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 10, 2019
Edition: Annual Edition 2019
Website: abstract.pugetsound.edu
©1997–2019 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 com-
ments and suggestions.
• 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 available as an open source textbook, a decision that I have
never regretted. With his assistance, the book has been rewritten in Pre-
TeXt (pretextbook.org), 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).
v
vi
Preface
This text is intended for a one or two-semester undergraduate course in
abstract algebra. Traditionally, these courses have covered the theoreti-
cal 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 signif-
icantly.
Until recently most abstract algebra texts included few if any applica-
tions. However, one of the major problems in teaching an abstract algebra
course is that for many students it is their first encounter with an envi-
ronment 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.)
vii
viii
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; computational, 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 problems require careful thought, so the student
should not be discouraged if the solution is not forthcoming after a few
minutes of work.
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.org) is a free, open source, software system for ad-
vanced 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.com). Robert Beezer has written a comprehensive
ix
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 8.8 (released 2019-07-02).
Thomas W. Judson
Nacogdoches, Texas 2019
x
Contents
Acknowledgements v
Preface vii
1 Preliminaries 1
1.1 A Short Note on Proofs . . . . . . . . . . . . . . . . . . . 1
1.2 Sets and Equivalence Relations . . . . . . . . . . . . . . . 3
1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.4 References and Suggested Readings . . . . . . . . . . . . . 16
2 The Integers 17
2.1 Mathematical Induction . . . . . . . . . . . . . . . . . . . 17
2.2 The Division Algorithm . . . . . . . . . . . . . . . . . . . 20
2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.4 Programming Exercises . . . . . . . . . . . . . . . . . . . 26
2.5 References and Suggested Readings . . . . . . . . . . . . . 27
3 Groups 29
3.1 Integer Equivalence Classes and Symmetries . . . . . . . . 29
3.2 Definitions and Examples . . . . . . . . . . . . . . . . . . 33
3.3 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.5 Additional Exercises: Detecting Errors . . . . . . . . . . . 44
3.6 References and Suggested Readings . . . . . . . . . . . . . 45
4 Cyclic Groups 47
4.1 Cyclic Subgroups . . . . . . . . . . . . . . . . . . . . . . . 47
4.2 Multiplicative Group of Complex Numbers . . . . . . . . 50
4.3 The Method of Repeated Squares . . . . . . . . . . . . . . 54
4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.5 Programming Exercises . . . . . . . . . . . . . . . . . . . 59
4.6 References and Suggested Readings . . . . . . . . . . . . . 59
5 Permutation Groups 61
5.1 Definitions and Notation . . . . . . . . . . . . . . . . . . . 61
5.2 Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . 68
5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
xi
xii
6 Cosets and Lagrange’s Theorem 75
6.1 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
6.2 Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . 77
6.3 Fermat’s and Euler’s Theorems . . . . . . . . . . . . . . . 79
6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
7 Introduction to Cryptography 83
7.1 Private Key Cryptography . . . . . . . . . . . . . . . . . . 84
7.2 Public Key Cryptography . . . . . . . . . . . . . . . . . . 86
7.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
7.4 Additional Exercises: Primality and Factoring . . . . . . . 90
7.5 References and Suggested Readings . . . . . . . . . . . . . 92
8 Algebraic Coding Theory 93
8.1 Error-Detecting and Correcting Codes . . . . . . . . . . . 93
8.2 Linear Codes . . . . . . . . . . . . . . . . . . . . . . . . . 100
8.3 Parity-Check and Generator Matrices . . . . . . . . . . . 103
8.4 Efficient Decoding . . . . . . . . . . . . . . . . . . . . . . 108
8.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
8.6 Programming Exercises . . . . . . . . . . . . . . . . . . . 115
8.7 References and Suggested Readings . . . . . . . . . . . . . 115
9 Isomorphisms 117
9.1 Definition and Examples . . . . . . . . . . . . . . . . . . . 117
9.2 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . 121
9.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
10 Normal Subgroups and Factor Groups 129
10.1 Factor Groups and Normal Subgroups . . . . . . . . . . . 129
10.2 The Simplicity of the Alternating Group . . . . . . . . . . 131
10.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
11 Homomorphisms 137
11.1 Group Homomorphisms . . . . . . . . . . . . . . . . . . . 137
11.2 The Isomorphism Theorems . . . . . . . . . . . . . . . . . 139
11.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
11.4 Additional Exercises: Automorphisms . . . . . . . . . . . 143
12 Matrix Groups and Symmetry 145
12.1 Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . 145
12.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
12.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
12.4 References and Suggested Readings . . . . . . . . . . . . . 160
13 The Structure of Groups 163
13.1 Finite Abelian Groups . . . . . . . . . . . . . . . . . . . . 163
13.2 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . 168
13.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
13.4 Programming Exercises . . . . . . . . . . . . . . . . . . . 172
13.5 References and Suggested Readings . . . . . . . . . . . . . 172
xiii
14 Group Actions 173
14.1 Groups Acting on Sets . . . . . . . . . . . . . . . . . . . . 173
14.2 The Class Equation . . . . . . . . . . . . . . . . . . . . . 176
14.3 Burnside’s Counting Theorem . . . . . . . . . . . . . . . . 177
14.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
14.5 Programming Exercise . . . . . . . . . . . . . . . . . . . . 186
14.6 References and Suggested Reading . . . . . . . . . . . . . 186
15 The Sylow Theorems 187
15.1 The Sylow Theorems . . . . . . . . . . . . . . . . . . . . . 187
15.2 Examples and Applications . . . . . . . . . . . . . . . . . 190
15.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
15.4 A Project . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
15.5 References and Suggested Readings . . . . . . . . . . . . . 195
16 Rings 197
16.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
16.2 Integral Domains and Fields . . . . . . . . . . . . . . . . . 201
16.3 Ring Homomorphisms and Ideals . . . . . . . . . . . . . . 202
16.4 Maximal and Prime Ideals . . . . . . . . . . . . . . . . . . 206
16.5 An Application to Software Design . . . . . . . . . . . . . 208
16.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
16.7 Programming Exercise . . . . . . . . . . . . . . . . . . . . 215
16.8 References and Suggested Readings . . . . . . . . . . . . . 215
17 Polynomials 217
17.1 Polynomial Rings . . . . . . . . . . . . . . . . . . . . . . . 217
17.2 The Division Algorithm . . . . . . . . . . . . . . . . . . . 220
17.3 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . 223
17.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
17.5 Additional Exercises: Solving the Cubic and Quartic Equa-
tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
18 Integral Domains 233
18.1 Fields of Fractions . . . . . . . . . . . . . . . . . . . . . . 233
18.2 Factorization in Integral Domains . . . . . . . . . . . . . . 236
18.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
18.4 References and Suggested Readings . . . . . . . . . . . . . 246
19 Lattices and Boolean Algebras 247
19.1 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
19.2 Boolean Algebras . . . . . . . . . . . . . . . . . . . . . . . 250
19.3 The Algebra of Electrical Circuits . . . . . . . . . . . . . . 255
19.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
19.5 Programming Exercises . . . . . . . . . . . . . . . . . . . 260
19.6 References and Suggested Readings . . . . . . . . . . . . . 260
20 Vector Spaces 261
20.1 Definitions and Examples . . . . . . . . . . . . . . . . . . 261
20.2 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
20.3 Linear Independence . . . . . . . . . . . . . . . . . . . . . 263
20.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
20.5 References and Suggested Readings . . . . . . . . . . . . . 268
xiv
21 Fields 269
21.1 Extension Fields . . . . . . . . . . . . . . . . . . . . . . . 269
21.2 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . 278
21.3 Geometric Constructions . . . . . . . . . . . . . . . . . . . 280
21.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
21.5 References and Suggested Readings . . . . . . . . . . . . . 287
22 Finite Fields 289
22.1 Structure of a Finite Field . . . . . . . . . . . . . . . . . . 289
22.2 Polynomial Codes . . . . . . . . . . . . . . . . . . . . . . 293
22.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 300
22.4 Additional Exercises: Error Correction for BCH Codes . . 302
22.5 References and Suggested Readings . . . . . . . . . . . . . 302
23 Galois Theory 305
23.1 Field Automorphisms . . . . . . . . . . . . . . . . . . . . 305
23.2 The Fundamental Theorem . . . . . . . . . . . . . . . . . 309
23.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 316
23.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
23.5 References and Suggested Readings . . . . . . . . . . . . . 322
A GNU Free Documentation License 325
B Hints and Answers to Selected Exercises 333
C Notation 347
Index 351
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 sci-
ences such as chemistry and physics, scientists perform experiments to
discover new principles and verify theories. Although mathematics is of-
ten motivated by physical experimentation or by computer simulations,
it is made rigorous through the use of logical arguments. In studying ab-
stract 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 arguments. 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 ̸= 0, then
√
−b ± b2 − 4ac
x= .
2a
• x3 − 4x2 + 5x − 6.
1
2 CHAPTER 1 PRELIMINARIES
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 instance, we can see that the
statement “2x = 6 exactly when x = 4” is false by evaluating 2 · 4 and
noting that 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 ̸= 0, then
√
−b ± b2 − 4ac
x= .
2a
The hypothesis is ax2 + bx + c = 0 and a ̸= 0; the conclusion is
√
−b ± b2 − 4ac
x= .
2a
Notice that the statement says nothing about whether or not the hypoth-
esis is true. However, if this entire statement is true and we can show
that ax2 + bx + c = 0 with a ̸= 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 proposition 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 support-
ing propositions, which are called lemmas, and use the results of these
1.2 SETS AND EQUIVALENCE RELATIONS 3
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 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 statement “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 sometimes 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 prov-
ing theorems. Theorems are tools that make new and productive appli-
cations 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.
4 CHAPTER 1 PRELIMINARIES
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}.
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 follow-
ing:
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 ̸= 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 oper-
ations 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 ̸= A. If A is not a subset of B, we write A ̸⊂ B; for
example, {4, 7, 9} ̸⊂ {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 opera-
tions: 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}.
1.2 SETS AND EQUIVALENCE RELATIONS 5
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}.
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}
6 CHAPTER 1 PRELIMINARIES
= {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 ′ .
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}.
1.2 SETS AND EQUIVALENCE RELATIONS 7
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 (a, b) ∈ f if for every element a ∈ A there exists a unique
element b ∈ B. Another way of saying this is that for every element in A,
f
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.
Example 1.6 Suppose A = {1, 2, 3} and B = {a, b, c}. In Figure 1.7, p. 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 de-
scribing what the function does to each specific element in the domain.
However, not all functions can be described in this manner. For exam-
ple, 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
8 CHAPTER 1 PRELIMINARIES
one-to-one or injective if a1 ̸= a2 implies f (a1 ) ̸= 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)).
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, p. 8 (top). The composition of these functions,
g ◦ f : A → C, is defined in Figure 1.9, p. 8 (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 ̸= g ◦ f .
□
Example 1.12
√ Sometimes it is the case that f ◦ g = g ◦ f . Let f (x) = x3
and g(x) = x. Then
3
√ √
(f ◦ g)(x) = f (g(x)) = f ( 3 x ) = ( 3 x )3 = x
and √
(g ◦ f )(x) = g(f (x)) = g(x3 ) =
3
x3 = x.
□
1.2 SETS AND EQUIVALENCE RELATIONS 9
Example 1.13 Given a 2 × 2 matrix
( )
a b
A= ,
c d
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.
■
10 CHAPTER 1 PRELIMINARIES
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 .
√
Example 1.16 The function f (x) = x3 has inverse f −1 (x) = 3 x by
Example 1.12, p. 8. □
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,
( )
2 −1
A−1 = ;
−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. □
1.2 SETS AND EQUIVALENCE RELATIONS 11
Example 1.19 Given the permutation
( )
1 2 3
π=
2 3 1
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 gen-
eralize 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 ̸= 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
12 CHAPTER 1 PRELIMINARIES
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 ma-
trices, by saying A ∼ B if there exists 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 P of a set X is a collection
∪ of nonempty sets X 1 , X 2 ..
, .
such that Xi ∩ Xj = ∅ for i ̸= 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 equiv-
alence 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 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
1.2 SETS AND EQUIVALENCE RELATIONS 13
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.
Example 1.27 In the equivalence relation in Example 1.21, p. 11, 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, p. 11, 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, p. 17.
□
14 CHAPTER 1 PRELIMINARIES
1.3 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
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 .
1.3 EXERCISES 15
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
(d) m ∼ n in Z if m ≡ n
(b) m ∼ n in Z if mn > 0 (mod 6)
26. Define a relation ∼ on R2 by stating that (a, b) ∼ (c, d) if and only
if a2 + b2 ≤ c2 + d2 . 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 counterex-
ample. “The reflexive property is redundant in the axioms for an
equivalence relation. If x ∼ y, then y ∼ x by the symmetric prop-
erty. Using the transitive property, we can deduce that x ∼ x.”
16 CHAPTER 1 PRELIMINARIES
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 ∼ 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.
1.4 References and Suggested Readings
[1] Artin, M. Abstract Algebra. 2nd ed. Pearson, Upper Saddle River,
NJ, 2011.
[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 num-
bers 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.
17
18 CHAPTER 2 THE INTEGERS
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 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 inte-
gers 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 ( )
∑
n n k n−k
(a + b) = a b ,
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
2.1 MATHEMATICAL INDUCTION 19
( n ( )
)
∑ n k n−k
= (a + b) a b
k
k=0
∑n ( ) n ( )
n k+1 n−k ∑ n k n+1−k
= a b + a b
k k
k=0 k=0
∑n ( ) ∑ n ( )
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
∑n [( ) ( )]
n n
= an+1 + + ak bn+1−k + bn+1
k−1 k
k=1
∑ 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 ele-
ment. 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 Math-
ematical 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 S = N. ■
Theorem 2.8 The Principle of Mathematical Induction implies the Prin-
ciple 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, p. 19. 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 in-
stance, there are two ways to define n!, the factorial of a positive integer
n.
• The explicit definition: n! = 1 · 2 · 3 · · · (n − 1) · n.
20 CHAPTER 2 THE INTEGERS
• 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 un-
derstanding of complex issues.
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 ̸= ∅. 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 ∈ ̸ b and so r < b.
/ S, r =
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
2.2 THE DIVISION ALGORITHM 21
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. 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, p. 20 allows us to compute the great-
est 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 numbers 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
22 CHAPTER 2 THE INTEGERS
= 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 sequence of positive integers r1 > r2 > · · · > rn = d; that is,
b = aq1 + r1
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 ̸= pi
that divides P . ■
2.2 THE DIVISION ALGORITHM 23
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.
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 = p1 p2 · · · pk = q1 q2 · · · ql ,
where p1 ≤ p2 ≤ · · · ≤ pk and q1 ≤ q2 ≤ · · · ≤ ql . By Lemma 2.13, p. 22,
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 impor-
tant results from antiquity 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
n
integer 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 conjecture 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
24 CHAPTER 2 THE INTEGERS
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.
2.3 Exercises
1. Prove that
n(n + 1)(2n + 1)
1 2 + 2 2 + · · · + 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
1∑
n
√n
a1 a2 · · · an ≤ ak .
n
k=1
(n)
8. Prove the Leibniz rule for f (x), where f (n) is the nth derivative
of f ; that is, show that
n ( )
∑ n
(f g)(n) (x) = f (k) (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.
2.3 EXERCISES 25
13. Prove that the two principles of mathematical induction stated in
Section 2.1, p. 17 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 limn→∞ fn /fn+1 = ( 5 − 1)/2.
(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
a 2 + b2 = r 2
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
26 CHAPTER 2 THE INTEGERS
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 p2 = 2q 2 . Demonstrate that therefore 2 cannot
be a rational number.
2.4 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 algo-
rithm. 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.
2.5 REFERENCES AND SUGGESTED READINGS 27
2.5 References and Suggested Readings
[1] Brookshear, J. G. Theory of Computation: Formal Languages, Au-
tomata, and Complexity. Benjamin/Cummings, Redwood City,
CA, 1989. Shows the relationships of the theoretical aspects of
computer science to set theory and the integers.
[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.
28 CHAPTER 2 THE INTEGERS
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 poly-
nomial in terms of its coefficients. Groups now play a central role in such
areas as coding theory, counting, and the study of symmetries; many ar-
eas of biology, chemistry, and physics have benefited from group theory.
3.1 Integer Equivalence Classes and Symme-
tries
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 appli-
cations 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, . . .}.
29
30 CHAPTER 3 GROUPS
When no confusion can arise, we will use 0, 1, . . . , 11 to indicate the equiv-
alence 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 n. Similarly, multipli-
cation 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 mod-
ulo 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 Table 3.3, p. 30. 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
Table 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).
3.1 INTEGER EQUIVALENCE CLASSES AND SYMMETRIES 31
4. Multiplication distributes over addition:
a(b + c) ≡ ab + ac (mod n).
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. ■
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 arrangement of its sides and vertices as well as its distances
32 CHAPTER 3 GROUPS
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, p. 31, 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.
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 △ABC. To
find a symmetry of △ABC, 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
3.2 DEFINITIONS AND EXAMPLES 33
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 sym-
metries are shown in Figure 3.6, p. 32.
A natural question to ask is what happens if one motion of the triangle
△ABC 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 ̸= µ1 ρ1 . A multiplication table for the
symmetries of an equilateral triangle △ABC is given in Table 3.7, p. 33.
Notice that in the multiplication table for the symmetries of an equi-
lateral 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 µ3 id ρ1 ρ2
µ2 µ2 µ3 µ1 ρ2 id ρ1
µ3 µ3 µ1 µ2 ρ1 ρ2 id
Table 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.
• The law of composition is associative. That is,
(a ◦ b) ◦ c = a ◦ (b ◦ c)
for a, b, c ∈ G.
34 CHAPTER 3 GROUPS
• 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 commutative. 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. Table 3.10, p. 34 is a
Cayley table for Z5 . By Proposition 3.4, p. 30, 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
Table 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
3.2 DEFINITIONS AND EXAMPLES 35
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 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, p. 30, 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 .
Table 3.12, p. 35 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
Table 3.12: Multiplication table for U (8)
□
Example 3.13 The symmetries of an equilateral triangle described in
Section 3.1, p. 29 form a nonabelian group. As we observed, it is not nec-
essarily true that αβ = βα for two symmetries α and β. Using Table 3.7,
p. 33, 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 ̸= 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 1 d −b
A = .
ad − bc −c a
The product of two invertible matrices is again invertible. Matrix multi-
plication is associative, 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. □
36 CHAPTER 3 GROUPS
Example 3.15 Let
( ) ( )
1 0 0 1
1= I=
0 1 −1 0
( ) ( )
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 =
a2 + 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 equation by a, we have
(a−1 )−1 = e(a−1 )−1 = aa−1 (a−1 )−1 = ae = a.
■
3.2 DEFINITIONS AND EXAMPLES 37
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.
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 ̸= 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 methods of solving polynomial equations. Later,
38 CHAPTER 3 GROUPS
Évariste Galois (1811–1832) succeeded in developing 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 pro-
posed that geometric spaces should be studied by examining those prop-
erties 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.
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 ma-
trices 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
3.3 SUBGROUPS 39
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 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). Table 3.29,
p. 39 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)
Table 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.
40 CHAPTER 3 GROUPS
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 ̸= ∅, 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 ̸= ∅ 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. We must show that their
product is also in H. However, h1 (h−12 )
−1
= h1 h2 ∈ H. Hence, H 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 pri-
marly for just group theory, and in continuous development since 1986.
Many of Sage’s computations for groups ultimately are performed by
GAP.
3.4 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 =
3.4 EXERCISES 41
{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 sym-
metries 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?
5. Describe the symmetries of a square and prove that the set of sym-
metries is a group. Give a Cayley table for the symmetries. How
many ways can the vertices of a square be permuted? Is each permu-
tation 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 ̸= 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 mul-
tiplication 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 ).
42 CHAPTER 3 GROUPS
Prove that Zn2 is a group under this operation. This group is impor-
tant in algebraic coding theory.
13. Show that R∗ = R \ {0} is a group under the operation of multipli-
cation.
14. Given the groups R∗ and Z, let G = R∗ × Z. Define a binary opera-
tion ◦ 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 ̸= 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).
22. Show that addition and multiplication mod n are well defined op-
erations. 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 opera-
tions.
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 ̸= 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, p. 37: 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, p. 37.
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.
3.4 EXERCISES 43
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, p. 39 for a
short description of the product of groups.)
35. Find all the subgroups of the symmetry group of an equilateral tri-
angle.
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.
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 sub-
group 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.
44 CHAPTER 3 GROUPS
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 cen-
tralizer 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.
3.5 Additional Exercises: Detecting Errors
1. UPC Symbols. Universal Product Code (upc) symbols are found
on most products 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, p. 44). 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, p. 44, is a valid upc number.
(b) Show that the number 0-50000-30043-6 is not a valid upc num-
ber.
(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 inter-
changed. 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.
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
3.6 REFERENCES AND SUGGESTED READINGS 45
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 detec-
tion 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.
3.6 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 Market-
place,” The American 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 Soci-
ety, Providence, 1959.
[5] Kurosh, A. E. The Theory of Groups, vols. I and II. American
Mathematical Society, Providence, 1979.
46 CHAPTER 3 GROUPS
[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 un-
derstood groups, are both examples of what are called cyclic groups. In
this chapter we will study the properties of cyclic groups and cyclic sub-
groups, 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 determined 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 multi-
plicative 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,
p. 40, 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
⟨a⟩ = {ak : k ∈ Z}
is a subgroup of G. Furthermore, ⟨a⟩ is the smallest subgroup of G that
contains a.
