abstract algebra theory and practice

7/24/2019 abstract algebra theory and practice

1/423

Abstract Algebra

Theory and Applications


2/423


3/423

Abstract Algebra

Theory and Applications

Thomas W. Judson

Stephen F. Austin State University

Sage Exercises for Abstract Algebra

Robert A. Beezer

University of Puget Sound

August 12, 2015


4/423

19972015 Thomas W. Judson, Robert A. Beezer

Permission is granted to copy, distribute and/or modify this doc-

ument under the terms of the GNU Free Documentation License,Version 1.2 or any later version published by the Free SoftwareFoundation; with no Invariant Sections, no Front-Cover Texts, andno Back-Cover Texts. A copy of the license is included in the ap-pendix entitled GNU Free Documentation License.


5/423

Acknowledgements

I would like to acknowledge the following reviewers for their helpfulcomments 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

Georey 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 Grin, Kelle Karshick, and the rest of the sta at PWSPublishing for their guidance throughout this project. It has beena pleasure to work with them.

Robert Beezer encouraged me to make Abstract Algebra: The-ory and Applicationsavailable as an open source textbook, a deci-sion that I have never regretted. With his assistance, the book hasbeen rewritten in MathBook XML (http://mathbook.pugetsound.edu),making it possible to quickly output print, web, PDF versions and

more from the same source. The open source version of this bookhas received support from the National Science Foundation (Award#DUE-1020957).

v


6/423

Preface

This text is intended for a one or two-semester undergraduatecourse in abstract algebra. Traditionally, these courses have cov-ered the theoretical aspects of groups, rings, and elds. However,with the development of computing in the last several decades, ap-plications that involve abstract algebra and discrete mathematicshave become increasingly important, and many science, engineer-

ing, and computer science students are now electing to minor inmathematics. Though theory still occupies a central role in thesubject of abstract algebra and no student should go through sucha course without a good notion of what a proof is, the importanceof applications such as coding theory and cryptography has grownsignicantly.

Until recently most abstract algebra texts included few if anyapplications. However, one of the major problems in teaching anabstract algebra course is that for many students it is their rstencounter with an environment that requires them to do rigorousproofs. Such students often nd it hard to see the use of learn-

ing to prove theorems and propositions; applied examples help theinstructor provide motivation.

This text contains more material than can possibly be coveredin a single semester. Certainly there is adequate material for a two-semester course, and perhaps more; however, for a one-semestercourse it would be quite easy to omit selected chapters and stillhave a useful text. The order of presentation of topics is standard:groups, then rings, and nally elds. Emphasis can be placed ei-ther on theory or on applications. A typical one-semester coursemight cover groups and rings while briey touching on eld theory,using Chapters 1 through 6, 9, 10, 11, 13 (the rst part), 16, 17,

18 (the rst part), 20, and 21. Parts of these chapters could bedeleted and applications substituted according to the interests ofthe students and the instructor. A two-semester course emphasiz-ing theory might cover Chapters 1 through 6, 9, 10, 11, 13 through18, 20, 21, 22 (the rst part), and 23. On the other hand, if appli-cations are to be emphasized, the course might cover Chapters 1through 14, and 16 through 22. In an applied course, some of themore theoretical results could be assumed or omitted. A chapter

vi


7/423

vii

dependency chart appears below. (A broken line indicates a partialdependency.)

Chapter 23

Chapter 22

Chapter 21

Chapter 18 Chapter 20 Chapter 19

Chapter 17 Chapter 15

Chapter 13 Chapter 16 Chapter 12 Chapter 14

Chapter 11

Chapter 10

Chapter 8 Chapter 9 Chapter 7

Chapters 16

Though there are no specic prerequisites for a course in ab-stract algebra, students who have had other higher-level courses inmathematics will generally be more prepared than those who havenot, because they will possess a bit more mathematical sophisti-cation. Occasionally, we shall assume some basic linear algebra;that is, we shall take for granted an elementary knowledge of ma-trices and determinants. This should present no great problem,since most students taking a course in abstract algebra have beenintroduced to matrices and determinants elsewhere in their career,if they have not already taken a sophomore or junior-level course

in linear algebra.Exercise sections are the heart of any mathematics text. Anexercise set appears at the end of each chapter. The nature of theexercises ranges over several categories; computational, conceptual,and theoretical problems are included. A section presenting hintsand solutions to many of the exercises appears at the end of thetext. Often in the solutions a proof is only sketched, and it is up tothe student to provide the details. The exercises range in diculty


8/423

viii

from very easy to very challenging. Many of the more substantialproblems require careful thought, so the student should not bediscouraged if the solution is not forthcoming after a few minutesof work.

There are additional exercises or computer projects at the endsof many of the chapters. The computer projects usually require aknowledge of programming. All of these exercises and projects aremore substantial in nature and allow the exploration of new resultsand theory.

Sage (sagemath.org) is a free, open source, software system foradvanced mathematics, which is ideal for assisting with a study ofabstract algebra. Sage can be used either on your own computer, alocal server, or on SageMathCloud (https://cloud.sagemath.com).Robert Beezer has written a comprehensive introduction to Sageand a selection of relevant exercises that appear at the end of eachchapter, including live Sage cells in the web version of the book.The Sage code has been tested for accuracy with the most recentversion available at this time: Sage Version 6.8 (released 2015-07-

26).

Thomas W. JudsonNacogdoches, Texas 2015


9/423


10/423

x CONTENTS

6 Cosets and Lagranges Theorem 876.1 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . 876.2 Lagranges Theorem . . . . . . . . . . . . . . . . . . 896.3 Fermats and Eulers Theorems . . . . . . . . . . . . 916.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 93

7 Introduction to Cryptography 95

7.1 Private Key Cryptography . . . . . . . . . . . . . . . 967.2 Public Key Cryptography . . . . . . . . . . . . . . . 987.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 1027.4 Additional Exercises: Primality and Factoring . . . . 1047.5 References and Suggested Readings . . . . . . . . . . 105

8 Algebraic Coding Theory 1078.1 Error-Detecting and Correcting Codes . . . . . . . . 1078.2 Linear Codes . . . . . . . . . . . . . . . . . . . . . . 1158.3 Parity-Check and Generator Matrices . . . . . . . . 1198.4 Ecient Decoding . . . . . . . . . . . . . . . . . . . 125

8.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 1288.6 Programming Exercises . . . . . . . . . . . . . . . . 1338.7 References and Suggested Readings . . . . . . . . . . 134

9 Isomorphisms 1359.1 Denition and Examples . . . . . . . . . . . . . . . . 1359.2 Direct Products . . . . . . . . . . . . . . . . . . . . . 1409.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 144

10 Normal Subgroups and Factor Groups 14910.1 Factor Groups and Normal Subgroups . . . . . . . . 14910.2 The Simplicity of the Alternating Group . . . . . . . 15210.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 155

11 Homomorphisms 15811.1 Group Homomorphisms . . . . . . . . . . . . . . . . 15811.2 The Isomorphism Theorems . . . . . . . . . . . . . . 16111.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 16411.4 Additional Exercises: Automorphisms . . . . . . . . 166

12 Matrix Groups and Symmetry 16812.1 Matrix Groups . . . . . . . . . . . . . . . . . . . . . 16812.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . 176

12.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 18312.4 References and Suggested Readings . . . . . . . . . . 185

13 The Structure of Groups 18713.1 Finite Abelian Groups . . . . . . . . . . . . . . . . . 18713.2 Solvable Groups . . . . . . . . . . . . . . . . . . . . 19113.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 19513.4 Programming Exercises . . . . . . . . . . . . . . . . 197


11/423

CONTENTS xi

13.5 References and Suggested Readings . . . . . . . . . . 197

14 Group Actions 198

14.1 Groups Acting on Sets . . . . . . . . . . . . . . . . . 19814.2 The Class Equation . . . . . . . . . . . . . . . . . . 20114.3 Burnsides Counting Theorem . . . . . . . . . . . . . 20314.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 21114.5 Programming Exercise . . . . . . . . . . . . . . . . . 21414.6 References and Suggested Reading . . . . . . . . . . 214

15 The Sylow Theorems 215

15.1 The Sylow Theorems . . . . . . . . . . . . . . . . . . 21515.2 Examples and Applications . . . . . . . . . . . . . . 21915.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 22315.4 A Project . . . . . . . . . . . . . . . . . . . . . . . . 22515.5 References and Suggested Readings . . . . . . . . . . 225

