Authors Thomas W. Judson
License GFDL-1.2-no-invariants-or-later
Abstract Algebra
Theory and Applications
Thomas W. Judson
Stephen F. Austin State University
August 27, 2010
ii
Copyright 1997 by Thomas W. Judson.
Permission is granted to copy, distribute and/or modify this document un-
der the terms of the GNU Free Documentation License, Version 1.2 or any
later version published by the Free Software Foundation; with no Invari-
ant 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”.
A current version can always be found via abstract.pugetsound.edu.
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 al-
gebra and discrete mathematics have become increasingly important, and
many science, engineering, and computer science students are now electing
to minor in mathematics. Though theory still occupies a central role in the
subject of abstract algebra and no student should go through such a course
without a good notion of what a proof is, the importance of applications
such as coding theory and cryptography has grown significantly.
Until recently most abstract algebra texts included few if any 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 environ-
ment 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 presen-
tation 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
iii
iv PREFACE
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 the-
oretical results could be assumed or omitted. A chapter dependency chart
appears below. (A broken line indicates a partial dependency.)
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 alge-
bra, students who have had other higher-level courses in mathematics will
generally be more prepared than those who have not, because they will pos-
sess a bit more mathematical sophistication. Occasionally, we shall assume
some basic linear algebra; that is, we shall take for granted an elemen-
tary 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.
PREFACE v
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 pro-
gramming. All of these exercises and projects are more substantial in nature
and allow the exploration of new results and theory.
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 for their guidance through-
out this project. It has been a pleasure to work with them.
Thomas W. Judson
Contents
Preface iii
1 Preliminaries 1
1.1 A Short Note on Proofs . . . . . . . . . . . . . . . . . . . . . 1
1.2 Sets and Equivalence Relations . . . . . . . . . . . . . . . . . 4
2 The Integers 22
2.1 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . 22
2.2 The Division Algorithm . . . . . . . . . . . . . . . . . . . . . 26
3 Groups 35
3.1 The Integers mod n and Symmetries . . . . . . . . . . . . . . 35
3.2 Definitions and Examples . . . . . . . . . . . . . . . . . . . . 40
3.3 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 Cyclic Groups 57
4.1 Cyclic Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.2 The Group C∗ . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4.3 The Method of Repeated Squares . . . . . . . . . . . . . . . . 66
5 Permutation Groups 74
5.1 Definitions and Notation . . . . . . . . . . . . . . . . . . . . . 75
5.2 The Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . 83
6 Cosets and Lagrange’s Theorem 92
6.1 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
6.2 Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 95
6.3 Fermat’s and Euler’s Theorems . . . . . . . . . . . . . . . . . 97
vi
CONTENTS vii
7 Introduction to Cryptography 100
7.1 Private Key Cryptography . . . . . . . . . . . . . . . . . . . . 101
7.2 Public Key Cryptography . . . . . . . . . . . . . . . . . . . . 104
8 Algebraic Coding Theory 111
8.1 Error-Detecting and Correcting Codes . . . . . . . . . . . . . 111
8.2 Linear Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
8.3 Parity-Check and Generator Matrices . . . . . . . . . . . . . 124
8.4 Efficient Decoding . . . . . . . . . . . . . . . . . . . . . . . . 131
9 Isomorphisms 141
9.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . 141
9.2 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . 146
10 Normal Subgroups and Factor Groups 155
10.1 Factor Groups and Normal Subgroups . . . . . . . . . . . . . 155
10.2 Simplicity of An . . . . . . . . . . . . . . . . . . . . . . . . . 158
11 Homomorphisms 165
11.1 Group Homomorphisms . . . . . . . . . . . . . . . . . . . . . 165
11.2 The Isomorphism Theorems . . . . . . . . . . . . . . . . . . . 168
12 Matrix Groups and Symmetry 175
12.1 Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 175
12.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
13 The Structure of Groups 196
13.1 Finite Abelian Groups . . . . . . . . . . . . . . . . . . . . . . 196
13.2 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . 201
14 Group Actions 209
14.1 Groups Acting on Sets . . . . . . . . . . . . . . . . . . . . . . 209
14.2 The Class Equation . . . . . . . . . . . . . . . . . . . . . . . 213
14.3 Burnside’s Counting Theorem . . . . . . . . . . . . . . . . . . 215
15 The Sylow Theorems 227
15.1 The Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . . 227
15.2 Examples and Applications . . . . . . . . . . . . . . . . . . . 231
viii CONTENTS
16 Rings 239
16.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
16.2 Integral Domains and Fields . . . . . . . . . . . . . . . . . . . 244
16.3 Ring Homomorphisms and Ideals . . . . . . . . . . . . . . . . 246
16.4 Maximal and Prime Ideals . . . . . . . . . . . . . . . . . . . . 250
16.5 An Application to Software Design . . . . . . . . . . . . . . . 253
17 Polynomials 263
17.1 Polynomial Rings . . . . . . . . . . . . . . . . . . . . . . . . . 264
17.2 The Division Algorithm . . . . . . . . . . . . . . . . . . . . . 268
17.3 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . 272
18 Integral Domains 283
18.1 Fields of Fractions . . . . . . . . . . . . . . . . . . . . . . . . 283
18.2 Factorization in Integral Domains . . . . . . . . . . . . . . . . 287
19 Lattices and Boolean Algebras 301
19.1 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
19.2 Boolean Algebras . . . . . . . . . . . . . . . . . . . . . . . . . 306
19.3 The Algebra of Electrical Circuits . . . . . . . . . . . . . . . . 312
20 Vector Spaces 319
20.1 Definitions and Examples . . . . . . . . . . . . . . . . . . . . 319
20.2 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
20.3 Linear Independence . . . . . . . . . . . . . . . . . . . . . . . 322
21 Fields 329
21.1 Extension Fields . . . . . . . . . . . . . . . . . . . . . . . . . 329
21.2 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 340
21.3 Geometric Constructions . . . . . . . . . . . . . . . . . . . . . 343
22 Finite Fields 353
22.1 Structure of a Finite Field . . . . . . . . . . . . . . . . . . . . 353
22.2 Polynomial Codes . . . . . . . . . . . . . . . . . . . . . . . . 358
23 Galois Theory 371
23.1 Field Automorphisms . . . . . . . . . . . . . . . . . . . . . . 371
23.2 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . 377
23.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
Hints and Solutions 395
CONTENTS ix
GNU Free Documentation License 410
Notation 418
Index 422
x CONTENTS
1
Preliminaries
A certain amount of mathematical maturity is necessary to find and study
applications of abstract algebra. A basic knowledge of set theory, mathe-
matical induction, equivalence relations, and matrices is a must. Even more
important is the ability to read and understand mathematical proofs. In
this chapter we will outline the background needed for a course in abstract
algebra.
1.1 A Short Note on Proofs
Abstract mathematics is different from other sciences. In laboratory sciences
such as chemistry and physics, scientists perform experiments to discover
new principles and verify theories. Although mathematics is often motivated
by physical experimentation or by computer simulations, it is made rigorous
through the use of logical arguments. In studying abstract mathematics, we
take what is called an axiomatic approach; that is, we take a collection of
objects S and assume some rules about their structure. These rules are called
axioms. Using the axioms for S, we wish to derive other information about
S by using logical 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.
1
2 CHAPTER 1 PRELIMINARIES
• 2x = 6 exactly when x = 4.
• If ax2 + bx + c = 0 and a 6= 0, then
√
−b ± b2 − 4ac
x= .
2a
• x3 − 4x2 + 5x − 6.
All but the first and last examples are statements, and must be either true
or false.
A mathematical proof is nothing more than a convincing argument
about the accuracy of a statement. Such an argument should contain enough
detail to convince the audience; for instance, we can see that the statement
“2x = 6 exactly when x = 4” is false by evaluating 2 · 4 and noting that
6 6= 8, an argument that would satisfy anyone. Of course, audiences may
vary widely: proofs can be addressed to another student, to a professor,
or to the reader of a text. If more detail than needed is presented in the
proof, then the explanation will be either long-winded or poorly written. If
too much detail is omitted, then the proof may not be convincing. Again
it is important to keep the audience in mind. High school students require
much more detail than do graduate students. A good rule of thumb for an
argument in an introductory abstract algebra course is that it should be
written to convince one’s peers, whether those peers be other students or
other readers of the text.
Let us examine different types of statements. A statement could be as
simple as “10/5 = 2”; however, mathematicians are usually interested in
more complex statements such as “If p, then q,” where p and q are both
statements. If certain statements are known or assumed to be true, we
wish to know what we can say about other statements. Here p is called
the hypothesis and q is known as the conclusion. Consider the following
statement: If ax2 + bx + c = 0 and a 6= 0, then
√
−b ± b2 − 4ac
x= .
2a
The hypothesis is ax2 + bx + c = 0 and a 6= 0; the conclusion is
√
−b ± b2 − 4ac
x= .
2a
Notice that the statement says nothing about whether or not the hypothesis
is true. However, if this entire statement is true and we can show that
1.1 A SHORT NOTE ON PROOFS 3
ax2 + bx + c = 0 with a 6= 0 is true, then the conclusion must be true. A
proof of this statement might simply be a series of equations:
ax2 + bx + c = 0
b c
x2 + x = −
a a
2 2
b b b c
x2 + x + = −
a 2a 2a a
2 2
b b − 4ac
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 propo-
sition. 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 supporting propositions, which
are called lemmas, and use the results of these propositions to prove the
main result. If we can prove a proposition or a theorem, we will often,
with very little effort, be able to derive other related propositions called
corollaries.
Some Cautions and Suggestions
There are several different strategies for proving propositions. In addition
to using different methods of proof, students often make some common mis-
takes 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.
4 CHAPTER 1 PRELIMINARIES
• Never assume any hypothesis that is not explicitly stated in the theo-
rem. 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. Prov-
ing 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
proving theorems. Theorems are tools that make new and productive ap-
plications of mathematics possible. We use examples to give insight into
existing theorems and to foster intuitions as to what new theorems might
be true. Applications, examples, and proofs are tightly interconnected—
much more so than they may seem at first appearance.
1.2 Sets and Equivalence Relations
Set Theory
A set is a well-defined collection of objects; that is, it is defined in such
a manner that we can determine for any given object x whether or not x
belongs to the set. The objects that belong to a set are called its elements
or members. We will denote sets by capital letters, such as A or X; if a is
an element of the set A, we write a ∈ A.
A set is usually specified either by listing all of its elements inside a
pair of braces or by stating the property that determines whether or not an
object x belongs to the set. We might write
X = {x1 , x2 , . . . , xn }
for a set containing elements x1 , x2 , . . . , xn or
X = {x : x satisfies P}
1.2 SETS AND EQUIVALENCE RELATIONS 5
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 following:
N = {n : n is a natural number} = {1, 2, 3, . . .};
Z = {n : n is an integer} = {. . . , −1, 0, 1, 2, . . .};
Q = {r : r is a rational number} = {p/q : p, q ∈ Z where q 6= 0};
R = {x : x is a real number};
C = {z : z is a complex number}.
We find various relations between sets and can perform operations on
sets. A set A is a subset of B, written A ⊂ B or B ⊃ A, if every element
of A is also an element of B. For example,
{4, 5, 8} ⊂ {2, 3, 4, 5, 6, 7, 8, 9}
and
N ⊂ Z ⊂ Q ⊂ R ⊂ C.
Trivially, every set is a subset of itself. A set B is a proper subset of a
set A if B ⊂ A but B 6= A. If A is not a subset of B, we write A 6⊂ B; for
example, {4, 7, 9} 6⊂ {2, 4, 5, 8, 9}. Two sets are equal, written A = B, if we
can show that A ⊂ B and B ⊂ A.
It is convenient to have a set with no elements in it. This set is called
the empty set and is denoted by ∅. Note that the empty set is a subset of
every set.
To construct new sets out of old sets, we can perform certain operations:
the union A ∪ B of two sets A and B is defined as
A ∪ B = {x : x ∈ A or x ∈ B};
the intersection of A and B is defined by
A ∩ B = {x : x ∈ A and x ∈ B}.
If A = {1, 3, 5} and B = {1, 2, 3, 9}, then
A ∪ B = {1, 2, 3, 5, 9} and A ∩ B = {1, 3}.
6 CHAPTER 1 PRELIMINARIES
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 collection of 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 A0 ,
to be the set
A0 = {x : x ∈ U and x ∈ / A}.
We define the difference of two sets A and B to be
A \ B = A ∩ B 0 = {x : x ∈ A and x ∈
/ B}.
Example 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}
A0 = {x ∈ R : x ≤ 0 or x > 3}.
Proposition 1.1 Let A, B, and C be sets. Then
1. A ∪ A = A, A ∩ A = A, and A \ A = ∅;
2. A ∪ ∅ = A and A ∩ ∅ = ∅;
1.2 SETS AND EQUIVALENCE RELATIONS 7
3. A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C;
4. A ∪ B = B ∪ A and A ∩ B = B ∩ A;
5. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C);
6. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
Proof. We will prove (1) and (3) and leave the remaining results to be
proven in the exercises.
(1) Observe that
A ∪ A = {x : x ∈ A or x ∈ A}
= {x : x ∈ A}
=A
and
A ∩ A = {x : x ∈ A and x ∈ A}
= {x : x ∈ A}
= A.
Also, A \ A = A ∩ A0 = ∅.
(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 1.2 (De Morgan’s Laws) Let A and B be sets. Then
1. (A ∪ B)0 = A0 ∩ B 0 ;
2. (A ∩ B)0 = A0 ∪ B 0 .
Proof. (1) We must show that (A ∪ B)0 ⊂ A0 ∩ B 0 and (A ∪ B)0 ⊃ A0 ∩ B 0 .
Let x ∈ (A ∪ B)0 . 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 ∈ A0
and x ∈ B 0 . Therefore, x ∈ A0 ∩ B 0 and we have (A ∪ B)0 ⊂ A0 ∩ B 0 .
8 CHAPTER 1 PRELIMINARIES
To show the reverse inclusion, suppose that x ∈ A0 ∩ B 0 . Then x ∈ A0
and x ∈ B 0 , and so x ∈/ A and x ∈ / A ∪ B and so x ∈ (A ∪ B)0 .
/ B. Thus x ∈
Hence, (A ∪ B) ⊃ A ∩ B and so (A ∪ B) = A0 ∩ B 0 .
0 0 0 0
The proof of (2) is left as an exercise.
Example 2. 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 0 ) ∩ (B ∩ A0 )
= A ∩ A0 ∩ B ∩ B 0
= ∅.
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 3. If A = {x, y}, B = {1, 2, 3}, and C = ∅, then A × B is the set
{(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}
and
A × C = ∅.
We define the Cartesian product of n sets to be
A1 × · · · × An = {(a1 , . . . , an ) : ai ∈ Ai for i = 1, . . . , n}.
If A = A1 = A2 = · · · = An , we often write An for A × · · · × A (where A
would be written n times). For example, the set R3 consists of all of 3-tuples
of real numbers.
Subsets of A × B are called relations. We will define a mapping or
function f ⊂ A × B from a set A to a set B to be the special type of
1.2 SETS AND EQUIVALENCE RELATIONS 9
relation in which for each element a ∈ A there is a unique element b ∈ B
such that (a, b) ∈ f ; another way of saying this is that for every element in
f
A, f assigns a unique element in B. We usually write f : A → B or A → B.
Instead of writing down ordered pairs (a, b) ∈ A × B, we write f (a) = b or
f : a 7→ b. The set A is called the domain of f and
f (A) = {f (a) : a ∈ A} ⊂ B
is called the range or image of f . We can think of the elements in the
function’s domain as input values and the elements in the function’s range
as output values.
A B
f
1 a
2 b
3 c
A g B
1 a
2 b
3 c
Figure 1.1. Mappings
Example 4. Suppose A = {1, 2, 3} and B = {a, b, c}. In Figure 1.1 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.
Given a function f : A → B, it is often possible to write a list describing
what the function does to each specific element in the domain. However,
not all functions can be described in this manner. For example, the function
f : R → R that sends each real number to its cube is a mapping that must
be described by writing f (x) = x3 or f : x 7→ x3 .
10 CHAPTER 1 PRELIMINARIES
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. A map is one-to-one or injective
if a1 6= a2 implies f (a1 ) 6= f (a2 ). Equivalently, a function is one-to-one if
f (a1 ) = f (a2 ) implies a1 = a2 . A map that is both one-to-one and onto is
called bijective.
Example 5. 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 1.2. Composition of maps
1.2 SETS AND EQUIVALENCE RELATIONS 11
Example 6. Consider the functions f : A → B and g : B → C that are
defined in Figure 1.2(a). The composition of these functions, g ◦ f : A → C,
is defined in Figure 1.2(b).
Example 7. Let f (x) = x2 and g(x) = 2x + 5. Then
(f ◦ g)(x) = f (g(x)) = (2x + 5)2 = 4x2 + 20x + 25
and
(g ◦ f )(x) = g(f (x)) = 2x2 + 5.
In general, order makes a difference; that is, in most cases f ◦ g 6= g ◦ f .
Example 8. Sometimes it is the case that f ◦ g = g ◦ f . Let f (x) = x3 and
√
g(x) = 3 x. Then
√ √
(f ◦ g)(x) = f (g(x)) = f ( 3 x ) = ( 3 x )3 = x
and √
3
(g ◦ f )(x) = g(f (x)) = g(x3 ) = x3 = x.
Example 9. 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 10. Suppose that S = {1, 2, 3}. Define a map π : S → S by
π(1) = 2, π(2) = 1, π(3) = 3.
12 CHAPTER 1 PRELIMINARIES
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 per-
mutation of S.
Theorem 1.3 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 a b ∈ B such that g(b) = c. Similarly, there is an
a ∈ A such that f (a) = b. Accordingly,
(g ◦ f )(a) = g(f (a)) = g(b) = c.
If S is any set, we will use idS or id to denote the identity mapping
from S to itself. Define this map by id(s) = s for all s ∈ S. A map g : B → A
1.2 SETS AND EQUIVALENCE RELATIONS 13
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 11. The function f (x) = x3 has inverse f −1 (x) = 3
x by Exam-
ple 8.
Example 12. 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 13. Suppose that
3 1
A= .
5 2
Then A defines a map from R2 to R2 by
TA (x, y) = (3x + y, 5x + 2y).
We can find an inverse map of TA by simply inverting the matrix A; that is,
TA−1 = TA−1 . In this example,
−1 2 −1
A = ;
−5 3
hence, the inverse map is given by
TA−1 (x, y) = (2x − y, −5x + 3y).
It is easy to check that
TA−1 ◦ TA (x, y) = TA ◦ TA−1 (x, y) = (x, y).
14 CHAPTER 1 PRELIMINARIES
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) = T ◦ TB−1 (x, y) = (3ax + 3by, 0)
for all x and y. Clearly this is impossible because y might not be 0.
Example 14. 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 1.4 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).
Now assume the converse; that is, let f be bijective. 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 .
1.2 SETS AND EQUIVALENCE RELATIONS 15
Equivalence Relations and Partitions
A fundamental notion in mathematics is that of equality. We can general-
ize equality with the introduction of 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 15. Let p, q, r, and s be integers, where q and s are nonzero.
Define p/q ∼ r/s if ps = qr. Clearly ∼ is reflexive and symmetric. To show
that it is also transitive, suppose that p/q ∼ r/s and r/s ∼ t/u, with q, s,
and u all nonzero. Then ps = qr and ru = st. Therefore,
psu = qru = qst.
Since s 6= 0, pu = qt. Consequently, p/q ∼ t/u.
Example 16. 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 0 (x) = g 0 (x). It is clear that ∼ is both reflexive and symmetric. To
demonstrate transitivity, suppose that f (x) ∼ g(x) and g(x) ∼ h(x). From
calculus we know that f (x) − g(x) = c1 and g(x) − h(x) = c2 , where c1 and
c2 are both constants. Hence,
f (x) − h(x) = (f (x) − g(x)) + (g(x) − h(x)) = c1 − c2
and f 0 (x) − h0 (x) = 0. Therefore, f (x) ∼ h(x).
Example 17. 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 18. Let A and B be 2 × 2 matrices with entries in the real
numbers. We can define an equivalence relation on the set of 2 × 2 matrices,
16 CHAPTER 1 PRELIMINARIES
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
S of nonempty sets X1 , X2 , . . .
such that Xi ∩ Xj = ∅ for i 6= j and k Xk = X. Let ∼ be an equivalence
relation on a set X and let x ∈ X. Then [x] = {y ∈ X : y ∼ x} is called the
equivalence class of x. We will see that an equivalence relation gives rise
to a partition via equivalence classes. Also, whenever a partition of a set
exists, there is some natural underlying equivalence relation, as the following
theorem demonstrates.
Theorem 1.5 Given an equivalence relation ∼ on a set X, the equivalence
classes of X form a partition of X. Conversely, if P = {Xi } is a partition of
a set X, then there is an equivalence relation on X with equivalence classes
Xi .
1.2 SETS AND EQUIVALENCE RELATIONS 17
Proof. Suppose there exists an equivalence relation ∼ on the set X. For
any x ∈ X, theSreflexive property shows that x ∈ [x] and so [x] is nonempty.
Clearly X = x∈X [x]. Now let x, y ∈ X. We need to show that either
[x] = [y] or [x] ∩ [y] = ∅. Suppose that the intersection of [x] and [y] is not
empty and that z ∈ [x] ∩ [y]. Then z ∼ x and z ∼ y. By symmetry and
transitivity x ∼ y; hence, [x] ⊂ [y]. Similarly, [y] ⊂ [x] and so [x] = [y].
Therefore, any two equivalence classes are either disjoint or exactly the same.
Conversely, suppose that P = {Xi } is a partition of a set X. Let two
elements be equivalent if they are in the same partition. Clearly, the relation
is reflexive. If x is in the same partition as y, then y is in the same partition
as x, so x ∼ y implies y ∼ x. Finally, if x is in the same partition as y and
y is in the same partition as z, then x must be in the same partition as z,
and transitivity holds.
Corollary 1.6 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 19. In the equivalence relation in Example 15, 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 20. In the equivalence relation in Example 16, two functions f (x)
and g(x) are in the same partition when they differ by a constant.
Example 21. 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 22. 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
18 CHAPTER 1 PRELIMINARIES
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 vari-
ous 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.
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.
EXERCISES 19
(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)0 = A0 ∪ B 0 .
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 .
20. (a) Define a function f : N → N that is one-to-one but not onto.
20 CHAPTER 1 PRELIMINARIES
(b) Define a function f : N → N that is onto but not one-to-one.
21. Prove the relation defined on R2 by (x1 , y1 ) ∼ (x2 , y2 ) if x21 + y12 = x22 + y22 is
an equivalence relation.
22. Let f : A → B and g : B → C be maps.
(a) If f and g are both one-to-one functions, show that g ◦ f is one-to-one.
(b) If g ◦ f is onto, show that g is onto.
(c) If g ◦ f is one-to-one, show that f is one-to-one.
(d) If g ◦ f is one-to-one and f is onto, show that g is one-to-one.
(e) If g ◦ f is onto and g is one-to-one, show that f is onto.
23. Define a function on the real numbers by
x+1
f (x) = .
x−1
What are the domain and range of f ? What is the inverse of f ? Compute
f ◦ f −1 and f −1 ◦ f .
24. Let f : X → Y be a map with A1 , A2 ⊂ X and B1 , B2 ⊂ Y .
(a) Prove f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ).
(b) Prove f (A1 ∩ A2 ) ⊂ f (A1 ) ∩ f (A2 ). Give an example in which equality
fails.
(c) Prove f −1 (B1 ∪ B2 ) = f −1 (B1 ) ∪ f −1 (B2 ), where
f −1 (B) = {x ∈ X : f (x) ∈ B}.
(d) Prove f −1 (B1 ∩ B2 ) = f −1 (B1 ) ∩ f −1 (B2 ).
(e) Prove f −1 (Y \ B1 ) = X \ f −1 (B1 ).
25. Determine whether or not the following relations are equivalence relations on
the given set. If the relation is an equivalence relation, describe the partition
given by it. If the relation is not an equivalence relation, state why it fails to
be one.
(a) x ∼ y in R if x ≥ y (c) x ∼ y in R if |x − y| ≤ 4
(b) m ∼ n in Z if mn > 0 (d) m ∼ n in Z if m ≡ n (mod 6)
26. Define a relation ∼ on 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 .
EXERCISES 21
28. Find the error in the following argument by providing a counterexample.
“The reflexive property is redundant in the axioms for an equivalence relation.
If x ∼ y, then y ∼ x by the symmetric property. Using the transitive
property, we can deduce that x ∼ x.”
29. Projective Real Line. Define a relation on R2 \ (0, 0) by letting (x1 , y1 ) ∼
(x2 , y2 ) if there exists a nonzero real number λ such that (x1 , y1 ) = (λx2 , λy2 ).
Prove that ∼ defines an equivalence relation on R2 \(0, 0). What are the corre-
sponding equivalence classes? This equivalence relation defines the projective
line, denoted by P(R), which is very important in geometry.
References and Suggested Readings
The following list contains references suitable for further reading. With the excep-
tion of [8] and [9] and perhaps [1] and [3], all of these books are more or less at the
same level as this text. Interesting applications of algebra can be found in [2], [5],
[10], and [11].
[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] Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper
Saddle River, NJ, 2003.
[5] Gallian, J. A. Contemporary Abstract Algebra. 7th ed. Brooks/Cole, Bel-
mont, CA, 2009.
[6] Halmos, P. Naive Set Theory. Springer, New York, 1991. One of the best
references for set theory.
[7] Herstein, I. N. Abstract Algebra. 3rd ed. Wiley, New York, 1996.
[8] Hungerford, T. W. Algebra. Springer, New York, 1974. One of the standard
graduate algebra texts.
[9] Lang, S. Algebra. 3rd ed. Springer, New York, 2002. Another standard
graduate text.
[10] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York,
1998.
[11] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[12] Nickelson, W. K. Introduction to Abstract Algebra. 3rd ed. Wiley, New York,
2006.
[13] Solow, D. How to Read and Do Proofs. 5th ed. Wiley, New York, 2009.
[14] 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 math-
ematical induction, the division algorithm, and the Fundamental Theorem
of Arithmetic.
2.1 Mathematical Induction
Suppose we wish to show that
n(n + 1)
1 + 2 + ··· + n =
2
for any natural number n. This formula is easily verified for small numbers
such as n = 1, 2, 3, or 4, but it is impossible to verify for all natural numbers
on a case-by-case basis. To prove the formula true in general, a more generic
method is required.
Suppose we have verified the equation for the first n cases. We will
attempt to show that we can generate the formula for the (n + 1)th case
from this knowledge. The formula is true for n = 1 since
1(1 + 1)
1= .
2
If we have verified the first n cases, then
n(n + 1)
1 + 2 + · · · + n + (n + 1) = +n+1
2
n2 + 3n + 2
=
2
(n + 1)[(n + 1) + 1]
= .
2
22
2.1 MATHEMATICAL INDUCTION 23
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.
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 n0 .
Example 1. For all integers n ≥ 3, 2n > n + 4. Since
8 = 23 > 3 + 4 = 7,
the statement is true for n0 = 3. Assume that 2k > k + 4 for k ≥ 3. Then
2k+1 = 2 · 2k > 2(k + 4). But
2(k + 4) = 2k + 8 > k + 5 = (k + 1) + 4
since k is positive. Hence, by induction, the statement holds for all integers
n ≥ 3.
Example 2. 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 3. We will prove the binomial theorem using mathematical in-
duction; that is,
n
n
X n k n−k
(a + b) = a b ,
k
k=0
24 CHAPTER 2 THE INTEGERS
where a and b are real numbers, n ∈ N, and
n n!
=
k k!(n − k)!
is the binomial coefficient. We first show that
n+1 n n
= + .
k k k−1
This result follows from
n n n! n!
+ = +
k k−1 k!(n − k)! (k − 1)!(n − k + 1)!
(n + 1)!
=
k!(n + 1 − k)!
n+1
= .
k
If n = 1, the binomial theorem is easy to verify. Now assume that the result
is true for n greater than or equal to 1. Then
(a + b)n+1 = (a + b)(a + b)n
n
!
X n k n−k
= (a + b) a b
k
k=0
n n
X n k+1 n−k X n k n+1−k
= a b + a b
k k
k=0 k=0
n n
n+1
X n k n+1−k
X n k n+1−k
=a + a b + a b + bn+1
k−1 k
k=1 k=1
n
X n n
= an+1 + + ak bn+1−k + bn+1
k−1 k
k=1
n+1
X n + 1
= ak bn+1−k .
k
k=0
We have an equivalent statement of the Principle of Mathematical In-
duction that is often very useful.
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
2.1 MATHEMATICAL INDUCTION 25
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 greater than n0 .
A nonempty subset S of Z is well-ordered if S contains a least element.
Notice that the set Z is not well-ordered since it does not contain a smallest
element. However, the natural numbers are well-ordered.
Principle of Well-Ordering. Every nonempty subset of the natural num-
bers is well-ordered.
The Principle of Well-Ordering is equivalent to the Principle of Mathe-
matical Induction.
Lemma 2.1 The Principle of Mathematical Induction implies that 1 is the
least positive natural number.
Proof. Let S = {n ∈ N : n ≥ 1}. Then 1 ∈ S. Now assume that n ∈ S;
that is, n ≥ 1. Since n + 1 ≥ 1, n + 1 ∈ S; hence, by induction, every natural
number is greater than or equal to 1.
Theorem 2.2 The Principle of Mathematical Induction implies that the
natural numbers are well-ordered.
Proof. We must show that if S is a nonempty subset of the natural num-
bers, then S contains a smallest element. If S contains 1, then the theorem
is true by Lemma 2.1. Assume that if S contains an integer k such that
1 ≤ k ≤ n, then S contains a smallest element. We will show that if a set S
contains an integer less than or equal to n+1, then S has a smallest 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 smallest integer.
Induction can also be very useful in formulating definitions. For instance,
there are two ways to define n!, the factorial of a positive integer n.
• The explicit definition: n! = 1 · 2 · 3 · · · (n − 1) · n.
• The inductive or recursive definition: 1! = 1 and n! = n(n − 1)! for
n > 1.
Every good mathematician or computer scientist knows that looking at prob-
lems recursively, as opposed to explicitly, often results in better understand-
ing of complex issues.
26 CHAPTER 2 THE INTEGERS
2.2 The Division Algorithm
An application of the Principle of Well-Ordering that we will use often is
the division algorithm.
Theorem 2.3 (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 0 and r0 are two other such numbers, then q = q 0 and
r = r0 .
Existence of q and r. Let
S = {a − bk : k ∈ Z and a − bk ≥ 0}.
If 0 ∈ S, then b divides a, and we can let q = a/b and r = 0. If 0 ∈
/ S, we can
use the Well-Ordering Principle. We must first show that S is nonempty.
If a > 0, then a − b · 0 ∈ S. If a < 0, then a − b(2a) = a(1 − 2b) ∈ S. In
either case S 6= ∅. By the Well-Ordering Principle, S must have a smallest
member, say r = a − bq. Therefore, a = bq + r, r ≥ 0. We now show that
r < b. Suppose that r > b. Then
a − b(q + 1) = a − bq − b = r − b > 0.
In this case we would have a − b(q + 1) in the set S. But then a − b(q + 1) <
a−bq, which would contradict the fact that r = a−bq is the smallest member
of S. So r ≤ b. Since 0 ∈
/ S, r 6= b and so r < b.
Uniqueness of q and r. Suppose there exist integers r, r0 , q, and q 0 such
that
a = bq + r, 0 ≤ r < b
and
a = bq 0 + r0 , 0 ≤ r0 < b.
Then bq + r = bq 0 + r0 . Assume that r0 ≥ r. From the last equation we have
b(q − q 0 ) = r0 − r; therefore, b must divide r0 − r and 0 ≤ r0 − r ≤ r0 < b.
This is possible only if r0 − r = 0. Hence, r = r0 and q = q 0 .
2.2 THE DIVISION ALGORITHM 27
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 d0 is any other common divisor
of a and b, then d0 | 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 2.4 Let a and b be nonzero integers. Then there exist integers r
and s such that
gcd(a, b) = ar + bs.
Furthermore, the greatest common divisor of a and b is unique.
Proof. Let
S = {am + bn : m, n ∈ Z and am + bn > 0}.
Clearly, the set S is nonempty; hence, by the Well-Ordering Principle S
must have a smallest member, say d = ar + bs. We claim that d = gcd(a, b).
Write a = dq + r where 0 ≤ r < d . 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 d0 is another common divisor of a and b, and we want to
show that d0 | d. If we let a = d0 h and b = d0 k, then
d = ar + bs = d0 hr + d0 ks = d0 (hr + ks).
So d0 must divide d. Hence, d must be the unique greatest common divisor
of a and b.
Corollary 2.5 Let a and b be two integers that are relatively prime. Then
there exist integers r and s such that ar + bs = 1.
28 CHAPTER 2 THE INTEGERS
The Euclidean Algorithm
Among other things, Theorem 2.4 allows us to compute the greatest common
divisor of two integers.
Example 4. 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
= 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 .
2.2 THE DIVISION ALGORITHM 29
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 2.6 (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 2.7 (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.
We will show that p must be a different prime number, which contradicts
the assumption that we have only n primes. If p is not prime, then it must
be divisible by some pi for 1 ≤ i ≤ n. In this case pi must divide p1 p2 · · · pn
and also divide 1. This is a contradiction, since p > 1.
30 CHAPTER 2 THE INTEGERS
Theorem 2.8 (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 2.6,
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,
n0 = 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
a 2 = q1 · · · qs .
Therefore,
a = a1 a2 = p1 · · · pr q1 · · · qs .
So a ∈
/ S, which is a contradiction.
Historical Note
EXERCISES 31
Prime numbers were first studied by the ancient Greeks. Two important 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 integer n. Pierre Fermat (1601?–1665) conjectured that
n
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 100 million, it has yet to be proven in general.
Since prime numbers play an important role in public key cryptography, there is
currently a great deal of interest in determining whether or not a large number is
prime.
Exercises
1. Prove that
n(n + 1)(2n + 1)
12 + 22 + · · · + n2 =
6
for n ∈ N.
2. Prove that
n2 (n + 1)2
1 3 + 2 3 + · · · + 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
n
√ 1X
n
a1 a2 · · · an ≤ ak .
n
k=1
8. Prove the Leibniz rule for f (n) (x), where f (n) is the nth derivative of f ; that
is, show that
n
X n (k)
(f g)(n) (x) = f (x)g (n−k) (x).
k
k=0
32 CHAPTER 2 THE INTEGERS
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.
13. Prove that the two principles of mathematical induction stated in Section 2.1
are equivalent.
14. Show that the Principle of Well-Ordering for the natural numbers implies
that 1 is the smallest natural number. Use this result to show that the
Principle of Well-Ordering implies the Principle of Mathematical Induction;
that is, show that if S ⊂ N such that 1 ∈ S and n + 1 ∈ S whenever n ∈ S,
then S = N.
15. For each of the following pairs of numbers a and b, calculate gcd(a, b) and
find integers r and s such that gcd(a, b) = ra + sb.
(a) 14 and 39 (d) 471 and 562
(b) 234 and 165 (e) 23,771 and 19,945
(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.
EXERCISES 33
(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 coprime and that
a2 + b2 = r2
a2 − b2 = s2 .
Prove that a, r, and s are odd and b is even.
22. Let n ∈ N. Use the division algorithm to prove that every integer is congruent
mod n to precisely one of the integers 0, 1, . . . , n − 1. Conclude that if r is
an integer, then there is exactly one s in Z such that 0 ≤ s < n and [r] = [s].
Hence, the integers are indeed partitioned by congruence mod n.
23. Define the least common multiple of two nonzero integers a and b,
denoted by lcm(a, b), to be the nonnegative integer m such that both a and
b divide m, and if a and b divide any other integer n, then m also divides n.
Prove that any two integers a and b have a unique least common multiple.
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 + 1.
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.
34 CHAPTER 2 THE INTEGERS
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 Ack-
ermann’s function. Modify the program to count the number of statements
executed in the program when Ackermann’s function is evaluated. How many
statements are executed in the evaluation of A(4, 1)? What about A(5, 1)?
3. Write a computer program that will implement the Euclidean algorithm.
The program should accept two positive integers a and b as input and should
output gcd(a, b) as well as integers r and s such that
gcd(a, b) = ra + sb.
References and Suggested Readings
References [2], [3], and [4] are good sources for elementary number theory.
[1] Brookshear, J. G. Theory of Computation: Formal Languages, Automata,
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.
3
Groups
We begin our study of algebraic structures by investigating sets associated
with single operations that satisfy certain reasonable axioms; that is, we
want to define an operation on a set in a way that will generalize such
familiar structures as the integers Z together with the single operation of
addition, or invertible 2 × 2 matrices together with the single operation of
matrix multiplication. The integers and the 2 × 2 matrices, together with
their respective single operations, are examples of algebraic structures known
as groups.
The theory of groups occupies a central position in mathematics. Modern
group theory arose from an attempt to find the roots of a polynomial in
terms of its coefficients. Groups now play a central role in such areas as
coding theory, counting, and the study of symmetries; many areas of biology,
chemistry, and physics have benefited from group theory.
3.1 The Integers mod n and Symmetries
Let us now investigate some mathematical structures that can be viewed as
sets with single operations.
The Integers mod n
The integers mod n have become indispensable in the theory and 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
35
36 CHAPTER 3 GROUPS
Zn . Consider the integers modulo 12 and the corresponding partition of the
integers:
[0] = {. . . , −12, 0, 12, 24, . . .},
[1] = {. . . , −11, 1, 13, 25, . . .},
..
.
[11] = {. . . , −1, 11, 23, 35, . . .}.
When no confusion can arise, we will use 0, 1, . . . , 11 to indicate the equiva-
lence 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, multiplication modulo
n is defined as (ab) (mod n), the remainder when ab is divided by n.
Table 3.1. Multiplication table for Z8
· 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
Example 1. The following examples illustrate integer arithmetic modulo n:
7 + 4 ≡ 1 (mod 5) 7 · 3 ≡ 1 (mod 5)
3 + 5 ≡ 0 (mod 8) 3 · 5 ≡ 7 (mod 8)
3 + 4 ≡ 7 (mod 12) 3 · 4 ≡ 0 (mod 12).
In particular, notice that it is possible that the product of two nonzero
numbers modulo n can be equivalent to 0 modulo n.
Example 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
3.1 THE INTEGERS MOD N AND SYMMETRIES 37
Z8 in Table 3.1. 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).
Proposition 3.1 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 an additive and a multiplicative identity:
a + 0 ≡ a (mod n)
a · 1 ≡ a (mod n).
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 ex-
ists 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.
38 CHAPTER 3 GROUPS
(6) Suppose that gcd(a, n) = 1. Then there exist integers r and s such
that ar + ns = 1. Since ns = 1 − ar, ra ≡ 1 (mod n). Letting b be the
equivalence class of r, ab ≡ 1 (mod n).
Conversely, suppose that there exists a 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
Figure 3.1. Rigid motions of a rectangle
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
A symmetry of a geometric figure is a rearrangement of the figure pre-
serving the arrangement of its sides and vertices as well as its distances and
angles. A map from the plane to itself preserving the symmetry of an object
is called a rigid motion. For example, if we look at the rectangle in Fig-
ure 3.1, 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
3.1 THE INTEGERS MOD N AND SYMMETRIES 39
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.
Figure 3.2. Symmetries of a triangle
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
Let us find the symmetries of the equilateral triangle 4ABC. To find a
symmetry of 4ABC, we must first examine the permutations of the vertices
A, B, and C and then ask if a permutation extends to a symmetry of the
triangle. Recall that a permutation of a set S is a one-to-one and onto
map π : S → S. The three vertices have 3! = 6 permutations, so the triangle
40 CHAPTER 3 GROUPS
has at most six symmetries. To see that there are six permutations, observe
there are three different possibilities for the first vertex, and two for the
second, and the remaining vertex is determined by the placement of the
first two. So we have 3 · 2 · 1 = 3! = 6 different arrangements. To denote the
permutation of the vertices of an equilateral triangle that sends A to B, B
to C, and C to A, we write the array
A B C
.
B C A
Notice that this particular permutation corresponds to the rigid motion
of rotating the triangle by 120◦ in a clockwise direction. In fact, every
permutation gives rise to a symmetry of the triangle. All of these symmetries
are shown in Figure 3.2.
A natural question to ask is what happens if one motion of the trian-
gle 4ABC is followed by another. Which symmetry is µ1 ρ1 ; that is, what
happens when we do the permutation ρ1 and then the permutation µ1 ? Re-
member that we are composing functions here. Although we usually multiply
left to right, we compose functions right to left. We have
(µ1 ρ1 )(A) = µ1 (ρ1 (A)) = µ1 (B) = C
(µ1 ρ1 )(B) = µ1 (ρ1 (B)) = µ1 (C) = B
(µ1 ρ1 )(C) = µ1 (ρ1 (C)) = µ1 (A) = A.
This is the same symmetry as µ2 . Suppose we do these motions in the
opposite order, ρ1 then µ1 . It is easy to determine that this is the same
as the symmetry µ3 ; hence, ρ1 µ1 6= µ1 ρ1 . A multiplication table for the
symmetries of an equilateral triangle 4ABC is given in Table 3.2.
Notice that in the multiplication table for the symmetries of an equilat-
eral triangle, for every motion of the triangle α there is another motion α0
such that αα0 = id; that is, for every motion there is another motion that
takes the triangle back to its original orientation.
3.2 Definitions and Examples
The integers mod n and the symmetries of a triangle or a rectangle are both
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.
3.2 DEFINITIONS AND EXAMPLES 41
Table 3.2. Symmetries of an equilateral triangle
◦ 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
• The law of composition is associative. That is,
(a ◦ b) ◦ c = a ◦ (b ◦ c)
for a, b, c ∈ G.
• There exists an element e ∈ G, called the identity element, such
that for any element a ∈ G
e ◦ a = a ◦ e = a.
• For each element a ∈ G, there exists an inverse element in G,
denoted by a−1 , such that
a ◦ a−1 = a−1 ◦ a = e.
A group G with the property that a ◦ b = b ◦ a for all a, b ∈ G is called
abelian or commutative. Groups not satisfying this property are said to
be nonabelian or noncommutative.
Example 3. 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 integers under addition have the additional
property that m + n = n + m and are therefore 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,
42 CHAPTER 3 GROUPS
Table 3.3. Cayley table for (Z5 , +)
+ 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
−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 4. 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 3.3 is a Cayley table for Z5 . By
Proposition 3.1, Zn = {0, 1, . . . , n − 1} is a group under the binary operation
of addition mod n.
Example 5. Not every set with a binary operation is a group. For example,
if we let modular multiplication be the binary operation on Zn , then Zn fails
to be a group. The element 1 acts as a group identity since 1 · k = k · 1 = k
for any k ∈ Zn ; however, a multiplicative inverse for 0 does not exist since
0 · k = k · 0 = 0 for every k in Zn . Even if we consider the set Zn \ {0},
we still may not have a group. For instance, let 2 ∈ Z6 . Then 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 3.1, 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 3.4 is
a Cayley table for the group U (8).
3.2 DEFINITIONS AND EXAMPLES 43
Table 3.4. Multiplication table for 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
Example 6. The symmetries of an equilateral triangle described in Sec-
tion 3.1 form a nonabelian group. As we observed, it is not necessarily true
that αβ = βα for two symmetries α and β. Using Table 3.2, 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 7. We use M2 (R) to denote the set of all 2 × 2 matrices. Let
GL2 (R) be the subset of M2 (R) consisting of invertible matrices; that is, a
matrix
a b
A=
c d
is in GL2 (R) if there exists a matrix A−1 such that AA−1 = A−1 A = I,
where I is the 2 × 2 identity matrix. For A to have an inverse is equivalent
to requiring that the determinant of A be nonzero; that is, det A = ad−bc 6=
0. The set of invertible matrices forms a group called the general linear
group. The identity of the group is the identity matrix
1 0
I= .
0 1
The inverse of A ∈ GL2 (R) is
−1 1 d −b
A = .
ad − bc −c a
The product of two invertible matrices is again invertible. Matrix multipli-
cation is associative, satisfying the other group axiom. For matrices it is
not true in general that AB 6= BA; hence, GL2 (R) is another example of a
nonabelian group.
44 CHAPTER 3 GROUPS
Example 8. 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 9. 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 con-
tains. 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 3.2 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 e0 are both identities in G. Then eg = ge = g
and e0 g = ge0 = g for all g ∈ G. We need to show that e = e0 . If we think
of e as the identity, then ee0 = e0 ; but if e0 is the identity, then ee0 = e.
Combining these two equations, we have e = ee0 = e0 .
Inverses in a group are also unique. If g 0 and g 00 are both inverses of an
element g in a group G, then gg 0 = g 0 g = e and gg 00 = g 00 g = e. We want
to show that g 0 = g 00 , but g 0 = g 0 e = g 0 (gg 00 ) = (g 0 g)g 00 = eg 00 = g 00 . We
summarize this fact in the following proposition.
3.2 DEFINITIONS AND EXAMPLES 45
Proposition 3.3 If g is any element in a group G, then the inverse of g,
g −1 , is unique.
Proposition 3.4 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 3.5 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.
It makes sense to write equations with group elements and group opera-
tions. 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 3.6 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. Multi-
plying both sides of ax = b by a−1 , we have 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 3.7 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
46 CHAPTER 3 GROUPS
Theorem 3.8 In a group, the usual laws of exponents hold; that is, for all
g, h ∈ G,
1. g m g n = g m+n for all m, n ∈ Z;
2. (g m )n = g mn for all m, n ∈ Z;
3. (gh)n = (h−1 g −1 )−n for all n ∈ Z. Furthermore, if G is abelian, then
(gh)n = g n hn .
We will leave the proof of this theorem as an exercise. Notice that
(gh)n 6= g n hn in general, since the group may not be abelian. If the group
is Z or Zn , we write the group operation additively and the exponential
operation multiplicatively; that is, we write ng instead of g n . The laws of
exponents now become
1. mg + ng = (m + n)g for all m, n ∈ Z;
2. m(ng) = (mn)g for all m, n ∈ Z;
3. m(g + h) = mg + mh for all n ∈ Z.
It is important to realize that the last statement can be made only because
Z and Zn are commutative groups.
Historical Note
Although the first clear axiomatic definition of a group was not given until the
late 1800s, group-theoretic methods had been employed before this time in the
development of many areas of mathematics, including geometry and the theory of
algebraic equations.
Joseph-Louis Lagrange used group-theoretic methods in a 1770–1771 memoir to
study methods of solving polynomial equations. Later, Évariste Galois (1811–1832)
succeeded in developing the mathematics necessary to determine exactly which
polynomial equations could be solved in terms of the polynomials’ coefficients.
Galois’ primary tool was group theory.
The study of geometry was revolutionized in 1872 when Felix Klein proposed
that geometric spaces should be studied by examining those properties that are
invariant under a transformation of the space. Sophus Lie, a contemporary of
Klein, used group theory to study solutions of partial differential equations. One of
the first modern treatments of group theory appeared in William Burnside’s The
Theory of Groups of Finite Order [1], first published in 1897.
3.3 SUBGROUPS 47
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 10. 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 11. 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 12. Let SL2 (R) be the subset of GL2 (R) consisting of matrices
of determinant one; that is, a matrix
a b
A=
c d
is in SL2 (R) exactly when ad − bc = 1. To show that SL2 (R) is a subgroup
of the general linear group, we must show that it is a group under matrix
48 CHAPTER 3 GROUPS
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 13. 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 G’s binary operation. 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 14. 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 3.5 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 3.5. Addition table for Z2 × Z2
EXERCISES 49
Some Subgroup Theorems
Let us examine some criteria for determining exactly when a subset of a
group is a subgroup.
Proposition 3.9 A subset H of G is a subgroup if and only if it satisfies
the following conditions.
1. The identity e of G is in H.
2. If h1 , h2 ∈ H, then h1 h2 ∈ H.
3. If h ∈ H, then h−1 ∈ H.
Proof. First suppose that H is a subgroup of G. We must show that the
three conditions hold. Since H is a group, it must have an identity eH .
We must show that eH = e, where e is the identity of G. We know that
eH eH = eH and that eeH = eH e = eH ; hence, eeH = eH eH . By right-hand
cancellation, e = eH . The second condition holds since a subgroup H is a
group. To prove the third condition, let h ∈ H. Since H is a group, there
is an element h0 ∈ H such that hh0 = h0 h = e. By the uniqueness of the
inverse in G, h0 = 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 associa-
tivity of the binary operation are exactly the axioms stated in the definition
of a group.
Proposition 3.10 Let H be a subset of a group G. Then H is a subgroup
of G if and only if H 6= ∅, and whenever g, h ∈ H then gh−1 is in H.
Proof. Let H be a nonempty subset of G. Then H contains some element
g. So gg −1 = e is in H. If g ∈ H, then eg −1 = g −1 is also in H. Finally,
let g, h ∈ H. We must show that their product is also in H. However,
g(h−1 )−1 = gh ∈ H. Hence, H is indeed a subgroup of G. Conversely, if
g and h are in H, we want to show that gh−1 ∈ 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.
Exercises
1. Find all x ∈ Z satisfying each of the following equations.
50 CHAPTER 3 GROUPS
(a) 3x ≡ 2 (mod 7) (d) 9x ≡ 3 (mod 5)
(b) 5x + 1 ≡ 13 (mod 23) (e) 5x ≡ 1 (mod 6)
(c) 5x + 1 ≡ 13 (mod 26) (f) 3x ≡ 1 (mod 6)
2. Which of the following multiplication tables defined on the set G = {a, b, c, d}
form a group? Support your answer in each case.
◦ a b c d ◦ a b c d
a a c d a a a b c d
(a) b b b c d (c) 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
◦ a b c d ◦ a b c d
a a b c d a a b c d
(b) b b a d c (d) b b a c d
c c d a b c c b a d
d d c b a d d d b c
3. Write out Cayley tables for groups formed by the symmetries of a rectangle
and for (Z4 , +). How many elements are in each group? Are the groups the
same? Why or why not?
4. Describe the symmetries of a rhombus and prove that the set of symmetries
forms a group. Give Cayley tables for both the symmetries of a rectangle
and the symmetries of a rhombus. Are the symmetries of a rectangle and
those of a rhombus the same?
5. Describe the symmetries of a square and prove that the set of symmetries
is a group. Give a Cayley table for the symmetries. How many ways can
the vertices of a square be permuted? Is each permutation necessarily a
symmetry of the square? The symmetry group of the square is denoted by
D4 .
6. Give a multiplication table for the group U (12).
7. Let S = R \ {−1} and define a binary operation on S by a ∗ b = a + b + ab.
Prove that (S, ∗) is an abelian group.
8. Give an example of two elements A and B in GL2 (R) with AB 6= BA.
9. Prove that the product of two matrices in SL2 (R) has determinant one.
10. Prove that the set of matrices of the form
1 x y
0 1 z
0 0 1
EXERCISES 51
is a group under matrix multiplication. This group, known as the Heisen-
berg group, is important in quantum physics. Matrix multiplication in the
Heisenberg group is defined by
1 x0 y 0 1 x + x0 y + y 0 + xz 0
1 x y
0 1 z 0 1 z 0 = 0 1 z + z0 .
0 0 1 0 0 1 0 0 1
11. Prove that det(AB) = det(A) det(B) in GL2 (R). Use this result to show
that the binary operation in the group GL2 (R) is closed; that is, if A and B
are in GL2 (R), then AB ∈ GL2 (R).
12. Let Zn2 = {(a1 , a2 , . . . , an ) : ai ∈ Z2 }. Define a binary operation on Zn2 by
(a1 , a2 , . . . , an ) + (b1 , b2 , . . . , bn ) = (a1 + b1 , a2 + b2 , . . . , an + bn ).
Prove that Zn2 is a group under this operation. This group is important in
algebraic coding theory.
13. Show that R∗ = R \ {0} is a group under the operation of multiplication.
14. Given the groups R∗ and Z, let G = R∗ × Z. Define a binary operation ◦
on G by (a, m) ◦ (b, n) = (ab, m + n). Show that G is a group under this
operation.
15. Prove or disprove that every group containing six elements is abelian.
16. Give a specific example of some group G and elements g, h ∈ G where (gh)n 6=
g n hn .
17. Give an example of three different groups with eight elements. Why are the
groups different?
18. Show that there are n! permutations of a set containing n items.
19. Show that
0 + a ≡ a + 0 ≡ a (mod n)
for all a ∈ Zn .
