ G(E/F ) is a transposition of two roots of f (x), show that σ(∆) = −∆.
(e) If σ ∈ G(E/F ) is an even permutation of the roots of f (x), show that σ(∆) = ∆.
(f) Prove that G(E/F ) is isomorphic to a subgroup of An if and only if ∆ ∈ F .
(g) Determine the Galois groups of x3 + 2x − 4 and x3 + x − 3.
23.6 References and Suggested Readings
[1] Artin, E. Theory: Lectures Delivered at the University of Notre Dame (Notre Dame
Mathematical Lectures, Number 2). Dover, Mineola, NY, 1997.
CHAPTER 23. GALOIS THEORY 310
[2] Edwards, H. M. Galois Theory. Springer-Verlag, New York, 1984.
[3] Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper Saddle
River, NJ, 2003.
[4] Gaal, L. Classical Galois Theory with Examples. American Mathematical Society,
Providence, 1979.
[5] Garling, D. J. H. A Course in Galois Theory. Cambridge University Press, Cambridge,
1986.
[6] Kaplansky, I. Fields and Rings. 2nd ed. University of Chicago Press, Chicago, 1972.
[7] Rothman, T. “The Short Life of Évariste Galois,” Scientific American, April 1982,
136–49.
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B
Hints and Answers to Selected Exercises
1 · Preliminaries
1.4 · Exercises
1.4.1.
Hint. (a) A ∩ B = {2}; (b) B ∩ C = {5}.
1.4.2.
Hint. (a) A×B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c, 1), (c, 2), (c, 3)}; (d) A×D = ∅.
1.4.6.
Hint. Observe that x ∈ A ∪ B if and only if x ∈ A or x ∈ B. Equivalently, x ∈ B or
x ∈ A, which is the same as x ∈ B ∪ A. Therefore, A ∪ B = B ∪ A.
1.4.10.
Hint. (A∩B)∪(A\B)∪(B \A) = (A∩B)∪(A∩B ′ )∪(B ∩A′ ) = [A∩(B ∪B ′ )]∪(B ∩A′ ) =
A ∪ (B ∩ A′ ) = (A ∪ B) ∩ (A ∪ A′ ) = A ∪ B.
1.4.14.
Hint. A\(B∪C) = A∩(B∪C)′ = (A∩A)∩(B ′ ∩C ′ ) = (A∩B ′ )∩(A∩C ′ ) = (A\B)∩(A\C).
1.4.17.
Hint. (a) Not a map since f (2/3) is undefined; (b) this is a map; (c) not a map, since
f (1/2) = 3/4 but f (2/4) = 3/8; (d) this is a map.
1.4.18.
Hint. (a) f is one-to-one but not onto. f (R) = {x ∈ R : x > 0}. (c) f is neither one-to-one
nor onto. f (R) = {x : −1 ≤ x ≤ 1}.
1.4.20.
Hint. (a) f (n) = n + 1.
1.4.22.
Hint. (a) Let x, y ∈ A. Then g(f (x)) = (g ◦ f )(x) = (g ◦ f )(y) = g(f (y)). Thus,
f (x) = f (y) and x = y, so g ◦ f is one-to-one. (b) Let c ∈ C, then c = (g ◦ f )(x) = g(f (x))
for some x ∈ A. Since f (x) ∈ B, g is onto.
1.4.23.
Hint. f −1 (x) = (x + 1)/(x − 1).
1.4.24.
Hint. (a) Let y ∈ f (A1 ∪ A2 ). Then there exists an x ∈ A1 ∪ A2 such that f (x) = y.
318
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 319
Hence, y ∈ f (A1 ) or f (A2 ). Therefore, y ∈ f (A1 ) ∪ f (A2 ). Consequently, f (A1 ∪ A2 ) ⊂
f (A1 ) ∪ f (A2 ). Conversely, if y ∈ f (A1 ) ∪ f (A2 ), then y ∈ f (A1 ) or f (A2 ). Hence, there
exists an x in A1 or A2 such that f (x) = y. Thus, there exists an x ∈ A1 ∪ A2 such that
f (x) = y. Therefore, f (A1 ) ∪ f (A2 ) ⊂ f (A1 ∪ A2 ), and f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ).
1.4.25.
Hint. (a) The relation fails to be symmetric. (b) The relation is not reflexive, since 0 is
not equivalent to itself. (c) The relation is not transitive.
1.4.28.
√
Hint. Let X = N ∪ { 2 } and define x ∼ y if x + y ∈ N.
2 · The Integers
2.4 · Exercises
2.4.1.
Hint. The base case, S(1) : [1(1 + 1)(2(1) + 1)]/6 = 1 = 12 is true. Assume that S(k) :
12 + 22 + · · · + k 2 = [k(k + 1)(2k + 1)]/6 is true. Then
12 + 22 + · · · + k 2 + (k + 1)2 = [k(k + 1)(2k + 1)]/6 + (k + 1)2
= [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6,
and so S(k + 1) is true. Thus, S(n) is true for all positive integers n.
2.4.3.
Hint. The base case, S(4) : 4! = 24 > 16 = 24 is true. Assume S(k) : k! > 2k is true.
Then (k + 1)! = k!(k + 1) > 2k · 2 = 2k+1 , so S(k + 1) is true. Thus, S(n) is true for all
positive integers n.
2.4.8.
Hint. Follow the proof in Example 2.4.
2.4.11.
Hint. The base case, S(0) : (1+x)0 −1 = 0 ≥ 0 = 0·x is true. Assume S(k) : (1+x)k −1 ≥
kx is true. Then
(1 + x)k+1 − 1 = (1 + x)(1 + x)k − 1
= (1 + x)k + x(1 + x)k − 1
≥ kx + x(1 + x)k
≥ kx + x
= (k + 1)x,
so S(k + 1) is true. Therefore, S(n) is true for all positive integers n.
2.4.17. Fibonacci Numbers.
Hint. For (a) and (b) use mathematical induction. (c) Show that f1 = 1, f2 = 1, and
fn+2 = fn+1 + fn . (e) Use part (b) and Exercise 2.4.16.
2.4.19.
Hint. Use the Fundamental Theorem of Arithmetic.
2.4.23.
Hint. Use the Principle of Well-Ordering and the division algorithm.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 320
2.4.27.
Hint. Since gcd(a, b) = 1, there exist integers r and s such that ar + bs = 1. Thus,
acr + bcs = c.
2.4.29.
Hint. Every prime must be of the form 2, 3, 6n + 1, or 6n + 5. Suppose there are only
finitely many primes of the form 6k + 5.
3 · Groups
3.5 · Exercises
3.5.1.