16 Rings 227

16.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . 22716.2 Integral Domains and Fields . . . . . . . . . . . . . . 23116.3 Ring Homomorphisms and Ideals . . . . . . . . . . . 23316.4 Maximal and Prime Ideals . . . . . . . . . . . . . . . 23716.5 An Application to Software Design . . . . . . . . . . 24016.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 24416.7 Programming Exercise . . . . . . . . . . . . . . . . . 24916.8 References and Suggested Readings . . . . . . . . . . 249

17 Polynomials 251

17.1 Polynomial Rings . . . . . . . . . . . . . . . . . . . . 251

17.2 The Division Algorithm . . . . . . . . . . . . . . . . 25517.3 Irreducible Polynomials . . . . . . . . . . . . . . . . 25917.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 26517.5 Additional Exercises: Solving the Cubic and Quartic

Equations . . . . . . . . . . . . . . . . . . . . . . . . 268

18 Integral Domains 271

18.1 Fields of Fractions . . . . . . . . . . . . . . . . . . . 27118.2 Factorization in Integral Domains . . . . . . . . . . . 27518.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 28418.4 References and Suggested Readings . . . . . . . . . . 287

19 Lattices and Boolean Algebras 288

19.1 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . 28819.2 Boolean Algebras . . . . . . . . . . . . . . . . . . . . 29219.3 The Algebra of Electrical Circuits . . . . . . . . . . . 29819.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 30119.5 Programming Exercises . . . . . . . . . . . . . . . . 30319.6 References and Suggested Readings . . . . . . . . . . 304


12/423

xii CONTENTS

20 Vector Spaces 30520.1 Denitions and Examples . . . . . . . . . . . . . . . 30520.2 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . 30720.3 Linear Independence . . . . . . . . . . . . . . . . . . 30820.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 31020.5 References and Suggested Readings . . . . . . . . . . 313

21 Fields 31521.1 Extension Fields . . . . . . . . . . . . . . . . . . . . 31521.2 Splitting Fields . . . . . . . . . . . . . . . . . . . . . 32621.3 Geometric Constructions . . . . . . . . . . . . . . . . 32821.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 33421.5 References and Suggested Readings . . . . . . . . . . 337

22 Finite Fields 33822.1 Structure of a Finite Field . . . . . . . . . . . . . . . 33822.2 Polynomial Codes . . . . . . . . . . . . . . . . . . . 34222.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 351

22.4 Additional Exercises: Error Correction for BCH Codes35322.5 References and Suggested Readings . . . . . . . . . . 354

23 Galois Theory 35623.1 Field Automorphisms . . . . . . . . . . . . . . . . . 35623.2 The Fundamental Theorem . . . . . . . . . . . . . . 36123.3 Applications . . . . . . . . . . . . . . . . . . . . . . . 36923.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 37523.5 References and Suggested Readings . . . . . . . . . . 377

A GNU Free Documentation License 379

Hints and Solutions to Selected Exercises 389

Notation 404


13/423

1

Preliminaries

A certain amount of mathematical maturity is necessary to ndand study applications of abstract algebra. A basic knowledgeof set theory, mathematical induction, equivalence relations, andmatrices is a must. Even more important is the ability to read andunderstand mathematical proofs. In this chapter we will outlinethe background needed for a course in abstract algebra.

1.1 A Short Note on Proofs

Abstract mathematics is dierent from other sciences. In labora-tory sciences such as chemistry and physics, scientists perform ex-periments to discover new principles and verify theories. Althoughmathematics is often motivated by physical experimentation or bycomputer simulations, it is made rigorous through the use of logi-cal arguments. In studying abstract mathematics, we take what iscalled an axiomatic approach; that is, we take a collection of ob-

jects

Sand assume some rules about their structure. These rules

are calledaxioms. Using the axioms for S, we wish to derive otherinformation aboutSby using logical arguments. We require thatour axioms be consistent; that is, they should not contradict oneanother. We also demand that there not be too many axioms. Ifa system of axioms is too restrictive, there will be few examples ofthe mathematical structure.

A statement in logic or mathematics is an assertion that iseither true or false. Consider the following examples:

3 + 56 13 + 8/2. All cats are black.

2 + 3 = 5.

2x= 6exactly when x= 4.

Ifax2 + bx + c= 0and a = 0, then

x=b b2 4ac

2a .

1


14/423

2 CHAPTER 1. PRELIMINARIES

x3 4x2 + 5x 6.

All but the rst and last examples are statements, and must beeither true or false.

A mathematical proof is nothing more than a convincingargument about the accuracy of a statement. Such an argumentshould contain enough detail to convince the audience; for instance,we can see that the statement 2x= 6exactly whenx = 4 is falseby evaluating2 4and noting that 6 = 8, an argument that wouldsatisfy anyone. Of course, audiences may vary widely: proofs canbe addressed to another student, to a professor, or to the reader ofa text. If more detail than needed is presented in the proof, thenthe explanation will be either long-winded or poorly written. Iftoo 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. Agood rule of thumb for an argument in an introductory abstractalgebra course is that it should be written to convince ones peers,whether those peers be other students or other readers of the text.

Let us examine dierent types of statements. A statement couldbe as simple as 10/5 = 2; however, mathematicians are usuallyinterested in more complex statements such as Ifp, thenq, wherepand qare both statements. If certain statements are known orassumed to be true, we wish to know what we can say about otherstatements. Herep is called the hypothesisand qis known as theconclusion. Consider the following statement: Ifax2 + bx + c= 0

and a = 0, then

x=b b2 4ac

2a .

The hypothesis is ax2 + bx + c= 0and a = 0; the conclusion is

x=b

b2

4ac

2a .

Notice that the statement says nothing about whether or not thehypothesis is true. However, if this entire statement is true andwe can show that ax2 +bx+ c = 0 with a= 0 is true, then theconclusion mustbe true. A proof of this statement might simply


15/423

1.1. A SHORT NOTE ON PROOFS 3

be a series of equations:

ax2 + bx + c= 0

x2 + b

ax= c

a

x2 + b

ax +

b

2a2

= b

2a2

c

ax +

b

2a

2=

b2 4ac4a2

x + b

2a=

b2 4ac2a

x=b b2 4ac

2a .

If we can prove a statement true, then that statement is calleda proposition. A proposition of major importance is called atheorem. Sometimes instead of proving a theorem or propositionall at once, we break the proof down into modules; that is, weprove 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, withvery little eort, be able to derive other related propositions calledcorollaries.

Some Cautions and Suggestions

There are several dierent strategies for proving propositions. Inaddition to using dierent methods of proof, students often makesome common mistakes when they are rst learning how to prove

theorems. To aid students who are studying abstract mathematicsfor the rst time, we list here some of the diculties that they mayencounter 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 thischapter and the remainder of the text.)

A theorem cannot be proved by example; however, the stan-dard way to show that a statement is not a theorem is toprovide a counterexample.

Quantiers are important. Words and phrases such as only,for all, for every, and for somepossess dierent meanings.

Never assume any hypothesis that is not explicitly stated inthe theorem. You cannot take things for granted.

Suppose you wish to show that an objectexistsand isunique.First show that there actually is such an object. To show thatit is unique, assume that there are two such objects, sayrands, and then show that r= s.


16/423


Sometimes it is easier to prove the contrapositive of a state-ment. Proving the statement If p, then q is exactly thesame as proving the statement If not q, then not p.

Although it is usually better to nd a direct proof of a the-orem, this task can sometimes be dicult. It may be easierto assume that the theorem that you are trying to prove isfalse, and to hope that in the course of your argument youare forced to make some statement that cannot possibly betrue.

Remember that one of the main objectives of higher mathemat-ics is proving theorems. Theorems are tools that make new andproductive applications of mathematics possible. We use examplesto give insight into existing theorems and to foster intuitions as towhat new theorems might be true. Applications, examples, andproofs are tightly interconnectedmuch more so than they mayseem at rst appearance.

1.2 Sets and Equivalence Relations

Set Theory

A set is a well-dened collection of objects; that is, it is denedin such a manner that we can determine for any given object xwhether or not xbelongs to the set. The objects that belong to aset are called its elements or members. We will denote sets bycapital letters, such as Aor X; ifais an element of the set A, we

write a A.A set is usually specied either by listing all of its elementsinside a pair of braces or by stating the property that determineswhether or not an object xbelongs to the set. We might write

X= {x1, x2, . . . , xn}

for a set containing elements x1, x2, . . . , xnor

X= {x: x satisesP}

if each x in Xsatises a certain propertyP. For example, if Eis the set of even positive integers, we can describe Eby writingeither

E= {2, 4, 6, . . .} or E= {x: x is an even integer and x >0}.