20. Prove that there is a multiplicative identity for the integers modulo n:
a · 1 ≡ a (mod n).
21. For each a ∈ Zn find a b ∈ Zn such that
a+b≡b+a≡0 (mod n).
22. Show that addition and multiplication mod n are associative operations.
23. Show that multiplication distributes over addition modulo n:
a(b + c) ≡ ab + ac (mod n).
52 CHAPTER 3 GROUPS
24. Let a and b be elements in a group G. Prove that abn a−1 = (aba−1 )n .
25. Let U (n) be the group of units in Zn . If n > 2, prove that there is an element
k ∈ U (n) such that k 2 = 1 and k 6= 1.
−1
26. Prove that the inverse of g1 g2 · · · gn is gn−1 gn−1 · · · g1−1 .
27. Prove the remainder of Theorem 3.6: if G is a group and a, b ∈ G, then the
equation xa = b has unique solutions in G.
28. Prove Theorem 3.8.
29. 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.
30. Show that if a2 = e for all a ∈ G, then G must be an abelian group.
31. 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.
32. 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.
33. Find all the subgroups of Z3 × Z3 . Use this information to show that Z3 × Z3
is not the same group as Z9 . (See Example 14 for a short description of the
product of groups.)
34. Find all the subgroups of the symmetry group of an equilateral triangle.
35. Compute the subgroups of the symmetry group of a square.
36. Let H = {2k : k ∈ Z}. Show that H is a subgroup of Q∗ .
37. 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.
38. Let T = {z ∈ C∗ : |z| = 1}. Prove that T is a subgroup of C∗ .
39. 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).
40. 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.
EXERCISES 53
41. 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.
42. Prove or disprove: SL2 (Z), the set of 2 × 2 matrices with integer entries and
determinant one, is a subgroup of SL2 (R).
43. List the subgroups of the quaternion group, Q8 .
44. Prove that the intersection of two subgroups of a group G is also a subgroup
of G.
45. Prove or disprove: If H and K are subgroups of a group G, then H ∪ K is a
subgroup of G.
46. 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?
47. 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.
48. Let a and b be elements of a group G. If a4 b = ba and a3 = e, prove that
ab = ba.
49. Give an example of an infinite group in which every nontrivial subgroup is
infinite.
50. Give an example of an infinite group in which every proper subgroup is finite.
51. If xy = x−1 y −1 for all x and y in G, prove that G must be abelian.
52. If (xy)2 = xy for all x and y in G, prove that G must be abelian.
53. Prove or disprove: Every nontrivial subgroup of an nonabelian group is non-
abelian.
54. Let H be a subgroup of G and
N (H) = {g ∈ G : gh = hg for all h ∈ H}.
Prove N (H) is a subgroup of G. This subgroup is called the normalizer
of H in G.
54 CHAPTER 3 GROUPS
0 50000 30042 6
Figure 3.3. A UPC code
Additional Exercises: Detecting Errors
Credit card companies, banks, book publishers, and supermarkets all take advan-
tage of the properties of integer arithmetic modulo n and group theory to obtain
error detection schemes for the identification codes that they use.
1. UPC Symbols. Universal Product Code (UPC) symbols are now 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 3.3). 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 Fig-
ure 3.3, is a valid UPC number.
(b) Show that the number 0-50000-30043-6 is not a valid UPC number.
(c) Write a formula to calculate the check digit, d12 , in the UPC number.
(d) The UPC error detection scheme can detect most transposition errors;
that is, it can determine if two digits have been interchanged. Show
that the transposition error 0-05000-30042-6 is detected. Find a trans-
position error that is not detected.
(e) Write a program that will determine whether or not a UPC number is
valid.
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)
EXERCISES 55
to mean
d1 w1 + d2 w2 + · · · + dk wk ≡ 0 (mod n).
Suppose that (d1 , d2 , . . . , dk )·(w1 , w2 , . . . , wk ) ≡ 0 (mod n) is an error detec-
tion scheme for the k-digit identification number d1 d2 · · · dk , where 0 ≤ di <
n. Prove that all single-digit errors are detected if and only if gcd(wi , n) = 1
for 1 ≤ i ≤ k.
3. Let (d1 , d2 , . . . , dk ) · (w1 , w2 , . . . , wk ) ≡ 0 (mod n) be an error detection
scheme for the k-digit identification number d1 d2 · · · dk , where 0 ≤ di < n.
Prove that all transposition errors of two digits di and dj are detected if and
only if gcd(wi − wj , n) = 1 for i and j between 1 and k.
4. ISBN Codes. Every book has an International Standard Book Number
(ISBN) code. This is a 10-digit code indicating the book’s publisher and
title. The tenth digit is a check digit satisfying
(d1 , d2 , . . . , d10 ) · (10, 9, . . . , 1) ≡ 0 (mod 11).
One problem is that d10 might have to be a 10 to make the inner product zero;
in this case, 11 digits would be needed to make this scheme work. Therefore,
the character X is used for the eleventh digit. So ISBN 3-540-96035-X is a
valid ISBN code.
(a) Is ISBN 0-534-91500-0 a valid ISBN code? What about ISBN 0-534-
91700-0 and ISBN 0-534-19500-0?
(b) Does this method detect all single-digit errors? What about all trans-
position 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.
References and Suggested Readings
References [2] and [3] show how group theory can be used in error detection schemes.
Other sources cover more advanced topics in group theory.
[1] Burnside, W. Theory of Groups of Finite Order. 2nd ed. Cambridge Uni-
versity Press, Cambridge, 1911; Dover, New York, 1953. A classic. Also
available at books.google.com.
56 CHAPTER 3 GROUPS
[2] Gallian, J. A. and Winters, S. “Modular Arithmetic in the Marketplace,”
The American Mathematical Monthly 95(1988): 548–51.
[3] Gallian, J. A. Contemporary Abstract Algebra. 7th ed. Brooks/Cole, Bel-
mont, CA, 2009.
[4] Hall, M. Theory of Groups. 2nd ed. American Mathematical Society, Provi-
dence, 1959.
[5] Kurosh, A. E. The Theory of Groups, vols. I and II. American Mathematical
Society, Providence, 1979.
[6] Rotman, J. J. An Introduction to the Theory of Groups. 4th ed. Springer,
New York, 1995.
4
Cyclic Groups
The groups Z and Zn , which are among the most familiar and easily under-
stood groups, are both examples of what are called cyclic groups. In this
chapter we will study the properties of cyclic groups and cyclic subgroups,
which play a fundamental part in the classification of all abelian groups.
4.1 Cyclic Subgroups
Often a subgroup will depend entirely on a single element of the group;
that is, knowing that particular element will allow us to compute any other
element in the subgroup.
Example 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 2. If H = {2n : n ∈ Z}, then H is a subgroup of the multiplicative
group of nonzero rational numbers, Q∗ . If a = 2m and b = 2n are in H, then
ab−1 = 2m 2−n = 2m−n is also in H. By Proposition 3.10, H is a subgroup
of Q∗ determined by the element 2.
57
58 CHAPTER 4 CYCLIC GROUPS
Theorem 4.1 Let G be a group and a be any element in G. Then the set
hai = {ak : k ∈ Z}
is a subgroup of G. Furthermore, hai is the smallest subgroup of G that
contains a.
Proof. The identity is in hai since a0 = e. If g and h are any two elements
in hai, then by the definition of hai we can write g = am and h = an for some
integers m and n. So gh = am an = am+n is again in hai. Finally, if g = an
in hai, then the inverse g −1 = a−n is also in hai. Clearly, any subgroup H
of G containing a must contain all the powers of a by closure; hence, H
contains hai. Therefore, hai is the smallest subgroup of G containing a.
Remark. If we are using the “+” notation, as in the case of the integers
under addition, we write hai = {na : n ∈ Z}.
For a ∈ G, we call hai the cyclic subgroup generated by a. If G contains
some element a such that G = hai, then G is a cyclic group. In this case a
is a generator of G. If a is an 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 3. Notice that a cyclic group can have more than a single gen-
erator. Both 1 and 5 generate Z6 ; hence, Z6 is a cyclic group. Not every
element in a cyclic group is necessarily a generator of the group. The order
of 2 ∈ Z6 is 3. The cyclic subgroup generated by 2 is h2i = {0, 2, 4}.
The groups Z and Zn are cyclic groups. The elements 1 and −1 are
generators for Z. We can certainly generate Zn with 1 although there may
be other generators of Zn , as in the case of Z6 .
Example 4. 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.
4.1 CYCLIC SUBGROUPS 59
Example 5. Not every group is a cyclic group. Consider the symmetry
group of an equilateral triangle S3 . The multiplication table for this group
is Table 3.2. The subgroups of S3 are shown in Figure 4.1. 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 4.1. Subgroups of S3
Theorem 4.2 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.
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 4.3 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. We can assume that n > 0.
60 CHAPTER 4 CYCLIC GROUPS
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
h0 ∈ H can be written as a power of h. Since h0 ∈ H and H is a subgroup
of G, h0 = ak for some positive 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,
h0 = ak = amq = hq
and H is generated by h.
Corollary 4.4 The subgroups of Z are exactly nZ for n = 0, 1, 2, . . ..
Proposition 4.5 Let G be a cyclic group of order n and suppose that a is
a generator for G. Then ak = e if and only if n divides k.
Proof. First suppose that ak = e. By the division algorithm, k = nq + r
where 0 ≤ r < n; hence,
e = ak = anq+r = anq ar = ear = ar .
Since the smallest positive integer m such that am = e is n, r = 0.
Conversely, if n divides k, then k = ns for some integer s. Consequently,
ak = ans = (an )s = es = e.
Theorem 4.6 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 4.5, 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.
4.2 THE GROUP C∗ 61
Corollary 4.7 The generators of Zn are the integers r such that 1 ≤ r < n
and gcd(r, n) = 1.
Example 6. 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 The Group C∗
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 polyno-
mials. 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
62 CHAPTER 4 CYCLIC GROUPS
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 7. 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.
y
z1 = 2 + 3i
z3 = −3 + 2i
0 x
z2 = 1 − 2i
Figure 4.2. 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 4.2.
Nonzero complex numbers can also be represented using polar coordi-
nates. To specify any nonzero point on the plane, it suffices to give an angle
4.2 THE GROUP C∗ 63
y
a + bi
r
θ
0 x
Figure 4.3. Polar coordinates of a complex number
θ from the positive x axis in the counterclockwise direction and a distance
r from the origin, as in Figure 4.3. We can see that
z = a + bi = r(cos θ + i sin θ).
Hence, p
r = |z| = a2 + b2
and
a = r cos θ
b = r sin θ.
We sometimes abbreviate r(cos θ + i sin θ) as r cis θ. To assure that the
representation of z is well-defined, we also require that 0◦ ≤ θ < 360◦ . If
the measurement is in radians, then 0 ≤ θ < 2π.
Example 8. 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.
64 CHAPTER 4 CYCLIC GROUPS
Conversely, if we are given a rectangular representation of a complex
number,
√ it is√ often useful to know the number’s polar representation. If
z = 3 2 − 3 2 i, then
p √
r = a2 + b2 = 36 = 6
and
b
θ = arctan = arctan(−1) = 315◦ ,
a
√ √
so 3 2 − 3 2 i = 6 cis 315◦ .
The polar representation of a complex number makes it easy to find prod-
ucts and powers of complex numbers. The proof of the following proposition
is straightforward and is left as an exercise.
Proposition 4.8 Let z = r cis θ and w = s cis φ be two nonzero complex
numbers. Then
zw = rs cis(θ + φ).
Example 9. If z = 3 cis(π/3) and w = 2 cis(π/6), then zw = 6 cis(π/2) =
6i.
Theorem 4.9 (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].
4.2 THE GROUP C∗ 65
Example 10. Suppose that z = 1 + i and we wish to compute z 10 . Rather
than computing (1 + i)10 directly, it is much easier to switch to polar coor-
dinates and calculate z 10 using DeMoivre’s Theorem:
z 10 = (1 + i)10
√ π 10
= 2 cis
4
√ 10 5π
= ( 2 ) cis
2
π
= 32 cis
2
= 32i.
The Circle Group and the Roots of Unity
The multiplicative group of the complex numbers, C∗ , possesses some in-
teresting 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 4.8.
Proposition 4.10 The circle group is a subgroup of C∗ .
Although the circle group has infinite order, it has many interesting finite
subgroups. Suppose that H = {1, −1, i, −i}. Then H is a subgroup of the
circle group. Also, 1, −1, i, and −i are exactly those complex numbers that
satisfy the equation z 4 = 1. The complex numbers satisfying the equation
z n = 1 are called the nth roots of unity.
Theorem 4.11 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.
66 CHAPTER 4 CYCLIC GROUPS
Proof. By DeMoivre’s Theorem,
n 2kπ
z = cis n = cis(2kπ) = 1.
n
The z’s are distinct since the numbers 2kπ/n are all distinct and are greater
than or equal to 0 but less than 2π. The fact that these are all of the roots
of the equation z n = 1 follows from from Corollary 17.6, which states that
a polynomial of degree n can have at most n roots. We will leave the proof
that the nth roots of unity form a cyclic subgroup of T as an exercise.
A generator for the group of the nth roots of unity is called a primitive
nth root of unity.
Example 11. The 8th roots of unity can be represented as eight equally
spaced points on the unit circle (Figure 4.4). 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
4.3 The Method of Repeated Squares1
Computing large powers can be very time-consuming. Just as anyone can
compute 22 or 28 , everyone knows how to compute
1000000
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
1
The results in this section are needed only in Chapter 7.
4.3 THE METHOD OF REPEATED SQUARES 67
y
i
ω3 ω
−1 0 1 x
ω5 ω7
−i
Figure 4.4. 8th roots of unity
reasonable. It could be thousands or even millions of digits long. However,
if we could compute something like 237398332 (mod 46389), 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.
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)
..
.
k
a2 (mod n).
68 CHAPTER 4 CYCLIC GROUPS
Each step involves squaring the answer obtained in the previous step, divid-
ing by n, and taking the remainder.
Example 12. 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).
So it will suffice to compute 271 2i (mod 481) where i = 0, 6, 8. It is very
easy to see that
1
2712 ≡ 73, 441 (mod 481)
≡ 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)
≡ 1, 082, 411 (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. 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.
EXERCISES 69
Exercises
1. Prove or disprove each of the following statements.
(a) U (8) is cyclic.
(b) All of the generators of Z60 are prime.
(c) Q is cyclic.
(d) If every 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 (c) 3 ∈ R∗ (e) 72 in Z240
√ ∗
(b) 3 ∈ R (d) −i ∈ C (f) 312 in 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 6 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 .
70 CHAPTER 4 CYCLIC GROUPS
7. What are all of the cyclic subgroups of the quaternion group, Q8 ?
8. List all of the cyclic subgroups of U (30).
9. List every generator of each subgroup of order 8 in Z32 .
10. Find all elements of finite order in each of the following groups. Here the “∗”
indicates the set with zero removed.
(a) Z (b) Q∗ (c) R∗
11. If a24 = e in a group G, what are the possible orders of a?
12. Find a cyclic group with exactly one generator. Can you find cyclic groups
with exactly two generators? Four generators? How about n generators?
13. For n ≤ 20, which groups U (n) are cyclic? Make a conjecture as to what is
true in general. Can you prove your conjecture?
14. Let
0 1 0 −1
A= and B=
−1 0 1 −1
be elements in GL2 (R). Show that A and B have finite orders but AB does
not.
15. Evaluate each of the following.
(a) (3 − 2i) + (5i − 6) (d) (9 − i)(9 − i)
(b) (4 − 5i) − (4i − 4) (e) i45
(c) (5 − 4i)(7 + 2i) (f) (1 + i) + (1 + i)
16. Convert the following complex numbers to the form a + bi.
(a) 2 cis(π/6) (c) 3 cis(π)
(b) 5 cis(9π/4) (d) cis(7π/4)/2
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.
EXERCISES 71
(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|
2
(b) zz = |z| (e) |z − w| ≥ ||z| − |w||
−1 2
(c) z = z/|z| (f) |zw| = |z||w|
20. List and graph the 6th roots of unity. What are the generators of this group?
What are the primitive 6th roots of unity?
21. List and graph the 5th roots of unity. What are the generators of this group?
What are the primitive 5th roots of unity?
22. Calculate each of the following.
(a) 2923171 (mod 582) (c) 20719521 (mod 4724)
(b) 2557341 (mod 5681) (d) 971321 (mod 765)
23. Let a, b ∈ G. Prove the following statements.
(a) The order of a is the same as the order of a−1 .
(b) For all g ∈ G, |a| = |g −1 ag|.
(c) The order of ab is the same as the order of ba.
24. Let p and q be distinct primes. How many generators does Zpq have?
25. Let p be prime and r be a positive integer. How many generators does Zpr
have?
26. Prove that Zp has no nontrivial subgroups if p is prime.
27. If g and h have orders 15 and 16 respectively in a group G, what is the order
of hgi ∩ hhi?
28. Let a be an element in a group G. What is a generator for the subgroup
ham i ∩ han i?
29. Prove that Zn has an even number of generators for n > 2.
30. Suppose that G is a group and let a, b ∈ G. Prove that if |a| = m and |b| = n
with gcd(m, n) = 1, then hai ∩ hbi = {e}.
31. Let G be an abelian group. Show that the elements of finite order in G form
a subgroup. This subgroup is called the torsion subgroup of G.
72 CHAPTER 4 CYCLIC GROUPS
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. For what integers n is −1 an nth root of unity?
40. 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)].
41. Prove that the circle group is a subgroup of C∗ .
42. Prove that the nth roots of unity form a cyclic subgroup of T of order n.
43. Prove that αm = 1 and αn = 1 if and only if αd = 1 for d = gcd(m, n).
44. Let z ∈ C∗ . If |z| =
6 1, prove that the order of z is infinite.
45. Let z = cos θ + i sin θ be in T where θ ∈ Q. Prove that the order of z is
infinite.
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 re-
peated squares. What are the largest values of n and x that your program
will accept?
EXERCISES 73
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 Intro-
duction,” in Cryptology and Computational Number Theory, Pomerance, C.,
ed. Proceedings of Symposia in Applied Mathematics, 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.
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 triangle
4ABC from Chapter 3. The symmetries actually consist of permutations
of the three vertices, where a permutation of the set S = {A, B, C} is a
one-to-one and onto map π : S → S. The three vertices have the following
six permutations.
A B C A B C A B C
A B C C A B B C A
A B C A B C A B C
A C B C B A B A C
We have used the array
A B C
B C A
to denote the permutation that sends A to B, B to C, and C to A. That is,
A 7→ B
B 7→ C
C 7→ A.
The symmetries of a triangle form a group. In this chapter we will study
groups of this type.
74
5.1 DEFINITIONS AND NOTATION 75
5.1 Definitions and Notation
In general, the permutations of a set X form a group SX . If X is a finite
set, we can assume X = {1, 2, . . . , n}. In this case we write Sn instead of
SX . The following theorem says that Sn is a group. We call this group the
symmetric group on n letters.
Theorem 5.1 The symmetric group on n letters, Sn , is a group with n!
elements, where the binary operation is the composition of maps.
Proof. The identity of Sn is just the identity map that sends 1 to 1, 2 to
2, . . ., n to n. If f : Sn → Sn is a permutation, then f −1 exists, since f is
one-to-one and onto; hence, every permutation has an inverse. Composition
of maps is associative, which makes the group operation associative. We
leave the proof that |Sn | = n! as an exercise.
A subgroup of Sn is called a permutation group.
Example 1. 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. 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) =
76 CHAPTER 5 PERMUTATION GROUPS
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 2. Permutation multiplication is not usually commutative. Let
1 2 3 4
σ=
4 1 2 3
1 2 3 4
τ= .
2 1 4 3
Then
1 2 3 4
στ = ,
1 4 3 2
but
1 2 3 4
τσ = .
3 2 1 4
Cycle Notation
The notation that we have used to represent permutations up to this point is
cumbersome, to say the least. To work effectively with permutation groups,
we need a more streamlined method of writing down and manipulating per-
mutations.
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
5.1 DEFINITIONS AND NOTATION 77
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 3. 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 4. It is very easy to compute products of cycles. Suppose that
σ = (1352)
τ = (256).
We can 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
Hence, στ = (1356). If µ = (1634), then σµ = (1652)(34).
78 CHAPTER 5 PERMUTATION GROUPS
Two cycles in SX , σ = (a1 , a2 , . . . , ak ) and τ = (b1 , b2 , . . . , bl ), are dis-
joint if ai 6= bj for all i and j.
Example 5. The cycles (135) and (27) are disjoint; however, the cycles
(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 5.2 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 {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.
5.1 DEFINITIONS AND NOTATION 79
Theorem 5.3 Every permutation in Sn can be written as the product of
disjoint cycles.
Proof. We can assume that X = {1, 2, . . . , n}. Let σ ∈ Sn , and define X1
to be {σ(1), σ 2 (1), . . .}. The set X1 is finite since X is finite. Now let i be
the first integer in X that is not in X1 and define X2 by {σ(i), σ 2 (i), . . .}.
Again, X2 is a finite set. Continuing in this manner, we can define finite
disjoint sets X3 , X4 , . . .. Since X is a finite set, we are guaranteed that this
process will end and there will be only a finite number of these sets, say r.
If σi is the cycle defined by
σ(x) x ∈ Xi
σi (x) =
x x∈ / Xi ,
then σ = σ1 σ2 · · · σr . Since the sets X1 , X2 , . . . , Xr are disjoint, the cycles
σ1 , σ2 , . . . , σr must also be disjoint.
Example 6. 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. 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 ),
80 CHAPTER 5 PERMUTATION GROUPS
any cycle can be written as the product of transpositions, leading to the
following proposition.
Proposition 5.4 Any permutation of a finite set containing at least two
elements can be written as the product of transpositions.
Example 7. 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 prod-
uct of transpositions. For instance, we can write the identity permutation 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 5.5 If the identity is written as the product of r transpositions,
id = τ1 τ2 · · · τr ,
then r is an even number.
Proof. We will employ induction on r. A transposition cannot be the
identity; hence, r > 1. If r = 2, then we are done. Suppose that r > 2. In
this case the product of the last two transpositions, τr−1 τr , must be one of
the following cases:
(ab)(ab) = id
(bc)(ab) = (ac)(bc)
(cd)(ab) = (ab)(cd)
(ac)(ab) = (ab)(bc),
where a, b, c, and d are distinct.
5.1 DEFINITIONS AND NOTATION 81
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 transpo-
sitions for the identity. In this new product the last occurrence of a will
be in the next-to-the-last transposition. We can continue this process with
τr−2 τr−1 to obtain either a product of r − 2 transpositions or a new product
of r transpositions where the last occurrence of a is in τr−2 . If the identity is
the product of r −2 transpositions, then again we are done, by our induction
hypothesis; otherwise, we will repeat the procedure with τr−3 τr−2 .
At some point either we will have two adjacent, identical transpositions
canceling each other out or a will be shuffled so that it will appear only in
the first transposition. However, the latter case cannot occur, because the
identity would not fix a in this instance. Therefore, the identity permutation
must be the product of r − 2 transpositions and, again by our induction
hypothesis, we are done.
Theorem 5.6 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 −1 is σm · · · σ1 . Since
id = σσm · · · σ1 = τ1 · · · τn σm · · · σ1 ,
n must be even by Lemma 5.5. 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 5.6, 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.
82 CHAPTER 5 PERMUTATION GROUPS
The Alternating Groups
One of the most important subgroups of Sn is the set of all even permuta-
tions, An . The group An is called the alternating group on n letters.
Theorem 5.7 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 transpo-
sition is itself,
σ −1 = σr σr−1 · · · σ1
is also in An .
Proposition 5.8 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
λ σ : An → B n
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 8. The group A4 is the subgroup of S4 consisting of even permu-
tations. There are twelve elements in A4 :
(1) (12)(34) (13)(24) (14)(23)
(123) (132) (124) (142)
(134) (143) (234) (243).
5.2 THE DIHEDRAL GROUPS 83
One of the end-of-chapter exercises will be to write down all the subgroups
of A4 . You will find that there is no subgroup of order 6. Does this surprise
you?
Historical Note
Lagrange first thought of permutations as functions from a set to itself, but it was
Cauchy who developed the basic theorems and notation for permutations. He was
the first to use cycle notation. Augustin-Louis Cauchy (1789–1857) was born in
Paris at the height of the French Revolution. His family soon left Paris for the
village of Arcueil to escape the Reign of Terror. One of the family’s neighbors there
was Pierre-Simon Laplace (1749–1827), who encouraged him to seek a career in
mathematics. Cauchy began his career as a mathematician by solving a problem
in geometry given to him by Lagrange. Over 800 papers were written by Cauchy
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.
1
n 2
n−1 3
4
Figure 5.1. A regular n-gon
5.2 The Dihedral Groups
Another special type of permutation group is the dihedral group. Recall the
symmetry group of an equilateral triangle in Chapter 3. Such groups consist
of the rigid motions of a regular n-sided polygon or n-gon. For n = 3, 4, . . .,
we define the nth dihedral group to be the group of rigid motions of a
regular n-gon. We will denote this group by Dn . We can number the vertices
of a regular n-gon by 1, 2, . . . , n (Figure 5.1). Notice that there are exactly
84 CHAPTER 5 PERMUTATION GROUPS
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.
Theorem 5.9 The dihedral group, Dn , is a subgroup of Sn of order 2n.
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 5.2. Rotations and reflections of a regular n-gon
Theorem 5.10 The group Dn , n ≥ 3, consists of all products of the two
elements r and s, satisfying the relations
rn = id
s2 = id
srs = r−1 .
Proof. The possible motions of a regular n-gon are either reflections or
rotations (Figure 5.2). There are exactly n possible rotations:
360◦ 360◦ 360◦
id, ,2 · , . . . , (n − 1) · .
n n n
5.2 THE DIHEDRAL GROUPS 85
1 1
6 2 2 6
5 3 3 5
4 4
1 1
5 2 2 5
4 3 3 4
Figure 5.3. Types of reflections of a regular n-gon
We will denote the rotation 360◦ /n by r. The rotation r generates all of the
other rotations. That is,
360◦
rk = k · .
n
Label the n reflections s1 , s2 , . . . , sn , where sk is the reflection that leaves
vertex k fixed. There are two cases of reflection, depending on whether n
is even or odd. If there are an even number of vertices, then 2 vertices are
left fixed by a reflection. If there are an odd number of vertices, then only
a single vertex is left fixed by a reflection (Figure 5.3). In either case, the
order of sk is two. Let s = s1 . Then s2 = id and rn = id. 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−1 . If it is replaced by k − 1, then t = rk−1 s.
Hence, r and s generate Dn ; that is, Dn consists of all finite products of r
and s. We will leave the proof that srs = r−1 as an exercise.
Example 9. The group of rigid motions of a square, D4 , consists of eight
elements. With the vertices numbered 1, 2, 3, 4 (Figure 5.4), the rotations
86 CHAPTER 5 PERMUTATION GROUPS
1 2
4 3
Figure 5.4. The group D4
are
r = (1234)
r2 = (13)(24)
r3 = (1432)
r4 = id
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).
The Motion Group of a Cube
We can investigate the groups of rigid motions of geometric objects other
than a regular n-sided polygon to obtain interesting examples of permutation
groups. Let us consider the group of rigid motions of a cube. 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.
5.2 THE DIHEDRAL GROUPS 87
1 2
4 3
3 4
2 1
Figure 5.5. The motion group of a cube
Proposition 5.11 The group of rigid motions of a cube contains 24 ele-
ments.
Theorem 5.12 The group of rigid motions of a cube is S4 .
1 2 2 1
4 3 4 3
3 4 3 4
2 1 1 2
Figure 5.6. Transpositions in the motion group of a cube
Proof. From Proposition 5.11, we already know that the motion group of
the cube has 24 elements, the same number of elements as there are in S4 .
There are exactly four diagonals in the cube. If we label these diagonals 1,
2, 3, and 4, we must show that the motion group of the cube will give us
any permutation of the diagonals (Figure 5.5). 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 5.6). There are six such
88 CHAPTER 5 PERMUTATION GROUPS
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 .
Exercises
1. Write the following permutations in cycle notation.
(a) (c)
1 2 3 4 5 1 2 3 4 5
2 4 1 5 3 3 5 1 4 2
(b) (d)
1 2 3 4 5 1 2 3 4 5
4 2 5 1 3 1 4 3 2 5
2. Compute each of the following.
(a) (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)
4. Find (a1 , a2 , . . . , an )−1 .
5. List all of the subgroups of S4 . Find each of the following sets.
(a) {σ ∈ S4 : σ(1) = 3}
(b) {σ ∈ S4 : σ(2) = 2}
(c) {σ ∈ S4 : σ(1) = 3 and σ(2) = 2}
EXERCISES 89
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 .
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 5.5, 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 . 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 fol-
lowing permutations.
(a) (12), (13), . . . , (1n)
90 CHAPTER 5 PERMUTATION GROUPS
(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 .
(c) Define the orbit of x ∈ X under σ ∈ SX to be the set
Ox,σ = {y : x ∼ y}.
Compute the orbits of α, β, γ where
α = (1254)
β = (123)(45)
γ = (13)(25).
(d) If Ox,σ ∩ Oy,σ 6= ∅, prove that Ox,σ = Oy,σ . The orbits under a permu-
tation 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 hσi is transitive if and only if
Ox,σ = X for some x ∈ X.
EXERCISES 91
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 5.10.
(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 the-
ory, 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 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}.
92
6.1 COSETS 93
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 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)}.
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 6.1 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;
94 CHAPTER 6 COSETS AND LAGRANGE’S THEOREM
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 6.2 Let H be a subgroup of a group G. Then the left cosets of
H in G partition G. That is, the group G is the disjoint union of the left
cosets of H in G.
Proof. Let g1 H and g2 H be two cosets of H in G. We must show that
either g1 H ∩ g2 H = ∅ or g1 H = g2 H. Suppose that g1 H ∩ g2 H 6= ∅ and
a ∈ g1 H ∩ g2 H. Then by the definition of a left coset, a = g1 h1 = g2 h2
for some elements h1 and h2 in H. Hence, g1 = g2 h2 h−1
1 or g1 ∈ g2 H. By
Lemma 6.1, g1 H = g2 H.
Remark. 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 3. Let G = Z6 and H = {0, 3}. Then [G : H] = 3.
Example 4. Suppose that G = S3 , H = {(1), (123), (132)}, and K =
{(1), (12)}. Then [G : H] = 2 and [G : K] = 3.
Theorem 6.3 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 6.1,
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 6.1, g1 H = g2 H. The map φ is onto since φ(g −1 H) = Hg.
6.2 LAGRANGE’S THEOREM 95
6.2 Lagrange’s Theorem
Proposition 6.4 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 6.5 (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 6.6 Suppose that G is a finite group and g ∈ G. Then the order
of g must divide the number of elements in G.
Corollary 6.7 Let |G| = p with p a prime number. Then G is cyclic and
any g ∈ G such that g 6= e is a generator.
Proof. Let g be in G such that g 6= e. Then by Corollary 6.6, the order of
g must divide the order of the group. Since |hgi| > 1, it must be p. Hence,
g generates G.
Corollary 6.7 suggests that groups of prime order p must somehow look
like Zp .
Corollary 6.8 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|
96 CHAPTER 6 COSETS AND LAGRANGE’S THEOREM
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 a subgroup H such that |H| = 6
and show that a contradiction must occur. The group A4 contains eight
3-cycles; hence, H must contain a 3-cycle. We will show that if H contains
one 3-cycle, then it must contain every 3-cycle, contradicting the assumption
that H has only 6 elements.
Theorem 6.9 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 ) = b0 , then µ(b) = b0 . 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 .
Corollary 6.10 The group A4 has no subgroup of order 6.
Proof. Since [A4 : H] = 2, there are only two cosets of H in A4 . Inasmuch
as one of the cosets is H itself, right and left cosets must coincide; therefore,
gH = Hg or gHg −1 = H for every g ∈ A4 . By Theorem 6.9, if H contains
one 3-cycle, then it must contain every 3-cycle, contradicting the order of H.
6.3 FERMAT’S AND EULER’S THEOREMS 97
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 3.1, 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 6.11 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 6.12 (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 6.11 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 6.13 (Fermat’s Little Theorem) Let p be any prime number
and suppose that p6 |a. Then
ap−1 ≡ 1 (mod p).
Furthermore, for any integer b, bp ≡ b (mod p).
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, communi-
cated 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
98 CHAPTER 6 COSETS AND LAGRANGE’S THEOREM
proclaiming that the “greatest king in Europe” should have the “greatest mathe-
matician in Europe” at his court. For 20 years Lagrange held the position vacated
by his mentor, Euler. His works include contributions to number theory, group
theory, physics and mechanics, the calculus of variations, the theory of equations,
and differential equations. Along with Laplace and Lavoisier, Lagrange was one of
the people responsible for designing the metric system. During his life Lagrange
profoundly influenced the development of mathematics, leaving much to the next
generation of mathematicians in the form of examples and new problems to be
solved.
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) h8i in Z24 (e) An in Sn
(b) h3i in U (8) (f) D4 in S4
(c) 3Z in Z (g) T in C∗
(d) A4 in S4 (h) H = {(1), (123), (132)} in S4
6. Describe the left cosets of SL2 (R) in GL2 (R). What is the index of SL2 (R)
in GL2 (R)?
7. Verify Euler’s Theorem for n = 15 and a = 4.
8. Use Fermat’s Little Theorem to show that if p = 4n + 3 is prime, there is no
solution to the equation x2 ≡ −1 (mod p).
9. Show that the integers have infinite index in the additive group of rational
numbers.
10. Show that the additive group of real numbers has infinite index in the additive
group of the complex numbers.
11. Let H be a subgroup of a group G and suppose that g1 , g2 ∈ G. Prove that
the following conditions are equivalent.
(a) g1 H = g2 H
EXERCISES 99
(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.
13. What fails in the proof of Theorem 6.3 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. Modify the proof of Theorem 6.9 to show that any two permutations α, β ∈
Sn have the same cycle structure if and only if there exists a permutation
γ such that β = γαγ −1 . If β = γαγ −1 for some γ ∈ Sn , then α and β are
conjugate.
16. If |G| = 2n, prove that the number of elements of order 2 is odd. Use this
result to show that G must contain a subgroup of order 2.
17. Suppose that [G : H] = 2. If a and b are not in H, show that ab ∈ H.
18. If [G : H] = 2, prove that gH = Hg.
19. Let H and K be subgroups of a group G. Prove that gH ∩ gK is a coset of
H ∩ K in G.
20. Let H and K be subgroups of a group G. Define a relation ∼ on G by a ∼ b
if there exists an h ∈ H and a k ∈ K such that hak = b. Show that this
relation is an equivalence relation. The corresponding equivalence classes are
called double cosets. Compute the double cosets of H = {(1), (123), (132)}
in A4 .
21. If G is a group of order pn where p is prime, 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 ?
22. Let G be a cyclic group of order n. Show that there are exactly φ(n) gener-
ators for G.
23. Let n = pe11 pe22 · · · pekk be the factorization of n into distinct primes. Prove
that
1 1 1
φ(n) = n 1 − 1− ··· 1 − .
p1 p2 pk
24. Show that X
n= φ(d)
d|n
for all positive integers n.
7
Introduction to
Cryptography
Cryptography is the study of sending and receiving secret messages. The aim
of cryptography is to send messages across a channel so only the intended
recipient of the message can read it. In addition, when a message is received,
the recipient usually requires some assurance that the message is authentic;
that is, that it has not been sent by someone who is trying to deceive the
recipient. Modern cryptography is heavily dependent on abstract algebra
and number theory.
The message to be sent is called the plaintext message. The disguised
message is called the ciphertext. The plaintext and the ciphertext are both
written in an alphabet, consisting of letters or characters. Characters can
include not only the familiar alphabetic characters A, . . ., Z and a, . . ., z but
also digits, punctuation marks, and blanks. A cryptosystem, or cipher,
has two parts: encryption, the process of transforming a plaintext message
to a ciphertext message, and decryption, the reverse transformation of
changing a ciphertext message into a plaintext message.
There are many different families of cryptosystems, each distinguished
by a particular encryption algorithm. Cryptosystems in a specified cryp-
tographic family are distinguished from one another by a parameter to the
encryption function called a key. A classical cryptosystem has a single key,
which must be kept secret, known only to the sender and the receiver of
the message. If person A wishes to send secret messages to two different
people B and C, and does not wish to have B understand C’s messages or
vice versa, A must use two separate keys, so one cryptosystem is used for
exchanging messages with B, and another is used for exchanging messages
with C.
100
7.1 PRIVATE KEY CRYPTOGRAPHY 101
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 en-
coding key can be made public. A public key cryptosystem allows A and B
to send messages to C using the same encoding key. Anyone is capable of
encoding a message to be sent to C, but only C knows how to decode such
a message.
7.1 Private Key Cryptography
In single or private key cryptosystems the same key is used for both
encrypting and 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 at if only examples of
coded messages are available.
Example 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.
102 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
Cryptanalysis is concerned with deciphering a received or intercepted
message. Methods from probability and statistics are great aids in deci-
phering an intercepted message; for example, the frequency analysis of the
characters appearing in the intercepted message often makes its decryption
possible.
Example 2. Suppose we receive a message that we know was encrypted by
using a shift transformation on single letters of the 26-letter alphabet. To
find out exactly what the shift transformation was, we must compute b in
the equation f (p) = p + b mod 26. We can do this using frequency analysis.
The letter E = 04 is the most commonly occurring letter in the English
language. Suppose that S = 18 is the most commonly occurring letter in
the ciphertext. Then we have good reason to suspect that 18 = 4+b mod 26,
or b = 14. Therefore, the most likely encrypting function is
f (p) = p + 14 mod 26.
The corresponding decrypting function is
f −1 (p) = p + 12 mod 26.
It is now easy to determine whether or not our guess is correct.
Simple shift codes are examples of monoalphabetic cryptosystems.
In these ciphers a character in the enciphered message represents exactly
one character in the original message. Such cryptosystems are not very
sophisticated and are quite easy to break. In fact, in a simple shift as
described in Example 1, there are only 26 possible keys. It would be quite
easy to try them all rather than to use frequency analysis.
Let us investigate a slightly more sophisticated cryptosystem. Suppose
that the encoding function is given by
f (p) = ap + b mod 26.
We first need to find out when a decoding function f −1 exists. Such a
decoding function exists when we can solve the equation
c = ap + b mod 26
for p. By Proposition 3.1, 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.
7.1 PRIVATE KEY CRYPTOGRAPHY 103
Such a cryptosystem is called an affine cryptosystem.
Example 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 rep-
resent 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 4. Suppose that we wish to encode the word HELP. The corre-
sponding digit string is 7, 4, 11, 15. If
3 5
A= ,
1 2
then
−1 2 21
A = .
25 3
104 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
If b = (2, 2)t , then our message is encrypted as RRCR. The encrypted letter
R represents more than one plaintext letter.
Frequency analysis can still be performed on a polyalphabetic cryptosys-
tem, 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 4.
7.2 Public Key Cryptography
If traditional cryptosystems are used, anyone who knows enough to encode
a message will also know enough to decode an intercepted message. In 1976,
W. Diffie and M. Hellman proposed public key cryptography, which is based
on the observation that the encryption and decryption procedures need not
have the same key. This removes the requirement that the encoding key be
kept secret. The encoding function f must be relatively easy to compute,
but f −1 must be extremely difficult to compute without some additional
information, so that someone who knows only the encrypting key cannot
find the decrypting key without prohibitive computation. It is interesting
to note that to date, no system has been proposed that has been proven to
be “one-way;” that is, for any existing public key cryptosystem, it has never
been shown to be computationally prohibitive to decode messages with only
knowledge of the encoding key.
The RSA Cryptosystem
The RSA cryptosystem introduced by R. Rivest, A. Shamir, and L. Adleman
in 1978, is based on the difficulty of factoring large numbers. Though it is not
a difficult task to find two large random primes and multiply them together,
factoring a 150-digit number that is the product of two large primes would
take 100 million computers operating at 10 million instructions per second
about 50 million years under the fastest algorithms currently known.
The RSA cryptosystem works as follows. Suppose that we choose two
random 150-digit prime numbers p and q. Next, we compute the prod-
uct 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.
7.2 PUBLIC KEY CRYPTOGRAPHY 105
Suppose now that person B (Bob) wishes to send person A (Alice) a
message over a public line. Since E and n are known to everyone, anyone can
encode messages. Bob first digitizes the message according to some scheme,
say A = 00, B = 02, . . . , Z = 25. If necessary, he will break the message into
pieces such that each piece is a positive integer less than n. Suppose x is
one of the pieces. Bob forms the number y = xE mod n and sends y to
Alice. For Alice to recover x, she need only compute x = y D mod n. Only
Alice knows D.
Example 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 23. 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
23487 mod 667 = 368.
This computation can be reasonably done by using the method of repeated
squares as described in Chapter 4. Using the Euclidean algorithm, we de-
termine that 191E = 1 + 151m; therefore, the decrypting key is (n, D) =
(667, 191). We can recover the original message by calculating
368191 mod 667 = 23.
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.
By Theorem 6.12,
y D = (xE )D = xDE = xkm+1 = (xφ(n) )k x = 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 5, we could have recovered D.
106 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
Message Verification
There is a problem of message verification in public key cryptosystems.
Since the encoding key is public knowledge, anyone has the ability to send
an encoded message. If Alice receives a message from Bob, she would like
to be able to verify that it was Bob who actually sent the message. Sup-
pose that Bob’s encrypting key is (n0 , E 0 ) and his decrypting key is (n0 , D0 ).
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:
0
x0 = xD mod n0 .
Anyone can change x0 back to x just by encryption, but only Bob has the
ability to form x0 . Now Bob encrypts x0 with Alice’s encryption key to form
E
y 0 = x0 mod n,
a message that only Alice can decode. Alice decodes the message and then
encodes the result with Bob’s key to read the original message, a message
that could have only been sent by Bob.
Historical Note
Encrypting secret messages goes as far back as ancient Greece and Rome. As we
know, Julius Caesar used a simple shift code to send and receive messages. However,
the formal study of encoding and decoding messages probably began with the Arabs
in the 1400s. In the fifteenth and sixteenth centuries mathematicians such as Alberti
and Viete discovered that monoalphabetic cryptosystems offered no real security.
In the 1800s, F. W. Kasiski established methods for breaking ciphers in which
a ciphertext letter can represent more than one plaintext letter, if the same key
was used several times. This discovery led to the use of cryptosystems with keys
that were used only a single time. Cryptography was placed on firm mathematical
foundations by such people as W. Friedman and L. Hill in the early part of the
twentieth century.
During World War II mathematicians were very active in cryptography. Efforts
to penetrate the cryptosystems of the Axis nations were organized in England and in
the United States by such notable mathematicians as Alan Turing and A. A. Albert.
The period after World War I saw the development of special-purpose machines for
encrypting and decrypting messages. 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.
EXERCISES 107
By the 1970s, interest in commercial cryptography had begun to take hold.
There was a growing need to protect banking transactions, computer data, and
electronic mail. In the early 1970s, IBM developed and implemented LUZIFER,
the forerunner of the National Bureau of Standards’ Data Encryption Standard
(DES).
The concept of a public key cryptosystem, due to Diffie and Hellman, is very
recent (1976). It was further developed by Rivest, Shamir, and Adleman with the
RSA cryptosystem (1978). It is not known how secure any of these systems are.
The trapdoor knapsack cryptosystem, developed by Merkle and Hellman, has been
broken. It is still an open question whether or not the RSA system can be broken.
At the time of the writing of this book, the largest number factored is 135 digits
long, and at the present moment a code is considered secure if the key is about
400 digits long and is the product of two 200-digit primes. There has been a great
deal of controversy about research in cryptography in recent times: the National
Security Agency would like to keep information about cryptography secret, whereas
the academic community has fought for the right to publish basic research.
Modern cryptography has come a long way since 1929, when Henry Stimson,
Secretary of State under Herbert Hoover, dismissed the Black Chamber (the State
Department’s cryptography division) in 1929 on the ethical grounds that “gentle-
men do not read each other’s mail.”
Exercises
1. Encode IXLOVEXMATH using the cryptosystem in Example 1.
2. Decode ZLOOA WKLVA EHARQ WKHA ILQDO, which was encoded using
the cryptosystem in Example 1.
3. Assuming that monoalphabetic code was used to encode the following secret
message, what was the original message?
NBQFRSMXZF YAWJUFHWFF ESKGQCFWDQ AFNBQFTILO FCWP
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 CRYP-
TOLOGY, 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.
108 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
(a) n = 3551, E = 629, x = 31
(b) n = 2257, E = 47, x = 23
(c) n = 120979, E = 13251, x = 142371
(d) n = 45629, E = 781, x = 231561
8. Compute the decoding key D for each of the encoding keys in Exercise 7.
9. Decrypt each of the following RSA messages y.
(a) n = 3551, D = 1997, y = 2791
(b) n = 5893, D = 81, y = 34
(c) n = 120979, D = 27331, y = 112135
(d) n = 79403, D = 671, y = 129381
10. For each of the following encryption keys (n, E) in the RSA cryptosystem,
compute D.
(a) (n, E) = (451, 231)
(b) (n, E) = (3053, 1921)
(c) (n, E) = (37986733, 12371)
(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.
Additional Exercises: Primality and Factoring
In the RSA cryptosystem it is important to be able to find large prime numbers
easily. Also, this cryptosystem is not secure if we can factor a composite number
that is the product of two large primes. The solutions to both of these problems
are quite easy. To find out if a number n is prime
√ or to factor n, we can use trial
division. We simply divide n by d = 2, 3, . . . , n. Either a factorization will be
obtained, or n is prime if no d divides n. The problem is that such a computation
is prohibitively time-consuming if n is very large.
EXERCISES 109
1. A better algorithm for factoring odd positive integers is Fermat’s factor-
ization 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 .
(b) Write a program to implement the following factorization algorithm
based on the observation in part (a).
√
x ← d ne
y←1
1: while x2 − y 2 > n do
y ←y+1
if x2 − y 2 < n then
x←x+1
y←1
goto 1
else if x2 − y 2 = 0 then
a←x−y
b←x+y
write n = a ∗ b
√
The expression d n e 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?
2. Primality Testing. Recall Fermat’s Little Theorem from Chapter 6. Let p
be prime with gcd(a, p) = 1. Then ap−1 ≡ 1 (mod p). We can use Fermat’s
Little Theorem as a screening test for primes. For example, 15 cannot be
prime since
215−1 ≡ 214 ≡ 4 (mod 15).
However, 17 is a potential prime since
217−1 ≡ 216 ≡ 1 (mod 17).
We say that an odd composite number n is a pseudoprime if
2n−1 ≡ 1 (mod n).
Which of the following numbers are primes and which are pseudoprimes?
110 CHAPTER 7 INTRODUCTION TO CRYPTOGRAPHY
(a) 342 (c) 601 (e) 771
(b) 811 (d) 561 (f) 631
3. Let n be an odd composite number and b be a positive integer such that
gcd(b, n) = 1. If bn−1 ≡ 1 (mod n), then n is a pseudoprime base b.
Show that 341 is a pseudoprime base 2 but not a pseudoprime base 3.
4. Write a program to determine all primes less than 2000 using trial division.
Write a second program that will determine all numbers less than 2000 that
are either primes or pseudoprimes. Compare the speed of the two programs.
How many pseudoprimes are there below 2000?
There exist composite numbers that are pseudoprimes for all bases to which
they are relatively prime. These numbers are called Carmichael num-
bers. The first Carmichael number is 561 = 3 · 11 · 17. In 1992, Al-
ford, Granville, and Pomerance proved that there are an infinite number of
Carmichael numbers [4]. However, Carmichael numbers are very rare. There
are only 2163 Carmichael numbers less than 25 × 109 . For more sophisticated
primality tests, see [1], [6], or [7].
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. “A New Kind of Cipher that Would Take a Million Years to
BREAK,” Scientific American 237 (1977), 120–24.
[4] Granville, A. “Primality Testing and Carmichael Numbers,” Notices of the
American Mathematical Society 39(1992), 696–700.
[5] Hellman, M. E. “The Mathematics of Public Key Cryptography,” Scientific
American 241 (1979), 130–39.
[6] Koblitz, N. A Course in Number Theory and Cryptography. 2nd ed. Springer,
New York, 1994.
[7] Pomerance, C., ed. Cryptology and Computational Number Theory. Proceed-
ings of Symposia in Applied Mathematics, vol. 42. American Mathematical
Society, Providence, RI, 1990.
[8] Rivest, R. L., Shamir, A., and Adleman, L., “A Method for Obtaining Sig-
natures and Public-key Cryptosystems,” Comm. ACM 21(1978), 120–26.
8
Algebraic Coding Theory
Coding theory is an application of algebra that has become increasingly
important over the last several decades. When we transmit data, we are
concerned about sending a message over a channel that could be affected
by “noise.” We wish to be able to encode and decode the information in a
manner that will allow the detection, and possibly the correction, of errors
caused by noise. This situation arises in many areas of communications,
including radio, telephone, television, computer communications, and even
compact disc player technology. Probability, combinatorics, group theory,
linear algebra, and polynomial rings over finite fields all play important roles
in coding theory.
8.1 Error-Detecting and Correcting Codes
Let us examine a simple model of a communications system for transmitting
and receiving coded messages (Figure 8.1).
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 occur-
rence of errors during transmission. An error occurs if there is a change in
one or more bits in the codeword. A decoding scheme is a method that
either converts an arbitrarily received n-tuple into a meaningful decoded
message or gives an error message for that n-tuple. If the received message
is a codeword (one of the special n-tuples allowed to be transmitted), then
the decoded message must be the unique message that was encoded into the
codeword. For received non-codewords, the decoding scheme will give an
error indication, or, if we are more clever, will actually try to correct the
111
112 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
error and reconstruct the original message. Our goal is to transmit error-free
messages as cheaply and quickly as possible.
Example 1. One possible coding scheme would be to send a message several
times and to compare the received copies with one another. Suppose that
the message to be encoded is a binary n-tuple (x1 , x2 , . . . , xn ). The message
is encoded into a binary 3n-tuple by simply repeating the message three
times:
(x1 , x2 , . . . , xn ) 7→ (x1 , x2 , . . . , xn , x1 , x2 , . . . , xn , x1 , x2 , . . . , xn ).
To decode the message, we choose as the ith digit the one that appears
in the ith place in at least two of the three transmissions. For exam-
ple, 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 de-
8.1 ERROR-DETECTING AND CORRECTING CODES 113
coded as (0110).1 This triple-repetition method will automatically detect
and correct all single errors, but it is slow and inefficient: to send a mes-
sage 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 2. 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, yield-
ing 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 char-
acters 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
1
We will adopt the convention that bits are numbered left to right in binary n-tuples.
114 CHAPTER 8 ALGEBRAIC CODING THEORY
1’s, 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 informa-
tion 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 1’s 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 1’s 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 1’s.)
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.
Received Word
000 001 010 011 100 101 110 111
Transmitted 000 0 1 1 2 1 2 2 3
Codeword 111 3 2 2 1 2 1 1 0
Table 8.1. A repetition code
Example 3. 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.
In Table 8.1, we present all possible words that might be received for the
transmitted codewords (000) and (111). Table 8.1 also shows the number of
bits by which each received 3-tuple differs from each original codeword.
8.1 ERROR-DETECTING AND CORRECTING CODES 115
Maximum-Likelihood Decoding
The coding scheme presented in Example 3 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.
p
0 0
q
q
1 p 1
Figure 8.2. Binary symmetric channel
A binary symmetric channel is a model that consists of a transmitter
capable of sending a binary signal, either a 0 or a 1, together with a receiver.
Let p be the probability that the signal is correctly received. Then q =
1 − p is the probability of an incorrect reception. If a 1 is sent, then the
probability that a 1 is received is p and the probability that a 0 is received is
q (Figure 8.2). The probability that no errors occur during the transmission
of a binary codeword of length n is pn . For example, if p = 0.999 and a
message consisting of 10,000 bits is sent, then the probability of a perfect
transmission is
(0.999)10,000 ≈ 0.00005.
116 CHAPTER 8 ALGEBRAIC CODING THEORY
Theorem 8.1 If a binary n-tuple (x1 , . . . , xn ) is transmitted across a binary
symmetric channel with probability p that no error will occur in each coor-
dinate, 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 4. 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
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.