Hint. (a) 3 + 7Z = {. . . , −4, 3, 10, . . .}; (c) 18 + 26Z; (e) 5 + 6Z.
3.5.2.
Hint. (a) Not a group; (c) a group.
3.5.6.
Hint.
· 1 5 7 11
1 1 5 7 11
5 5 1 11 7
7 7 11 1 5
11 11 7 5 1
3.5.8.
Hint. Pick two matrices. Almost any pair will work.
3.5.15.
Hint. There is a nonabelian group containing six elements.
3.5.16.
Hint. Look at the symmetry group of an equilateral triangle or a square.
3.5.17.
Hint. The are five different groups of order 8.
3.5.18.
Hint. Let
1 2 ··· n
σ=
a1 a2 · · · an
be in Sn . All of the ai s must be distinct. There are n ways to choose a1 , n − 1 ways to
choose a2 , . . ., 2 ways to choose an−1 , and only one way to choose an . Therefore, we can
form σ in n(n − 1) · · · 2 · 1 = n! ways.
3.5.25.
Hint.
(aba−1 )n = (aba−1 )(aba−1 ) · · · (aba−1 )
= ab(aa−1 )b(aa−1 )b · · · b(aa−1 )ba−1
= abn a−1 .
3.5.31.
Hint. Since abab = (ab)2 = e = a2 b2 = aabb, we know that ba = ab.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 321
3.5.35.
Hint. H1 = {id}, H2 = {id, ρ1 , ρ2 }, H3 = {id, µ1 }, H4 = {id, µ2 }, H5 = {id, µ3 }, S3 .
3.5.41.
√ √ √ √
Hint. The identity of G is 1 = 1+0 2. Since (a+b √ −12 )(c+d 22 ) = 2(ac+2bd)+(ad+bc)
√ 2,
G is closed under multiplication. Finally, (a + b 2 ) = a/(a − 2b ) − b 2/(a2 − 2b2 ).
3.5.46.
Hint. Look at S3 .
3.5.49.
Hint. ba = a4 b = a3 ab = ab
4 · Cyclic Groups
4.5 · Exercises
4.5.1.
Hint. (a) False; (c) false; (e) true.
4.5.2.
Hint. (a) 12; (c) infinite; (e) 10.
4.5.3.
Hint. (a) 7Z = {. . . , −7, 0, 7, 14, . . .}; (b) {0, 3, 6, 9, 12, 15, 18, 21}; (c) {0}, {0, 6}, {0, 4, 8},
{0, 3, 6, 9}, {0, 2, 4, 6, 8, 10}; (g) {1, 3, 7, 9}; (j) {1, −1, i, −i}.
4.5.4.
Hint. (a)
1 0 −1 0 0 −1 0 1
, , , .
0 1 0 −1 1 0 −1 0
(c)
1 0 1 −1 −1 1 0 1 0 −1 −1 0
, , , , , .
0 1 1 0 −1 0 −1 1 1 −1 0 −1
4.5.10.
Hint. (a) 0; (b) 1, −1.
4.5.11.
Hint. 1, 2, 3, 4, 6, 8, 12, 24.
4.5.15.
Hint. (a) −3 + 3i; (c) 43 − 18i; (e) i
4.5.16.
√
Hint. (a) 3 + i; (c) −3.
4.5.17.
√ √
Hint. (a) 2 cis(7π/4); (c) 2 2 cis(π/4); (e) 3 cis(3π/2).
4.5.18.
√
Hint. (a) (1 − i)/2; (c) 16(i − 3 ); (e) −1/4.
4.5.22.
Hint. (a) 292; (c) 1523.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 322
4.5.27.
Hint. |hgi ∩ hhi| = 1.
4.5.31.
Hint. The identity element in any group has finite order. Let g, h ∈ G have orders m and
n, respectively. Since (g −1 )m = e and (gh)mn = e, the elements of finite order in G form a
subgroup of G.
4.5.37.
Hint. If g is an element distinct from the identity in G, g must generate G; otherwise, hgi
is a nontrivial proper subgroup of G.
5 · Permutation Groups
5.4 · Exercises
5.4.1.
Hint. (a) (1 2 4 5 3); (c) (1 3)(2 5).
5.4.2.
Hint. (a) (1 3 5)(2 4); (c) (1 4)(2 3); (e) (1 3 2 4); (g) (1 3 4)(2 5); (n) (1 7 3 5 2).
5.4.3.
Hint. (a) (1 6)(1 5)(1 3)(1 4); (c) (1 6)(1 4)(1 2).
5.4.4.
Hint. (a1 , a2 , . . . , an )−1 = (a1 , an , an−1 , . . . , a2 )
5.4.5.
Hint. (a) {(1 3), (1 3)(2 4), (1 3 2), (1 3 4), (1 3 2 4), (1 3 4 2)} is not a subgroup.
5.4.8.
Hint. (1 2 3 4 5)(6 7 8).
5.4.11.
Hint. Permutations of the form
(1), (a1 , a2 )(a3 , a4 ), (a1 , a2 , a3 ), (a1 , a2 , a3 , a4 , a5 )
are possible for A5 .
5.4.17.
Hint. Calculate (1 2 3)(1 2) and (1 2)(1 2 3).
5.4.25.
Hint. Consider the cases (a, b)(b, c) and (a, b)(c, d).
5.4.29.
Hint. Show that the center of Dn consists of the identity if n is odd and consists of the
identity and a 180◦ rotation if n is even.
5.4.30.
Hint. For (a), show that στ σ −1 (σ(ai )) = σ(ai+1 ).
6 · Cosets and Lagrange’s Theorem
6.5 · Exercises
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 323
6.5.1.
Hint. The order of g and the order h must both divide the order of G.
6.5.2.
Hint. The possible orders must divide 60.
6.5.3.
Hint. This is true for every proper nontrivial subgroup.
6.5.4.
Hint. False.
6.5.5.
Hint. (a) h8i, 1 + h8i, 2 + h8i, 3 + h8i, 4 + h8i, 5 + h8i, 6 + h8i, and 7 + h8i; (c) 3Z, 1 + 3Z,
and 2 + 3Z.
6.5.7.
Hint. 4ϕ(15) ≡ 48 ≡ 1 (mod 15).
6.5.12.
Hint. Let g1 ∈ gH. Show that g1 ∈ Hg and thus gH ⊂ Hg.
6.5.19.
Hint. Show that g(H ∩ K) = gH ∩ gK.
6.5.22.
Hint. If gcd(m, n) = 1, then ϕ(mn) = ϕ(m)ϕ(n) (Exercise 2.4.26 in Chapter 2).
7 · Introduction to Cryptography
7.4 · Exercises
7.4.1.