Proof. The identity is in ⟨a⟩ since a0 = e. If g and h are any two elements
in ⟨a⟩, then by the definition of ⟨a⟩ we can write g = am and h = an for
some integers m and n. So gh = am an = am+n is again in ⟨a⟩. Finally,
if g = an in ⟨a⟩, then the inverse g −1 = a−n is also in ⟨a⟩. Clearly,
any subgroup H of G containing a must contain all the powers of a by
closure; hence, H contains ⟨a⟩. Therefore, ⟨a⟩ is the smallest subgroup of
47
48 CHAPTER 4 CYCLIC GROUPS
G containing a. ■
Remark 4.4 If we are using the “+” notation, as in the case of the
integers under addition, we write ⟨a⟩ = {na : n ∈ Z}.
For a ∈ G, we call ⟨a⟩ the cyclic subgroup generated by a. If G
contains some element a such that G = ⟨a⟩, then G is a cyclic group.
In this case a is a generator of G. If a is an 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 ⟨2⟩ = {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 Table 3.7, p. 33. The subgroups of S3 are shown in Figure 4.8,
p. 48. 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. ■
4.1 CYCLIC SUBGROUPS 49
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. Conse-
quently,
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 Proposition 4.12, p. 49, 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.
50 CHAPTER 4 CYCLIC GROUPS
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
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 Num-
bers
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 poly-
nomials. 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 = 2 .
a + 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.
□
4.2 MULTIPLICATIVE GROUP OF COMPLEX NUMBERS 51
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, p. 51.
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 coor-
dinates. To specify 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, p. 51. We can see that
z = a + bi = r(cos θ + i sin θ).
Hence, √
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.
52 CHAPTER 4 CYCLIC GROUPS
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
√ √
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)]
= rn+1 [cos(n + 1)θ + i sin(n + 1)θ].
■
10
Example 4.23 Suppose that z = 1 + i and we wish to compute z .
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
4.2 MULTIPLICATIVE GROUP OF COMPLEX NUMBERS 53
= 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, p. 52.
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 num-
bers 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,
( )
2kπ
z n = 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, p. 222,
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 prim-
itive nth root of unity.
Example 4.26 The 8th roots of unity can be represented as eight equally
spaced points on the unit circle (Figure 4.27, p. 54). 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
54 CHAPTER 4 CYCLIC GROUPS
y
i
ω3 ω
−1 0 1 x
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 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.1
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
k
c ≡ ay (mod n), 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)
..
.
1 The results in this section are needed only in Chapter 7, p. 83
4.3 THE METHOD OF REPEATED SQUARES 55
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)
20 26 8
≡ 271 · 271 · 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, p. 83. 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.
56 CHAPTER 4 CYCLIC GROUPS
4.4 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
(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).
4.4 EXERCISES 57
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∗
24
11. If a = 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
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|.
58 CHAPTER 4 CYCLIC GROUPS
(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 ⟨g⟩ ∩ ⟨h⟩?
28. Let a be an element in a group G. What is a generator for the
subgroup ⟨am ⟩ ∩ ⟨an ⟩?
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 ⟨a⟩ ∩ ⟨b⟩ = {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, . . ..
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| ̸= 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.
4.5 PROGRAMMING EXERCISES 59
4.5 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?
4.6 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 The-
ory, Pomerance, C., ed. Proceedings of Symposia in Applied Math-
ematics, vol. 42, American Mathematical Society, Providence, RI,
1990. This book gives an excellent account of how the method of
repeated squares is used in cryptography.
60 CHAPTER 4 CYCLIC GROUPS
5
Permutation Groups
Permutation groups are central to the study of geometric symmetries and
to Galois theory, 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 trian-
gle △ABC from Chapter 3, p. 29. 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.
61
62 CHAPTER 5 PERMUTATION GROUPS
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 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
□
5.1 DEFINITIONS AND NOTATION 63
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 permu-
tation groups, we need a more streamlined 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= = (162354)
6 3 5 1 4 2 7
is a cycle of length 6, whereas
( )
1 2 3 4 5 6
τ= = (243)
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
= (1243)(56).
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 = (1352) and τ = (256).
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 = (1356). If µ = (1634), then σµ = (1652)(34). □
Two cycles in SX , σ = (a1 , a2 , . . . , ak ) and τ = (b1 , b2 , . . . , bl ), are
disjoint if ai ̸= bj for all i and j.
Example 5.7 The cycles (135) and (27) are disjoint; however, the cycles
64 CHAPTER 5 PERMUTATION GROUPS
(135) and (347) are not. Calculating their products, we find that
(135)(27) = (135)(27)
(135)(347) = (13475).
The product of two cycles that are not disjoint may reduce to something
less complicated; the product of disjoint cycles cannot be simplified. □
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. ■
5.1 DEFINITIONS AND NOTATION 65
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
Using cycle notation, we can write
σ = (1624)
τ = (13)(456)
στ = (136)(245)
τ σ = (143)(256).
□
Remark 5.11 From this point forward we will find it convenient to use
cycle notation to represent 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 proposition.
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
(16)(253) = (16)(23)(25) = (16)(45)(23)(45)(25).
As we can see, there is no unique way to represent permutation as the
product of transpositions. For instance, we can write the identity permu-
tation as (12)(12), as (13)(24)(13)(24), 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 (16) by
(23)(16)(23)
or by
(35)(16)(13)(16)(13)(35)(56),
but (16) 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.
66 CHAPTER 5 PERMUTATION GROUPS
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:
(ab)(ab) = id
(bc)(ab) = (ac)(bc)
(cd)(ab) = (ab)(cd)
(ac)(ab) = (ab)(bc),
where a, b, c, and d are distinct.
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 transposi-
tions 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, be-
cause 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, p. 65. 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, p. 66, 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.
5.1 DEFINITIONS AND NOTATION 67
The Alternating Groups
One of the most important subgroups of Sn is the set of all even per-
mutations, 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 trans-
position is itself,
σ −1 = σr σr−1 · · · σ1
is also in An . ■
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) (12)(34) (13)(24) (14)(23)
(123) (132) (124) (142)
(134) (143) (234) (243).
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 Revo-
lution. 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 mathemat-
ics. Cauchy began his career as a mathematician by solving a problem
68 CHAPTER 5 PERMUTATION GROUPS
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, p. 29. 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,
p. 68). 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.
1
n 2
n−1 3
4
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, satisfying the relations
rn = 1
s2 = 1
srs = r−1 .
Proof. The possible motions of a regular n-gon are either reflections or
rotations (Figure 5.22, p. 69). 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
5.2 DIHEDRAL GROUPS 69
1 2
8 2 1 3
rotation
7 3 8 4
6 4 7 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, p. 69). 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 = srk . 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, sr, sr2 , . . . , srn−1 }.
We will leave the proof that srs = r−1 as an exercise.
1 1
6 2 2 6
5 3 3 5
4 4
1 1
5 2 2 5
4 3 3 4
Figure 5.23: Types of reflections of a regular n-gon
70 CHAPTER 5 PERMUTATION GROUPS
■
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, p. 70),
the rotations are
r = (1234)
r2 = (13)(24)
r3 = (1432)
r4 = (1)
and the reflections are
s1 = (24)
s2 = (13).
The order of D4 is 8. The remaining two elements are
rs1 = (12)(34)
r3 s1 = (14)(23).
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 permu-
tation 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, p. 70, 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
5.3 EXERCISES 71
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, p. 71).
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, p. 71). 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 .
1 2
4 3
3 4
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 Exercises
72 CHAPTER 5 PERMUTATION GROUPS
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) (1345)(234) (i) (123)(45)(1254)−2
(b) (12)(1253) (j) (1254)100
(c) (143)(23)(24) (k) |(1254)|
(d) (1423)(34)(56)(1324) (l) |(1254)2 |
(e) (1254)(13)(25) (m) (12)−1
(f) (1254)(13)(25)2 (n) (12537)−1
(g) (1254)−1 (123)(45)(1254) (o) [(12)(34)(12)(47)]−1
(h) (1254)2 (123)(45) (p) [(1235)(467)]−1
3. Express the following permutations as products of transpositions and
identify them as even or odd.
(a) (14356) (d) (17254)(1423)(154632)
(b) (156)(234)
(c) (1426)(142) (e) (142637)
−1
4. Find (a1 , a2 , . . . , an ) .
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 ?
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 .
5.3 EXERCISES 73
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, p. 71, 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) (12), (13), . . . , (1n)
(b) (12), (23), . . . , (n − 1, n)
(c) (12), (12 . . . 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.
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, we will say that x ∼ y.
(a) Show that ∼ is an equivalence relation on X.
(b) If σ ∈ An and τ ∈ Sn , show that τ −1 στ ∈ An .
74 CHAPTER 5 PERMUTATION GROUPS
(c) Define the orbit of x ∈ X under σ ∈ SX to be the set
Ox,σ = {y : x ∼ y}.
Compute the orbits of each of the following elements in S5 :
α = (1254)
β = (123)(45)
γ = (13)(25).
(d) If Ox,σ ∩ Oy,σ ̸= ∅, prove that Ox,σ = Oy,σ . The orbits under
a permutation σ are the equivalence classes corresponding to
the equivalence relation ∼.
(e) 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 ⟨σ⟩ 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. Show that α−1 β −1 αβ is even for α, β ∈ Sn .
36. Let r and s be the elements in Dn described in Theorem 5.21, p. 68
(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), (123), (132)}. The left cosets of H are
(1)H = (123)H = (132)H = {(1), (123), (132)}
(12)H = (13)H = (23)H = {(12), (13), (23)}.
75
76 CHAPTER 6 COSETS AND LAGRANGE’S THEOREM
The right cosets of H are exactly the same as the left cosets:
H(1) = H(123) = H(132) = {(1), (123), (132)}
H(12) = H(13) = H(23) = {(12), (13), (23)}.
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), (12)}.
Then the left cosets of K are
(1)K = (12)K = {(1), (12)}
(13)K = (123)K = {(13), (123)}
(23)K = (132)K = {(23), (132)};
however, the right cosets of K are
K(1) = K(12) = {(1), (12)}
K(13) = K(132) = {(13), (132)}
K(23) = K(123) = {(23), (123)}.
□
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 ̸= ∅ 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, p. 76, 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. □
6.2 LAGRANGE’S THEOREM 77
Example 6.7 Suppose that G = S3 , H = {(1), (123), (132)}, and K =
{(1), (12)}. 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, p. 76, the map ϕ is well-defined; that is, if g1 H = g2 H, then
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, p. 76, 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 ̸= e is a generator.
Proof. Let g be in G such that g ̸= e. Then by Corollary 6.11, p. 77, the
order of g must divide the order of the group. Since |⟨g⟩| > 1, it must be
p. Hence, g generates G. ■
Corollary 6.12, p. 77 suggests that groups of prime order p must some-
how 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|
■
78 CHAPTER 6 COSETS AND LAGRANGE’S THEOREM
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 . Inas-
much 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 (123) is in H. Then (123)−1 = (132) must also
be in H. Since ghg −1 ∈ H for all g ∈ A4 and all h ∈ H and
(124)(123)(124)−1 = (124)(123)(142) = (243)
(243)(123)(243)−1 = (243)(123)(234) = (142)
we can conclude that H must have at least seven elements
(1), (123), (132), (243), (243)−1 = (234), (142), (142)−1 = (124).
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 79
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, p. 30, 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, p. 79 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.
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 va-
cated 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 design-
ing 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.
80 CHAPTER 6 COSETS AND LAGRANGE’S THEOREM
6.4 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.
5. List the left and right cosets of the subgroups in each of the following.
(a) ⟨8⟩ in Z24 (e) An in Sn
(b) ⟨3⟩ in U (8) (f) D4 in S4
(g) T in C∗
(c) 3Z in Z
(h) H = {(1), (123), (132)} in
(d) A4 in S4 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, p. 77 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 = (12)(345)(78)(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.
6.4 EXERCISES 81
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), (123), (132)} in A4 .
21. Let G be a cyclic group of order n. Show that there are exactly ϕ(n)
generators for G.
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 ∑
n= ϕ(d)
d|n
for all positive integers n.
82 CHAPTER 6 COSETS AND LAGRANGE’S THEOREM
7
Introduction to Cryptography
Cryptography is the study of sending and receiving secret messages. The
aim of cryptography 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. Char-
acters can include not only the familiar alphabetic characters A, . . ., Z and
a, . . ., z but also digits, punctuation marks, and blanks. A cryptosys-
tem, or cipher, has two parts: encryption, the process of transforming
a plaintext message to a ciphertext message, and decryption, the re-
verse 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 cryp-
tographic family are distinguished from one another by a parameter to
the encryption function called a key. A classical cryptosystem has a sin-
gle 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 cryp-
tosystem 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.
83
84 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
7.1 Private Key Cryptography
In single or private key cryptosystems the same key is used for both
encrypting and decrypting 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 encrypted 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.
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 mes-
sage, 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 inter-
cepted message. Methods from probability and statistics are great aids
in deciphering an intercepted message; for example, the frequency analy-
sis 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
7.1 PRIVATE KEY CRYPTOGRAPHY 85
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, p. 84, 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. Sup-
pose 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, p. 30, this is possible exactly when a has an
inverse or, equivalently, 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
86 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
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 cryp-
tosystem, 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, p. 85.
7.2 Public Key Cryptography
If traditional cryptosystems are used, anyone who knows enough to en-
code 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 encoding key be kept secret. The encoding function f must be
relatively easy to compute, but f −1 must be extremely difficult to com-
pute without some additional information, so that someone who knows
only the encrypting key cannot find the decrypting key without prohib-
itive 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 com-
putationally 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. Adle-
man 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 in-
structions 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 compu-
tationally 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
7.2 PUBLIC KEY CRYPTOGRAPHY 87
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 re-
peated squares as described in Chapter 4, p. 47. Using the Euclidean
algorithm, we determine that 191E = 1 + 151m; therefore, the decrypt-
ing 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); 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, p. 79,
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) ̸= 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, p. 79, 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 cryptosys-
tem. 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, p. 87, 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
88 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
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 mes-
sages probably began with the Arabs in the 1400s. In the fifteenth and
sixteenth centuries mathematicians such as Alberti and Viete discovered
that monoalphabetic cryptosystems offered no real security. In the 1800s,
F. W. Kasiski established methods for breaking ciphers in which a cipher-
text 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 encrypting and decrypting messages, and mathematicians
were very active in cryptography during World War II. Efforts to pene-
trate 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, com-
puter data, and electronic mail. In the early 1970s, ibm developed and
implemented luzifer, the forerunner of the National Bureau of Stan-
dards’ 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 se-
cure any of these systems are. The trapdoor knapsack cryptosystem,
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-
7.3 EXERCISES 89
rsa-challenge-numbers.htm). 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 Depart-
ment’s cryptography division) on the ethical grounds that “gentlemen do
not read each other’s mail.” During the last two decades of the twenti-
eth 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 mathemati-
cal cryptography and computational number theory is very active, and
mathematicians are free to publish their results in these areas.
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 Exercises
1. Encode IXLOVEXMATH using the cryptosystem in Example 7.1, p. 84.
2. Decode ZLOOA WKLVA EHARQ WKHA ILQDO, which was encoded using
the cryptosystem in Example 7.1, p. 84.
3. Assuming that monoalphabetic code was used to encode the follow-
ing 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 cryptogra-
phy?
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
90 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
8. Compute the decoding key D for each of the encoding keys in Exer-
cise 7.3.7, p. 89.
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 cryp-
tosystem, compute D.
(a) (n, E) = (451, 231)
(b) (n, E) = (3053, 1921)
(c) (n, E) = (37986733, 12371)
(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.
7.4 Additional Exercises: Primality and Fac-
toring
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 algorithm.
(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 .
7.4 ADDITIONAL EXERCISES: PRIMALITY AND FACTORING 91
(b) Write a program to implement the following factorization al-
gorithm 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
2. Primality Testing. Recall Fermat’s Little Theorem from Chap-
ter 6, p. 75. 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 pseudo-
primes?
(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 pseudo-
prime 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
92 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
numbers less than 25 × 109 . For more sophisticated primality tests,
see [1], [6], or [7].
7.5 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. Inform. 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 The-
ory”, Proceedings of Symposia in Applied Mathematics 42 (1990)
American Mathematical Society, Providence, RI.
[8] Rivest, R. L., Shamir, A., and Adleman, L., “A Method for Ob-
taining 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 communica-
tions, 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 transmit-
ting and receiving coded messages (Figure 8.1, p. 94).
93
94 CHAPTER 8 ALGEBRAIC CODING THEORY
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 typ-
ically 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 oc-
currence 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 mean-
ingful 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 mes-
sage. 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. Sup-
pose 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), 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).2 This triple-repetition method will automatically de-
tect and correct all single errors, but it is slow and inefficient: to send a
8.1 ERROR-DETECTING AND CORRECTING CODES 95
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.) □
2 We will adopt the convention that bits are numbered left to right in binary n-
tuples.
96 CHAPTER 8 ALGEBRAIC CODING THEORY
The even parity system is easy to implement, but has two drawbacks.
First, multiple 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.
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
Table 8.5: A repetition code
In Table 8.5, p. 96, we present all possible words that might be received
for the transmitted codewords (000) and (111). Table 8.5, p. 96 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, p. 96 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. 3
3 This section requires a knowledge of probability, but can be skipped without loss
of continuity.
8.1 ERROR-DETECTING AND CORRECTING CODES 97
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 trans-
mitter 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, p. 97). The probability that no errors occur
during the transmission of a binary codeword of length n is pn . For ex-
ample, 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
98 CHAPTER 8 ALGEBRAIC CODING THEORY
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 : Zm2 → 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 num-
ber 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 code-
words. 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.
□
8.1 ERROR-DETECTING AND CORRECTING CODES 99
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);
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, p. 99 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.
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
Table 8.12: Distances between 4-bit codewords
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
100 CHAPTER 8 ALGEBRAIC CODING THEORY
transmission of a single bit gives y with d(x, y) = 1, but d(z, y) ≥ 2 for
any other codeword z ̸= 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).
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, p. 100, the codewords c1 = (00000), c2 =
(00111), c3 = (11100), and c4 = (11011) determine a single 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
Table 8.15: Hamming distances for an error-correcting code
□
Historical Note
Modern coding theory began in 1948 with C. Shannon’s paper, “A Math-
ematical 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, al-
ready 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 encoding, decoding, and error detection, we need to add
additional structure to our codes. One way to accomplish this is to require
8.2 LINEAR CODES 101
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 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 ̸= 0}.
Proof. Observe that
dmin = min{d(x, y) : x ̸= y}
= min{d(x, y) : x + y ̸= 0}
= min{w(x + y) : x + y ̸= 0}
= min{w(z) : z ̸= 0}.
■
102 CHAPTER 8 ALGEBRAIC CODING THEORY
Linear Codes
From Example 8.16, p. 101, 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.4
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
= x1 y1 + x2 y2 + · · · + xn yn .
Example 8.19 Suppose that the words to be encoded consist of all bi-
nary 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
1 = 0.
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
4 Since we will be working with matrices, we will write binary n-tuples as column
vectors for the remainder of this chapter.
8.3 PARITY-CHECK AND GENERATOR MATRICES 103
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
Suppose that the 6-tuple x = (010011)t is received. It is a simple matter
of matrix multiplication 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
104 CHAPTER 8 ALGEBRAIC CODING THEORY
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
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
8.3 PARITY-CHECK AND GENERATOR MATRICES 105
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, p. 105). □
Message Word x Codeword Gx
000 000000
001 001101
010 010110
011 011011
100 100011
101 101110
110 110101
111 111000
Table 8.24: A matrix-generated code
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, p. 104, 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 specif-
ically, 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 standard generating matrices, we need to prove a lemma.
Lemma 8.27 ) H = (A | Im ) be an m×n canonical parity-check matrix
( Let
and G = In−mA be the corresponding n × (n − m) standard generator
matrix. Then HG = 0.
106 CHAPTER 8 ALGEBRAIC CODING THEORY
Proof. Let C = HG. The ijth entry in C is
∑
n
cij = hik gkj
k=1
∑
n−m ∑
n
= hik gkj + hik gkj
k=1 k=n−m+1
∑
n−m ∑
n
= aik δkj + δi−(m−n),k akj
k=1 k=n−m+1
= aij + aij
= 0,
where {
1 i=j
δij =
0 i ̸= j
is the Kronecker delta. ■
Theorem 8.28 Let( H =)(A | Im ) be an m × n canonical parity-check
matrix and let G = In−m 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.
Proof. First suppose that y ∈ C. Then Gx = y for some x ∈ Zm 2 . By
Lemma 8.27, p. 105, 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−m2 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+1 = 0
..
.
am1 y1 + am2 y2 + · · · + am,n−m yn−m + yn−m+1 = 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+1 = a21 y1 + a22 y2 + · · · + a2,n−m yn−m
..
.
yn−m+1 = 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
8.3 PARITY-CHECK AND GENERATOR MATRICES 107
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 coor-
dinate 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
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, p. 107, Hei ̸= 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, p. 107. 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
108 CHAPTER 8 ALGEBRAIC CODING THEORY
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 ̸= 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
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 information 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 pos-
sible 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
8.4 EFFICIENT DECODING 109
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
n
of x is Hx. The 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, p. 109
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.