We write 2 Ewhen we want to say that 2 is in the set E, and3 / Eto say that3is not in the set E.


17/423

1.2. SETS AND EQUIVALENCE RELATIONS 5

Some of the more important sets that we will consider are thefollowing:

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 Zwhere q= 0};R = {x: x is a real number};

C = {z: z is a complex number}.

We can nd various relations between sets as well as performoperations on sets. A set A is a subsetof B, written A B orB A, if every element ofA is also an element ofB . For example,

{4, 5, 8} {2, 3, 4, 5, 6, 7, 8, 9}

andN

Z

Q

R

C.

Trivially, every set is a subset of itself. A set B is aproper subsetof a set A ifB Abut B =A. IfA is not a subset ofB , we writeA B; for example,{4, 7, 9} {2, 4, 5, 8, 9}. Two sets are equal,written A= B , if we can show that A B and B A.

It is convenient to have a set with no elements in it. This setis called theempty setand is denoted by. Note that the emptyset is a subset of every set.

To construct new sets out of old sets, we can perform certainoperations: the unionA B of two sets Aand B is dened as

A

B={

x: x

Aor x

B}

;

the intersectionofAand B is dened by

A B= {x: x Aand x B}.

IfA= {1, 3, 5}and B= {1, 2, 3, 9}, then

A B= {1, 2, 3, 5, 9} and A B= {1, 3}.

We can consider the union and the intersection of more than twosets. In this case we write

ni=1

Ai= A1 . . . An

andn

i=1

Ai= A1 . . . An

for the union and intersection, respectively, of the sets A1, . . . , An.


18/423


When two sets have no elements in common, they are said tobe disjoint; for example, ifEis the set of even integers and Oisthe set of odd integers, then Eand O are disjoint. Two sets A andB are disjoint exactly when A B = .

Sometimes we will work within one xed set U, called theuni-versal set. For any set A U, we dene the complementofA,denoted by A, to be the set

A = {x: x U and x / A}.We dene the dierenceof two sets Aand B to be

A \ B= A B = {x: x Aand x / B}.Example 1.1. LetRbe the universal set and suppose that

A= {x R : 0< x 3} and B= {x R : 2 x


19/423


Also, A \ A= A A = .(3) For sets A, B, and C,

A (B C) =A {x: x B or x C}= {x: x Aor x B, or x C}= {x: x Aor x B} C= (A B) C.

A similar argument proves that A (B C) = (A B) C.Theorem 1.3 (De Morgans Laws). LetA andB be sets. Then

1. (A B) =A B;2. (A B) =A B.

Proof. (1) We must show that(A B) A B and(A B) A

B. Let x

(A

B). Then x /

A

B. So xis neither in A

nor in B, by the denition of the union of sets. By the denitionof the complement, x A and x B. Therefore,x A B andwe have (A B) A B.

To show the reverse inclusion, suppose that x A B. Thenx A and x B, and so x / Aand x / B. Thus x / A B andsox (AB). Hence,(AB) AB and so(AB) =AB.

The proof of (2) is left as an exercise.

Example 1.4. Other relations between sets often hold true. Forexample,

(A \ B) (B \ A) = .

To see that this is true, observe that

(A \ B) (B \ A) = (A B) (B A)=A A B B= .

Cartesian Products and Mappings

Given sets A and B, we can dene a new set A B, called theCartesian productofAand B, as a set of ordered pairs. Thatis,

A B= {(a, b) :a Aand b B}.Example 1.5. IfA = {x, y},B = {1, 2, 3}, andC= , thenABis the set

{(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}and

A C= .


20/423


We dene the Cartesian product of n setsto be

A1 An= {(a1, . . . , an) :ai Aifor 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 ofA B are called relations. We will dene a map-pingor function f A B from a set A to a set B to be thespecial type of relation where (a, b) f if for every element a Athere exists a unique element b B. Another way of saying thisis that for every element in A, fassigns a unique element in B.

We usually write f : A B or A f B. Instead of writing downordered pairs (a, b) A B, we write f(a) =b or f : a b. ThesetAis called the domainoffand

f(A) = {f(a) :a A} B

is called the rangeor imageoff. We can think of the elementsin the functions domain as input values and the elements in thefunctions range as output values.

1

2

3

a

b

c

1

2

3

a

b

c

A B

A Bg

f

Figure 1.6: Mappings and relations

Example 1.7. Suppose A =

{1, 2, 3

}and B =

{a,b,c

}. In Fig-

ure 1.6 we dene relationsfand gfromAto B . The relation f isa mapping, but gis not because 1 Ais not assigned to a uniqueelement in B; that is, g(1) =a and g(1) =b.

Given a function f : A B, it is often possible to write alist describing what the function does to each specic element inthe domain. However, not all functions can be described in thismanner. For example, the function f : R Rthat sends each real


21/423


number to its cube is a mapping that must be described by writingf(x) =x3 or f :x x3.

Consider the relationf : Q Z given byf(p/q) =p. We knowthat 1/2 = 2/4, but is f(1/2) = 1or 2? This relation cannot be amapping because it is not well-dened. A relation iswell-denedif each element in the domain is assigned to a uniqueelement inthe range.

Iff :A B is a map and the image off is B, i.e., f(A) =B ,then f is said to be ontoor surjective. In other words, if thereexists ana Afor eachb Bsuch thatf(a) =b, thenfis onto. Amap is one-to-oneor injectiveifa1=a2 implies f(a1) =f(a2).Equivalently, a function is one-to-one iff(a1) =f(a2)impliesa1=a2. A map that is both one-to-one and onto is called bijective.

Example 1.8. Let f : Z Q be dened by f(n) = n/1. Thenf is one-to-one but not onto. Dene g : Q Z by g(p/q) = pwhere p/qis a rational number expressed in its lowest terms witha positive denominator. The function g is onto but not one-to-one.

Given two functions, we can construct a new function by usingthe range of the rst function as the domain of the second function.Let f : A B and g : B Cbe mappings. Dene a new map,thecompositionoffandg fromA to C, by(g f)(x) =g(f(x)).

A B C

1

2

3

a

b

c

X

Y

Z

f g

A C

1

2

3

X

Y

Z

g f

Figure 1.9: Composition of maps

Example 1.10. Consider the functions f :A Band g : B Cthat are dened in Figure 1.9 (top). The composition of thesefunctions,g f :A C, is dened in Figure 1.9 (bottom).Example 1.11. Letf(x) =x2 and g(x) = 2x + 5. Then

(f g)(x) =f(g(x)) = (2x + 5)2 = 4x2 + 20x + 25


22/423


and(g f)(x) =g(f(x)) = 2x2 + 5.

In general, order makes a dierence; that is, in most cases f g=g f.Example 1.12. Sometimes it is the case that f g = g f. Letf(x) =x

3

and g(x) = 3

x. Then(f g)(x) =f(g(x)) =f( 3x ) = ( 3x )3 =x

and(g f)(x) =g(f(x)) =g(x3) = 3

x3 =x.

Example 1.13. Given a 2 2matrix

A=

a b

c d

,

we can dene 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

c d

x

y

=

ax + by

cx + dy

.

Maps fromRn toRm given by matrices are called linear mapsorlinear transformations.

Example 1.14. Suppose thatS=

{1, 2, 3

}. Dene a map : S

Sby(1) = 2, (2) = 1, (3) = 3.

This is a bijective map. An alternative way to write is 1 2 3

(1) (2) (3)

=

1 2 3

2 1 3

.

For any set S, a one-to-one and onto mapping : S Sis calleda permutationofS.

Theorem 1.15. Let f : A B, g : B C, and h : C D.Then

1. The composition of mappings is associative; that is,(h g) f=h (g f);

2. If f and g are both one-to-one, then the mapping g f isone-to-one;

3. Iff andg are both onto, then the mappingg f is onto;


23/423


4. Iff andg are bijective, then so isg 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 Awe have

(h (g f))(a) =h((g f)(a))=h(g(f(a)))

= (h g)(f(a))= ((h g) f)(a).

(3) Assume thatfandg are both onto functions. Given c C,we must show that there exists an a

A such that (g

f)(a) =

g(f(a)) = c. However, since g is onto, there is an element b Bsuch thatg(b) =c. Similarly, there is an a Asuch 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 identitymapping from S to itself. Dene this map by id(s) = s for alls S. A map g: B Ais an inverse mappingoff :A B ifg f = idA and f g = idB; in other words, the inverse functionof a function simply undoes the function. A map is said to be

invertibleif it has an inverse. We usually write f1 for the inverseoff.