Hint. LAORYHAPDWK
7.4.3.
Hint. Hint: V = E, E = X (also used for spaces and punctuation), K = R.
7.4.4.
Hint. 26! − 1
7.4.7.
Hint. (a) 2791; (c) 11213525032442.
7.4.9.
Hint. (a) 31 (c) 14.
7.4.10.
Hint. (a) n = 11 · 41; (c) n = 8779 · 4327.
8 · Algebraic Coding Theory
8.6 · Exercises
8.6.2.
Hint. This cannot be a group code since (0000) ∈
/ C.
8.6.3.
Hint. (a) 2; (c) 2.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 324
8.6.4.
Hint. (a) 3; (c) 4.
8.6.6.
Hint. (a) dmin = 2; (c) dmin = 1.
8.6.7.
Hint.
(a) (00000), (00101), (10011), (10110)
0 1
0 0
G = 1 0
0 1
1 1
(b) (000000), (010111), (101101), (111010)
1 0
0 1
1 0
G=
1 1
0 1
1 1
8.6.9.
Hint. Multiple errors occur in one of the received words.
8.6.11.
Hint. (a) A canonical parity-check matrix with standard generator matrix
1
1
G = 0 .
0
1
(c) A canonical parity-check matrix with standard generator matrix
1 0
0 1
G= .
1 1
1 0
8.6.12.
Hint. (a) All possible syndromes occur.
8.6.15.
Hint. (a) C, (10000) + C, (01000) + C, (00100)+ C, (00010) + C, (11000)+ C, (01100) + C,
(01010) + C. A decoding table does not exist for C since this is only a single error-detecting
code.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 325
8.6.19.
Hint. Let x ∈ C have odd weight and define a map from the set of odd codewords to the
set of even codewords by y 7→ x + y. Show that this map is a bijection.
8.6.23.
Hint. For 20 information positions, at least 6 check bits are needed to ensure an error-
correcting code.
9 · Isomorphisms
9.4 · Exercises
9.4.1.
Hint. Every infinite cyclic group is isomorphic to Z by Theorem 9.7.
9.4.2.
Hint. Define ϕ : C∗ → GL2 (R) by
a b
ϕ(a + bi) = .
−b a
9.4.3.
Hint. False.
9.4.6.
Hint. Define a map from Zn into the nth roots of unity by k 7→ cis(2kπ/n).
9.4.8.
Hint. Assume that Q is cyclic and try to find a generator.
9.4.11.
Hint. There are two nonabelian and three abelian groups that are not isomorphic.
9.4.16.
Hint. (a) 12; (c) 5.
9.4.19.
Hint. Draw the picture.
9.4.20.
Hint. True.
9.4.25.
Hint. True.
9.4.27.
Hint. Let a be a generator for G. If ϕ : G → H is an isomorphism, show that ϕ(a) is a
generator for H.
9.4.38.
Hint. Any automorphism of Z6 must send 1 to another generator of Z6 .
9.4.45.
Hint. To show that ϕ is one-to-one, let g1 = h1 k1 and g2 = h2 k2 and consider ϕ(g1 ) =
ϕ(g2 ).
10 · Normal Subgroups and Factor Groups
10.4 · Exercises
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 326
10.4.1.
Hint. (a)
A4 (1 2)A4
A4 A4 (1 2)A4
(1 2)A4 (1 2)A4 A4
(c) D4 is not normal in S4 .
10.4.8.
Hint. If a ∈ G is a generator for G, then aH is a generator for G/H.
10.4.11.
Hint. For any g ∈ G, show that the map ig : G → G defined by ig : x 7→ gxg −1 is an
isomorphism of G with itself. Then consider ig (H).
10.4.12.
Hint. Suppose that hgi is normal in G and let y be an arbitrary element of G. If x ∈ C(g),
we must show that yxy −1 is also in C(g). Show that (yxy −1 )g = g(yxy −1 ).
10.4.14.
Hint. (a) Let g ∈ G and h ∈ G′ . If h = aba−1 b−1 , then
ghg −1 = gaba−1 b−1 g −1
= (gag −1 )(gbg −1 )(ga−1 g −1 )(gb−1 g −1 )
= (gag −1 )(gbg −1 )(gag −1 )−1 (gbg −1 )−1 .
We also need to show that if h = h1 · · · hn with hi = ai bi a−1 −1
i bi , then ghg
−1 is a product of el-
ements of the same type. However, ghg = gh1 · · · hn g = (gh1 g )(gh2 g −1 ) · · · (ghn g −1 ).
−1 −1 −1
11 · Homomorphisms
11.4 · Exercises
11.4.2.
Hint. (a) is a homomorphism with kernel {1}; (c) is not a homomorphism.
11.4.4.
Hint. Since ϕ(m + n) = 7(m + n) = 7m + 7n = ϕ(m) + ϕ(n), ϕ is a homomorphism.
11.4.5.
Hint. For any homomorphism ϕ : Z24 → Z18 , the kernel of ϕ must be a subgroup of Z24
and the image of ϕ must be a subgroup of Z18 . Now use the fact that a generator must map
to a generator.
11.4.9.
Hint. Let a, b ∈ G. Then ϕ(a)ϕ(b) = ϕ(ab) = ϕ(ba) = ϕ(b)ϕ(a).
11.4.17.
Hint. Find a counterexample.
12 · Matrix Groups and Symmetry
12.4 · Exercises
12.4.1.
Hint.
1 1
kx + yk2 + kxk2 − kyk2 = hx + y, x + yi − kxk2 − kyk2
2 2
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 327
1
= kxk2 + 2hx, yi + kyk2 − kxk2 − kyk2
2
= hx, yi.
12.4.3.
Hint. (a) is in SO(2); (c) is not in O(3).
12.4.5.
Hint. (a) hx, yi = hy, xi.
12.4.7.
Hint. Use the unimodular matrix
5 2
.
2 1
12.4.10.
Hint. Show that the kernel of the map det : O(n) → R∗ is SO(n).
12.4.13.
Hint. True.
12.4.17.
Hint. p6m
13 · The Structure of Groups
13.4 · Exercises
13.4.1.
Hint. There are three possible groups.
13.4.4.
Hint. (a) {0} ⊂ h6i ⊂ h3i ⊂ Z12 ; (e) {(1)} × {0} ⊂ {(1), (1 2 3), (1 3 2)} × {0} ⊂ S3 × {0} ⊂
S3 × h2i ⊂ S3 × Z4 .
13.4.7.
Hint. Use the Fundamental Theorem of Finitely Generated Abelian Groups.
13.4.12.