8.1 ERROR-DETECTING AND CORRECTING CODES 117
Block Codes
If we are to develop efficient error-detecting and error-correcting codes, we
will need more sophisticated mathematical tools. Group theory will allow
faster methods of encoding and decoding messages. A code is an (n, m)-
block code if the information that is to be coded can be divided into blocks
of m binary digits, each of which can be encoded into n binary digits. More
specifically, an (n, m)-block code consists of an encoding function
E : Zm n
2 → Z2
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 5. The even-parity coding system developed to detect single
errors in ASCII characters is an (8, 7)-block code. The encoding function is
E(x7 , x6 , . . . , x1 ) = (x8 , x7 , . . . , x1 ),
where x8 = x7 + x6 + · · · + x1 with addition in Z2 .
Let x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) be binary n-tuples. The
Hamming distance or distance, d(x, y), between x and y is the number
of bits in which x and y differ. The distance between two codewords is the
minimum number of transmission errors required to change one codeword
into the other. The minimum distance for a code, dmin , is the minimum
of all distances d(x, y), where x and y are distinct codewords. The weight,
w(x), of a binary codeword x is the number of 1’s in x. Clearly, w(x) =
d(x, 0), where 0 = (00 · · · 0).
Example 6. 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.
118 CHAPTER 8 ALGEBRAIC CODING THEORY
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 8.2 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 8.2 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-correcting 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 8.2. Distances between 4-bit codewords
8.1 ERROR-DETECTING AND CORRECTING CODES 119
To determine exactly what the error-detecting and error-correcting ca-
pabilities 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 cor-
rects all single errors. Starting with a codeword x, an error in the transmis-
sion of a single bit gives y with d(x, y) = 1, but d(z, y) ≥ 2 for any other
codeword z 6= x. If we do not require the correction of errors, then we can
detect multiple errors when a code has a minimum distance that is greater
than 3.
Theorem 8.3 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 7. In Table 8.3, the codewords c1 = (00000), c2 = (00111),
c3 = (11100), and c4 = (11011) determine a single error-correcting code.
Historical Note
Modern coding theory began in 1948 with C. Shannon’s paper, “A Mathematical
Theory of Information” [7]. This paper offered an example of an algebraic code, and
120 CHAPTER 8 ALGEBRAIC CODING THEORY
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 8.3. Hamming distances for an error-correcting code
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 years. The Theory of Error-Correcting Codes,
by MacWilliams and Sloane [5], published in 1977, already contained over 1500
references. Linear codes (Reed-Muller (32, 6)-block codes) were used on NASA’s
Mariner space probes. More recent space probes such as Voyager have used what
are called convolution codes. Currently, very active research is being done with
Goppa codes, which are heavily dependent on algebraic geometry.
8.2 Linear Codes
To gain more knowledge of a particular code and develop more efficient tech-
niques of encoding, decoding, and error detection, we need to add additional
structure to our codes. One way to accomplish this is to require that the
code also be a group. A group code is a code that is also a subgroup of
Zn2 .
To check that a code is a group code, we need only verify one thing. If
we add any two elements in the code, the result must be an n-tuple that is
again in the code. It is not necessary to check that the inverse of the n-tuple
is in the code, since every codeword is its own inverse, nor is it necessary to
check that 0 is a codeword. For instance,
(11000101) + (11000101) = (00000000).
Example 8. 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).
8.2 LINEAR CODES 121
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 8.4 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 8.5 Let dmin be the minimum distance for a group code C. Then
dmin is the minimum of all the nonzero weights of the nonzero codewords in
C. That is,
dmin = min{w(x) : x 6= 0}.
Proof. Observe that
dmin = min{d(x, y) : x 6= y}
= min{d(x, y) : x + y 6= 0}
= min{w(x + y) : x + y 6= 0}
= min{w(z) : z 6= 0}.
122 CHAPTER 8 ALGEBRAIC CODING THEORY
Linear Codes
From Example 8, 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.2
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 9. Suppose that the words to be encoded consist of all binary
3-tuples and that our encoding scheme is even-parity. To encode an arbitrary
3-tuple, we add a fourth bit to obtain an even number of 1’s. Notice that
an arbitrary n-tuple x = (x1 , x2 , . . . , xn )t has an even number of 1’s exactly
when x1 + x2 + · · · + xn = 0; hence, a 4-tuple x = (x1 , x2 , x3 , x4 )t has an
even number of 1’s if x1 + x2 + x3 + x4 = 0, or
1
1
x · 1 = xt 1 = x 1 x 2 x 3 x 4
1 = 0.
1
This example leads us to hope that there is a connection between matrices
and coding theory.
2
Since we will be working with matrices, we will write binary n-tuples as column vectors
for the remainder of this chapter.
8.2 LINEAR CODES 123
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 10. Suppose that
0 1 0 1 0
H = 1 1 1 1 0 .
0 0 1 1 1
For a 5-tuple x = (x1 , x2 , x3 , x4 , x5 )t to be in the null space of H, Hx = 0.
Equivalently, the following system of equations must be satisfied:
x2 + x4 = 0
x1 + x2 + x3 + x4 = 0
x3 + x4 + x5 = 0.
The set of binary 5-tuples satisfying these equations is
(00000) (11110) (10101) (01011).
This code is easily determined to be a group code.
Theorem 8.6 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) = 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 11. 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
124 CHAPTER 8 ALGEBRAIC CODING THEORY
Suppose that the 7-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
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
8.3 PARITY-CHECK AND GENERATOR MATRICES 125
Our goal will be to show that Gx = y if and only if Hy = 0. Given a
message block x to be encoded, G will allow us to quickly encode it into a
linear codeword y.
Example 12. Suppose that we have the following eight words to be en-
coded:
(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 1’s in the identity matrix serve as parity checks
for the 1’s 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
126 CHAPTER 8 ALGEBRAIC CODING THEORY
from having to check on each other. Hence, x1 , x2 , and x3 can be arbitrary
but x4 , x5 , and x6 must be chosen to ensure parity. The null space of H is
easily computed to be
(000000) (001101) (010110) (011011)
(100011) (101110) (110101) (111000).
An even easier way to compute the null space is with the generator matrix
G (Table 8.4).
Message Word Codeword
x Gx
000 000000
001 001101
010 010110
011 011011
100 100011
101 101110
110 110101
111 111000
Table 8.4. A matrix-generated code
Theorem 8.7 Let H ∈ Mm×n (Z2 ) be 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 12, the first three bits are the information
bits and the last three are the check bits.
Theorem 8.8 Suppose that G is an n × k standard generator matrix. Then
C = y : Gx = y for x ∈ Zk2 is an (n, k)-block code. More specifically, C
is a group code.
Proof. Let Gx1 = y1 and Gx2 = y2 be two codewords. Then y1 + y2 is
in C since
G(x1 + x2 ) = Gx1 + Gx2 = y1 + y2 .
8.3 PARITY-CHECK AND GENERATOR MATRICES 127
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 ma-
trices and standard generating matrices, we need to prove a lemma.
H = (A | Im ) be an m × n canonical parity-check matrix
Lemma 8.9 Let
In−m
and G = A be the corresponding n×(n−m) standard generator matrix.
Then HG = 0.
Proof. Let C = HG. The ijth entry in C is
n
X
cij = hik gkj
k=1
n−m
X n
X
= hik gkj + hik gkj
k=1 k=n−m+1
n−m
X Xn
= aik δkj + δi−(m−n),k akj
k=1 k=n−m+1
= aij + aij
= 0,
where (
1, i=j
δij =
0, i 6= j
is the Kronecker delta.
Theorem 8.10 Let H = (A | Im ) be an m×n canonical parity-check matrix
In−m
and let G = A be the n × (n − m) standard generator matrix associated
with H. Let C be the code generated by G. Then y is in C if and only if
Hy = 0. In particular, C is a linear code with canonical parity-check matrix
H.
128 CHAPTER 8 ALGEBRAIC CODING THEORY
Proof. First suppose that y ∈ C. Then Gx = y for some x ∈ Zm 2 . By
Lemma 8.9, Hy = HGx = 0.
Conversely, suppose that y = (y1 , . . . , yn )t is in the null space of H. We
need to find an x in Zn−m
2 such that Gxt = y. Since Hy = 0, the following
set of equations must be satisfied:
a11 y1 + a12 y2 + · · · + a1,n−m yn−m + yn−m+1 = 0
a21 y1 + a22 y2 + · · · + a2,n−m yn−m + yn−m+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
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 13. Observe that
0
1
1 1 1 0 0 1
1 0 0 1 0 0 = 0 .
1 1 0 0 1 0 1
0
8.3 PARITY-CHECK AND GENERATOR MATRICES 129
We state this result in the following proposition and leave the proof as
an exercise.
Proposition 8.11 Let ei be the binary n-tuple with a 1 in the ith coordinate
and 0’s elsewhere and suppose that H ∈ Mm×n (Z2 ). Then Hei is the ith
column of the matrix H.
Theorem 8.12 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 Propo-
sition 8.11, Hei 6= 0.
Example 14. 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 8.12. This theorem gives us con-
ditions on a matrix H that tell us when the minimum weight of the code
130 CHAPTER 8 ALGEBRAIC CODING THEORY
formed by the null space of H is 2. We can also determine when the mini-
mum distance of a linear code is 3 by examining the corresponding matrix.
Example 15. If we let
1 1 1 0
H = 1 0 0 1
1 1 0 0
and want to determine whether or not H is the canonical parity-check matrix
for an error-correcting code, it is necessary to make certain that Null(H)
does not contain any 4-tuples of weight 2. That is, (1100), (1010), (1001),
(0110), (0101), and (0011) must not be in Null(H). The next theorem
states that we can indeed determine that the code generated by H is error-
correcting by examining the columns of H. Notice in this example that not
only does H have no zero columns, but also that no two columns are the
same.
Theorem 8.13 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 1’s in the ith and jth entries and 0’s
elsewhere, and w(ei + ej ) = 2 for i 6= j. Since
0 = H(ei + ej ) = Hei + Hej
can only occur if the ith and jth columns are identical, the null space of H
is a single error-correcting code.
Suppose now that we have a canonical parity-check matrix H with three
rows. Then we might ask how many more columns we can add to the
matrix and still have a null space that is a single error-detecting and single
error-correcting code. Since each column has three entries, there are 23 = 8
possible distinct columns. We cannot add the columns
0 1 0 0
0 , 0 , 1 , 0 .
0 0 0 1
So we can add as many as four columns and still maintain a minimum
distance of 3.
8.4 EFFICIENT DECODING 131
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 , . . . , en be excluded, leaving 2m − (1 + n) remaining columns
for information if we are still to maintain the ability not only to detect but
also to correct single errors.
8.4 Efficient Decoding
We are now at the stage where we are able to generate linear codes that
detect and correct errors fairly easily, but it is still a time-consuming process
to decode a received n-tuple and determine which is the closest codeword,
because the received n-tuple must be compared to each possible codeword
to determine the proper decoding. This can be a serious impediment if the
code is very large.
Example 16. Given the binary matrix
1 1 1 0 0
H = 0 1 0 1 0
1 0 0 0 1
and the 5-tuples x = (11011)t and y = (01011)t , we can compute
0 1
Hx = 0 and Hy = 0 .
0 1
Hence, x is a codeword and y is not, since x is in the null space and y is
not. Notice that Hx 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 ∈ Zn2 , then we say that the syndrome of
x is Hx. The following proposition allows the quick detection and correction
of errors.
Proposition 8.14 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.
132 CHAPTER 8 ALGEBRAIC CODING THEORY
Proof. 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 8.14 and from the
fact that He is the ith column of the matrix H.
Theorem 8.15 Let H ∈ Mm×n (Z2 ) and suppose that the linear code cor-
responding 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 17. Consider the matrix
1 0 1 1 0 0
H = 0 1 1 0 1 0
1 1 1 0 0 1
and suppose that the 6-tuples x = (111110)t , y = (111111)t , and z =
(010111)t have been received. Then
1 1 1
Hx = 1 , Hy = 1 , Hz = 0 .
1 0 0
Hence, x has an error in the third bit and z has an error in the fourth bit. The
transmitted codewords for x and z must have been (110110) and (010011),
respectively. The syndrome of y does not occur in any of the columns of the
matrix H, so multiple errors must have occurred to produce y.
Coset Decoding
We can use group theory to obtain another way of decoding messages. A
linear code C is a subgroup of Zn2 . Coset or standard decoding uses
the cosets of C in Zn2 to implement maximum-likelihood decoding. Suppose
that C is an (n, m)-linear code. A coset of C in Zn2 is written in the form
8.4 EFFICIENT DECODING 133
Cosets
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 8.5. Cosets of C
x + C, where x ∈ Zn2 . By Lagrange’s Theorem (Theorem 6.5), there are
2n−m distinct cosets of C in Zn2 .
Example 18. 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 8.5.
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 19. In Table 8.5, notice that we have chosen a representative
of the least possible weight for each coset. These representatives are coset
leaders. Now suppose that r = (01111) is the received word. To decode r,
134 CHAPTER 8 ALGEBRAIC CODING THEORY
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 proposi-
tion 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 des-
ignates a coset leader corresponding to each syndrome. 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 8.6. Syndromes for each coset
Proposition 8.16 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 20. Table 8.6 is a decoding table for the code C given in Exam-
ple 18. 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
EXERCISES 135
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.
Exercises
1. Why is the following encoding scheme not acceptable?
Information: 0 1 2 3 4 5 6 7 8
Codeword: 000 001 010 011 101 110 111 000 001
2. Without doing any addition, explain why the following set of 4-tuples in Z42
cannot be a group code.
(0110) (1001) (1010) (1100)
3. Compute the Hamming distances between the following pairs of n-tuples.
(a) (011010), (011100) (c) (00110), (01111)
(b) (11110101), (01010100) (d) (1001), (0111)
4. Compute the weights of the following n-tuples.
(a) (011010) (c) (01111)
(b) (11110101) (d) (1011)
5. Suppose that a linear code C has a minimum weight of 7. What are the
error-detection and error-correction capabilities of C?
6. In each of the following codes, what is the minimum distance for the code?
What is the best situation we might hope for in connection with error detec-
tion 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?
136 CHAPTER 8 ALGEBRAIC CODING THEORY
(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
(d)
(b)
0 0 0 1 1 1 1
1 0 1 0 0 0 0
1 1 1 0 0 1 1
1 0 1 0 0
1
0
0 1 0 1 0 1
1 0 0 1 0
0 1 1 0 0 1 1
1 1 0 0 0 1
8. Construct a (5, 2)-block code. Discuss the error-detection and error-correction
capabilities of your code.
9. Let C be the code obtained from the null space of the matrix
0 1 0 0 1
H = 1 0 1 0 1 .
0 0 1 1 1
Decode the message
01111 10101 01110 00011
if possible.
10. Suppose that a 1000-bit binary message is transmitted. Assume that the
probability of a single error is p and that the errors occurring in different
bits are independent of one another. If p = 0.01, what is the probability of
more than one error occurring? What is the probability of exactly two errors
occurring? Repeat this problem for p = 0.0001.
11. Which matrices are canonical parity-check matrices? For those matrices that
are canonical parity-check matrices, what are the corresponding standard
generator matrices? What are the error-detection and error-correction capa-
bilities 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 0 0 1 0 0 0
0 1 1 0 0 0 0 1 1 0 1 0 0
1 1 0 1 0 0
1 0 1 0 0 1 0
0 1 0 0 1 0 0 1 1 0 0 0 1
1 1 0 0 0 1
EXERCISES 137
12. List all possible syndromes for the codes generated by each of the matrices
in the previous exercise.
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 H in Z32 . Why was there no need to specify right or
left cosets? Give the single transmission error, if any, to which each coset
corresponds.
15. For each of the following matrices, find the cosets of the corresponding code
C. Give a decoding table for each code if possible.
(a) (c)
0 1 0 0 0 1 0 0 1 1
1 0 1 0 1 0 1 0 1 1
1 0 0 1 0
(d)
(b)
0 0 1 0 0 1 0 0 1 1 1 1
1 1 0 1 0 1 1 1 0 0 1 1
0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 1 1 0 0 1 0
16. Let x, y, and z be binary n-tuples. Prove each of the following statements.
(a) w(x) = d(x, 0)
(b) d(x, y) = d(x + z, y + z)
(c) d(x, y) = w(x − y)
17. A metric on a set X is a map d : X × X → R satisfying the following
conditions.
(a) d(x, y) ≥ 0 for all x, y ∈ X;
(b) d(x, y) = 0 exactly when x = y;
(c) d(x, y) = d(y, x);
138 CHAPTER 8 ALGEBRAIC CODING THEORY
(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 matrix for seven infor-
mation 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.
EXERCISES 139
(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 information
positions and four check positions. What are the check positions and
what are the information positions? Encode the messages (101101) and
(001001). Decode the received words (0010000101) and (0000101100).
What are the possible syndromes for this code?
(d) What is the number of check bits and the number of information bits
in an (m, n)-block Hamming code? Give both an upper and a lower
bound on the number of information bits in terms of the number of
check bits. Hamming codes having the maximum possible number of
information bits with k check bits are called 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.
Programming Exercises
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.
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.
140 CHAPTER 8 ALGEBRAIC CODING THEORY
[3] Levinson, N. “Coding Theory: A Counterexample to G. H. Hardy’s Concep-
tion of Applied Mathematics,” American Mathematical Monthly 77 (1970),
249–58.
[4] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York,
1998.
[5] MacWilliams, F. J. and Sloane, N. J. A. The Theory of Error-Correcting
Codes. North-Holland Mathematical Library, 16, Elsevier, Amsterdam, 1983.
[6] Roman, S. Coding and Information Theory. Springer-Verlag, New York,
1992.
[7] Shannon, C. E. “A Mathematical Theory of Communication,” Bell System
Technical Journal 27 (1948), 379–423, 623–56.
[8] Thompson, T. M. From Error-Correcting Codes through Sphere Packing to
Simple Groups. Carus Monograph Series, No. 21. Mathematical Association
of America, Washington, DC, 1983.
[9] van Lint, J. H. Introduction to Coding Theory. Springer, New York, 1999.
9
Isomorphisms
Many groups may appear to be different at first glance, but can be shown
to be the same by a simple renaming of the group elements. For example,
Z4 and the subgroup of the circle group T generated by i can be shown
to be the same by demonstrating a one-to-one correspondence between the
elements of the two groups and between the group operations. In such a
case we say that the groups are isomorphic.
9.1 Definition and Examples
Two groups (G, ·) and (H, ◦) are isomorphic if there exists a one-to-one
and onto map φ : G → H such that the group operation is preserved; that is,
φ(a · b) = φ(a) ◦ φ(b)
for all a and b in G. If G is isomorphic to H, we write G ∼
= H. The map φ
is called an isomorphism.
Example 1. To show that Z4 ∼ = hii, define a map φ : Z4 → hii by φ(n) = in .
We must show that φ is bijective and preserves the group operation. The
map φ is one-to-one and onto because
φ(0) = 1
φ(1) = i
φ(2) = −1
φ(3) = −i.
Since
φ(m + n) = im+n = im in = φ(m)φ(n),
141
142 CHAPTER 9 ISOMORPHISMS
the group operation is preserved.
Example 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.
Example 3. The integers are isomorphic to the subgroup of Q∗ consisting
of elements of the form 2n . Define a map φ : Z → Q∗ by φ(n) = 2n . Then
φ(m + n) = 2m+n = 2m 2n = φ(m)φ(n).
By definition the map φ is onto the subset {2n : n ∈ Z} of Q∗ . To show that
the map is injective, assume that m 6= n. If we can show that φ(m) 6= φ(n),
then we are done. Suppose that m > n and assume that φ(m) = φ(n). Then
2m = 2n or 2m−n = 1, which is impossible since m − n > 0.
Example 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 14 in Chapter 3).
Example 5. Even though S3 and Z6 possess the same number of elements,
we would suspect that they are not isomorphic, because Z6 is abelian and
9.1 DEFINITION AND EXAMPLES 143
S3 is nonabelian. To demonstrate that this is indeed the case, suppose that
φ : Z6 → S3 is an isomorphism. Let a, b ∈ S3 be two elements such that
ab 6= ba. Since φ is an isomorphism, there exist elements m and n in Z6
such that
φ(m) = a and φ(n) = b.
However,
ab = φ(m)φ(n) = φ(m + n) = φ(n + m) = φ(n)φ(m) = ba,
which contradicts the fact that a and b do not commute.
Theorem 9.1 Let φ : G → H be an isomorphism of two groups. Then the
following statements are true.
1. φ−1 : H → G is an isomorphism.
2. |G| = |H|.
3. If G is abelian, then H is abelian.
4. If G is cyclic, then H is cyclic.
5. If G has a subgroup of order n, then H has a subgroup of order n.
Proof. Assertions (1) and (2) follow from the fact that φ is a bijection.
We will prove (3) here and leave the remainder of the theorem to be proved
in the exercises.
(3) Suppose that h1 and h2 are elements of H. Since φ is onto, there
exist elements 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 9.2 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).
144 CHAPTER 9 ISOMORPHISMS
To show that φ is injective, suppose that m and n are two elements in Z,
where m 6= n. We can assume that m > n. We must show that am 6= an .
Let us suppose the contrary; that is, am = an . In this case am−n = e, where
m − n > 0, which contradicts the fact that a has infinite order. Our map
is onto since any element in G can be written as an for some integer n and
φ(n) = an .
Theorem 9.3 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 9.4 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 6.7.
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 9.5 The isomorphism of groups determines an equivalence rela-
tion 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 6. Consider the group Z3 . The Cayley table for Z3 is as follows.
+ 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1
9.1 DEFINITION AND EXAMPLES 145
The addition table of Z3 suggests that it is the same as the permutation
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 9.6 (Cayley) Every group is isomorphic to a group of permu-
tations.
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).
146 CHAPTER 9 ISOMORPHISMS
It is also one-to-one, because if φ(g)(a) = φ(h)(a), then
ga = λg a = λh a = ha.
Hence, g = h. That φ is onto follows from the fact that φ(g) = λg for any
λg ∈ G.
The isomorphism g 7→ λg is known as the left regular representation
of G.
Historical Note
Arthur Cayley was born in England in 1821, though he spent much of the first
part of his life in Russia, where his father was a merchant. Cayley was educated
at Cambridge, where he took the first Smith’s Prize in mathematics. A lawyer
for much of his adult life, he wrote several papers in his early twenties before
entering the legal profession at the age of 25. While practicing law he continued his
mathematical research, writing more than 300 papers during this period of his life.
These included some of his best work. In 1863 he left 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 some-
times 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 );
9.2 DIRECT PRODUCTS 147
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 9.7 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 7. Let R be the group of real numbers under addition. The
Cartesian product of R with itself, R × R = R2 , is also a group, in which the
group operation is just addition in each coordinate; that is, (a, b) + (c, d) =
(a + c, b + d). The identity is (0, 0) and the inverse of (a, b) is (−a, −b).
Example 8. Consider
Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)}.
Although Z2 × Z2 and Z4 both contain four elements, it is easy to see that
they are not isomorphic since for every element (a, b) in Z2 × Z2 , (a, b) +
(a, b) = (0, 0), but 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
Y
Gi = G1 × G2 × · · · × Gn
i=1
of the groups G1 , G2 , . . . , Gn is defined in exactly the same manner. If
G = G1 = G2 = · · · = Gn , we often write Gn instead of G1 × G2 × · · · × Gn .
Example 9. 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.
148 CHAPTER 9 ISOMORPHISMS
Theorem 9.8 Let (g, h) ∈ G × H. If g and h have finite orders r and s
respectively, then the order of (g, h) in G × H is the least common multiple
of r and s.
Proof. Suppose that m is the least common multiple of r and s and let
n = |(g, h)|. Then
(g, h)m = (g m , hm ) = (eG , eH )
(g n , hn ) = (g, h)n = (eG , eH ).
Hence, n must divide m, and n ≤ m. However, by the second equation,
both r and s must divide n; therefore, n is a common multiple of r and s.
Since m is the least common multiple of r and s, m ≤ n. Consequently, m
must be equal to n.
Q
Corollary 9.9 Let (g1 , . . . ,Q 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 10. 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 11. The group Z2 × Z3 consists of the pairs
(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2).
In this case, unlike that of Z2 × Z2 and Z4 , it is true that Z2 × Z3 ∼
= Z6 .
We need only show that Z2 × Z3 is cyclic. It is easy to see that (1, 1) is a
generator for Z2 × Z3 .
The next theorem tells us exactly when the direct product of two cyclic
groups is cyclic.
Theorem 9.10 The group Zm × Zn is isomorphic to Zmn if and only if
gcd(m, n) = 1.
Proof. Assume first that if Zm × Zn ∼ = Zmn , then gcd(m, n) = 1. To
show this, 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
9.2 DIRECT PRODUCTS 149
Therefore, no (a, b) can generate all of Zm × Zn .
The converse follows directly from Theorem 9.8 since lcm(m, n) = mn if
and only if gcd(m, n) = 1.
Corollary 9.11 Let n1 , . . . , nk be positive integers. Then
k
Zni ∼
Y
= Zn1 ···nk
i=1
if and only if gcd(ni , nj ) = 1 for i 6= j.
Corollary 9.12 If
m = pe11 · · · pekk ,
where the pi s are distinct primes, then
Zm ∼
= Zpe11 × · · · × Zpek .
k
e
Proof. Since the greatest common divisor of pei i and pj j is 1 for i 6= j, the
proof follows from Corollary 9.11.
In Chapter 13, we will prove that all finite abelian groups are isomorphic
to direct products of the form
Zpe1 × · · · × Zpek
1 k
where p1 , . . . , pk are (not necessarily distinct) primes.
Internal Direct Products
The external direct product of two groups builds a large group out of two
smaller groups. We would like to be able to reverse this process and con-
veniently 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.
150 CHAPTER 9 ISOMORPHISMS
Then G is the internal direct product of H and K.
Example 12. The group U (8) is the internal direct product of
H = {1, 3} and K = {1, 5}.
Example 13. 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 14. Not every group can be written as the internal direct product
of two of its proper subgroups. If the group S3 were an internal direct
product of its proper subgroups H and K, then one of the subgroups, say H,
would have to have order 3. In this case H is the subgroup {(1), (123), (132)}.
The subgroup K must have order 2, but no matter which subgroup we
choose for K, the condition that hk = kh will never be satisfied for h ∈ H
and k ∈ K.
Theorem 9.13 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 = h0 k 0 . Then h−1 h0 = k(k 0 )−1 is in both H and K, so
it must be the identity. Therefore, h = h0 and k = k 0 , 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 ).
EXERCISES 151
We will leave the proof that φ is one-to-one and onto as an exercise.
Example 15. 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 ∩ h∪j6=i Hj i = {e};
• hi hj = hj hi for all hi ∈ Hi and hj ∈ Hj .
We will leave the proof of the following theorem as an exercise.
Theorem 9.14 Let G be the internal direct Q product of subgroups Hi , where
i = 1, 2, . . . , n. Then G is isomorphic to i Hi .
Exercises
1. Prove that Z ∼
= nZ for n 6= 0.
2. Prove that C∗ is isomorphic to the subgroup of GL2 (R) consisting of matrices
of the form
a b
−b a
3. Prove or disprove: U (8) ∼
= Z4 .
4. Prove that U (8) is isomorphic to the group of matrices
1 0 1 0 −1 0 −1 0
, , , .
0 1 0 −1 0 1 0 −1
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.
152 CHAPTER 9 ISOMORPHISMS
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 n
B= ,
0 1
where n ∈ Zn , is a group isomorphic to Dn .
15. List all of the elements of Z4 × Z2 .
16. Find the order of each of the following elements.
(a) (3, 4) in Z4 × Z6
(b) (6, 15, 4) in Z30 × Z45 × Z24
(c) (5, 10, 15) in Z25 × Z25 × Z25
(d) (8, 8, 8) in Z10 × Z24 × Z80
17. Prove that D4 cannot be the internal direct product of two of its proper
subgroups.
18. Prove that the subgroup of Q∗ consisting of elements of the form 2m 3n for
m, n ∈ Z is an internal direct product isomorphic to Z × Z.
19. Prove that S3 × Z2 is isomorphic to D6 . Can you make a conjecture about
D2n ? Prove your conjecture. [Hint: Draw the picture.]
20. Prove or disprove: Every abelian group of order divisible by 3 contains a
subgroup of order 3.
EXERCISES 153
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 : G1 → G2 be a group isomorphism. Show that φ(x) = e if and only if
x = e.
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.
∼ Z4 . Can you generalize this result to show that U (p) =
32. Prove U (5) = ∼ Zp−1 ?
33. Write out the permutations associated with each element of S3 in the proof
of Cayley’s Theorem.
34. An automorphism of a group G is an isomorphism with itself. Prove that
complex conjugation is an automorphism of the additive group of complex
numbers; that is, show that the map φ(a + bi) = a − bi is an isomorphism
from C to C.
35. Prove that a + ib 7→ a − ib is an automorphism of C∗ .
36. Prove that A 7→ B −1 AB is an automorphism of SL2 (R) for all B in GL2 (R).
37. We will denote the set of all automorphisms of G by Aut(G). Prove that
Aut(G) is a subgroup of SG , the group of permutations of G.
38. Find Aut(Z6 ).
39. Find Aut(Z).
40. Find two nonisomorphic groups G and H such that Aut(G) ∼
= Aut(H).
41. Let G be a group and g ∈ G. Define a map ig : G → G by ig (x) = gxg −1 .
Prove that ig defines an automorphism of G. Such an automorphism is called
an inner automorphism. The set of all inner automorphisms is denoted
by Inn(G).
154 CHAPTER 9 ISOMORPHISMS
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 map ρg : G → G 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.
= G and H ∼
47. If G ∼ = 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 ∼
Y
= Zn1 ···nk
i=1
if and only if gcd(ni , nj ) = 1 for i 6= j.
50. Prove that A × B is abelian if and only if A and B are abelian.
51. If G is the
Q internal direct product of H1 , H2 , . . . , Hn , prove that G is isomor-
phic to i Hi .
52. Let H1 and H2 be subgroups of G1 and G2 , respectively. Prove that H1 × H2
is a subgroup of G1 × G2 .
53. Let m, n ∈ Z. Prove that hm, ni ∼
= hdi if and only if d = gcd(m, n).
54. Let m, n ∈ Z. Prove that hmi ∩ hni ∼
= hli if and only if d = lcm(m, n).
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 by using homomorphisms,
a generalization of isomorphisms.
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 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 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 , consisting
of the permutations (1), (123), and (132), is normal since the cosets of N
155
156 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
are
N = {(1), (123), (132)}
(12)N = N (12) = {(12), (13), (23)}.
The following theorem is fundamental to our understanding of normal
subgroups.
Theorem 10.1 Let G be a group and N be a subgroup of G. Then the
following statements are equivalent.
1. The subgroup N is normal in G.
2. For all g ∈ G, gN g −1 ⊂ N .
3. For all g ∈ G, gN g −1 = N .
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 n0 in N such that gn = n0 g.
Therefore, gng −1 = n0 ∈ 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 = n0 for
some n0 ∈ N . Therefore, n = gn0 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 n0 ∈ N such that gng −1 = n0 . Consequently, gn = n0 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 10.2 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 ).
10.1 FACTOR GROUPS AND NORMAL SUBGROUPS 157
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 3. 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 com-
plicated 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 captured in G/N is parity;
that is, multiplying two even or two odd permutations results in an even per-
mutation, whereas multiplying an odd permutation by an even permutation
yields an odd permutation.
Example 4. 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 multiplication table below.
158 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
+ 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 + Z and l + Z is k + l + Z. Notice that we have
written our cosets additively, because the group operation is integer addition.
Example 5. 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 Simplicity of An
Of special interest are groups with no nontrivial normal subgroups. Such
groups are called simple groups. Of course, we already have a whole
class of examples of simple groups, Zp , where p is prime. These groups are
trivially simple since they have no proper subgroups other than the subgroup
consisting solely of the identity. Other examples of simple groups are not
so easily found. We can, however, show that the alternating group, An , is
simple for n ≥ 5. The proof of this result requires several lemmas.
Lemma 10.3 The alternating group An is generated by 3-cycles for n ≥ 3.
10.2 SIMPLICITY OF AN 159
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 10.4 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 10.3, these 3-cycles generate An ; hence, N = An .
Lemma 10.5 For n ≥ 5, every 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 ).
160 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
• 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 dis-
joint 2-cycles.
If σ is a 3-cycle, then we are done. If N contains a product of disjoint
cycles, σ, and at least one of these cycles has length greater than 3, say
σ = τ (a1 a2 · · · ar ), then
(a1 a2 a3 )σ(a1 a2 a3 )−1
is in N since N is normal; hence,
σ −1 (a1 a2 a3 )σ(a1 a2 a3 )−1
is also in N . Since
σ −1 (a1 a2 a3 )σ(a1 a2 a3 )−1
= σ −1 (a1 a2 a3 )σ(a1 a3 a2 )
= (a1 a2 · · · ar )−1 τ −1 (a1 a2 a3 )τ (a1 a2 · · · ar )(a1 a3 a2 )
= (a1 ar ar−1 · · · a2 )(a1 a2 a3 )(a1 a2 · · · ar )(a1 a3 a2 )
= (a1 a3 ar ),
N must contain a 3-cycle; hence, N = An .
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 ).
10.2 SIMPLICITY OF AN 161
So N contains a disjoint cycle of length greater than 3, and we can apply
the previous case.
Suppose N contains a disjoint product of the form σ = τ (a1 a2 a3 ), where
τ is the product of disjoint 2-cycles. Since σ ∈ N , σ 2 ∈ N , and
σ 2 = τ (a1 a2 a3 )τ (a1 a2 a3 )
= (a1 a3 a2 ).
So N contains a 3-cycle.
The only remaining possible case is a disjoint product of the form
σ = τ (a1 a2 )(a3 a4 ),
where τ is the product of an even number of disjoint 2-cycles. But
σ −1 (a1 a2 a3 )σ(a1 a2 a3 )−1
is in N since (a1 a2 a3 )σ(a1 a2 a3 )−1 is in N ; and so
σ −1 (a1 a2 a3 )σ(a1 a2 a3 )−1
= τ −1 (a1 a2 )(a3 a4 )(a1 a2 a3 )τ (a1 a2 )(a3 a4 )(a1 a2 a3 )−1
= (a1 a3 )(a2 a4 ).
Since n ≥ 5, we can find b ∈ {1, 2, . . . , n} such that b 6= a1 , a2 , a3 , a4 . Let
µ = (a1 a3 b). Then
µ−1 (a1 a3 )(a2 a4 )µ(a1 a3 )(a2 a4 ) ∈ N
and
µ−1 (a1 a3 )(a2 a4 )µ(a1 a3 )(a2 a4 )
= (a1 ba3 )(a1 a3 )(a2 a4 )(a1 a3 b)(a1 a3 )(a2 a4 )
= (a1 a3 b).
Therefore, N contains a 3-cycle. This completes the proof of the lemma.
Theorem 10.6 The alternating group, An , is simple for n ≥ 5.
Proof. Let N be a normal subgroup of An . By Lemma 10.5, N contains
a 3-cycle. By Lemma 10.4, N = An ; therefore, An contains no proper
nontrivial normal subgroups for n ≥ 5.
162 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
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 years. In a sense, finite simple groups are the building blocks of all finite
groups. The first nonabelian simple groups to be discovered were the alternating
groups. Galois was the first to prove that A5 was simple. Later mathematicians,
such as C. Jordan and L. E. Dickson, found several infinite families of matrix
groups that were simple. Other families of simple groups were discovered in the
1950s. At the turn of the century, William Burnside conjectured that all nonabelian
simple groups must have even order. In 1963, W. Feit and J. Thompson proved
Burnside’s conjecture and published their results in the paper “Solvability of Groups
of Odd Order,” which appeared in the Pacific Journal of Mathematics. Their
proof, running over 250 pages, gave impetus to a program in the 1960s and 1970s
to classify all finite simple groups. Daniel Gorenstein was the organizer of this
remarkable effort. One of the last simple groups was the “Monster,” discovered
by R. Greiss. The Monster, a 196,833 × 196,833 matrix group, is one of the 26
sporadic, or special, simple groups. These sporadic simple groups are groups that
fit into no infinite family of simple groups.
Exercises
1. For each of the following groups G, determine whether H is a normal sub-
group 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 subgroups are
normal? What are all the factor groups of Q8 up to isomorphism?
4. Let T be the group of nonsingular upper triangular 2×2 matrices with entries
in R; that is, matrices of the form
a b
,
0 c
EXERCISES 163
where a, b, c ∈ R and ac 6= 0. Let U consist of matrices of the form
1 x
,
0 1
where x ∈ R.
(a) Show that U is a subgroup of T .
(b) Prove that U is abelian.
(c) Prove that U is normal in T .
(d) Show that T /U is abelian.
(e) Is T normal in GL2 (R)?
5. Show that the intersection of two normal subgroups is a normal subgroup.
6. If G is abelian, prove that G/H must also be abelian.
7. Prove or disprove: If H is a normal subgroup of G such that H and G/H
are abelian, then G is abelian.
8. If G is cyclic, prove that G/H must also be cyclic.
9. Prove or disprove: If H and G/H are cyclic, then G is cyclic.
10. Let H be a subgroup of index 2 of a group G. Prove that H must be a normal
subgroup of G. Conclude that Sn is not simple.
11. Let G be a group of order p2 , where p is a prime number. If H is a subgroup
of G of order p, show that H is normal in G. Prove that G must be abelian.
12. If a group G has exactly one subgroup H of order k, prove that H is normal
in G.
13. 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.
14. 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.
164 CHAPTER 10 NORMAL SUBGROUPS AND FACTOR GROUPS
15. Let G be a group and let G0 = haba−1 b−1 i; that is, G0 is the subgroup of all
finite products of elements in G of the form aba−1 b−1 . The subgroup G0 is
called the commutator subgroup of G.
(a) Show that G0 is a normal subgroup of G.
(b) Let N be a normal subgroup of G. Prove that G/N is abelian if and
only if N contains the commutator subgroup of G.
11
Homomorphisms
11.1 Group Homomorphisms
One of the basic ideas of algebra is the concept of a homomorphism, a nat-
ural generalization of an isomorphism. If we relax the requirement that an
isomorphism of groups be bijective, we have a homomorphism. A homo-
morphism 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 isomor-
phic; 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 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.
165
166 CHAPTER 11 HOMOMORPHISMS
Example 2. Let G = GL2 (R). If
a b
A=
c d
is in G, then the determinant is nonzero; that is, det(A) = ad − bc 6= 0.
Also, for any two elements A and B in G, det(AB) = det(A) det(B). Using
the determinant, we can define a homomorphism φ : GL2 (R) → R∗ by
A 7→ det(A).
Example 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 homomor-
phisms.
Proposition 11.1 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 ∈ G : 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 e0 are the identities of G1 and G2 , respec-
tively; then
e0 φ(e) = φ(e) = φ(ee) = φ(e)φ(e).
By cancellation, φ(e) = e0 .
11.1 GROUP HOMOMORPHISMS 167
(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 H2 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 3.10.
(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 11.1, φ−1 ({e}) is a subgroup of G. This
subgroup is called the kernel of φ and will be denoted by ker φ. In fact, this
subgroup is a normal subgroup of G since the trivial subgroup is normal in
H. We state this result in the following theorem, which says that with every
homomorphism of groups we can naturally associate a normal subgroup.
Theorem 11.2 Let φ : G → H be a group homomorphism. Then the kernel
of φ is a normal subgroup of G.
Example 4. 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 homomor-
phism is all 2×2 matrices having determinant one. That is, ker φ = SL2 (R).
Example 5. 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 6. Suppose that we wish to determine all possible homomor-
phisms φ from Z7 to Z12 . Since the kernel of φ must be a subgroup of
168 CHAPTER 11 HOMOMORPHISMS
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 7. Let G be a group. Suppose that g ∈ G and φ is the homo-
morphism 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
Though at first it is not evident that factor groups correspond exactly to
homomorphic images, we can use factor groups to study homomorphisms.
We already know that with every group homomorphism φ : G → H we can
associate a normal subgroup of G, ker φ; the converse is also true. 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 among group homomorphisms, normal subgroups, and factor
groups.
Theorem 11.3 (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.
11.2 THE ISOMORPHISM THEOREMS 169
Proof. We already know that K is normal in G. Define η : G/K → ψ(G)
by η(gK) = ψ(g). We must first show that this is a well-defined map.
Suppose that g1 K = g2 K. 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).
Since η(g1 K) = η(g2 K), η does not depend on the choice of coset represen-
tative. 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.
Finally, we must show that η is a homomorphism, but
η(g1 Kg2 K) = η(g1 g2 K)
= ψ(g1 g2 )
= ψ(g1 )ψ(g2 )
= η(g1 K)η(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 8. 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 .
170 CHAPTER 11 HOMOMORPHISMS
Theorem 11.4 (Second Isomorphism Theorem) Let H be a subgroup
of a group G (not necessarily normal in G) and N a normal subgroup of G.
Then HN is a subgroup of G, H ∩ N is a normal subgroup of H, and
H/H ∩ N ∼
= HN/N.
Proof. We will first show that HN = {hn : h ∈ H, n ∈ N } is a subgroup
of G. Suppose that h1 n1 , h2 n2 ∈ HN . Since N is normal, (h2 )−1 n1 h2 ∈ N .
So
(h1 n1 )(h2 n2 ) = h1 h2 ((h2 )−1 n1 h2 )n2
is in HN . The inverse of hn ∈ HN is in HN since
(hn)−1 = n−1 h−1 = h−1 (hn−1 h−1 ).
Next, we prove that H ∩ N is normal in H. Let h ∈ H and n ∈ H ∩ N .
Then h−1 nh ∈ H since each element is in H. Also, h−1 nh ∈ N since N is
normal in G; therefore, h−1 nh ∈ H ∩ N .
Now define a map φ from H to HN/N by h 7→ hN . The map φ is onto,
since any coset hnN = hN is the image of h in H. We also know that φ is
a homomorphism because
φ(hh0 ) = hh0 N = hN h0 N = φ(h)φ(h0 ).
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 11.5 (Correspondence Theorem) Let N be a normal sub-
group of a group G. Then H 7→ H/N is a one-to-one correspondence be-
tween the set of subgroups H containing N and the set of subgroups of G/N .
Furthermore, the normal subgroups of H 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 sense. 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 .
EXERCISES 171
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 ) =
hh0 N ∈ S and h−11 N ∈ S. Therefore, H must be a subgroup of G. Clearly,
H contains N . Therefore, S = H/N . Consequently, the map H 7→ H/H 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/H is one-to-one.
Suppose that H is normal in G and N is a subgroup of H. Then it
is easy to verify that the map G/N → G/H defined by gN 7→ gH is a
homomorphism. The kernel of this homomorphism is H/N , which proves
that H/N is normal in G/N .
Conversely, suppose that H/N is normal in G/N . The homomorphism
given by
G/N
G → G/N →
H/N
has kernel H. Hence, H must be normal in G.
Notice that in the course of the proof of Theorem 11.5, we have also
proved the following theorem.
Theorem 11.6 (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 9. By the Third Isomorphism Theorem,
Z/mZ ∼
= (Z/mnZ)/(mZ/mnZ).
Since |Z/mnZ| = mn and |Z/mZ| = m, we have |mZ/mnZ| = n.
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∗ .
172 CHAPTER 11 HOMOMORPHISMS
2. Which of the following maps are homomorphisms? If the map is a homomor-
phism, 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 homomor-
phism. 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 = h4i and N = h6i.
(a) List the elements in HN (we usually write H + N for these additive
groups) and H ∩ N .
(b) List the cosets in HN/N , showing the elements in each coset.
(c) List the cosets in H/(H ∩ N ), showing the elements in each coset.
(d) Give the correspondence between HN/N and H/(H ∩ N ) described in
the proof of the Second Isomorphism Theorem.
EXERCISES 173
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 deter-
mined by its action on the generator of the group.
12. Let G be a group of order p2 , where p is a prime number. If H is a subgroup
of G of order p, show that H is normal in G. Prove that G must be abelian.
13. If a group G has exactly one subgroup H of order k, prove that H is normal
in G.
14. Prove or disprove: Q/Z ∼
= Q.
15. 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.
16. 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 natural homomorphism φ : (G1 /H1 ) → (G2 /H2 ) if φ(H1 ) ⊆ H2 .
17. 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.
18. 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 .
19. Let φ : G → H be a group homomorphism. Show that φ is one-to-one if and
only if φ−1 (e) = {e}.
Additional Exercises: Automorphisms
1. Let Aut(G) be the set of all automorphisms of G; that is, isomorphisms from
G to itself. Prove this set forms a group and is a subgroup of the group of
permutations of G; that is, Aut(G) ≤ SG .
2. An inner automorphism of G,
ig : G → G,
is defined by the map
ig (x) = gxg −1 ,
for g ∈ G. Show that ig ∈ Aut(G).
174 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 automorphism.
5. Let G be a group and ig be an inner automorphism of G, and define a map
G → Aut(G)
by
g 7→ ig .
Prove that this map is a homomorphism with image Inn(G) and kernel Z(G).
Use this result to conclude that
G/Z(G) ∼
= Inn(G).
6. Compute Aut(S3 ) and Inn(S3 ). Do the same thing for D4 .
7. Find all of the homomorphisms φ : Z → Z. What is Aut(Z)?
8. Find all of the automorphisms of Z8 . Prove that Aut(Z8 ) ∼
= U (8).
9. For k ∈ Zn , define a map φk : Zn → Zn by a 7→ ka. Prove that φk is a
homomorphism.
10. Prove that φk is an isomorphism if and only if k is a generator of Zn .
11. Show that every automorphism of Zn is of the form φk , where k is a generator
of Zn .
12. Prove that ψ : U (n) → Aut(Zn ) is an isomorphism, where ψ : k 7→ φk .
12
Matrix Groups and
Symmetry
When Felix Klein (1849–1925) accepted a chair at the University of Er-
langen, he outlined in his inaugural address a program to classify different
geometries. Central to Klein’s program was the theory of groups: he con-
sidered geometry to be the study of properties that are left invariant under
transformation groups. Groups, especially matrix groups, have now become
important in the study of symmetry and have found applications in such
disciplines as chemistry and physics. In the first part of this chapter, we
will examine some of the classical matrix groups, such as the general linear
group, the special linear group, and the orthogonal group. We will then
use these matrix groups to investigate some of the ideas behind geometric
symmetry.
12.1 Matrix Groups
Some Facts from Linear Algebra
Before we study matrix groups, we must recall some basic facts from linear
algebra. One of the most fundamental ideas of linear algebra is that of a
linear transformation. A linear transformation or linear map T : Rn →
Rm is a map that preserves vector addition and scalar multiplication; that
is, for vectors x and y in Rn and a scalar α ∈ R,
T (x + y) = T (x) + T (y)
T (αy) = αT (y).
An m × n matrix with entries in R represents a linear transformation from
Rn to Rm . If we write vectors x = (x1 , . . . , xn )t and y = (y1 , . . . , yn )t in Rn
175
176 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
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),
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 ,
12.1 MATRIX GROUPS 177
then
T (x) = T (x1 e1 + x2 e2 + · · · + xn en )
= x1 T (e1 ) + x2 T (e2 ) + · · · + xn T (en )
n n
!t
X X
= a1k xk , . . . , amk xk
k=1 k=1
= Ax.
Example 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
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 2. If A is the matrix
2 1
,
5 3
then the inverse of A is
−1 3 −1
A = .
−5 2
178 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
We are guaranteed that A−1 exists, since det(A) = 2 · 3 − 5 · 1 = 1 is nonzero.
Some other facts about determinants will also prove useful in the course
of this chapter. Let A and B be n × n matrices. From linear algebra we
have the following properties of determinants.
• The determinant is a homomorphism into the multiplicative group of
real numbers; that is, det(AB) = (det A)(det B).
• If A is an invertible matrix, then det(A−1 ) = 1/ det A.
• If we define the transpose of a matrix A = (aij ) to be At = (aji ), then
det(At ) = det A.
• Let T be the linear transformation associated with an n × n matrix A.
Then T multiplies volumes by a factor of | det A|. In the case of R2 ,
this means that T multiplies areas by | det A|.
Linear maps, matrices, and determinants are covered in any elementary
linear algebra text; however, if you have not had a course in linear algebra,
it is a straightforward process to verify these properties directly for 2 × 2
matrices, the case with which we are most concerned.
The General and Special Linear Groups
The set of all n × n invertible matrices forms a group called the general
linear group. We will denote this group by GLn (R). The general linear
group has several important subgroups. The multiplicative properties of
the determinant imply that the set of matrices with determinant one is a
subgroup of the general linear group. Stated another way, suppose that
det(A) = 1 and det(B) = 1. Then det(AB) = det(A) det(B) = 1 and
det(A−1 ) = 1/ det A = 1. This subgroup is called the special linear group
and is denoted by SLn (R).
Example 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 matri-
ces in which ad − bc 6= 0. The inverse of A is
−1 1 d −b
A = .
ad − bc −c a
12.1 MATRIX GROUPS 179
If A is in SL2 (R), then
d −b
A−1 = .
−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 12.1, 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 12.1. 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 or-
thogonal 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 4. 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
180 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
There is a more geometric way of viewing the group O(n). The orthog-
onal matrices are exactly those matrices that preserve the length of vectors.
We can define the length of a vector using the Euclidean inner product,
or dot product, of two vectors. The Euclidean inner product of two vectors
x = (x1 , . . . , xn )t and y = (y1 , . . . , yn )t is
y1
y2
hx, yi = xt y = (x1 , x2 , . . . , xn ) . = x1 y1 + · · · + xn yn .
..
yn
We define the length of a vector x = (x1 , . . . , xn )t to be
p q
kxk = hx, xi = x21 + · · · + x2n .
Associated with the notion of the length of a vector is the idea of the distance
between two vectors. We define the distance between two vectors x and y
to be kx − yk. We leave as an exercise the proof of the following proposition
about the properties of Euclidean inner products.
Proposition 12.1 Let x, y, and w be vectors in Rn and α ∈ R. Then
1. hx, yi = hy, xi.
2. hx, y + wi = hx, yi + hx, wi.
3. hαx, yi = hx, αyi = αhx, yi.
4. hx, xi ≥ 0 with equality exactly when x = 0.
5. If hx, yi = 0 for all x in Rn , then y = 0.
√
Example 5. 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.
12.1 MATRIX GROUPS 181
Since det(AAt ) = det(I) = 1 and det(A) = det(At ), the determinant of
any orthogonal matrix is either 1 or −1. Consider the column vectors
a1j
a2j
aj = .
..
anj
of the orthogonal matrix A = (aij ). Since AAt = I, har , as i = δrs , where
1 r=s
δrs =
0 r 6= s
is the Kronecker delta. Accordingly, column vectors of an orthogonal ma-
trix 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, A−1 = At .
We say that a matrix A is distance-preserving, length-preserving,
or inner product-preserving when kT x−T yk = kx−yk, kT xk = kxk, or
hT x, T yi = hx, yi, respectively. The following theorem, which characterizes
the orthogonal group, says that these notions are the same.
Theorem 12.2 Let A be an n × n matrix. The following statements are
equivalent.
1. The columns of the matrix A form an orthonormal set.
2. A−1 = At .
3. For vectors x and y, hAx, Ayi = hx, yi.
4. For vectors x and y, kAx − Ayk = kx − yk.
5. For any vector x, kAxk = kxk.
Proof. We have already shown (1) and (2) to be equivalent.
(2) ⇒ (3).
hAx, Ayi = (Ax)t Ay
= xt At Ay
= xt y
= hx, yi.
182 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
(3) ⇒ (2). Since
hx, xi = hAx, Axi
= xt At Ay
= hx, At Axi,
we know that hx, (At A − I)xi = 0 for all x. Therefore, At A − I = 0 or
A−1 = At .
(3) ⇒ (4). If A is inner product-preserving, then A is distance-preserving,
since
kAx − Ayk2 = kA(x − y)k2
= hA(x − y), A(x − y)i
= hx − y, x − yi
= kx − yk2 .
(4) ⇒ (5). If A is distance-preserving, then A is length-preserving.
Letting y = 0, we have
kAxk = kAx − Ayk = kx − yk = kxk.
(5) ⇒ (3). We use the following identity to show that length-preserving
implies inner product-preserving:
1
kx + yk2 − kxk2 − kyk2 .
hx, yi =
2
Observe that
1
kAx + Ayk2 − kAxk2 − kAyk2
hAx, Ayi =
2
1
kA(x + y)k2 − kAxk2 − kAyk2
=
2
1
kx + yk2 − kxk2 − kyk2
=
2
= hx, yi.
Example 6. Let us examine the orthogonal group on R2 a bit more closely.