Example 1.16. The function f(x) =x3 has inverse f1(x) = 3

x

by Example 1.12.

Example 1.17. The natural logarithm and the exponential func-tions, f(x) = ln x and f1(x) = ex, are inverses of each otherprovided that we are careful about choosing domains. Observethat

f(f1(x)) =f(ex) =ln ex =x

and f1(f(x)) =f1(ln x) =elnx =x

whenever composition makes sense.

Example 1.18. Suppose that

A=

3 1

5 2

.


24/423


Then Adenes a map from R2 to R2 by

TA(x, y) = (3x + y, 5x + 2y).

We can nd an inverse map ofTA by simply inverting the matrixA; that is, T1A =TA1 . In this example,

A1 = 2 1

5 3

;

hence, the inverse map is given by

T1A (x, y) = (2x y, 5x + 3y).

It is easy to check that

T1A TA(x, y) =TA T1A (x, y) = (x, y).

Not every map has an inverse. If we consider the map

TB(x, y) = (3x, 0)

given by the matrix

B=

3 0

0 0

,

then an inverse map would have to be of the form

T1B (x, y) = (ax + by, cx + dy)

and(x, y) =T T1B (x, y) = (3ax + 3by, 0)

for all xand y. Clearly this is impossible because y might not be0.

Example 1.19. Given the permutation

=

1 2 3

2 3 1

on S= {1, 2, 3}, it is easy to see that the permutation dened by

1 = 1 2 33 1 2

is the inverse of . In fact, any bijective mapping possesses aninverse, as we will see in the next theorem.

Theorem 1.20. A mapping is invertible if and only if it is bothone-to-one and onto.


25/423


Proof. Suppose rst that f :A Bis invertible with inverse g :B A. Theng f=idAis the identity map; that is, g(f(a)) =a.Ifa1, a2 Awith f(a1) =f(a2), then a1=g(f(a1)) =g(f(a2)) =a2. Consequently, f is one-to-one. Now suppose that b B. Toshow that f is onto, it is necessary to nd an a A such thatf

(a

) =b

, butf

(g

(b

)) =b

withg

(b

) A

. Leta

=g

(b

).Conversely, let fbe bijective and let b B. Since f is onto,there exists an a Asuch that f(a) =b. Because f is one-to-one,a must be unique. Dene g by letting g(b) = a. We have nowconstructed the inverse off.

Equivalence Relations and Partitions

A fundamental notion in mathematics is that of equality. Wecan generalize equality with equivalence relations and equivalenceclasses. An equivalence relationon a set X is a relation RX

Xsuch that

(x, x) Rfor all x X(reexive property);

(x, y) Rimplies (y, x) R(symmetric property);

(x, y)and(y, z) Rimply(x, z) R(transitive property).

Given an equivalence relationRon a setX, we usually writex yinstead of (x, y) R. If the equivalence relation already has anassociated notation such as=,, or=, we will use that notation.

Example 1.21. Let p, q, r, and s be integers, where q and s

are nonzero. Dene p/q r/s if ps = qr. Clearly is reexiveand symmetric. To show that it is also transitive, suppose thatp/q r/s and r/s t/u, with q, s, and u all nonzero. Thenps= qrand ru= st. Therefore,

psu= qru= qst.

Since s = 0, pu= qt. Consequently, p/q t/u.

Example 1.22. Suppose thatfand g are dierentiable functionson R. We can dene an equivalence relation on such functionsby letting f(x) g(x)if f(x) = g(x). It is clear that is bothreexive and symmetric. To demonstrate transitivity, suppose thatf(x) g(x)and g(x) h(x). From calculus we know that f(x) g(x) =c1and g(x)h(x) =c2, wherec1and c2are both constants.Hence,

f(x) h(x) = (f(x) g(x)) + (g(x) h(x)) =c1 c2

and f(x) h(x) = 0. Therefore, f(x) h(x).


26/423


Example 1.23. For (x1, y1) and (x2, y2) in R2, dene (x1, y1)(x2, y2)ifx21+ y

21 =x

22+ y

22. Thenis an equivalence relation on

R2.

Example 1.24. LetA and B be2 2matrices with entries in thereal numbers. We can dene an equivalence relation on the set of2

2matrices, by saying A

Bif there exists an invertible matrix

Psuch that P AP1 =B . For example, if

A=

1 2

1 1

and B=18 33

11 20

,

then A B since P AP1 =B for

P =

2 5

1 3

.

LetIbe the 2 2identity matrix; that is,

I=1 0

0 1

.

Then IAI1 = IAI = A; therefore, the relation is reexive. Toshow symmetry, suppose that A B. Then there exists an invert-ible matrix Psuch that P AP1 =B . So

A= P1BP =P1B(P1)1.

Finally, suppose that A B and B C. Then there exist invert-ible matrices P and Qsuch that P AP1 = B and QBQ1 = C.

SinceC=QBQ1 =QPAP1Q1 = (QP)A(QP)1,

the relation is transitive. Two matrices that are equivalent in thismanner are said to be similar.

A partitionP of a set X is a collection of nonempty setsX1, X2, . . .such that Xi Xj =for i= j and

kXk = X. Let

be an equivalence relation on a set X and let x X. Then[x] ={y X : y x} is called the equivalence classof x. Wewill see that an equivalence relation gives rise to a partition via

equivalence classes. Also, whenever a partition of a set exists, thereis some natural underlying equivalence relation, as the followingtheorem demonstrates.

Theorem 1.25. Given an equivalence relation on a setX, theequivalence classes of X form a partition of X. Conversely, ifP ={Xi} is a partition of a set X, then there is an equivalencerelation onXwith equivalence classesXi.


27/423


Proof. Suppose there exists an equivalence relationon the setX. For any x X, the reexive property shows that x [x]andso [x] is nonempty. Clearly X=

xX[x]. Now let x, y X. We

need to show that either[x] = [y]or [x] [y] = . Suppose that theintersection of[x]and [y]is not empty and thatz [x] [y]. Thenz

x and z

y. By symmetry and transitivity x

y; hence,[x][y]. Similarly, [y][x]and so [x] = [y]. Therefore, any twoequivalence classes are either disjoint or exactly the same.

Conversely, suppose thatP ={Xi} is a partition of a set X.Let two elements be equivalent if they are in the same partition.Clearly, the relation is reexive. If x is in the same partition asy, 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 samepartition as z, then x must be in the same partition as z, andtransitivity holds.

Corollary 1.26. Two equivalence classes of an equivalence relation

are either disjoint or equal.

Let us examine some of the partitions given by the equivalenceclasses in the last set of examples.

Example 1.27. In the equivalence relation in Example 1.21, twopairs of integers,(p, q)and (r, s), are in the same equivalence classwhen they reduce to the same fraction in its lowest terms.

Example 1.28. In the equivalence relation in Example 1.22, twofunctionsf(x)and g(x)are in the same partition when they dierby a constant.

Example 1.29.We dened an equivalence class on R2 by(x1, y1) (x2, y2)ifx21+ y

21 =x

22+ y

22. Two pairs of real numbers are in the

same partition when they lie on the same circle about the origin.

Example 1.30. Let r and s be two integers and suppose thatn N. We say that r is congruent to s modulo n, or r iscongruent 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) since41 17 = 24is divisible by 8.We claim that congruence modulo n forms an equivalence relationofZ. Certainly any integer ris equivalent to itself since r r= 0is divisible byn. We will now show that the relation is symmetric.If r s (mod n), then r s =(s r) is divisible by n. Sos ris divisible byn and s r (mod n). Now suppose thatr s(mod n)ands t (mod n). Then there exist integers k and l suchthatr s= knand s t= ln. To show transitivity, it is necessaryto prove that r tis divisible by n. However,

r t= r s + s t= kn + ln= (k+ l)n,


28/423


and so r tis divisible by n.If we consider the equivalence relation established by the inte-

gers 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 thestudy of abstract algebra and will become quite useful in our inves-tigation of various algebraic structures such as groups and rings. Inour discussion of the integers modulo nwe have actually assumeda result known as the division algorithm, which will be stated andproved in Chapter 2.

1.3 Exercises

1. Suppose that

A= {x: x Nand xis even},B= {x: x Nand xis prime},C= {x: x Nand xis a multiple of 5}.

Describe each of the following sets.