Hint. If N and G/N are solvable, then they have solvable series
N = Nn ⊃ Nn−1 ⊃ · · · ⊃ N1 ⊃ N0 = {e}
G/N = Gn /N ⊃ Gn−1 /N ⊃ · · · G1 /N ⊃ G0 /N = {N }.
13.4.16.
Hint. Use the fact that Dn has a cyclic subgroup of index 2.
13.4.21.
Hint. G/G′ is abelian.
14 · Group Actions
14.5 · Exercises
14.5.1.
Hint. Example 14.1: 0, R2 \ {0}. Example 14.2: X = {1, 2, 3, 4}.
14.5.2.
Hint. (a) X(1) = {1, 2, 3}, X(1 2) = {3}, X(1 3) = {2}, X(2 3) = {1}, X(1 2 3) = X(1 3 2) = ∅.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 328
G1 = {(1), (2 3)}, G2 = {(1), (1 3)}, G3 = {(1), (1 2)}.
14.5.3.
Hint. (a) O1 = O2 = O3 = {1, 2, 3}.
14.5.6.
Hint. The conjugacy classes for S4 are
O(1) = {(1)},
O(12) = {(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)},
O(1 2)(3 4) = {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)},
O(123) = {(1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3)},
O(1234) = {(1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2)}.
The class equation is 1 + 3 + 6 + 6 + 8 = 24.
14.5.8.
Hint. (34 + 31 + 32 + 31 + 32 + 32 + 33 + 33 )/8 = 21.
14.5.11.
Hint. The group of rigid motions of the cube can be described by the allowable permuta-
tions of the six faces and is isomorphic to S4 . There are the identity cycle, 6 permutations
with the structure (abcd) that correspond to the quarter turns, 3 permutations with the
structure (ab)(cd) that correspond to the half turns, 6 permutations with the structure
(ab)(cd)(ef ) that correspond to rotating the cube about the centers of opposite edges, and
8 permutations with the structure (abc)(def ) that correspond to rotating the cube about
opposite vertices.
14.5.15.
Hint. (1 · 26 + 3 · 24 + 4 · 23 + 2 · 22 + 2 · 21 )/12 = 13.
14.5.17.
Hint. (1 · 28 + 3 · 26 + 2 · 24 )/6 = 80.
14.5.22.
Hint. Use the fact that x ∈ gC(a)g −1 if and only if g −1 xg ∈ C(a).
15 · The Sylow Theorems
15.4 · Exercises
15.4.1.
Hint. If |G| = 18 = 2 · 32 , then the order of a Sylow 2-subgroup is 2, and the order of a
Sylow 3-subgroup is 9.
15.4.2.
Hint. The four Sylow 3-subgroups of S4 are P1 = {(1), (1 2 3), (1 3 2)}, P2 = {(1), (1 2 4), (1 4 2)},
P3 = {(1), (1 3 4), (1 4 3)}, P4 = {(1), (2 3 4), (2 4 3)}.
15.4.5.
Hint. Since |G| = 96 = 25 · 3, G has either one or three Sylow 2-subgroups by the Third
Sylow Theorem. If there is only one subgroup, we are done. If there are three Sylow 2-
subgroups, let H and K be two of them. Therefore, |H ∩ K| ≥ 16; otherwise, HK would
have (32 · 32)/8 = 128 elements, which is impossible. Thus, H ∩ K is normal in both H and
K since it has index 2 in both groups.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 329
15.4.8.
Hint. Show that G has a normal Sylow p-subgroup of order p2 and a normal Sylow q-
subgroup of order q 2 .
15.4.10.
Hint. False.
15.4.17.
Hint. If G is abelian, then G is cyclic, since |G| = 3 · 5 · 17. Now look at Example 15.14.
15.4.23.
Hint. Define a mapping between the right cosets of N (H) in G and the conjugates of H
in G by N (H)g 7→ g −1 Hg. Prove that this map is a bijection.
15.4.26.
Hint. Let aG′ , bG′ ∈ G/G′ . Then (aG′ )(bG′ ) = abG′ = ab(b−1 a−1 ba)G′ = (abb−1 a−1 )baG′ =
baG′ .
16 · Rings
16.7 · Exercises
16.7.1.
√
Hint. (a) 7Z is a ring but not a field; (c) Q( 2 ) is a field; (f) R is not a ring.
16.7.3.
Hint. (a) {1, 3, 7, 9}; (c) {1, 2, 3, 4, 5, 6}; (e)
1 0 1 1 1 0 0 1 1 1 0 1
, , , , , , .
0 1 0 1 1 1 1 0 1 0 1 1
16.7.4.
Hint. (a) {0}, {0, 9}, {0, 6, 12}, {0, 3, 6, 9, 12, 15}, {0, 2, 4, 6, 8, 10, 12, 14, 16}; (c) there are
no nontrivial ideals.
16.7.7.
Hint. Assume there is an isomorphism ϕ : C → R with ϕ(i) = a.
16.7.8.
√ √ √
Hint. False. Assume there is an isomorphism ϕ : Q( 2 ) → Q( 3 ) such that ϕ( 2 ) = a.
16.7.13.
Hint. (a) x ≡ 17 (mod 55); (c) x ≡ 214 (mod 2772).
16.7.16.
Hint. If I 6= {0}, show that 1 ∈ I.
16.7.18.
Hint. (a) ϕ(a)ϕ(b) = ϕ(ab) = ϕ(ba) = ϕ(b)ϕ(a).
16.7.26.
Hint. Let a ∈ R with a 6= 0. Then the principal ideal generated by a is R. Thus, there
exists a b ∈ R such that ab = 1.
16.7.28.
Hint. Compute (a + b)2 and (−ab)2 .
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 330
16.7.33.
Hint. Let a/b, c/d ∈ Z(p) . Then a/b + c/d = (ad + bc)/bd and (a/b) · (c/d) = (ac)/(bd) are
both in Z(p) , since gcd(bd, p) = 1.
16.7.37.
Hint. Suppose that x2 = x and x 6= 0. Since R is an integral domain, x = 1. To find a
nontrivial idempotent, look in M2 (R).
17 · Polynomials
17.5 · Exercises
17.5.2.
Hint. (a) 9x2 + 2x + 5; (b) 8x4 + 7x3 + 2x2 + 7x.
17.5.3.
Hint. (a) 5x3 + 6x2 − 3x + 4 = (5x2 + 2x + 1)(x − 2) + 6; (c) 4x5 − x3 + x2 + 4 =
(4x2 + 4)(x3 + 3) + 4x2 + 2.
17.5.5.
Hint. (a) No zeros in Z12 ; (c) 3, 4.
17.5.7.
Hint. Look at (2x + 1).
17.5.8.