An element T ∈ O(2) is determined by its action on e1 = (1, 0)t and e2 =
12.1 MATRIX GROUPS 183
y y
(sin θ, − cos θ)
(cos θ, sin θ)
(a, b)
θ
x x
(a, −b)
Figure 12.2. O(2) acting on R2
(0, 1)t . If T (e1 ) = (a, b)t , then a2 + b2 = 1 and T (e2 ) = (−b, a)t . Hence, T
can be represented by
a −b cos θ − sin θ
A= = ,
b a sin θ cos θ
where 0 ≤ θ < 2π. A matrix T in O(2) either reflects or rotates a vector in
R2 (Figure 12.2). A reflection is given by the matrix
1 0
,
0 −1
whereas a rotation by an angle θ in a counterclockwise direction must come
from a matrix of the form
cos θ sin θ
.
sin θ − cos θ
If det A = −1, then A gives a reflection.
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 Eu-
clidean 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 6, you are asked to check that E(n) is indeed a group under this
operation.
184 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
y y
x+y
x
x x
Figure 12.3. Translations in R2
12.2 Symmetry
An isometry or rigid motion in Rn is a distance-preserving function f
from Rn to Rn . This means that f must satisfy
kf (x) − f (y)k = kx − yk
for all x, y ∈ Rn . It is not difficult to show that f must be a one-to-one
map. By Theorem 12.2, 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 12.3); 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 translation
followed by a reflection (Figure 12.4). In Rn all isometries are given in the
same manner. The proof is very easy to generalize.
Lemma 12.3 An isometry f that fixes the origin in R2 is a linear trans-
formation. In particular, f is given by an element in O(2).
12.2 SYMMETRY 185
y y
x
x x
T (x)
Figure 12.4. Glide reflections
Proof. Let f be an isometry in R2 fixing the origin. We will first show
that f preserves inner products. Since f (0) = 0, kf (x)k = kxk; therefore,
kxk2 − 2hf (x), f (y)i + kyk2 = kf (x)k2 − 2hf (x), f (y)i + kf (y)k2
= hf (x) − f (y), f (x) − f (y)i
= kf (x) − f (y)k2
= kx − yk2
= hx − y, x − yi
= kxk2 − 2hx, yi + kyk2 .
Consequently,
hf (x), f (y)i = hx, yi.
Now let e1 and e2 be (1, 0)t and (0, 1)t , respectively. If
x = (x1 , x2 ) = x1 e1 + x2 e2 ,
then
f (x) = hf (x), f (e1 )if (e1 ) + hf (x), f (e2 )if (e2 ) = x1 f (e1 ) + x2 f (e2 ).
The linearity of f easily follows.
For any arbitrary isometry, f , Tx f will fix the origin for some vector
x in R2 ; hence, Tx f (y) = Ay for some matrix A ∈ O(2). Consequently,
186 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
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 12.4 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 ⊂ R2 . 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 12.5 The only finite symmetry groups in R2 are Zn and Dn .
Proof. Any finite symmetry group G in R2 must be a finite subgroup of
O(2); otherwise, G would have an element in E(2) of the form (A, x), where
x 6= 0. Such an element must have infinite order.
By Example 6, 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 θ − sin θ
Tθ = .
sin θ cos θ
Notice that det(Rθ ) = 1, det(Tθ ) = −1, and Tθ2 = 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,
12.2 SYMMETRY 187
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 Tθ . 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.
Figure 12.5. A wallpaper pattern in R2
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 12.5). The analogs of wallpaper patterns in R3 are crystals,
which we can think of as repeating patterns of molecules in three dimensions
(Figure 12.6). The mathematical equivalent of a wallpaper or crystal pattern
is called a lattice.
188 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
Figure 12.6. 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.
(−1, 1) (1, 1)
(2, 0)
(−1, −1)
Figure 12.7. A lattice in R2
Notice that a lattice can have several bases. For example, the vectors
(1, 1)t and (2, 0)t have the same lattice as the vectors (−1, 1)t and (−1, −1)t
(Figure 12.7). However, any lattice is completely determined by a basis.
Given two bases for the same lattice, say {x1 , x2 } and {y1 , y2 }, we can
12.2 SYMMETRY 189
write
y1 = α1 x1 + α2 x2
y2 = β1 x1 + β2 x2 ,
where α1 , α2 , β1 , and β2 are integers. The matrix corresponding to this
transformation is
α1 α2
U= .
β1 β2
If we wish to give x1 and x2 in terms of y1 and y2 , we need only calculate
U −1 ; that is,
−1 y1 x1
U = .
y2 x2
Since U has integer entries, U −1 must also have integer entries; hence the
determinants of both U and U −1 must be integers. Because U U −1 = I,
det(U U −1 ) = det(U ) det(U −1 ) = 1;
consequently, det(U ) = ±1. A matrix with determinant ±1 and integer
entries is called unimodular. For example, the matrix
3 1
5 2
is unimodular. It should be clear that there is a minimum length for vectors
in a lattice.
We can classify lattices by studying their symmetry groups. The sym-
metry 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 group. 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]).
190 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
Theorem 12.6 Every translation group in R2 is isomorphic to Z × Z.
Rectangular
Square Rhombic
Parallelogram
Hexagonal
Figure 12.8. Types of lattices in R2
The point group of G is G0 = {A : (A, b) ∈ G for some b}. In particular,
G0 must be a subgroup of O(2). Suppose that x is a vector in a lattice
L with space group G, translation group H, and point group G0 . For any
element (A, y) in G,
(A, y)(I, x)(A, y)−1 = (A, Ax + y)(A−1 , −A−1 y)
= (AA−1 , −AA−1 y + Ax + y)
= (I, Ax);
hence, (I, Ax) is in the translation group of G. More specifically, Ax must
be in the lattice L. It is important to note that G0 is not usually a subgroup
of the space group G; however, if T is the translation subgroup of G, then
G/T ∼ = G0 . The proof of the following theorem can be found in [2], [5],
or [6].
Theorem 12.7 The point group in the wallpaper groups is isomorphic to
Zn or Dn , where n = 1, 2, 3, 4, 6.
12.2 SYMMETRY 191
To answer the question of how the point groups and the translation
groups can be combined, we must look at the different types of lattices.
Lattices can be classified by the structure of a single lattice cell. The possible
cell shapes are parallelogram, rectangular, square, rhombic, and hexagonal
(Figure 12.8). The wallpaper groups can now be classified according to the
types of reflections that occur in each group: these are ordinarily reflections,
glide reflections, both, or none.
Table 12.1. The 17 wallpaper groups
Notation and Reflections
Space Groups Point Group Lattice Type or Glide Reflections?
p1 Z1 parallelogram none
p2 Z2 parallelogram none
p3 Z3 hexagonal none
p4 Z4 square none
p6 Z6 hexagonal none
pm D1 rectangular reflections
pg D1 rectangular glide reflections
cm D1 rhombic both
pmm D2 rectangular reflections
pmg D2 rectangular glide reflections
pgg D2 rectangular both
c2mm D2 rhombic both
p3m1, p31m D3 hexagonal both
p4m, p4g D4 square both
p6m D6 hexagonal both
Theorem 12.8 There are exactly 17 wallpaper groups.
The 17 wallpaper groups are listed in Table 12.1. The groups p3m1 and
p31m can be distinguished by whether or not all of their threefold centers
lie on the reflection axes: those of p3m1 must, whereas those of p31m may
not. Similarly, the fourfold centers of p4m must lie on the reflection axes
whereas those of p4g need not (Figure 12.9). 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].
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 Inter-
192 CHAPTER 12 MATRIX GROUPS AND SYMMETRY
p4m p4g
Figure 12.9. The wallpaper groups p4m and p4g
national Congress of Mathematicians in 1900, David Hilbert gave a now-famous
address outlining 23 problems to guide mathematics in the twentieth century.
Hilbert’s eighteenth problem asked whether or not crystallographic groups in n
dimensions were always finite. In 1910, L. Bieberbach proved that crystallographic
groups are finite in every dimension. Finding out how many of these groups there
are in each dimension is another matter. In R3 there are 230 different space groups;
in R4 there are 4783. No one has been able to compute the number of space groups
for R5 and beyond. It is interesting to note that the crystallographic groups were
found mathematically for R3 before the 230 different types of crystals were actually
discovered in nature.
Exercises
1. Prove the identity
1
kx + yk2 − kxk2 − kyk2 .
hx, yi =
2
2. Show that O(n) is a group.
3. Prove that the following matrices are orthogonal. Are any of these matrices
in SO(n)?
(a) (b)
√ √ √ √
1/√2 −1/√ 2 1/ √5 2/√5
1/ 2 1/ 2 −2/ 5 1/ 5
EXERCISES 193
(c) (d)
√ √
4/ √5 0 3/√5 1/3 2/3 −2/3
−3/ 5 0 4/ 5 −2/3 2/3 1/3
0 −1 0 −2/3 1/3 2/3
4. Determine the symmetry group of each of the figures in Figure 12.10.
(a)
(c)
(b)
Figure 12.10.
5. Let x, y, and w be vectors in Rn and α ∈ R. Prove each of the following
properties of inner products.
(a) hx, yi = hy, xi.
(b) hx, y + wi = hx, yi + hx, wi.
(c) hαx, yi = hx, αyi = αhx, yi.
(d) hx, xi ≥ 0 with equality exactly when x = 0.
(e) If hx, yi = 0 for all x in Rn , then y = 0.
6. Verify that
E(n) = {(A, x) : A ∈ O(n) and x ∈ Rn }
is a group.
7. Prove that {(2, 1), (1, 1)} and {(12, 5), (7, 3)} are bases for the same lattice.
8. Let G be a subgroup of E(2) and suppose that T is the translation subgroup
of G. Prove that the point group of G is isomorphic to G/T .
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.
194 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. Show that an element in E(2) of the form (A, x), where x 6= 0, has infinite
order.
13. Prove or disprove: There exists an infinite abelian subgroup of O(n).
14. Let x = (x1 , x2 ) be a point on the unit circle in R2 ; that is, x21 + x22 = 1. If
A ∈ O(2), show that Ax is also a point on the unit circle.
15. Let G be a group with a subgroup H (not necessarily normal) and a normal
subgroup N . Then G is a semidirect product of N by H if
• H ∩ N = {id};
• HN = G.
Show that each of the following is true.
(a) S3 is the semidirect product of A3 by H = {(1), (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 12.5.
Figure 12.11.
17. Determine which of the 17 wallpaper groups preserves the symmetry of the
pattern in Figure 12.11.
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.
EXERCISES 195
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 Mathe-
matical Monthly 93 (1986), 765–79.
[4] Lockwood, E. H. and Macmillan, R. H. Geometric Symmetry. Cambridge
University Press, Cambridge, 1978.
[5] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[6] Martin, G. Transformation Groups: An Introduction to Symmetry. Springer-
Verlag, New York, 1982.
[7] Milnor, J. “Hilbert’s Problem 18: On Crystallographic Groups, Fundamental
Domains, and Sphere Packing,” Proceedings of Symposia in Pure Mathemat-
ics 18, American Mathematical Society, 1976.
[8] Phillips, F. C. An Introduction to Crystallography. 4th ed. Wiley, New York,
1971.
[9] Rose, B. I. and Stafford, R. D. “An Elementary Course in Mathematical
Symmetry,” American Mathematical Monthly 88 (1980), 54–64.
[10] Schattschneider, D. “The Plane Symmetry Groups: Their Recognition and
Their Notation,” American Mathematical Monthly 85 (1978), 439–50.
[11] Schwarzenberger, R. L. “The 17 Plane Symmetry Groups,” Mathematical
Gazette 58 (1974), 123–31.
[12] Weyl, H. Symmetry. Princeton University Press, Princeton, NJ, 1952.
13
The Structure of Groups
The ultimate goal of group theory is to classify all groups up to isomorphism;
that is, given a particular group, we should be able to match it up with a
known group via an isomorphism. For example, we have already proved that
any finite cyclic group of order n is isomorphic to Zn ; hence, we “know” all
finite cyclic groups. It is probably not reasonable to expect that we will ever
know all groups; however, we can often classify certain types of groups or
distinguish between groups in special cases.
In this chapter we will characterize all finite abelian groups. We shall also
investigate groups with sequences of subgroups. If a group has a sequence
of subgroups, say
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e},
where each subgroup Hi is normal in Hi+1 and each of the factor groups
Hi+1 /Hi is abelian, then G is a solvable group. In addition to allowing us
to distinguish between certain classes of groups, solvable groups turn out to
be central to the study of solutions to polynomial equations.
13.1 Finite Abelian Groups
In our investigation of cyclic groups we found that every group of prime order
was isomorphic to Zp , where p was a prime number. We also determined
that Zmn ∼ = Zm × Zn when gcd(m, n) = 1. In fact, much more is true.
Every finite abelian group is isomorphic to a direct product of cyclic groups
of prime power order; that is, every finite abelian group is isomorphic to a
group of the type
Zpα1 × · · · × Zpαnn .
1
196
13.1 FINITE ABELIAN GROUPS 197
First, let us examine a slight generalization of finite abelian groups.
Suppose that G is a group and let {gi } be a set of elements in G, where i
is in some index set I (not necessarily finite). The smallest subgroup of G
containing all of the gi ’s is the subgroup of G generated by the gi ’s. If this
subgroup of G is in fact all of G, then G is generated by the set {gi : i ∈ I}.
In this case the gi ’s are said to be the generators of G. If there is a finite
set {gi : i ∈ I} that generates G, then G is finitely generated.
Example 1. Obviously, all finite groups are finitely generated. For example,
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 2. Not all groups are finitely generated. Consider the rational
numbers Q under the operation of addition. Suppose that Q is finitely
generated with generators p1 /q1 , . . . , pn /qn , where each pi /qi is a fraction
expressed in its lowest terms. Let p be some prime that does not divide
any of the denominators q1 , . . . , qn . We claim that 1/p cannot be in the
subgroup of Q that is generated by p1 /q1 , . . . , pn /qn , since p does not divide
the denominator of any element in this subgroup. This fact is easy to see
since the sum of any two generators is
pi /qi + pj /qj = (pi qj + pj qi )/(qi qj ).
Theorem 13.1 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.
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 could always be simplified (in this case, to a4 b5 ).
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.
198 CHAPTER 13 THE STRUCTURE OF GROUPS
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 = g1k1 · · · giknn in K must also be in K. However,
g −1 = (g1k1 · · · giknn )−1 = (g1−kn · · · gi−k
n
1
).
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 that every finite
abelian group is isomorphic to a direct product of cyclic p-groups. Before we
state the main theorem concerning finite abelian groups, we shall consider
a special case.
Theorem 13.2 Every finite abelian group G is the direct product of p-
groups.
Proof. If |G| = 1, then the theorem is trivial. Suppose that the order of
G is greater than 1, say
|G| = pα1 1 · · · pαnn ,
where p1 , . . . , pn are all prime, and define Gi to be the set of elements in G of
order pki for some integer k. Since G is an abelian group, we are guaranteed
that Gi is a subgroup of G for i = 1, . . . , n. We must show that
G = G 1 × · · · × Gn .
That is, we must be able to write every g ∈ G as a unique product gp1 · · · gpn
where gpi is of the order of some power of pi . Since the order of g divides
the order of G, we know that
|g| = pβ1 1 pβ2 2 · · · pβnn
for integers β1 , . . . , βn . Letting ai = |g|/pβi i , the ai ’s are relatively prime;
hence, there exist integers b1 , . . . , bn such that a1 b1 + · · · + an bn = 1. Con-
sequently,
g = g a1 b1 +···+an bn = g a1 b1 · · · g an bn .
Since βi
g (ai bi )pi = g bi |g| = e,
13.1 FINITE ABELIAN GROUPS 199
it follows that g ai bi must be in Gi . Let gi = g ai bi . Then g = g1 · · · gn and
Gi ∩ Gj = {e} for i 6= j.
To show uniqueness, suppose that g = g1 · · · gn = h1 · · · hn , with hi ∈ Gi .
Then
e = (g1 · · · gn )(h1 · · · hn )−1 = g1 h−1 −1
1 · · · gn hn .
The order of gi h−1 −1 −1
i is a power of pi ; hence, the order of g1 h1 · · · gn hn is the
−1
least common multiple of the orders of the gi hi . This must be 1, since the
order of the identity is 1. Therefore, |gi h−1
i | = 1 or gi = hi for i = 1, . . . , n.
We shall now state the Fundamental Theorem of Finite Abelian Groups.
Theorem 13.3 (Fundamental Theorem of Finite Abelian Groups)
Every finite abelian group G is isomorphic to a direct product of cyclic groups
of the form
Zpα1 × Zpα2 × · · · × Zpαnn
1 2
where the pi ’s are primes (not necessarily distinct).
Example 3. Suppose that we wish to classify all abelian groups of order
540 = 22 · 33 · 5. The Fundamental Theorem of Finite Abelian Groups tells
us that we have the following six possibilities.
• Z2 × Z2 × Z3 × Z3 × Z3 × Z5 ;
• Z2 × Z2 × Z3 × Z9 × Z5 ;
• Z2 × Z2 × Z27 × Z5 ;
• Z4 × Z3 × Z3 × Z3 × Z5 ;
• Z4 × Z3 × Z9 × Z5 ;
• Z4 × Z27 × Z5 .
The proof of the Fundamental Theorem relies on the following lemma.
Lemma 13.4 Let G be a finite abelian p-group and suppose that g ∈ G has
maximal order. Then G can be written as hgi×H for some subgroup H of G.
200 CHAPTER 13 THE STRUCTURE OF GROUPS
Proof. Suppose that the order of G is pn . We shall induct on n. If n = 1,
then G is cyclic of order p and must be generated by g. Suppose now that
the statement of the lemma holds for all integers k with 1 ≤ k < n and let
m
g be of maximal order in G, say |g| = pm . Then ap = e for all a ∈ G.
Now choose h in G such that h ∈ / hgi, where h has the smallest possible
order. Certainly such an h exists; otherwise, G = hgi and we are done. Let
H = hhi.
We claim that hgi ∩ H = {e}. It suffices to show that |H| = p. Since
|h | = |h|/p, the order of hp is smaller than the order of h and must be in
p
hgi by the minimality of h; that is, hp = g r for some number r. Hence,
m−1 m−1 m
(g r )p = (hp )p = hp = e,
and the order of g r must be less than or equal to pm−1 . Therefore, g r cannot
generate hgi. Notice that p must occur as a factor of r, say r = ps, and
hp = g r = g ps . Define a to be g −s h. Then a cannot be in hgi; otherwise, h
would also have to be in hgi. Also,
ap = g −sp hp = g −r hp = h−p hp = e.
We have now formed an element a with order p such that a ∈ / hgi. Since h
was chosen to have the smallest order of all of the elements that are not in
hgi, |H| = p.
Now we will show that the order of gH in the factor group G/H must
be the same as the order of g in G. If |gH| < |g| = pm , then
m−1 m−1
H = (gH)p = gp H;
m−1
hence, g p must be in hgi ∩ H = {e}, which contradicts the fact that the
order of g is pm . Therefore, gH must have maximal order in G/H. By the
Correspondence Theorem and our induction hypothesis,
G/H ∼
= hgHi × K/H
for some subgroup K of G containing H. We claim that hgi ∩ K = {e}. If
b ∈ hgi ∩ K, then bH ∈ hgHi ∩ K/H = {H} and b ∈ hgi ∩ H = {e}. It
follows that G = hgiK implies that G ∼
= hgi × K.
The proof of the Fundamental Theorem of Finite Abelian Groups follows
very quickly from Lemma 13.4. Suppose that G is a finite abelian group and
let g be an element of maximal order in G. If hgi = G, then we are done;
otherwise, G ∼= Z|g| × H for some subgroup H contained in G by the lemma.
Since |H| < |G|, we can apply mathematical induction.
13.2 SOLVABLE GROUPS 201
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 13.5 (Fundamental Theorem of Finitely Generated Abelian
Groups) Every finitely generated abelian group G is isomorphic to a direct
product of cyclic groups of the form
Zpα1 × Zpα2 × · · · × Zpαnn × Z × · · · × Z,
1 2
where the pi ’s are primes (not necessarily distinct).
13.2 Solvable Groups
A subnormal series of a group G is a finite sequence of subgroups
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e},
where Hi is a normal subgroup of Hi+1 . If each subgroup Hi is normal in
G, then the series is called a normal series. The length of a subnormal
or normal series is the number of proper inclusions.
Example 4. Any series of subgroups of an abelian group is a normal series.
Consider the following series of groups:
Z ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0},
Z24 ⊃ h2i ⊃ h6i ⊃ h12i ⊃ {0}.
Example 5. A subnormal series need not be a normal series. Consider 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 6. The series
Z ⊃ 3Z ⊃ 9Z ⊃ 45Z ⊃ 90Z ⊃ 180Z ⊃ {0}
202 CHAPTER 13 THE STRUCTURE OF GROUPS
is a refinement of the series
Z ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0}.
The correct way to study a subnormal or normal series of subgroups,
{Hi } of G, is actually to study the factor groups Hi+1 /Hi . We say that two
subnormal (normal) series {Hi } and {Kj } of a group G are isomorphic if
there is a one-to-one correspondence between the collections of factor groups
{Hi+1 /Hi } and {Kj+1 /Kj }.
Example 7. The two normal series
Z60 ⊃ h3i ⊃ h15i ⊃ {0}
Z60 ⊃ h4i ⊃ h20i ⊃ {0}
of the group Z60 are isomorphic since
Z60 /h3i ∼
= h20i/{0} ∼
= Z3
∼
h3i/h15i = h4i/h20i ∼
= Z5
h15i/{0} ∼
= Z60 /h4i ∼
= Z4 .
A subnormal series {Hi } of a group G is a composition series if all the
factor groups are simple; that is, if none of the factor groups of the series
contains a normal subgroup. A normal series {Hi } of G is a principal
series if all the factor groups are simple.
Example 8. The group Z60 has a composition series
Z60 ⊃ h3i ⊃ h15i ⊃ h30i ⊃ {0}
with factor groups
Z60 /h3i ∼
= Z3
h3i/h15i =∼ Z5
h15i/h30i ∼
= Z2
h30i/{0} ∼
= Z2 .
13.2 SOLVABLE GROUPS 203
Since Z60 is an abelian group, this series is automatically a principal series.
Notice that a composition series need not be unique. The series
Z60 ⊃ h2i ⊃ h4i ⊃ h20i ⊃ {0}
is also a composition series.
Example 9. or n ≥ 5, the series
Sn ⊃ An ⊃ {(1)}
is a composition series for Sn since Sn /An ∼
= Z2 and An is simple.
Example 10. 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 nZ for some n ∈ N. In this case H1 /H0 ∼ = nZ 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 13.6 (Jordan-Hölder) Any two composition series of G are
isomorphic.
Proof. We shall employ mathematical induction on the length of the com-
position series. If the length of a composition series is 1, then G must be a
simple group. In this case any two composition series are isomorphic.
Suppose now that the theorem is true for all groups having a composition
series of length k, where 1 ≤ k < n. Let
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}
G = Km ⊃ Km−1 ⊃ · · · ⊃ K1 ⊃ K0 = {e}
204 CHAPTER 13 THE STRUCTURE OF GROUPS
be two composition series for G. We can form two new subnormal series for
G since Hi ∩ Km−1 is normal in Hi+1 ∩ Km−1 and Kj ∩ Hn−1 is normal in
Kj+1 ∩ Hn−1 :
G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e}
G = Km ⊃ Km−1 ⊃ Km−1 ∩ Hn−1 ⊃ · · · ⊃ K0 ∩ Hn−1 = {e}.
Since Hi ∩Km−1 is normal in Hi+1 ∩Km−1 , the Second Isomorphism Theorem
(Theorem 11.4) 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 se-
ries, 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}
EXERCISES 205
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 composition series {Hi } such that all
of the factor groups Hi+1 /Hi are abelian. Solvable groups will play a fun-
damental role when we study Galois theory and the solution of polynomial
equations.
Example 11. 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.
Exercises
1. Find all of the abelian groups of order less than or equal to 40 up to isomor-
phism.
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.
206 CHAPTER 13 THE STRUCTURE OF GROUPS
7. A group G is a torsion group if every element of G has finite order. Prove
that a finitely generated torsion group must be finite.
8. Let G, H, and K be finitely generated abelian groups. Show that if G × H ∼
=
G × K, then H ∼ = K. Give a counterexample to show that this cannot be
true in general.
9. Let G and H be solvable groups. Show that G × H is also solvable.
10. If G has a composition (principal) series and if N is a proper normal subgroup
of G, show there exists a composition (principal) series containing N .
11. Prove or disprove: Let N be a normal subgroup of G. If N and G/N have
composition series, then G must also have a composition series.
12. Let N be a normal subgroup of G. If N and G/N are solvable groups, show
that G is also a solvable group.
13. Prove that G is a solvable group if and only if G has a series of subgroups
G = Pn ⊃ Pn−1 ⊃ · · · ⊃ P1 ⊃ P0 = {e}
where Pi is normal in Pi+1 and the order of Pi+1 /Pi is prime.
14. Let G be a solvable group. Prove that any subgroup of G is also solvable.
15. Let G be a solvable group and N a normal subgroup of G. Prove that G/N
is solvable.
16. Prove that Dn is solvable for all integers n.
17. Suppose that G has a composition series. If N is a normal subgroup of G,
show that N and G/N also have composition series.
18. Let G be a cyclic p-group with subgroups H and K. Prove that either H is
contained in K or K is contained in H.
19. Suppose that G is a solvable group with order n ≥ 2. Show that G contains
a normal nontrivial abelian subgroup.
20. Recall that the commutator subgroup G0 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) = G0 , and
G(i+1) = (G(i) )0 .
(a) Prove that G(i+1) is normal in (G(i) )0 . 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.
EXERCISES 207
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 ∗ ).
=
[Hint: Use the diagram in Figure 13.1. The Zassenhaus Lemma is often
referred to as the Butterfly Lemma because of this diagram.]
H K
H ∩K
H ∗ (H ∩ K) K ∗ (H ∩ K)
H ∗ (H ∩ K ∗ ) K ∗ (H ∗ ∩ K)
H∗ K∗
(H ∗ ∩ K)(H ∩ K ∗)
H∗ ∩ K H ∩ K∗
Figure 13.1. The Zassenhaus Lemma
23. Schreier’s Theorem. Use the Zassenhaus Lemma to prove that two sub-
normal (normal) series of a group G have isomorphic refinements.
24. Use Schreier’s Theorem to prove the Jordan-Hölder Theorem.
Programming Exercises
Write a program that will compute all possible abelian groups of order n. What is
the largest n for which your program will work?
208 CHAPTER 13 THE STRUCTURE OF GROUPS
References and Suggested Readings
Each of the following references contains a proof of the Fundamental Theorem of
Finitely Generated Abelian Groups.
[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 may 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 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 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
209
210 CHAPTER 14 GROUP ACTIONS
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)
for σ ∈ G and x ∈ X.
Example 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 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.
14.1 GROUPS ACTING ON SETS 211
Example 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 0 )xH = g(g 0 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 14.1 Let X be a G-set. Then G-equivalence is an equivalence
relation on X.
Proof. The relation ∼ is reflexive since ex = x. Suppose that x ∼ y for
x, y ∈ X. Then there exists a g such that gx = y. In this case g −1 y = x;
hence, y ∼ x. To show that the relation is transitive, suppose that x ∼ y
and y ∼ z. Then there must exist group elements g and h such that gx = y
and hy = z. So z = hy = (hg)x, and x is equivalent to z.
If X is a G-set, then each partition of X associated with G-equivalence
is called an orbit of X under G. We will denote the orbit that contains an
element x of X by Ox .
Example 6. 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}.
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. It is important to remember that Xg ⊂ X and Gx ⊂ G.
Example 7. Let X = {1, 2, 3, 4, 5, 6} and suppose that G is the permutation
group given by the permutations
{(1), (12)(3456), (35)(46), (12)(3654)}.
212 CHAPTER 14 GROUP ACTIONS
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 14.2 Let G be a group acting on a set X and x ∈ X. The
stabilizer group, Gx , of x 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 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 14.3 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. 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 . First we must
show that this map is well-defined and does not depend on our selection of
g. Suppose that h is another element in G such that hx = y. Then gx = hx
or x = g −1 hx; hence, g −1 h is in the stabilizer subgroup of x. Therefore,
h ∈ gGx or gGx = hGx . Thus, y gets mapped to the same coset regardless
of the choice of the representative from that coset.
To show that φ is one-to-one, assume that φ(x1 ) = φ(x2 ). Then there
exist g1 , g2 ∈ G such that x1 = g1 x and x2 = g2 x. Since there exists a
g ∈ Gx such that g2 = g1 g,
x2 = g2 x = g1 gx = g1 x = x1 ;
14.2 THE CLASS EQUATION 213
consequently, the map φ is one-to-one. Finally, we must show that the map
φ is onto. Let gGx be a left coset. If gx = y, then φ(y) = gGx .
14.2 The Class Equation
Let X be a finite G-set and XG be the set of fixed points in X; that is,
XG = {x ∈ X : gx = x for all g ∈ G}.
Since the orbits of the action partition X,
n
X
|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 repre-
sentatives 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 12.3, 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 8. 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.
214 CHAPTER 14 GROUP ACTIONS
Example 9. The conjugacy classes for D4 are
{(1)}, {(13), (24)}, {(1432), (1234)}, {(12)(34), (14)(23), (13)(24)}.
The class equation is 8 = 1 + 2 + 2 + 3.
Example 10. 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 6.9,
τ στ −1 = (τ (a1 ), . . . , τ (ak )).
Consequently, any two cycles of the same length are conjugate. Now let
σ = σ1 σ2 · · · σr be a cycle decomposition, where the length of each cycle σi
is ri . Then σ is conjugate to every other τ ∈ Sn whose cycle decomposition
has the same lengths.
The number of conjugate classes in Sn is the number of ways in which
n can be partitioned into sums of positive integers. 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. The problem of finding the
number of such partitions for any positive integer n is what computer scien-
tists call NP-complete. This effectively means that the problem cannot be
solved for a large n because the computations would be too time-consuming
for even the largest computer.
Theorem 14.4 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, p must divide each ni . Also, p | |G|; hence, p
must divide |Z(G)|. Since the identity is always in the center of G, |Z(G)| ≥
1. Therefore, |Z(G)| ≥ p and there exists some g ∈ Z(G) such that g 6= 1.
14.3 BURNSIDE’S COUNTING THEOREM 215
Corollary 14.5 Let G be a group of order p2 where p is prime. Then G is
abelian.
Proof. By Theorem 14.4, |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 are 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 14.1).
B W W B
W W W W
W W W W
B W W B
Figure 14.1. Equivalent colorings of square
216 CHAPTER 14 GROUP ACTIONS
Burnside’s Counting Theorem offers a method of computing the number
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 Counting Theorem
depends on the following lemma.
Lemma 14.6 Let X be a G-set and suppose that x ∼ y. Then Gx is iso-
morphic 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(a).
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 14.7 (Burnside) Let G be a finite group acting on a set X and
let k denote the number of orbits of X. Then
1 X
k= |Xg |.
|G|
g∈G
Proof. We look at all the fixed points x of all the elements in g ∈ G; that
is, we look at all g’s and all x’s such that gx = x. If viewed in terms of fixed
point sets, the number of all g’s fixing x’s is
X
|Xg |.
g∈G
However, if viewed in terms of the stabilizer subgroups, this number is
X
|Gx |;
x∈X
14.3 BURNSIDE’S COUNTING THEOREM 217
P P
hence, g∈G |Xg | = x∈X |Gx |. By Lemma 14.6,
X
|Gy | = |Ox | · |Gx |.
y∈Ox
By Theorem 14.3 and Lagrange’s Theorem, this expression is equal to |G|.
Summing over all of the k distinct orbits, we conclude that
X X
|Xg | = |Gx | = k · |G|.
g∈G x∈X
Example 11. Let X = {1, 2, 3, 4, 5} and suppose that G is the permutation
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 X 1
k= |Xg | = (5 + 3 + 1 + 3) = 3.
|G| 4
g∈G
A Geometric Example
Before we apply Burnside’s Theorem to switching-theory problems, let us
examine the number of ways in which the vertices of a square can be colored
black or white. Notice that we can sometimes obtain equivalent colorings
by simply applying a rigid motion to the square. For instance, as we have
pointed out, if we color one of the vertices black and the remaining three
white, it does not matter which vertex was colored black since a rotation
will give an equivalent coloring.
The symmetry group of a square, D4 , is given by the following permu-
tations:
(1) (13) (24) (1432)
(1234) (12)(34) (14)(23) (13)(24)
218 CHAPTER 14 GROUP ACTIONS
The group G acts on the set of vertices {1, 2, 3, 4} in the usual manner. We
can describe the different colorings by mappings from X into Y = {B, W }
where B and W represent the colors black and white, respectively. Each map
f : X → Y describes a way to color the corners of the square. Every σ ∈ D4
induces a permutation σ e of the possible colorings given by σe(f ) = f ◦ σ for
f : X → Y . For example, suppose that f is defined by
f (1) = B
f (2) = W
f (3) = W
f (4) = W
and σ = (12)(34). Then σ e(f ) = f ◦ σ sends vertex 2 to B and the remaining
vertices to W . The set of all such σ e is a permutation group G e on the set
of possible colorings. Let X denote the set of all possible colorings; that is,
e
Xe is the set of all possible maps from X to Y . Now we must compute the
number of G-equivalence
e classes.
1. X e(1) = Xe since the identity fixes every possible coloring. |X| e =
24 = 16.
2. Xe(1234) consists of all f ∈ X
e such that f is unchanged by the permuta-
tion (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 |X e(1234) | = 2.
3. |X
e(1432) | = 2.
e(13)(24) | = 22 = 4.
e(13)(24) , f (1) = f (3) and f (2) = f (4). Thus, |X
4. For X
5. |X
e(12)(34) | = 4.
6. |X
e(14)(23) | = 4.
7. For X
e(13) , f (1) = f (3) and the other corners can be of any color;
e(13) | = 23 = 8.
hence, |X
8. |X
e(24) | = 8.
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.
14.3 BURNSIDE’S COUNTING THEOREM 219
Proposition 14.8 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 G e acting
on X, where σ
e e ∈ G is defined by σ
e e(f ) = f ◦ σ for σ ∈ G and f ∈ X. e
Furthermore, if n is the number of cycles in the cycle decomposition of σ,
then |Xeσ | = |Y |n .
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 σ e(f ) = σe(g). Then for each
x ∈ X,
f (σ(x)) = σ
e(f )(x) = σ
e(g)(x) = g(σ(x)).
Since σ is a permutation of X, every element x0 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 12. 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 ∈ Fg must have the same value on each cycle in g. There are |Y | = 3
such choices for any value, so |Fg | = 34 = 81.
Example 13. Suppose that we wish to color the vertices of a square using
four different colors. By Proposition 12.8, we can immediately decide that
there are
1 4
(4 + 41 + 42 + 41 + 42 + 42 + 43 + 43 ) = 55
8
possible ways.
x1
x2
.. f f (x1 , x2 , . . . , xn )
.
xn
Figure 14.2. A switching function of n variables
220 CHAPTER 14 GROUP ACTIONS
Switching Functions
In switching theory we are concerned with the design of electronic circuits
with binary inputs and outputs. The simplest of these circuits is a switching
function that has n inputs and a single output (Figure 14.2). 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 14.3).
a a
f f (a, b) f f (b, a) = g(a, b)
b b
Figure 14.3. A switching function of two 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
n
values for each binary n-tuple and there are 2n binary n-tuples, 22 switching
functions are possible for n variables. In general, allowing permutations of
the inputs greatly reduces the number of different kinds of modules that are
needed to build a large circuit.
The possible switching functions with two input variables a and b are
listed in Table 14.1. 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
f10 ∼ f12
f11 ∼ f13 .
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 neces-
sary to employ Burnside’s Theorem.
14.3 BURNSIDE’S COUNTING THEOREM 221
Table 14.1. Switching functions in two variables
Inputs Outputs
f0 f1 f2 f3 f4 f5 f6 f7
0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1
Inputs Outputs
f8 f9 f10 f11 f12 f13 f14 f15
0 0 1 1 1 1 1 1 1 1
0 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1
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).
222 CHAPTER 14 GROUP ACTIONS
Let X be the set of output values for a switching function in n variables.
Then |X| = 2n . We can enumerate these values as follows:
(0, . . . , 0, 1) 7→ 0
(0, . . . , 1, 0) 7→ 1
(0, . . . , 1, 1) 7→ 2
..
.
(1, . . . , 1, 1) 7→ 2n − 1.
Now let us consider a circuit with four input variables and a single out-
put. Suppose that we can permute the leads of any circuit according to the
following permutation group:
(a) (ac) (bd) (adcb)
(abcd) (ab)(cd) (ad)(bc) (ac)(bd).
The permutations of the four possible input variables induce the permuta-
tions of the output values in Table 14.2.
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.
Table 14.2. Permutations of switching functions in four variables
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
Historical Note
EXERCISES 223
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, differ-
ential 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.
Exercises
1. Examples 1–5 in the first section each describe an action of a group G on a set
X, which will give rise to the equivalence relation defined by G-equivalence.
For each example, compute the equivalence classes of the equivalence relation,
the G-equivalence classes .
2. Compute all Xg and all Gx for each of the following permutation groups.
(a) X = {1, 2, 3},
G = S3 = {(1), (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 2.
For each x ∈ X verify that |G| = |Ox | · |Gx |.
4. Let G be the additive group of real numbers. Let the action of θ ∈ G on
the real plane R2 be given by rotating the plane counterclockwise about the
origin through θ radians. Let P be a point on the plane other than the origin.
(a) Show that R2 is a G-set.
(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.
224 CHAPTER 14 GROUP ACTIONS
(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 (c) Z9
(b) D5 (d) Q8
7. Write the class equation for S5 and for A5 .
8. If a square remains fixed in the plane, how many different ways can the
corners of the square be colored if three colors are used?
9. How many ways can the vertices of an equilateral triangle be colored using
three different colors?
10. Find the number of ways a six-sided die can be constructed if each side is
marked differently with 1, . . . , 6 dots.
11. Up to a rotation, how many ways can the faces of a cube be colored with
three different colors?
12. Consider 12 straight wires of equal lengths with their ends soldered together
to form the edges of a cube. Either silver or copper wire can be used for each
edge. How many different ways can the cube be constructed?
13. Suppose that we color each of the eight corners of a cube. Using three
different colors, how many ways can the corners be colored up to a rotation
of the cube?
14. Each of the faces of a regular tetrahedron can be painted either red or white.
Up to a rotation, how many different ways can the tetrahedron be painted?
15. Suppose that the vertices of a regular hexagon are to be colored either red or
white. How many ways can this be done up to a symmetry of the hexagon?
16. A molecule of benzene is made up of six carbon atoms and six hydrogen
atoms, linked together in a hexagonal shape as in Figure 14.4.
(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.
17. How many equivalence classes of switching functions are there if the input
variables x1 , x2 , and x3 can be permuted by any permutation in S3 ? What if
the input variables x1 , x2 , x3 , and x4 can be permuted by any permutation
in S4 ?
EXERCISES 225
H
H H
H H
H
Figure 14.4. A benzene ring
18. How many equivalence classes of switching functions are there if the input
variables x1 , x2 , x3 , and x4 can be permuted by any permutation in the
subgroup of S4 generated by the permutation (x1 x2 x3 x4 )?
19. A striped necktie has 12 bands of color. Each band can be colored by one of
four possible colors. How many possible different-colored neckties are there?
20. A group acts faithfully on a G-set X if the identity is the only element of
G that leaves every element of X fixed. Show that G acts faithfully on X
if and only if no two distinct elements of G have the same action on each
element of X.
21. Let p be prime. Show that the number of different abelian groups of order pn
(up to isomorphism) is the same as the number of conjugacy classes in Sn .
22. Let a ∈ G. Show that for any g ∈ G, gC(a)g −1 = C(gag −1 ).
23. Let |G| = pn and suppose that |Z(G)| = pn−1 for p prime. Prove that G is
abelian.
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).
Programming Exercise
Write a program to compute the number of conjugacy classes in Sn . What is the
largest n for which your program will work?
226 CHAPTER 14 GROUP ACTIONS
References and Suggested Reading
[1] De Bruijin, N. G. “Pólya’s Theory of Counting,” in Applied Combinatorial
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 Contributions to
Chemical Enumeration,” in Chemical Applications of Graph Theory, Bala-
ban, A. T., ed. Academic Press, London, 1976.
[4] Gåding, 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,” Mathe-
matics 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 sub-
groups 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 the idea of group actions to prove the Sylow Theorems. Recall
for a moment what it means for G to act on itself by conjugation and
how conjugacy classes are distributed in the group according to the class
equation, discussed in Chapter 14. A group G acts on itself by conjugation
via the map (g, x) 7→ gxg −1 . Let x1 , . . . , xk be representatives from each of
the distinct conjugacy classes of G that consist of more than one element.
Then the class equation can be written as
|G| = |Z(G)| + [G : C(x1 )] + · · · + [G : C(xk )],
where Z(G) = {g ∈ G : gx = xg for all x ∈ G} is the center of G and
C(xi ) = {g ∈ G : gxi = xi g} is the centralizer subgroup of xi .
We now begin our investigation of the Sylow Theorems by examining
subgroups of order p, where p is prime. A group G is a p-group if every
element in G has as its order a power of p, where p is a prime number. A
subgroup of a group G is a p-subgroup if it is a p-group.
Theorem 15.1 (Cauchy) Let G be a finite group and p a prime such that
p divides the order of G. Then G contains a subgroup of order p.
227
228 CHAPTER 15 THE SYLOW THEOREMS
Proof. We will use induction on the order of G. If |G| = p, then clearly G
must have an element of order p. 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. The order of one of the centralizer subgroups, C(xi ), is divisible
by p for some i, i = 1, . . . , k. In this case, by our induction hypothesis, we
are done. Since C(xi ) is a proper subgroup of G and p divides |C(xi )|, C(xi )
must contain an element of order p. Hence, G must contain an element of
order p.
Case 2. 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 1. 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 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 15.3 (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. 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 ap-
ply the induction hypothesis to C(xi ).
15.1 THE SYLOW THEOREMS 229
Hence, we may assume that p divides [G : C(xi )] for all i. Since p divides
|G|, the class equation says that p must divide |Z(G)|; hence, by Cauchy’s
Theorem, Z(G) has an element of order p, say g. Let N be the group
generated by g. Clearly, N is a normal subgroup of Z(G) since Z(G) is
abelian; therefore, N is normal in G since every element in Z(G) commutes
with every element in G. Now consider the factor group G/N of order |G|/p.
By the induction hypothesis, G/N contains a subgroup H of order pr−1 . The
inverse image of H under the canonical homomorphism φ : G → G/N is a
subgroup of order pr in G.
A Sylow p-subgroup P of a group G is a maximal p-subgroup of G.
To prove the other two Sylow Theorems, we need to consider conjugate
subgroups as opposed to conjugate elements in a group. For a group G, let
S be the collection of all subgroups of G. For any subgroup H, S is a H-set,
where H acts on S by conjugation. That is, we have an action
H ×S →S
defined by
h · K 7→ hKh−1
for K in S.
The set
N (H) = {g ∈ G : gHg −1 = H}
is a subgroup of G. 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. We call N (H) the
normalizer of H in G.
Lemma 15.4 Let P be a Sylow p-subgroup of a finite group G and let x
have as its order a power of p. If x−1 P x = P . Then x ∈ P .
Proof. Certainly x ∈ N (P ), and the cyclic subgroup, hxP i ⊂ N (P )/P ,
has as its order a power of p. By the Correspondence Theorem there exists
a subgroup H of N (P ) such that H/P = hxP i. Since |H| = |P | · |hxP i|,
the order of H must be a power of p. However, P is a Sylow p-subgroup
contained in H. Since the order of P is the largest power of p dividing |G|,
H = P . Therefore, H/P is the trivial subgroup and xP = P , or x ∈ P .
Lemma 15.5 Let H and K be subgroups of G. The number of distinct
H-conjugates of K is [H : N (K) ∩ H].
230 CHAPTER 15 THE SYLOW THEOREMS
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
1 ∈ N (K). Therefore, K = h2 h1 Kh1 h2 or
h−1 −1
1 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 15.6 (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
and |P | = pr . Let
P = {P = P1 , P2 , . . . , Pk }
consist of the distinct conjugates of P in G. By Lemma 15.5, 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 ∈ P. Consider the Q-conjugacy classes of
each Pi . Clearly, these conjugacy classes partition P. The size of the parti-
tion containing Pi is [Q : N (Pi ) ∩ Q]. Lagrange’s Theorem tells us that this
number is 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 . Therefore, for some Pj , x−1 Pj x = Pj for all x ∈ Q. By
Lemma 15.4, Pj = Q.
Theorem 15.7 (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,
P = {P = P1 , P2 , . . . , Pk },
by conjugation. From the proof of the Second Sylow Theorem, the only
P -conjugate of P is itself and the order of the other P -conjugacy classes is a
15.2 EXAMPLES AND APPLICATIONS 231
power of p. Each P -conjugacy class contributes a positive power of p toward
|P| except the equivalence class {P }. Since |P| is the sum of positive powers
of p and 1, |P| ≡ 1 (mod p).
Now suppose that G acts on P by conjugation. Since all Sylow p-
subgroups are conjugate, there can be only one orbit under this action.
For P ∈ P,
|P| = |orbit of P| = [G : N (P )].
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 ob-
tained a temporary appointment at Christiania University. Even though his ap-
pointment 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 2. 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.
232 CHAPTER 15 THE SYLOW THEOREMS
Theorem 15.8 If p and q are distinct primes with p < q, then every group
G of order pq has a single subgroup of order q and this subgroup is normal
in G. Hence, G cannot be simple. Furthermore, if q ≡ 6 1 (mod p), then G
is cyclic.
Proof. We know that G contains a subgroup H of order q. The number of
conjugates of H divides pq and is equal to 1 + kq for k = 0, 1, . . .. However,
1 + q is already too large to divide the order of the group; hence, H can only
be conjugate to itself. That is, H must be normal in G.
The group G also has a Sylow p-subgroup, say K. The number of con-
jugates of K must divide q and be equal to 1 + kp for k = 0, 1, . . .. Since q
is prime, either 1 + kp = q or 1 + kp = 1. If 1 + kp = 1, then K is normal
in G. In this case, we can easily show that G satisfies the criteria, given in
Chapter 8, for the internal direct product of H and K. Since H is isomor-
phic to Zq and K is isomorphic to Zp , G ∼ = Zp × Zq ∼= Zpq by Theorem 9.10.
Example 3. Every group of order 15 is cyclic. This is true because 15 = 5·3
and 5 6≡ 1 (mod 3).
Example 4. 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 14.5, 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 15.9 Let G0 = haba−1 b−1 : a, b ∈ Gi be the subgroup consisting
of all finite products of elements of the form aba−1 b−1 in a group G. Then
G0 is a normal subgroup of G and G/G0 is abelian.
15.2 EXAMPLES AND APPLICATIONS 233
The subgroup G0 of G is called the commutator subgroup of G. We
leave the proof of this theorem as an exercise.
Example 5. We will now show that every group of order 5 · 7 · 47 = 1645
is abelian, and cyclic by Corollary 9.11. 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 15.8. Hence, the commutator subgroup of G is
contained in H which tells us that |G0 | is either 1 or 47. If |G0 | = 1, we
are done. Suppose that |G0 | = 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, G0 must be a subgroup of
Hi , i = 1, 2. Therefore, the order of G0 is 1, 5, or 7. However, we already
have determined that |G0 | = 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 6. Let us show that no group G of order 20 can be simple. By
the Third Sylow Theorem, G contains one or more Sylow 5-subgroups. The
number of such subgroups is congruent to 1 (mod 5) and must also divide
20. The only possible such number is 1. Since there is only a single Sylow
5-subgroup and all Sylow 5-subgroups are conjugate, this subgroup must be
normal.
Example 7. Let G be a finite group of order pn , n > 1 and p prime. By
Theorem 14.4, 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 14.5.
Example 8. 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
234 CHAPTER 15 THE SYLOW THEOREMS
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 15.10 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; conse-
quently,
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
EXERCISES 235
the form hi ki for hi ∈ H and ki ∈ K, as many times as there are elements
in H ∩ K; that is, |H ∩ K| times. Therefore, |HK| = (|H| · |K|)/|H ∩ K|.
Example 9. 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 13.10,
16 · 16
|HK| = = 64,
4
which is impossible. So H ∩ K is normal in both H and K since it has index
2. The normalizer of H ∩ K contains both H and K, and |H ∩ K| must
both be a multiple of 16 greater than 1 and divide 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 15.11 (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.
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.
236 CHAPTER 15 THE SYLOW THEOREMS
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
q6 |p2 − 1 and p6 |q 2 − 1. Prove that G must be abelian. List three pairs of
primes satisfying these conditions.
9. Show directly that a group of order 33 has only one Sylow 3-subgroup.
10. Let H be a subgroup of a group G. Prove or disprove that the normalizer of
H is normal in G.
11. Let G be a finite group 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 automor-
phism 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.
EXERCISES 237
(a) Suppose p is prime and p does not divide m. Show that
k
p m
p6 | .
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 p6 ||OT | for some T ∈ S.
(e) Let {T1 , . . . , Tu } be an orbit such that p6 |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 G0 = haba−1 b−1 : a, b ∈ Gi is a normal subgroup
of G and G/G0 is abelian. Find an example to show that {aba−1 b−1 : a, b ∈
G} is not necessarily a group.
A Project
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 15.1. Numbers of distinct groups G, |G| ≤ 60
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
238 CHAPTER 15 THE SYLOW THEOREMS
of order less than or equal to 60. However, we can break the problem down into
several intermediate problems.
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.
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 15.2 as a guide.
References and Suggested Readings
[1] Edwards, H. “A Short History of the Fields Medal,” Mathematical Intelli-
gencer 1 (1978), 127–29.
[2] Feit, W. and Thompson, J. G. “Solvability of Groups of Odd Order,” Pacific
Journal of Mathematics 13 (1963), 775–1029.
[3] Gallian, J. A. “The Search for Finite Simple Groups,” Mathematics Magazine
49(1976), 163–79.
[4] Gorenstein, D. “Classifying the Finite Simple Groups,” Bulletin of the Amer-
ican Mathematical Society 14 (1986), 1–98.
[5] Gorenstein, D. Finite Groups. AMS Chelsea Publishing, Providence RI, 1968.
[6] Gorenstein, D., Lyons, R., and Solomon, R. The Classification of Finite
Simple Groups. American Mathematical Society, Providence RI, 1994.
16
Rings
Up to this point we have studied sets with a single binary operation satis-
fying certain axioms, but often we are more interested in working with sets
that have two binary operations. For example, one of the most natural alge-
braic structures to study is the integers with the operations of addition and
multiplication. These operations are related to one another by the distribu-
tive property. If we consider a set with two such related binary operations
satisfying certain axioms, we have an algebraic structure called a ring. In a
ring we add and multiply such elements 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.
239
240 CHAPTER 16 RINGS
This last condition, the distributive axiom, relates the binary operations of
addition and multiplication. Notice that the first four axioms simply require
that a ring be an abelian group under addition, so we could also have defined
a ring to be an abelian group (R, +) together with a second binary operation
satisfying the fifth and sixth conditions given above.
If there is an element 1 ∈ R such that 1 6= 0 and 1a = a1 = a for
each element a ∈ R, we say that R is a ring with unity or identity . A
ring R for which ab = ba for all a, b in R is called a commutative ring.
A commutative ring R with identity is called an integral domain if, for
every a, b ∈ R such that ab = 0, either a = 0 or b = 0. A division ring
is a ring R, with an identity, in which every nonzero element in R is a
unit; that is, for each a ∈ R with a 6= 0, there exists a unique element a−1
such that a−1 a = aa−1 = 1. A commutative division ring is called a field.
The relationship among rings, integral domains, division rings, and fields is
shown in Figure 16.1.
Rings
Commutative Rings with
Rings Identity
Integral Division
Domains Rings
Fields
Figure 16.1. Types of rings
Example 1. 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 2. Under the ordinary operations of addition and multiplication,
all of the familiar number systems are rings: the rationals, Q; the real
numbers, R; and the complex numbers, C. Each of these rings is a field.
16.1 RINGS 241
Example 3. 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 4. 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 5. The 2 × 2 matrices with entries in R form a ring under
the usual operations of matrix addition and multiplication. This ring is
noncommutative, since it is usually the case that AB 6= BA. Also, notice
that we can have AB = 0 when neither A nor B is zero.