(a) A B(b) B C (c) A B(d) A (B C)

2. If A ={a,b,c}, B ={1, 2, 3}, C ={x}, and D =, list all ofthe elements in each of the following sets.

(a) A B(b) B A

(c) A B C(d) A D

3. Find an example of two nonempty sets A and B for which A B= B Ais true.

4. Prove A =A and A = .

5. Prove A B=B Aand A B= B A.

6. Prove A (B C) = (A B) (A C).


29/423

1.3. EXERCISES 17

7. Prove A (B C) = (A B) (A C).

8. Prove A B if and only ifA B= A.

9. Prove (A B) =A B.

10. Prove A B= (A B) (A \ B) (B \ A).11. Prove (A B) C= (A C) (B C).

12. Prove (A B) \ B= .

13. Prove (A B) \ B=A \ B.

14. Prove A \ (B C) = (A \ B) (A \ C).

15. Prove A (B \ C) = (A B) \ (A C).

16. Prove (A \ B) (B \ A) = (A B) \ (A B).17. Which of the following relations f : Q Qdene a mapping?In each case, supply a reason why fis or is not a mapping.

(a) f(p/q) = p + 1

p 2(b) f(p/q) =

3p

3q

(c) f(p/q) = p + q

q2

(d) f(p/q) =3p2

7q2 p

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 Rdened by f(x) =ex(b) f : Z Zdened by f(n) =n2 + 3(c) f : R Rdened by f(x) =sin x(d) f : Z Zdened by f(x) =x2

19. Let f : A B and g : B Cbe invertible mappings; thatis, mappings such that f1 and g1 exist. Show that (g f)1 =f1 g1.

20. (a) Dene a function f : N Nthat is one-to-one but notonto.(b) Dene a function f : N Nthat is onto but not one-to-one.

21. Prove the relation dened on R2 by(x1, y1) (x2, y2)ifx21+y21 =x

22+ y

22 is an equivalence relation.

22. Letf :A B and g: B Cbe maps.


30/423


(a) If f and g are both one-to-one functions, show that g f isone-to-one.

(b) Ifg fis onto, show that gis onto.(c) Ifg f is one-to-one, show that f is one-to-one.(d) Ifg f is one-to-one and fis onto, show thatg is one-to-one.(e) Ifg

fis onto and g is one-to-one, show that f is onto.

23. Dene a function on the real numbers by

f(x) = x + 1

x 1 .

What are the domain and range of f? What is the inverse of f?Compute f f1 and f1 f.

24. Letf :X Ybe a map with A1, A2 Xand B1, B2 Y.(a) Prove f(A1 A2) =f(A1) f(A2).

(b) Provef(A1A2) f(A1)f(A2). Give an example in whichequality fails.(c) Prove f1(B1 B2) =f1(B1) f1(B2), where

f1(B) = {x X :f(x) B}.

(d) Prove f1(B1 B2) =f1(B1) f1(B2).(e) Prove f1(Y\ B1) =X \ f1(B1).

25. Determine whether or not the following relations are equiva-lence relations on the given set. If the relation is an equivalencerelation, describe the partition given by it. If the relation is not anequivalence relation, state why it fails to be one.

(a) x yin Rifx y(b) m nin Z ifmn >0(c) x yin Rif|x y| 4

(d) m n in Z if m n(mod 6)

26. Dene a relationon R2 by stating that (a, b) (c, d)if andonly if a2 +b2 c2 +d2. Show that is reexive and transitivebut not symmetric.

27. Show that an m n matrix gives rise to a well-dened mapfromRn to Rm.

28. Find the error in the following argument by providing a coun-terexample. The reexive property is redundant in the axiomsfor an equivalence relation. Ifx y, then y xby the symmet-ric property. Using the transitive property, we can deduce thatx x.


31/423

1.4. REFERENCES AND SUGGESTED READINGS 19

29. (Projective Real Line) Dene a relation on R2\{(0, 0)} by let-ting(x1, y1) (x2, y2)if there exists a nonzero real number suchthat (x1, y1) = (x2, y2). Prove that denes an equivalencerelation on R2 \ (0, 0). What are the corresponding equivalenceclasses? This equivalence relation denes the projective line, de-noted by P(R), which is very important in geometry.

1.4 References and Suggested Readings

[1] Artin, M. Abstract Algebra. 2nd ed. Pearson, Upper SaddleRiver, NJ, 2011.

[2] Childs, L. A Concrete Introduction to Higher Algebra. 2nded. Springer-Verlag, New York, 1995.

[3] Dummit, D. and Foote, R. Abstract Algebra. 3rd ed. Wiley,New York, 2003.

[4] Ehrlich, G.Fundamental Concepts of Algebra. PWS-KENT,Boston, 1991.

[5] Fraleigh, J. B.A First Course in Abstract Algebra. 7th ed.Pearson, Upper Saddle River, NJ, 2003.

[6] Gallian, J. A.Contemporary Abstract Algebra. 7th ed. Brook-s/Cole, Belmont, CA, 2009.

[7] Halmos, P. Naive Set Theory. Springer, New York, 1991.One of the best references for set theory.

[8] Herstein, I. N. Abstract Algebra. 3rd ed. Wiley, New York,1996.

[9] Hungerford, T. W. Algebra. Springer, New York, 1974. One

of the standard graduate algebra texts.

[10] Lang, S.Algebra. 3rd ed. Springer, New York, 2002. Anotherstandard graduate text.

[11] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed.Springer, New York, 1998.


32/423


[12] Mackiw, G. Applications of Abstract Algebra. Wiley, NewYork, 1985.

[13] Nickelson, W. K. Introduction to Abstract Algebra. 3rd ed.Wiley, New York, 2006.

[14] Solow, D. How to Read and Do Proofs. 5th ed. Wiley, NewYork, 2009.

[15] van der Waerden, B. L. A History of Algebra. Springer-Verlag, New York, 1985. An account of the historical de-velopment of algebra.


33/423

2

The Integers

The integers are the building blocks of mathematics. In this chap-ter we will investigate the fundamental properties of the integers,including mathematical induction, the division algorithm, and theFundamental Theorem of Arithmetic.

2.1 Mathematical InductionSuppose we wish to show that

1 + 2 + + n= n(n + 1)2

for any natural number n. This formula is easily veried for smallnumbers such as n = 1, 2, 3, or 4, but it is impossible to verify forall natural numbers on a case-by-case basis. To prove the formulatrue in general, a more generic method is required.

Suppose we have veried the equation for the rst 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= 1since

1 =1(1 + 1)

2 .

If we have veried the rst ncases, then

1 + 2 + + n + (n + 1) = n(n + 1)2

+ n + 1

= n2 + 3n + 2

2

=

(n + 1)[(n + 1) + 1]

2 .

This is exactly the formula for the (n + 1)th case.This method of proof is known as mathematical induction.

Instead of attempting to verify a statement about some subset Sof the positive integers N on a case-by-case basis, an impossibletask ifSis an innite set, we give a specic proof for the smallestinteger being considered, followed by a generic argument showing

21


34/423

22 CHAPTER 2. THE INTEGERS

that if the statement holds for a given case, then it must also holdfor the next case in the sequence. We summarize mathematicalinduction in the following axiom.

Principle 2.1 (First Principle of Mathematical Induction). LetS(n) be a statement about integers forn N and supposeS(n0) istrue 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 ngreater than or equal to n0.

Example 2.2. For all integers n 3, 2n > n + 4. Since

8 = 23 >3 + 4 = 7,

the statement is true for n0= 3. Assume that2k > k + 4for k 3.Then2k+1 = 2 2k >2(k+ 4). But

2(k+ 4) = 2k+ 8 > k+ 5 = (k+ 1) + 4

sincek is positive. Hence, by induction, the statement holds for allintegersn 3.

Example 2.3. Every integer 10n+1 + 3 10n + 5 is divisible by 9for n N. For n= 1,

101+1 + 3 10 + 5 = 135 = 9 15

is divisible by 9. Suppose that 10k+1 + 3 10k + 5is divisible by 9for k 1. Then

10(k+1)+1

+ 3 10k+1

+ 5 = 10k+2

+ 3 10k+1

+ 50 45= 10(10k+1 + 3 10k + 5) 45

is divisible by 9.

Example 2.4. We will prove the binomial theorem using mathe-matical induction; that is,

(a + b)n =n

k=0

n

k

akbnk,

wherea

andb

are real numbers,n

N, andn

k

=

n!

k!(n k)!

is the binomial coecient. We rst show thatn + 1

k

=

n

k

+

n

k 1

.


35/423

2.1. MATHEMATICAL INDUCTION 23

This result follows fromn

k

+

n

k 1

= n!

k!(n k)!+ n!

(k 1)!(n k+ 1)!=

(n + 1)!

k!(n + 1 k)!