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. □
110 CHAPTER 8 ALGEBRAIC CODING THEORY
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,
p. 77), 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, p. 110. □
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)
Table 8.40: Cosets of C
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 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, p. 110, notice that we have chosen a repre-
sentative 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.
8.5 EXERCISES 111
Such a list is called a decoding table.
Syndrome Coset Leader
(000) (00000)
(001) (00001)
(010) (00010)
(011) (10000)
(100) (00100)
(101) (01000)
(110) (00110)
(111) (10100)
Table 8.42: Syndromes for each coset
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, p. 111 is a decoding table for the code C given
in Example 8.39, p. 110. If x = (01111) is received, then its syndrome
can be computed to be
0
Hx = 1 .
1
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.
8.5 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)
112 CHAPTER 8 ALGEBRAIC CODING THEORY
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)
(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.
8.5 EXERCISES 113
11. Which matrices are canonical parity-check matrices? For those ma-
trices that are canonical parity-check matrices, what are the corre-
sponding 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 Exercise 8.5.11, p. 113.
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 correspond-
ing 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)
114 CHAPTER 8 ALGEBRAIC CODING THEORY
(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.
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 ma-
trix 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 information 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.
8.6 PROGRAMMING EXERCISES 115
(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 Hamming 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 in-
formation positions and four check positions. What are the
check positions and what are the information positions? En-
code 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 infor-
mation 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 hav-
ing the maximum possible number of information bits with k
check bits are called 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.
8.6 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.
8.7 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.
116 CHAPTER 8 ALGEBRAIC CODING THEORY
[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, Else-
vier, 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. Math-
ematical 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 ∼ = ⟨i⟩, define a map ϕ : Z4 → ⟨i⟩ 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. □
117
118 CHAPTER 9 ISOMORPHISMS
Example 9.3 The integers are isomorphic to the subgroup of Q∗ consist-
ing 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 ̸= n. If we can show that
ϕ(m) ̸= 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, p. 39 in Chapter 3, p. 29). □
Example 9.5 Even though S3 and Z6 possess the same number of el-
ements, 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 ̸= 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.
9.1 DEFINITION AND EXAMPLES 119
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 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 ̸= n. We can assume that m > n. We must show that am ̸= 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, p. 77. ■
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 permu-
tations on some set; hence, every group is a permutation group. Cayley’s
Theorem is what we call a representation theorem. The aim of represen-
tation 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.
120 CHAPTER 9 ISOMORPHISMS
+ 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 permuta-
tion group G = {(0), (012), (021)}. The isomorphism here is
( )
0 1 2
0 7→ = (0)
0 1 2
( )
0 1 2
1 7→ = (012)
1 2 0
( )
0 1 2
2 7→ = (021).
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 representa-
9.2 DIRECT PRODUCTS 121
tion 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 mathe-
matics. 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 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). □
122 CHAPTER 9 ISOMORPHISMS
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 iso-
morphic. Every element (a, b) in Z2 ×Z2 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
∏
n
Gi = G1 × G2 × · · · × G n
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 .
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. ■
∏
Corollary 9.18 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.
9.2 DIRECT PRODUCTS 123
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, p. 122 since lcm(m, n) =
mn if and only if gcd(m, n) = 1. ■
Corollary 9.22 Let n1 , . . . , nk be positive integers. Then
∏
k
Zni ∼
= Zn1 ···nk
i=1
if and only if gcd(ni , nj ) = 1 for i ̸= j.
Corollary 9.23 If
m = pe11 · · · pekk ,
where the pi s are distinct primes, then
Zm ∼
= Zpe11 × · · · × Zpekk .
e
Proof. Since the greatest common divisor of pei i and pj j is 1 for i ̸= j,
the proof follows from Corollary 9.22, p. 123. ■
In Chapter 13, p. 163, we will prove that all finite abelian groups are
isomorphic to direct products of the form
Zpe11 × · · · × Zpek
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}.
□
124 CHAPTER 9 ISOMORPHISMS
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 sub-
groups, 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 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 subgroups H1 , H2 , . . . , Hn of G, by requiring that
• G = H1 H2 · · · Hn = {h1 h2 · · · hn : hi ∈ Hi };
• Hi ∩ ⟨∪j̸=i Hj ⟩ = {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 be the 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 iso-
morphic, even though this should, in theory, be a very difficult compu-
tation.
9.3 EXERCISES 125
9.3 Exercises
1. Prove that Z ∼
= nZ for n ̸= 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
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 .
126 CHAPTER 9 ISOMORPHISMS
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.
20. Prove or disprove: Every abelian group of order divisible by 3 con-
tains 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.
27. Let G ∼
= H. Show that if G is cyclic, then so is H.
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).
9.3 EXERCISES 127
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.
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
∏
k
Zni ∼
= Zn1 ···nk
i=1
if and only if gcd(ni , nj ) = 1 for i ̸= j.
50. Prove that A × B is abelian if and only if A and B are abelian.
51. If 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 ⟨m, n⟩ = ⟨d⟩ if and only if d = gcd(m, n).
54. Let m, n ∈ Z. Prove that ⟨m⟩∩⟨n⟩ = ⟨l⟩ 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.
128 CHAPTER 9 ISOMORPHISMS
(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 = ⟨z⟩ 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 = ⟨z⟩ 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 = ⟨z⟩ 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, p. 137.
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 = {(123), (13)} and H(123) = {(123), (23)},
H cannot be a normal subgroup of S3 . However, the subgroup N , consist-
ing of the permutations (1), (123), and (132), is normal since the cosets
of N are
N = {(1), (123), (132)}
(12)N = N (12) = {(12), (13), (23)}.
□
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.
129
130 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
2. For all g ∈ G, gN g −1 ⊂ N .
3. For all g ∈ G, gN g −1 = N .
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), (123), (132)}.
The cosets of N in S3 are N and (12)N . The factor group S3 /N has the
following multiplication table.
N (12)N
N N (12)N
(12)N (12)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 (12)N =
{(12), (13), (23)} is the set of odd permutations. The information cap-
10.2 THE SIMPLICITY OF THE ALTERNATING GROUP 131
tured in G/N is parity; that is, multiplying two even or two odd per-
mutations results in an even permutation, whereas multiplying an odd
permutation by an even permutation yields an odd permutation. □
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 sub-
group 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.
132 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
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
(ab) = (ba), every pair of transpositions must be one of the following:
(ab)(ab) = id
(ab)(cd) = (acb)(acd)
(ab)(ac) = (acb).
■
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 (ijk), 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
(iaj) = (ija)2
(iab) = (ijb)(ija)2
(jab) = (ijb)2 (ija)
(abc) = (ija)2 (ijc)(ijb)2 (ija).
Now suppose that N is a nontrivial normal subgroup of An for n ≥ 3
such that N contains a 3-cycle of the form (ija). Using the normality of
N , we see that
[(ij)(ak)](ija)2 [(ij)(ak)]−1 = (ijk)
is in N . Hence, N must contain all of the 3-cycles (ijk) for 1 ≤ k ≤ n.
By Lemma 10.8, p. 131, 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 )
10.2 THE SIMPLICITY OF THE ALTERNATING GROUP 133
= (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 .
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 ̸= 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.
■
134 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
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, p. 132,
N contains a 3-cycle. By Lemma 10.9, p. 132, N = An ; therefore, An
contains no proper nontrivial normal subgroups for n ≥ 5. ■
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 pub-
lished their results in the paper “Solvability of Groups of Odd Order,”
which appeared in the Pacific Journal of Mathematics. Their proof, run-
ning 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 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), (123), (132)}
(c) G = S4 and H = D4
(d) G = Q8 and H = {1, −1, I, −I}
(e) G = Z and H = 5Z
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 sub-
groups are normal? What are all the factor groups of Q8 up to
isomorphism?
10.3 EXERCISES 135
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 ̸= 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.
14. Let G be a group and let G′ = ⟨aba−1 b−1 ⟩; that is, G′ is the subgroup
of all finite products 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.
136 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
(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 iso-
morphic; 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 permutations 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 ̸= 0.
Also, for any two elements A and B in G, det(AB) = det(A) det(B).
137
138 CHAPTER 11 HOMOMORPHISMS
Using the determinant, we can define a homomorphism ϕ : GL2 (R) → R∗
by A 7→ det(A). □
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 homo-
morphisms.
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 , respec-
tively; 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, p. 40.
(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, p. 138, ϕ−1 ({e}) is a subgroup of G.
This subgroup is called the kernel of ϕ and will be denoted by ker ϕ. In
11.2 THE ISOMORPHISM THEOREMS 139
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.
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 homo-
morphisms 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 homo-
morphisms. We already know that with every group homomorphism
ϕ : G → H we can associate a normal subgroup of G, ker ϕ. The converse
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.
140 CHAPTER 11 HOMOMORPHISMS
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)
= 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 theorems. 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 sub-
group of a group G (not necessarily normal in G) and N a normal sub-
group 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 ).
11.2 THE ISOMORPHISM THEOREMS 141
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.
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 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 makes 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 homomor-
phism 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, p. 141, 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).
142 CHAPTER 11 HOMOMORPHISMS
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 relation-
ships between groups, normal subgroups, quotient groups and properties
of homomorphisms.
11.3 Exercises
1. Prove that det(AB) = det(A) det(B) for A, B ∈ GL2 (R). This
shows that the determinant 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 .
7. In the group Z24 , let H = ⟨4⟩ and N = ⟨6⟩.
(a) List the elements in HN (we usually write H + N for these
additive groups) and H ∩ N .
11.4 ADDITIONAL EXERCISES: AUTOMORPHISMS 143
(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 ) de-
scribed 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 respectively. 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.
11.4 Additional Exercises: Automorphisms
1. Let Aut(G) be the set of all automorphisms of G; that is, isomor-
phisms 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).
144 CHAPTER 11 HOMOMORPHISMS
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 automor-
phism.
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 Erlan-
gen, 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 appli-
cations 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),
145
146 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
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 )
( n )t
∑ ∑
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
12.1 MATRIX GROUPS 147
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
( )
3 −1
A−1 = .
−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 elemen-
tary 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 ̸= 0. The inverse of A is
( )
1 d −b
A−1 = .
ad − bc −c a
148 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
If A is in SL2 (R), then
( )
−1 d −b
A = .
−c a
Geometrically, SL2 (R) is the group that preserves the areas of parallelo-
grams. Let ( )
1 1
A=
0 1
be in SL2 (R). In Figure 12.4, p. 148, 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 or-
thogonal 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
⟨x, y⟩ = xt y = (x1 , x2 , . . . , xn ) . = x1 y1 + · · · + xn yn .
..
yn
We define the length of a vector x = (x1 , . . . , xn )t to be
√ √
∥x∥ = ⟨x, x⟩ = x21 + · · · + x2n .
12.1 MATRIX GROUPS 149
Associated with the notion of the length of a vector is the idea of the dis-
tance between two vectors. We define the distance between two vectors
x and y to be ∥x − y∥. 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. ⟨x, y⟩ = ⟨y, x⟩.
2. ⟨x, y + w⟩ = ⟨x, y⟩ + ⟨x, w⟩.
3. ⟨αx, y⟩ = ⟨x, αy⟩ = α⟨x, y⟩.
4. ⟨x, x⟩ ≥ 0 with equality exactly when x = 0.
5. If ⟨x, y⟩ = 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, ⟨ar , as ⟩ = δrs , where
{
1 r=s
δrs =
0 r ̸= 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 ∥Ax − Ay∥ = ∥x − y∥, ∥Ax∥ =
∥x∥, or ⟨Ax, Ay⟩ = ⟨x, y⟩, 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, ⟨Ax, Ay⟩ = ⟨x, y⟩.
4. For vectors x and y, ∥Ax − Ay∥ = ∥x − y∥.
5. For any vector x, ∥Ax∥ = ∥x∥.
150 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
Proof. We have already shown (1) and (2) to be equivalent.
(2) ⇒ (3).
⟨Ax, Ay⟩ = (Ax)t Ay
= xt At Ay
= xt y
= ⟨x, y⟩.
(3) ⇒ (2). Since
⟨x, x⟩ = ⟨Ax, Ax⟩
= xt At Ax
= ⟨x, At Ax⟩,
we know that ⟨x, (At A − I)x⟩ = 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
∥Ax − Ay∥2 = ∥A(x − y)∥2
= ⟨A(x − y), A(x − y)⟩
= ⟨x − y, x − y⟩
= ∥x − y∥2 .
(4) ⇒ (5). If A is distance-preserving, then A is length-preserving.
Letting y = 0, we have
∥Ax∥ = ∥Ax − Ay∥ = ∥x − y∥ = ∥x∥.
(5) ⇒ (3). We use the following identity to show that length-preserving
implies inner product-preserving:
1[ ]
⟨x, y⟩ = ∥x + y∥2 − ∥x∥2 − ∥y∥2 .
2
Observe that
1[ ]
⟨Ax, Ay⟩ = ∥Ax + Ay∥2 − ∥Ax∥2 − ∥Ay∥2
2
1[ ]
= ∥A(x + y)∥2 − ∥Ax∥2 − ∥Ay∥2
2
1[ ]
= ∥x + y∥2 − ∥x∥2 − ∥y∥2
2
= ⟨x, y⟩.
■
12.1 MATRIX GROUPS 151
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 mul-
tiplication 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, p. 151).
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 ( )
2 1 0
B = .
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, p. 151). 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 12.3.6, p. 159, you are asked to check that E(n) is indeed a
group under this operation.
152 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
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
∥f (x) − f (y)∥ = ∥x − y∥
for all x, y ∈ Rn . It is not difficult to show that f must be a one-to-one
map. By Theorem 12.8, p. 149, 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, p. 152);
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 isome-
tries in R2 are rotations and reflections about the origin, translations,
and combinations of the two. For example, a glide reflection is a trans-
lation followed by a reflection (Figure 12.12, p. 152). 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
12.2 SYMMETRY 153
that f preserves inner products. Since f (0) = 0, ∥f (x)∥ = ∥x∥; therefore,
∥x∥2 − 2⟨f (x), f (y)⟩ + ∥y∥2 = ∥f (x)∥2 − 2⟨f (x), f (y)⟩ + ∥f (y)∥2
= ⟨f (x) − f (y), f (x) − f (y)⟩
= ∥f (x) − f (y)∥2
= ∥x − y∥2
= ⟨x − y, x − y⟩
= ∥x∥2 − 2⟨x, y⟩ + ∥y∥2 .
Consequently,
⟨f (x), f (y)⟩ = ⟨x, y⟩.
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) = ⟨f (x), f (e1 )⟩f (e1 ) + ⟨f (x), f (e2 )⟩f (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 R2 ; hence, 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, p. 151, 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
154 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
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 .
Case 2. The group G contains a reflection T . The kernel of the ho-
momorphism 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 dimensions. Wallpaper patterns are simply repeating patterns
in the plane (Figure 12.16, p. 154). The analogs of wallpaper patterns in
R3 are crystals, which we can think of as repeating patterns of molecules
in three dimensions (Figure 12.17, p. 155). The mathematical equivalent
of a wallpaper or crystal pattern is called a lattice.
Figure 12.16: A wallpaper pattern in R2
12.2 SYMMETRY 155
Figure 12.17: A crystal structure in R3
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 vec-
tors (1, 1)t and (2, 0)t have the same lattice as the vectors (−1, 1)t and
(−1, −1)t (Figure 12.18, p. 156). However, any lattice is completely de-
termined 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,
( ) ( )
y1 x1
U −1 = .
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.
156 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
(−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 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 rotations 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 par-
ticular, 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
12.2 SYMMETRY 157
possible cell shapes are parallelogram, rectangular, square, rhombic, and
hexagonal (Figure 12.21, p. 157). The wallpaper groups can now be clas-
sified according to the types of reflections that occur in each group: these
are ordinarily reflections, glide reflections, both, or none.
Rectangular
Square Rhombic
Parallelogram
Hexagonal
Figure 12.21: Types of lattices in R2
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
Table 12.22: The 17 wallpaper groups
Theorem 12.23 There are exactly 17 wallpaper groups.
158 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
p4m p4g
Figure 12.24: The wallpaper groups p4m and p4g
The 17 wallpaper groups are listed in Table 12.22, p. 157. The groups
p3m1 and p31m can be distinguished by whether or not all of their three-
fold 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, p. 158). 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 yet included any Sage material related to this chap-
ter.
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 math-
ematics in the twentieth century. Hilbert’s eighteenth problem asked
whether or not crystallographic groups in n dimensions were always fi-
nite. 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 num-
ber 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 Exercises
1. Prove the identity
1[ ]
⟨x, y⟩ = ∥x + y∥2 − ∥x∥2 − ∥y∥2 .
2
2. Show that O(n) is a group.
3. Prove that the following matrices are orthogonal. Are any of these
12.3 EXERCISES 159
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 in Figure 12.25,
p. 159.
(a)
(c)
(b)
Figure 12.25
5. Let x, y, and w be vectors in Rn and α ∈ R. Prove each of the
following properties of inner products.
(a) ⟨x, y⟩ = ⟨y, x⟩.
(b) ⟨x, y + w⟩ = ⟨x, y⟩ + ⟨x, w⟩.
(c) ⟨αx, y⟩ = ⟨x, αy⟩ = α⟨x, y⟩.
(d) ⟨x, x⟩ ≥ 0 with equality exactly when x = 0.
(e) If ⟨x, y⟩ = 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 .
9. Let A ∈ SL2 (R) and suppose that the vectors x and y form two sides
of a parallelogram in R2 . Prove that the area of this parallelogram
is the same as the area of the parallelogram with sides Ax and Ay.
160 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
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 ̸= 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), (12)}.
(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, p. 154.
17. Determine which of the 17 wallpaper groups preserves the symmetry
of the pattern in Figure 12.26, p. 160.
Figure 12.26
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.
12.4 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.
12.4 REFERENCES AND SUGGESTED READINGS 161
[4] Lockwood, E. H. and Macmillan, R. H. Geometric Symmetry. Cam-
bridge 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 Sym-
posia 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 Math-
ematical Symmetry,” American Mathematical Monthly 88 (1980),
54–64.
[10] Schattschneider, D. “The Plane Symmetry Groups: Their Recog-
nition and Their Notation,” American Mathematical Monthly 85
(1978), 439–50.
[11] Schwarzenberger, R. L. “The 17 Plane Symmetry Groups,” Mathe-
matical Gazette 58 (1974), 123–31.
[12] Weyl, H. Symmetry. Princeton University Press, Princeton, NJ,
1952.
162 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
13
The Structure of Groups
The ultimate goal of group theory is to classify all groups up to isomor-
phism; 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 1 × · · · × Zpαnn ,
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.
163
164 CHAPTER 13 THE STRUCTURE OF GROUPS
Example 13.1 Obviously, all finite groups are finitely generated. For ex-
ample, the group S3 is generated by the permutations (12) and (123). The
group Z×Zn is an infinite group but is finitely generated by {(1, 0), (0, 1)}.
□
Example 13.2 Not all groups are finitely generated. Consider the ra-
tional 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 1 × Zpα2 2 × · · · × Zpαnn
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
13.1 FINITE ABELIAN GROUPS 165
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 ;
• 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 = ⟨a⟩, where a is any
element other than the identity. By Exercise 4.4.39, p. 58, 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 ̸= 1. Suppose
r
a = 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 ̸= 1 is an element of order p in G. ■
Lemma 13.6, p. 165 is a special case of Cauchy’s Theorem (Theo-
rem 15.1, p. 187), 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, p. 187.
166 CHAPTER 13 THE STRUCTURE OF GROUPS
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, p. 165, G has an element of order q and thus is not a
p-group. ■
1 · · · pk ,
αk
Lemma 13.8 Let G be a finite abelian group of order n = pα 1
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 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 = G1 G2 · · · Gk
and Gi ∩ Gj = {1} for i ̸= j. Suppose that g1 ∈ G1 is in the subgroup
generated by G2 , G3 , . . . , Gk . Then g1 = g2 g3 · · · gk for gi ∈ Gi . Since gi
α2 αk
αi p ···p
has order pαi , we know that gip = 1 for i = 2, 3, . . . , k, and g1 2 k
= 1.
Since the order of g1 is a power of p1 and gcd(p1 , p2 · · · pk ) = 1, it
α2 αk
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 ̸= 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 · · · pβkk
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, p. 166.
Lemma 13.9 Let G be a finite abelian p-group and suppose that g ∈ G
has maximal order. Then G is isomorphic to ⟨g⟩ × H for some subgroup
H of G.
Proof. By Lemma 13.7, p. 166, 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
13.1 FINITE ABELIAN GROUPS 167
all integers k with 1 ≤ k < n and let g be of maximal order in G, say
m
|g| = pm . Then ap = e for all a ∈ G. Now choose h in G such that
h ∈/ ⟨g⟩, where h has the smallest possible order. Certainly such an h
exists; otherwise, G = ⟨g⟩ and we are done. Let H = ⟨h⟩.
We claim that ⟨g⟩ ∩ 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 ⟨g⟩ 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 ⟨g⟩. 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
⟨g⟩; otherwise, h would also have to be in ⟨g⟩. 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 ∈ / ⟨g⟩. Since
h was chosen to have the smallest order of all of the elements that are
not in ⟨g⟩, |H| = p.
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 ⟨g⟩ ∩ 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 ∼
= ⟨gH⟩ × K/H
for some subgroup K of G containing H. We claim that ⟨g⟩ ∩ K = {e}.