Example 6. 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
242 CHAPTER 16 RINGS
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 + a1 b1 + c1 d2 − d1 c2
γ = a1 c2 − b1 d2 + c1 a2 − d1 b2
δ = a1 d2 + b1 c2 − c1 b2 − d1 a2 .
Though multiplication looks complicated, it is actually a straightforward
computation if we remember that we just add and multiply elements in H
like polynomials and keep in mind the relationships between the generators
i, j, and k. The ring H is called the ring of quaternions.
To show that the quaternions are a division ring, we must be able to find
an inverse for each nonzero element. Notice that
(a + bi + cj + dk)(a − bi − cj − dk) = a2 + b2 + c2 + d2 .
This element can be zero only if a, b, c, and d are all zero. So if a + bi + cj +
dk 6= 0,
a − bi − cj − dk
(a + bi + cj + dk) = 1.
a2 + b2 + c2 + d2
Proposition 16.1 Let R be a ring with a, b ∈ R. Then
1. a0 = 0a = 0;
2. a(−b) = (−a)b = −ab;
3. (−a)(−b) = ab.
16.1 RINGS 243
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 7. 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 16.2 Let R be a ring and S a subset of R. Then S is a subring
of R if and only if the following conditions are satisfied.
1. S 6= ∅.
2. rs ∈ S for all r, s ∈ S.
3. r − s ∈ S for all r, s ∈ S.
Example 8. 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
a0 b0
a b
A= and B =
0 c 0 c0
are in T , then clearly A − B is also in T . Also,
0
aa ab0 + bc0
AB =
0 cc0
is in T .
244 CHAPTER 16 RINGS
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 9. 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 10. 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 11. 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 16.3 (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.
16.2 INTEGRAL DOMAINS AND FIELDS 245
Proof. Let D be an integral domain. Then D has no zero divisors. Let
ab = ac with a 6= 0. Then a(b − c) = 0. Hence, b − c = 0 and b = c.
Conversely, let us suppose that cancellation is possible in D. That is,
suppose that ab = ac implies b = c. Let ab = 0. If a 6= 0, then ab = a0 or
b = 0. Therefore, a cannot be a zero divisor.
The following surprising theorem is due to Wedderburn.
Theorem 16.4 Every finite integral domain is a field.
Proof. Let D be a finite integral domain and D∗ be the set of nonzero
elements of D. We must show that every element in D∗ has an inverse. For
each a ∈ D∗ we can define a map λa : D∗ → D∗ by λa (d) = ad. This map
makes sense, because if a 6= 0 and d 6= 0, then ad 6= 0. The map λa is
one-to-one, since for d1 , d2 ∈ D∗ ,
ad1 = λa (d1 ) = λa (d2 ) = ad2
implies d1 = d2 by left cancellation. Since D∗ is a finite set, the map λa
must also be onto; hence, for some d ∈ D∗ , λa (d) = ad = 1. Therefore, a
has a left inverse. Since D is commutative, d must also be a right inverse
for a. Consequently, D is a field.
For any nonnegative integer n and any element r in a ring R we write
r + · · · + r (n times) as nr. We define the characteristic of a ring R to be
the least positive integer n such that nr = 0 for all r ∈ R. If no such integer
exists, then the characteristic of R is defined to be 0.
Example 12. For every prime p, Zp is a field of characteristic p. By
Proposition 3.1, 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.
Theorem 16.5 The characteristic of an integral domain is either prime
or zero.
Proof. Let D be an integral domain and suppose that the characteristic
of D is n with n 6= 0. If n is not prime, then n = ab, where 1 < a < n and
1 < b < n. 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.
246 CHAPTER 16 RINGS
16.3 Ring Homomorphisms and Ideals
In the study of groups, a homomorphism is a map that preserves the op-
eration of the group. Similarly, a homomorphism between rings preserves
the operations of addition and multiplication in the ring. More specifically,
if R and S are rings, then a ring homomorphism is a map φ : R → S
satisfying
φ(a + b) = φ(a) + φ(b)
φ(ab) = φ(a)φ(b)
for all a, b ∈ R. If φ : R → S is a one-to-one and onto homomorphism, then
φ is called an isomorphism of rings.
The set of elements that a ring homomorphism maps to 0 plays a funda-
mental 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 13. For any integer n we can define a ring homomorphism
φ : Z → Zn by a 7→ a (mod n). This is indeed a ring homomorphism,
since
φ(a + b) = (a + b) (mod n)
=a (mod n) + b (mod n)
= φ(a) + φ(b)
and
φ(ab) = ab (mod n)
= a (mod n) · b (mod n)
= φ(a)φ(b).
The kernel of the homomorphism φ is nZ.
Example 14. Let C[a, b] be the ring of continuous real-valued functions
on an interval [a, b] as in Example 4. 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).
16.3 RING HOMOMORPHISMS AND IDEALS 247
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 16.6 Let φ : R → S be a ring homomorphism.
1. If R is a commutative ring, then φ(R) is a commutative ring.
2. φ(0) = 0.
3. Let 1R and 1S be the identities for R and S, respectively. If φ is onto,
then φ(1R ) = 1S .
4. If R is a field and φ(R) 6= 0, then φ(R) is a field.
In group theory we found that normal subgroups play a special role.
These subgroups have nice characteristics that make them more interesting
to study than arbitrary subgroups. In ring theory the objects corresponding
to normal subgroups are a special class of subrings called ideals. An ideal
in a ring R is a subring I of R such that if a is in I and r is in R, then both
ar and ra are in I; that is, rI ⊂ I and Ir ⊂ I for all r ∈ R.
Example 15. 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 R. Since for any r ∈ R, r1 = r ∈ I by the definition of an ideal,
I = R.
Example 16. If a is any element in a commutative ring R with identity,
then the set
hai = {ar : r ∈ R}
is an ideal in R. Certainly, hai is nonempty since both 0 = a0 and a = a1 are
in hai. The sum of two elements in hai is again in hai since ar+ar0 = a(r+r0 ).
The inverse of ar is −ar = a(−r) ∈ hai. Finally, if we multiply an element
ar ∈ hai by an arbitrary element s ∈ R, we have s(ar) = a(sr). Therefore,
hai satisfies the definition of an ideal.
If R is a commutative ring with identity, then an ideal of the form hai =
{ar : r ∈ R} is called a principal ideal.
Theorem 16.7 Every ideal in the ring of integers Z is a principal ideal.
248 CHAPTER 16 RINGS
Proof. The zero ideal {0} is a principal ideal since h0i = {0}. If I is any
nonzero ideal in Z, then I must contain some positive integer m. There
exists at least one such positive integer n in I by the Principle of Well-
Ordering. Now let a be any element in I. Using the division algorithm, we
know that there exist integers q and r such that
a = nq + r
where 0 ≤ r < n. This equation tells us that r = a − nq ∈ I, but r must be
0 since n is the least positive element in I. Therefore, a = nq and I = hni.
Example 17. 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 16.7, these are
the only ideals of Z.
Proposition 16.8 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. 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 16.9 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.
16.3 RING HOMOMORPHISMS AND IDEALS 249
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 r0 ∈ r +I and s0 ∈ s+I,
then r0 s0 must be in rs + I. Since r0 ∈ r + I, there exists an element a in
I such that r0 = r + a. Similarly, there exists a b ∈ I such that s0 = s + b.
Notice that
r0 s0 = (r + a)(s + b) = rs + as + rb + ab
and as + rb + ab ∈ I since I is an ideal; consequently, r0 s0 ∈ rs + I. We will
leave as an exercise the verification of the associative law for multiplication
and the distributive laws.
The ring R/I in Theorem 16.9 is called the factor or quotient ring.
Just as with group homomorphisms and normal subgroups, there is a rela-
tionship between ring homomorphisms and ideals.
Theorem 16.10 Let I be an ideal of R. The map ψ : R → R/I defined by
ψ(r) = r + I is a ring homomorphism of R onto R/I with kernel I.
Proof. Certainly ψ : R → R/I is a surjective abelian group homomor-
phism. It remains to show that ψ works correctly under ring multiplication.
Let r and s be in R. Then
ψ(r)ψ(s) = (r + I)(s + I) = rs + I = ψ(rs),
which completes the proof of the theorem.
The map ψ : R → R/I is often called the natural or canonical homo-
morphism. In ring theory we have isomorphism theorems relating ideals
and ring homomorphisms similar to the isomorphism theorems for groups
that relate normal subgroups and homomorphisms in Chapter 11. We will
prove only the First Isomorphism Theorem for rings in this chapter and
leave the proofs of the other two theorems as exercises. All of the proofs are
similar to the proofs of the isomorphism theorems for groups.
Theorem 16.11 (First Isomorphism Theorem) Let φ : R → S be a
ring homomorphism. Then ker φ is an ideal of R. If ψ : R → R/ ker φ
is the canonical homomorphism, then there exists a unique isomorphism
η : R/ ker φ → φ(R) such that φ = ηψ.
Proof. Let K = ker φ. 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
250 CHAPTER 16 RINGS
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 16.12 (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 16.13 (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 16.14 (Correspondence Theorem) Let I be a 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
containing I correspond to ideals of R/I.
16.4 Maximal and Prime Ideals
In this particular section we are especially interested in certain ideals of
commutative rings. These ideals give us special types of factor rings. More
specifically, we would like to characterize those ideals I of a commutative
ring R such that R/I is an integral domain or a field.
A proper ideal M of a ring R is a maximal ideal of R if the ideal
M is not a proper subset of any ideal of R except R itself. That is, M
is a maximal ideal if for any ideal I properly containing M , I = R. The
following theorem completely characterizes maximal ideals for commutative
rings with identity in terms of their corresponding factor rings.
Theorem 16.15 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.
16.4 MAXIMAL AND PRIME IDEALS 251
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 14.2 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 a 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 18. Let pZ be an ideal in Z, where p is prime. Then pZ is a
maximal ideal since Z/pZ ∼
= Zp is a field.
An ideal P in a commutative ring R is called a prime ideal if whenever
ab ∈ P , then either a ∈ P or b ∈ P .
Example 19. 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 16.16 Let R be a commutative ring with identity. Then P is
a prime ideal in R if and only if R/P is an integral domain.
Proof. First let us assume that P is an ideal in R and R/P is an integral
domain. Suppose that ab ∈ P . If a + P and b + P are two elements of R/P
252 CHAPTER 16 RINGS
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 20. 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 16.17 Every maximal ideal in a commutative ring with identity
is also a prime ideal.
Historical Note
Amalie Emmy Noether, one of the outstanding mathematicians of this century, was
born in Erlangen, Germany in 1882. She was the daughter of Max Noether (1844–
1921), a distinguished mathematician at the University of Erlangen. Together with
Paul Gordon (1837–1912), Emmy Noether’s father strongly influenced her early
education. She entered the University of Erlangen at the age of 18. Although
women had been admitted to universities in England, France, and Italy for decades,
there was great resistance to their presence at universities in Germany. Noether
was one of only two women among the university’s 986 students. After completing
her doctorate under Gordon in 1907, she continued to do research at Erlangen,
occasionally lecturing when her father was ill.
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.
16.5 AN APPLICATION TO SOFTWARE DESIGN 253
In 1922, Noether became a Privatdozent at Göttingen. Over the next 11 years
she used axiomatic methods to develop an abstract theory of rings and ideals.
Though she was not good at lecturing, Noether was an inspiring teacher. One of her
many students was B. L. van der Waerden, author of the first text treating abstract
algebra from a modern point of view. Some of the other mathematicians Noether
influenced or closely worked with were Alexandroff, Artin, Brauer, Courant, Hasse,
Hopf, Pontryagin, von Neumann, and Weyl. One of the high points of her career
was an invitation to address the International Congress of Mathematicians in Zurich
in 1932. In spite of all the recognition she received from her colleagues, Noether’s
abilities were never recognized as they should have been during her lifetime. She
was never promoted to full professor by the Prussian academic bureaucracy.
In 1933, Noether, a Jew, was banned from participation in all academic activi-
ties 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-tsı̈ wrote about the theorem in the first century A.D.
This theorem has some interesting consequences in the design of software
for parallel processors.
Lemma 16.18 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)
254 CHAPTER 16 RINGS
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).
Example 21. 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 = −1 and t = 1. Consequently,
x = a + k1 m = 3 + 4k1 = 3 + 4[(5 − 4)4] = 19.
Theorem 16.19 (Chinese Remainder Theorem) Let n1 , n2 , . . . , nk be
positive integers such that gcd(ni , nj ) = 1 for i 6= j. Then for any integers
a1 , . . . , ak , the system
x ≡ a1 (mod n1 )
x ≡ a2 (mod n2 )
..
.
x ≡ ak (mod nk )
has a solution. Furthermore, any two solutions of the system are congruent
modulo n1 n2 · · · nk .
16.5 AN APPLICATION TO SOFTWARE DESIGN 255
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 16.18. 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 mod-
ulo 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 22. Let us solve the system
x≡3 (mod 4)
x≡4 (mod 5)
x≡1 (mod 9)
x≡5 (mod 7).
From Example 21 we know that 19 is a solution of the first two congruences
and any other solution of the system is congruent to 19 (mod 20). Hence,
we can reduce the system to a system of three congruences:
x ≡ 19 (mod 20)
x≡1 (mod 9)
x≡5 (mod 7).
Solving the next two equations, we can reduce the system to
x ≡ 19 (mod 180)
x≡5 (mod 7).
Solving this last system, we find that 19 is a solution for the system that is
unique up to modulo 1260.
256 CHAPTER 16 RINGS
One interesting application of the Chinese Remainder Theorem in the
design of computer software is that the theorem allows us to break up a
calculation involving large integers into several less formidable calculations.
Most computers will handle integer calculations only up to a certain size.
For example, the largest integer available on many workstations is 231 − 1 =
2,147,483,647. Special software is required for calculations involving larger
integers which cannot be added directly by the machine. However, by using
the Chinese Remainder Theorem we can break down large integer additions
and multiplications into calculations that the computer can handle directly.
This is especially useful on parallel processing computers which have the
ability to run several programs concurrently.
Most computers have a single central processing unit (CPU), which can
only add two numbers at a time. To add a list of ten numbers, the CPU must
do nine additions in sequence. However, a parallel processing computer has
more than one CPU. A computer with 10 CPUs, for example, can perform 10
different additions at the same time. If we can take a large integer and break
it down into parts, sending each part to a different CPU, then by performing
several additions or multiplications simultaneously on those parts, we can
work with an integer that the computer would not be able to handle as a
whole.
Example 23. 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)
1531 ≡ 76 (mod 97)
1531 ≡ 61 (mod 98)
1531 ≡ 46 (mod 99).
EXERCISES 257
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 transform the
problem as we have done above and to perform the necessary calculations
simultaneously.
Exercises
1. Which of the following sets are rings with respect to the usual operations of
addition and multiplication? If the set is a ring, is it also a field?
(a) 7Z
(b) Z18
√ √
(c) Q( 2 ) = {a + b 2 : a, b ∈ Q}
√ √ √ √ √
(d) Q( 2, 3 ) = {a + b 2 + c 3 + d 6 : a, b, c, d ∈ Q}
√ √
(e) Z[ 3 ] = {a + b 3 : a, b ∈ Z}
√
(f) R = {a + b 3 3 : a, b ∈ Q}
(g) Z[i] = {a + bi : a, b ∈ Z and i2 = −1}
258 CHAPTER 16 RINGS
√ √ √
(h) Q( 3 3 ) = {a + b 3 3 + c 3 9 : a, b, c ∈ Q}
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. Prove or disprove:
√ √ Q( 2 ) = {a + b 2 : a, b ∈ Q} is isomorphic to
The ring
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 ?
EXERCISES 259
10. Define a map φ : C → M2 (R) by
a b
φ(a + bi) = .
−b a
Show that φ is an isomorphism of C with its image in M2 (R).
11. Prove that the Gaussian integers, Z[i], are an integral domain.
√ √
12. Prove that Z[ 3 i] = {a + b 3 i : a, b ∈ Z} is an integral domain.
13. Solve each of the following systems of congruences.
(a) (c)
x≡2 (mod 4)
x≡2 (mod 5) x≡4 (mod 7)
x≡6 (mod 11) x≡7 (mod 9)
x≡5 (mod 11)
(b) (d)
x≡3 (mod 5)
x≡3 (mod 7) x≡0 (mod 8)
x≡0 (mod 8) x≡1 (mod 11)
x≡5 (mod 15) x≡5 (mod 13)
14. Use the method of parallel computation outlined in the text to calculate
2234 + 4121 by dividing the calculation into four separate additions modulo
95, 97, 98, and 99.
15. Explain why the method of parallel computation outlined in the text fails
for 2134 · 1531 if we attempt to break the calculation down into two smaller
calculations modulo 98 and 99.
16. If R is a field, show that the only two ideals of R are {0} and R itself.
17. Let a be any element in a ring R with identity. Show that (−1)a = −a.
18. Prove that (−a)(−b) = ab for any elements a and b in a ring R.
19. Let φ : R → S be a ring homomorphism. Prove each of the following state-
ments.
(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 .
260 CHAPTER 16 RINGS
(d) If R is a field and φ(R) 6= 0, then φ(R) is a field.
20. Prove that the associative law for multiplication and the distributive laws
hold in R/I.
21. 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.
22. 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
23. Prove the Correspondence Theorem: Let I be a 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.
24. Let R be a ring and S a subset of R. Show that S is a subring of R if and
only if each of the following conditions is satisfied.
(a) S 6= ∅.
(b) rs ∈ S for all r, s ∈ S.
(c) r − s ∈ S for all r, s ∈ S.
T
25. 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 cannot
be a subring.
T
26. 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.
27. Let R be an integral domain. Show that if the only ideals in R are {0} and
R itself, R must be a field.
28. 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.
29. A ring R is a Boolean ring if for every a ∈ R, a2 = a. Show that every
Boolean ring is a commutative ring.
30. Let R be a ring, where a3 = a for all a ∈ R. Prove that R must be a
commutative ring.
31. Let R be a ring with identity 1R and S a subring of R with identity 1S .
Prove or disprove that 1R = 1S .
EXERCISES 261
32. 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}.
33. Let S be a subset of a ring R. Prove that there is a subring R0 of R that
contains S.
34. 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.
35. 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.
36. Prove or disprove: Every finite integral domain is isomorphic to Zp .
37. Let R be a ring.
(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).
38. 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) + (r0 , s0 ) = (r + r0 , s + s0 )
(b) (r, s)(r0 , s0 ) = (rr0 , ss0 )
39. An element a 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.
40. Let gcd(a, n) = d and gcd(b, d) = 1. Prove that ax ≡ b (mod n) does not
have a solution.
41. The Chinese Remainder Theorem for Rings. Let R be a ring and I
and J be ideals in R such that I + J = R.
262 CHAPTER 16 RINGS
(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.
Programming Exercise
Write a computer program implementing fast addition and multiplication
using the Chinese Remainder Theorem and the method outlined in the text.
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 Amer-
ica, 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 Algo-
rithms, vol. 2. 3rd ed. Addison-Wesley Professional, Boston, 1997.
[6] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York,
1998. A good source for applications.
[7] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[8] McCoy, N. H. Rings and Ideals. Carus Monograph Series, No. 8. Mathemat-
ical 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 polyno-
mial rings. We can prove many results for polynomial rings that are similar
to the theorems we proved for the integers. Analogs of prime numbers, of
the division algorithm, and of the Euclidean algorithm exist for polynomials.
263
264 CHAPTER 17 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
X
f (x) = ai xi = a0 + a1 x + a2 x2 + · · · + an xn ,
i=0
where ai ∈ R and an 6= 0, is called a polynomial over R with indeter-
minate x. The elements a0 , a1 , . . . , an are called the coefficients of f .
The coefficient an is called the leading coefficient. A polynomial is called
monic if the leading coefficient is 1. If n is the largest nonnegative number
for which an 6= 0, we say that the degree of f is n and write deg f (x) = n.
If no such n exists—that is, if f = 0 is the zero polynomial—then the degree
of f is defined to be −∞. We will denote the set of all polynomials with
coefficients in a ring R by R[x]. Two polynomials are equal exactly when
their corresponding coefficients are equal; that is, if we let
p(x) = a0 + a1 x + · · · + an xn
q(x) = b0 + b1 x + · · · + bm xm ,
then p(x) = q(x) if and only if ai = bi for all i ≥ 0.
To show that the set of all polynomials forms a ring, we must first de-
fine 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
X
ci = ak bi−k = a0 bi + a1 bi−1 + · · · + ai−1 b1 + ai b0
k=0
17.1 POLYNOMIAL RINGS 265
for each i. Notice that in each case some of the coefficients may be zero.
Example 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 2. Let
p(x) = 3 + 3x3 and q(x) = 4 + 4x2 + 4x4
be polynomials in Z12 [x]. The sum of p(x) and q(x) is 7 + 4x2 + 3x3 + 4x4 .
The product of the two polynomials is the zero polynomial. This example
tells us that R[x] cannot be an integral domain if R is not an integral domain.
Theorem 17.1 Let R be a commutative ring with identity. Then R[x] is a
commutative ring with identity.
Proof. Our first task is to show that R[x] is an abelian group under
polynomial addition. The zero
Pnpolynomial, f (x) = 0, is the additive identity.
Given a polynomial p(x) = i=0 i
Pai x , the inverse of p(x) is easily verified to
be −p(x) = ni=0 (−ai )xi = − ni=0 ai xi . Commutativity and associativity
P
follow immediately from the definition of polynomial addition and from the
fact that addition in R is both commutative and associative.
266 CHAPTER 17 POLYNOMIALS
To show that polynomial multiplication is associative, let
m
X
p(x) = ai xi ,
i=0
Xn
q(x) = bi xi ,
i=0
Xp
r(x) = ci xi .
i=0
Then
m n p
" ! !# !
X X X
[p(x)q(x)]r(x) = ai xi bi xi ci xi
i=0 i=0 i=0
m+n i p
!
X X X
= aj bi−j xi ci xi
i=0 j=0 i=0
!
m+n+p
X X i j
X
= ak bj−k cj xi
i=0 j=0 k=0
m+n+p
X X
= aj bk cr xi
i=0 j+k+l=i
!
m+n+p
X X i i−j
X
= aj bk ci−j−k xi
i=0 j=0 k=0
m
! n+p i
X X X
= ai xi bj ci−j xi
i=0 i=0 j=0
m n p
!" ! !#
X X X
i i i
= ai x bi x ci x
i=0 i=0 i=0
= p(x)[q(x)r(x)]
The commutativity and distribution properties of polynomial multiplication
are proved in a similar manner. We shall leave the proofs of these properties
as an exercise.
Proposition 17.2 Let p(x) and q(x) be polynomials in R[x], where R is an
integral domain. Then deg p(x) + deg q(x) = deg(p(x)q(x)). Furthermore,
R[x] is an integral domain.
17.1 POLYNOMIAL RINGS 267
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 6= 0 and bn 6= 0. The degrees of p and q are m and n, respectively.
The leading term of p(x)q(x) is am bn xm+n , which cannot be zero since R is
an integral domain; hence, the degree of p(x)q(x) is m+n, and p(x)q(x) 6= 0.
Since p(x) 6= 0 and q(x) 6= 0 imply that p(x)q(x) 6= 0, we know that R[x]
must also be an integral domain.
We also want to consider polynomials in two or more variables, such
as 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 17.3 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 .
Proof. Let p(x) = ni=0 ai xi and q(x) = m i
P P
i=0 bi x . It is easy to show
that φα (p(x) + q(x)) = φα (p(x)) + φα (q(x)). To show that multiplication is
preserved under the map φα , observe that
φα (p(x))φα (q(x)) = p(α)q(α)
n m
! !
X X
i i
= ai α bi α
i=0 i=0
m+n i
!
X X
= 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.
268 CHAPTER 17 POLYNOMIALS
17.2 The Division Algorithm
Recall that the division algorithm for integers (Theorem 2.2) says that if
a and b are integers with b > 0, then there exist unique integers q and r
such that a = bq + r, where 0 ≤ r < b. The algorithm by which q and r
are found is just long division. A similar theorem exists for polynomials.
The division algorithm for polynomials has several important consequences.
Since its proof is very similar to the corresponding proof for integers, it is
worthwhile to review Theorem 2.2 at this point.
Theorem 17.4 (Division Algorithm) Let f (x) and g(x) be two nonzero
polynomials in F [x], where F is a field and g(x) is a nonconstant 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). Let S =
{f (x) − g(x)h(x) : h(x) ∈ F [x]} and assume that
g(x) = a0 + a1 x + · · · + an xn
is a polynomial of degree n. This set is nonempty since f (x) ∈ S. If f (x) is
the zero polynomial, then
0 = f (x) = 0 · g(x) + 0;
hence, both q and r must also be the zero polynomial.
Now suppose that the zero polynomial is not in S. In this case the
degree of every polynomial in S is nonnegative. Choose a polynomial r(x)
of smallest degree in S; hence, there must exist a q(x) ∈ F [x] such that
r(x) = f (x) − g(x)q(x),
or
f (x) = g(x)q(x) + r(x).
We need to show that the degree of r(x) is less than the degree of g(x).
Assume that deg g(x) ≤ deg r(x). Say r(x) = b0 + b1 x + · · · + bm xm and
m ≥ n. Then
f (x) − g(x)[q(x) − (bm /an )xm−n ] = f (x) − g(x)q(x) + (bm /an )xm−n g(x)
= r(x) + (bm /an )xm−n g(x)
= r(x) + bm xm + terms of lower degree
17.2 THE DIVISION ALGORITHM 269
is in S. This is a polynomial of lower degree than r(x), which contradicts
the fact that r(x) is a polynomial of smallest degree in S; hence, deg r(x) <
deg g(x).
To show that q(x) and r(x) are unique, suppose that there exist two
other polynomials q 0 (x) and r0 (x) such that f (x) = g(x)q 0 (x) + r0 (x) and
deg r0 (x) < deg g(x) or r0 (x) = 0, so that
f (x) = g(x)q(x) + r(x) = g(x)q 0 (x) + r0 (x),
and
g(x)[q(x) − q 0 (x)] = r0 (x) − r(x).
If g is not the zero polynomial, then
deg(g(x)[q(x) − q 0 (x)]) = deg(r0 (x) − r(x)) ≥ deg g(x).
However, the degrees of both r(x) and r0 (x) are strictly less than the degree
of g(x); therefore, r(x) = r0 (x) and q(x) = q 0 (x).
Example 3. The division algorithm merely formalizes long division of poly-
nomials, a task we have been familiar with since high school. For example,
suppose that we divide x3 − x2 + 2x − 3 by x − 2.
x2 + x + 4
x−2 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 17.5 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)
270 CHAPTER 17 POLYNOMIALS
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(x) + 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(x) = 0.
Corollary 17.6 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 17.5. The degree of q(x) is n − 1 by
Proposition 17.2. Let β be some other zero of p(x) that is distinct from α.
Then p(β) = (β − α)q(β) = 0. Since α 6= β and F is a field, q(β) = 0. By
our induction hypothesis, p(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 d0 (x) dividing both p(x) and q(x),
d0 (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 17.7 Let F be a field and suppose that d(x) is the greatest
common divisor of two polynomials p(x) and q(x) in F [x]. Then there exist
polynomials r(x) and s(x) such that
d(x) = r(x)p(x) + s(x)q(x).
Furthermore, the greatest common divisor of two polynomials is unique.
17.2 THE DIVISION ALGORITHM 271
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 d0 (x) is another common divisor of p(x) and q(x). We will show that
d0 (x) | d(x). Since d0 (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)d0 (x) and q(x) = v(x)d0 (x).
Therefore,
d(x) = r(x)p(x) + s(x)q(x)
= r(x)u(x)d0 (x) + s(x)v(x)d0 (x)
= d0 (x)[r(x)u(x) + s(x)v(x)].
Since d0 (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 d0 (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) = d0 (x)[r(x)u(x) + s(x)v(x)]. Since
deg d(x) = deg d0 (x) + deg[r(x)u(x) + s(x)v(x)]
and d(x) and d0 (x) are both greatest common divisors, deg d(x) = deg d0 (x).
Since d(x) and d0 (x) are both monic polynomials of the same degree, it must
be the case that d(x) = d0 (x).
Notice the similarity between the proof of Proposition 17.7 and the proof
of Theorem 2.4.
272 CHAPTER 17 POLYNOMIALS
17.3 Irreducible Polynomials
A nonconstant polynomial f (x) ∈ F [x] is irreducible over a field F if
f (x) cannot be expressed as a product of two polynomials g(x) and h(x)
in F [x], where the degrees of g(x) and h(x) are both smaller than the de-
gree of f (x). Irreducible polynomials function as the “prime numbers” of
polynomial rings.
Example 4. 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 5. 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 17.8 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
17.3 IRREDUCIBLE POLYNOMIALS 273
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 17.9 (Gauss’s Lemma) Let p(x) ∈ Z[x] be a monic polynomial
such that p(x) factors into a product of two polynomials α(x) and β(x) in
Q[x], where the degrees of both α(x) and β(x) are less than the degree of
p(x). Then p(x) = a(x)b(x), where a(x) and b(x) are monic polynomials in
Z[x] with deg α(x) = deg a(x) and deg β(x) = deg b(x).
Proof. By Lemma 17.8, we can assume that
c1 c1
α(x) = (a0 + a1 x + · · · + am xm ) = α1 (x)
d1 d1
c2 c2
β(x) = (b0 + b1 x + · · · + bn xn ) = β1 (x),
d2 d2
where the ai ’s are relatively prime and the bi ’s are relatively prime. Conse-
quently,
c1 c2 c
p(x) = α(x)β(x) = α1 (x)β1 (x) = α1 (x)β1 (x),
d1 d2 d
where c/d is the product of c1 /d1 and c2 /d2 expressed in lowest terms.
Hence, dp(x) = cα1 (x)β1 (x).
If d = 1, then cam bn = 1 since p(x) is a monic polynomial. Hence, either
c = 1 or c = −1. If c = 1, then either am = bn = 1 or am = bn = −1. In the
first case p(x) = α1 (x)β1 (x), where α1 (x) and β1 (x) are monic polynomials
with deg α(x) = deg α1 (x) and deg β(x) = deg β1 (x). In the second case
a(x) = −α1 (x) and b(x) = −β1 (x) are the correct monic polynomials since
p(x) = (−α1 (x))(−β1 (x)) = a(x)b(x). The case in which c = −1 can be
handled similarly.
Now suppose that d 6= 1. Since gcd(c, d) = 1, there exists a prime p such
that p | d and p6 |c. Also, since the coefficients of α1 (x) are relatively prime,
there exists a coefficient ai such that p6 |ai . Similarly, there exists a coefficient
bj of β1 (x) such that p6 |bj . Let α10 (x) and β10 (x) be the polynomials in Zp [x]
obtained by reducing the coefficients of α1 (x) and β1 (x) modulo p. Since
p | d, α10 (x)β10 (x) = 0 in Zp [x]. However, this is impossible since neither
α10 (x) nor β10 (x) is the zero polynomial and Zp [x] is an integral domain.
Therefore, d = 1 and the theorem is proven.
274 CHAPTER 17 POLYNOMIALS
Corollary 17.10 Let p(x) = xn + an−1 xn−1 + · · · + a0 be a polynomial with
coefficients in Z and a0 6= 0. If p(x) has a zero in Q, then p(x) also has a
zero α in Z. Furthermore, α divides a0 .
Proof. Let p(x) have a zero a ∈ Q. Then p(x) must have a linear factor
x − a. By Gauss’s Lemma, p(x) has a factorization with a linear factor in
Z[x]. Hence, for some α ∈ Z
p(x) = (x − α)(xn−1 + · · · − a0 /α).
Thus a0 /α ∈ Z and so α | a0 .
Example 6. Let p(x) = x4 − 2x3 + x + 1. We shall show that p(x) is
irreducible over Q[x]. Assume that p(x) is reducible. Then either p(x) has
a linear factor, say p(x) = (x − α)q(x), where q(x) is a polynomial of degree
three, or p(x) has two quadratic factors.
If p(x) has a linear factor in Q[x], then it has a zero in Z. By Corol-
lary 17.10, 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 polyno-
mials, say
p(x) = (x2 + ax + b)(x2 + cx + d)
= x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd,
where each factor is in Z[x] by Gauss’s Lemma. Hence,
a + c = −2
ac + b + d = 0
ad + bc = 1
bd = 1.
Since bd = 1, either b = d = 1 or b = d = −1. In either case b = d and so
ad + bc = b(a + c) = 1.
Since a + c = −2, we know that −2b = 1. This is impossible since b is an
integer. Therefore, p(x) must be irreducible over Q.
17.3 IRREDUCIBLE POLYNOMIALS 275
Theorem 17.11 (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 p6 |an and p2 6 |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 p6 |b0 and p | c0 . Since p6 |an and an = br cs , neither br nor cs is
divisible by p. Let m be the smallest value of k such that p6 |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 7. The polynomial
p(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 polyno-
mials 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 The-
orem 17.11 is that we now have an easy method of generating irreducible
polynomials of any degree.
Ideals in F [x]
Let F be a field. Recall that a principal ideal in F [x] is an ideal hp(x)i
generated by some polynomial p(x); that is,
hp(x)i = {p(x)q(x) : q(x) ∈ F [x]}.
276 CHAPTER 17 POLYNOMIALS
Example 8. The polynomial x2 in F [x] generates the ideal hx2 i consisting
of all polynomials with no constant term or term of degree 1.
Theorem 17.12 If F is a field, then every ideal in F [x] is a principal ideal.
Proof. Let I be an ideal of F [x]. If I is the zero ideal, the theorem is
easily true. Suppose that I is a nontrivial ideal in F [x], and let p(x) ∈ I be
a nonzero element of minimal degree. If deg p(x) = 0, then p(x) is a nonzero
constant and 1 must be in I. Since 1 generates all of F [x], h1i = I = F [x]
and I is again a principal ideal.
Now assume that deg p(x) ≥ 1 and let f (x) be any element in I. By
the division algorithm there exist q(x) and r(x) in F [x] such that f (x) =
p(x)q(x) + r(x) and deg r(x) < deg p(x). Since f (x), p(x) ∈ I and I is an
ideal, r(x) = f (x) − p(x)q(x) is also in I. However, since we chose p(x) to
be of minimal degree, r(x) must be the zero polynomial. Since we can write
any element f (x) in I as p(x)q(x) for some q(x) ∈ F [x], it must be the case
that I = hp(x)i.
Example 9. 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 17.13 Let F be a field and suppose that p(x) ∈ F [x]. Then the
ideal generated by p(x) is maximal if and only if p(x) is irreducible.
Proof. Suppose that p(x) generates a maximal ideal of F [x]. Then hp(x)i is
also a prime ideal of F [x]. Since a maximal ideal must be properly contained
inside F [x], p(x) cannot be a constant polynomial. Let us assume that p(x)
factors into two polynomials of lesser degree, say p(x) = f (x)g(x). Since
hp(x)i is a prime ideal one of these factors, say f (x), is in hp(x)i and therefore
be a multiple of p(x). But this would imply that hp(x)i ⊂ hf (x)i, which is
impossible since hp(x)i is maximal.
Conversely, suppose that p(x) is irreducible over F [x]. Let I be an ideal
in F [x] containing hp(x)i. By Theorem 17.12, I is a principal ideal; hence,
I = hf (x)i for some f (x) ∈ F [x]. Since p(x) ∈ I, it must be the case
that p(x) = f (x)g(x) for some g(x) ∈ F [x]. However, p(x) is irreducible;
hence, either f (x) or g(x) is a constant polynomial. If f (x) is constant,
then I = F [x] and we are done. If g(x) is constant, then f (x) is a constant
17.3 IRREDUCIBLE POLYNOMIALS 277
multiple of I and I = hp(x)i. Thus, there are no proper ideals of F [x] that
properly contain hp(x)i.
Historical Note
Throughout history, the solution of polynomial equations has been a challenging
problem. The Babylonians knew how to solve the equation ax2 + bx + c = 0.
Omar Khayyam (1048–1131) devised methods of solving cubic equations through
the use of geometric constructions and conic sections. The algebraic solution of
the general cubic equation ax3 + bx2 + cx + d = 0 was not discovered until the
sixteenth century. An Italian mathematician, Luca Paciola (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 “de-
pressed cubic,”
ax3 + cx + d = 0.
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 chal-
lenge to Niccolo Fontana (1499–1557). Fontana was known as Tartaglia (the Stam-
merer). 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 perma-
nently impaired. Tartaglia sent Fior a list of 30 various mathematical problems;
Fior countered by sending Tartaglia a list of 30 depressed cubics. Tartaglia would
either solve all 30 of the problems or absolutely fail. After much effort Tartaglia
finally succeeded in solving the depressed cubic and defeated Fior, who faded into
obscurity.
At this point another mathematician, Gerolamo Cardano (1501–1576), entered
the story. Cardano wrote to Tartaglia, begging him for the solution to the depressed
cubic. Tartaglia refused several of his requests, then finally revealed the solution to
Cardano after the latter swore an oath not to publish the secret or to pass it on to
anyone else. Using the knowledge that he had obtained from Tartaglia, Cardano
eventually solved the general cubic
ax3 + bx2 + cx + d = 0.
278 CHAPTER 17 POLYNOMIALS
Cardano shared the secret with his student, Ludovico Ferrari (1522–1565), who
solved the general quartic equation,
ax4 + bx3 + cx2 + dx + e = 0.
In 1543, Cardano and Ferrari examined del Ferro’s papers and discovered that he
had also solved the depressed cubic. Cardano felt that this relieved him of his
obligation to Tartaglia, so he proceeded to publish the solutions in Ars Magna
(1545), in which he gave credit to del Ferro for solving the special case of the cubic.
This resulted in a bitter dispute between Cardano and Tartaglia, who published
the story of the oath a year later.
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
(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) p(x) = 5x3 + 6x2 − 3x + 4 and q(x) = x − 2 in Z7 [x]
(b) p(x) = 6x4 − 2x3 + x2 − 3x + 1 and q(x) = x2 + x − 2 in Z7 [x]
(c) p(x) = 4x5 − x3 + x2 + 4 and q(x) = x3 − 2 in Z5 [x]
(d) p(x) = x5 + x3 − x2 − x and q(x) = x3 + x in Z2 [x]
4. Find the greatest common divisor of each of the following pairs p(x) and q(x)
of polynomials. If d(x) = gcd(p(x), q(x)), find two polynomials a(x) and b(x)
such that a(x)p(x) + b(x)q(x) = d(x).
(a) p(x) = 7x3 + 6x2 − 8x + 4 and q(x) = x3 + x − 2, 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.
EXERCISES 279
(a) 5x3 + 4x2 − x + 9 in Z12 (c) 5x4 + 2x2 − 3 in Z7
3 2
(b) 3x − 4x − 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. If f (x) | p(x)q(x), prove that either f (x) | p(x) or
f (x) | q(x).
16. Suppose that R and S are isomorphic rings. Prove that R[x] ∼
= S[x].
17. Let F be a field and a ∈ F . If p(x) ∈ F [x], show that p(a) is the remainder
obtained when p(x) is divided by x − a.
18. Let Q∗ be the multiplicative group of positive rational numbers. Prove that
Q∗ is isomorphic to (Z[x], +).
19. Cyclotomic Polynomials. The polynomial
xn − 1
Φn (x) = = xn−1 + xn−2 + · · · + x + 1
x−1
is called the cyclotomic polynomial. Show that Φp (x) is irreducible over
Q for any prime p.
20. If F is a field, show that there are infinitely many irreducible polynomials in
F [x].
21. Let R be a commutative ring with identity. Prove that multiplication is
commutative in R[x].
22. Let R be a commutative ring with identity. Prove that multiplication is
distributive in R[x].
280 CHAPTER 17 POLYNOMIALS
23. Show that xp − x has p distinct zeros in Zp [x], for any prime p. Conclude
that therefore
xp − x = x(x − 1)(x − 2) · · · (x − (p − 1)).
24. Let F be a ring and f (x) = a0 + a1 x + · · · + an xn be in F [x]. Define
f 0 (x) = a1 + 2a2 x + · · · + nan xn−1 to be the derivative of f (x).
(a) Prove that
(f + g)0 (x) = f 0 (x) + g 0 (x).
Conclude that we can define a homomorphism of abelian groups D :
F [x] → F [x] by (D(f (x)) = f 0 (x).
(b) Calculate the kernel of D if charF = 0.
(c) Calculate the kernel of D if charF = p.
(d) Prove that
(f g)0 (x) = f 0 (x)g(x) + f (x)g 0 (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 0 (x)
are relatively prime.
25. Let F be a field. Show that F [x] is never a field.
26. Let R be an integral domain. Prove that R[x1 , . . . , xn ] is an integral domain.
27. Let R be a commutative ring with identity. Show that R[x] has a subring R0
isomorphic to R.
28. 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)).
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 determines the
nature of the solutions of the equation. If ∆ > 0, the equation has two
distinct real solutions. If ∆ = 0, the equation has a single repeated real root.
If ∆ < 0, there are two distinct imaginary solutions.
EXERCISES 281
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. Show√that the product of the solutions obtained in (4) is −p3 /27, deducing
that 3 AB = −p/3.
6. Prove that the possible solutions for z in (4) are given by
√ √ √ √ √ √
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
s r s r
3 q p 3 q 2 q p3 q2
i 2i 3
ω − + + +ω − − + ,
2 27 4 2 27 4
where i = 0, 1, 2.
7. The discriminant of the cubic equation is
p3 q2
∆= + .
27 4
Show that y 3 + py + q = 0
(a) has three real roots, at least two of which are equal, if ∆ = 0.
(b) has one real root and two conjugate imaginary roots if ∆ > 0.
(c) has three distinct real roots if ∆ < 0.
8. Solve the following cubic equations.
282 CHAPTER 17 POLYNOMIALS
(a) x3 − 4x2 + 11x + 30 = 0 (c) x3 − 3x + 2 = 0
3
(b) x − 3x + 5 = 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
y2 + z = (z − p)y 2 − qy + z −r .
2 4
11. Show that the right-hand side of (10) can be put in the form (my + k)2 if
and only if
2 1 2
q − 4(z − p) z − r = 0.
4
12. From (11) obtain the resolvent cubic equation
z 3 − pz 2 − 4rz + (4pr − q 2 ) = 0.
Solving the resolvent cubic equation, put the equation found in (10) in the
form 2
2 1
y + z = (my + k)2
2
to obtain the solution of the quartic equation.
13. Use this method to solve the following quartic equations.
(a) x4 − x2 − 3x + 2 = 0 (c) x4 − 2x2 + 4x − 3 = 0
(b) x4 + x3 − 7x2 − x + 6 = 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, an-
swering questions about the ideal structure of integral domains, polynomial
rings over integral domains, and whether or not an integral domain can be
embedded in a field.
18.1 Fields of Fractions
Every field is also an integral domain; however, there are many integral
domains that are not fields. For example, the integers Z are an integral
domain but not a field. A question that naturally arises is how we might
associate an integral domain with a field. There is a natural way to construct
the rationals Q from the integers: the rationals can be represented as formal
quotients of two integers. The rational numbers are certainly a field. In fact,
it can be shown that the rationals are the smallest field that contains the
integers. Given an integral domain D, our question now becomes how to
construct a smallest field F containing D. We will do this in the same way
as we constructed the rationals from the integers.
An element p/q ∈ Q is the quotient of two integers p and q; however,
different pairs of integers can represent the same rational number. For in-
stance, 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
283
284 CHAPTER 18 INTEGRAL DOMAINS
elements in Q as ordered pairs in Z × Z. A quotient p/q can be written
as (p, q). For instance, (3, 7) would represent the fraction 3/7. However,
there are problems if we consider all possible pairs in Z × Z. There is no
fraction 5/0 corresponding to the pair (5, 0). Also, the pairs (3, 6) and (2, 4)
both represent the fraction 1/2. The first problem is easily solved if we
require the second coordinate to be nonzero. The second problem is solved
by considering two pairs (a, b) and (c, d) to be equivalent if ad = bc.
If we use the approach of ordered pairs instead of fractions, then we can
study integral domains in general. Let D be any integral domain and let
S = {(a, b) : a, b ∈ D and b 6= 0}.
Define a relation on S by (a, b) ∼ (c, d) if ad = bc.
Lemma 18.1 The relation ∼ between elements of S is an equivalence rela-
tion.
Proof. Since D is commutative, ab = ba; hence, ∼ is reflexive on D.
Now suppose that (a, b) ∼ (c, d). Then ad = bc or cb = da. Therefore,
(c, d) ∼ (a, b) and the relation is symmetric. Finally, to show that the
relation is transitive, let (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). In this case
ad = bc and cf = de. Multiplying both sides of ad = bc by f yields
af d = adf = bcf = bde = bed.
Since D is an integral domain, we can deduce that af = be or (a, b) ∼ (e, f ).
We will denote the set of equivalence classes on S by FD . We now need
to define the operations of addition and multiplication on FD . Recall how
fractions are added and multiplied in Q:
a c ad + bc
+ = ;
b d bd
a c ac
· = .
b d bd
It seems reasonable to define the operations of addition and multiplication
on FD in a similar manner. If we denote the equivalence class of (a, b) ∈ S by
[a, b], then we are led to define the operations of addition and multiplication
on FD by
[a, b] + [c, d] = [ad + bc, bd]
18.1 FIELDS OF FRACTIONS 285
and
[a, b] · [c, d] = [ac, bd],
respectively. The next lemma demonstrates that these operations are inde-
pendent of the choice of representatives from each equivalence class.
Lemma 18.2 The operations of addition and multiplication on FD are well-
defined.
Proof. We will prove that the operation of addition is well-defined. The
proof that multiplication is well-defined is left as an exercise. Let [a1 , b1 ] =
[a2 , b2 ] and [c1 , d1 ] = [c2 , d2 ]. We must show that
[a1 d1 + b1 c1 , b1 d1 ] = [a2 d2 + b2 c2 , b2 d2 ]
or, equivalently, that
(a1 d1 + b1 c1 )(b2 d2 ) = (b1 d1 )(a2 d2 + b2 c2 ).
Since [a1 , b1 ] = [a2 , b2 ] and [c1 , d1 ] = [c2 , d2 ], we know that a1 b2 = b1 a2 and
c1 d2 = d1 c2 . Therefore,
(a1 d1 + b1 c1 )(b2 d2 ) = a1 d1 b2 d2 + b1 c1 b2 d2
= a1 b2 d1 d2 + b1 b2 c1 d2
= b1 a2 d1 d2 + b1 b2 d1 c2
= (b1 d1 )(a2 d2 + b2 c2 ).
Lemma 18.3 The set of equivalence classes of S, FD , under the equiva-
lence relation ∼, together with the operations of addition and multiplication
defined by
[a, b] + [c, d] = [ad + bc, bd]
[a, b] · [c, d] = [ac, bd],
is a field.
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].
286 CHAPTER 18 INTEGRAL DOMAINS
It is easy to show that [1, 1] is the multiplicative identity. Let [a, b] ∈ FD
such that a 6= 0. Then [b, a] is also in FD and [a, b] · [b, a] = [1, 1]; hence,
[b, a] is the multiplicative inverse for [a, b]. Similarly, [−a, b] is the additive
inverse of [a, b]. We leave as exercises the verification of the associative and
commutative properties of multiplication in FD . We also leave it to the
reader to show that FD is an abelian group under addition.
It remains to show that the distributive property holds in FD ; however,
[a, b][e, f ] + [c, d][e, f ] = [ae, bf ] + [ce, df ]
= [aedf + bf ce, bdf 2 ]
= [aed + bce, bdf ]
= [ade + bce, bdf ]
= ([a, b] + [c, d])[e, f ]
and the lemma is proved.
The field FD in Lemma 18.3 is called the field of fractions or field of
quotients of the integral domain D.
Theorem 18.4 Let D be an integral domain. Then D can be embedded in
a field of fractions FD , where any element in FD can be expressed as the
quotient of two elements in D. Furthermore, the field of fractions FD is
unique in the sense that if E is any field containing D, then there exists
a map ψ : FD → E giving an isomorphism with a subfield of E such that
ψ(a) = a for all elements a ∈ D.
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 expressed as the quotient of two elements in D, since
φ(a)[φ(b)]−1 = [a, 1][b, 1]−1 = [a, 1] · [1, b] = [a, b].
Now let E be a field containing D and define a map ψ : FD → E by
ψ([a, b]) = ab−1 . To show that ψ is well-defined, let [a1 , b1 ] = [a2 , b2 ]. Then
a1 b2 = b1 a2 . Therefore, a1 b−1 −1
1 = a2 b2 and ψ([a1 , b1 ]) = ψ([a2 , b2 ]).
18.2 FACTORIZATION IN INTEGRAL DOMAINS 287
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 1. Since Q is a field, Q[x] is an integral domain. The field
of fractions of Q[x] is the set of all rational expressions p(x)/q(x), where
p(x) and q(x) are polynomials over the rationals and q(x) is not the zero
polynomial. We will denote this field by Q(x).
We will leave the proofs of the following corollaries of Theorem 18.4 as
exercises.
Corollary 18.5 Let F be a field of characteristic zero. Then F contains a
subfield isomorphic to Q.
Corollary 18.6 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.
288 CHAPTER 18 INTEGRAL DOMAINS
Let R be a commutative ring with identity, and let a and b be elements
in R. We say that a divides b, and write a | b, if there exists an element
c ∈ R such that b = ac. A unit in R is an element that has a multiplicative
inverse. Two elements a and b in R are said to be associates if there exists
a unit u in R such that a = ub.
Let D be an integral domain. A nonzero element p ∈ D that is not a
unit is said to be irreducible provided that whenever p = ab, either a or b
is a unit. Furthermore, p is prime if whenever p | ab either p | a or p | b.
Example 2. It is important to notice that prime and irreducible elements
do not always coincide. Let R be the subring 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 in-
teger n > 1 can be factored into a product of prime numbers p1 · · · pk , where
the pi ’s are not necessarily distinct. We also know that such factorizations
are unique up to the order of the pi ’s. We can easily extend this result
to the integers. The question arises of whether or not such factorizations
are possible in other rings. Generalizing this definition, we say an integral
domain D is a unique factorization domain, or UFD, if D satisfies the
following criteria.
1. Let a ∈ D such that a 6= 0 and a is not a unit. Then a can be written
as the product of irreducible elements in D.
2. Let a = p1 · · · pr = q1 · · · qs , where the pi ’s and the qi ’s are irre-
ducible. Then r = s and there is a π ∈ Sk such that pi = qπ(j)
for j = 1, . . . , r = s.
Example 3. The integers are a unique factorization domain by the Funda-
mental Theorem of Arithmetic.
Example 4. Not √ every integral√domain is a unique factorization domain.
The subring Z[ 3 i] = {a + b 3 i} of the complex numbers √ is an inte-
gral domain
√ (Exercise 12, Chapter 16). Let z = a + b 3 i and define
2 2 2
ν : Z[ 3 i] → N ∪ {0} by ν(z) = |z| = a + 3b . 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.
18.2 FACTORIZATION IN INTEGRAL DOMAINS 289
We claim that 4 has two distinct factorizations into irreducible elements:
√ √
4 = 2 · 2 = (1 − 3 i)(1 + 3 i).
√
We must show that each of these factors is an irreducible element√in Z[ 3 i].
If 2 is not irreducible, then 2 = zw for elements z, w in Z[ 3 i] where √
ν(z) = ν(w) = 2. However, there does not exist an element in z in Z[ 3 i]
such that ν(z) = 2 because the equation a2 +3b2 = 2 has no integer solutions.
√ be irreducible. A similar argument shows that both 1 −
Therefore, 2 must
√
3 i√and 1 + √3 i are irreducible. Since 2 is not a unit multiple of either
1 − 3 i or 1 + 3 i, 4 has at least two distinct factorizations into irreducible
elements.
Principal Ideal Domains
Let R be a commutative ring with identity. Recall that a principal ideal
generated by a ∈ R is an ideal of the form hai = {ra : r ∈ R}. An integral
domain in which every ideal is principal is called a principal ideal domain,
or PID.
Lemma 18.7 Let D be an integral domain and let a, b ∈ D. Then
1. a | b ⇔ hbi ⊂ hai.
2. a and b are associates ⇔ hbi = hai.
3. a is a unit in D ⇔ hai = D.
Proof. (1) Suppose that a | b. Then b = ax for some x ∈ D. Hence, for
every r in D, br = (ax)r = a(xr) and hbi ⊂ hai. Conversely, suppose that
hbi ⊂ hai. Then b ∈ hai. Consequently, b = ax for some x ∈ D. Thus, a | b.
(2) Since a and b are associates, there exists a unit u such that a = ub.