= n + 1k .Ifn= 1, the binomial theorem is easy to verify. Now assume thatthe result is true for ngreater than or equal to 1. Then

(a + b)n+1 = (a + b)(a + b)n

= (a + b)

nk=0

n

k

akbnk

=n

k=0n

kak+1bnk +

n

k=0n

kakbn+1k

=an+1 +n

k=1

n

k 1

akbn+1k +n

k=1

n

k

akbn+1k + bn+1

=an+1 +n

k=1

n

k 1

+

n

k

akbn+1k + bn+1

=n+1k=0

n + 1

k

akbn+1k.

We have an equivalent statement of the Principle of Mathemat-ical Induction that is often very useful.

Principle 2.5(Second Principle of Mathematical Induction). LetS(n) be a statement about integers forn N and supposeS(n0) istrue for some integer n0. If S(n0), S(n0+ 1), . . . , S (k) imply thatS(k + 1)fork n0, then the statementS(n)is true for all integersn n0.

A nonempty subsetSofZ iswell-orderedifScontains a leastelement. Notice that the setZ is not well-ordered since it does notcontain a smallest element. However, the natural numbers are well-ordered.

Principle 2.6(Principle of Well-Ordering). Every nonempty sub-set of the natural numbers is well-ordered.

The Principle of Well-Ordering is equivalent to the Principle ofMathematical Induction.

Lemma 2.7. The Principle of Mathematical Induction implies that1 is the least positive natural number.


36/423


Proof. Let S ={n N : n 1}. Then 1 S. Now assumethat nS; that is, n1. Since n+ 11, n+ 1S; hence, byinduction, every natural number is greater than or equal to 1.

Theorem 2.8. The Principle of Mathematical Induction impliesthe Principle of Well-Ordering. That is, every nonempty subset of

N contains a least element.

Proof. We must show that ifSis a nonempty subset of the natu-ral numbers, thenScontains a least element. IfScontains 1, thenthe theorem is true by Lemma 2.7. Assume that ifScontains anintegerk such that1 k n, thenScontains a least element. Wewill show that if a set Scontains an integer less than or equal ton + 1, then Shas a least element. IfSdoes not contain an integerless than n + 1, then n + 1is the smallest integer in S. Otherwise,since Sis nonempty, Smust contain an integer less than or equal

to n. In this case, by induction, Scontains a least element.Induction can also be very useful in formulating denitions. For

instance, there are two ways to denen!, the factorial of a positiveintegern.

The explicitdenition: n! = 1 2 3 (n 1) n. The inductiveor recursivedenition: 1! = 1and n! =n(n

1)!for n >1.

Every good mathematician or computer scientist knows that look-ing at problems recursively, as opposed to explicitly, often resultsin better understanding of complex issues.

2.2 The Division Algorithm

An application of the Principle of Well-Ordering that we will useoften is the division algorithm.

Theorem 2.9(Division Algorithm). Letaandb be integers, withb >0. Then there exist unique integersqandr such that

a= bq+ r

where0 r < b.

Proof. This is a perfect example of the existence-and-uniquenesstype of proof. We must rst prove that the numbers qand r actu-ally exist. Then we must show that ifq and r are two other suchnumbers, then q= q and r= r .


37/423

2.2. THE DIVISION ALGORITHM 25

Existence ofqandr. Let

S= {a bk: k Zand a bk 0}.

If 0 S, then b divides a, and we can let q = a/b and r = 0.If0 / S, we can use the Well-Ordering Principle. We must rstshow that S is nonempty. If a > 0, then a

b

0

S. If a < 0,

thena b(2a) =a(1 2b) S. In either case S= . By the Well-Ordering Principle,Smust have a smallest member, say r = abq.Therefore, a = bq+ r, r 0. We now show that r < b. Supposethat 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 thenab(q+1) < abq, which would contradict the fact that r = abqis the smallest member ofS. So rb. Since 0 / S, r=b and sor < b.

Uniqueness of q and r. Suppose there exist integers r, r, q,and q such that

a= bq+ r, 0 r < b and a= bq + r, 0 r < b.

Thenbq+ r= bq + r. Assume thatr r. From the last equationwe have b(q q) = r r; therefore, b must divide r r and0 r r r < b. This is possible only if r r = 0. Hence,r= r and q= q.

Letaand bbe integers. Ifb= ak for some integer k, we writea| b. An integer d is called a common divisor of a and b ifd|

a and d|

b. The greatest common divisorof integers aand b is a positive integer d such that d is a common divisor ofa and b and if d is any other common divisor of a and b, thend | d. We write d= gcd(a, b); for example, gcd(24, 36) = 12andgcd(120, 102) = 6. We say that two integers a and b are relativelyprimeif gcd(a, b) = 1.

Theorem 2.10. Letaandbbe nonzero integers. Then there existintegersr ands such that

gcd(a, b) =ar + bs.

Furthermore, the greatest common divisor ofa andb is unique.

Proof. Let

S= {am + bn: m, n Zand am + bn >0}.

Clearly, the set Sis nonempty; hence, by the Well-Ordering Prin-ciple Smust have a smallest member, say d = ar+ bs. We claim


38/423


that d= gcd(a, b). Write a= dq+ r where 0r < d. Ifr >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 thesmallest member of S. Hence, r = 0and d divides a. A similarargument shows thatd dividesb. Therefore,d is a common divisorofaand b.

Suppose thatd is another common divisor ofaand b, and wewant to show that d | d. If we let a= dhand b= dk, then

d= ar+ bs= dhr+ dks = d(hr+ ks).

Sod must divided. Hence,dmust be the unique greatest common

divisor ofaand b.Corollary 2.11. Let a and b be two integers that are relativelyprime. Then there exist integersr ands such thatar+ bs= 1.

The Euclidean Algorithm

Among other things, Theorem 2.10 allows us to compute the great-est common divisor of two integers.

Example 2.12. Let us compute the greatest common divisor of945and 2415. First observe that

2415 = 945 2 + 525945 = 525 1 + 420525 = 420 1 + 105420 = 105 4 + 0.

Reversing our steps, 105 divides 420, 105 divides 525, 105 divides945, and 105 divides 2415. Hence, 105 divides both 945 and 2415.If dwere another common divisor of 945 and 2415, then d wouldalso 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 that105 = 525 + (1) 420

= 525 + (1) [945 + (1) 525]= 2 525 + (1) 945= 2 [2415 + (2) 945] + (1) 945= 2 2415 + (5) 945.


39/423

2.2. THE DIVISION ALGORITHM 27

So r =5and s = 2. Notice that r and s are not unique, sincer= 41and s= 16would also work.

To compute gcd(a, b) = d, we are using repeated divisions toobtain a decreasing sequence of positive integers r1 > r2 > >rn= d; that is,

b= aq1+ r1

a= r1q2+ r2

r1 = r2q3+ r3...

rn2 = rn1qn+ rn

rn1 = rnqn+1.

To ndr and s such thatar + bs= d, we begin with this last equa-tion and substitute results obtained from the previous equations:

d= rn

=rn2 rn1qn=rn2 qn(rn3 qn1rn2)= qnrn3+ (1 + qnqn1)rn2

...

=ra + sb.

The algorithm that we have just used to nd the greatest com-mon divisor dof two integers aand band to write das the linearcombination ofaand bis known as the Euclidean algorithm.

Prime Numbers

Let p be an integer such that p > 1. We say that p is a primenumber, or simply pis prime, if the only positive numbers thatdivide pare 1 and p itself. An integer n > 1that is not prime issaid to be composite.

Lemma 2.13 (Euclid). Leta and b be integers and p be a primenumber. Ifp | ab, then eitherp | a orp | b.

Proof. Suppose that p does not divide a. We must show thatp| b. Since gcd(a, p) = 1, there exist integers r and s such thatar+ps= 1. So

b= b(ar+ ps) = (ab)r+ p(bs).

Sincepdivides both aband itself, pmust divide b = (ab)r +p(bs).


40/423


Theorem 2.14(Euclid). There exist an innite number of primes.

Proof. We will prove this theorem by contradiction. Supposethat there are only a nite number of primes, say p1, p2, . . . , pn.Let P = p1p2 pn + 1. Then Pmust be divisible by some pifor 1

i

n. In this case, pi must divide P

p1p2

pn = 1,

which is a contradiction. Hence, either P is prime or there existsan additional prime number p =pithat divides P.

Theorem 2.15 (Fundamental Theorem of Arithmetic). Let n bean integer such thatn >1. Then

n= p1p2 pk,wherep1, . . . , pkare primes (not necessarily distinct). Furthermore,this factorization is unique; that is, if

n= q1q2 ql,

thenk = l and theqis are just thepis rearranged.