Hint. (a) Reducible; (c) irreducible.
17.5.10.
Hint. One factorization is x2 + x + 8 = (x + 2)(x + 9).
17.5.13.
Hint. The integers Z do not form a field.
17.5.14.
Hint. False.
17.5.16.
Hint. Let ϕ : R → S be an isomorphism. Define ϕ : R[x] → S[x] by ϕ(a0 + a1 x + · · · +
an xn ) = ϕ(a0 ) + ϕ(a1 )x + · · · + ϕ(an )xn .
17.5.20. Cyclotomic Polynomials.
Hint. The polynomial
xn − 1
Φn (x) = = xn−1 + xn−2 + · · · + x + 1
x−1
is called the cyclotomic polynomial. Show that Φp (x) is irreducible over Q for any prime p.
17.5.26.
Hint. Find a nontrivial proper ideal in F [x].
18 · Integral Domains
18.4 · Exercises
18.4.1.
√ √ √
Hint. Note that z −1 = 1/(a + b 3 i) = (a − b 3 i)/(a2 + 3b2 ) is in Z[ 3 i] if and only if
a2 + 3b2 = 1. The only integer solutions to the equation are a = ±1, b = 0.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 331
18.4.2.
Hint. (a) 5 = −i(1 + 2i)(2 + i); (c) 6 + 8i = −i(1 + i)2 (2 + i)2 .
18.4.4.
Hint. True.
18.4.9.
Hint. Let z = a + bi and w = c + di 6= 0 be in Z[i]. Prove that z/w ∈ Q(i).
18.4.15.
Hint. Let a = ub with u a unit. Then ν(b) ≤ ν(ub) ≤ ν(a). Similarly, ν(a) ≤ ν(b).
18.4.16.
Hint. Show that 21 can be factored in two different ways.
19 · Lattices and Boolean Algebras
19.5 · Exercises
19.5.2.
Hint.
30
10 15
2 5 3
1
19.5.4.
Hint. What are the atoms of B?
19.5.5.
Hint. False.
19.5.6.
Hint. (a) (a ∨ b ∨ a′ ) ∧ a
a
b a
a′
(c) a ∨ (a ∧ b)
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 332
a b
a
19.5.8.
Hint. Not equivalent.
19.5.10.
Hint. (a) a′ ∧ [(a ∧ b′ ) ∨ b] = a ∧ (a ∨ b).
19.5.14.
Hint. Let I, J be ideals in R. We need to show that I + J = {r + s : r ∈ I and s ∈ J}
is the smallest ideal in R containing both I and J. If r1 , r2 ∈ I and s1 , s2 ∈ J, then
(r1 +s1 )+(r2 +s2 ) = (r1 +r2 )+(s1 +s2 ) is in I +J. For a ∈ R, a(r1 +s1 ) = ar1 +as1 ∈ I +J;
hence, I + J is an ideal in R.
19.5.18.
Hint. (a) No.
19.5.20.
Hint. (⇒). a = b ⇒ (a ∧ b′ ) ∨ (a′ ∧ b) = (a ∧ a′ ) ∨ (a′ ∧ a) = O ∨ O = O. (⇐).
(a∧b′ )∨(a′ ∧b) = O ⇒ a∨b = (a∨a)∨b = a∨(a∨b) = a∨[I ∧(a∨b)] = a∨[(a∨a′ )∧(a∨b)] =
[a ∨ (a ∧ b′ )] ∨ [a ∨ (a′ ∧ b)] = a ∨ [(a ∧ b′ ) ∨ (a′ ∧ b)] = a ∨ 0 = a. A symmetric argument
shows that a ∨ b = b.
20 · Vector Spaces
20.5 · Exercises
20.5.3.
√ √ √ √ √
Hint. Q( 2, 3 ) has basis {1, 2, 3, 6 } over Q.
20.5.5.
Hint. The set {1, x, x2 , . . . , xn−1 } is a basis for Pn .
20.5.7.
Hint. (a) Subspace of dimension 2 with basis {(1, 0, −3), (0, 1, 2)}; (d) not a subspace
20.5.10.
Hint. Since 0 = α0 = α(−v + v) = α(−v) + αv, it follows that −αv = α(−v).
20.5.12.
Hint. Let v0 = 0, v1 , . . . , vn ∈ V and α0 6= 0, α1 , . . . , αn ∈ F . Then α0 v0 + · · · + αn vn = 0.
20.5.15. Linear Transformations.
Hint. (a) Let u, v ∈ ker(T ) and α ∈ F . Then
T (u + v) = T (u) + T (v) = 0
T (αv) = αT (v) = α0 = 0.
Hence, u + v, αv ∈ ker(T ), and ker(T ) is a subspace of V .
(c) The statement that T (u) = T (v) is equivalent to T (u − v) = T (u) − T (v) = 0, which
is true if and only if u − v = 0 or u = v.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 333
20.5.17. Direct Sums.
Hint. (a) Let u, u′ ∈ U and v, v ′ ∈ V . Then
(u + v) + (u′ + v ′ ) = (u + u′ ) + (v + v ′ ) ∈ U + V
α(u + v) = αu + αv ∈ U + V .
21 · Fields
21.5 · Exercises
21.5.1.
Hint. (a) x4 − (2/3)x2 − 62/9; (c) x4 − 2x2 + 25.
21.5.2.
√ √ √ √ √
Hint. (a) {1, 2, 3, 6 }; (c) {1, i, 2, 2 i}; (e) {1, 21/6 , 21/3 , 21/2 , 22/3 , 25/6 }.
21.5.3.
√ √
Hint. (a) Q( 3, 7 ).
21.5.5.
Hint. Use the fact that the elements of Z2 [x]/hx3 + x + 1i are 0, 1, α, 1 + α, α2 , 1 + α2 ,
α + α2 , 1 + α + α2 and the fact that α3 + α + 1 = 0.
21.5.8.
Hint. False.
21.5.14.
Hint. Suppose that E is algebraic over F and K is algebraic over E. Let α ∈ K. It
suffices to show that α is algebraic over some finite extension of F . Since α is algebraic over
E, it must be the zero of some polynomial p(x) = β0 + β1 x + · · · + βn xn in E[x]. Hence α
is algebraic over F (β0 , . . . , βn ).
21.5.22.
√ √ √ √ √ √ √ √ √
Hint. Since
√ √ {1, 3, 7, 21 } is √ for Q( 3, 7 ) over Q, Q( 3, 7 ) ⊃ Q( 3 + 7 ).
√ a basis
Since [Q( 3, √7 ) : Q] √ 7 ) : Q]
√ = 4, [Q(√ 3 + √ = 2√or 4. Since the degree of the minimal
polynomial of 3 + 7 is 4, Q( 3, 7 ) = Q( 3 + 7 ).