Therefore, b | a and hai ⊂ hbi. Similarly, hbi ⊂ hai. It follows that hai = hbi.
Conversely, suppose that hai = hbi. By part (1), a | b and b | a. Then
a = bx and b = ay for some x, y ∈ D. Therefore, a = bx = ayx. Since D
is an integral domain, xy = 1; that is, x and y are units and a and b are
associates.
(3) An element a ∈ D is a unit if and only if a is an associate of 1.
However, a is an associate of 1 if and only if hai = h1i = D.
Theorem 18.8 Let D be a PID and hpi be a nonzero ideal in D. Then hpi
is a maximal ideal if and only if p is irreducible.
290 CHAPTER 18 INTEGRAL DOMAINS
Proof. Suppose that hpi is a maximal ideal. If some element a in D
divides p, then hpi ⊂ hai. Since hpi is maximal, either D = hai or hpi = hai.
Consequently, either a and p are associates or a is a unit. Therefore, p is
irreducible.
Conversely, let p be irreducible. If hai is an ideal in D such that hpi ⊂
hai ⊂ D, then a | p. Since p is irreducible, either a must be a unit or a
and p are associates. Therefore, either D = hai or hpi = hai. Thus, hpi is a
maximal ideal.
Corollary 18.9 Let D be a PID. If p is irreducible, then p is prime.
Proof. Let p be irreducible and suppose that p | ab. Then habi ⊂ hpi. By
Corollary 16.17, since hpi is a maximal ideal, hpi must also be a prime ideal.
Thus, either a ∈ hpi or b ∈ hpi. Hence, either p | a or p | b.
Lemma 18.10 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 = ∞
S
i=1 is an ideal of D. Certainly I is not empty,
since I1 ⊂ I and 0 ∈ I. If a, b ∈ I, then a ∈ Ii and b ∈ Ij for some i and j
in N. Without loss of generality we can assume that i ≤ j. Hence, a and b
are both in Ij and so a − b is also in Ij . Now let r ∈ D and a ∈ I. Again,
we note that a ∈ Ii for some positive integer i. Since Ii is an ideal, ra ∈ Ii
and hence must be in I. Therefore, we have shown that I is an ideal in D.
Since D is a principal ideal domain, there exists an element a ∈ D that
generates I. Since a is in IN for some N ∈ N, we know that IN = I = hai.
Consequently, In = IN for n ≥ N .
Any commutative ring satisfying the condition in Lemma 18.10 is said
to satisfy the ascending chain condition, or ACC. Such rings are called
Noetherian rings, after Emmy Noether.
Theorem 18.11 Every PID is a UFD.
Proof. Existence of a factorization. Let D be a PID and a be a nonzero
element in D that is not a unit. If a is irreducible, then we are done. If not,
then there exists a factorization a = a1 b1 , where neither a1 nor b1 is a unit.
Hence, hai ⊂ ha1 i. By Lemma 18.7, we know that hai 6= ha1 i; otherwise, a
and a1 would be associates and b1 would be a unit, which would contradict
our assumption. Now suppose that a1 = a2 b2 , where neither a2 nor b2 is a
18.2 FACTORIZATION IN INTEGRAL DOMAINS 291
unit. By the same argument as before, ha1 i ⊂ ha2 i. We can continue with
this construction to obtain an ascending chain of ideals
hai ⊂ ha1 i ⊂ ha2 i ⊂ · · · .
By Lemma 16.10, there exists a positive integer N such that han i = haN i
for all n ≥ N . Consequently, aN must be irreducible. We have now shown
that a is the product of two elements, one of which must be irreducible.
Now suppose that a = c1 p1 , where p1 is irreducible. If c1 is not a unit,
we can repeat the preceding argument to conclude that hai ⊂ hc1 i. Either
c1 is irreducible or c1 = c2 p2 , where p2 is irreducible and c2 is not a unit.
Continuing in this manner, we obtain another chain of ideals
hai ⊂ hc1 i ⊂ hc2 i ⊂ · · · .
This chain must satisfy the ascending chain condition; therefore,
a = p1 p2 · · · pr
for irreducible elements p1 , . . . , pr .
Uniqueness of the factorization. To show uniqueness, let
a = p1 p2 · · · pr = q1 q2 · · · qs ,
where each pi and each qi is irreducible. Without loss of generality, we can
assume that r < s. Since p1 divides q1 q2 · · · qs , by Corollary 16.9 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 18.12 Let F be a field. Then F [x] is a UFD.
292 CHAPTER 18 INTEGRAL DOMAINS
Example 5. Every PID is a UFD, but it is not the case that every UFD
is a PID. In Corollary 18.22, we will prove that Z[x] is a UFD. However,
Z[x] is not a PID. Let I = {5f (x) + xg(x) : f (x), g(x) ∈ Z[x]}. We can
easily show that I is an ideal of Z[x]. Suppose that I = hp(x)i. Since 5 ∈ I,
5 = f (x)p(x). In this case p(x) = p must be a constant. Since x ∈ I,
x = pg(x); consequently, p = ±1. However, it follows from this fact that
hp(x)i = Z[x]. But this would mean that 3 is in I. Therefore, we can write
3 = 5f (x) + xg(x) for some f (x) and g(x) in Z[x]. Examining the constant
term of this polynomial, we see that 3 = 5f (x), which is impossible.
Euclidean Domains
We have repeatedly used the division algorithm when proving results about
either Z or F [x], where F is a field. We should now ask when a division
algorithm is available for an integral domain.
Let D be an integral domain such that for each a ∈ D there is a non-
negative 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 6= 0. Then there exist elements
q, r ∈ D such that a = bq + r and either r = 0 or ν(r) < ν(b).
Then D is called a Euclidean domain and ν is called a Euclidean val-
uation.
Example 6. Absolute value on Z is a Euclidean valuation.
Example 7. Let F be a field. Then the degree of a polynomial in F [x] is
a Euclidean valuation.
Example 8. Recall that the Gaussian integers in Example 9 of Chapter 16
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).
18.2 FACTORIZATION IN INTEGRAL DOMAINS 293
Next, we must show that for any z = a + bi and w = c + di in Z[i]
with w 6= 0, there exist elements q and r in Z[i] such that z = qw + r
with either r = 0 or ν(r) < ν(w). We can view z and w as elements in
Q(i) = {p + qi : p, q ∈ Q}, the field of fractions of Z[i]. Observe that
c − di
zw−1 = (a + bi)
c2 + d2
ac + bd bc − ad
= 2 + 2 i
c + d2
c +d2
n1 n2
= m1 + 2 + m2 + 2 i
c + d2 c + d2
n1 n2
= (m1 + m2 i) + + 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).
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 18.13 Every Euclidean domain is a principal ideal domain.
Proof. Let D be a Euclidean domain and let ν be a Euclidean valuation
on D. Suppose I is a nontrivial ideal in D and choose a nonzero element
b ∈ I such that ν(b) is minimal for all a ∈ I. Since D is a Euclidean domain,
there exist elements q and r in D such that a = bq + r and either r = 0 or
ν(r) < ν(b). But r = a − bq is in I since I is an ideal; therefore, r = 0 by
the minimality of b. It follows that a = bq and I = hbi.
294 CHAPTER 18 INTEGRAL DOMAINS
Corollary 18.14 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 ques-
tions 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 17.9).
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 , . . . , a1 . We say that p(x) is primitive if gcd(a0 , . . . , an ) = 1.
Example 9. 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; how-
ever, the polynomial q(x) = 4x2 − 6x + 8 is not primitive since the content
of q(x) is 2.
Theorem 18.15 (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.
Proof. Let f (x) = m
P i
Pn i
i=0 ai x and g(x) = i=0 bi x . Suppose that p is
a prime dividing the coefficients of f (x)g(x). Let r be the smallest integer
such that p6 |ar and s be the smallest integer such that p6 |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 18.16 Let D be a UFD, and let p(x) and q(x) be in D[x]. Then the
content of p(x)q(x) is equal to the product of the contents of p(x) and q(x).
Proof. Let p(x) = cp1 (x) and q(x) = dq1 (x), where c and d are the
contents of p(x) and q(x), respectively. Then p1 (x) and q1 (x) are primitive.
We can now write p(x)q(x) = cdp1 (x)q1 (x). Since p1 (x)q1 (x) is primitive,
the content of p(x)q(x) must be cd.
18.2 FACTORIZATION IN INTEGRAL DOMAINS 295
Lemma 18.17 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 polynomi-
als, 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 18.17.
Corollary 18.18 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 18.19 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 18.20 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 18.18, 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. There-
fore, 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)
296 CHAPTER 18 INTEGRAL DOMAINS
be two factorizations of p(x), where all of the factors are irreducible in
D[x]. By Corollary 18.18, each of the fi ’s and gi ’s is irreducible in F [x].
The ai ’s and the bi ’s are units in F . Since F [x] is a PID, it is a UFD;
therefore, n = s. Now rearrange the gi (x)’s so that fi (x) and gi (x) are
associates for i = 1, . . . , n. Then there exist c1 , . . . , cn and d1 , . . . , dn in
D such that (ci /di )fi (x) = gi (x) or ci fi (x) = di gi (x). The polynomials
fi (x) and gi (x) are primitive; hence, ci and di are associates in D. Thus,
a1 · · · am = ub1 · · · br in D, where u is a unit in D. Since D is a unique
factorization domain, m = s. Finally, we can reorder the bi ’s so that ai and
bi are associates for each i. This completes the uniqueness part of the proof.
The theorem that we have just proven has several obvious but important
corollaries.
Corollary 18.21 Let F be a field. Then F [x] is a UFD.
Corollary 18.22 Z[x] is a UFD.
Corollary 18.23 Let D be a UFD. Then D[x1 , . . . , xn ] is a UFD.
Remark. It is important to notice that every Euclidean domain is a PID
and every PID is a UFD. However, as demonstrated by our examples, the
converse of each of these statements fails. There are principal ideal domains
that are not Euclidean domains, and there are unique factorization domains
that are not principal ideal domains (Z[x]).
Historical Note
Karl Friedrich Gauss, born in Brunswick, Germany on April 30, 1777, is
considered to be one of the greatest mathematicians who ever lived. Gauss
was truly a child prodigy. At the age of three he was able to detect errors
in the books of his father’s business. Gauss entered college at the age of 15.
Before the age of 20, Gauss was able to construct a regular 17-sided polygon
with a ruler and compass. This was the first new construction of a regular
n-sided polygon since the time of the ancient Greeks. Gauss succeeded in
n
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
EXERCISES 297
with real coefficients can be factored into linear factors over the complex
numbers. The acceptance of complex numbers was brought √ about by Gauss,
who was the first person to use the notation of i for −1.
Gauss then turned his attention toward number theory; in 1801, he
published his famous book on number theory, Disquisitiones Arithmeticae.
Throughout his life Gauss was 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 Univer-
sity 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, mechanics, optics, geodesy,
and magnetism. Along with Wilhelm Weber, he coinvented the first prac-
tical electric telegraph some years before a better version was invented by
Samuel F. B. Morse.
Gauss was clearly the most prominent mathematician in the world in the
early nineteenth century. His status naturally made his discoveries subject
to intense scrutiny. Gauss’s cold and distant personality many times led him
to ignore the work of his contemporaries, making him many enemies. He
did not enjoy teaching very much, and young mathematicians who sought
him out for encouragement were often rebuffed. Nevertheless, he had many
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 mathematician. Gauss died at the age
of 78 in Göttingen on February 23, 1855.
Exercises
√ √
1. Let z = a + b 3 i be in Z[ 3 i].√ If a2 + 3b2 = 1, show that z must be a unit.
Show that the only units of Z[ 3 i] are 1 and −1.
2. The Gaussian integers, Z[i], are a UFD. Factor each of the following elements
in Z[i] into a product of irreducibles.
(a) 5 (c) 6 + 8i
(b) 1 + 3i (d) 2
3. Let D be an integral domain.
(a) Prove that FD is an abelian group under the operation of addition.
298 CHAPTER 18 INTEGRAL DOMAINS
(b) Show that the operation of multiplication is well-defined in the field of
fractions, FD .
(c) Verify the associative and commutative properties for multiplication in
FD .
4. Prove or disprove: Any subring of a field F containing 1 is an integral domain.
5. Let F be a field of characteristic zero. Prove that F contains a subfield
isomorphic to Q.
6. Let F be a field.
(a) Prove that the field of fractions of F [x], denoted by F (x), is isomorphic
to the set all rational expressions p(x)/q(x), where q(x) is not the zero
polynomial.
(b) Let p(x1 , . . . , xn ) and q(x1 , . . . , xn ) be polynomials in F [x1 , . . . , xn ].
Show that the set of all rational expressions p(x1 , . . . , xn )/q(x1 , . . . , xn )
is isomorphic to the field of fractions of F [x1 , . . . , xn ]. We denote the
field of fractions of F [x1 , . . . , xn ] by F (x1 , . . . , xn ).
7. 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.
8. Prove that the field of fractions of the Gaussian integers, Z[i], is
Q(i) = {p + qi : p, q ∈ Q}.
9. 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 .
√ √
10. 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 .
11. 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.
EXERCISES 299
(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. 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.
12. 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.
13. Let D be a Euclidean domain with Euclidean valuation ν. If u is a unit in
D, show that ν(u) = ν(1).
14. Let D be a Euclidean domain with Euclidean valuation ν. If a and b are
associates in D, prove that ν(a) = ν(b).
√
15. Show that Z[ 5 i] is not a unique factorization domain.
16. Prove or disprove: Every subdomain of a UFD is also a UFD.
17. 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.
18. Let D be an integral domain with a descending chain of ideals I1 ⊃ I2 ⊃ · · · .
Show 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.
19. Let R be a commutative ring with identity. We define a multiplicative
subset of R to be a subset S such that 1 ∈ S and ab ∈ S if a, b ∈ S.
(a) Define a relation ∼ on R × S by (a, s) ∼ (a0 , s0 ) if there exists an s∗ ∈ S
such that s∗ (s0 a − sa0 ) = 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 operations. The ring
S −1 R is called the ring of quotients of R with respect to S.
(c) Show that the map ψ : R → S −1 R defined by ψ(a) = a/1 is a ring
homomorphism.
300 CHAPTER 18 INTEGRAL DOMAINS
(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.
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 multi-
plication 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 intersec-
tion. 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 by 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 .
301
302 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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 X ⊆ Y with sets X and Y . A set X together with a
partial order is called a partially ordered set, or poset.
Example 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 2. Let X be any set. We will define the power set of X to be
the set of all subsets of X. We denote the power set of X by P(X). For
example, let X = {a, b, c}. Then P(X) is the set of all subsets of the set
{a, b, c}:
∅ {a} {b} {c}
{a, b} {a, c} {b, c} {a, b, c}.
On any power set of a set X, set inclusion, ⊆, is a partial order. We can
represent the order on {a, b, c} schematically by a diagram such as the one
in Figure 19.1.
{a, b, c}
{a, b} {a, c} {b, c}
{a} {b} {c}
∅
Figure 19.1. Partial order on P({a, b, c})
Example 3. Let G be a group. The set of subgroups of G is a poset, where
the partial order is set inclusion.
Example 4. 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
19.1 LATTICES 303
relation is also antisymmetric. The relation is transitive, because if m | n
and n | p, then m | p.
Example 5. Let X = {1, 2, 3, 4, 6, 8, 12, 24} be the set of divisors of 24 with
the partial order defined in Example 4. Figure 19.2 shows the partial order
on X.
24
8 12
4 6
2 3
1
Figure 19.2. 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 .
Example 6. Let Y = {2, 3, 4, 6} be contained in the set X of Example 5.
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 19.1 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.
304 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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 7. 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 8. 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)0 must
also be true about A0 ∩ B 0 . We also have a duality principle for lattices.
Principle of Duality. Any statement that is true for all lattices remains
true when is replaced by and ∨ and ∧ are interchanged throughout the
statement.
The following theorem tells us that a lattice is an algebraic structure
with two binary operations that satisfy certain axioms.
Theorem 19.2 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.
19.1 LATTICES 305
3. Idempotent laws: a ∨ a = a and a ∧ a = a.
4. Absorption laws: a ∨ (a ∧ b) = a and a ∧ (a ∨ b) = a.
Proof. By the Principle of Duality, we need only prove the first statement
in each part.
(1) By definition a ∨ b is the least upper bound of {a, b}, and b ∨ a is the
least upper bound of {b, a}; however, {a, b} = {b, a}.
(2) We will show that a ∨ (b ∨ c) and (a ∨ b) ∨ c are both least upper
bounds of {a, b, c}. Let d = a ∨ b. Then c d ∨ c = (a ∨ b) ∨ c. We also
know that
a a ∨ b = d d ∨ c = (a ∨ b) ∨ c.
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 con-
ditions 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 19.3 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,
306 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
or a c.
To show that L is a lattice, we must prove that a ∨ b and a ∧ b are,
respectively, the least upper and greatest lower bounds of a and b. Since
a = (a ∨ b) ∧ a = a ∧ (a ∨ b), it follows that a a ∨ b. Similarly, b a ∨ b.
Therefore, a ∨ b is an upper bound for a and b. Let u be any other upper
bound of both a and b. Then a u and b u. But a ∨ b u since
(a ∨ b) ∨ u = a ∨ (b ∨ u) = a ∨ u = u.
The proof that a ∧ b is the greatest lower bound of a and b is left as an
exercise.
19.2 Boolean Algebras
Let us investigate the example of the power set, P(X), of a set X more
closely. The power set is a lattice that is ordered by inclusion. By the
definition of the power set, the largest element in P(X) is X itself and the
smallest element is ∅, the empty set. For any set A in P(X), we know that
A ∩ X = A and A ∪ ∅ = A. This suggests the following definition for lattices.
An element I in a poset X is a largest element if a I for all a ∈ X. An
element O is a smallest element of X if O a for all a ∈ X.
Let A be in P(X). Recall that the complement of A is
A0 = X \ A = {x : x ∈ X and x ∈
/ A}.
We know that A ∪ A0 = X and A ∩ A0 = ∅. 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 ∈ X, there exists an a0 such that a ∨ a0 = I
and a ∧ a0 = 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.
19.2 BOOLEAN ALGEBRAS 307
Theorem 19.4 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 19.5 A set B is a Boolean algebra if and only if there exist binary
operations ∨ and ∧ on B satisfying the following axioms.
1. a ∨ b = b ∨ a and a ∧ b = b ∧ a for a, b ∈ B.
2. a ∨ (b ∨ c) = (a ∨ b) ∨ c and a ∧ (b ∧ c) = (a ∧ b) ∧ c for a, b, c ∈ B.
3. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) and a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) for
a, b, c ∈ B.
4. There exist elements I and O such that a ∨ O = a and a ∧ I = a for
all a ∈ B.
5. For every a ∈ B there exists an a0 ∈ B such that a ∨ a0 = I and
a ∧ a0 = O.
308 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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 ∧ a0 )
= (a ∨ a) ∧ (a ∨ a0 )
= (a ∨ a) ∧ I
= a ∨ a.
Observe that
I ∨ b = (I ∨ b) ∧ I = (I ∧ I) ∨ (b ∧ I) = I ∨ I = 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 19.3 are met; therefore, B
must be a lattice. Condition (4) tells us that B is a distributive lattice.
For a ∈ B, O ∨ a = a; hence, O a and O is the smallest element in B.
To show that I is the largest element in B, we will first show that a ∨ b = b
is equivalent to a ∧ b = a. Since a ∨ I = a for all a ∈ B, using the absorption
laws we can determine that
a ∨ I = (a ∧ I) ∨ I = I ∨ (I ∧ a) = I
or a I for all a in B. Finally, since we know that B is complemented by
(5), B must be a Boolean algebra.
Conversely, suppose that B is a Boolean algebra. Let I and O be the
greatest and least elements in B, respectively. If we define a ∨ b and a ∧ b as
least upper and greatest lower bounds of {a, b}, then B is a Boolean algebra
by Theorem 19.3 , Theorem 19.4, and our hypothesis.
Many other identities hold in Boolean algebras. Some of these identities
are listed in the following theorem.
Theorem 19.6 Let B be a Boolean algebra. Then
19.2 BOOLEAN ALGEBRAS 309
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 = a0 .
4. (a0 )0 = a for all a ∈ B.
5. I 0 = O and O0 = I.
6. (a ∨ b)0 = a0 ∧ b0 and (a ∧ b)0 = a0 ∨ b0 (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.
310 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
We will show that any finite Boolean algebra is isomorphic to the Boolean
algebra obtained by taking the power set of some finite set X. We will need
a few lemmas and definitions before we prove this result. Let B be a finite
Boolean algebra. An element a ∈ B is an atom of B if a 6= O and a ∧ b = a
for all 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 19.7 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 19.8 Let a and b be atoms in a finite Boolean algebra B such that
a 6= b. Then a ∧ b = O.
Proof. Since a ∧ b is the greatest lower bound of a and b, we know that
a ∧ b a. Hence, either a ∧ b = a or a ∧ b = O. However, if a ∧ b = a, then
either a b or a = O. In either case we have a contradiction because a and
b are both atoms; therefore, a ∧ b = O.
Lemma 19.9 Let B be a Boolean algebra and a, b ∈ B. The following
statements are equivalent.
1. a b.
2. a ∧ b0 = O.
3. a0 ∨ b = I.
Proof. (1) ⇒ (2). If a b, then a ∨ b = b. Therefore,
a ∧ b0 = a ∧ (a ∨ b)0
= a ∧ (a0 ∧ b0 )
= (a ∧ a0 ) ∧ b0
= O ∧ b0
= O.
19.2 BOOLEAN ALGEBRAS 311
(2) ⇒ (3). If a ∧ b0 = O, then a0 ∨ b = (a ∧ b0 )0 = O0 = I.
(3) ⇒ (1). If a0 ∨ b = I, then
a = a ∧ (a0 ∨ b)
= (a ∧ a0 ) ∨ (a ∧ b)
= O ∨ (a ∧ b)
= a ∧ b.
Thus, a b.
Lemma 19.10 Let B be a Boolean algebra and b and c be elements in B
such that b 6 c. Then there exists an atom a ∈ B such that a b and a 6 c.
Proof. By Lemma 19.9, b ∧ c0 6= O. Hence, there exists an atom a such
that a b ∧ c0 . Consequently, a b and a 6 c.
Lemma 19.11 Let b ∈ B and a1 , . . . , an be the atoms of B such that ai b.
Then b = a1 ∨· · ·∨an . Furthermore, if a, a1 , . . . , an are atoms of B such that
a b, ai b, and b = a ∨ a1 ∨ · · · ∨ an , then a = ai for some i = 1, . . . , n.
Proof. Let b1 = a1 ∨ · · · ∨ an . Since ai b for each i, we know that b1 b.
If we can show that b b1 , then the lemma is true by antisymmetry. Assume
b 6 b1 . Then there exists an atom a such that a b and a 6 b1 . Since a is
an atom and a b, we can deduce that a = ai for some ai . However, this is
impossible since a b1 . Therefore, b b1 .
Now suppose that b = a1 ∨ · · · ∨ an . If a is an atom less than b,
a = a ∧ b = a ∧ (a1 ∨ · · · ∨ an ) = (a ∧ a1 ) ∨ · · · ∨ (a ∧ an ).
But each term is O or a with a ∧ ai occurring for only one ai . Hence, by
Lemma 19.8, a = ai for some i.
Theorem 19.12 Let B be a finite Boolean algebra. Then there exists a set
X such that B is isomorphic to P(X).
Proof. We will show that B is isomorphic to P(X), where X is the set
of atoms of B. Let a ∈ B. By Lemma 19.11, 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 }.
312 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
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 19.13 The order of any finite Boolean algebra must be 2n for
some positive integer n.
19.3 The Algebra of Electrical Circuits
The usefulness of Boolean algebras has become increasingly apparent over
the past several decades with the development of the modern computer.
The circuit design of computer chips can be expressed in terms of Boolean
algebras. In this section we will develop the Boolean algebra of electrical
circuits and switches; however, these results can easily be generalized to the
design of integrated computer circuitry.
A switch is a device, located at some point in an electrical circuit, that
controls the flow of current through the circuit. Each switch has two possible
states: it can be open, and not allow the passage of current through the
circuit, or a it can be closed, and allow the passage of current. These states
are mutually exclusive. We require that every switch be in one state or the
other: a switch cannot be open and closed at the same time. Also, if one
switch is always in the same state as another, we will denote both by the
same letter; that is, two switches that are both labeled with the same letter
a will always be open at the same time and closed at the same time.
Given two switches, we can construct two fundamental types of circuits.
Two switches a and b are in series if they make up a circuit of the type
that is illustrated in Figure 19.3. 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
19.3 THE ALGEBRA OF ELECTRICAL CIRCUITS 313
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 19.4.
In the case of a parallel circuit, current can pass between A and B if either
one of the switches is closed. We denote a parallel combination of circuits a
and b by a ∨ b.
A a b B
Figure 19.3. a ∧ b
a
A B
b
Figure 19.4. a ∨ b
We can build more complicated electrical circuits out of series and par-
allel 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 19.5. If a is a switch, then a0 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 ∧ a0 = O and a ∨ a0 = I are shown in Figure 19.6.
Example 9. Every Boolean expression represents a switching circuit. For
example, given the expression (a ∨ b) ∧ (a ∨ b0 ) ∧ (a ∨ b), we can construct
the circuit in Figure 19.7.
Theorem 19.14 The set of all circuits is a Boolean algebra.
314 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
b a b
a
c a c
Figure 19.5. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
a
a a0
a0
Figure 19.6. a ∧ a0 = O and a ∨ a0 = I
We leave as an exercise the proof of this theorem for the Boolean alge-
bra axioms not yet verified. We can now apply the techniques of Boolean
algebras to switching theory.
Example 10. Given a complex circuit, we can now apply the techniques
of Boolean algebra to reduce it to a simpler one. Consider the circuit in
Figure 19.7. Since
(a ∨ b) ∧ (a ∨ b0 ) ∧ (a ∨ b) = (a ∨ b) ∧ (a ∨ b) ∧ (a ∨ b0 )
= (a ∨ b) ∧ (a ∨ b0 )
= a ∨ (b ∧ b0 )
=a∨O
= a,
a a a
b b0 b
Figure 19.7. (a ∨ b) ∧ (a ∨ b0 ) ∧ (a ∨ b)
EXERCISES 315
we can replace the more complicated circuit with a circuit containing the
single switch a and achieve the same function.
Historical Note
George Boole (1815–1864) was the first person to study lattices. In 1847, he pub-
lished The Investigation of the Laws of Thought, a book in which he used lattices to
formalize logic and the calculus of propositions. Boole believed that mathematics
was the study of form rather than of content; that is, he was not so much concerned
with what he was calculating as with how he was calculating it. Boole’s work was
carried on by his friend Augustus De Morgan (1806–1871). De Morgan observed
that the principle of duality often held in set theory, as is illustrated by De Morgan’s
laws for set theory. He believed, as did Boole, that mathematics was the study of
symbols and abstract operations.
Set theory and logic were further advanced by such mathematicians as Alfred
North 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 indis-
pensable 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 com-
puter chips. Sociologists have used lattices and Boolean algebras to model social
hierarchies; biologists have used them to describe biosystems.
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).
316 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
5. Prove or disprove: Z is a poset under the relation a b if a | b.
6. Draw the switching circuit for each of the following Boolean expressions.
(a) (a ∨ b ∨ a0 ) ∧ a (c) a ∨ (a ∧ b)
0
(b) (a ∨ b) ∧ (a ∨ b) (d) (c ∨ a ∨ b) ∧ c0 ∧ (a ∨ b)0
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
a0 b
a c0 a c0
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 b0
a0
b
a a b
a0
b a0 b
a b c
a0 b0 c
a b0 c0
11. Prove or disprove: The set of all nonzero integers is a lattice, where a b is
defined by a | b.
12. Prove that a ∧ b is the greatest lower bound of a and b in Theorem 19.3.
EXERCISES 317
13. 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, as in Theorem 19.3, by a b if a ∨ b = b. Prove that the
greatest lower bound of a and b is a ∧ b.
14. 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.
15. 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.
16. 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 = a0 .
(c) (a0 )0 = a for all a ∈ B.
(d) I 0 = O and O0 = I.
(e) (a ∨ b)0 = a0 ∧ b0 and (a ∧ b)0 = a0 ∨ b0 (De Morgan’s laws).
17. By drawing the appropriate diagrams, complete the proof of Theorem 19.14
to show that the switching functions form a Boolean algebra.
18. Let B be a Boolean algebra. Define binary operations + and · on B by
a + b = (a ∧ b0 ) ∨ (a0 ∧ b)
a · b = a ∧ b.
Prove that B is a commutative ring under these operations satisfying a2 = a
for all a ∈ B.
19. 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 ≤.
20. 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.
21. Let B be a Boolean algebra. Prove that a = b if and only if (a∧b0 )∨(a0 ∧b) = O
for a, b ∈ B.
318 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS
Table 19.1. Boolean polynomials
x y x0 x∨y x∧y
0 0 1 0 0
0 1 1 1 0
1 0 0 1 0
1 1 0 1 1
22. Let B be a Boolean algebra. Prove that a = 0 if and only if (a∧b0 )∨(a0 ∧b) = b
for all b ∈ B.
23. Let L and M be lattices. Define an order relation on L × M by (a, b) (c, d)
if a c and b d. Show that L × M is a lattice under this partial order.
Programming Exercises
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 0 . The values of the functions
are defined in Table 19.1. Write a program to evaluate Boolean polynomials.
References and Suggested Readings
[1] Donnellan, T. Lattice Theory. Pergamon Press, Oxford, 1968.
[2] Halmos, P. R. “The Basic Concepts of Algebraic Logic,” American Mathe-
matical Monthly 53 (1956), 363–87.
[3] Hohn, F. “Some Mathematical Aspects of Switching,” American Mathemat-
ical 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
319
320 CHAPTER 20 VECTOR SPACES
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 1. The n-tuples of real numbers, denoted by Rn , form a vector
space over R. Given vectors u = (u1 , . . . , un ) and v = (v1 , . . . , vn ) in Rn and
α in R, we can define vector addition by
u + v = (u1 , . . . , un ) + (v1 , . . . , vn ) = (u1 + v1 , . . . , un + vn )
and scalar multiplication by
αu = α(u1 , . . . , un ) = (αu1 , . . . , αun ).
Example 2. If F is a field, then F [x] is a vector space over F . The vectors
in F [x] are simply polynomials. Vector addition is just polynomial addition.
If α ∈ F and p(x) ∈ F [x], then scalar multiplication is defined by αp(x).
Example 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 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 20.1 Let V be a vector space over F . Then each of the fol-
lowing 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.
20.2 SUBSPACES 321
4. (−1)v = −v for all v ∈ V .
5. −(αv) = (−α)v = α(−v) for all α ∈ F and all v ∈ V .
Proof. To prove (1), observe that
0v = (0 + 0)v = 0v + 0v;
consequently, 0 + 0v = 0v + 0v. Since V is an abelian group, 0 = 0v.
The proof of (2) is almost identical to the proof of (1). For (3), we are
done if α = 0. Suppose that α 6= 0. Multiplying both sides of αv = 0 by
1/α, we have v = 0.
To show (4), observe that
v + (−1)v = 1v + (−1)v = (1 − 1)v = 0v = 0,
and so −v = (−1)v. We will leave the proof of (5) as an exercise.
20.2 Subspaces
Just as groups have subgroups and rings have subrings, vector spaces also
have substructures. Let V be a vector space over a field F , and W a subset
of V . Then W is a subspace of V if it is closed under vector addition and
scalar multiplication; that is, if u, v ∈ W and α ∈ F , it will always be the
case that u + v and αv are also in W .
Example 5. 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 6. 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 .
322 CHAPTER 20 VECTOR SPACES
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
Xn
w= αi vi = α1 v1 + α2 v2 + · · · + αn vn
i=1
is called a linear combination of the vectors v1 , v2 , . . . , vn . The spanning
set of vectors v1 , v2 , . . . , vn is the set of vectors obtained from all possible lin-
ear combinations of v1 , v2 , . . . , vn . If W is the spanning set of v1 , v2 , . . . , vn ,
then we often say that W is spanned by v1 , v2 , . . . , vn .
Proposition 20.2 Let S = {v1 , v2 , . . . , vn } be vectors in a vector space V .
Then the span of S is a subspace of V .
Proof. Let u and v be in S. We can write both of these vectors as linear
combinations of the vi ’s:
u = α1 v1 + α2 v2 + · · · + αn vn
v = β1 v1 + β2 v2 + · · · + βn vn .
Then
u + v = (α1 + β1 )v1 + (α2 + β2 )v2 + · · · + (αn + βn )vn
is a linear combination of the vi ’s. For α ∈ F ,
αu = (αα1 )v1 + (αα2 )v2 + · · · + (ααn )vn
is in the span of S.
20.3 Linear Independence
Let S = {v1 , v2 , . . . , vn } be a set of vectors in a vector space V . If there
exist scalars α1 , α2 . . . αn ∈ F such that not all of the αi ’s are zero and
α1 v1 + α2 v2 + · · · + αn vn = 0,
then S is said to be linearly dependent. If the set S is not linearly depen-
dent, 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 }.
20.3 LINEAR INDEPENDENCE 323
Proposition 20.3 Let {v1 , v2 , . . . , vn } be a set of linearly independent vec-
tors 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 20.4 A set {v1 , v2 , . . . , vn } of vectors in a vector space V is
linearly dependent if and only if one of the vi ’s is a linear combination of
the rest.
Proof. Suppose that {v1 , v2 , . . . , vn } is a set of linearly dependent vectors.
Then there exist scalars α1 , . . . , αn such that
α1 v1 + α2 v2 + · · · + αn vn = 0,
with at least one of the αi ’s not equal to zero. Suppose that αk 6= 0. Then
α1 αk−1 αk+1 αn
vk = − v1 − · · · − vk−1 − vk+1 − · · · − vn .
αk αk αk αk
Conversely, suppose that
vk = β1 v1 + · · · + βk−1 vk−1 + βk+1 vk+1 + · · · + βn vn .
Then
β1 v1 + · · · + βk−1 vk−1 − vk + βk+1 vk+1 + · · · + βn vn = 0.
The following proposition is a consequence of the fact that any system of
homogeneous linear equations with more unknowns than equations will have
a nontrivial solution. We leave the details of the proof for the end-of-chapter
exercises.
324 CHAPTER 20 VECTOR SPACES
Proposition 20.5 Suppose that a vector space V is spanned by n vectors.
If m > n, then any set of m vectors in V must be linearly dependent.
A set {e1 , e2 , . . . , en } of vectors in a vector space V is called a basis for
V if {e1 , e2 , . . . , en } is a linearly independent set that spans V .
Example 7. The vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1)
form a basis for R3 . The set certainly spans R3 , since any arbitrary vector
(x1 , x2 , x3 ) in R3 can be written as x1 e1 + x2 e2 + x3 e3 . Also, none of the
vectors e1 , e2 , e3 can be written as a linear combination of the other two;
hence, they are linearly independent. The vectors e1 , e2 , e3 are not the only
basis of R3 : the set {(3, 2, 1), (3, 2, 0), (1, 1, 1)} is also a basis for R3 .
√ √ √
Example
√ 8.√Let Q( 2 ) = {a + b 2√: a, b ∈ Q}. The sets {1, 2 } and
{1 + 2, 1 − 2 } are both bases of Q( 2 ).
From the last two examples it should be clear that a given vector space
has several bases. In fact, there are an infinite number of bases for both
of these examples. In general, there is no unique basis for a vector space.
However, 3
√ every basis of R consists of exactly three vectors, and every basis
of Q( 2 ) consists of exactly two vectors. This is a consequence of the next
proposition.
Proposition 20.6 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 18.5, 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 20.7 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 .
EXERCISES 325
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 .
Exercises
1. If F is a field, show that F [x] is a vector space over F , where the vectors
in F [x] are polynomials. Vector addition is polynomial addition, and scalar
multiplication is defined by αp(x) for α ∈ F .
√
2. Prove that Q( 2 ) is a vector space.
√ √ √ √
3. Let Q( 2, 3 ) be the field generated√ by elements
√ of the form a + b 2 + c 3,
where a, b, c 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 operations.
7. Which of the following sets are subspaces of R3 ? If the set is indeed a
subspace, find a basis for the subspace and compute its dimension.
(a) {(x1 , x2 , x3 ) : 3x1 − 2x2 + x3 = 0}
(b) {(x1 , x2 , x3 ) : 3x1 + 4x3 = 0, 2x1 − x2 + x3 = 0}
(c) {(x1 , x2 , x3 ) : x1 − 2x2 + 2x3 = 2}
(d) {(x1 , x2 , x3 ) : 3x1 − 2x22 = 0}
326 CHAPTER 20 VECTOR SPACES
8. Show that the set of all possible solutions (x, y, z) ∈ R3 of the equations
Ax + By + Cz = 0
Dx + Ey + Cz = 0
forms 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 state-
ments.
(a) If S = {v1 , . . . , vn } is a set of linearly independent vectors for V , then
S is a basis for V .
(b) If S = {v1 , . . . , vn } spans V , then S is a basis for V .
(c) If S = {v1 , . . . , vk } is a set of linearly independent vectors for V with
k < n, then there exist vectors vk+1 , . . . , vn such that
{v1 , . . . , vk , vk+1 , . . . , vn }
is a basis for V .
12. Prove that any set of vectors containing 0 is linearly dependent.
13. Let V be a vector space. Show that {0} is a subspace of V of dimension zero.
14. If a vector space V is spanned by n vectors, show that any set of m vectors
in V must be linearly dependent for m > n.
15. Linear Transformations. Let V and W be vector spaces over a field F , of
dimensions m and n, respectively. If T : V → W is a map satisfying
T (u + v) = T (u) + T (v)
T (αv) = αT (v)
for all α ∈ F and all u, v ∈ V , then T is called a linear transformation
from V into W .
(a) Prove that the kernel of T , ker(T ) = {v ∈ V : T (v) = 0}, is a
subspace of V . The kernel of T is sometimes called the null space of
T.
(b) Prove that the range or range space of T , R(V ) = {w ∈ W :
T (v) = w for some v ∈ V }, is a subspace of W .
(c) Show that T : V → W is injective if and only if ker(T ) = {0}.
EXERCISES 327
(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
of U and V and we write W = U ⊕ V . Show that every element w ∈ W
can be written uniquely as w = u + v, where u ∈ U and v ∈ V .
(c) Let U be a subspace of dimension k of a vector space W of dimension
n. Prove that there exists a subspace V of dimension n − k such that
W = U ⊕ V . Is the subspace V unique?
(d) If U and V are arbitrary subspaces of a vector space W , show that
dim(U + V ) = dim U + dim V − dim(U ∩ V ).
18. Dual Spaces. Let V and W be finite dimensional vector spaces over a
field F .
(a) Show that the set of all linear transformations from V into W , denoted
by Hom(V, W ), is a vector space over F , where we define vector addition
as 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 lin-
ear 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).
328 CHAPTER 20 VECTOR SPACES
(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 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 ∗∗ .
References and Suggested Readings
[1] Beezer, R. A First Course in Linear Algebra. Available online at
http://linear.ups.edu/. 2004.
[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, Engle-
wood Cliffs, NJ, 1971.
[5] Johnson, L. W., Riess, R. D., and Arnold, J. T. Introduction to Linear Alge-
bra. 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 polyno-
mials 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 1. For example, let
√ √
F = Q( 2 ) = {a + b 2 : a, b ∈ Q}
329
330 CHAPTER 21 FIELDS
√ √
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.
Example 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 17.13, the ideal hp(x)i generated by p(x) is maximal;
hence, Z2 [x]/hp(x)i is a field. Let f (x) + hp(x)i be an arbitrary element of
Z2 [x]/hp(x)i. By the division algorithm,
f (x) = (x2 + x + 1)q(x) + r(x),
where the degree of r(x) is less than the degree of x2 + x + 1. Therefore,
f (x) + hx2 + x + 1i = r(x) + hx2 + x + 1i.
The only possibilities for r(x) are then 0, 1, x, and 1 + x. Consequently,
E = Z2 [x]/hx2 + x + 1i is a field with four elements and must be a field
extension of Z2 , containing a zero α of p(x). The field Z2 (α) consists of
elements
0 + 0α = 0
1 + 0α = 1
0 + 1α = α
1 + 1α = 1 + α.
Notice that α2 + α + 1 = 0; hence, if we compute (1 + α)2 ,
(1 + α)(1 + α) = 1 + α + α + (α)2 = α.
Other calculations are accomplished in a similar manner. We summarize
these computations in the following tables, which tell us how to add and
multiply elements in E.
+ 0 1 α 1+α
0 0 1 α 1+α
1 1 0 1+α α
α α 1+α 0 1
1+α 1+α α 1 0
21.1 EXTENSION FIELDS 331
· 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
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.
Theorem 21.1 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 2. Clearly, we can assume that p(x) is an irreducible
polynomial. We wish to find an extension field E of F containing an element
α such that p(α) = 0. The ideal hp(x)i generated by p(x) is a maximal ideal
in F [x] by Theorem 17.13; hence, F [x]/hp(x)i is a field. We claim that
E = F [x]/hp(x)i is the desired field.
We first show that E is a field extension of F . We can define a ho-
momorphism of commutative rings by the map ψ : F → F [x]/hp(x)i, where
ψ(a) = a + hp(x)i for a ∈ F . It is easy to check that ψ is indeed a ring
homomorphism. Observe that
ψ(a) + ψ(b) = (a + hp(x)i) + (b + hp(x)i) = (a + b) + hp(x)i = ψ(a + b)
and
ψ(a)ψ(b) = (a + hp(x)i)(b + hp(x)i) = ab + hp(x)i = ψ(ab).
To prove that ψ is one-to-one, assume that
a + hp(x)i = ψ(a) = ψ(b) = b + hp(x)i.
Then a − b is a multiple of p(x), since it lives in the ideal hp(x)i. Since
p(x) is a nonconstant polynomial, the only possibility is that a − b = 0.
Consequently, a = b and ψ is injective. Since ψ is one-to-one, we can
identify F with the subfield {a + hp(x)i : a ∈ F } of E and view E as an
extension field of F .
332 CHAPTER 21 FIELDS
It remains for us to prove that p(x) has a zero α ∈ F . Set α = x+hp(x)i.
Then α is in E. If p(x) = a0 + a1 x + · · · + an xn , then
p(α) = a0 + a1 (x + hp(x)i) + · · · + an (x + hp(x)i)n
= a0 + (a1 x + hp(x)i) + · · · + (an xn + hp(x)i)
= a0 + a1 x + · · · + an xn + hp(x)i
= 0 + hp(x)i.
Therefore, we have found an element α ∈ E = F [x]/hp(x)i such that α is a
zero of p(x).
Example 3. Let p(x) = x5 + x4 + 1 ∈ Z2 [x]. Then p(x) has irreducible
factors x2 + x + 1 and x3 + x + 1. For a field extension E of Z2 such that p(x)
has a root in E, we can let E be either Z2 [x]/hx2 +x+1i or Z2 [x]/hx3 +x+1i.
We will leave it as an exercise to show that Z2 [x]/hx3 + x + 1i is a field with
23 = 8 elements.
Algebraic Elements
An element α in an extension field E over F is algebraic over F if f (α) = 0
for some nonzero polynomial f (x) ∈ F [x]. An element in E that is not
algebraic over F is transcendental over F . An extension field E of a field
F is an algebraic extension of F if every element in E is algebraic over
F . If E is a field extension of F and α1 , . . . , αn are contained in E, we
denote the smallest field containing F and α1 , . . . , αn by F (α1 , . . . , αn ). If
E = F (α) for some α ∈ E, then E is a simple extension of F .
√
Example 4. 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.1 (In many
cases we do not know whether or not a particular number is transcendental;
for example, it is not known whether π + e is transcendental or algebraic.)
1
If we choose a number in R, then there is a probability of 1 that the number will be
transcendental over Q.
21.1 EXTENSION FIELDS 333
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.
p √
Example 5. We will show that 2 + 3 is algebraic over Q. If α =
p √ √ √
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 21.2 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) 6= 0 for all noncon-
stant polynomials p(x) ∈ F [x]. This is true if and only if ker φα = {0}; that
is, it is true exactly when φα is one-to-one. Hence, E must contain a copy
of F [x]. The smallest field containing F [x] is the field of fractions F (x). By
Theorem 18.4, E must contain a copy of this field.
We have a more interesting situation in the case of algebraic extensions.
Theorem 21.3 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 monic
polynomial in F [x] such that f (α) = 0, then p(x) divides f (x).
Proof. Let φα : F [x] → E be the evaluation homomorphism. The kernel of
φα is a principal ideal generated by some p(x) ∈ F [x] with deg p(x) ≥ 1. We
know that such a polynomial exists, since F [x] is a principal ideal domain
and α is algebraic. The ideal hp(x)i consists exactly of those elements of
F [x] having α as a zero. If f (α) = 0 and f (x) is not the zero polynomial,
then f (x) ∈ hp(x)i and p(x) divides f (x). So p(x) is a polynomial of minimal
degree having α as a zero. Any other polynomial of the same degree having
α as a zero must have the form βp(x) for some β ∈ F .
Suppose now that p(x) = r(x)s(x) is a factorization of p into polyno-
mials 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.
334 CHAPTER 21 FIELDS
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 6. Let f (x) = x2 − 2 and g(x)p
= x4 − 4x2 + 1. These polynomials
√ √
are the minimal polynomials of 2 and 2 + 3, respectively.
Proposition 21.4 Let E be a field extension of F and α ∈ E be algebraic
over F . Then F (α) ∼
= F [x]/hp(x)i, where p(x) is the minimal polynomial of
α over F .
Proof. Let φα : F [x] → E be the evaluation homomorphism. The kernel
of this map is the minimal polynomial p(x) 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 21.5 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 .
21.1 EXTENSION FIELDS 335
To show uniqueness, suppose that
β = b0 + b1 α + · · · + bn−1 αn−1 = c0 + c1 α + · · · + cn−1 αn−1
for bi and ci in F . Then
g(x) = (b0 − c0 ) + (b1 − c1 )x + · · · + (bn−1 − cn−1 )xn−1
is in F [x] and g(α) = 0. Since the degree of g(x) is less than the degree
of p(x), the irreducible polynomial of α, g(x) must be the zero polynomial.
Consequently,
b0 − c0 = b1 − c1 = · · · = bn−1 − cn−1 = 0,
or bi = ci for i = 0, 1, . . . , n − 1. Therefore, we have shown uniqueness.
Example 7. Since x2 + 1 is irreducible over R, hx2 + 1i is a maximal ideal
in R[x]. So E = R[x]/hx2 + 1i is a field extension of R that contains a root of
x2 + 1. Let α = x + hx2 + 1i. We can identify E with the complex numbers.
By Proposition 21.4, E is isomorphic to R(α) = {a + bα : a, b ∈ R}. We
know that α2 = −1 in E, since
α2 + 1 = (x + hx2 + 1i)2 + (1 + hx2 + 1i)
= (x2 + 1) + hx2 + 1i
= 0.
Hence, we have an isomorphism of R(α) with C defined by the map that
takes a + bα to a + bi.
Let E be a field extension of a field F . If we regard E as a vector space
over F , then we can bring the machinery of linear algebra to bear on the
problems that we will encounter in our study of fields. The elements in the
field E are vectors; the elements in the field F are scalars. We can think
of addition in E as adding vectors. When we multiply an element in E
by an element of F , we are multiplying a vector by a scalar. This view of
field extensions is especially fruitful if a field extension E of F is a finite
dimensional vector space over F , and Theorem 21.5 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 .
336 CHAPTER 21 FIELDS
Theorem 21.6 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. Theorem 21.6 says that every finite extension of a field F is an
algebraic extension. The converse is false, however. We will leave it as an
exercise to show that the set of all elements in R that are algebraic over Q
forms an infinite field extension of Q.
The next theorem is a counting theorem, similar to Lagrange’s Theorem
in group theory. Theorem 21.6 will prove to be an extremely useful tool in
our investigation of finite field extensions.
Theorem 21.7 If E is a finite extension of F and K is a finite extension
of E, then K is a finite extension of F and
[K : F ] = [K : E][E : F ].
Proof. Let {α1 , . . . , αn } be a basis for E as a vector space over F and
{β1 , . . . , βm } be a basis for K as a vector space over E. We claim that
{αi βj } is a basis for K overP F . We will first showPnthat 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
m n
!
X X X
u= aij αi βj = aij (αi βj ).
j=1 i=1 i,j
So the mn vectors αi βj must span K over F .
We must show that {αi βj } are linearly independent. Recall that a set
of vectors v1 , v2 , . . . , vn in a vector space V are linearly independent if
c1 v1 + c2 v2 + · · · + cn vn = 0
21.1 EXTENSION FIELDS 337
implies that
c1 = c2 = · · · = cn = 0.
Let X
u= cij (αi βj ) = 0
i,j
for cij ∈ F . We need to prove that all of the cij ’s are zero. We can rewrite
u as
m n
!
X X
cij αi βj = 0,
j=1 i=1
P
where i cij αi ∈ E. Since the βj ’s are linearly independent over E, it must
be the case that
Xn
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 21.8 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 21.9 Let E be an extension field of F . If α ∈ E is algebraic over
F with minimal polynomial p(x) and β ∈ F (α) with minimal polynomial
q(x), then deg q(x) divides deg p(x).
Proof. We know that deg p(x) = [F (α) : F ] and deg q(x) = [F (β) : F ].
Since F ⊂ F (β) ⊂ F (α),
[F (α) : F ] = [F (α) : F (β)][F (β) : F ].
√ √
Example 8. Let us determine an extension field of Q√containing
√ 3+ 5. It
4
is easy to determine that the minimal polynomial of 3 + 5 is x − 16x + 4.
It follows that √ √
[Q( 3 + 5 ) : Q] = 4.
338 CHAPTER 21 FIELDS
√ √ √ √
We know that √ {1, 3 } is a basis for
√ Q( 3 ) over Q. Hence,
√ 3 + 5 can-
not be in√Q( 3 ). It follows that √ √5 cannot be √ in Q(√ 3 ) either.√ There-
} is
fore,√ {1,√ 5√ √ a basis
√ for Q( 3, 5 ) = (Q(
√ √ 3 ))( 5 )√over √
Q( 3 ) and
{1, 3, 5, 3 5 = 15 } is a basis for Q( 3, 5 ) = Q( 3 + 5 ) over Q.
This example shows that it is possible that some extension F (α1 , . . . , αn ) is
actually a simple extension of F even though n > 1.
√ √ √
Example 9. Let us compute √ a basis for Q( 3 5, 5 i), where 5 is the
positive square
√ √ root of 5 and 3 5 is the real cube root of 5. We know that
5i ∈ / Q( 3 5 ), so √ √ √
3 3
[Q( 5, 5 i) : Q( 5 )] = 2.
√ √
3
√ √
3
It is easy to determine that
√ {1,
√ 5i } is a basis for Q( √ 5, 5 i) over Q( 5 ).
3 3 2 3
We also know √ {1, 5, ( 5 ) } is a basis for Q( 5 ) over Q. Hence, a
√ that
basis for Q( 5, 3 5 ) 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. Conse-
quently, √ √ √
6 3
Q ⊂ Q( 5 ) ⊂ Q( 5, 5 i).
√ √ √
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 21.10 Let E be a field extension of F . Then the following state-
ments 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 ).
21.1 EXTENSION FIELDS 339
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 21.6.
(2) ⇒ (3). Suppose that E = F (α1 , . . . , αn ), where every αi is algebraic
over F . Then
E = F (α1 , . . . , αn ) ⊃ F (α1 , . . . , αn−1 ) ⊃ · · · ⊃ F (α1 ) ⊃ F,
where each field F (α1 , . . . , αi ) is algebraic over F (α1 , . . . , αi−1 ).
(3) ⇒ (1). Let
E = F (α1 , . . . , αn ) ⊃ F (α1 , . . . , αn−1 ) ⊃ · · · ⊃ F (α1 ) ⊃ F,
where each field F (α1 , . . . , αi ) is algebraic over F (α1 , . . . , αi−1 ). Since
F (α1 , . . . , αi ) = F (α1 , . . . , αi−1 )(αi )
is simple extension and αi is algebraic over F (α1 , . . . , αi−1 ), it follows that
[F (α1 , . . . , αi ) : F (α1 , . . . , αi−1 )]
is finite for each i. Therefore, [E : F ] is finite.
Algebraic Closure
Given a field F , the question arises as to whether or not we can find a field
E such that every polynomial p(x) has a root in E. This leads us to the
following theorem.