Proof. Uniqueness. To show uniqueness we will use induction onn. The theorem is certainly true for n = 2since in this case nisprime. Now assume that the result holds for all integers m suchthat1 m < n, and

n= p1p2 pk =q1q2 ql,wherep1 p2 pk and q1 q2 ql. By Lemma 2.13,p1| qifor some i= 1, . . . , land q1| pj for some j= 1, . . . , k. Sinceall of the pis and qis are prime, p1 = qi and q1 = pj. Hence,

p1= q1since p1 pj =q1 qi= p1. By the induction hypothesis,n =p2 pk =q2 ql

has a unique factorization. Hence, k = l and qi = pi for i =1, . . . , k.

Existence. To show existence, suppose that there is some inte-ger that cannot be written as the product of primes. Let Sbe theset of all such numbers. By the Principle of Well-Ordering,Shasa smallest number, say a. If the only positive factors of a are aand 1, then ais prime, which is a contradiction. Hence, a= a1a2where 1 < a1 < a and 1 < a2 < a. Neither a1 Snor a2 S,since ais the smallest element in S. So

a1= p1 pra2= q1 qs.

Therefore,a= a1a2= p1 prq1 qs.

Soa / S, which is a contradiction.


41/423

2.3. EXERCISES 29

Historical Note

Prime numbers were rst studied by the ancient Greeks. Twoimportant results from antiquity are Euclids proof that an innitenumber of primes exist and the Sieve of Eratosthenes, a methodof computing all of the prime numbers less than a xed positiveinteger n. One problem in number theory is to nd a function f

such that f(n)is prime for each integer n. Pierre Fermat (1601?1665) conjectured that 22

n+ 1was prime for all n, but later it was

shown by Leonhard Euler (17071783) that

225

+ 1 =4,294,967,297

is a composite number. One of the many unproven conjecturesabout prime numbers is Goldbachs Conjecture. In a letter to Eulerin 1742, Christian Goldbach stated the conjecture that every eveninteger 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 hasbeen veried for the numbers up through 4

1018, it has yet to be

proven in general. Since prime numbers play an important role inpublic key cryptography, there is currently a great deal of interestin determining whether or not a large number is prime.

Sage Sages original purpose was to support research in num-ber theory, so it is perfect for the types of computations with theintegers that we have in this chapter.

2.3 Exercises

1. Prove that

12 + 22 + + n2 = n(n + 1)(2n + 1)6

for n N.

2. Prove that

13 + 23 + + n3 = n2(n + 1)2

4

for n N.

3. Prove that n!> 2n for n 4.

4. Prove that

x + 4x + 7x + + (3n 2)x= n(3n 1)x2

for n N.


42/423


5. Prove that 10n+1 + 10n + 1is divisible by 3 for n N.

6. Prove that 4 102n + 9 102n1 + 5is divisible by 99 for n N.

7. Show thatn

a1a2 an 1

n

n

k=1

ak.

8. Prove the Leibniz rule for f(n)(x), wheref(n) is thenth deriva-tive off; that is, show that

(f g)(n)(x) =n

k=0

n

k

f(k)(x)g(nk)(x).

9. Use induction to prove that 1 + 2 + 22 + + 2n = 2n+1 1forn N.

10. Prove that1

2+

1

6+ + 1

n(n + 1)=

n

n + 1

for n N.

11. Ifxis a nonnegative real number, then show that(1+x)n1 nxfor n= 0, 1, 2, . . ..

12. (Power Sets) Let Xbe a set. Dene the power set of X,denotedP(X), to be the set of all subsets ofX. For example,

P({a, b}) = {, {a}, {b}, {a, b}}.

For every positive integer n, show that a set with exactlynelementshas a power set with exactly 2n elements.

13. Prove that the two principles of mathematical induction statedin Section 2.1 are equivalent.

14. Show that the Principle of Well-Ordering for the natural num-bers implies that 1 is the smallest natural number. Use this resultto show that the Principle of Well-Ordering implies the Principle

of Mathematical Induction; that is, show that ifS N such that1 Sand n + 1 Swhenever n S, then S= N.

15. For each of the following pairs of numbers aand b, calculategcd(a, b)and nd integers rand ssuch that gcd(a, b) =ra + sb.


43/423

2.3. EXERCISES 31

(a) 14 and 39

(b) 234 and 165

(c) 1739 and 9923

(d) 471 and 562

(e) 23,771 and 19,945

(f)4357and 3754

16. Let aand bbe nonzero integers. If there exist integers rand

ssuch that ar+ bs= 1, show that aand bare relatively prime.17. (Fibonacci Numbers) The Fibonacci numbers are

1, 1, 2, 3, 5, 8, 13, 21, . . . .

We can dene 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+1fn1= f2n+ (1)n, n 2.(c) Prove that fn= [(1 +

5 )n (1 5 )n]/2n5.

(d) Show that limn fn/fn+1= (

5

1)/2.

(e) Prove that fn and fn+1 are relatively prime.

18. Let aand b be integers such that gcd(a, b) = 1. Let r and sbe 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 xand ymust both be perfect squares.

20. Using the division algorithm, show that every perfect square

is of the form4kor 4k+ 1 for some nonnegative integer k .

21. Suppose that a,b,r,sare pairwise relatively prime and that

a2 + b2 =r2

a2 b2 =s2.Prove that a, r, and sare odd and bis even.

22. Letn N. Use the division algorithm to prove that every inte-ger is congruent mod n to precisely one of the integers0, 1, . . . , n1.Conclude that if r is an integer, then there is exactly one s in Z

such that 0s < n

and [r

] = [s

]. Hence, the integers are indeedpartitioned by congruence mod n.

23. Dene theleast common multipleof two nonzero integersaand b, denoted by lcm(a, b), to be the nonnegative integer msuchthat botha and b dividem, and ifa and b divide any other integern, thenm also dividesn. Prove that any two integers a and b havea unique least common multiple.


44/423


24. Ifd=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) = 1if and only if gcd(ab,c) = 1for integers a, b, and c.

27. Let a, b, c Z. Prove that if gcd(a, b) = 1 and a| bc, thena | c.

28. Let p2. Prove that if2p 1is prime, then pmust also beprime.

29. Prove that there are an innite number of primes of the form6n + 5.

30. Prove that there are an innite number of primes of the form4n

1.

31. Using the fact that 2 is prime, show that there do not existintegers pand qsuch that p2 = 2q2. Demonstrate that therefore

2cannot be a rational number.

2.4 Programming Exercises

1. (The Sieve of Eratosthenes) One method of computing all of theprime numbers less than a certain xed positive integer Nis to list

all of the numbersnsuch that1 < n < N. Begin by eliminating allof the multiples of 2. Next eliminate all of the multiples of 3. Noweliminate all of the multiples of 5. Notice that 4 has already beencrossed out. Continue in this manner, noticing that we do not haveto go all the way toN; it suces to stop at

N. Using this method,

compute all of the prime numbers less than N= 250. We can alsouse this method to nd all of the integers that are relatively primeto an integer N. Simply eliminate the prime factors ofNand allof their multiples. Using this method, nd all of the numbers thatare relatively prime to N= 120. Using the Sieve of Eratosthenes,write a program that will compute all of the primes less than anintegerN.

2. Let N0 = N {0}. Ackermanns function is the function A :N0 N0 N0 dened 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)).


45/423

2.5. REFERENCES AND SUGGESTED READINGS 33

Use this denition to computeA(3, 1). Write a program to evaluateAckermanns function. Modify the program to count the numberof statements executed in the program when Ackermanns functionis evaluated. How many statements are executed in the evaluationofA(4, 1)? What about A(5, 1)?

3. Write a computer program that will implement the Euclideanalgorithm. The program should accept two positive integers a andbas input and should output gcd(a, b)as well as integers r and ssuch that

gcd(a, b) =ra + sb.

2.5 References and Suggested Readings

[1] Brookshear, J. G. Theory of Computation: Formal Lan-guages, Automata, and Complexity. Benjamin/Cummings,

Redwood City, CA, 1989. Shows the relationships of thetheoretical aspects of computer science to set theory and theintegers.

[2] Hardy, G. H. and Wright, E. M.An Introduction to the The-ory of Numbers. 6th ed. Oxford University Press, New York,2008.

[3] Niven, I. and Zuckerman, H. S.An Introduction to the Theoryof Numbers. 5th ed. Wiley, New York, 1991.

[4] Vanden Eynden, C. Elementary Number Theory. 2nd ed.Waveland Press, Long Grove IL, 2001.