If b ∈ ⟨g⟩ ∩ K, then bH ∈ ⟨gH⟩ ∩ K/H = {H} and b ∈ ⟨g⟩ ∩ H = {e}. It
follows that G = ⟨g⟩K implies that G ∼
= ⟨g⟩ × K. ■
The proof of the Fundamental Theorem of Finite Abelian Groups fol-
lows very quickly from Lemma 13.9, p. 166. Suppose that G is a finite
abelian group and let g be an element of maximal order in G. If ⟨g⟩ = G,
then we are done; otherwise, G ∼ = Z|g| × H for some subgroup H con-
tained 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 Gen-
erated Abelian Groups. Every finitely generated abelian group G is
isomorphic to a direct product of cyclic groups of the form
Zpα1 1 × Zpα2 2 × · · · × Zpαnn × Z × · · · × Z,
where the pi ’s are primes (not necessarily distinct).
168 CHAPTER 13 THE STRUCTURE OF GROUPS
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 ⊃ ⟨2⟩ ⊃ ⟨6⟩ ⊃ ⟨12⟩ ⊃ {0}.
□
Example 13.12 A subnormal series need not be a normal series. Con-
sider the following subnormal series of the group D4 :
D4 ⊃ {(1), (12)(34), (13)(24), (14)(23)} ⊃ {(1), (12)(34)} ⊃ {(1)}.
The subgroup {(1), (12)(34)} is not normal in D4 ; consequently, this series
is not a normal series. □
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 isomor-
phic 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 ⊃ ⟨3⟩ ⊃ ⟨15⟩ ⊃ {0}
Z60 ⊃ ⟨4⟩ ⊃ ⟨20⟩ ⊃ {0}
of the group Z60 are isomorphic since
Z60 /⟨3⟩ ∼
= ⟨20⟩/{0} ∼
= Z3
⟨3⟩/⟨15⟩ = ⟨4⟩/⟨20⟩ ∼
∼ = Z5
⟨15⟩/{0} ∼
= Z60 /⟨4⟩ ∼
= 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
13.2 SOLVABLE GROUPS 169
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 ⊃ ⟨3⟩ ⊃ ⟨15⟩ ⊃ ⟨30⟩ ⊃ {0}
with factor groups
Z60 /⟨3⟩ ∼
= Z3
∼
⟨3⟩/⟨15⟩ = Z5
⟨15⟩/⟨30⟩ ∼= Z2
⟨30⟩/{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 ⊃ ⟨2⟩ ⊃ ⟨4⟩ ⊃ ⟨20⟩ ⊃ {0}
is also a composition series. □
Example 13.16 For n ≥ 5, the series
Sn ⊃ An ⊃ {(1)}
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 compo-
sition 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}
170 CHAPTER 13 THE STRUCTURE OF GROUPS
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 (Theorem 11.12, p. 140) 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 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), (12)(34), (13)(24), (14)(23)} ⊃ {(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. □
13.3 EXERCISES 171
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 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.
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.
172 CHAPTER 13 THE STRUCTURE OF GROUPS
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.
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 refine-
ments.
24. Use Schreier’s Theorem to prove the Jordan-Hölder Theorem.
13.4 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?
13.5 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), (13), (24), (1432), (1234), (12)(34), (14)(23), (13)(24)}.
The elements of D4 act on X as functions. The permutation (13)(24)
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)
173
174 CHAPTER 14 GROUP ACTIONS
for σ ∈ G and x ∈ X.
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 equiva-
lence 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), (123), (132), (45), (123)(45), (132)(45)}
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}. □
14.1 GROUPS ACTING ON SETS 175
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.
Example 14.9 Let X = {1, 2, 3, 4, 5, 6} and suppose that G is the per-
mutation group given by the permutations
{(1), (12)(3456), (35)(46), (12)(3654)}.
Then the fixed point sets of X under the action of G are
X(1) = X,
X(35)(46) = {1, 2},
X(12)(3456) = X(12)(3654) = ∅,
and the stabilizer subgroups are
G1 = G2 = {(1), (35)(46)},
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, p. 77). 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 . ■
176 CHAPTER 14 GROUP ACTIONS
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}.
Since the orbits of the action partition 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,
p. 175, 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)}, {(123), (132)}, {(12), (13), (23)}.
The class equation is 6 = 1 + 2 + 3. □
Example 14.13 The center of D4 is {(1), (13)(24)}, and the conjugacy
classes are
{(13), (24)}, {(1432), (1234)}, {(12)(34), (14)(23)}.
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, p. 78,
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
14.3 BURNSIDE’S COUNTING THEOREM 177
n can be partitioned 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 prob-
lem 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 com-
putations would be too time-consuming for even the largest computer.
□
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 ̸= 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, p. 177, |Z(G)| = p or p2 . If |Z(G)| = p2 , then
we are done. 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. ■
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, p. 178).
178 CHAPTER 14 GROUP ACTIONS
B W W B
W W W W
W W W W
B W W B
Figure 14.17: Equivalent colorings of square
Burnside’s Counting Theorem offers a method of computing the num-
ber of distinguishable 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 Count-
ing 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 ∑
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
∑
|Xg |.
g∈G
However, if viewed in terms of the stabilizer subgroups, this number is
∑
|Gx |;
x∈X
14.3 BURNSIDE’S COUNTING THEOREM 179
∑ ∑
hence, g∈G |Xg | = x∈X |Gx |. By Lemma 14.18, p. 178,
∑
|Gy | = |Ox | · |Gx |.
y∈Ox
By Theorem 14.11, p. 175 and Lagrange’s Theorem, this expression is
equal to |G|. Summing over all of the k distinct orbits, we conclude that
∑ ∑
|Xg | = |Gx | = k · |G|.
g∈G x∈X
■
Example 14.20 Let X = {1, 2, 3, 4, 5} and suppose that G is the permu-
tation group G = {(1), (13), (13)(25), (25)}. The orbits of X are {1, 3},
{2, 5}, and {4}. The fixed point sets are
X(1) = X
X(13) = {2, 4, 5}
X(13)(25) = {4}
X(25) = {1, 3, 4}.
Burnside’s Theorem says that
1 ∑ 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 per-
mutations:
(1) (13) (24) (1432)
(1234) (12)(34) (14)(23) (13)(24)
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
e(f ) = f ◦ σ for f : X → Y . For example, suppose that f is defined by
σ
f (1) = B
f (2) = W
f (3) = W
f (4) = W
180 CHAPTER 14 GROUP ACTIONS
and σ = (12)(34). Then σ e(f ) = f ◦ σ sends vertex 2 to B and the
remaining vertices to W . The set of all such σ e is a permutation group
Ge on the set of possible colorings. Let X e denote the set of all possible
e is the set of all possible maps from X to Y . Now we
colorings; that is, X
e
must compute the number of G-equivalence classes.
1. Xe(1) = X
e since the identity fixes every possible coloring. |X|
e =
4
2 = 16.
2. Xe(1234) consists of all f ∈ Xe such that f is unchanged by the
permutation (1234). 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 |Xe(1234) | = 2.
e(1432) | = 2.
3. |X
4. For Xe(13)(24) , f (1) = f (3) and f (2) = f (4). Thus, |X
e(13)(24) | =
22 = 4.
e(12)(34) | = 4.
5. |X
e(14)(23) | = 4.
6. |X
e(13) , f (1) = f (3) and the other corners can be of any color;
7. For X
e(13) | = 23 = 8.
hence, |X
e(24) | = 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 to Y . Then there exists a permutation group Ge acting
e
on X, where σ e
e ∈ G is defined by σ
e(f ) = f ◦ σ for σ ∈ G and f ∈ X. e
Furthermore, if n is the number of cycles in the cycle decomposition of
eσ | = |Y |n .
σ, then |X
Proof. Let σ ∈ G and f ∈ X. e Clearly, f ◦ σ is also in X.
e Suppose that
g is another function from X to Y such that σe(f ) = σ
e(g). Then for each
x ∈ X,
e(f )(x) = σ
f (σ(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 X eσ must have the same value on each cycle of σ.
eσ | = |Y |n .
Since there are n cycles and |Y | 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 permutation of X given by (13)(245) = (13)(245)(6)(7), then
n = 4. Any f ∈ X eg must have the same value on each cycle in g. There
eg | = 34 = 81.
are |Y | = 3 such choices for any value, so |X □
14.3 BURNSIDE’S COUNTING THEOREM 181
Example 14.23 Suppose that we wish to color the vertices of a square
using four different colors. By Proposition 14.21, p. 180, we can immedi-
ately 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 cir-
cuits 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,
p. 181). 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, p. 181).
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 and there are 2n binary n-tuples,
n
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: A switching function of two variables
The possible switching functions with two input variables a and b
are listed in Table 14.26, p. 182. 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 (acb). In the case of switching functions of two variables, the
permutation (ab) reduces 16 possible switching functions to 12 equivalent
functions since
f2 ∼ f4
f3 ∼ f5
182 CHAPTER 14 GROUP ACTIONS
f10 ∼ f12
f11 ∼ f13 .
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
Table 14.26: Switching functions in two variables
3
For three input variables there are 22 = 256 possible switching func-
4
tions; in the case 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.
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 vari-
ables. 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
14.3 BURNSIDE’S COUNTING THEOREM 183
following permutation group:
(a), (ac), (bd), (adcb),
(abcd), (ab)(cd), (ad)(bc), (ac)(bd).
The permutations of the four possible input variables induce the permu-
tations of the output values in Table 14.27, p. 183.
Hence, there are
1 16
(2 + 2 · 212 + 2 · 26 + 3 · 210 ) = 9616
8
possible switching functions of four variables under this group of permuta-
tions. This number will be even smaller if we consider the full symmetric
group on four letters.
Group Number
Permutation Switching Function Permutation of Cycles
(a) (0) 16
(ac) (2, 8)(3, 9)(6, 12)(7, 13) 12
(bd) (1, 4)(3, 6)(9, 12)(11, 14) 12
(adcb) (1, 2, 4, 8)(3, 6.12, 9)(5, 10)(7, 14, 13, 11) 6
(abcd) (1, 8, 4, 2)(3, 9, 12, 6)(5, 10)(7, 11, 13, 14) 6
(ab)(cd) (1, 2)(4, 8)(5, 10)(6, 9)(7, 11)(13, 14) 10
(ad)(bc) (1, 8)(2, 4)(3, 12)(5, 10)(7, 14)(11, 13) 10
(ac)(bd) (1, 4)(2, 8)(3, 12)(6, 9)(7, 13)(11, 14) 10
Table 14.27: Permutations of switching functions in four variables
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.
184 CHAPTER 14 GROUP ACTIONS
14.4 Exercises
1. Examples 14.1, p. 173–14.5, p. 174 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), (12), (13), (23), (123), (132)}
(b) X = {1, 2, 3, 4, 5, 6}, G = {(1), (12), (345), (354), (12)(345), (12)(354)}
3. Compute the G-equivalence classes of X for each of the G-sets in
Exercise 14.4.2, p. 184. 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.
(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?
14.4 EXERCISES 185
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 sym-
metry of the hexagon?
16. A molecule of benzene is made up of six carbon atoms and six hydro-
gen atoms, linked together in a hexagonal shape as in Figure 14.28,
p. 185.
(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
17. How many equivalence classes of switching functions are there if the
input variables x1 , x2 , and x3 can be permuted by any permuta-
tion 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 permuta-
tion 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).
186 CHAPTER 14 GROUP ACTIONS
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 ?
14.5 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?
14.6 References and Suggested Reading
[1] De Bruijin, N. G. “Pólya’s Theory of Counting,” in Applied Com-
binatorial Mathematics, 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 Con-
tributions to Chemical 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 Chemical Compounds. Springer-Verlag, New York,
1985.
[7] Shapiro, L. W. “Finite Groups Acting on Sets with Applications,”
Mathematics Magazine, 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, p. 173. 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 sub-
groups 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
187
188 CHAPTER 15 THE SYLOW THEOREMS
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.
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}
15.1 THE SYLOW THEOREMS 189
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, ⟨xP ⟩ ⊂ 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 = ⟨xP ⟩. Since
|H| = |P | · |⟨xP ⟩|, 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 . ■
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
1 ∈ N (K). Therefore, K = h2 h1 Kh1 h2 or
−1
−1 −1
h1 Kh1 = h2 Kh2 , 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, p. 189,
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, p. 189, 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, p. 189,
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
190 CHAPTER 15 THE SYLOW THEOREMS
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, p. 189. 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. ■
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 ̸≡ 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,
15.2 EXAMPLES AND APPLICATIONS 191
given in Chapter 9, p. 117, 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, p. 123. ■
Example 15.11 Every group of order 15 is cyclic. This is true because
15 = 5 · 3 and 5 ̸≡ 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, p. 177, 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.
Theorem 15.13 Let G′ = ⟨aba−1 b−1 : a, b ∈ G⟩ be the subgroup consist-
ing 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 10.3.14, p. 135 in
Chapter 10, p. 129).
Example 15.14 We will now show that every group of order 5 · 7 · 47 =
1645 is abelian, and cyclic by Theorem 9.21, p. 123. 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, p. 190. 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
192 CHAPTER 15 THE SYLOW THEOREMS
must be normal. □
n
Example 15.16 Let G be a finite group of order p , n > 1 and p
prime. By Theorem 14.15, p. 177, 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, p. 177. □
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.
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|.
■
15.3 EXERCISES 193
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, p. 192,
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 di-
visor 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 nor-
malizer of a subgroup.
15.3 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.
194 CHAPTER 15 THE SYLOW THEOREMS
11. Let G be a finite group 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?
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′ = ⟨aba−1 b−1 : a, b ∈ G⟩ is a normal
subgroup of G and G/G′ is abelian. Find an example to show that
15.4 A PROJECT 195
{aba−1 b−1 : a, b ∈ G} is not necessarily a group.
15.4 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,
p. 195 as a guide.
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
Table 15.21: Numbers of distinct groups G, |G| ≤ 60
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.
15.5 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 Or-
der,” 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.
196 CHAPTER 15 THE SYLOW THEOREMS
[5] Gorenstein, D. Finite Groups. AMS Chelsea Publishing, Provi-
dence 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 sat-
isfying certain axioms, but we are often more interested in working with
sets that have two binary operations. For example, one of the most nat-
ural 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 struc-
ture 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 ̸= 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
197
198 CHAPTER 16 RINGS
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 ̸= 0, there exists a unique element
a−1 such that a−1 a = aa−1 = 1. A commutative division ring is called
a field. The relationship among rings, integral domains, division rings,
and fields is shown in Figure 16.1, p. 198.
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 multipli-
cation, 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 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 ̸= BA. Also, notice
that we can have AB = 0 when neither A nor B is zero. □
16.1 RINGS 199
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
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 ̸= 0,
( )
a − bi − cj − dk
(a + bi + cj + dk) = 1.
a2 + b2 + c2 + d2
□
200 CHAPTER 16 RINGS
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.
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 ̸= ∅.
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 201
16.2 Integral Domains and Fields
Let us briefly recall some definitions. If R is a 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. 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 16.14 The set Q( √ 2 ) =√{a + b 2 : a, b ∈ Q} is a field. The
inverse of an element a + b 2 in Q( 2 ) is
a −b √
+ 2 2.
a2 − 2b2 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 ̸= 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 ̸= 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 ̸= 0 and d ̸= 0, then ad ̸= 0. The map λa
is one-to-one, since for d1 , d2 ∈ D∗ ,
ad1 = λa (d1 ) = λa (d2 ) = ad2
202 CHAPTER 16 RINGS
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, p. 30, every nonzero element in Zp has an inverse; hence,
Zp is a field. If a is any nonzero element 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 ̸= 0. If n is not prime, then n = ab, where 1 < a < n
and 1 < b < n. By Lemma 16.18, p. 202, 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 pre-
serves 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,
16.3 RING HOMOMORPHISMS AND IDEALS 203
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 func-
tions on an interval [a, b] as in Example 16.5, p. 198. For a fixed α ∈ [a, b],
we can define a ring homomorphism 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 homomor-
phisms. □
In the next proposition we will examine some fundamental properties
of ring homomorphisms. 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) ̸= {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 inter-
esting 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 iden-
tity, then the set
⟨a⟩ = {ar : r ∈ R}
is an ideal in R. Certainly, ⟨a⟩ is nonempty since both 0 = a0 and
a = a1 are in ⟨a⟩. The sum of two elements in ⟨a⟩ is again in ⟨a⟩ since
ar + ar′ = a(r + r′ ). The inverse of ar is −ar = a(−r) ∈ ⟨a⟩. Finally, if
204 CHAPTER 16 RINGS
we multiply an element ar ∈ ⟨a⟩ by an arbitrary element s ∈ R, we have
s(ar) = a(sr). Therefore, ⟨a⟩ satisfies the definition of an ideal. □
If R is a commutative ring with identity, then an ideal of the form
⟨a⟩ = {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 ⟨0⟩ = {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 = ⟨n⟩. ■
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,
p. 204, these are the only ideals of Z. □
Proposition 16.27 The kernel of any ring homomorphism ϕ : R → S is
an ideal in R.
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. ■
16.3 RING HOMOMORPHISMS AND IDEALS 205
The ring R/I in Theorem 16.29, p. 204 is called the factor or quo-
tient ring. Just as with group homomorphisms and normal subgroups,
there is a relationship between ring homomorphisms 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 homomor-
phism. It remains to show that ϕ works correctly under ring multiplica-
tion. 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,
p. 137. 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.
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.
206 CHAPTER 16 RINGS
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 characterize those ideals I of a com-
mutative 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, p. 200 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 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 .5
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≠ 0. Then P is a prime ideal in R if and only if R/P is an integral
domain.
5 It is possible to define prime ideals in a noncommutative ring. See [1] or [3].
16.4 MAXIMAL AND PRIME IDEALS 207
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 iden-
tity 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 en-
tered 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.
Noether went to Göttingen to study in 1916. David Hilbert and Felix
Klein tried unsuccessfully 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 inspir-
ing 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, Pontrya-
gin, 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
208 CHAPTER 16 RINGS
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 cen-
tury 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,
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).
■
16.5 AN APPLICATION TO SOFTWARE DESIGN 209
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 ̸= 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 congru-
ent 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, p. 208. 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. ■
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, p. 209 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
210 CHAPTER 16 RINGS
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 calcula-
tions. 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 mul-
tiplications 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) contain-
ing 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 com-
puter 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 four parts:
2134 ≡ 44 (mod 95)
2134 ≡ 0 (mod 97)
2134 ≡ 76 (mod 98)
2134 ≡ 55 (mod 99)
and
1531 ≡ 11 (mod 95)
16.6 EXERCISES 211
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 computing 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 trans-
form 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 possibilities for computing with rings and fields. Ideals, quo-
tients, and homomorphisms are all available.
16.6 Exercises
1. Which of the following sets are rings with respect to the usual oper-
ations 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 3 : a, b ∈ Q}
(g) Z[i] = {a + bi : a, b ∈ Z and i2 = −1}
√ √ √
(h) Q( 3 3 ) = {a + b 3 3 + c 3 9 : a, b, c ∈ Q}
212 CHAPTER 16 RINGS
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
(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. √ Q( 2 ) =
Prove or disprove: The ring √ {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.
16.6 EXERCISES 213
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 cal-
culate 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) ̸= 0, then ϕ(R) is a field.
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 ̸= ∅.
214 CHAPTER 16 RINGS
(b) rs ∈ S for all r, s ∈ S.
(c) r − s ∈ S for all r, s ∈ S.
∩
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.
∩
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 S be a nonempty subset of a ring R. Prove that there is a
subring R′ of R that contains S.
33. 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.
34. 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.
35. Prove or disprove: Every finite integral domain is isomorphic to Zp .
36. 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).
16.7 PROGRAMMING EXERCISE 215
37. 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′ )
38. An element x in a ring is called an idempotent if x2 = x. Prove
that the only idempotents in an integral domain are 0 and 1. Find
a ring with a idempotent x not equal to 0 or 1.
39. Let gcd(a, n) = d and gcd(b, d) ̸= 1. Prove that ax ≡ b (mod n)
does not have a solution.
40. 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.
16.7 Programming Exercise
1. Write a computer program implementing fast addition and multi-
plication using the Chinese Remainder Theorem and the method
outlined in the text.
16.8 References and Suggested Readings
[1] Anderson, F. W. and Fuller, K. R. Rings and Categories of Modules.
2nd ed. Springer, New York, 1992.
[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, Washington, 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.
216 CHAPTER 16 RINGS
[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
∑
n
f (x) = ai xi = a0 + a1 x + a2 x2 + · · · + an xn ,
i=0
where ai ∈ R and an ̸= 0, is called a polynomial over R with inde-
terminate x. The elements a0 , a1 , . . . , an are called the coefficients of
217
218 CHAPTER 17 POLYNOMIALS
f . The coefficient an is called the leading coefficient. A polynomial
is called monic if the leading coefficient is 1. If n is the largest non-
negative number for which an ̸= 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
∑
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
17.1 POLYNOMIAL RINGS 219
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 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 poly-
nomial addition. The zero polynomial,
∑n f (x) = 0, is the additive identity.
i
Given a polynomial
∑n p(x) = i=0 a
∑n
i x , the inverse of p(x) is easily verified
to be −p(x) = i=0 (−ai )xi = − i=0 ai xi . Commutativity and associa-
tivity follow immediately from the definition of polynomial addition and
from the fact that addition in R is both commutative and associative.