Theorem 21.11 Let E be an extension field of F . The set of elements in
E that are algebraic over F form a field.
Proof. Let α, β ∈ E be algebraic over F . Then F (α, β) is a finite extension
of F . Since every element of F (α, β) is algebraic over F , α ± β, α/β, and
α/β (β 6= 0) are all algebraic over F . Consequently, the set of elements in
E that are algebraic over F forms a field.
Corollary 21.12 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 .
340 CHAPTER 21 FIELDS
Theorem 21.13 A field F is algebraically closed if and only if every non-
constant 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 noncon-
stant 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] fac-
tors into linear factors. Let ax − b be such a factor. Then p(b/a) = 0.
Consequently, F is algebraically closed.
Corollary 21.14 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 21.15 Every field F has a unique algebraic closure.
It is a nontrivial fact that every field has a unique algebraic closure. The
proof is not extremely difficult, but requires some rather sophisticated set
theory. We refer the reader to [3], [4], or [8] for a proof of this result.
We now state the Fundamental Theorem of Algebra, first proven by
Gauss at the age of 22 in his doctoral thesis. This theorem states that
every polynomial with coefficients in the complex numbers has a root in the
complex numbers. The proof of this theorem will be given in Chapter 23.
Theorem 21.16 (Fundamental Theorem of Algebra) The field of com-
plex numbers is algebraically closed.
21.2 Splitting Fields
Let F be a field and p(x) be a nonconstant polynomial in F [x]. We already
know that we can find a field extension of F that contains a root of p(x).
However, we would like to know whether an extension E of F containing all
of the roots of p(x) exists. In other words, can we find a field extension of
21.2 SPLITTING FIELDS 341
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 10. 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 √ 11. Let p(x) = x3 − 3 be in Q[x]. Then p(x) has a root in the
3
field Q( 3 ). However, this field is not a splitting field for p(x) since the
complex cube roots of 3, √ √
− 3 3 ± ( 6 3 )5 i
,
2
√
are not in Q( 3 3 ).
Theorem 21.17 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 21.1, 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 question
is answered in the affirmative. Given two splitting fields K and L of a
polynomial p(x) ∈ F [x], there exists a field isomorphism φ : K → L that
preserves F . In order to prove this result, we must first prove a lemma.
342 CHAPTER 21 FIELDS
Lemma 21.18 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 ψ : E(α) → F (β) such that ψ(α) = β and ψ agrees
with φ on E.
Proof. If p(x) has degree n, then by Theorem 21.5 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
ψ(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 ψ is an isomorphism could be checked by
direct computation; however, it is easier to observe that ψ is a composition
of maps that we already know to be isomorphisms.
We can extend φ to be an isomorphism from E[x] to F [x], which we will
also denote by φ, by letting
φ(a0 + a1 x + · · · + an xn ) = φ(a0 ) + φ(a1 )x + · · · + φ(an )xn .
This extension agrees with the original isomorphism φ : E → F , since
constant polynomials get mapped to constant polynomials. By assumption,
φ(p(x)) = q(x); hence, φ maps hp(x)i onto hq(x)i. Consequently, we have
an isomorphism φ : E[x]/h p(x)i → F [x]/h q(x)i. By Theorem 21.4, we have
isomorphisms σ : E[x]/hp(x)i → F (α) and τ : F [x]/hq(x)i → F (β), defined
by evaluation at α and β, respectively. Therefore, ψ = τ −1 φσ is the required
isomorphism.
ψ
E(α)
−→ F
(β)
yσ yτ
φ
E[x]/hp(x)i
−→ F [x]/hq(x)i
y y
φ
E −→ F
We leave the proof of uniqueness as a exercise.
21.3 GEOMETRIC CONSTRUCTIONS 343
Theorem 21.19 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 E, 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 21.18, there exists
an isomorphism φ : E(α) → F (β) such that φ(α) = β and φ agrees with φ
on E.
ψ
K −→ L
y y
φ
E(α)
−→ F
(β)
y y
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 21.20 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 ge-
ometric 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.
344 CHAPTER 21 FIELDS
1. Given an arbitrary angle, can one trisect the angle into three equal
subangles using only a straightedge and compass?
2. Given an arbitrary circle, can one construct a square with the same
area using only a straightedge and compass?
3. Given a cube, can one construct the edge of another cube having
twice the volume of the original? Again, we are only allowed to use a
straightedge and compass to do the construction.
After puzzling mathematicians for over two thousand years, each of these
constructions was finally shown to be impossible. We will use the theory of
fields to provide a proof that the solutions do not exist. It is quite remarkable
that the long-sought solution to each of these three geometric problems came
from abstract algebra.
First, let us determine more specifically what we mean by a straightedge
and compass, and also examine the nature of these problems in a bit more
depth. To begin with, a straightedge is not a ruler. We cannot measure
arbitrary lengths with a straightedge. It is merely a tool for drawing a line
through two points. The statement that the trisection of an arbitrary angle is
impossible means that there is at least one angle that is impossible to trisect
with a straightedge-and-compass construction. Certainly it is possible to
trisect an angle in special cases. We can construct a 30◦ angle; hence, it is
possible to trisect a 90◦ 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 21.21 The set of all constructible real numbers forms a subfield
F of the field of real numbers.
Proof. Let α and β be constructible numbers. We must show that α + β,
α − β, αβ, and α/β (β 6= 0) are also constructible numbers. We can assume
that both α and β are positive with α > β. It is quite obvious how to
construct α+β and α−β. To find a line segment with length αβ, we assume
that β > 1 and construct the triangle in Figure 21.1 such that triangles
4ABC and 4ADE are similar. Since α/1 = x/β, the line segment x has
21.3 GEOMETRIC CONSTRUCTIONS 345
length αβ. A similar construction can be made if β < 1. We will leave it as
an exercise to show that the same triangle can be used to construct α/β for
β 6= 0.
D
β B
1
α C
A E
x
Figure 21.1. Construction of products
√
Lemma 21.22 If α is a constructible number, then α is a constructible
number.
Proof. In Figure 21.2 the triangles 4ABD, 4BCD, and 4ABC are
similar; hence, 1/x = x/α, or x2 = α.
B
x
1 α
A D C
Figure 21.2. Construction of roots
By Theorem 21.21, 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.
346 CHAPTER 21 FIELDS
Lemma 21.23 Let F be a subfield of R.
1. If a line contains two points in F , then it has the equation ax+by+c =
0, where a, b, and c are in F .
2. If a circle has a center at a point with coordinates in F and a radius
that is also in F , then it has the equation x2 + y 2 + dx + ey + f = 0,
where d, e, and f are in F .
Proof. Let (x1 , y1 ) and (x2 , y2 ) be points on a line whose coordinates are
in F . If x1 = x2 , then the equation of the line through the two points is
x−x1 = 0, which has the form ax+by +c = 0. If x1 6= x2 , then the equation
of the line through the two points is given by
y2 − y1
y − y1 = (x − x1 ),
x2 − x1
which can also be put into the proper form.
To prove the second part of the lemma, suppose that (x1 , y1 ) is the center
of a circle of radius r. Then the circle has the equation
(x − x1 )2 + (y − y1 )2 − r2 = 0.
This equation can easily be put into the appropriate form.
Starting with a field of constructible numbers F , we have three possible
ways of 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 x + f1 = 0
x2 + y 2 + d2 x + e2 x + f2 = 0
21.3 GEOMETRIC CONSTRUCTIONS 347
be the equations of two circles, where di , ei , and fi are in F for i = 1, 2.
These circles have the same intersection as the circle
x2 + y 2 + d1 x + e1 x + f1 = 0
and the line
(d1 − d2 )x + b(e2 − e1 )y + (f2 − f1 ) = 0.
The last equation is that of the chord passing through the intersection points
of the two circles. Hence, the intersection of two circles can be reduced to
the case of an intersection of a line with a circle.
Considering the case of the intersection of a line and a circle, we must
determine the nature of the solutions of the equations
ax + by + c = 0
2 2
x + y + dx + ey + f = 0.
If we eliminate y from these equations, we obtain an equation of the form
Ax2 + Bx + C = 0, where A, B, and C are in F . The x coordinate of the
intersection points is given by
√
−B ± B 2 − 4AC
x=
2A
√
and is in F ( α ), where α = B 2 − 4AC > 0. We have proven the following
lemma.
Lemma 21.24 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 21.25 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 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 21.24 and of the fact that
[Fk : Q] = [Fk : Fk−1 ][Fk−1 : Fk−2 ] · · · [F1 : Q] = 2k .
348 CHAPTER 21 FIELDS
Corollary 21.26 The field of all constructible numbers is an algebraic ex-
tension of Q.
As we can see by the field of constructible numbers, not every algebraic
extension of a field is a finite extension.
Doubling the Cube and Squaring the Circle
We are now ready to investigate the classical problems of doubling the cube
and squaring the circle. We can use the field of constructible numbers to
show exactly when a particular geometric construction can be accomplished.
Doubling the cube is impossible. Given the edge of the cube, it is im-
possible 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,
√3
[Q( 2 ) : Q] = 3
This is impossible, since 3 is not a power of 2.
Squaring the circle is impossible. Suppose that we have a circle of radius
1. The area of the circle is π; therefore, we must be able to construct a
√ √
square with side π. This is impossible since π and consequently π are
both transcendental. Therefore, using a straightedge and compass, it is not
possible to construct a square with the same area as the circle.
Trisecting an Angle
Trisecting an arbitrary angle is impossible. We will show that it is impossible
to construct a 20◦ angle. Consequently, a 60◦ angle cannot be trisected. We
first need to calculate the triple-angle formula for the cosine:
cos 3θ = cos(2θ + θ)
= cos 2θ cos θ − sin 2θ sin θ
= (2 cos2 θ − 1) cos θ − 2 sin2 θ cos θ
= (2 cos2 θ − 1) cos θ − 2(1 − cos2 θ) cos θ
= 4 cos3 θ − 3 cos θ.
The angle θ can be constructed if and only if α = cos θ is constructible. Let
θ = 20◦ . Then cos 3θ = cos 60◦ = 1/2. By the triple-angle formula for the
EXERCISES 349
cosine,
1
4α3 − 3α = .
2
Therefore, α is a zero of 8x3 −6x−1. This polynomial has no factors in Z[x],
and hence is irreducible over Q[x]. Thus, [Q(α) : Q] = 3. Consequently, α
cannot be a constructible number.
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 contribu-
tions to algebraic number theory by such notable mathematicians as Leonhard
Euler (1707–1783). Significant advances in the understanding of Fermat’s Last
Theorem were made by Ernst Kummer (1810–1893). Kummer’s student, Leopold
Kronecker (1823–1891), became one of the leading algebraists of the nineteenth
century. Kronecker’s theory of ideals and his study of algebraic number theory
added much to the understanding of fields.
David Hilbert (1862–1943) and Hermann Minkowski (1864–1909) were among
the mathematicians who led the way in this subject at the beginning of the twentieth
century. Hilbert and Minkowski were both mathematicians at Göttingen University
in Germany. Göttingen was truly one the most important centers of mathematical
research during the last two centuries. The large number of exceptional mathemati-
cians who studied there included Gauss, Dirichlet, Riemann, Dedekind, Noether,
and Weyl.
André Weil answered questions in number theory using algebraic geometry, a
field of mathematics that studies geometry by studying commutative rings. From
about 1955 to 1970, A. Grothendieck dominated the field of algebraic geometry.
Pierre Deligne, a student of Grothendieck, solved several of Weil’s number-theoretic
conjectures. One of the most recent contributions to algebra and number theory is
Gerd Falting’s proof of the Mordell-Weil conjecture. This conjecture of Mordell and
Weil essentially says that certain polynomials p(x, y) in Z[x, y] have only a finite
number of integral solutions.
350 CHAPTER 21 FIELDS
Exercises
1. Show that each of the following numbers is algebraic over Q by finding the
minimal polynomial of the number over Q.
q √
(a) 1/3 + 7
√ √
(b) 3 + 3 5
√ √
(c) 3 + 2 i
(d) cos θ + i sin θ for θ = 2π/n with n ∈ N
p√
3
(e) 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
√
4. Determine all of the subfields of Q( 4 3, i).
5. Show that Z2 [x]/hx3 + x + 1i is a field with eight elements. Construct a
multiplication table for the multiplicative group of the field.
6. Show that the regular 9-gon is not constructible with a straightedge and
compass, but that the regular 20-gon is constructible.
7. Prove that the cosine of one degree (cos 1◦ ) is algebraic over Q but not con-
structible.
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.
EXERCISES 351
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 exten-
sions are finite.]
15. Prove or disprove: Z[x]/hx3 − 2i is a field.
16. Let F be a field of characteristic p. Prove that p(x) = xp − a either is
irreducible over F or splits in F .
17. Let E be the algebraic closure of a field F . Prove that every polynomial p(x)
in F [x] splits in E.
18. If every irreducible polynomial p(x) in F [x] is linear, show that F is an
algebraically closed field.
19. Prove that if α and β are constructible numbers such that β 6= 0, then so is
α/β.
20. Show that the set of all elements in R that are algebraic over Q form a field
extension of Q that is not finite.
21. Let E be an algebraic extension of a field F , and let σ be an automorphism
of E leaving F fixed. Let α ∈ E. Show that σ induces a permutation of the
set of all zeros of the minimal polynomial of α that are in E.
√ √ √ √
22. Show
√ that
√ Q( 3, √ 7 )√= Q( 3 + 7 ). Extend your proof to show that
Q( a, b ) = Q( a + b ).
23. Let E be a finite extension of a field F . If [E : F ] = 2, show that E is a
splitting field of F .
24. Prove or disprove: Given a polynomial p(x) in Z6 [x], it is possible to construct
a ring R such that p(x) has a root in R.
25. Let E be a field extension of F and α ∈ E. Determine [F (α) : F (α3 )].
26. Let α, β be transcendental over Q. Prove that either αβ or α + β is also
transcendental.
27. Let E be an extension field of F and α ∈ E be transcendental over F . Prove
that every element in F (α) that is not in F is also transcendental over F .
352 CHAPTER 21 FIELDS
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, Mine-
ola, NY, 2010.
[8] Walker, E. A. Introduction to Abstract Algebra. Random House, New York,
1987. This work contains a proof showing that every field has an algebraic
closure.
22
Finite Fields
Finite fields appear in many applications of algebra, including coding theory
and 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 16.5 we know
that p must be prime. Suppose that F is a finite field with n elements.
Then nα = 0 for all α in F . Consequently, the characteristic of F must
be p, where p is a prime dividing n. This discussion is summarized in the
following proposition.
Proposition 22.1 If F is a finite field, then the characteristic of F is p,
where p is prime.
Throughout this chapter we will assume that p is a prime number unless
otherwise stated.
Proposition 22.2 If F is a finite field of characteristic p, then the order
of F is pn for some n ∈ N.
Proof. Let φ : Z → F be the ring homomorphism defined by φ(n) = n · 1.
Since the characteristic of F is p, the kernel of φ must be pZ and the image of
353
354 CHAPTER 22 FINITE FIELDS
φ must be a subfield of F isomorphic to Zp . We will denote this subfield by
K. Since F is a finite field, it must be a finite extension of K and, therefore,
an algebraic extension of K. Suppose that [F : K] = n is the dimension of
F , where F is a K vector space. There must exist elements α1 , . . . , αn ∈ F
such that any element α in F can be written uniquely in the form
α = a1 α1 + · · · + an αn ,
where the ai ’s are in K. Since there are p elements in K, there are pn
possible linear combinations of the αi ’s. Therefore, the order of F must be
pn .
Lemma 22.3 (Freshman’s Dream) Let p be prime and D be an integral
domain of characteristic p. Then
n n n
ap + bp = (a + b)p
for all positive integers n.
Proof. We will prove this lemma using mathematical induction on n. We
can use the binomial formula (see Chapter 2, Example 3) to verify the case
for n = 1; that is,
p
p
X p k p−k
(a + b) = a b .
k
k=0
If 0 < k < p, then
p p!
=
k k!(p − k)!
must be divisible by p, since p cannot divide k!(p − k)!. Note that D is an
integral domain of characteristic p, so all but the first and last terms in the
sum must be zero. Therefore, (a + b)p = ap + bp .
Now suppose that the result holds for all k, where 1 ≤ k ≤ n. By the
induction hypothesis,
n+1 n n n n n+1 n+1
(a + b)p = ((a + b)p )p = (ap + bp )p = (ap )p + (bp )p = ap + bp .
Therefore, the lemma is true for n + 1 and the proof is complete.
Let F be a field. A polynomial f (x) ∈ F [x] of degree n is separable if
it has n distinct roots in the splitting field of f (x); that is, f (x) is separable
when it factors into distinct linear factors over the splitting field of F . An
22.1 STRUCTURE OF A FINITE FIELD 355
extension E of F is a separable extension of F if every element in E is
the root of a separable polynomial in F [x].
2
Example √ 1. The √ polynomial x − √ 2 is separable over Q since it factors
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. If b = 0, then α is a root of x − a. If
b 6= 0, then α is the root of the separable polynomial
√ √
x2 − 2ax + a2 − 2b2 = (x − (a + b 2 ))(x − (a − b 2 )).
Fortunately, we have an easy test to determine the separability of any
polynomial. Let
f (x) = a0 + a1 x + · · · + an xn
be any polynomial in F [x]. Define the derivative of f (x) to be
f 0 (x) = a1 + 2a2 x + · · · + nan xn−1 .
Lemma 22.4 Let F be a field and f (x) ∈ F [x]. Then f (x) is separable if
and only if f (x) and f 0 (x) are relatively prime.
Proof. Let f (x) be separable. Then f (x) factors over some extension field
of F as f (x) = (x − α1 )(x − α2 ) · · · (x − αn ), where αi 6= αj for i 6= j. Taking
the derivative of f (x), we see that
f 0 (x) = (x − α2 ) · · · (x − αn )
+ (x − α1 )(x − α3 ) · · · (x − αn )
+ · · · + (x − α1 ) · · · (x − αn−1 ).
Hence, f (x) and f 0 (x) can have no common factors.
To prove the converse, we will show that the contrapositive of the state-
ment is true. Suppose that f (x) = (x − α)k g(x), where k > 1. Differentiat-
ing, we have
f 0 (x) = k(x − α)k−1 g(x) + (x − α)k g 0 (x).
Therefore, f (x) and f 0 (x) have a common factor.
Theorem 22.5 For every prime p and every positive integer n, there exists
a finite field F with pn elements. Furthermore, any field of order pn is
n
isomorphic to the splitting field of xp − x over Zp .
356 CHAPTER 22 FINITE FIELDS
n
Proof. Let f (x) = xp − x and let F be the splitting field of f (x). Then by
n
Lemma 20.4, f (x) has pn distinct zeros in F , since f 0 (x) = pn xp −1 −1 = −1
is relatively prime to f (x). We claim that the roots of f (x) form a subfield
of F . Certainly 0 and 1 are zeros of f (x). If α and β are zeros of f (x),
n n n
then α + β and αβ are also zeros of f (x), since αp + β p = (α + β)p
n n n
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), −α = (p − 1)α is also a zero of f (x). If α 6= 0, then
n n
(α−1 )p = (αp )−1 = α−1 . Since the zeros of f (x) form a 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
n
multiplicative group of nonzero elements of E is pn − 1; hence, αp −1 = 1
n
or αp − α = 0. Since E contains pn elements, E must be a splitting field
of f (x); however, by Corollary 21.20, the splitting field of any polynomial is
unique up to isomorphism.
The unique finite field with pn elements is called the Galois field of
order pn . We will denote this field by GF(pn ).
Theorem 22.6 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
m n m
divides pn −1. Consequently, xp −1 −1 divides xp −1 −1. Therefore, xp −x
n m n
must divide xp − x, and every zero of xp − x is also a zero of xp − x.
m
Thus, GF(pn ) contains, as a subfield, a splitting field of xp − x, which must
be isomorphic to GF(pm ).
Example 2. The lattice of subfields of GF(p24 ) is given in Figure 22.1.
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.
22.1 STRUCTURE OF A FINITE FIELD 357
GF(p24 )
GF(p8 ) GF(p12 )
GF(p4 ) GF(p6 )
GF(p2 ) GF(p3 )
GF(p)
Figure 22.1. Subfields of GF(p24 )
Theorem 22.7 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 ∗ with n = pe11 · · · pekk elements,
where pi ’s are (not necessarily distinct) primes. By the Fundamental Theo-
rem of Finite Abelian Groups (Theorem 13.3),
G∼
= Zpe1 × · · · × Zpek .
1 k
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 22.8 The multiplicative group of all nonzero elements of a finite
field is cyclic.
Corollary 22.9 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 3. The finite field GF(24 ) is isomorphic to the field Z2 /h1+x+x4 i.
Therefore, the elements of GF(24 ) can be taken to be
{a0 + a1 α + a2 α2 + a3 α3 : ai ∈ Z2 and 1 + α + α4 = 0}.
358 CHAPTER 22 FINITE FIELDS
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
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 7. 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 4. Consider the (6, 3)-linear codes generated by the two matrices
1 0 0 1 0 0
0 1 0 1 1 0
0 0 1 1 1 1
G1 =
and G2 = 1 1 1 .
1 0 0
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).
22.2 POLYNOMIAL CODES 359
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 ,
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 5. 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 15, Chapter 20). Let
360 CHAPTER 22 FINITE FIELDS
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 )x3 = x2 + x5 .
We obtain the code corresponding to the generator matrix G1 in Example 4.
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 imme-
diate goal is to establish a link between polynomial codes and ring theory.
Recall that xn − 1 = (x − 1)(xn−1 + · · · + x + 1). The factor ring
Rn = Z2 [x]/hxn − 1i
can be considered to be the ring of polynomials of the form
f (t) = a0 + a1 t + · · · + an−1 tn−1
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]/hxn − 1i. In this manner we can interpret a linear
code as a subset of Z[x]/hxn − 1i.
The additional ring structure on polynomial codes is very powerful in
describing cyclic codes. A cyclic shift of an n-tuple can be described by
22.2 POLYNOMIAL CODES 361
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 22.10 A linear code C in Zn2 is cyclic if and only if it is an ideal
in Rn = Z[x]/hxn − 1i.
Proof. Let C be a linear cyclic code and suppose that f (t) is in C.
Then tf (t) must also be in C. Consequently, tk f (t) is in C for all k ∈
N. Since C is a linear code, any linear combination of the codewords
f (t), tf (t), t2 f (t), . . . , tn−1 f (t) is also a codeword; therefore, for every poly-
nomial p(t), p(t)f (t) is in C. Hence, C is an ideal.
Conversely, let C be an ideal in Z2 [x]/hxn + 1i. Suppose that f (t) =
a0 + a1 t + · · · + an−1 tn−1 is a codeword in C. Then tf (t) is a codeword in
C; that is, (a1 , . . . , an−1 , a0 ) is in C.
Theorem 22.10 tells us that knowing the ideals of Rn is equivalent to
knowing the linear cyclic codes in Zn2 . Fortunately, the ideals in Rn are easy
to describe. The natural ring homomorphism φ : Z2 [x] → Rn defined by
φ[f (x)] = f (t) is a surjective homomorphism. The kernel of φ is the ideal
generated by xn − 1. By Theorem 16.14, every ideal C in Rn is of the form
φ(I), where I is an ideal in Z2 [x] that contains hxn − 1i. By Theorem 17.12,
we know that every ideal I in Z2 [x] is a principal ideal, since Z2 is a field.
Therefore, I = hg(x)i for some unique monic polynomial in Z2 [x]. Since
hxn − 1i is contained in I, it must be the case that g(x) divides xn − 1.
Consequently, every ideal C in Rn is of the form
C = hg(t)i = {f (t)g(t) : f (t) ∈ Rn and g(x) | (xn − 1) in Z2 [x]}.
The unique monic polynomial of the smallest degree that generates C is
called the minimal generator polynomial of C.
Example 6. 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 5, it is easy to calculate a generator matrix
362 CHAPTER 22 FINITE FIELDS
by examining what g(t) does to the polynomials 1, t, t2 , and t3 . A generator
matrix for C is
1 0 0 0
1 1 0 0
0 1 1 0
G= 1 0 1 1.
0 1 0 1
0 0 1 0
0 0 0 1
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 · · · g 1
.. .. .. .
..
. . .
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 22.11 Let C = hg(t)i be a cyclic code in Rn and suppose that
xn − 1 = g(x)h(x). Then G and H are generator and parity-check matrices
for C, respectively. Furthermore, HG = 0.
Example 7. In Example 6,
x7 − 1 = g(x)h(x) = (1 + x + x3 )(1 + x + x2 + x4 ).
22.2 POLYNOMIAL CODES 363
Therefore, a parity-check matrix for this code is
0 0 1 0 1 1 1
H = 0 1 0 1 1 1 0 .
1 0 1 1 1 0 0
To determine the error-detecting and error-correcting capabilities of a
cyclic code, we need to know something about determinants. If α1 , . . . , αn
are elements in a field F , then the n × n matrix
1 1 ··· 1
α1
α2 · · · αn
α2 α 2 · · · α 2
1 2 n
.. .. .. ..
. . . .
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 22.12 Let α1 , . . . , αn be elements in a field F with n ≥ 2. Then
1 1 ··· 1
α1
α2 ··· αn
α2 α 2 ··· αn2
Y
det 1 2 = (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
2 α2 · · · αn−1 x2
2 2
p(x) = det α1 .
.. .. . . .. ..
. . . . .
n−1 n−1 n−1 n−1
α1 α2 · · · αn−1 x
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
364 CHAPTER 22 FINITE FIELDS
α1 , . . . , αn−1 , since the substitution of any one of these elements in the last
column will produce a column identical to the last column in the matrix.
Remember that the determinant of a matrix is zero if it has two identical
columns. Therefore,
p(x) = (x − α1 )(x − α2 ) · · · (x − αn−1 )β,
where
1 1 ··· 1
α1 α2 ··· αn−1
β = (−1)n+n det
α12 α22 ··· 2
αn−1 .
.. .. .. ..
. . . .
α1n−2 α2n−2 · · · n−2
αn−1
By our induction hypothesis,
Y
β = (−1)n+n (αi − αj ).
1≤j<i≤n−1
If we let x = αn , the result now follows immediately.
The following theorem gives us an estimate on the error detection and
correction capabilities for a particular generator polynomial.
Theorem 22.13 Let C = hg(t)i be a cyclic code in Rn and suppose that ω
is a primitive nth root of unity over Z2 . If s consecutive powers of ω are
roots of g(x), then the minimum distance of C is at least s + 1.
Proof. Suppose that
g(ω r ) = g(ω r+1 ) = · · · = g(ω r+s−1 ) = 0.
Let f (x) be some polynomial in C with s or fewer nonzero coefficients. We
can assume that
f (x) = ai0 xi0 + ai1 xi1 + · · · + ais−1 xis−1
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.
22.2 POLYNOMIAL CODES 365
Equivalently, we have the following system of equations:
ai0 (ω r )i0 + ai1 (ω r )i1 + · · · + ais−1 (ω r )is−1 = 0
ai0 (ω r+1 )i0 + ai1 (ω r+1 )i2 + · · · + ais−1 (ω r+1 )is−1 = 0
..
.
ai0 (ω r+s−1 )i0 + ai1 (ω r+s−1 )i1 + · · · + ais−1 (ω r+s−1 )is−1 = 0.
Therefore, (ai0 , ai1 , . . . , ais−1 ) is a solution to the homogeneous system of
linear equations
(ω i0 )r x0 + (ω i1 )r x1 + · · · + (ω is−1 )r xn−1 = 0
(ω i0 )r+1 x0 + (ω i1 )r+1 x1 + · · · + (ω is−1 )r+1 xn−1 = 0
..
.
(ω i0 )r+s−1 x0 + (ω i1 )r+s−1 x1 + · · · + (ω is−1 )r+s−1 xn−1 = 0.
However, this system has a unique solution, since the determinant of the
matrix
(ω i0 )r (ω i1 )r (ω is−1 )r
···
(ω i0 )r+1 (ω i1 )r+1 · · · (ω is−1 )r+1
.. .. . . ..
. . . .
i
(ω )0 r+s−1 i
(ω )1 r+s−1 · · · (ω i s−1 ) r+s−1
can be shown to be nonzero using Lemma 22.12 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 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 small-
est degree that has the largest error detection and error correction capabil-
ities. Let d = 2r + 1 for some r ≥ 0. Suppose that ω is a primitive nth root
366 CHAPTER 22 FINITE FIELDS
of unity over Z2 , and let mi (x) be the minimal polynomial over Z2 of ω i . If
g(x) = lcm[m1 (x), m2 (x), . . . , m2r (x)],
then the cyclic code hg(t)i in Rn is called the BCH code of length n and
distance d. By Theorem 22.13, the minimum distance of C is at least d.
Theorem 22.14 Let C = hg(t)i be a cyclic code in Rn . The following
statements are equivalent.
1. The code C is a BCH code whose minimum distance is at least d.
2. A code polynomial f (t) is in C if and only if f (ω i ) = 0 for 1 ≤ i < d.
3. The matrix
ω2 ··· ω n−1
1 ω
1
ω2 ω4 ··· ω (n−1)(2)
H = 1
ω3 ω6 ··· ω (n−1)(3)
. .. .. .. ..
..
. . . .
1 ω 2r ω 4r · · · ω (n−1)(2r)
is a parity-check matrix for C.
Proof. (1) ⇒ (2). If f (t) is in C, then g(x) | f (x) in Z2 [x]. Hence, for
i = 1, . . . , 2r, f (ω i ) = 0 since g(ω i ) = 0. Conversely, suppose that f (ω i ) = 0
for 1 ≤ i ≤ d. Then f (x) is divisible by each mi (x), since mi (x) is the
minimal polynomial of ω i . Therefore, g(x) | f (x) by the definition of g(x).
Consequently, f (x) is a codeword.
(2) ⇒ (3). Let f (t) = a0 + a1 t + · · · + an−1 vtn−1 be in Rn . The corre-
sponding n-tuple in Zn2 is x = (a0 a1 · · · an−1 )t . By (2),
a0 + a1 ω + · · · + an−1 ω n−1
f (ω)
a0 + a1 ω 2 + · · · + an−1 (ω 2 )n−1 f (ω 2 )
Hx = = .. = 0
..
. .
a0 + a1 ω 2r + · · · + an−1 (ω 2r )n−1 f (ω 2r )
exactly when f (t) is in C. Thus, H is a parity-check matrix for C.
(3) ⇒ (1). By (3), a code polynomial f (t) = a0 + a1 t + · · · + an−1 tn−1 is
in C exactly when f (ω i ) = 0 for i = 1, . . . , 2r. The smallest such polynomial
is g(t) = lcm[m1 (t), . . . , m2r (t)]. Therefore, C = hg(t)i.
Example 8. 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),
EXERCISES 367
where each of the factors is an irreducible polynomial. Let ω be a root of
1 + x + x4 . The Galois field GF(24 ) is
{a0 + a1 ω + a2 ω 2 + a3 ω 3 : ai ∈ Z2 and 1 + ω + ω 4 = 0}.
By Example 3, 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
Exercises
1. Calculate each of the following.
(a) [GF(36 ) : GF(33 )] (c) [GF(625) : GF(25)]
(b) [GF(128) : GF(16)] (d) [GF(p12 ) : GF(p2 )]
2. Calculate [GF(pm ) : GF(pn )], where n | m.
3. What is the lattice of subfields for GF(p30 )?
4. Let α be a zero of x3 + x2 + 1 over Z2 . Construct a finite field of order 8.
Show that x3 + x2 + 1 splits in Z2 (α).
5. Construct a finite field of order 27.
6. Prove or disprove: Q∗ is cyclic.
7. Factor each of the following polynomials in Z2 [x].
368 CHAPTER 22 FINITE FIELDS
(a) x5 − 1 (c) x9 − 1
6 5 4 3 2
(b) x + x + x + x + x + x + 1 (d) x4 + x3 + x2 + x + 1
8. Prove or disprove: Z2 [x]/hx3 + x + 1i ∼
= Z2 [x]/hx3 + x2 + 1i.
9. Determine the number of cyclic codes of length n for n = 6, 7, 8, 10.
10. Prove that the ideal ht + 1i in Rn is the code in Zn2 consisting of all words of
even parity.
11. Construct all BCH codes of
(a) length 7. (b) length 15.
12. Prove or disprove: There exists a finite field that is algebraically closed.
13. Let p be prime. Prove that the field of rational functions Zp (x) is an infinite
field of characteristic p.
n
14. Let D be an integral domain of characteristic p. Prove that (a − b)p =
n n
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 separable over F , show that K is also
separable over 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).
EXERCISES 369
25. If g(t) is the minimal generator polynomial for a cyclic code C in Rn , prove
that the constant term of g(x) is 1.
26. Often it is conceivable that a burst of errors might occur during transmission,
as in the case of a power surge. Such a momentary burst of interference
might alter several consecutive bits in a codeword. Cyclic codes permit the
detection of such error bursts. Let C be an (n, k)-cyclic code. Prove that
any error burst up to n − k digits can be detected.
27. Prove that the rings Rn and Zn2 are isomorphic as vector spaces.
28. Let C be a code in Rn that is generated by g(t). If hf (t)i is another code in
Rn , show that hg(t)i ⊂ hf (t)i if and only if f (x) divides g(x) in Z2 [x].
29. Let C = hg(t)i be a cyclic code in Rn and suppose that xn − 1 = g(x)h(x),
where g(x) = g0 + g1 x + · · · + gn−k xn−k and h(x) = h0 + h1 x + · · · + hk xk .
Define G to be the n × k matrix
g0 0 ··· 0
g1 g0 ··· 0
.. .. ..
. .
.
. . .
G = gn−k gn−k−1 · · ·
g0
0
gn−k ··· g1
. .. .. ..
.. . . .
0 0 ··· gn−k
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.
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 .
370 CHAPTER 22 FINITE FIELDS
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 7. By Theorem 8.3, this code
is capable of correcting two errors. Suppose that these errors occur in bits
a1 and a2 . The error-locator polynomial is s(x) = (x + ω a1 )(x + ω a2 ). Show
that
2 2 s3
s(x) = x + s1 x + s1 + .
s1
4. Let w(t) = 1 + t2 + t4 + t5 + t7 + t12 + t13 . Determine what the originally
transmitted code polynomial was.
References and Suggested Readings
[1] Childs, L. A Concrete Introduction to Higher Algebra. 2nd ed. Springer-
Verlag, New York, 1995. .
[2] Gåding, L. and Tambour, T. Algebra for Computer Science. Springer-Verlag,
New York, 1988.
[3] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York,
1998. An excellent presentation of finite fields and their applications.
[4] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985.
[5] Roman, S. Coding and Information Theory. Springer-Verlag, New York,
1992.
[6] van Lint, J. H. Introduction to Coding Theory. Springer, New York, 1999.
23
Galois Theory
A classic problem of algebra has been to find the solutions of a polynomial
equation. The solution to the quadratic equation was known in antiquity.
Italian mathematicians found general solutions to the general cubic and
quartic equations in the sixteenth century; however, attempts to solve the
general fifth-degree, or quintic, polynomial were repulsed for the next three
hundred years. Certainly, equations such as x5 − 1 = 0 or x6 − x3 − 6 = 0
could be solved, but no solution like the quadratic formula was found for
the general quintic,
ax5 + bx4 + cx3 + dx2 + ex + f = 0.
Finally, at the beginning of the nineteenth century, Ruffini and Abel both
found quintics that could not be solved with any formula. It was Galois,
however, who provided the full explanation by showing which polynomials
could and could not be solved by formulas. He discovered the connection
between groups and field extensions. Galois theory demonstrates the strong
interdependence of group and field theory, and has had far-reaching impli-
cations 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.
371
372 CHAPTER 23 GALOIS THEORY
Proof. If σ and τ are automorphisms of E, then so are στ and σ −1 . The
identity is certainly an automorphism; hence, the set of all automorphisms
of a field F is indeed a group.
Proposition 23.2 Let E be a field extension of F . Then the set of all
automorphisms of E that fix F elementwise is a group; that is, the set of all
automorphisms σ : E → E such that σ(α) = α for all α ∈ F is a group.
Proof. We need only show that the set of automorphisms of E that fix F
elementwise is a subgroup of the group of all automorphisms of E. Let σ
and τ be two automorphisms of E such that σ(α) = α and τ (α) = α for all
α ∈ F . Then στ (α) = σ(α) = α and σ −1 (α) = α. Since the identity fixes
every element of E, the set of automorphisms of E that leave elements of F
fixed is a subgroup of the entire group of automorphisms of E.
Let E be a field extension of F . We will denote the full group of auto-
morphisms 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 1. 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 √ 2. Consider the fields Q ⊂ Q( 5 ) ⊂ Q( 3, 5 ). Then for
a, b ∈ Q( 5 ), √ √
σ(a + b 3 ) = a − b 3
√ √ √
is an automorphism of Q( 3, 5 ) leaving Q( 5 ) fixed. Similarly,
√ √
τ (a + b 5 ) = a − b 5
√ √ √
is an automorphism of√Q( 3, √5 ) leaving Q( 3 ) fixed. The automorphism
µ = στ moves both √ √ 5. It will soon be clear that {id, σ, τ, µ} is
3 and
the Galois group of Q( 3, 5 ) over Q. The following table shows that this
group is isomorphic to Z2 × Z2 .
23.1 FIELD AUTOMORPHISMS 373
id σ τ µ
id id σ τ µ
σ σ id µ τ
τ τ µ id σ
µ µ τ σ id
√ √
We may also√regard
√ √the field Q( 3, 5 ) as a vector space√ 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 23.3 Let E be a field extension of F and f (x) be a polynomial
in F [x]. Then any automorphism in G(E/F ) defines a permutation of the
roots of f (x) that lie in E.
Proof. Let
f (x) = a0 + a1 x + a2 x2 + · · · + an xn
and suppose that α ∈ E is a zero of f (x). Then for σ ∈ G(E/F ),
0 = σ(0)
= σ(f (α))
= σ(a0 + a1 α + a2 α2 + · · · + an αn )
= a0 + a1 σ(α) + a2 [σ(α)]2 + · · · + an [σ(α)]n ;
therefore, σ(α) is also a zero of f (x).
Let E be an algebraic extension of a field F . Two elements α, β ∈ E are
conjugate over√F if they have the√same minimal √ polynomial. For example,
in the field Q( 2 ) the elements 2 and − 2 are conjugate over Q since
they are both roots of the irreducible polynomial x2 − 2.
A converse of the last proposition exists. The proof follows directly from
Lemma 21.18.
Proposition 23.4 If α and β are conjugate over F , there exists an isomor-
phism σ : F (α) → F (β) such that σ is the identity when restricted to F .
Theorem 23.5 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 ].
374 CHAPTER 23 GALOIS THEORY
Proof. The proof is similar to the proof of Theorem 21.19. 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. 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. If β is any other
root of p(x), then F ⊂ F (β) ⊂ E. By Lemma 21.18, 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 F (β), there are exactly r such isomorphisms.
We can factor p(x) in F (α) as p(x) = (x − α)p1 (x). The degrees of p1 (x)
and q1 (x) are both less than r. Since we know that E is the splitting field
of p1 (x) over F (α), we can apply the induction hypothesis to conclude that
|G(E/F (α))| = [E : F (α)].
Consequently, there are
[E : F ] = [E : F (α)][F (α) : F ]
possible automorphisms of E that fix F , or |G(E/F )| = [E : F ].
Corollary 23.6 Let F be a finite field with a finite extension E such that
[E : F ] = k. T hen G(E/F ) is cyclic.
Proof. Let p be the characteristic of E and F and assume that the orders
of E and F are pm and pn , respectively. Then nk = m. We can also assume
m
that E is the splitting field of xp − x over a subfield of order p. Therefore,
m
E must also be the splitting field of xp −x over F . Applying Theorem 23.5,
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. Also, it is easy to show that σ(αβ) = σ(α)σ(β). Since σ
is a nonzero homomorphism of fields, it must be injective. It must also be
onto, since E is a finite field. We know that σ must be in G(E/F ), since
n
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
k
of σ is k. By Theorem 23.5, we know that σ k (α) = αp = α is the identity
23.1 FIELD AUTOMORPHISMS 375
of G(E/F ). However, σ r cannot be the identity for 1 ≤ r < k; otherwise,
rk
xp − x would have pm roots, which is impossible.
√ √
Example 3. We can now confirm that the Galois group of Q( 3, 5 )
over Q in Example 2 is indeed isomorphic√ √ to Z2 × 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 4. 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 19 in Chapter 17.
Furthermore, since (x − 1)f (x) = x5 − 1, we can use DeMoivre’s Theorem
to determine that the roots of f (x) are ω i , where i = 1, . . . , 4 and
ω = cos(2π/5) + i sin(2π/5).
Hence, the splitting field of f (x) must be Q(ω). We can define 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
Y
f (x) = (x − α1 )n1 (x − α2 )n2 · · · (x − αr )nr = (x − αi )ni .
i=1
376 CHAPTER 23 GALOIS THEORY
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 0 (x)) = 1 (Lemma 22.4).
Proposition 23.7 Let f (x) be an irreducible polynomial over F . If the
characteristic of F is 0, then f (x) is separable. If the characteristic of F is
p and f (x) 6= g(xp ) for some g(x) in F [x], then f (x) is also separable.
Proof. First assume that charF = 0. Since deg f 0 (x) < deg f (x) and
f (x) is irreducible, the only way gcd(f (x), f 0 (x)) 6= 1 is if f 0 (x) is the zero
polynomial; however, this is impossible in a field of characteristic zero. If
charF = p, then f 0 (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 √ √ √
3 6
Q( 5, 5 i) = Q( 5 i).
Corollary 22.9 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 23.8 (Primitive Element Theorem) Let E be a finite sepa-
rable 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 mathemat-
ical 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
23.2 THE FUNDAMENTAL THEOREM 377
that f (x) has zeros α = α1 , . . . , αn in K and g(x) has zeros β = β1 , . . . , βm
in K. All of these zeros have multiplicity 1, since E is separable over F .
Since F is infinite, we can find an a in F such that
αi − α
a 6=
β − βj
for all i and j with j 6= 1. Therefore, a(β − βj ) 6= αi − α. Let γ = α + aβ.
Then
γ = α + aβ 6= αi + aβj ;
hence, γ − aβj 6= αi for all i, j with j 6= 1. Define h(x) ∈ F (γ)[x] by
h(x) = f (γ − ax). Then h(β) = f (α) = 0. However, h(βj ) 6= 0 for j 6= 1.
Hence, h(x) and g(x) have a single common factor in F (γ)[x]; that is, the
irreducible polynomial of β over F (γ) must be linear, since β is the only
zero common to both g(x) and h(x). So β ∈ F (γ) and α = γ − aβ is in
F (γ). Hence, F (α, β) = F (γ).
23.2 The Fundamental Theorem
The goal of this section is to prove the Fundamental Theorem of Galois
Theory. This theorem explains the connection between the subgroups of
G(E/F ) and the intermediate fields between E and F .
Proposition 23.9 Let {σi : i ∈ I} be a collection of automorphisms of a
field F . Then
F{σi } = {a ∈ F : σi (a) = a for all σi }
is a subfield of F .
Proof. Let σi (a) = a and σi (b) = b. Then
σi (a ± b) = σi (a) ± σi (b) = a ± b
and
σi (ab) = σi (a)σi (b) = ab.
If a 6= 0, then σi(a−1 )
= [σi (a)]−1 = a−1 . Finally, σi (0) = 0 and σi (1) = 1
since σi is an automorphism.
Corollary 23.10 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 .
378 CHAPTER 23 GALOIS THEORY
The subfield F{σi } of F is called the fixed field of {σi }. The field fixed
for a subgroup G of Aut(F ) will be denoted by FG .
√ √ √ √
Example√ 5. 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 23.11 Let E be a splitting field over F of a separable polyno-
mial. 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 23.5,
|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 23.12 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 homo-
geneous 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 equations than unknowns. From linear algebra we know that this
system has a nontrivial solution, say xi = ai for i = 1, 2, . . . , n + 1. Since σ1
is the identity, the first equation translates to
a1 α1 + a2 α2 + · · · + an+1 αn+1 = 0.
23.2 THE FUNDAMENTAL THEOREM 379
The problem is that some of the ai ’s may be in E but not in F . We must
show that this is impossible.
Suppose that at least one of the ai ’s is in E but not in F . By rearranging
the αi ’s we may assume that a1 is nonzero. Since any nonzero multiple of a
solution is also a solution, we can also assume that a1 = 1. Of all possible
solutions fitting this description, we choose the one with the smallest number
of nonzero terms. Again, by rearranging α2 , . . . , αn+1 if necessary, we can
assume that a2 is in E but not in F . Since F is the subfield of E that is fixed
elementwise by G, there exists a σi in G such that σi (a2 ) 6= a2 . Applying
σi to each equation in the system, we end up with the same homogeneous
system, since G is a group. Therefore, x1 = σi (a1 ) = 1, x2 = σi (a2 ), . . .,
xn+1 = σi (an+1 ) is also a solution of the original system. We know that
a linear combination of two solutions of a homogeneous system is also a
solution; consequently,
x1 = 1 − 1 = 0
x2 = a2 − σi (a2 )
..
.
xn+1 = an+1 − σi (an+1 )
must be another solution of the system. This is a nontrivial solution because
σi (a2 ) 6= a2 , and has fewer nonzero entries than our original solution. This
is a contradiction, since the number of nonzero solutions to our original
solution was assumed to be minimal. We can therefore conclude that a1 =
· · · = an+1 = 0.
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 23.13 Let E be a field extension of F . Then the following state-
ments 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 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).
380 CHAPTER 23 GALOIS THEORY
Let f (x) be the minimal polynomial of α over F . The field E must contain
all of the roots of f (x) since it is a normal extension F ; hence, E is a splitting
field for f (x).
(2) ⇒ (3). Let E be the splitting field over F of a separable polynomial.
By Proposition 23.11, 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 23.3, automorphisms in G permute
the roots of f (x) lying in E. Hence, if we let G Q act on α, we can obtain
distinct roots α1 = α, α2 , . . . , αn in E. Let g(x) = ni=1 (x − αi ). Then g(x)
is separable over F and g(α) = 0. Any automorphism σ in G permutes the
factors of g(x) since it permutes these roots; hence, when σ acts on g(x), it
must fix the coefficients of g(x). Therefore, the coefficients of g(x) must be
in F . Since deg g(x) ≤ deg f (x) and f (x) is the minimal polynomial of α,
f (x) = g(x).
Corollary 23.14 Let K be a field extension of F such that F = KG for
some finite group of automorphisms G of K. Then G = G(K/F ).
Proof. Since F = KG , G is a subgroup of G(K/F ). Hence,
[K : F ] ≤ |G| ≤ |G(K/F )| = [K : F ].
It follows that G = G(K/F ), since they must have the same order.
Before we determine the exact correspondence between field extensions
and automorphisms of fields, let us return to a familiar example.
√ √
Example 6. In Example 2 we examined the automorphisms of Q( 3, 5 )
fixing Q. Figure 23.1 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 relationship is between the two lattices.
We are now ready to state and prove the Fundamental Theorem of Galois
Theory.
Theorem 23.15 (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.
23.2 THE FUNDAMENTAL THEOREM 381
√ √
{id, σ, τ, µ} Q( 3, 5 )
√ √ √
{id, σ} {id, τ } {id, µ} Q( 3 ) Q( 5 ) Q( 15 )
{id} Q
√ √
Figure 23.1. G(Q( 3, 5 )/Q)
1. The map K 7→ G(E/K) is a bijection of subfields K of E containing
F with the subgroups of G(E/F ).
2. If F ⊂ K ⊂ E, then
[E : K] = |G(E/K)| and [K : F ] = [G(E/F ) : G(E/K)].
3. F ⊂ K ⊂ L ⊂ E if and only if {id} ⊂ G(E/L) ⊂ G(E/K) ⊂ G(E/F ).
4. K is a normal extension of F if and only if G(E/K) is a normal
subgroup of G(E/F ). In this case
G(K/F ) ∼
= G(E/F )/G(E/K).
Proof. (1) Suppose that G(E/K) = G(E/L) = G. Both K and L are
fixed fields of G; hence, K = L and the map defined by K 7→ G(E/K) is
one-to-one. To show that the map is onto, let G be a subgroup of G(E/F )
and K be the field fixed by G. Then F ⊂ K ⊂ E; consequently, E is a
normal extension of K. Thus, G(E/K) = G and the map K 7→ G(E/K) is
a bijection.
(2) By Theorem 23.5, |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)].
(3) Statement (3) is illustrated in Figure 23.2. 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
382 CHAPTER 23 GALOIS THEORY
E {id}
L G(E/L)
K G(E/K)
F G(E/F )
Figure 23.2. Subgroups of G(E/F ) and subfields of E
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 restrict-
ing σ to K. Since K is a normal extension, the argument in the preced-
ing 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 homomor-
phism since
φ(στ ) = (στ )K = σK τK = φ(σ)φ(τ ).
23.2 THE FUNDAMENTAL THEOREM 383
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 Isomor-
phism Theorem, we have
G(K/F ) ∼
= G(E/F )/G(E/K).
Example 7. In this example we will illustrate the Fundamental Theorem of
Galois Theory by determining the lattice of subgroups of the Galois group of
f (x) = x4 − 2. We will compare this lattice to the lattice of field extensions
of Q that √ are contained in the splitting field of x4 − 2. The splitting √ field
√ of
f (x) is Q( 4 2, i). To see this, notice that f (x) factors as (x 2 + 2 )(x2 − 2 );
√ √ √
hence, the roots of f (x) are ± 4 2 and ± 4 2 i. We √ first adjoin the root 4
2
to Q and then adjoin the root i of x 2 + 1 to Q( 4 2 ). The splitting field of
√
4
√
4
f (x) is then Q(
√ 2 )(i) = Q( 2, i). √
4
Since [Q( √2 ) : Q] = 4 and i is not
√ √ in Q( 4 2 ), it must be the case that
[Q( 4 2, i) : Q( 4 2 )] = 2. Hence, [Q( 4 2, i) : Q] = 8. The set
√ √ √ √ √ √
{1, 2, ( 2 )2 , ( 2 )3 , i, i 2, i( 2 )2 , i( 2 )3 }
4 4 4 4 4 4
√4
is a basis
√ of Q( 2, i) over Q. The lattice of field extensions of Q contained
4
in Q( 2, i) is illustrated in Figure 23.3(a).
The Galois group√G of f (x) √ must be of order 8. Let σ be the automor-
4 4
phism defined by σ( 2 ) = 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 23.3(b).
Historical Note
Solutions for the cubic and quartic equations were discovered in the 1500s. At-
tempts to find solutions for the quintic equations puzzled some of history’s best
mathematicians. In 1798, P. Ruffini submitted a paper that claimed no such so-
lution could be found; however, the paper was not well received. In 1826, Niels
384 CHAPTER 23 GALOIS THEORY
√
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 23.3. Galois group of x4 − 2
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 for-
mal 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
23.3 APPLICATIONS 385
he could read the paper. Another paper was presented, but was not published
until 1846.
Galois’s democratic sympathies led him into the Revolution of 1830. He was
expelled from school and sent to prison for his part in the turmoil. After his release
in 1832, he was drawn into a duel 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 21.
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 in a finite number of operations on the coefficients
of f (x). The allowable operations are addition, subtraction, multiplication,
division, and the extraction of nth roots. Certainly the solution to the
quadratic equation, ax2 + bx + c = 0, illustrates this process:
√
−b ± b2 − 4ac
x= .
2a
The only one of these operations that might demand a larger field is the
taking of nth roots. We are led to the following definition.
An extension field E of a field F is an extension by radicals if there
are elements α1 , . . . , αr ∈ K and positive integers n1 , . . . , nr such that
E = F (α1 , . . . , αr ),
where α1n1 ∈ F and
αini ∈ F (α1 , . . . , αi−1 )
for i = 2, . . . , r. A polynomial f (x) is solvable by radicals over F if the
splitting field K of f (x) over F is contained in an extension of F by radicals.
Our goal is to arrive at criteria that will tell us whether or not a polynomial
f (x) is solvable by radicals by examining the Galois group f (x).
The easiest polynomial to solve by radicals is one of the form xn − a. As
we discussed in Chapter 4, the roots of xn − 1 are called the nth roots of
unity. These roots are a finite subgroup of the splitting field of xn − 1. By
386 CHAPTER 23 GALOIS THEORY
Theorem 22.7, the nth roots of unity form a cyclic group. Any generator of
this group is called a primitive nth root of unity.