21.5.27.
Hint. Let β ∈ F (α) not in F . Then β = p(α)/q(α), where p and q are polynomials in α
with q(α) 6= 0 and coefficients in F . If β is algebraic over F , then there exists a polynomial
f (x) ∈ F [x] such that f (β) = 0. Let f (x) = a0 + a1 x + · · · + an xn . Then
p(α) p(α) p(α) n
0 = f (β) = f = a0 + a1 + · · · + an .
q(α) q(α) q(α)
Now multiply both sides by q(α)n to show that there is a polynomial in F [x] that has α as
a zero.
21.5.28.
Hint. See the comments following Theorem 21.13.
22 · Finite Fields
22.4 · Exercises
22.4.1.
Hint. Make sure that you have a field extension.
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 334
22.4.4.
Hint. There are eight elements in Z2 (α). Exhibit two more zeros of x3 + x2 + 1 other than
α in these eight elements.
22.4.5.
Hint. Find an irreducible polynomial p(x) in Z3 [x] of degree 3 and show that Z3 [x]/hp(x)i
has 27 elements.
22.4.7.
Hint. (a) x5 − 1 = (x + 1)(x4 + x3 + x2 + x + 1); (c) x9 − 1 = (x + 1)(x2 + x + 1)(x6 + x3 + 1).
22.4.8.
Hint. True.
22.4.11.
Hint. (a) Use the fact that x7 − 1 = (x + 1)(x3 + x + 1)(x3 + x2 + 1).
22.4.12.
Hint. False.
22.4.17.
Hint. If p(x) ∈ F [x], then p(x) ∈ E[x].
22.4.18.
Hint. Since α is algebraic over F of degree n, we can write any element β ∈ F (α)
uniquely as β = a0 + a1 α + · · · + an−1 αn−1 with ai ∈ F . There are q n possible n-tuples
(a0 , a1 , . . . , an−1 ).
22.4.24. Wilson’s Theorem.
Hint. Factor xp−1 − 1 over Zp .
23 · Galois Theory
23.5 · Exercises
23.5.1.
Hint. (a) Z2 ; (c) Z2 × Z2 × Z2 .
23.5.2.
Hint. (a) Separable over Q since x3 + 2x2 − x − 2 = (x − 1)(x + 1)(x + 2); (c) not separable
over Z3 since x4 + x2 + 1 = (x + 1)2 (x + 2)2 .
23.5.3.
Hint. If
[GF(729) : GF(9)] = [GF(729) : GF(3)]/[GF(9) : GF(3)] = 6/2 = 3,
then G(GF(729)/ GF(9)) ∼= Z3 . A generator for G(GF(729)/ GF(9)) is σ, where σ36 (α) =
6
3
α =α 729 for α ∈ GF(729).
23.5.4.
Hint. (a) S5 ; (c) S3 ; (g) see Example 23.11.
23.5.5.
Hint. (a) Q(i)
23.5.7.
Hint. Let E be the splitting field of a cubic polynomial in F [x]. Show that [E : F ] is less
than or equal to 6 and is divisible by 3. Since G(E/F ) is a subgroup of S3 whose order is
APPENDIX B. HINTS AND ANSWERS TO SELECTED EXERCISES 335
divisible by 3, conclude that this group must be isomorphic to Z3 or S3 .
23.5.9.
Hint. G is a subgroup of Sn .
23.5.16.
Hint. True.
23.5.20.
Hint.
(a) Clearly ω, ω 2 , . . . , ω p−1 are distinct since ω 6= 1 or 0. To show that ω i is a zero of Φp ,
calculate Φp (ω i ).
(b) The conjugates of ω are ω, ω 2 , . . . , ω p−1 . Define a map ϕi : Q(ω) → Q(ω i ) by
ϕi (a0 + a1 ω + · · · + ap−2 ω p−2 ) = a0 + a1 ω i + · · · + cp−2 (ω i )p−2 ,
where ai ∈ Q. Prove that ϕi is an isomorphism of fields. Show that ϕ2 generates
G(Q(ω)/Q).
(c) Show that {ω, ω 2 , . . . , ω p−1 } is a basis for Q(ω) over Q, and consider which linear
combinations of ω, ω 2 , . . . , ω p−1 are left fixed by all elements of G(Q(ω)/Q).
C
Notation
The following table defines the notation used in this book. Page numbers or references refer
to the first appearance of each symbol.
Symbol Description Page
a∈A a is in the set A 3
N the natural numbers 4
Z the integers 4
Q the rational numbers 4
R the real numbers 4
C the complex numbers 4
A⊂B A is a subset of B 4
∅ the empty set 4
A∪B the union of sets A and B 4
A∩B the intersection of sets A and B 4
A′ complement of the set A 4
A\B difference between sets A and B 4
A×B Cartesian product of sets A and B 6
An A × · · · × A (n times) 6
id identity mapping 9
f −1 inverse of the function f 9
a ≡ b (mod n) a is congruent to b modulo n 13
n! n factorial 18
n
k binomial coefficient n!/(k!(n − k)!) 18
a|b a divides b 20
gcd(a, b) greatest common divisor of a and b 20
P(X) power set of X 24
lcm(m, n) the least common multiple of m and n 25
Zn the integers modulo n 28
U (n) group of units in Zn 35
Mn (R) the n × n matrices with entries in R 35
det A the determinant of A 35
GLn (R) the general linear group 35
Q8 the group of quaternions 36
(Continued on next page)
336
APPENDIX C. NOTATION 337
Symbol Description Page
C∗ the multiplicative group of complex numbers 36
|G| the order of a group 36
R∗ the multiplicative group of real numbers 38
Q∗ the multiplicative group of rational numbers 38
SLn (R) the special linear group 38
Z(G) the center of a group 43
hai cyclic group generated by a 46
|a| the order of an element a 47
cis θ cos θ + i sin θ 51
T the circle group 52
Sn the symmetric group on n letters 59
(a1 , a2 , . . . , ak ) cycle of length k 61
An the alternating group on n letters 64
Dn the dihedral group 65
[G : H] index of a subgroup H in a group G 75
LH the set of left cosets of a subgroup H in a group G 75
RH the set of right cosets of a subgroup H in a group G 75
a∤b a does not divide b 78
d(x, y) Hamming distance between x and y 96
dmin the minimum distance of a code 96
w(x) the weight of x 96
Mm×n (Z2 ) the set of m × n matrices with entries in Z2 100
Null(H) null space of a matrix H 100
δij Kronecker delta 103
G∼ =H G is isomorphic to a group H 114
Aut(G) automorphism group of a group G 123
ig ig (x) = gxg −1 123
Inn(G) inner automorphism group of a group G 123
ρg right regular representation 123
G/N factor group of G mod N 126
G′ commutator subgroup of G 132
ker ϕ kernel of ϕ 134
(aij ) matrix 142
O(n) orthogonal group 144
kxk length of a vector x 144
SO(n) special orthogonal group 147
E(n) Euclidean group 147