To show that polynomial multiplication is associative, let
∑
m
p(x) = ai xi ,
i=0
∑
n
q(x) = bi xi ,
i=0
∑
p
r(x) = ci xi .
i=0
Then
[( m )( )] ( )
∑ ∑
n ∑
p
i i i
[p(x)q(x)]r(x) = ai x bi x ci x
i=0 i=0 i=0
( )
∑
m+n ∑
i ∑p
= aj bi−j xi ci x i
i=0 j=0 i=0
( j )
∑
m+n+p ∑i ∑
= ak bj−k ci−j xi
i=0 j=0 k=0
∑
m+n+p ∑
= aj bk cl xi
i=0 j+k+l=i
( i−j )
∑
m+n+p ∑i ∑
= aj bk ci−j−k xi
i=0 j=0 k=0
(m ) n+p i
∑ ∑ ∑
= ai xi bj ci−j xi
i=0 i=0 j=0
(m ) [( )( )]
∑ ∑
n ∑
p
= ai xi bi xi ci x i
i=0 i=0 i=0
= p(x)[q(x)r(x)]
The commutativity and distribution properties of polynomial multiplica-
tion are proved in a similar manner. We shall leave the proofs of these
properties as an exercise. ■
220 CHAPTER 17 POLYNOMIALS
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)). Further-
more, R[x] is an integral domain.
Proof. Suppose that we have two nonzero polynomials
p(x) = am xm + · · · + a1 x + a0
and
q(x) = bn xn + · · · + b1 x + b0
with am ̸= 0 and bn ̸= 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) ̸= 0. Since p(x) ̸= 0 and q(x) ̸= 0 imply that p(x)q(x) ̸= 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 x2 − 3xy + 2y 3 . 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 .
∑n ∑m
Proof. Let p(x) = i=0 ai xi and q(x) = i=0 bi xi . 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)
( n )( m )
∑ ∑
i i
= ai α bi α
i=0 i=0
( )
∑
m+n ∑
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, p. 20) 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 poly-
nomials. The division algorithm for polynomials has several important
17.2 THE DIVISION ALGORITHM 221
consequences. Since its proof is very similar to the corresponding proof
for integers, it is worthwhile to review Theorem 2.9, p. 20 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.
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); therefore, 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.
222 CHAPTER 17 POLYNOMIALS
x2 + x + 4
x − 2 x3 − x2 + 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. □
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 polynomials 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, p. 222. The degree
of q(x) is n − 1 by Proposition 17.4, p. 220. Let β be some other zero of
p(x) that is distinct from α. Then p(β) = (β − α)q(β) = 0. Since α ̸= 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
17.3 IRREDUCIBLE POLYNOMIALS 223
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)
= 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, p. 222
and the proof of Theorem 2.10, p. 20.
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)
224 CHAPTER 17 POLYNOMIALS
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. □
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 poly-
nomial 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, p. 224, we can assume that
c1 c1
α(x) = (a0 + a1 x + · · · + am xm ) = α1 (x)
d1 d1
17.3 IRREDUCIBLE POLYNOMIALS 225
c2 c2
β(x) = (b0 + b1 x + · · · + bn xn ) = β1 (x),
d2 d2
where the ai ’s are relatively prime and the bi ’s are relatively prime. Con-
sequently,
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 ̸= 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. ■
Corollary 17.15 Let p(x) = xn + an−1 xn−1 + · · · + a0 be a polynomial
with coefficients in Z and a0 ̸= 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) = x − 2x + x + 1. We shall show that p(x)
4 3
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 Corol-
lary 17.15, p. 225, 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 poly-
nomials, 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
226 CHAPTER 17 POLYNOMIALS
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. ■
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 poly-
nomials 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 Criterion. The real value of
Theorem 17.17, p. 226 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 ⟨p(x)⟩
generated by some polynomial p(x); that is,
⟨p(x)⟩ = {p(x)q(x) : q(x) ∈ F [x]}.
Example 17.19 The polynomial x2 in F [x] generates the ideal ⟨x2 ⟩
consisting of all polynomials 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.
17.3 IRREDUCIBLE POLYNOMIALS 227
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],
⟨1⟩ = 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 = ⟨p(x)⟩. ■
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 ⟨p(x)⟩
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 ⟨p(x)⟩ is a prime ideal one of these factors, say f (x), is
in ⟨p(x)⟩ and therefore be a multiple of p(x). But this would imply that
⟨p(x)⟩ ⊂ ⟨f (x)⟩, which is impossible since ⟨p(x)⟩ is maximal.
Conversely, suppose that p(x) is irreducible over F [x]. Let I be an
ideal in F [x] containing ⟨p(x)⟩. By Theorem 17.20, p. 226, I is a principal
ideal; hence, I = ⟨f (x)⟩ 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 = ⟨p(x)⟩. Thus, there are no proper ideals
of F [x] that properly contain ⟨p(x)⟩. ■
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 equa-
tion 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.
228 CHAPTER 17 POLYNOMIALS
He kept his solution an absolute secret. This may seem surprising today,
when mathematicians 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 so-
lution 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 re-
lieved 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.
17.4 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
(b) (5x2 + 3x − 4)(4x2 − x + 9) in Z12
(c) (7x3 + 3x2 − x) + (6x2 − 8x + 4) in Z9
(d) (3x2 + 2x − 4) + (4x2 + 2) in Z5
17.4 EXERCISES 229
(e) (3x2 + 2x − 4)(4x2 + 2) in Z5
(f) (5x2 + 3x − 2)2 in Z12
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 polyno-
mials 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 (c) 5x4 + 2x2 − 3 in Z7
(b) 3x3 − 4x2 − x + 4 in Z5 (d) x3 + x + 1 in Z2
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.
230 CHAPTER 17 POLYNOMIALS
18. The Rational Root Theorem. Let
p(x) = an xn + an−1 xn−1 + · · · + a0 ∈ Z[x],
where an ̸= 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 irre-
ducible over Q for any prime p.
21. If F is a field, show that there are infinitely many irreducible poly-
nomials 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).
(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.
17.5 ADDITIONAL EXERCISES: SOLVING THE CUBIC AND QUARTIC EQUATIONS231
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)).
17.5 Additional Exercises: Solving the Cubic
and Quartic Equations
1. Solve the general quadratic equation
ax2 + bx + c = 0
to obtain √
−b ± b2 − 4ac
x= .
2a
The discriminant of the quadratic equation ∆ = b2 − 4ac deter-
mines 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
ω 3 = 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 .
5. product of the solutions obtained in (4) is −p3 /27,
Show that the √
deducing that AB = −p/3.
3
6. Prove that the possible solutions for z in (4) are given by
√ √ √ √ √ √
A, ω A, ω 2 A, B, ω B, ω 2 B
3 3 3 3 3 3
and use this result to show that the three possible solutions for y are
√ √ √ √
3 q p3 q 2 q p3 q2
2i 3
ω − +
i
+ +ω − − + ,
2 27 4 2 27 4
where i = 0, 1, 2.
232 CHAPTER 17 POLYNOMIALS
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
( )2 ( )
1 1 2
2
y + z = (z − p)y − qy +
2
z −r .
2 4
11. Show that the right-hand side of Exercise 17.5.10, p. 232 can be put
in the form (my + k)2 if and only if
( )
1 2
q 2 − 4(z − p) z − r = 0.
4
12. From Exercise 17.5.11, p. 232 obtain the resolvent cubic equation
z 3 − pz 2 − 4rz + (4pr − q 2 ) = 0.
Solving the resolvent cubic equation, put the equation found in Ex-
ercise 17.5.10, p. 232 in the form
( )2
1
y2 + 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 do-
mains, 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 con-
struct 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.
233
234 CHAPTER 18 INTEGRAL DOMAINS
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 ̸= 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.
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 multiplica-
tion 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
18.1 FIELDS OF FRACTIONS 235
= (b1 d1 )(a2 d2 + b2 c2 ).
■
Lemma 18.3 The set of equivalence classes of S, FD , under the equiva-
lence relation ∼, together with the operations of addition and multiplica-
tion defined by
[a, b] + [c, d] = [ad + bc, bd]
[a, b] · [c, d] = [ac, bd],
is a field.
Proof. The additive and multiplicative identities are [0, 1] and [1, 1], re-
spectively. 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 ̸= 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 ; how-
ever,
[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, p. 235 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].
236 CHAPTER 18 INTEGRAL DOMAINS
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
1 = a2 b2 and ψ([a1 , b1 ]) = ψ([a2 , b2 ]).
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])
= (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,
p. 235 as exercises.
Corollary 18.6 Let F be a field of characteristic zero. Then F contains
a subfield isomorphic 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 el-
ements 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.
18.2 FACTORIZATION IN INTEGRAL DOMAINS 237
Example 18.8 It is important to notice that prime and irreducible el-
ements 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 fac-
torizations 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 ̸= 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 irre-
ducible. 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 Theorem of Arithmetic. □
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√ 16.6.12, p. 212, Chapter 16, p. 197). Let
z = a+b 3 i and define ν : Z[ 3 i] → N∪{0} by ν(z) = |z|2 = a2 +3b2 . 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 ele-
ments: √ √
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 + are irreducible. Since 2 is not a
3i √
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 ⟨a⟩ = {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 ⟨b⟩ ⊂ ⟨a⟩.
2. a and b are associates if and only if ⟨b⟩ = ⟨a⟩.
3. a is a unit in D if and only if ⟨a⟩ = D.
238 CHAPTER 18 INTEGRAL DOMAINS
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 ⟨b⟩ ⊂ ⟨a⟩. Conversely, suppose that
⟨b⟩ ⊂ ⟨a⟩. Then b ∈ ⟨a⟩. 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 ⟨a⟩ ⊂ ⟨b⟩. Similarly, ⟨b⟩ ⊂ ⟨a⟩. It follows that
⟨a⟩ = ⟨b⟩. Conversely, suppose that ⟨a⟩ = ⟨b⟩. 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 ⟨a⟩ = ⟨1⟩ = D. ■
Theorem 18.12 Let D be a pid and ⟨p⟩ be a nonzero ideal in D. Then
⟨p⟩ is a maximal ideal if and only if p is irreducible.
Proof. Suppose that ⟨p⟩ is a maximal ideal. If some element a in D
divides p, then ⟨p⟩ ⊂ ⟨a⟩. Since ⟨p⟩ is maximal, either D = ⟨a⟩ or
⟨p⟩ = ⟨a⟩. Consequently, either a and p are associates or a is a unit.
Therefore, p is irreducible.
Conversely, let p be irreducible. If ⟨a⟩ is an ideal in D such that
⟨p⟩ ⊂ ⟨a⟩ ⊂ D, then a | p. Since p is irreducible, either a must be a unit
or a and p are associates. Therefore, either D = ⟨a⟩ or ⟨p⟩ = ⟨a⟩. Thus,
⟨p⟩ 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 ⟨ab⟩ ⊂ ⟨p⟩. By
Corollary 16.40, p. 207, since ⟨p⟩ is a maximal ideal, ⟨p⟩ must also be a
prime ideal. Thus, either a ∈ ⟨p⟩ or b ∈ ⟨p⟩. Hence, either p | a or p | b.
■
Lemma 18.14 Let D be a pid. Let I1 , I2 , . . . be a set of ideals such that
I1 ⊂ I2 ⊂ · · ·. Then there exists an integer N such that In = IN for all
n ≥ N.
∪∞
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 = ⟨a⟩.
Consequently, In = IN for n ≥ N . ■
Any commutative ring satisfying the condition in Lemma 18.14, p. 238
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, ⟨a⟩ ⊂ ⟨a1 ⟩. By Lemma 18.11, p. 237, we know
that ⟨a⟩ ̸= ⟨a1 ⟩; 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
18.2 FACTORIZATION IN INTEGRAL DOMAINS 239
before, ⟨a1 ⟩ ⊂ ⟨a2 ⟩. We can continue with this construction to obtain an
ascending chain of ideals
⟨a⟩ ⊂ ⟨a1 ⟩ ⊂ ⟨a2 ⟩ ⊂ · · · .
By Lemma 18.14, p. 238, there exists a positive integer N such that ⟨an ⟩ =
⟨aN ⟩ 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 ⟨a⟩ ⊂ ⟨c1 ⟩.
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
⟨a⟩ ⊂ ⟨c1 ⟩ ⊂ ⟨c2 ⟩ ⊂ · · · .
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,
p. 238 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. ■
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, p. 243, 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 = ⟨p(x)⟩.
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 ⟨p(x)⟩ = 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. □
240 CHAPTER 18 INTEGRAL DOMAINS
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 for each a ∈ D there is a
nonnegative integer ν(a) 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 = ̸ 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,
p. 201 of Chapter 16, p. 197 are defined by
Z[i] = {a + bi : a, b ∈ Z}.
We usually measure
√ the size of a complex
√ number a + bi by its absolute
value, |a + bi| = a2 + b2 ; however, a2 + b2 may not be an integer. For
our valuation we will let ν(a + bi) = a2 + b2 to ensure that we have an
integer.
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 ̸= 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 + d 2
ac + bd bc − ad
= 2 + 2 i
c + d2 c + d2
( ) ( )
n1 n2
= m1 + 2 + m 2 + i
c + d2 c2 + d2
( )
n1 n2
= (m1 + m2 i) + + i
c2 + d2 c2 + d2
= (m1 + m2 i) + (s + ti)
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).
18.2 FACTORIZATION IN INTEGRAL DOMAINS 241
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 = ⟨b⟩. ■
Corollary 18.22 Every Euclidean domain is a unique factorization do-
main.
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, p. 224).
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.
∑m ∑n
Proof. Let f (x) = i=0 ai xi and g(x) = i=0 bi xi . 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. ■
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).
242 CHAPTER 18 INTEGRAL DOMAINS
Proof. Let p(x) = cp1 (x) and q(x) = dq1 (x), where c and d are the con-
tents 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 , b2 ∈ 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,
p. 242.
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,
p. 242, 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 Corollary 18.27, p. 242, 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
18.2 FACTORIZATION IN INTEGRAL DOMAINS 243
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 impor-
tant corollaries.
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 factoriza-
tion domains that are not principal ideal domains (Z[x]).
Sage. Sage supports distinctions between “plain” rings, domains, prin-
cipal ideal domains and fields. Support is often very good for construc-
tions and computations with PID’s, but sometimes problems get signifi-
cantly 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
n
succeeded in showing that if 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 pub-
lished his famous book on number theory, Disquisitiones Arithmeticae.
Throughout his life Gauss was intrigued 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 Uni-
versity of Göttingen, a position he held until his death. This position
required him to study applications of mathematics to the sciences. He
succeeded in making contributions to fields such as astronomy, mechan-
ics, optics, geodesy, and magnetism. Along with Wilhelm Weber, he
coinvented the first practical electric telegraph some years before a bet-
ter version was invented by Samuel F. B. Morse.
244 CHAPTER 18 INTEGRAL DOMAINS
Gauss was clearly the most prominent mathematician in the world in the
early nineteenth century. His status naturally made his discoveries sub-
ject to intense scrutiny. Gauss’s cold and distant personality many times
led him to ignore the work of his contemporaries, making him many en-
emies. He did not enjoy teaching very much, and young mathematicians
who sought him out for encouragement were often rebuffed. Neverthe-
less, he had many outstanding 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 math-
ematician. Gauss died at the age of 78 in Göttingen on February 23,
1855.
18.3 Exercises
√ √
1. Let z = a + b 3 i be in Z[ 3 i]. If a2 + 3b 2
√ = 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 ad-
dition.
(b) Show that the operation of multiplication is well-defined in the
field of fractions, FD .
(c) Verify the associative and commutative properties for multi-
plication 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 de-
note 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}.
18.3 EXERCISES 245
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.
(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 ∈ R
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.
246 CHAPTER 18 INTEGRAL DOMAINS
20. Let R be a commutative ring with identity. We define a multiplica-
tive 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 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 oper-
ations. 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 homomorphism.
(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.
18.4 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}
247
248 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
{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, p. 248. □
{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, p. 248. Figure 19.7,
p. 248 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 .
19.1 LATTICES 249
Example 19.8 Let Y = {2, 3, 4, 6} be contained in the set X of Exam-
ple 19.6, p. 248. 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 inter-
changed 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
250 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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.
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
19.2 BOOLEAN ALGEBRAS 251
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; 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
252 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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′ )
= (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, p. 250 are met;
therefore, B must be a lattice. Condition (4) tells us that B is a distrib-
utive 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, p. 250, Theorem 19.15, p. 251, and
our hypothesis. ■
Many other identities hold in Boolean algebras. Some of these iden-
tities 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.
19.2 BOOLEAN ALGEBRAS 253
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)
= (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 ̸= O and a ∧ b = a for all nonzero b ∈ B. Equivalently, a is an atom
of B if there is no nonzero b ∈ B distinct from a such that O ⪯ b ⪯ a.
Lemma 19.18 Let B be a finite Boolean algebra. If b is a nonzero
element of B, 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 ̸= b. Then a ∧ b = O.
254 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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.
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 ̸⪯ c. Then there exists an atom a ∈ B such that a ⪯ b and
a ̸⪯ c.
Proof. By Lemma 19.20, p. 254, b ∧ c′ ̸= O. Hence, there exists an atom
a such that a ⪯ b ∧ c′ . Consequently, a ⪯ b and a ̸⪯ 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 ̸⪯ b1 . Then there exists an atom a such that a ⪯ b and a ̸⪯ 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, p. 253, 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).
19.3 THE ALGEBRA OF ELECTRICAL CIRCUITS 255
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, p. 254, 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 }
= {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 cir-
cuits. Two switches a and b are in series if they make up a circuit of the
type that is illustrated in Figure 19.25, p. 256. 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, p. 256. 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.
256 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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, 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, p. 256. 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, p. 256.
b a b
a
c a c
Figure 19.27: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
a
a a′
a′
Figure 19.28: a ∧ a′ = O and a ∨ a′ = I
Example 19.29 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 19.32, p. 257. □
Theorem 19.30 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.
19.3 THE ALGEBRA OF ELECTRICAL CIRCUITS 257
Example 19.31 Given a complex circuit, we can now apply the tech-
niques of Boolean algebra to reduce it to a simpler one. Consider the
circuit in Figure 19.32, p. 257. 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. □
a a a
b b′ b
Figure 19.32: (a ∨ b) ∧ (a ∨ b′ ) ∧ (a ∨ b)
Sage. Sage has a full suite of functionality for both posets and lattices,
all as part of its excellent 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 princi-
ple 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 Whitehead (1861–1947), Bertrand Russell (1872–1970), and
David Hilbert (1862–1943). In Principia Mathematica, Whitehead and
Russell attempted to show the connection between mathematics 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
258 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
Boolean algebras to model social hierarchies; biologists have used them
to describe biosystems.
19.4 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 36. 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 expres-
sions.
(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.
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.
19.4 EXERCISES 259
12. Let L be a nonempty set with two binary operations ∨ and ∧ satisfy-
ing the commutative, associative, idempotent, and absorption laws.
We can define a partial order on L, as in Theorem 19.14, p. 250, 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 The-
orem 19.30, p. 256 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 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 homomorphism
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
260 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
this partial order.
19.5 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 19.33, p. 260. Write a program to evaluate 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
Table 19.33: Boolean polynomials
19.6 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 )
261
262 CHAPTER 20 VECTOR SPACES
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 ).
□
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 poly-
nomial 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 ̸= 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
20.3 LINEAR INDEPENDENCE 263
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, 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
∑
n
w= αi vi = α1 v1 + α2 v2 + · · · + αn vn
i=1
is called a linear combination of the vectors v1 , v2 , . . . , vn . The span-
ning 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,
264 CHAPTER 20 VECTOR SPACES
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 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 proposition.
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 ̸= 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.
20.4 EXERCISES 265
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 arbi-
trary 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.
√ basis of R consists of exactly three vectors, and every
3
However, 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 Proposition 20.11, p. 264, 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 set-
tings, 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 Exercises
1. If F is a field, show that F [x] is a vector space over F , where the vec-
tors 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.
266 CHAPTER 20 VECTOR SPACES
√ √
3. √ Q( √2, 3 √
Let ) 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 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 opera-
tions.
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 equa-
tions
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.
20.4 EXERCISES 267
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.
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 transfor-
mation 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 .
268 CHAPTER 20 VECTOR SPACES
(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 follows:
(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 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 ∗∗ .
20.5 References and Suggested Readings
[1] Beezer, R. A First Course in Linear Algebra . Available online at
http://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 poly-
nomials 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. □
269
270 CHAPTER 21 FIELDS
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, p. 227, the ideal ⟨p(x)⟩ generated by p(x) is
maximal; hence, Z2 [x]/⟨p(x)⟩ is a field. Let f (x) + ⟨p(x)⟩ be an arbitrary
element of Z2 [x]/⟨p(x)⟩. 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) + ⟨x2 + x + 1⟩ = r(x) + ⟨x2 + x + 1⟩.
The only possibilities for r(x) are then 0, 1, x, and 1 + x. Consequently,
E = Z2 [x]/⟨x2 + x + 1⟩ 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
Table 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
Table 21.4: Multiplication Table for Z2 (α)
The following theorem, due to Kronecker, is so important and so basic
to our understanding of fields that it is often known as the Fundamental
Theorem of Field Theory.