Example 8. 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).
Recall that a subnormal series of a group G is a finite sequence of sub-
groups
G = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e},
where Hi is normal in Hi+1 . A subnormal series is a composition series
if all the factor groups are simple; that is, if none of the factor groups
of the series contains a normal subgroup. A group G is solvable if it has
a composition series {Hi } such that all of the factor groups Hi+1 /Hi are
abelian. For example, if we examine the series {id} ⊂ A3 ⊂ S3 , we see that
A3 is solvable. On the other hand, S5 is not solvable, by Theorem 10.6.
Lemma 23.16 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. First suppose that F contains all of its nth roots of unity. 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.
If ζ is one of these roots, then distinct roots of xn − 1 are ζ, ωζ, . . . , ω n−1 ζ,
and E = F (ζ). Since G(E/F ) permutes the roots xn − 1, 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 ij 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 ) is solvable.
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 . Any automorphism σ in
G(F (ω)/F ) is determined by σ(ω). It must be the case that σ(ω) = ω i for
23.3 APPLICATIONS 387
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. Since G(E/F (ω)) and
G(E/F )/G(E/F (ω)) ∼
= G(F (ω)/F )
are both abelian, G(E/F ) is solvable.
Lemma 23.17 Let F be a field of characteristic zero and let E be a radical
extension of F . Then there exists a normal radical extension K of F that
contains E.
Proof. Since E is a radical extension of F , there exist elements α1 , . . . , αr ∈
K and positive integers n1 , . . . , nr such that
E = F (α1 , . . . , αr ),
where α1n1 ∈ F and
αini ∈ F (α1 , . . . , αi−1 )
for i = 2, . . . , r. Let f (x) = f1 (x) · · · fr (x), where fi is the minimal poly-
nomial of αi over F , and let K be the splitting field of K over F . Every
root of f (x) in K is of the form σ(αi ), where σ ∈ G(K/F ). Therefore, for
any σ ∈ G(K/F ), we have [σ(α1 )]n1 ∈ F and [σ(αi )]ni ∈ F (α1 , . . . , αi−1 ) for
i = 2, . . . , r. Hence, if G(K/F ) = {σ1 = id, σ2 , . . . , σk }, then K = F (σ1 (αj ))
is a radical extension of F .
We will now prove the main theorem about solvability by radicals.
Theorem 23.18 Let f (x) be in F [x], where charF = 0. If f (x) is solvable
by radicals, then the Galois group of f (x) over F is solvable.
Proof. Let K be a splitting field of f (x) over F . Since f (x) is solvable,
there exists an extension E of radicals F = F0 ⊂ F1 ⊂ · · · Fn = E. Since Fi
is normal over Fi−1 , we know by Lemma 23.17 that E is a normal extension
of each Fi . By the Fundamental Theorem of Galois Theory, G(E/Fi ) is a
388 CHAPTER 23 GALOIS THEORY
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 23.16, G(Fi /Fi−1 ) is solvable; hence, G(E/F ) is also solvable.
The converse of Theorem 23.18 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 23.19 Any subgroup of Sn that contains a transposition and a cycle
of length n must be all of Sn .
Proof. Let G be a subgroup of Sn that contains a transposition σ and
a cycle τ of length n. We may assume that σ = (12) and τ = (12 . . . n).
Since (12)(1 . . . n) = (2 . . . n) and (2 . . . n)k (1, 2)(2 . . . n)−k = (1k), we can
obtain all the transpositions of the form (1, n + 1 − k). However, these
transpositions generate all transpositions in Sn , since (1j)(1i)(1j) = (ij).
The transpositions generate Sn .
Example 9. We will show that f (x) = x5 − 6x3 − 27x − 3 ∈ Q[x] is
not solvable. We claim that the Galois group of f (x) over Q is S5 . By
Eisenstein’s Criterion, f (x) is irreducible and, therefore, must be separable.
The derivative of f (x) is f 0 (x) = 5x4 − 18x2 − 27; hence, setting f 0 (x) = 0
and solving, we find that the only real roots of f 0 (x) are
s √
6 6+9
x=± .
5
Therefore, f (x) can have at most one maximum and one minimum. It is
easy to show that f (x) changes sign between −3 and −2, between −2 and 0,
23.3 APPLICATIONS 389
y
f (x) = x5 − 6x3 − 27x − 3
40
-4 -2 2 4
x
-40
Figure 23.4. The graph of f (x) = x5 − 6x3 − 27x − 3
and once again between 0 and 4 (Figure 23.4). 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) ⊂ S5 . By Lemma 23.19, S5 is generated by
a transposition and an element of order 5; therefore, G(K/F ) must be all
of S5 . By Theorem 10.6, S5 is not solvable. Consequently, f (x) cannot be
solved by radicals.
The Fundamental Theorem of Algebra
It seems fitting that the last theorem that we will state and prove is the
Fundamental Theorem of Algebra. This theorem was first proven by Gauss
390 CHAPTER 23 GALOIS THEORY
in his doctoral thesis. Prior to Gauss’s proof, mathematicians suspected
that there might exist polynomials over the real and complex numbers hav-
ing no solutions. The Fundamental Theorem of Algebra states that every
polynomial over the complex numbers factors into distinct linear factors.
Theorem 23.20 (Fundamental Theorem of Algebra) The field of com-
plex numbers is algebraically closed; that is, every polynomial in C[x] has a
root in C.
For our proof we shall assume two facts from calculus. We need the
results that every polynomial of odd degree over R has a real root and that
every positive real number has a square root.
Proof. Suppose that E is a proper finite field extension of the complex
numbers. Since any finite extension of a field of characteristic zero is a
simple extension, there exists an α ∈ E such that E = C(α) with α the root
of an irreducible polynomial f (x) in C[x]. The splitting field L of f (x) is a
finite normal separable extension of C that contains E. We must show that
it is impossible for L to be a proper extension of C.
Suppose that L is a proper extension of C. Since L is the splitting field
of f (x)(x2 + 1) over R, L is a finite normal separable extension of R. Let
K be the fixed field of a Sylow 2-subgroup G of G(L/R). Then L ⊃ K ⊃ R
and |G(L/K)| = [L : K]. Since [L : R] = [L : K][K : R], we know that
[K : R] must be odd. Consequently, K = R(β) with β having a minimal
polynomial f (x) of odd degree. Therefore, K = R.
We now know that G(L/R) must be a 2-group. It follows that G(L/C)
is a 2-group. We have assumed that L 6= C; therefore, |G(L/C)| ≥ 2. By the
first Sylow Theorem and the Fundamental Theorem of Galois Theory, there
exists a subgroup G of G(L/C) of index 2 and a field E fixed elementwise
by G. Then [E : C] = 2 and there exists an element γ ∈ E with √ minimal
polynomial x2 +bx+c in C[x]. This polynomial has roots (−b± b2 − 4c )/2
that are in C, since b2 − 4c is in C. This is impossible; hence, L = C.
Although our proof was strictly algebraic, we were forced to rely on
results from 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.
EXERCISES 391
Exercises
1. Compute each of the following Galois groups. Which of these field extensions
are normal field extensions? If the extension is not normal, find a normal
extension of Q in which the extension field is contained.
√ √ √
(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) x3 − 5
(d) x4 − x2 − 6 (h) x8 + 1
(e) x5 + 1 (i) x4 − 3x2 − 10
5. Find a primitive element in the splitting field of each of the following poly-
nomials 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 iso-
morphic 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!.
392 CHAPTER 23 GALOIS THEORY
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.
16. Let K be the splitting field of x3 + x2 + 1 ∈ Z2 [x]. Prove or disprove that K
is an extension by radicals.
17. Let F be a field such that √ char F 6= 2. Prove that the splitting field of
f (x) = ax2 + bx + c is F ( α ), where α = b2 − 4ac.
18. Prove or disprove: Two different subgroups of a Galois group will have dif-
ferent fixed fields.
19. Let K be the splitting field of a polynomial over F . If E is a field extension
of F contained in K and [E : F ] = 2, then E is the splitting field of some
polynomial in F [x].
20. We know that the cyclotomic polynomial
xp − 1
Φp (x) = = xp−1 + xp−2 + · · · + x + 1
x−1
is irreducible over Q for every prime p. Let ω be a zero of Φp (x), and consider
the field Q(ω).
(a) Show that ω, ω 2 , . . . , ω p−1 are distinct zeros of Φp (x), and conclude that
they are all the zeros of Φp (x).
(b) Show that G(Q(ω)/Q) is abelian of order p − 1.
(c) Show that the fixed field of G(Q(ω)/Q) is Q.
21. Let F be a finite field or a field of characteristic zero. Let E be a finite normal
extension of F with Galois group G(E/F ). Prove that F ⊂ K ⊂ L ⊂ E if
and only if {id} ⊂ G(E/L) ⊂ G(E/K) ⊂ G(E/F ).
EXERCISES 393
22. Let F be a field of characteristic zero and let f (x) ∈ F [x] be a separable
polynomial of degree n. If E isQthe splitting field of f (x), let α1 , . . . , αn be the
roots of f (x) in E. Let ∆ = i6=j (αi − αj ). We define the discriminant
of f (x) to be ∆2 .
(a) If f (x) = ax2 + bx + c, show that ∆2 = b2 − 4ac.
(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.
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.
[3] Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper
Saddle River, NJ, 2003.
[4] Gaal, L. Classical Galois Theory with Examples. American Mathematical
Society, Providence, 1979.
[5] Garling, D. J. H. A Course in Galois Theory. Cambridge University Press,
Cambridge, 1986.
[6] Kaplansky, I. Fields and Rings. 2nd ed. University of Chicago Press, Chicago,
1972.
[7] Rothman, T. “The Short Life of Évariste Galois,” Scientific American, April
1982, 136–49.
394 CHAPTER 23 GALOIS THEORY
Hints and Solutions
Chapter 1. Preliminaries
1. (a) {2}. (b) {5}.
2. (a) {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c, 1), (c, 2), (c, 3)}.
(d) ∅.
6. If x ∈ A ∪ (B ∩ C), then either x ∈ A or x ∈ B ∩ C ⇒ x ∈ A ∪ B and A ∪ C ⇒
x ∈ (A ∪ B) ∩ (A ∪ C) ⇒ A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C).
Conversely, x ∈ (A ∪ B) ∩ (A ∪ C) ⇒ x ∈ A ∪ B and A ∪ C ⇒ x ∈
A or x is in both B and C⇒ x ∈ A∪(B∩C) ⇒ (A∪B)∩(A∪C) ⊂ A∪(B∩C).
Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
10. (A ∩ B) ∪ (A \ B) ∪ (B \ A) = (A ∩ B) ∪ (A ∩ B 0 ) ∪ (B ∩ A0 ) = [A ∩ (B ∪
B 0 )] ∪ (B ∩ A0 ) = A ∪ (B ∩ A0 ) = (A ∪ B) ∩ (A ∪ A0 ) = A ∪ B.
14. A \ (B ∪ C) = A ∩ (B ∪ C)0 = (A ∩ A) ∩ (B 0 ∩ C 0 ) = (A ∩ B 0 ) ∩ (A ∩ C 0 ) =
(A \ B) ∩ (A \ C).
17. (a) Not a map. f (2/3) is undefined.
(c) Not a map. f (1/2) = 3/4 and f (2/4) = 3/8.
18. (a) One-to-one but not onto. f (R) = {x ∈ R : x > 0}.
(c) Neither one-to-one nor onto.
20. (a) f (n) = n + 1.
22. (a) Let x, y ∈ A. Then g(f (x)) = (g ◦ f )(x) = (g ◦ f )(y) = g(f (y)) ⇒ f (x) =
f (y) ⇒ 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.
23. f −1 (x) = (x + 1)/(x − 1).
24. (a) Let y ∈ f (A1 ∪ A2 ) ⇒ there exists an x ∈ A1 ∪ A2 such that f (x) = y ⇒
y ∈ f (A1 ) or f (A2 ) ⇒ y ∈ f (A1 ) ∪ f (A2 ) ⇒ f (A1 ∪ A2 ) ⊂ f (A1 ) ∪ f (A2 ).
395
396 HINTS AND SOLUTIONS
Conversely, let y ∈ f (A1 ) ∪ f (A2 ) ⇒ y ∈ f (A1 ) or f (A2 ) ⇒ there exists
an x ∈ A1 or there exists an x ∈ A2 such that f (x) = y ⇒ there exists an
x ∈ A1 ∪ A2 such that f (x) = y ⇒ f (A1 ) ∪ f (A2 ) ⊂ f (A1 ∪ A2 ). Hence,
f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ).
25. (a) Not an equivalence relation. Fails to be symmetric.
(c) Not an equivalence relation. Fails to be transitive.
√
28. Let X = N ∪ { 2 } and define x ∼ y if x + y ∈ N.
Chapter 2. The Integers
1. S(1) : [1(1 + 1)(2(1) + 1)]/6 = 1 = 12 is true. Assume 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, so
S(k + 1) is true. Thus S(n) is true for all positive integers n.
3. 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.
8. Look at the proof in Example 3.
11. 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. Thus S(n) is true
for all positive integers n.
15. (a) (14)14 + (−5)39 = 1.
(c) (3709)1739 + (−650)9923 = 1.
(e) (881)23771 + (−1050)19945 = 1.
17. (b) Use mathematical induction. (c) Show that f1 = 1, f2 = 1, and fn+2 =
fn+1 + fn . (d) Use part (c). (e) Use part (b) and Problem 16.
19. Use the Fundamental Theorem of Arithmetic.
23. Let S = {s ∈ N : a | s, b | s}. S 6= ∅, since |ab| ∈ S. By the Principle of
Well-Ordering, S contains a least element m. To show uniqueness, suppose
that a | n and b | n for some n ∈ N. By the division algorithm, there exist
unique integers q and r such that n = mq + r, where 0 ≤ r < m. a | m, b | m,
a | n, b | n ⇒ a | r, b | r ⇒ r = 0 by the minimality of m. Therefore, m | n.
27. Since gcd(a, b) = 1, there exist integers r and s such that ar + bs = 1 ⇒
acr + bcs = c. Since a | a and a | bc, a | c.
29. Let p = p1 p2 · · · pk + 1, where p1 = 2, p2 = 3, . . . , pk are the first k primes.
Show that p is prime.
HINTS AND SOLUTIONS 397
Chapter 3. Groups
1. (a) {. . . , −4, 3, 10, . . .}. (c) {. . . , −8, 18, 44, . . .}. (e) {. . . , −1, 5, 11, . . .}.
2. (a) Not a group. (c) A group.
6. · 1 5 7 11
1 1 5 7 11
5 5 1 11 7
7 7 11 1 5
11 11 7 5 1
8. Pick two matrices. Almost any pair will work.
15. There is a group of order 6 that is nonabelian.
16. Look at the symmetry group of an equilateral triangle or a square.
17. There are actually five different groups of order 8.
18. 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.
24. (aba−1 )n = (aba−1 )(aba−1 ) · · · (aba−1 ) = ab(aa−1 )b(aa−1 )b · · · (aa−1 )ba−1 =
abn a−1 .
30. abab = (ab)2 = e = a2 b2 = aabb ⇒ ba = ab.
34. H1 = {id}, H2 = {id, ρ1 , ρ2 }, H3 = {id, µ1 }, H4 = {id, µ2 }, H5 = {id, µ3 },
S3 .
√ √ √ √
40. id = 1√= 1 + 0 2, (a + b 2 )(c√+ d 2 ) = (ac + 2bd) + (ad + bc) 2, and
(a + b 2 )−1 = a/(a2 − 2b2 ) − b 2/(a2 − 2b2 ).
45. Not a subgroup. Look at S3 .
48. a4 b = ba ⇒ b = a6 b = a2 ba ⇒ ab = a3 ba = ba.
Chapter 4. Cyclic Groups
1. (a) False. (c) False. (e) True.
2. (a) 12. (c) Infinite. (e) 10.
3. (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}.
398 HINTS AND SOLUTIONS
4. (a) 1 0 −1 0 0 −1 0 1
, , , .
0 1 0 −1 1 0 −1 0
(c) 1 0 1 −1 −1 1
, , ,
0 1 1 0 −1 0
0 1 0 −1 −1 0
, , .
−1 1 1 −1 0 −1
10. (a) 0, 1, −1. (b) 1, −1.
11. 1, 2, 3, 4, 6, 8, 12, 24.
15. (a) 3i − 3. (c) 43 − 18i. (e) i.
√
16. (a) 3 + i. (c) −3.
√ √
17. (a) 2 cis(7π/4). (c) 2 2 cis(π/4). (e) 3 cis(3π/2).
√
18. (a) (1 − i)/2. (c) 16(i − 3 ). (e) −1/4.
22. (a) 292. (c) 1523.
27. |hgi ∩ hhi| = 1.
31. 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.
37. If g is an element distinct from the identity in G, g must generate G; other-
wise, hgi is a nontrivial proper subgroup of G.
Chapter 5. Permutation Groups
1. (a) (12453). (c) (13)(25).
2. (a) (135)(24). (c) (14)(23). (e) (1324). (g) (134)(25). (n) (17352).
3. (a) (16)(15)(13)(14). (c) (16)(14)(12).
4. (a1 , an , an−1 , . . . , a2 ).
5. (a) {(13), (13)(24), (132), (134), (1324), (1342)}. Not a subgroup.
8. (12345)(678).
11. Permutations of the form (1), (a1 , a2 )(a3 , a4 ), (a1 , a2 , a3 ), (a1 , a2 , a3 , a4 , a5 )
are possible for A5 .
17. (123)(12) = (13) 6= (23) = (12)(123).
25. Use the fact that (ab)(bc) = (abc) and (ab)(cd) = (abc)(bcd).
30. (a) Show that στ σ −1 (i) = (σ(a1 ), σ(a2 ), . . . , σ(ak ))(i) for 1 ≤ i ≤ n.
HINTS AND SOLUTIONS 399
Chapter 6. Cosets and Lagrange’s Theorem
1. The order of g and the order h must both divide the order of G. The smallest
number that 5 and 7 both divide is lcm(5, 7) = 35.
2. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
3. False.
4. False.
5. (a) H = {0, 8, 16} 4+H = {4, 12, 20}
1+H = {1, 9, 17} 5+H = {5, 13, 21}
2+H = {2, 10, 18} 6+H = {6, 14, 22}
3+H = {3, 11, 19} 7+H = {7, 15, 23}.
(c) 3Z = {. . . , −3, 0, 3, 6, . . .}
1 + 3Z = {. . . , −2, 1, 4, 7, . . .}
2 + 3Z = {. . . , −1, 2, 5, 8, . . .}.
7. 4φ(15) ≡ 48 ≡ 1 (mod 15).
12. Let g1 ∈ gH. Then there exists an h ∈ H such that g1 = gh = ghg −1 g ⇒
g1 ∈ Hg ⇒ gH ⊂ Hg. Similarly, Hg ⊂ gH. Therefore, gH = Hg.
/ H, then a−1 ∈
17. If a ∈ / H ⇒ a−1 ∈ aH = a−1 H = bH ⇒ there exist h1 , h2 ∈ H
such that a h1 = bh2 ⇒ ab = h1 h−1
−1
2 ∈ H.
Chapter 7. Introduction to Cryptography
1. LAORYHAPDWK.
3. Hint: Q = E, F = X, A = R.
4. 26! − 1.
7. (a) 2791. (c) 112135 25032 442.
9. (a) 31. (c) 14.
10. (a) n = 11 · 41. (c) n = 8779 · 4327.
Chapter 8. Algebraic Coding Theory
2. (0000) ∈
/ C.
3. (a) 2. (c) 2.
4. (a) 3. (c) 4.
6. (a) dmin = 2. (c) dmin = 1.
400 HINTS AND SOLUTIONS
7. (a) (00000), (00101), (10011), (10110)
0 1
0 0
1
G= 0
.
0 1
1 1
(b) (00000), (010111), (101101), (111010)
1 0
0 1
1 0
G= 1
.
1
0 1
1 1
9. Multiple errors occur in one of the received words.
11. (a) A canonical parity-check matrix with standard generator matrix
1
1
G= 0 .
0
1
(c) A canonical parity-check matrix with standard generator matrix
1 0
0 1
G= 1 1 .
1 0
12. (a) All possible syndromes occur.
15. (a) The cosets of C are
Cosets
C (00000) (00101) (10011) (10110)
(10000) + C (10000) (10101) (00011) (00110)
(01000) + C (01000) (01101) (11011) (11110)
(00100) + C (00100) (00001) (10111) (10010)
(00010) + C (00010) (00111) (10001) (10100)
(11000) + C (11000) (11101) (01011) (01110)
(01100) + C (01100) (01001) (11111) (11010)
(01010) + C (01010) (01111) (11001) (11100)
HINTS AND SOLUTIONS 401
A decoding table does not exist for C since it is only single error-detecting.
19. 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.
23. For 20 information positions, at least six check bits are needed to ensure an
error-correcting code.
Chapter 9. Isomorphisms
1. The group nZ is an infinite cyclic group generated by n. Every infinite cyclic
group is isomorphic to Z.
2. Define φ : C∗ → GL2 (R) by
a b
φ(a + bi) = .
−b a
3. False.
6. Define a map from Zn into the nth roots of unity by k 7→ cis(2kπ/n).
8. Assume that Q is cyclic and try to find a generator.
11. D4 , Q8 , Z8 , Z2 × Z4 , Z2 × Z2 × Z2 .
16. (a) 12. (c) 5.
20. True.
25. Z2 × Z2 × Z13 is not cyclic.
27. Let a be a generator for G. If φ : G → H is an isomorphism, show that φ(a)
is a generator for H.
38. Any automorphism of Z6 must send 1 to another generator of Z6 .
45. To show that φ is one-to-one, let g1 = h1 k1 and g2 = h2 k2 . Then φ(g1 ) =
φ(g2 ) ⇒ φ(h1 k1 ) = φ(h2 k2 ) ⇒ (h1 , k1 ) = (h2 , k2 ) ⇒ h1 = h2 , k1 = k2 ⇒
g1 = g2 .
Chapter 10. Normal Subgroups and Factor Groups
1. (a) A4 (12)A4
A4 A4 (12)A4
(12) A4 (12)A4 A4
(c) D4 is not normal in S4 .
8. If a ∈ G is a generator for G, then aH is a generator for G/H.
402 HINTS AND SOLUTIONS
13. Since eg = ge for all g ∈ G, the identity is in C(g). If x, y ∈ C(g), then xyg =
xgy = gxy ⇒ xy ∈ C(g). If xg = gx, then x−1 g = gx−1 ⇒ x−1 ∈ C(g) ⇒
C(g) is a subgroup of G. If hgi is normal in G, then g1 xg1−1 g = gg1 xg1−1 for
all g1 ∈ G.
15. (a) Let g ∈ G and h ∈ G0 . If h = aba−1 b−1 , then ghg −1 = gaba−1 b−1 g −1 =
(gag −1 )(gbg −1 )(ga−1 g −1 )(gb−1 g −1 ) = (gag −1 )(gbg −1 )(gag −1 )−1 (gbg −1 )−1 .
We also need to show that if h = h1 · · · hn with hi = ai bi a−1 −1
i bi , then ghg
−1
−1 −1
is a product of elements of the same type. However, ghg = gh1 · · · hn g =
(gh1 g −1 )(gh2 g −1 ) · · · (ghn g −1 ).
Chapter 11. Homomorphisms
2. (a) A homomorphism. (c) Not a homomorphism.
4. φ(m + n) = 7(m + n) = 7m + 7n = φ(m) + φ(n). The kernel of φ is {0} and
the image of φ is 7Z.
5. 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 .
9. Let a, b ∈ G. Then φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a).
Chapter 12. Matrix Groups and Symmetry
1 1
kx + yk2 + kxk2 − kyk2 = hx + y, x + yi − kxk2 − kyk2
1.
2 2
1
kxk2 + 2hx, yi + kyk2 − kxk2 − kyk2
=
2
= hx, yi.
3. (a) An element of SO(2). (c) Not in O(3).
5. (a) hx, yi = x1 y1 + · · · + xn yn = y1 x1 + · · · + yn xn = hy, xi.
7. Use the unimodular matrix
5 2
.
2 1
10. Show that the kernel of the map det : O(n) → R∗ is SO(n).
13. True.
17. p6m.
HINTS AND SOLUTIONS 403
Chapter 13. The Structure of Groups
1. Since 40 = 23 · 5, the possible abelian groups of order 40 are Z40 ∼
= Z8 × Z5 ,
Z5 × Z4 × Z2 , and Z5 × Z2 × Z2 × Z2 .
4. (a) {0} ⊂ h6i ⊂ h3i ⊂ Z12 .
(e) {((1), 0)} ⊂ {(1), (123), (132)} × {0} ⊂ S3 × {0} ⊂ S3 × h2i ⊂ S3 × Z4 .
7. Use the Fundamental Theorem of Finitely Generated Abelian Groups.
12. 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 }.
The series
G = Gn ⊃ Gn−1 ⊃ · · · ⊃ G0 = N = Nn ⊃ Nn−1 ⊃ · · · ⊃ N1 ⊃ N0 = {e}
is a subnormal series. The factors of this series are abelian since Gi+1 /Gi ∼
=
(Gi+1 /N )/(Gi /N ).
16. Use the fact that Dn has a cyclic subgroup of index 2.
21. G/G0 is abelian.
Chapter 14. Group Actions
1. Example 1. 0, R2 \ {0}.
Example 2. X = {1, 2, 3, 4}.
2. (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)}.
3. (a) O1 = O2 = O3 = {1, 2, 3}.
6. (a) 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.
8. (34 + 31 + 32 + 31 + 32 + 32 + 33 + 33 )/8 = 21.
11. (1 · 34 + 6 · 33 + 11 · 32 + 6 · 31 )/24 = 15.
15. (1 · 26 + 3 · 24 + 4 · 23 + 2 · 22 + 2 · 21 )/12 = 13.
17. (1 · 28 + 3 · 26 + 2 · 24 )/6 = 80.
22. x ∈ gC(a)g −1 ⇔ g −1 xg ∈ C(a) ⇔ ag −1 xg = g −1 xga ⇔ gag −1 x = xgag −1 ⇔
x ∈ C(gag −1 ).
404 HINTS AND SOLUTIONS
Chapter 15. The Sylow Theorems
1. 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.
If |G| = 54 = 2 · 33 , then the order of a Sylow 2-subgroup is 2, and the order
of a Sylow 3-subgroup is 27.
2. The four Sylow 3-subgroups of S4 are
P1 = {(1), (123), (132)},
P2 = {(1), (124), (142)},
P3 = {(1), (134), (143)},
P4 = {(1), (234), (243)}.
5. 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. |H ∩ K| ≥ 16;
otherwise, HK would have (32 · 32)/8 = 128 elements, which is impossible.
H ∩K is normal in both H and K since it has index 2 in both groups. Hence,
N (H ∩K) contains both H and K. Therefore, |N (H ∩K)| must be a multiple
of 32 greater than 1 and still divide 96, so N (H ∩ K) = G.
8. G has a Sylow q-subgroup of order q 2 . Since the number of such subgroups
is congruent to 1 modulo q and divides p2 q 2 , there must be either 1, p, or p2
Sylow q-subgroups. Since q6 |p2 − 1 = (p − 1)(p + 1), there can be only one
Sylow q-subgroup, say Q. Similarly, we can show that there is a single Sylow
p-subgroup P . Every element in Q other than the identity has order q or q 2 ,
so P ∩ Q = {e}. Now show that hk = kh for h ∈ P and k ∈ Q. Deduce that
G = P × Q is abelian.
10. False.
17. If G is abelian, then G is cyclic, since |G| = 3 · 5 · 17. Now look at Example 5.
23. 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.
26. Let aG0 , bG0 ∈ G/G0 . Then (aG0 )(bG0 ) = abG0 = ab(b−1 a−1 ba)G0 =
(abb−1 a−1 )baG0 = baG0 .
Chapter 16. Rings
√
1. (a) 7Z is a ring but not a field. (c) Q( 2 ) is a field. (f) R is not a ring.
3. (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
4. (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.
HINTS AND SOLUTIONS 405
7. Assume there is an isomorphism φ : C → R with φ(i) = a.
√ √
8. False.
√ Assume there is an isomorphism φ : Q( 2 ) → Q( 3 ) such that
φ( 2 ) = a.
13. (a) x ≡ 17 (mod 55). (c) x ≡ 214 (mod 2772).
16. If I 6= {0}, show that 1 ∈ I.
19. (a) φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a).
27. Let a ∈ R with a 6= 0. The principal ideal generated by a is R ⇒ there exists
a b ∈ R such that ab = 1.
29. Compute (a + b)2 and (−ab)2 .
35. 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.
39. Suppose that x2 = x and x 6= 0. Since R is an integral domain, x = 1. To
find a nontrivial idempotent, look in M2 (R).
Chapter 17. Polynomials
2. (a) 9x2 + 2x + 5. (b) 8x4 + 7x3 + 2x2 + 7x.
3. (a) 5x3 + 6x2 − 3x + 4 = (5x2 2x + 1)(x − 2) + 6.
(c) 4x5 − x3 + x2 + 4 = (4x2 + 4)(x3 + 3) + 4x2 + 2.
5. (a) No zeros in Z12 . (c) 3, 4.
7. (2x + 1)2 = 1.
8. (a) Reducible. (c) Irreducible.
10. x2 + x + 8 = (x + 2)(x + 9) = (x + 7)(x + 4).
13. Z is not a field.
14. False. x2 + 1 = (x + 1)(x + 1).
16. 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 .
19. Define g(x) by g(x) = Φp (x + 1) and show that g(x) is irreducible over Q.
25. Find a nontrivial proper ideal in F [x].
406 HINTS AND SOLUTIONS
Chapter 18. Integral Domains
√ √ √
1. z −1 = 1/(a + b 3 i) = (a − b 3 i)/(a2 + 3b2 ) is in Z[ 3 i] if and only if
a2 + 3b2 = 1. The only integer solutions to the equation are a = ±1, b = 0.
2. (a) 5 = 1 + 2i)(1 − 2i). (c) 6 + 8i = (−1 + 7i)(1 − i).
4. True.
8. Let z = a + bi and w = c + di 6= 0 be in Z[i]. Prove that z/w ∈ Q(i).
14. Let a = ub with u a unit. Then ν(b) ≤ ν(ub) ≤ ν(a). Similarly, ν(a) ≤ ν(b).
15. Show that 21 can be factored in two different ways.
Chapter 19. Lattices and Boolean Algebras
2.
30
10 15
2 5 3
1
5. False.
6. (a) (a ∨ b ∨ a0 ) ∧ a.
a
b a
a0
(c) a ∨ (a ∧ b).
a b
a
8. Not equivalent.
HINTS AND SOLUTIONS 407
10. a0 ∧ [(a ∧ b0 ) ∨ b] = a ∧ (a ∨ b).
15. 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. (a) No.
21. (⇒). a = b ⇒ (a ∧ b0 ) ∨ (a0 ∧ b) = (a ∧ a0 ) ∨ (a0 ∧ a) = O ∨ O = O.
(⇐). (a∧b0 )∨(a0 ∧b) = O ⇒ a∨b = (a∨a)∨b = a∨(a∨b) = a∨[I ∧(a∨b)] =
a∨[(a∨a0 )∧(a∨b)] = [a∨(a∧b0 )]∨[a∨(a0 ∧b)] = a∨[(a∧b0 )∨(a0 ∧b)] = a∨0 = a.
A symmetric argument shows that a ∨ b = b.
Chapter 20. Vector Spaces
√ √ √ √ √
3. Q( 2, 3 ) has basis {1, 2, 3, 6 } over Q.
5. Pn has basis {1, x, x2 , . . . , xn−1 }.
7. (a) Subspace of dimension 2 with basis {(1, 0, −3), (0, 1, 2)}.
(d) Not a subspace.
10. 0 = α0 = α(−v + v) = α(−v) + αv ⇒ −αv = α(−v).
12. Let v0 = 0, v1 , . . . , vn ∈ V and α0 6= 0, α1 , . . . , αn ∈ F . Then α0 v0 + · · · +
αn vn = 0.
15. (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 ) ⇒ ker(T ) is a subspace of V .
(c) T (u) = T (v) ⇔ T (u − v) = T (u) − T (v) = 0 ⇔ u − v = 0 ⇔ u = v.
17. (a) Let u, u0 ∈ U and v, v 0 ∈ V . Then
(u + v) + (u0 + v 0 ) = (u + u0 ) + (v + v 0 ) ∈ U + V
α(u + v) = αu + αv ∈ U + V.
Chapter 21. Fields
1. (a) x4 − 32 x2 − 62 4 2
9 . (c) x − 2x + 25.
√ √ √ √ √
2. (a) {1, 2, 3, 6 }. (c) {1, i, 2, 2 i}. (e) {1, 21/6 , 21/3 , 21/2 , 22/3 , 25/6 }.
√ √
3. (a) Q( 3, 7 ).
5. Use the fact that the elements of Z2 [x]/hx3 + x + 1i are 0, 1, α, 1 + α, α2 ,
1 + α2 , α + α2 , 1 + α + α2 and the fact that α3 + α + 1 = 0.
408 HINTS AND SOLUTIONS
8. False.
14. 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 ).
√ √ √ √ √ √ √ √ √
22. Q( 3, 7 ) ⊃ Q( √ 3 +√ 7 ) since {1, 3,√ 7, √ 21 } is a basis for Q( 3, 7 )
over Q. 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 ).
27. Let β ∈ F (α) not in F . Then β = p(α)/q(α), where p and q are polynomials
in α with q(α) 6= 0 and coefficients in F . If β is algebraic over F , then
there exists a polynomial f (x) ∈ F [x] such that f (β) = 0. Let f (x) =
a0 + a1 x + · · · + an xn . Then
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.
Chapter 22. Finite Fields
1. (a) 2. (c) 2.
4. There are eight elements in Z2 (α). Exhibit two more zeros of x3 + x2 + 1
other than α in these eight elements.
5. Find an irreducible polynomial p(x) in Z3 [x] of degree 3 and show that
Z3 [x]/hp(x)i has 27 elements.
7. (a) x5 − 1 = (x + 1)(x4 + x3 + x2 + x + 1).
(c) x9 − 1 = (x + 1)(x2 + x + 1)(x6 + x3 + 1).
8. True.
11. (a) Use the fact that x7 − 1 = (x + 1)(x3 + x + 1)(x3 + x2 + 1).
12. False.
17. If p(x) ∈ F [x], then p(x) ∈ E[x].
18. 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 ).
24. Factor xp−1 − 1 over Zp .
HINTS AND SOLUTIONS 409
Chapter 23. Galois Theory
1. (a) Z2 . (c) Z2 × Z2 × Z2 .
2. (a) Separable. (c) Not separable.
3. [GF(729) : GF(9)] = [GF(729) : GF(3)]/[GF(9) : GF(3)] = 6/2 = 3 ⇒
G(GF(729)/GF(9)) ∼ = Z3 . A generator for G(GF(729)/GF(9)) is σ, where
6
σ36 (α) = α3 = α729 for α ∈ GF(729).
4. (a) S5 . (c) S3 .
5. (a) Q(i).
7. Let E be the splitting field of a cubic polynomial in F [x]. Show that [E : F ]
is less than or equal to 6 and is divisible by 3. Since G(E/F ) is a subgroup of
S3 whose order is divisible by 3, conclude that this group must be isomorphic
to Z3 or S3 .
9. G is a subgroup of Sn .
16. True.
20. (a) Clearly ω, ω 2 , . . . , ω p−1 are distinct since ω 6= 1 or 0. To show that ω i is
a zero of Φp , calculate Φp (ω i ).
(b) The conjugates of ω are ω, ω 2 , . . . , ω p−1 . Define a map φi : Q(ω) → Q(ω i )
by
φi (a0 + a1 ω + · · · + ap−2 ω p−2 ) = a0 + a1 ω i + · · · + cp−2 (ω i )p−2 ,
where ai ∈ Q. Prove that φi is an isomorphism of fields. Show that φ2
generates G(Q(ω)/Q).
(c) Show that {ω, ω 2 , . . . , ω p−1 } is a basis for Q(ω) over Q, and consider
which linear combinations of ω, ω 2 , . . . , ω p−1 are left fixed by all elements of
G(Q(ω)/Q).
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Notation
The following table defines the notation used in this book. Page numbers
refer to the first appearance of each symbol.
Symbol Description Page
a∈A a is in the set A 4
N the natural numbers 5
Z the integers 5
Q the rational numbers 5
R the real numbers 5
C the complex numbers 5
A⊂B A is a subset of B 5
∅ the empty set 5
A∪B union of sets A and B 5
A∩B intersection of sets A and B 5
A0 complement of the set A 6
A\B difference between sets A and B 6
A×B Cartesian product of sets A and B 8
An A × · · · × A (n times) 8
id identity mapping 12
f −1 inverse of the function f 13
a ≡ b (mod n) a is congruent to b modulo n 17
n!
n factorial 24
n
binomial coefficient n!/(k!(n − k)!) 24
k
m|n m divides n 27
gcd(m, n) greatest common divisor of m and n 27
P(X) power set of X 32
418
NOTATION 419
Symbol Description Page
Zn the integers modulo n 36
lcm(m, n) least common multiple of m and n 33
U (n) group of units in Zn 42
Mn (R) the n × n matrices with entries in R 43
det A determinant of A 43
GLn (R) general linear group 43
Q8 the group of quaternions 44
C∗ the multiplicative group of complex numbers 44
|G| order of a group G 44
R∗ the multiplicative group of real numbers 47
Q∗ the multiplicative group of rational numbers 47
SLn (R) special linear group 47
Z(G) center of a group G 53
hai cyclic subgroup generated by a 58
|a| order of an element a 58
cis θ cos θ + i sin θ 63
T the circle group 65
Sn symmetric group on n letters 75
(a1 , a2 , . . . , ak ) cycle of length k 77
An alternating group on n letters 82
Dn dihedral group 83
[G : H] index of a subgroup H in a group G 94
LH set of left cosets of H in a group G 94
RH set of right cosets of H in a group G 94
d(x, y) Hamming distance between x and y 117
dmin minimum distance of a code 117
w(x) weight of x 117
Mm×n (Z2 ) set of m by n matrices with entries in Z2 123
Null(H) null space of a matrix H 123
δij Kronecker delta 127
G∼ =H G is isomorphic to H 141
Aut(G) automorphism group of G 153
ig ig (x) = gxg −1 153
Inn(G) inner automorphism group of G 153
ρg right regular representation 154
G/N factor group of G mod N 156
ker φ kernel of φ 167
G0 commutator subgroup of G 164
420 NOTATION
Symbol Description Page
(aij ) matrix 176
O(n) orthogonal group 179
kxk length of a vector x 180
SO(n) special orthogonal group 183
E(n) Euclidean group 183
Ox orbit of x 211
Xg fixed point set of g 211
Gx isotropy subgroup of x 211
XG set of fixed points in a G-set X 213
N (H) normalizer of a subgroup H 229
H the ring of quaternions 241
char R characteristic of a ring R 245
Z[i] the Gaussian integers 244
Z(p) ring of integers localized at p 261
R[x] ring of polynomials over R 264
deg p(x) degree of p(x) 264
R[x1 , x2 , . . . , xn ] ring of polynomials in n variables 267
φα evaluation homomorphism at α 267
Q(x) field of rational functions over Q 287
ν(a) Euclidean valuation of a 292
F (x) field of rational functions in x 298
F (x1 , . . . , xn ) field of rational functions in x1 , . . . , xn 298
ab a is less than b 302
a∧b meet of a and b 304
a∨b join of a and b 304
I largest element in a lattice 306
O smallest element in a lattice 306
a0 complement of a in a lattice 306
dim V dimension of a vector space V 324
U ⊕V direct sum of vector spaces U and V 327
Hom(V, W ) set of all linear transformations from U to V 327
V∗ dual of a vector space V 327
F (α1 , . . . , αn ) smallest field containing F and α1 , . . . , αn 332
[E : F ] dimension of a field extension of E over F 335
GF(pn ) Galois field of order pn 356
F∗ multiplicative group of a field F 356
G(E/F ) Galois group of E over F 372
F{σi } field fixed by automorphisms σi 377
NOTATION 421
Symbol Description Page
FG field fixed by automorphism group G 377
∆2 discriminant of a polynomial 393
Index
G-equivalence classes, 223 Boolean function, 220, 318
G-equivalent, 211 Boolean ring, 260
G-set, 209 Burnside’s Counting Theorem, 216
nth root of unity, 65, 385 Burnside, William, 46, 162, 223
Abel, Niels Henrik, 384 Cancellation law
Abelian group, 41 for groups, 45
Ackermann’s function, 34 for integral domains, 244
Adleman, L., 104 Cardano, Gerolamo, 277
Algebraic closure, 339 Carmichael numbers, 110
Algebraic extension, 332 Cauchy’s Theorem, 227
Algebraic number, 333 Cauchy, Augustin-Louis, 83
Algorithm Cayley table, 42
division, 268 Cayley’s Theorem, 145
Euclidean, 29 Cayley, Arthur, 146
Artin, Emil, 299 Center
Ascending chain condition, 290 of a group, 53
Associate elements, 288 of a ring, 261
Atom, 310 Centralizer
Automorphism of a subgroup, 213
inner, 153, 173 of an element, 163
of a group, 153 Characteristic of a ring, 245
Chinese Remainder Theorem
Basis of a lattice, 188 for integers, 254
Bieberbach, L., 192 for rings, 261
Binary operation, 40 Cipher, 100
Binary symmetric channel, 115 Ciphertext, 100
Boole, George, 315 Circuit
Boolean algebra parallel, 313
atom in a, 310 series, 312
definition of, 307 series-parallel, 313
finite, 309 Class equation, 213
isomorphism, 309 Code
422
INDEX 423
BCH, 366 disjoint, 78
cyclic, 358
dual, 138 De Morgan’s laws
group, 120 for Boolean algebras, 309
Hamming for sets, 7
definition of, 138 De Morgan, Augustus, 315
perfect, 139 Decoding table, 134
shortened, 139 Deligne, Pierre, 349
linear, 123 DeMoivre’s Theorem, 64
minimum distance of, 117 Derivative, 280, 355
polynomial, 359 Derived series, 206
Commutative diagrams, 169 Descending chain condition, 299
Commutative rings, 240 Determinant, Vandermonde, 363
Composite integer, 29 Dickson, L. E., 162
Composition series, 202 Diffie, W., 104
Congruence modulo n, 17 Direct product of groups
Conjugacy classes, 213 external, 147
Conjugate elements, 373 internal, 150
Conjugate fields, 392 Direct sum of vector spaces, 327
Conjugate permutations, 99 Discriminant
Conjugate, complex, 62 of a separable polynomial, 393
Conjugation, 210 of the cubic equation, 281
Constructible number, 344 of the quadratic equation, 280
Correspondence Theorem Division algorithm
for groups, 170 for integers, 26
for rings, 250 for polynomials, 268
Coset Division ring, 240
double, 99 Domain
leader, 133 Euclidean, 292
left, 92 principal ideal, 289
representative, 92 unique factorization, 288
right, 92 Doubling the cube, 348
Coset decoding, 132
Cryptanalysis, 102 Eisenstein’s Criterion, 275
Cryptosystem Element
affine, 103 associate, 288
definition of, 100 centralizer of, 163
monoalphabetic, 102 idempotent, 261
polyalphabetic, 103 identity, 41
private key, 101 inverse, 41
public key, 101 irreducible, 288
RSA, 104 nilpotent, 260
single key, 101 order of, 58
Cycle prime, 288
definition of, 76 primitive, 376
424 INDEX
transcendental, 332 Frobenius map, 369
Equivalence class, 16 Function
Equivalence relation, 15 bijective, 10
Euclidean algorithm, 29 Boolean, 220, 318
Euclidean domain, 292 composition of, 10
Euclidean group, 183 definition of, 8
Euclidean inner product, 180 domain of, 9
Euclidean valuation, 292 identity, 12
Euler φ-function, 97 injective, 10
Euler, Leonhard, 97, 349 invertible, 13
Extension one-to-one, 10
algebraic, 332 onto, 10
field, 329 order-preserving, 317
finite, 335 range of, 9
normal, 379 surjective, 10
radical, 385 switching, 220, 318
separable, 355, 376 Fundamental Theorem
simple, 332 of Algebra, 340, 390
External direct product, 147 of Arithmetic, 30
of Finite Abelian Groups, 199
Faltings, Gerd, 349 of Galois Theory, 380
Feit, W., 162, 223
Fermat’s factorization algorithm, 109 Gödel, Kurt, 315
Fermat’s Little Theorem, 97 Galois field, 356
Fermat, Pierre de, 97, 349 Galois group, 372
Ferrari, Ludovico, 278 Galois, Évariste, 46, 384
Ferro, Scipione del, 277 Gauss’s Lemma, 294
Field, 240 Gauss, Karl Friedrich, 296
algebraically closed, 339 Gaussian integers, 244
base, 329 Generator of a cyclic subgroup, 58
conjugate, 392 Generators for a group, 197
extension, 329 Glide reflection, 184
fixed, 378 Gorenstein, Daniel, 162
Galois, 356 Greatest common divisor
of fractions, 286 of elements in a UFD, 298
of quotients, 286 of two integers, 27
prime, 298 of two polynomials, 270
splitting, 341 Greatest lower bound, 303
Finitely generated group, 197 Greiss, R., 162
Fior, Antonio, 277 Grothendieck, A., 349
First Isomorphism Theorem Group
for groups, 168 p-group, 198, 227
for rings, 249 abelian, 41
Fixed point set, 211 action, 209
Freshman’s Dream, 354 alternating, 82
INDEX 425
automorphism of, 153 kernel of a group, 167
center of, 90, 163, 213 kernel of a ring, 246
circle, 65 lattice, 317
commutative, 41 natural, 168, 249
cyclic, 58 of groups, 165
definition of, 40 ring, 246
dihedral, 83
Euclidean, 183 Ideal
factor, 156 definition of, 247
finite, 44 maximal, 250
finitely generated, 197 one-sided, 248
Galois, 372 prime, 251
general linear, 43, 178 principal, 247
generators of, 197 trivial, 247
Heisenberg, 51 two-sided, 248
homomorphism of, 165 Idempotent, 261
infinite, 44 Indeterminate, 264
isomorphic, 141 Index of a subgroup, 94
isomorphism of, 141 Induction
nonabelian, 41 first principle of, 23
noncommutative, 41 second principle of, 24
of units, 42 Infimum, 303
order of, 44 Inner product, 122
orthogonal, 179 Integral domain, 240
permutation, 75 Internal direct product, 150
point, 189 International standard book number, 55
quaternion, 44 Irreducible element, 288
quotient, 156 Irreducible polynomial, 272
simple, 158, 162 Isometry, 184
solvable, 205 Isomorphism
space, 189 of Boolean algebras, 309
special linear, 48, 178 of groups, 141
special orthogonal, 183 ring, 246
symmetric, 75
symmetry, 186 Join, 304
torsion, 206 Jordan, C., 162
Jordan-Hölder Theorem, 203
Hamming distance, 117
Hamming, R., 120 Kernel
Hellman, M., 104 of a group homomorphism, 167
Hilbert, David, 192, 252, 315, 349 of a linear transformation, 326
Homomorphic image, 165 of a ring homomorphism, 246
Homomorphism Key
canonical, 168, 249 definition of, 100
evaluation, 247, 267 private, 101
426 INDEX
public, 101 Matrix, Vandermonde, 363
single, 101 Maximal ideal, 250
Klein, Felix, 46, 175, 252 Maximum-likelihood decoding, 115
Kronecker delta, 127, 181 Meet, 304
Kronecker, Leopold, 349 Metric, 137
Kummer, Ernst, 349 Minimal generator polynomial, 361
Minimal polynomial, 334
Lagrange’s Theorem, 95 Minkowski, Hermann, 349
Lagrange, Joseph-Louis, 46, 83, 97 Monic polynomial, 264
Laplace, Pierre-Simon, 83 Mordell-Weil conjecture, 349
Lattice Multiplicative subset, 299
completed, 306 Multiplicity of a root, 376
definition of, 304
distributive, 306 Nilpotent element, 260
homomorphism, 317 Noether, A. Emmy, 252
Lattice of points, 188 Noether, Max, 252
Lattices, Principle of Duality for, 304 Normal extension, 379
Least upper bound, 303 Normal series of a group, 201
Left regular representation, 146 Normal subgroup, 155
Lie, Sophus, 46, 231 Normalizer, 53, 229
Linear combination, 322 Null space
Linear dependence, 322 of a linear transformation, 326
Linear functionals, 327 of a matrix, 123
Linear independence, 322
Linear map, 175 Odd Order Theorem, 235
Linear transformation Orbit, 90, 211
definition of, 11, 175, 326 Orthogonal group, 179
kernel of, 326 Orthogonal matrix, 179
null space of, 326 Orthonormal set, 181
range of, 326
Lower bound, 303 Partial order, 301
Partially ordered set, 302
Mapping, see Function Partitions, 16
Matrix Permutation
distance-preserving, 181 conjugate, 99
generator, 124 definition of, 12, 74
inner product-preserving, 181 even, 81
invertible, 177 odd, 81
length-preserving, 181 Permutation group, 75
nonsingular, 177 Plaintext, 100
null space of, 123 Polynomial
orthogonal, 179 code, 359
parity-check, 124 content of, 294
similar, 16 cyclotomic, 279
unimodular, 189 definition of, 264
INDEX 427
degree of, 264 commutative, 240
error, 370 definition of, 239
error-locator, 370 division, 240
greatest common divisor of, 270 factor, 249
in n indeterminates, 267 finitely generated, 299
irreducible, 272 homomorphism, 246
leading coefficient of, 264 isomorphism, 246
minimal, 334 local, 300
minimal generator, 361 Noetherian, 290
monic, 264 of integers localized at p, 261
primitive, 294 of quotients, 299
root of, 269 quotient, 249
separable, 376 with identity, 240
zero of, 269 with unity, 240
Polynomial separable, 354 Rivest, R., 104
Poset RSA cryptosystem, 104
definition of, 302 Ruffini, P., 383
largest element in, 306 Russell, Bertrand, 315
smallest element in, 306
Power set, 32, 302 Scalar product, 319
Prime element, 288 Schreier’s Theorem, 207
Prime field, 298 Second Isomorphism Theorem
Prime ideal, 251 for groups, 170
Prime integer, 29 for rings, 250
Prime subfield, 298 Semidirect product, 194
Primitive nth root of unity, 66, 386 Shamir, A., 104
Primitive element, 376 Shannon, C., 119
Primitive Element Theorem, 376 Sieve of Eratosthenes, 34
Primitive polynomial, 294 Simple extension, 332
Principal ideal, 247 Simple group, 158
Principal ideal domain (PID), 289 Simple root, 376
Principal series, 202 Solvability by radicals, 385
Pseudoprime, 109 Spanning set, 322
Splitting field, 341
Quaternions, 44, 242 Squaring the circle, 348
Standard decoding, 132
Repeated squares, 66 Subfield
Resolvent cubic equation, 282 prime, 298
Right regular representation, 154 Subgroup
Rigid motion, 38, 184 p-subgroup, 227
Ring centralizer, 213
Artinian, 299 commutator, 164, 206, 233
Boolean, 260 cyclic, 58
center of, 261 definition of, 47
characteristic of, 245 index of, 94
428 INDEX
isotropy, 211 dual of, 327
normal, 155 subspace of, 321
normalizer of, 229
proper, 47 Weight of a codeword, 117
stabilizer, 211 Weil, André, 349
Sylow p-subgroup, 229 Well-defined map, 10
torsion, 71 Well-ordered set, 25
transitive, 90 Whitehead, Alfred North, 315
translation, 189 Wilson’s Theorem, 369
trivial, 47
Subnormal series of a group, 201 Zassenhaus Lemma, 207
Subring, 243 Zero
Supremum, 303 multiplicity of, 376
Switch of a polynomial, 269
closed, 312 Zero divisor, 241
definition of, 312
open, 312
Switching function, 220, 318
Sylow p-subgroup, 229
Sylow, Ludvig, 231
Syndrome of a code, 131, 370
Tartaglia, 277
Third Isomorphism Theorem
for groups, 171
for rings, 250
Thompson, J., 162, 223
Totally ordered set, 317
Transcendental element, 332
Transcendental number, 333
Transposition, 79
Trisection of an angle, 348
Unique factorization domain (UFD), 288
Unit, 240, 288
Universal Product Code, 54
Upper bound, 303
Vandermonde determinant, 363
Vandermonde matrix, 363
Vector space
basis of, 324
definition of, 319
dimension of, 324
direct sum of, 327