Ox orbit of x 169
Xg fixed point set of g 169
Gx isotropy subgroup of x 169
N (H) normalizer of s subgroup H 182
H the ring of quaternions 192
Z[i] the Gaussian integers 194
char R characteristic of a ring R 194
(Continued on next page)
APPENDIX C. NOTATION 338
Symbol Description Page
Z(p) ring of integers localized at p 206
deg f (x) degree of a polynomial 210
R[x] ring of polynomials over a ring R 210
R[x1 , x2 , . . . , xn ] ring of polynomials in n indeterminants 212
ϕα evaluation homomorphism at α 212
Q(x) field of rational functions over Q 228
ν(a) Euclidean valuation of a 231
F (x) field of rational functions in x 235
F (x1 , . . . , xn ) field of rational functions in x1 , . . . , xn 235
ab a is less than b 238
a∨b join of a and b 240
a∧b meet of a and b 240
I largest element in a lattice 241
O smallest element in a lattice 241
a′ complement of a in a lattice 241
dim V dimension of a vector space V 256
U ⊕V direct sum of vector spaces U and V 258
Hom(V, W ) set of all linear transformations from U into V 258
V∗ dual of a vector space V 258
F (α1 , . . . , αn ) smallest field containing F and α1 , . . . , αn 262
[E : F ] dimension of a field extension of E over F 265
GF(pn ) Galois field of order pn 280
F∗ multiplicative group of a field F 281
G(E/F ) Galois group of E over F 294
F{σi } field fixed by the automorphism σi 297
FG field fixed by the automorphism group G 297
∆2 discriminant of a polynomial 309
Index
G-equivalent, 169 for integral domains, 194
G-set, 168 Cardano, Gerolamo, 219
nth root of unity, 52, 303 Carmichael numbers, 89
rsa cryptosystem, 84 Cauchy’s Theorem, 181
Cauchy, Augustin-Louis, 65
Abel, Niels Henrik, 302 Cayley table, 34
Abelian group, 34 Cayley’s Theorem, 117
Ackermann’s function, 26 Cayley, Arthur, 117
Adleman, L., 84 Centralizer
Algebraic closure, 267 of a subgroup, 171
Algebraic extension, 262 Characteristic of a ring, 194
Algebraic number, 262 Chinese Remainder Theorem
Algorithm for integers, 201
division, 212 Cipher, 81
Euclidean, 22 Ciphertext, 81
Ascending chain condition, 230 Circuit
Associate elements, 228 parallel, 246
Atom, 244 series, 246
Automorphism series-parallel, 246
inner, 139 Class equation, 171
Basis of a lattice, 151 Code
Bieberbach, L., 154 bch, 287
Binary operation, 33 cyclic, 282
Binary symmetric channel, 95 group, 98
Boole, George, 248 linear, 100
Boolean algebra minimum distance of, 96
atom in a, 244 polynomial, 283
definition of, 242 Commutative diagrams, 136
finite, 244 Commutative rings, 190
isomorphism, 244 Composite integer, 22
Boolean function, 176, 251 Composition series, 163
Burnside’s Counting Theorem, 173 Congruence modulo n, 13
Burnside, William, 37, 130, 178 Conjugacy classes, 171
Conjugate elements, 294
Cancellation law Conjugate, complex, 49
for groups, 37 Conjugation, 169
339
INDEX 340
Constructible number, 271 Eisenstein’s Criterion, 217
Correspondence Theorem Element
for groups, 137 associate, 228
for rings, 198 identity, 34
Coset inverse, 34
leader, 108 irreducible, 228
left, 74 order of, 47
representative, 74 prime, 228
right, 74 primitive, 297
Coset decoding, 107 transcendental, 262
Cryptanalysis, 82 Equivalence class, 12
Cryptosystem Equivalence relation, 11
rsa, 84 Euclidean algorithm, 22
affine, 83 Euclidean domain, 231
definition of, 81 Euclidean group, 147
monoalphabetic, 82 Euclidean inner product, 144
polyalphabetic, 83 Euclidean valuation, 231
private key, 81 Euler ϕ-function, 77
public key, 81 Euler, Leonhard, 78, 275
single key, 81 Extension
Cycle algebraic, 262
definition of, 61 field, 260
disjoint, 61 finite, 264
normal, 299
De Morgan’s laws radical, 303
for Boolean algebras, 243 separable, 279, 296
for sets, 5 simple, 262
De Morgan, Augustus, 248 External direct product, 118
Decoding table, 108 Faltings, Gerd, 275
Deligne, Pierre, 275 Feit, W., 130, 178
DeMoivre’s Theorem, 51 Fermat’s factorizationalgorithm, 88
Derivative, 279 Fermat’s Little Theorem, 78
Determinant, Vandermonde, 285 Fermat, Pierre de, 77, 274
Dickson, L. E., 130 Ferrari, Ludovico, 219
Diffie, W., 83 Ferro, Scipione del, 219
Direct product of groups Field, 191
external, 118 algebraically closed, 267
internal, 120 base, 260
Discriminant extension, 260
of the cubic equation, 223 fixed, 298
of the quadratic equation, 222 Galois, 280
Division algorithm of fractions, 227
for integers, 20 of quotients, 227
for polynomials, 212 splitting, 268
Division ring, 190 Finitely generated group, 158
Domain Fior, Antonio, 219
Euclidean, 231 First Isomorphism Theorem
principal ideal, 229 for groups, 135
unique factorization, 228 for rings, 197
Doubling the cube, 274 Fixed point set, 169
INDEX 341
Freshman’s Dream, 278 finite, 36
Function finitely generated, 158
bijective, 7 Galois, 294
Boolean, 176, 251 general linear, 35, 143
composition of, 7 generators of, 158
definition of, 6 homomorphism of, 133
domain of, 6 infinite, 36
identity, 9 isomorphic, 114
injective, 7 isomorphism of, 114
invertible, 9 nonabelian, 34
one-to-one, 7 noncommutative, 34
onto, 7 of units, 35
range of, 6 order of, 36
surjective, 7 orthogonal, 144
switching, 176, 251 permutation, 60
Fundamental Theorem point, 152
of Algebra, 268, 307 quaternion, 36
of Arithmetic, 22 quotient, 126
of Finite Abelian Groups, 159 simple, 127, 130
Fundamental Theorem of Galois Theory, solvable, 165
300 space, 152
special linear, 38, 143
Galois field, 280 special orthogonal, 147