21.1 EXTENSION FIELDS 271
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, p. 270. 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 ⟨p(x)⟩ generated
by p(x) is a maximal ideal in F [x] by Theorem 17.22, p. 227; hence,
F [x]/⟨p(x)⟩ is a field. We claim that E = F [x]/⟨p(x)⟩ 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]/⟨p(x)⟩,
where ψ(a) = a + ⟨p(x)⟩ for a ∈ F . It is easy to check that ψ is indeed a
ring homomorphism. Observe that
ψ(a) + ψ(b) = (a + ⟨p(x)⟩) + (b + ⟨p(x)⟩) = (a + b) + ⟨p(x)⟩ = ψ(a + b)
and
ψ(a)ψ(b) = (a + ⟨p(x)⟩)(b + ⟨p(x)⟩) = ab + ⟨p(x)⟩ = ψ(ab).
To prove that ψ is one-to-one, assume that
a + ⟨p(x)⟩ = ψ(a) = ψ(b) = b + ⟨p(x)⟩.
Then a − b is a multiple of p(x), since it lives in the ideal ⟨p(x)⟩. 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 + ⟨p(x)⟩ : 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 + ⟨p(x)⟩. Then α is in E. If p(x) = a0 + a1 x + · · · + an xn , then
p(α) = a0 + a1 (x + ⟨p(x)⟩) + · · · + an (x + ⟨p(x)⟩)n
= a0 + (a1 x + ⟨p(x)⟩) + · · · + (an xn + ⟨p(x)⟩)
= a0 + a1 x + · · · + an xn + ⟨p(x)⟩
= 0 + ⟨p(x)⟩.
Therefore, we have found an element α ∈ E = F [x]/⟨p(x)⟩ 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]/⟨x2 + x + 1⟩ or
Z2 [x]/⟨x3 +x+1⟩. We will leave it as an exercise to show that Z2 [x]/⟨x3 +
x + 1⟩ 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
272 CHAPTER 21 FIELDS
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.6 (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.
√ √
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 transcendental over F if and only if ϕα (p(x)) = p(α) ̸= 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, p. 235, E must contain a copy
of this field. ■
We have a more interesting situation in the case of algebraic exten-
sions.
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 ⟨p(x)⟩ 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) ∈ ⟨p(x)⟩ 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 poly-
nomials 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. ■
6 The probability that a real number chosen at random from the interval [0, 1] will
be transcendental over the rational numbers is one.
21.1 EXTENSION FIELDS 273
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.
Example 21.11 Let f (x) = x2 − 2 and g(x) = x4√
− 4x2 + 1. These poly-
√ √
nomials are the 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]/⟨p(x)⟩, 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 ⟨p(x)⟩, 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
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 combination 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.
■
274 CHAPTER 21 FIELDS
Example 21.14 Since x2 + 1 is irreducible over R, ⟨x2 + 1⟩ is a maximal
ideal in R[x]. So E = R[x]/⟨x2 + 1⟩ is a field extension of R that contains
a root of x2 +1. Let α = x+⟨x2 +1⟩. We can identify E with the complex
numbers. By Proposition 21.12, p. 273, E is isomorphic to R(α) = {a +
bα : a, b ∈ R}. We know that α2 = −1 in E, since
α2 + 1 = (x + ⟨x2 + 1⟩)2 + (1 + ⟨x2 + 1⟩)
= (x2 + 1) + ⟨x2 + 1⟩
= 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, p. 273
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.
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, p. 274 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 Theo-
rem in group theory. Theorem 21.17, p. 274 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 ].
21.1 EXTENSION FIELDS 275
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 a basis for K over∑F . We will first show
∑nthat these vectors span
m
K. Let u ∈ K. Then u = j=1 bj βj and bj = i=1 aij αi , where bj ∈ E
and aij ∈ F . Then
( n )
∑m ∑ ∑
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 ∑
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 ( n )
∑
m ∑
cij αi βj = 0,
j=1 i=1
∑
where i cij αi ∈ E. Since the βj ’s are linearly independent over E, it
must be the case that
∑n
cij αi = 0
i=1
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 alge-
braic 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
276 CHAPTER 21 FIELDS
√ √ √
cannot be in √ Q( 3 ). It follows that √ √ 5 cannot√be in √ Q( 3 ) either.
√
Therefore, √ {1, √ √ 5 }√ √ for Q( 3, 5 ) = (Q(
is a basis √ Q(√3 )
√ 3√))( 5 ) over
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( 5, 5 i), where 5 is 3
the positive
√ square
√ root of 5 and 3 5 is the real cube root of 5. We know
that 5 i ∈ / 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, 3 5, ( 3 5 )2 } is a basis for Q( 3 5 ) over
Q. Hence, a basis for Q( 3 5, 5 i) over Q is
√ √ √ √ √ √ √
{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 ,
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, p. 274.
(2) ⇒ (3). Suppose that E = F (α1 , . . . , αn ), where every αi is alge-
braic 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 )]
21.1 EXTENSION FIELDS 277
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 ex-
tension of F . Since every element of F (α, β) is algebraic over F , α ± β,
αβ, and α/β (β ̸= 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.
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 sophisti-
cated 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, p. 305.
Theorem 21.28 Fundamental Theorem of Algebra. The field of
complex numbers is algebraically closed.
278 CHAPTER 21 FIELDS
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 x2 − 2 and x2 + 4. Therefore, the field Q( 2, i) is a
splitting field for p(x). □
Example 21.30
√ Let p(x) = x3 − 3 be in Q[x]. Then p(x) has a root in
the field Q( 3 ). However, this field is not a splitting field for p(x) since
3
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, p. 271, 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). ■
The question of uniqueness now arises for splitting fields. This ques-
tion 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
21.2 SPLITTING FIELDS 279
ϕ(α) = β and ϕ agrees with ϕ on E.
Proof. If p(x) has degree n, then by Theorem 21.13, p. 273 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 assump-
tion, ϕ(p(x)) = q(x); hence, ϕ maps ⟨p(x)⟩ onto ⟨q(x)⟩. Consequently,
we have an isomorphism ψ : E[x]/⟨p(x)⟩ → F [x]/⟨q(x)⟩. By Proposi-
tion 21.12, p. 273, we have isomorphisms σ : E[x]/⟨p(x)⟩ → E(α) and
τ : F [x]/⟨q(x)⟩ → F (β), defined by evaluation at α and β, respectively.
Therefore, ϕ = τ ψσ −1 is the required isomorphism.
ψ
E[x]/⟨p(x)⟩ F [x]/⟨q(x)⟩
σ τ
ϕ
E(α) F (β)
ϕ
E F
We leave the proof of uniqueness as a exercise. ■
Theorem 21.33 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.
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,
p. 278, there exists an isomorphism ϕ : E(α) → F (β) such that ϕ(α) = β
and ϕ agrees with ϕ on E.
280 CHAPTER 21 FIELDS
ψ
K L
ϕ
E(α) F (β)
ϕ
E F
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 21.34 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 straight-
edge 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
21.3 GEOMETRIC CONSTRUCTIONS 281
that is impossible to trisect with a straightedge-and-compass construc-
tion. 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◦ 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 21.35 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 ̸= 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 21.36, p. 281
such that triangles △ABC and △ADE 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 ̸= 0. ■
D
β B
1
α C
A E
x
Figure 21.36: Construction of products
√
Lemma 21.37 If α is a constructible number, then α is a constructible
number.
Proof. In Figure 21.38, p. 281 the triangles △ABD, △BCD, and △ABC
are similar; hence, 1/x = x/α, or x2 = α. ■
B
x
1 α
A D C
Figure 21.38: Construction of roots
282 CHAPTER 21 FIELDS
By Theorem 21.35, p. 281, 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 21.39 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 ̸= 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 pos-
sible ways of constructing 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 coordinates in F and whose radii are of lengths in F .
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.
21.3 GEOMETRIC CONSTRUCTIONS 283
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 −4AC > 0. We have proven the following
2
lemma.
Lemma 21.40 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 21.41 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 21.40, p. 283 and of the fact that
[Fk : Q] = [Fk : Fk−1 ][Fk−1 : Fk−2 ] · · · [F1 : Q] = 2k .
■
Corollary 21.42 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 num-
bers to show exactly when a particular geometric construction can be
accomplished.
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.
284 CHAPTER 21 FIELDS
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 im-
possible to construct a 20◦ angle. Consequently, a 60◦ angle cannot be
trisected. We first need to calculate the triple-angle formula for the co-
sine:
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 8x − 6x − 1. This polynomial has no factors in
3
Z[x], and hence is irreducible over Q[x]. Thus, [Q(α) : Q] = 3. Conse-
quently, 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.
Attempts to prove Fermat’s Last Theorem have led to important contri-
butions 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 understand-
21.4 EXERCISES 285
ing of fields.
David Hilbert (1862–1943) and Hermann Minkowski (1864–1909) were
among the mathematicians who led the way in this subject at the begin-
ning of the twentieth century. Hilbert and Minkowski were both mathe-
maticians at Göttingen University in Germany. Göttingen was truly one
the most important centers of mathematical research during the last two
centuries. 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 geome-
try, a field of mathematics that studies geometry by studying commuta-
tive rings. From about 1955 to 1970, Alexander Grothendieck dominated
the field of algebraic geometry. Pierre Deligne, a student of Grothen-
dieck, solved several of Weil’s number-theoretic conjectures. One of the
most recent contributions to algebra and number theory is Gerd Falt-
ing’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 Exercises
1. Show that each of the following numbers is algebraic over Q by
finding the minimal polynomial of the number over Q.
√ √
(a) 1/3 + 7
√ √
3
(b) 3+ 5
√ √
(c) 3+ 2i
(d) cos θ + i sin θ for θ = 2π/n with n ∈ N
√√
(e) 3
2−i
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
√
(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
286 CHAPTER 21 FIELDS
√
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]/⟨x3 +x+1⟩ 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]/⟨x3 − 2⟩ 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 poly-
nomial 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 ̸= 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.
21. Let E be an algebraic extension of a field F , and let σ be an auto-
morphism 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.
√ √ √ √
√ Q(
22. Show that √ 3, 7 )√= Q(√ 3 + 7 ). Extend your proof to show
that Q( a, b ) = Q( a + b ), where gcd(a, b) = 1.
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.
21.5 REFERENCES AND SUGGESTED READINGS 287
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.
21.5 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.
288 CHAPTER 21 FIELDS
22
Finite Fields
Finite fields appear in many applications of algebra, including coding
theory and cryptography. 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,
p. 202 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 charac-
teristic 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 . ■
289
290 CHAPTER 22 FINITE FIELDS
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.
Proof. We will prove this lemma using mathematical induction on n. We
can use the binomial formula (see Chapter 2, p. 17, Example 2.4, p. 18)
to verify the case for n = 1; that is,
p ( )
∑ p
(a + b)p = ak bp−k .
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].
√ polynomial x√− 2 is separable over Q since it factors
2
Example √ 22.4 The
as (x − √ 2 )(x + 2 ). 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 ̸= 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 ̸= αj for i ̸= 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 ).
22.1 STRUCTURE OF A FINITE FIELD 291
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. ■
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
n
pn is isomorphic to the splitting field of 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, p. 290, f (x) has pn distinct zeros in F , since f ′ (x) =
pn xp −1 − 1 = −1 is relatively prime to f (x). We claim that the roots
n
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 ̸= 0, then (α−1 )p = (αp )−1 = α−1 . Since the zeros of f (x) form a
n n
subfield of F and 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
n
pn
or α − α = 0. Since E contains p elements, E must be a splitting
n
field of f (x); however, by Corollary 21.34, p. 280, the splitting field of
any polynomial is unique up to isomorphism. ■
n
The unique finite field with p elements is called the Galois field of
order pn . 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,
m n
m n m
xp − x must divide xp − x, and every zero of xp − x is also a zero of
n m
xp − x. Thus, GF(pn ) contains, as a subfield, a 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,
p. 292. □
292 CHAPTER 22 FINITE FIELDS
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, p. 164),
G∼
= Zpe11 × · · · × Zpek ,
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 /⟨1+
x + x4 ⟩. 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
22.2 POLYNOMIAL CODES 293
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, p. 93. 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 1 0
0 1
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 ,
294 CHAPTER 22 FINITE FIELDS
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 20.4.15,
p. 267, Chapter 20, p. 261). 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 polynomials 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 Exam-
ple 22.14, p. 293. 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]/⟨xn − 1⟩
can be considered to be the ring of polynomials of the form
f (t) = a0 + a1 t + · · · + an−1 tn−1
22.2 POLYNOMIAL CODES 295
that satisfy the condition tn = 1. It is an easy exercise to show that Zn2
and Rn are isomorphic as vector spaces. We will often identify elements
in Zn2 with elements in Z[x]/⟨xn − 1⟩. In this manner we can interpret a
linear code as a subset of Z[x]/⟨xn − 1⟩.
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]/⟨xn − 1⟩.
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 code-
words 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]/⟨xn + 1⟩. 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, p. 295 tells us that knowing the ideals of Rn is equiv-
alent 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, p. 205, every ideal
C in Rn is of the form ϕ(I), where I is an ideal in Z2 [x] that contains
⟨xn − 1⟩. By Theorem 17.20, p. 226, we know that every ideal I in Z2 [x]
is a principal ideal, since Z2 is a field. Therefore, I = ⟨g(x)⟩ for some
unique monic polynomial in Z2 [x]. Since ⟨xn − 1⟩ 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 = ⟨g(t)⟩ = {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, p. 294, it is easy to calculate
a generator matrix by examining what g(t) does to the polynomials 1, t,
296 CHAPTER 22 FINITE FIELDS
t2 , and t3 . A generator matrix for C is
1 0 0 0
1 0
1 0
0 0
1 1
G = 1 0 1 1 .
0 1 0 1
0 0 1 0
0 0 0 1
□
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 = ⟨g(t)⟩ 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, p. 295,
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
22.2 POLYNOMIAL CODES 297
are elements in a field F , then the n × n matrix
1 1 ··· 1
α1 α2 ··· αn
α12 α22 ··· αn2
.. .. .. ..
. . . .
α1n−1 α2n−1 ··· αnn−1
is called the Vandermonde matrix. The determinant of this matrix
is called the Vandermonde determinant. We will need the following
lemma in our investigation of cyclic codes.
Lemma 22.20 Let α1 , . . . , αn be elements in a field F with n ≥ 2. Then
1 1 ··· 1
α1 α2 ··· αn
∏
α12 α22 ··· αn2
det = (αi − αj ).
.. .. .. ..
. . . . 1≤j<i≤n
α1n−1 α2n−1 ··· αnn−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
α2 α2 2
· · · αn−1
2
x2
p(x) = det 1 .
. .. .. .. ..
.. . . . .
α1n−1 α2n−1 ··· n−1
αn−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
α12 α22 ··· 2
αn−1
β = (−1) n+n
det .
.. .. .. ..
. . . .
α1n−2 α2n−2 ··· n−2
αn−1
By our induction hypothesis,
∏
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 capabilities for a particular generator polynomial.
298 CHAPTER 22 FINITE FIELDS
Theorem 22.21 Let C = ⟨g(t)⟩ 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
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
i0 r+1
(ω i1 )r+1 ··· (ω is−1 )r+1
(ω )
.. .. .. ..
. . . .
i0 r+s−1
(ω ) i1 r+s−1
(ω ) · · · (ω is−1 r+s−1
)
can be shown to be nonzero using Lemma 22.20, p. 297 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. Hoc-
quenghem 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
22.2 POLYNOMIAL CODES 299
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 ⟨g(t)⟩ in Rn is called the bch code of length n and
distance d. By Theorem 22.21, p. 298, the minimum distance of C is at
least d.
Theorem 22.22 Let C = ⟨g(t)⟩ 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)
ω3 ω6 ··· ω (n−1)(3)
H = 1
. .. .. .. ..
.. . . . .
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 = ⟨g(t)⟩. ■
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}.
300 CHAPTER 22 FINITE FIELDS
By Example 22.8, p. 291, 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 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 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(p ) : GF(p )], where n | m.
m n
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]/⟨x3 + x + 1⟩ ∼
= Z2 [x]/⟨x3 + x2 + 1⟩.
9. Determine the number of cyclic codes of length n for n = 6, 7, 8, 10.
10. Prove that the ideal ⟨t + 1⟩ 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.
12. Prove or disprove: There exists a finite field that is algebraically
closed.
22.3 EXERCISES 301
13. Let p be prime. Prove that the field of rational functions Zp (x) is
an infinite field of characteristic p.
14. Let D be an integral domain of characteristic p. Prove that (a −
n n n
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 code-
word. 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 ⟨f (t)⟩ is another
code in Rn , show that ⟨g(t)⟩ ⊂ ⟨f (t)⟩ if and only if f (x) divides g(x)
in Z2 [x].
29. Let C = ⟨g(t)⟩ 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
g ··· 0
1 g0
. .. .. ..
.. . .
.
G = gn−k gn−k−1 · · · g0
0 gn−k ··· g1
.. .. .. ..
. . . .
0 0 · · · gn−k
302 CHAPTER 22 FINITE FIELDS
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.
22.4 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, p. 296. By The-
orem 8.13, p. 100, this code is capable of correcting two errors. Sup-
pose that these errors occur in bits a1 and a2 . The error-locator
polynomial is s(x) = (x + ω a1 )(x + ω a2 ). Show that
( )
s3
s(x) = x2 + s1 x + s21 + .
s1
4. Let w(t) = 1 + t2 + t4 + t5 + t7 + t12 + t13 . Determine what the
originally transmitted code polynomial was.
22.5 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.
22.5 REFERENCES AND SUGGESTED READINGS 303
[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.
[6] van Lint, J. H. Introduction to Coding Theory. Springer, New York,
1999.
304 CHAPTER 22 FINITE FIELDS
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 polyno-
mials could and could not be solved by formulas. He discovered the
connection between groups and field extensions. Galois theory demon-
strates 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.
305
306 CHAPTER 23 GALOIS THEORY
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. ■
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 automor-
phism 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 √ √ the field Q( 3, 5 ) as a vector space √
√ regard over√Q that has
basis
√ √{1, 3, 5, 15 }. 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 polyno-
mial 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 )
23.1 FIELD AUTOMORPHISMS 307
= 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.
A converse of the last proposition exists. The proof follows directly
from Lemma 21.32, p. 278.
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 the degree of f (x). If the
degree of f (x) is 0 or 1, then E = F and there is nothing to show. Assume
that the result holds for all polynomials of degree k with 0 ≤ k < n.
Suppose that the degree of f (x) is n. Let p(x) be an irreducible factor of
f (x) of degree r. Since all of the roots of p(x) are in E, we can choose
one of these roots, say α, so that F ⊂ F (α) ⊂ E. Then
[E : F (α)] = n/r and [F (α) : F ] = r.
If β is any other root of p(x), then F ⊂ F (β) ⊂ E. By Lemma 21.32,
p. 278, there exists a unique isomorphism σ : F (α) → F (β) for each such
β that fixes F elementwise. Since E is a splitting field of p(x), there are
exactly r such isomorphisms. For each of these automorphisms, we can
use our induction hypothesis on [E : F (α)] = n/r < n to conclude that
|G(E/F (α))| = [E : F (α)].
Consequently, there are
[E : F ] = [E : F (α)][F (α) : F ] = n
possible automorphisms of E that fix F , or |G(E/F )| = [E : F ]. ■
Corollary 23.8 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
m
assume that E is the splitting field of xp − x over a subfield of order p.
m
Therefore, E must also be the splitting field of xp − x over F . Applying
Theorem 23.7, p. 307, we find that |G(E/F )| = k.
To prove that G(E/F ) is cyclic, we must find a generator for G(E/F ).
n
Let σ : E → E 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, p. 290. 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
308 CHAPTER 23 GALOIS THEORY
must also be onto, since E is a finite field. We know that σ must be in
n
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, p. 307, 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 ≤
nr
r < k; otherwise, xp − x would have pm roots, which is impossible. ■
√ √
Example 23.9 We can now confirm that the Galois group of Q( 3, 5 )
over Q in Example 23.4, p. 306 is indeed isomorphic√to Z√2 × Z2 . Certainly
the group H = {id, σ, τ, µ}
√ √ is a subgroup 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 23.10 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 17.4.20, p. 230 in
Chapter 17, p. 217. 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 automor-
phisms 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
∏
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,
p. 290).
23.2 THE FUNDAMENTAL THEOREM 309
Proposition 23.11 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) ̸= 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)) ̸= 1 is if f ′ (x) is the
zero polynomial; however, this is impossible 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 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, p. 292 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 23.12 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 math-
ematical 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 ̸=
β − βj
for all i and j with j ̸= 1. Therefore, a(β − βj ) ̸= αi − α. Let γ = α + aβ.
Then
γ = α + aβ ̸= αi + aβj ;
hence, γ − aβj ̸= αi for all i, j with j ̸= 1. Define h(x) ∈ F (γ)[x] by
h(x) = f (γ − ax). Then h(β) = f (α) = 0. However, h(βj ) ̸= 0 for j ̸= 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 .
310 CHAPTER 23 GALOIS THEORY
Proposition 23.13 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 ̸= 0, then σi (a−1 ) = [σi (a)]−1 = a−1 . Finally, σi (0) = 0 and σi (1) = 1
since σi is an automorphism. ■
Corollary 23.14 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}
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 23.15
√ 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 23.16 Let E be a splitting field over F of a separable poly-
nomial. 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, p. 307,
|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 monograph on the subject [1]. The very clever proof of the
following lemma is due to Artin.
Lemma 23.17 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 ho-
mogeneous 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.
23.2 THE FUNDAMENTAL THEOREM 311
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 rear-
ranging 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 ) ̸= 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 )
..
.
xn+1 = an+1 − σi (an+1 )
must be another solution of the system. This is a nontrivial solution
because σi (a2 ) ̸= 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 23.18 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 polyno-
mial. By Proposition 23.16, p. 310, 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, p. 306,
312 CHAPTER 23 GALOIS THEORY
automorphisms in G permute the roots of f (x) lying in E. Hence, if we
let G act on∏α, we can obtain distinct roots α1 = α, α2 , . . . , αn in E.
n
Let g(x) = i=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 23.19 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 exten-
sions and automorphisms of fields, let us return to a familiar example.
Example 23.20 √ √In Example 23.4, p. 306 we examined the automor-
phisms of Q( 3, 5 ) fixing Q. Figure 23.21, p. 312 compares √ the
√ lattice
of field extensions of Q with the lattice of subgroups of G(Q( 3, 5 )/Q).
The Fundamental Theorem of Galois Theory tells us what the relation-
ship is between the two lattices. □
√ √
{id, σ, τ, µ} Q( 3, 5 )
√ √ √
{id, σ} {id, τ } {id, µ} Q( 3 ) Q( 5 ) Q( 15 )
{id} Q
√ √
Figure 23.21: G(Q( 3, 5 )/Q)
We are now ready to state and prove the Fundamental Theorem of
Galois Theory.
Theorem 23.22 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).
23.2 THE FUNDAMENTAL THEOREM 313
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, p. 307, |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 23.23, p. 314. 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);
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 re-
stricting 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).
■
314 CHAPTER 23 GALOIS THEORY
E {id}
L G(E/L)
K G(E/K)
F G(E/F )
Figure 23.23: Subgroups of G(E/F ) and subfields of E
Example 23.24 In this example we will illustrate the Fundamental The-
orem of Galois Theory by determining the lattice of subgroups of the Ga-
lois 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
√ is Q( 2, i). To see this, notice that
4
splitting field
√ of2 f (x) √ f (x) factors√
as (x + 2 )(x − 2 );√hence, the roots of f (x) are ± 4 2 and ± 4 2 i.
2
We√first adjoin the root 4 2 to Q and then adjoin √ the root√i of x2 + 1 to
Q( 2 ). The √
4
splitting field of f (x) is then Q( √2 )(i) = Q( 4 2, i).
4
Since√[Q( 4 2 ) :√Q] = 4 and i is not √ in Q( 4 2 ), it must be the case
that [Q( 2, i) : Q( 2 )] = 2. Hence, [Q( 2, i) : Q] = 8. The set
4 4 4
√ √ √ √ √ √
{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 23.25, p. 315(a).
The Galois group√G of f (x) √ must be of order 8. Let σ be the automor-
phism defined by 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 ele-
ment 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 23.25, p. 315(b). □
23.2 THE FUNDAMENTAL THEOREM 315
√
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 23.25: 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 his-
tory’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 extraordinary mathematical talent at the age of 14. He applied for
entrance to the École Polytechnique several times; however, he had great
difficulty meeting the formal entrance requirements, 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.
316 CHAPTER 23 GALOIS THEORY
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, subtrac-
tion, 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, p. 47, 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, p. 292, the nth roots of unity form a
cyclic group. Any generator of this group is called a primitive nth root
of unity.
Example 23.26 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},
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
23.3 APPLICATIONS 317
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, p. 134.
Lemma 23.27 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 solv-
able.
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 23.28 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
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
318 CHAPTER 23 GALOIS THEORY
a primitive n1 th root of unity. If F contains all of its n1 roots of unity,
then K1 = F (α! ). 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 23.29 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 23.28,
p. 317, 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 23.27, p. 317, G(Fi /Fi−1 ) is solvable; hence, G(E/F ) is also
solvable. ■
The converse of Theorem 23.29, p. 318 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 23.30 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 = (12). The order of
τ is p and τ n must be a cycle of length p for 1 ≤ n < p. There-
fore, we may assume that µ = τ n = (12i3 . . . ip ) for some n, where
1 ≤ n < p (see Exercise 5.3.13, p. 72 in Chapter 5, p. 61). Noting that
(12)(12i3 . . . ip ) = (2i3 . . . ip ) and (2i3 . . . ip )k (12)(2i3 . . . ip )−k = (1ik ),
we can obtain all the transpositions of the form (1n) for 1 ≤ n < p.
However, these transpositions generate all transpositions in Sp , since
(1j)(1i)(1j) = (ij). The transpositions generate Sp . ■
23.3 APPLICATIONS 319
y
60
f(x) = x5 −6x3 −27x−3
40
20
x
-3 -2 -1 1 2 3
-20
-40
-60
Figure 23.31: The graph of f (x) = x5 − 6x3 − 27x − 3
Example 23.32 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 sepa-
rable. 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
√ √
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 23.31, p. 319). 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
transpostion. If α is one of the real roots of f (x), then [Q(α) : Q] = 5
by Exercise 21.4.28, p. 287. Since Q(α) is a subfield of K, it must be
the case the [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 23.30, p. 318, S5 is generated by a transposition and an element of
order 5; therefore, G(K/Q) must be all of S5 . By Theorem 10.11, p. 134,
S5 is not solvable. Consequently, f (x) cannot be solved by radicals. □
320 CHAPTER 23 GALOIS THEORY
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 dis-
tinct linear factors.
Theorem 23.33 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 ̸= 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 calculus. 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 Exercises
1. Compute each of the following Galois groups. Which of these field
extensions are normal field extensions? If the extension is not nor-
mal, find a normal extension of Q in which the extension field is
23.4 EXERCISES 321
contained. √ √ √
(a) G(Q( 30 )/Q) (d) G(Q( 2, 3 2, i)/Q)
√
(b) G(Q( 4 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, determine 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) x − 5
3
(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.
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.
322 CHAPTER 23 GALOIS THEORY
16. Let K be the splitting field of x3 + x2 + 1 ∈ Z2 [x]. Prove or disprove
that K is an extension by radicals.
√ ) ̸= 2. Prove that the splitting field
17. Let F be a field such that char(F
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 con-
clude 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 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.
23.5 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.
[2] Edwards, H. M. Galois Theory. Springer-Verlag, New York, 1984.
23.5 REFERENCES AND SUGGESTED READINGS 323
[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 Math-
ematical 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 Amer-
ican, April 1982, 136–49.
324 CHAPTER 23 GALOIS THEORY
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B
Hints and Answers to Selected Exercises
1 · Preliminaries
1.3 · Exercises
1.3.1. Hint. (a) A ∩ B = {2}; (b) B ∩ C = {5}.
1.3.2. 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 = ∅.
1.3.6. Hint. If x ∈ A ∪ (B ∩ C), then either x ∈ A or x ∈ B ∩ C.
Thus, x ∈ A ∪ B and A ∪ C. Hence, x ∈ (A ∪ B) ∩ (A ∪ C). Therefore,
A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C). Conversely, if x ∈ (A ∪ B) ∩ (A ∪ C),
then x ∈ A ∪ B and A ∪ C. Thus, x ∈ A or x is in both B and C. So
x ∈ A ∪ (B ∩ C) and therefore (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C). Hence,
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
1.3.10. 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.
1.3.14. Hint. A \ (B ∪ C) = A ∩ (B ∪ C)′ = (A ∩ A) ∩ (B ′ ∩ C ′ ) =
(A ∩ B ′ ) ∩ (A ∩ C ′ ) = (A \ B) ∩ (A \ C).
1.3.17. 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.
1.3.18. 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}.
1.3.20. Hint. (a) f (n) = n + 1.
1.3.22. 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.
1.3.23. Hint. f −1 (x) = (x + 1)/(x − 1).
1.3.24. Hint. (a) Let y ∈ f (A1 ∪ A2 ). Then there exists an x ∈
A1 ∪ A2 such that f (x) = y. 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 ) =
333
334APPENDIX B HINTS AND ANSWERS TO SELECTED EXERCISES
f (A1 ) ∪ f (A2 ).
1.3.25. 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.
√
1.3.28. Hint. Let X = N ∪ { 2 } and define x ∼ y if x + y ∈ N.
2 · The Integers
2.3 · Exercises
2.3.1. 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.
2.3.3. 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.
2.3.8. Hint. Follow the proof in Example 2.4, p. 18.
2.3.11. 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.
2.3.17. Fibonacci Numbers. Hint. For (a) and (b) use mathemati-
cal induction. (c) Show that f1 = 1, f2 = 1, and fn+2 = fn+1 + fn . (d)
Use part (c). (e) Use part (b) and Exercise 2.3.16, p. 25.
2.3.19. Hint. Use the Fundamental Theorem of Arithmetic.
2.3.23. Hint. Use the Principle of Well-Ordering and the division al-
gorithm.
2.3.27. Hint. Since gcd(a, b) = 1, there exist integers r and s such
that ar + bs = 1. Thus, acr + bcs = c.
2.3.29. 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
3.4 · Exercises
3.4.1. Hint. (a) 3 + 7Z = {. . . , −4, 3, 10, . . .}; (c) 18 + 26Z; (e) 5 + 6Z.
3.4.2. Hint. (a) Not a group; (c) a group.
335
3.4.6. 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
3.4.8. Hint. Pick two matrices. Almost any pair will work.
3.4.15. Hint. There is a nonabelian group containing six elements.
3.4.16. Hint. Look at the symmetry group of an equilateral triangle
or a square.
3.4.17. Hint. The are five different groups of order 8.
3.4.18. 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.
3.4.25. 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 .
3.4.31. Hint. Since abab = (ab)2 = e = a2 b2 = aabb, we know that
ba = ab.
3.4.35. Hint. H1 = {id}, H2 = {id, ρ1 , ρ2 }, H3 = {id, µ1 }, H4 =
{id, µ2 }, H5 = {id, µ3 }, S3 .
√ √
3.4.41.
√ Hint. The identity √ of G is 1 = 1 + 0 2. Since (a + b 2 )(c +
d 2)√ = (ac+2bd)+(ad+bc) 2, √ G is closed under multiplication. Finally,
(a + b 2 )−1 = a/(a2 − 2b2 ) − b 2/(a2 − 2b2 ).
3.4.46. Hint. Look at S3 .
3.4.49. Hint. ba = a4 b = a3 ab = ab
4 · Cyclic Groups
4.4 · Exercises
4.4.1. Hint. (a) False; (c) false; (e) true.
4.4.2. Hint. (a) 12; (c) infinite; (e) 10.
4.4.3. 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}.
4.4.4. Hint. (a)
( ) ( ) ( ) ( )
1 0 −1 0 0 −1 0 1
, , , .
0 1 0 −1 1 0 −1 0
336APPENDIX B HINTS AND ANSWERS TO SELECTED EXERCISES
(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
4.4.10. Hint. (a) 0; (b) 1, −1.
4.4.11. Hint. 1, 2, 3, 4, 6, 8, 12, 24.
4.4.15. Hint. (a) −3 + 3i; (c) 43 − 18i; (e) i
√
4.4.16. Hint. (a) 3 + i; (c) −3.
√ √
4.4.17. Hint. (a) 2 cis(7π/4); (c) 2 2 cis(π/4); (e) 3 cis(3π/2).
√
4.4.18. Hint. (a) (1 − i)/2; (c) 16(i − 3 ); (e) −1/4.
4.4.22. Hint. (a) 292; (c) 1523.
4.4.27. Hint. |⟨g⟩ ∩ ⟨h⟩| = 1.
4.4.31. 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.
4.4.37. Hint. If g is an element distinct from the identity in G, g must
generate G; otherwise, ⟨g⟩ is a nontrivial proper subgroup of G.
5 · Permutation Groups
5.3 · Exercises
5.3.1. Hint. (a) (12453); (c) (13)(25).
5.3.2. Hint. (a) (135)(24); (c) (14)(23); (e) (1324); (g) (134)(25); (n)
(17352).
5.3.3. Hint. (a) (16)(15)(13)(14); (c) (16)(14)(12).
5.3.4. Hint. (a1 , a2 , . . . , an )−1 = (a1 , an , an−1 , . . . , a2 )
5.3.5. Hint. (a) {(13), (13)(24), (132), (134), (1324), (1342)} is not a
subgroup.
5.3.8. Hint. (12345)(678).
5.3.11. Hint. Permutations of the form
(1), (a1 , a2 )(a3 , a4 ), (a1 , a2 , a3 ), (a1 , a2 , a3 , a4 , a5 )
are possible for A5 .
5.3.17. Hint. Calculate (123)(12) and (12)(123).
5.3.25. Hint. Consider the cases (ab)(bc) and (ab)(cd).
5.3.30. Hint. For (a), show that στ σ −1 (σ(ai )) = σ(ai+1 ).
6 · Cosets and Lagrange’s Theorem
6.4 · Exercises
6.4.1. Hint. The order of g and the order h must both divide the order
of G.
6.4.2. Hint. The possible orders must divide 60.
337
6.4.3. Hint. This is true for every proper nontrivial subgroup.
6.4.4. Hint. False.
6.4.5. Hint. (a) ⟨8⟩, 1 + ⟨8⟩, 2 + ⟨8⟩, 3 + ⟨8⟩, 4 + ⟨8⟩, 5 + ⟨8⟩, 6 + ⟨8⟩,
and 7 + ⟨8⟩; (c) 3Z, 1 + 3Z, and 2 + 3Z.
6.4.7. Hint. 4ϕ(15) ≡ 48 ≡ 1 (mod 15).
6.4.12. Hint. Let g1 ∈ gH. Show that g1 ∈ Hg and thus gH ⊂ Hg.
6.4.19. Hint. Show that g(H ∩ K) = gH ∩ gK.
6.4.22. Hint. If gcd(m, n) = 1, then ϕ(mn) = ϕ(m)ϕ(n) (Exer-
cise 2.3.26, p. 26 in Chapter 2, p. 17).
7 · Introduction to Cryptography
7.3 · Exercises
7.3.1. Hint. LAORYHAPDWK
7.3.3. Hint. Hint: V = E, E = X (also used for spaces and punctuation),
K = R.
7.3.4. Hint. 26! − 1
7.3.7. Hint. (a) 2791; (c) 11213525032442.
7.3.9. Hint. (a) 31 (c) 14.
7.3.10. Hint. (a) n = 11 · 41; (c) n = 8779 · 4327.
8 · Algebraic Coding Theory
8.5 · Exercises
8.5.2. Hint. This cannot be a group code since (0000) ∈
/ C.
8.5.3. Hint. (a) 2; (c) 2.
8.5.4. Hint. (a) 3; (c) 4.
8.5.6. Hint. (a) dmin = 2; (c) dmin = 1.
8.5.7. 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
8.5.9. Hint. Multiple errors occur in one of the received words.
338APPENDIX B HINTS AND ANSWERS TO SELECTED EXERCISES
8.5.11. Hint. (a) A canonical parity-check matrix with standard gen-
erator 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
8.5.12. Hint. (a) All possible syndromes occur.
8.5.15. 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.
8.5.19. 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.
8.5.23. Hint. For 20 information positions, at least 6 check bits are
needed to ensure an error-correcting code.
9 · Isomorphisms
9.3 · Exercises
9.3.1. Hint. Every infinite cyclic group is isomorphic to Z by Theo-
rem 9.7, p. 119.
9.3.2. Hint. Define ϕ : C∗ → GL2 (R) by
( )
a b
ϕ(a + bi) = .
−b a
9.3.3. Hint. False.
9.3.6. Hint. Define a map from Zn into the nth roots of unity by
k 7→ cis(2kπ/n).
9.3.8. Hint. Assume that Q is cyclic and try to find a generator.
9.3.11. Hint. There are two nonabelian and three abelian groups that
are not isomorphic.
9.3.16. Hint. (a) 12; (c) 5.
9.3.19. Hint. Draw the picture.
9.3.20. Hint. True.
9.3.25. Hint. True.
9.3.27. Hint. Let a be a generator for G. If ϕ : G → H is an isomor-
phism, show that ϕ(a) is a generator for H.
9.3.38. Hint. Any automorphism of Z6 must send 1 to another gener-
ator of Z6 .
339
9.3.45. 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
10.3 · Exercises
10.3.1. Hint. (a)
A4 (12)A4
A4 A4 (12)A4
(12)A4 (12)A4 A4
(c) D4 is not normal in S4 .
10.3.8. Hint. If a ∈ G is a generator for G, then aH is a generator for
G/H.
10.3.11. 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).
10.3.12. Hint. Suppose that ⟨g⟩ is normal in G and let y be an ar-
bitrary 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 ).
10.3.14. 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
−1
ghg is a product of elements of the same type. However, ghg −1 =
gh1 · · · hn g −1 = (gh1 g −1 )(gh2 g −1 ) · · · (ghn g −1 ).
11 · Homomorphisms
11.3 · Exercises
11.3.2. Hint. (a) is a homomorphism with kernel {1}; (c) is not a
homomorphism.
11.3.4. Hint. Since ϕ(m + n) = 7(m + n) = 7m + 7n = ϕ(m) + ϕ(n),
ϕ is a homomorphism.
11.3.5. 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.
11.3.9. Hint. Let a, b ∈ G. Then ϕ(a)ϕ(b) = ϕ(ab) = ϕ(ba) =
ϕ(b)ϕ(a).
11.3.17. Hint. Find a counterexample.
12 · Matrix Groups and Symmetry
12.3 · Exercises
12.3.1. Hint.
1[ ] 1[ ]
∥x + y∥2 + ∥x∥2 − ∥y∥2 = ⟨x + y, x + y⟩ − ∥x∥2 − ∥y∥2
2 2
340APPENDIX B HINTS AND ANSWERS TO SELECTED EXERCISES
1[ ]
= ∥x∥2 + 2⟨x, y⟩ + ∥y∥2 − ∥x∥2 − ∥y∥2
2
= ⟨x, y⟩.
12.3.3. Hint. (a) is in SO(2); (c) is not in O(3).
12.3.5. Hint. (a) ⟨x, y⟩ = ⟨y, x⟩.
12.3.7. Hint. Use the unimodular matrix
( )
5 2
.
2 1
12.3.10. Hint. Show that the kernel of the map det : O(n) → R∗ is
SO(n).
12.3.13. Hint. True.
12.3.17. Hint. p6m
13 · The Structure of Groups
13.3 · Exercises
13.3.1. Hint. There are three possible groups.
13.3.4. Hint. (a) {0} ⊂ ⟨6⟩ ⊂ ⟨3⟩ ⊂ Z12 ; (e) {(1)}×{0} ⊂ {(1), (123), (132)}×
{0} ⊂ S3 × {0} ⊂ S3 × ⟨2⟩ ⊂ S3 × Z4 .
13.3.7. Hint. Use the Fundamental Theorem of Finitely Generated
Abelian Groups.
13.3.12. 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 }.
13.3.16. Hint. Use the fact that Dn has a cyclic subgroup of index 2.
13.3.21. Hint. G/G′ is abelian.
14 · Group Actions
14.4 · Exercises
14.4.1. Hint. Example 14.1, p. 173: 0, R2 \ {0}. Example 14.2, p. 173:
X = {1, 2, 3, 4}.
14.4.2. Hint. (a) X(1) = {1, 2, 3}, X(12) = {3}, X(13) = {2}, X(23) =
{1}, X(123) = X(132) = ∅. G1 = {(1), (23)}, G2 = {(1), (13)}, G3 =
{(1), (12)}.
14.4.3. Hint. (a) O1 = O2 = O3 = {1, 2, 3}.
14.4.6. Hint. The conjugacy classes for S4 are
O(1) = {(1)},
O(12) = {(12), (13), (14), (23), (24), (34)},
O(12)(34) = {(12)(34), (13)(24), (14)(23)},
O(123) = {(123), (132), (124), (142), (134), (143), (234), (243)},
O(1234) = {(1234), (1243), (1324), (1342), (1423), (1432)}.
The class equation is 1 + 3 + 6 + 6 + 8 = 24.
341
14.4.8. Hint. (34 + 31 + 32 + 31 + 32 + 32 + 33 + 33 )/8 = 21.
14.4.11. Hint. The group of rigid motions of the cube can be described
by the allowable permutations 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 struc-
ture (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.
14.4.15. Hint. (1 · 26 + 3 · 24 + 4 · 23 + 2 · 22 + 2 · 21 )/12 = 13.
14.4.17. Hint. (1 · 28 + 3 · 26 + 2 · 24 )/6 = 80.
14.4.22. Hint. Use the fact that x ∈ gC(a)g −1 if and only if g −1 xg ∈
C(a).
15 · The Sylow Theorems
15.3 · Exercises
15.3.1. 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.
15.3.2. Hint. The four Sylow 3-subgroups of S4 are P1 = {(1), (123), (132)},
P2 = {(1), (124), (142)}, P3 = {(1), (134), (143)}, P4 = {(1), (234), (243)}.
15.3.5. 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.
15.3.8. Hint. Show that G has a normal Sylow p-subgroup of order p2
and a normal Sylow q-subgroup of order q 2 .
15.3.10. Hint. False.
15.3.17. Hint. If G is abelian, then G is cyclic, since |G| = 3 · 5 · 17.
Now look at Example 15.14, p. 191.
15.3.23. 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.
15.3.26. 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
16.6 · Exercises
√
16.6.1. Hint. (a) 7Z is a ring but not a field; (c) Q( 2 ) is a field; (f)
R is not a ring.
16.6.3. 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
342APPENDIX B HINTS AND ANSWERS TO SELECTED EXERCISES
16.6.4. 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.