Galois group, 294 symmetric, 59
Galois, Évariste, 37, 302 symmetry, 149
Gauss’s Lemma, 232 Gödel, Kurt, 248
Gauss, Karl Friedrich, 234
Gaussian integers, 194 Hamming distance, 96
Generator of a cyclic subgroup, 46 Hamming, R., 98
Generators for a group, 158 Hellman, M., 83
Glide reflection, 148 Hilbert, David, 154, 200, 248, 275
Gorenstein, Daniel, 130 Homomorphic image, 133
Greatest common divisor Homomorphism
of two integers, 20 canonical, 135, 197
of two polynomials, 214 evaluation, 196, 212
Greatest lower bound, 239 kernel of a group, 134
Greiss, R., 130 kernel of a ring, 195
Grothendieck, Alexander, 275 natural, 135, 197
Group of groups, 133
p-group, 159, 181 ring, 195
abelian, 34
action, 168 Ideal
alternating, 64 definition of, 196
center of, 171 maximal, 198
circle, 52 one-sided, 197
commutative, 34 prime, 199
cyclic, 46 principal, 196
definition of, 33 trivial, 196
dihedral, 65 two-sided, 197
Euclidean, 147 Indeterminate, 209
factor, 126 Index of a subgroup, 75
INDEX 342
Induction Lower bound, 239
first principle of, 17
second principle of, 19 Mapping, see Function
Infimum, 239 Matrix
Inner product, 99 distance-preserving, 145
Integral domain, 190 generator, 101
Internal direct product, 120 inner product-preserving, 145
International standard book number, 44 invertible, 142
Irreducible element, 228 length-preserving, 145
Irreducible polynomial, 215 nonsingular, 143
Isometry, 148 null space of, 100
Isomorphism orthogonal, 144
of Boolean algebras, 244 parity-check, 101
of groups, 114 similar, 12
ring, 195 unimodular, 151
Matrix, Vandermonde, 285
Join, 240 Maximal ideal, 198
Jordan, C., 130 Maximum-likelihood decoding, 94
Jordan-Hölder Theorem, 164 Meet, 240
Minimal generator polynomial, 284
Kernel Minimal polynomial, 263
of a group homomorphism, 134 Minkowski, Hermann, 275
of a ring homomorphism, 195 Monic polynomial, 209
Key Mordell-Weil conjecture, 275
definition of, 81 Multiplicity of a root, 296
private, 81
Noether, A. Emmy, 199
public, 81
Noether, Max, 199
single, 81
Normal extension, 299
Klein, Felix, 37, 141, 200
Normal series of a group, 162
Kronecker delta, 103, 145
Normal subgroup, 125
Kronecker, Leopold, 275
Normalizer, 182
Kummer, Ernst, 275
Null space
Lagrange’s Theorem, 76 of a matrix, 100
Lagrange, Joseph-Louis, 37, 65, 78
Odd Order Theorem, 186
Laplace, Pierre-Simon, 65
Orbit, 169
Lattice Orthogonal group, 144
completed, 241 Orthogonal matrix, 144
definition of, 240 Orthonormal set, 145
distributive, 242
Lattice of points, 151 Partial order, 238
Lattices, Principle of Duality for, 240 Partially ordered set, 238
Least upper bound, 239 Partitions, 12
Left regular representation, 117 Permutation
Lie, Sophus, 37, 184 cycle structure of, 79
Linear combination, 254 definition of, 9, 59
Linear dependence, 254 even, 64
Linear independence, 254 odd, 64
Linear map, 141 Permutation group, 60
Linear transformation Plaintext, 81
definition of, 9, 141 Polynomial
INDEX 343
code, 283 Rivest, R., 84
content of, 232 Ruffini, P., 302
definition of, 209 Russell, Bertrand, 248
degree of, 210
error, 291 Scalar product, 252
error-locator, 291 Second Isomorphism Theorem
greatest common divisor of, 214 for groups, 136
in n indeterminates, 212 for rings, 198
irreducible, 215 Shamir, A., 84
leading coefficient of, 209 Shannon, C., 98
minimal, 263 Sieve of Eratosthenes, 26
minimal generator, 284 Simple extension, 262
monic, 209 Simple group, 127
primitive, 232 Simple root, 296
root of, 214 Solvability by radicals, 303
separable, 296 Spanning set, 254
zero of, 214 Splitting field, 268
Polynomial separable, 279 Squaring the circle is impossible, 274
Poset Standard decoding, 107
definition of, 238 Subgroup
largest element in, 241 p-subgroup, 181
smallest element in, 241 centralizer, 171
Power set, 238 commutator, 185
Prime element, 228 cyclic, 46
Prime ideal, 199 definition of, 38
Prime integer, 22 index of, 75
Primitive nth root of unity, 52, 303 isotropy, 169
Primitive element, 297 normal, 125
Primitive Element Theorem, 297 normalizer of, 182
Primitive polynomial, 232 proper, 38
Principal ideal, 196 stabilizer, 169
Principal ideal domain (pid), 229 Sylowp-subgroup, 182
Principal series, 163 translation, 152
Pseudoprime, 89 trivial, 38
Subnormal series of a group, 162
Quaternions, 36, 192 Subring, 193
Supremum, 239
Resolvent cubic equation, 223 Switch
Rigid motion, 31, 148 closed, 246
Ring definition of, 246
characteristic of, 194 open, 246
commutative, 190 Switching function, 176, 251
definition of, 190 Sylow p-subgroup, 182
division, 190 Sylow, Ludvig, 184
factor, 197 Syndrome of a code, 106, 291
homomorphism, 195
isomorphism, 195 Tartaglia, 219
Noetherian, 230 Third Isomorphism Theorem
quotient, 197 for groups, 137
with identity, 190 for rings, 198
with unity, 190 Thompson, J., 130, 178
INDEX 344
Transcendental element, 262 definition of, 252
Transcendental number, 262 dimension of, 256
Transposition, 63 subspace of, 253
Trisection of an angle, 274
Weight of a codeword, 96
Unique factorization domain (ufd), 228 Weil, André, 275
Unit, 190, 228 Well-defined map, 7
Universal Product Code, 43 Well-ordered set, 19
Upper bound, 239 Whitehead, Alfred North, 248
Vandermonde determinant, 285 Zero
Vandermonde matrix, 285 multiplicity of, 296
Vector space of a polynomial, 214
basis of, 255 Zero divisor, 191
Colophon
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