16.6.7. Hint. Assume there is an isomorphism ϕ : C → R with ϕ(i) =
a.
√
16.6.8.
√ √ Assume there is an isomorphism ϕ : Q( 2 ) →
Hint. False.
Q( 3 ) such that ϕ( 2 ) = a.
16.6.13. Hint. (a) x ≡ 17 (mod 55); (c) x ≡ 214 (mod 2772).
16.6.16. Hint. If I ̸= {0}, show that 1 ∈ I.
16.6.18. Hint. (a) ϕ(a)ϕ(b) = ϕ(ab) = ϕ(ba) = ϕ(b)ϕ(a).
16.6.26. Hint. Let a ∈ R with a ̸= 0. Then the principal ideal gener-
ated by a is R. Thus, there exists a b ∈ R such that ab = 1.
16.6.28. Hint. Compute (a + b)2 and (−ab)2 .
16.6.34. 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.
16.6.38. Hint. Suppose that x2 = x and x ̸= 0. Since R is an integral
domain, x = 1. To find a nontrivial idempotent, look in M2 (R).
17 · Polynomials
17.4 · Exercises
17.4.2. Hint. (a) 9x2 + 2x + 5; (b) 8x4 + 7x3 + 2x2 + 7x.
17.4.3. Hint. (a) 5x3 + 6x2 − 3x + 4 = (5x2 + 2x + 1)(x − 2) + 6; (c)
4x5 − x3 + x2 + 4 = (4x2 + 4)(x3 + 3) + 4x2 + 2.
17.4.5. Hint. (a) No zeros in Z12 ; (c) 3, 4.
17.4.7. Hint. Look at (2x + 1).
17.4.8. Hint. (a) Reducible; (c) irreducible.
17.4.10. Hint. One factorization is x2 + x + 8 = (x + 2)(x + 9).
17.4.13. Hint. The integers Z do not form a field.
17.4.14. Hint. False.
17.4.16. 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 .
17.4.20. 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.
17.4.26. Hint. Find a nontrivial proper ideal in F [x].
18 · Integral Domains
18.3 · Exercises
√ √
18.3.1.√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
343
equation are a = ±1, b = 0.
18.3.2. Hint. (a) 5 = −i(1 + 2i)(2 + i); (c) 6 + 8i = −i(1 + i)2 (2 + i)2 .
18.3.4. Hint. True.
18.3.9. Hint. Let z = a + bi and w = c + di ̸= 0 be in Z[i]. Prove that
z/w ∈ Q(i).
18.3.15. Hint. Let a = ub with u a unit. Then ν(b) ≤ ν(ub) ≤ ν(a).
Similarly, ν(a) ≤ ν(b).
18.3.16. Hint. Show that 21 can be factored in two different ways.
19 · Lattices and Boolean Algebras
19.4 · Exercises
19.4.2. Hint.
30
10 15
2 5 3
1
19.4.5. Hint. False.
19.4.6. Hint. (a) (a ∨ b ∨ a′ ) ∧ a
a
b a
a′
(c) a ∨ (a ∧ b)
a b
a
19.4.8. Hint. Not equivalent.
19.4.10. Hint. (a) a′ ∧ [(a ∧ b′ ) ∨ b] = a ∧ (a ∨ b).
19.4.14. 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.
19.4.18. Hint. (a) No.
344APPENDIX B HINTS AND ANSWERS TO SELECTED EXERCISES
19.4.20. 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
20.4 · Exercises
√ √ √ √ √
20.4.3. Hint. Q( 2, 3 ) has basis {1, 2, 3, 6 } over Q.
20.4.5. Hint. The set {1, x, x2 , . . . , xn−1 } is a basis for Pn .
20.4.7. Hint. (a) Subspace of dimension 2 with basis {(1, 0, −3), (0, 1, 2)};
(d) not a subspace
20.4.10. Hint. Since 0 = α0 = α(−v + v) = α(−v) + αv, it follows
that −αv = α(−v).
20.4.12. Hint. Let v0 = 0, v1 , . . . , vn ∈ V and α0 ̸= 0, α1 , . . . , αn ∈ F .
Then α0 v0 + · · · + αn vn = 0.
20.4.15. 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.
20.4.17. 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
21.4 · Exercises
21.4.1. Hint. (a) x4 − (2/3)x2 − 62/9; (c) x4 − 2x2 + 25.
√ √ √ √ √
21.4.2. Hint. (a) {1, 2, 3, 6 }; (c) {1, i, 2, 2 i}; (e) {1, 21/6 , 21/3 , 21/2 , 22/3 , 25/6 }.
√ √
21.4.3. Hint. (a) Q( 3, 7 ).
21.4.5. Hint. Use the fact that the elements of Z2 [x]/⟨x3 +x+1⟩ are 0,
1, α, 1 + α, α2 , 1 + α2 , α + α2 , 1 + α + α2 and the fact that α3 + α + 1 = 0.
21.4.8. Hint. False.
21.4.14. 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 ).
√ √ √ √ √
21.4.22.
√ √ Hint. Since √ {1,√ 3, 7, 21√ } is √a basis for Q( 3,√ 7 )√over
Q, Q( 3, 7 ) ⊃ Q( 3+ 7 ). Since [Q( 3, 7 ) : Q] = 4, [Q(√ 3+ √7 ) :
Q] = √2 or√4. Since √ the degree
√ of the minimal polynomial of 3 + 7 is
4, Q( 3, 7 ) = Q( 3 + 7 ).
345
21.4.27. 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) ̸= 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
( ) ( ) ( )n
p(α) p(α) p(α)
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.
21.4.28. Hint. See the comments following Theorem 21.13, p. 273.
22 · Finite Fields
22.3 · Exercises
22.3.1. Hint. Make sure that you have a field extension.
22.3.4. Hint. There are eight elements in Z2 (α). Exhibit two more
zeros of x3 + x2 + 1 other than α in these eight elements.
22.3.5. Hint. Find an irreducible polynomial p(x) in Z3 [x] of degree 3
and show that Z3 [x]/⟨p(x)⟩ has 27 elements.
22.3.7. Hint. (a) x5 − 1 = (x + 1)(x4 + x3 + x2 + x + 1); (c) x9 − 1 =
(x + 1)(x2 + x + 1)(x6 + x3 + 1).
22.3.8. Hint. True.
22.3.11. Hint. (a) Use the fact that x7 − 1 = (x + 1)(x3 + x + 1)(x3 +
x2 + 1).
22.3.12. Hint. False.
22.3.17. Hint. If p(x) ∈ F [x], then p(x) ∈ E[x].
22.3.18. 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 ).
22.3.24. Wilson’s Theorem. Hint. Factor xp−1 − 1 over Zp .
23 · Galois Theory
23.4 · Exercises
23.4.1. Hint. (a) Z2 ; (c) Z2 × Z2 × Z2 .
23.4.2. 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 .
23.4.3. 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
6
σ, where σ36 (α) = α3 = α729 for α ∈ GF(729).
23.4.4. Hint. (a) S5 ; (c) S3 ; (g) see Example 23.10, p. 308.
23.4.5. Hint. (a) Q(i)
23.4.7. 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
346APPENDIX B HINTS AND ANSWERS TO SELECTED EXERCISES
G(E/F ) is a subgroup of S3 whose order is divisible by 3, conclude that
this group must be isomorphic to Z3 or S3 .
23.4.9. Hint. G is a subgroup of Sn .
23.4.16. Hint. True.
23.4.20. Hint.
(a) Clearly ω, ω 2 , . . . , ω p−1 are distinct since ω ̸= 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 5
A\B difference between sets A and B 5
A×B Cartesian product of sets A and B 6
An A × · · · × A (n times) 7
id identity mapping 10
f −1 inverse of the function f 10
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 29
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
C∗ the multiplicative group of complex num- 36
bers
(Continued on next page)
347
348 APPENDIX C NOTATION
Symbol Description Page
|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
⟨a⟩ cyclic group generated by a 47
|a| the order of an element a 48
cis θ cos θ + i sin θ 51
T the circle group 53
Sn the symmetric group on n letters 61
(a1 , a2 , . . . , ak ) cycle of length k 63
An the alternating group on n letters 67
Dn the dihedral group 68
[G : H] index of a subgroup H in a group G 76
LH the set of left cosets of a subgroup H in a 77
group G
RH the set of right cosets of a subgroup H in a 77
group G
a∤b a does not divide b 79
d(x, y) Hamming distance between x and y 98
dmin the minimum distance of a code 98
w(x) the weight of x 98
Mm×n (Z2 ) the set of m × n matrices with entries in Z2 102
Null(H) null space of a matrix H 102
δij Kronecker delta 106
G∼ =H G is isomorphic to a group H 117
Aut(G) automorphism group of a group G 127
−1
ig ig (x) = gxg 127
Inn(G) inner automorphism group of a group G 127
ρg right regular representation 127
G/N factor group of G mod N 130
G′ commutator subgroup of G 135
ker ϕ kernel of ϕ 138
(aij ) matrix 146
O(n) orthogonal group 148
∥x∥ length of a vector x 148
SO(n) special orthogonal group 151
E(n) Euclidean group 151
Ox orbit of x 174
Xg fixed point set of g 175
Gx isotropy subgroup of x 175
N (H) normalizer of s subgroup H 188
H the ring of quaternions 199
Z[i] the Gaussian integers 201
char R characteristic of a ring R 202
Z(p) ring of integers localized at p 214
deg f (x) degree of a polynomial 218
R[x] ring of polynomials over a ring R 218
R[x1 , x2 , . . . , xn ] ring of polynomials in n indeterminants 220
(Continued on next page)
349
Symbol Description Page
ϕα evaluation homomorphism at α 220
Q(x) field of rational functions over Q 236
ν(a) Euclidean valuation of a 240
F (x) field of rational functions in x 244
F (x1 , . . . , xn ) field of rational functions in x1 , . . . , xn 244
a⪯b a is less than b 247
a∨b join of a and b 249
a∧b meet of a and b 249
I largest element in a lattice 251
O smallest element in a lattice 251
a′ complement of a in a lattice 251
dim V dimension of a vector space V 265
U ⊕V direct sum of vector spaces U and V 267
Hom(V, W ) set of all linear transformations from U into 268
V
V∗ dual of a vector space V 268
F (α1 , . . . , αn ) smallest field containing F and α1 , . . . , αn 272
[E : F ] dimension of a field extension of E over F 274
GF(pn ) Galois field of order pn 291
F∗ multiplicative group of a field F 292
G(E/F ) Galois group of E over F 306
F{σi } field fixed by the automorphism σi 310
FG field fixed by the automorphism group G 310
∆2 discriminant of a polynomial 322
350 APPENDIX C NOTATION
Index
G-equivalent, 174 Cardano, Gerolamo, 228
G-set, 173 Carmichael numbers, 91
nth root of unity, 53, 316 Cauchy’s Theorem, 187
rsa cryptosystem, 86 Cauchy, Augustin-Louis, 67
Cayley table, 34
Abel, Niels Henrik, 315 Cayley’s Theorem, 120
Abelian group, 34 Cayley, Arthur, 121
Adleman, L., 86 Centralizer
Algebraic closure, 277 of a subgroup, 176
Algebraic extension, 271 Characteristic of a ring, 202
Algebraic number, 272 Chinese Remainder Theorem
Algorithm for integers, 209
division, 221 Cipher, 83
Euclidean, 22 Ciphertext, 83
Ascending chain condition, 238 Circuit
Associate elements, 236 parallel, 255
Atom, 253 series, 255
Automorphism series-parallel, 256
inner, 143 Class equation, 176
Code
Basis of a lattice, 155 bch, 299
Bieberbach, L., 158 cyclic, 293
Binary operation, 33 group, 101
Binary symmetric channel, 97 linear, 103
Boole, George, 257 minimum distance of, 98
Boolean algebra polynomial, 294
atom in a, 253 Commutative diagrams, 140
definition of, 251 Commutative rings, 197
finite, 253 Composite integer, 22
isomorphism, 253 Composition series, 168
Boolean function, 181, 260 Congruence modulo n, 13
Burnside’s Counting Theorem, Conjugacy classes, 176
178 Conjugate elements, 307
Burnside, William, 38, 134, 183 Conjugate, complex, 50
Conjugation, 174
Cancellation law Constructible number, 281
for groups, 37 Correspondence Theorem
for integral domains, 201 for groups, 141
351
352
for rings, 205 inverse, 34
Coset irreducible, 236
leader, 110 order of, 48
left, 75 prime, 236
representative, 75 primitive, 309
right, 75 transcendental, 271
Coset decoding, 110 Equivalence class, 12
Cryptanalysis, 84 Equivalence relation, 11
Cryptosystem Euclidean algorithm, 22
rsa, 86 Euclidean domain, 240
affine, 85 Euclidean group, 151
definition of, 83 Euclidean inner product, 148
monoalphabetic, 84 Euclidean valuation, 240
polyalphabetic, 85 Euler ϕ-function, 79
private key, 84 Euler, Leonhard, 79, 284
public key, 83 Extension
single key, 84 algebraic, 271
Cycle field, 269
definition of, 63 finite, 274
disjoint, 63 normal, 311
radical, 316
De Morgan’s laws separable, 290, 308
for Boolean algebras, 253 simple, 272
for sets, 6 External direct product, 122
De Morgan, Augustus, 257
Decoding table, 111 Faltings, Gerd, 285
Deligne, Pierre, 285 Feit, W., 134, 183
DeMoivre’s Theorem, 52 Fermat’s
Derivative, 290 factorizationalgorithm,
Determinant, Vandermonde, 90
297 Fermat’s Little Theorem, 79
Dickson, L. E., 134 Fermat, Pierre de, 79, 284
Diffie, W., 86 Ferrari, Ludovico, 228
Direct product of groups Ferro, Scipione del, 227
external, 122 Field, 198
internal, 123 algebraically closed, 277
Discriminant base, 269
of the cubic equation, 232 extension, 269
of the quadratic equation, fixed, 310
231 Galois, 291
Division algorithm of fractions, 235
for integers, 20 of quotients, 235
for polynomials, 221 splitting, 278
Division ring, 198 Finitely generated group, 163
Domain Fior, Antonio, 228
Euclidean, 240 First Isomorphism Theorem
principal ideal, 237 for groups, 139
unique factorization, 237 for rings, 205
Doubling the cube, 283 Fixed point set, 175
Freshman’s Dream, 290
Eisenstein’s Criterion, 226 Function
Element bijective, 8
associate, 236 Boolean, 181, 260
identity, 34 composition of, 8
353
definition of, 7 infinite, 36
domain of, 7 isomorphic, 117
identity, 10 isomorphism of, 117
injective, 8 nonabelian, 34
invertible, 10 noncommutative, 34
one-to-one, 8 of units, 35
onto, 7 order of, 36
range of, 7 orthogonal, 148
surjective, 7 permutation, 62
switching, 181, 260 point, 156
Fundamental Theorem quaternion, 36
of Algebra, 277, 320 quotient, 130
of Arithmetic, 23 simple, 131, 134
of Finite Abelian Groups, solvable, 170
164 space, 156
Fundamental Theorem of special linear, 39, 147
Galois Theory, 312 special orthogonal, 151
symmetric, 61
Galois field, 291 symmetry, 153
Galois group, 306 Groupp-group, 164
Galois, Évariste, 38, 315 Gödel, Kurt, 257
Gauss’s Lemma, 241
Gauss, Karl Friedrich, 243 Hamming distance, 98
Gaussian integers, 201 Hamming, R., 100
Generator of a cyclic subgroup, Hellman, M., 86
48 Hilbert, David, 158, 207, 257,
Generators for a group, 163 285
Glide reflection, 152 Homomorphic image, 137
Gorenstein, Daniel, 134 Homomorphism
Greatest common divisor canonical, 139, 205
of two integers, 20 evaluation, 203, 220
of two polynomials, 222 kernel of a group, 138
Greatest lower bound, 248 kernel of a ring, 202
Greiss, R., 134 natural, 139, 205
Grothendieck, Alexander, 285 of groups, 137
Group ring, 202
p-group, 187
abelian, 34 Ideal
action, 173 definition of, 203
alternating, 67 maximal, 206
center of, 176 one-sided, 204
circle, 53 prime, 206
commutative, 34 principal, 204
cyclic, 48 trivial, 203
definition of, 33 two-sided, 204
dihedral, 68 Indeterminate, 217
Euclidean, 151 Index of a subgroup, 76
factor, 130 Induction
finite, 36 first principle of, 18
finitely generated, 163 second principle of, 19
Galois, 306 Infimum, 248
general linear, 35, 147 Inner product, 102
generators of, 163 Integral domain, 198
homomorphism of, 137 Internal direct product, 123
354
International standard book inner product-preserving,
number, 45 149
Irreducible element, 236 invertible, 146
Irreducible polynomial, 223 length-preserving, 149
Isometry, 152 nonsingular, 147
Isomorphism null space of, 102
of Boolean algebras, 253 orthogonal, 148
of groups, 117 parity-check, 103
ring, 202 similar, 12
unimodular, 155
Join, 249 Matrix, Vandermonde, 297
Jordan, C., 134 Maximal ideal, 206
Jordan-Hölder Theorem, 169 Maximum-likelihood decoding,
96
Kernel Meet, 249
of a group homomorphism, Minimal generator polynomial,
138 295
of a ring homomorphism, Minimal polynomial, 273
202 Minkowski, Hermann, 285
Key Monic polynomial, 218
definition of, 83 Mordell-Weil conjecture, 285
private, 84 Multiplicity of a root, 308
public, 83
single, 84 Noether, A. Emmy, 207
Klein, Felix, 38, 145, 207 Noether, Max, 207
Kronecker delta, 106, 149 Normal extension, 311
Kronecker, Leopold, 284 Normal series of a group, 168
Kummer, Ernst, 284 Normal subgroup, 129
Normalizer, 189
Lagrange’s Theorem, 77 Null space
Lagrange, Joseph-Louis, 37, 67, of a matrix, 102
79
Laplace, Pierre-Simon, 67 Odd Order Theorem, 193
Lattice Orbit, 174
completed, 251 Orthogonal group, 148
definition of, 249 Orthogonal matrix, 148
distributive, 251 Orthonormal set, 149
Lattice of points, 155
Lattices, Principle of Duality Partial order, 247
for, 249 Partially ordered set, 247
Least upper bound, 248 Partitions, 12
Left regular representation, 121 Permutation
Lie, Sophus, 38, 190 cycle structure of, 80
Linear combination, 263 definition of, 9, 61
Linear dependence, 264 even, 66
Linear independence, 264 odd, 66
Linear map, 145 Permutation group, 62
Linear transformation Plaintext, 83
definition of, 9, 145 Polynomial
Lower bound, 248 code, 294
content of, 241
Mapping, see Function definition of, 217
Matrix degree of, 218
distance-preserving, 149 error, 302
generator, 104 error-locator, 302
355
greatest common divisor for groups, 140
of, 222 for rings, 205
in n indeterminates, 220 Shamir, A., 86
irreducible, 223 Shannon, C., 100
leading coefficient of, 218 Simple extension, 272
minimal, 273 Simple group, 131
minimal generator, 295 Simple root, 308
monic, 218 Solvability by radicals, 316
primitive, 241 Spanning set, 263
root of, 222 Splitting field, 278
separable, 308 Squaring the circle is
zero of, 222 impossible, 284
Polynomial separable, 290 Standard decoding, 110
Poset Subgroup
definition of, 247 p-subgroup, 187
largest element in, 250 centralizer, 176
smallest element in, 251 commutator, 191
Power set, 247 cyclic, 48
Prime element, 236 definition of, 38
Prime ideal, 206 index of, 76
Prime integer, 22 isotropy, 175
Primitive nth root of unity, 53, normal, 129
316 normalizer of, 189
Primitive element, 309 proper, 38
Primitive Element Theorem, stabilizer, 175
309 Sylowp-subgroup, 188
Primitive polynomial, 241 translation, 156
Principal ideal, 204 trivial, 38
Principal ideal domain (pid), Subnormal series of a group,
237 168
Principal series, 169 Subring, 200
Pseudoprime, 91 Supremum, 248
Switch
Quaternions, 36, 199
closed, 255
Resolvent cubic equation, 232 definition of, 255
Rigid motion, 32, 152 open, 255
Ring Switching function, 181, 260
characteristic of, 202 Sylow p-subgroup, 188
commutative, 197 Sylow, Ludvig, 190
definition of, 197 Syndrome of a code, 109, 302
division, 198
factor, 205 Tartaglia, 228
homomorphism, 202 Third Isomorphism Theorem
isomorphism, 202 for groups, 141
Noetherian, 238 for rings, 205
quotient, 205 Thompson, J., 134, 183
with identity, 197 Transcendental element, 271
with unity, 197 Transcendental number, 272
Rivest, R., 86 Transposition, 65
Ruffini, P., 315 Trisection of an angle, 284
Russell, Bertrand, 257
Unique factorization domain
Scalar product, 261 (ufd), 237
Second Isomorphism Theorem Unit, 198, 236
356
Universal Product Code, 44 Weight of a codeword, 98
Upper bound, 248 Weil, André, 285
Well-defined map, 7
Vandermonde determinant, 297 Well-ordered set, 19
Vandermonde matrix, 297 Whitehead, Alfred North, 257
Vector space
basis of, 265 Zero
definition of, 261 multiplicity of, 308
dimension of, 265 of a polynomial, 222
subspace of, 262 Zero divisor, 198
Colophon
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