46/423

3

Groups

We begin our study of algebraic structures by investigating sets as-sociated with single operations that satisfy certain reasonable ax-ioms; that is, we want to dene an operation on a set in a way thatwill generalize such familiar structures as the integers Z togetherwith the single operation of addition, or invertible 2 2matricestogether with the single operation of matrix multiplication. Theintegers and the22matrices, together with their respective singleoperations, are examples of algebraic structures known as groups.

The theory of groups occupies a central position in mathemat-ics. Modern group theory arose from an attempt to nd the rootsof a polynomial in terms of its coecients. Groups now play a cen-tral role in such areas as coding theory, counting, and the studyof symmetries; many areas of biology, chemistry, and physics havebeneted from group theory.

3.1 Integer Equivalence Classes and Sym-

metries

Let us now investigate some mathematical structures that can beviewed as sets with single operations.

The Integers mod n

The integers modn

have become indispensable in the theory andapplications of algebra. In mathematics they are used in cryptog-raphy, coding theory, and the detection of errors in identicationcodes.

We have already seen that two integers aand bare equivalentmod n if n divides a b. The integers mod n also partition Zinto ndierent equivalence classes; we will denote the set of theseequivalence classes by Zn. Consider the integers modulo 12 and

34


47/423


48/423

36 CHAPTER 3. GROUPS

1. Addition and multiplication are commutative:

a + b b + a (mod n)ab ba (mod n).

2. Addition and multiplication are associative:

(a + b) + c a + (b + c) (mod n)(ab)c a(bc) (mod n).

3. There are both additive and multiplicative identities:

a + 0 a (mod n)a 1 a (mod n).

4. Multiplication distributes over addition:

a(b + c) ab + ac (mod n).

5. For every integera there is an additive inversea:

a + (a) 0 (mod n).

6. Leta be a nonzero integer. Thengcd(a, n) = 1 if and only ifthere exists a multiplicative inverseb fora (mod n); that is,

a nonzero integerb such that

ab 1 (mod n).

Proof. We will prove (1) and (6) and leave the remaining prop-erties to be proven in the exercises.

(1) Addition and multiplication are commutative modulo nsince the remainder of a+ b divided by n is the same as the re-mainder ofb + adivided by n.

(6) Suppose that gcd(a, n) = 1. Then there exist integersr andssuch thatar + ns= 1. Sincens = 1ar, it must be the case thatar 1 (mod n). Letting b be the equivalence class of r, ab 1(mod n).

Conversely, suppose that there exists an integer b such thatab 1 (mod n). Then n divides ab 1, so there is an integer ksuch thatab nk= 1. Let d=gcd(a, n). Since ddivides ab nk,dmust also divide 1; hence, d= 1.


49/423

3.1. INTEGER EQUIVALENCE CLASSES AND SYMMETRIES37

Symmetries

reection

horizontal axis

A

D

B

C

C

B

D

A

reection

vertical axis

A

D

B

C

A

D

B

C

180

rotation

A

D

B

C

D

A

C

B

identityA

D

B

C

B

C

A

D

Figure 3.5: Rigid motions of a rectangle

Asymmetryof a geometric gure is a rearrangement of the gurepreserving the arrangement of its sides and vertices as well as itsdistances and angles. A map from the plane to itself preserving

the symmetry of an object is called a rigid motion. For example,if we look at the rectangle in Figure 3.5, it is easy to see that arotation of180 or 360 returns a rectangle in the plane with thesame orientation as the original rectangle and the same relation-ship among the vertices. A reection of the rectangle across eitherthe vertical axis or the horizontal axis can also be seen to be asymmetry. However, a90 rotation in either direction cannot be asymmetry unless the rectangle is a square.


50/423


A

B

C

reection

B C

A

3=

A B C

B A C

A

B

C

reection

C A

B

2=

A B C

C B A

A

B

C

reection

A B

C

1=

A B C

A C B

A

B

C

rotation

B A

C

2=

A B C

C A B

A

B

C

rotation

C B

A

1= A B CB C A

A

B

C

identity

A C

B

id=

A B C

A B C

Figure 3.6: Symmetries of a triangle

Let us nd the symmetries of the equilateral triangleABC.To nd a symmetry ofABC, we must rst examine the permu-tations of the vertices A, B, and Cand then ask if a permutationextends to a symmetry of the triangle. Recall that a permuta-tionof a setSis a one-to-one and onto map : S S. The threevertices have 3! = 6 permutations, so the triangle has at most sixsymmetries. To see that there are six permutations, observe thereare three dierent possibilities for the rst vertex, and two for thesecond, and the remaining vertex is determined by the placementof the rst two. So we have 3 2 1 = 3! = 6dierent arrangements.To denote the permutation of the vertices of an equilateral triangle

that sendsA

toB

,B

toC

, andC

toA

, we write the arrayA B C

B C A

.

Notice that this particular permutation corresponds to the rigidmotion of rotating the triangle by 120 in a clockwise direction. Infact, every permutation gives rise to a symmetry of the triangle.All of these symmetries are shown in Figure 3.6.


51/423

3.2. DEFINITIONS AND EXAMPLES 39

A natural question to ask is what happens if one motion of thetriangleABCis followed by another. Which symmetry is 11;that is, what happens when we do the permutation1and then thepermutation 1? Remember that we are composing functions here.Although we usually multiply left to right, we compose functionsright to left. We have

(11)(A) =1(1(A)) =1(B) =C

(11)(B) =1(1(B)) =1(C) =B

(11)(C) =1(1(C)) =1(A) =A.

This is the same symmetry as 2. Suppose we do these motionsin the opposite order, 1 then 1. It is easy to determine thatthis is the same as the symmetry 3; hence, 11 = 11. Amultiplication table for the symmetries of an equilateral triangleABCis given in Table 3.7.

Notice that in the multiplication table for the symmetries ofan equilateral triangle, for every motion of the triangle there isanother motion such that = id; that is, for every motionthere is another motion that takes the triangle back to its originalorientation.

id 1 2 1 2 3id id 1 2 1 2 31 1 2 id 3 1 22 2 id 1 2 3 11 1 2 3 id 1 22 2 3 1 2 id 13 3 1 2 1 2 id

Table 3.7: Symmetries of an equilateral triangle

3.2 Denitions and Examples

The integers modnand the symmetries of a triangle or a rectangleare examples of groups. A binary operationor law of compo-sitionon a set G is a function G G G that assigns to eachpair (a, b)

G

Ga unique element a

b, or abin G, called the

composition ofaand b. Agroup(G, )is a set G together with alaw of composition(a, b) abthat satises the following axioms.

The law of composition is associative. That is,

(a b) c= a (b c)

for a, b, c G.


52/423


There exists an elemente 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 elementin

G, denoted by a1, such that

a a1

=a1

a= e.A group Gwith the property that a b= b afor all a, b G

is called abelian or commutative. Groups not satisfying thisproperty are said to be nonabelianor noncommutative.

Example 3.8. The integers Z = {. . . , 1, 0, 1, 2, . . .} form a groupunder the operation of addition. The binary operation on two inte-gers m, nZ is just their sum. Since the integers under additionalready have a well-established notation, we will use the operator+instead of; that is, we shall write m + ninstead ofm n. Theidentity is 0, and the inverse ofnZ is written asninstead ofn1

. Notice that the set of integers under addition have the addi-tional property that m + n= n + mand therefore form an abeliangroup.

Most of the time we will write abinstead ofa b; however, ifthe group already has a natural operation such as addition in theintegers, we will use that operation. That is, if we are adding twointegers, we still write m+ n,n for the inverse, and 0 for theidentity as usual. We also write m ninstead ofm + (n).

It is often convenient to describe a group in terms of an additionor multiplication table. Such a table is called a Cayley table.

Example 3.9. The integers mod n form a group under addition

modulo n. Consider Z5, consisting of the equivalence classes ofthe integers 0, 1, 2, 3, and 4. We dene the group operation onZ5 by modular addition. We write the binary operation on thegroup additively; that is, we write m+ n. The element 0 is theidentity of the group and each element in Z5 has an inverse. Forinstance,2+ 3 = 3+ 2 = 0. Table 3.10 is a Cayley table forZ5. ByProposition 3.4,Zn= {0, 1, . . . , n 1} is a group under the binaryoperation of addition mod n.

+ 0 1 2 3 4

0 0 1 2 3 41 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3

Table 3.10: Cayley table for (Z5, +)


53/423


Example 3.11. Not every set with a binary operation is a group.For example, if we let modular multiplication be the binary oper-ation on Zn, then Zn fails to be a group. The element 1 acts asa group identity since 1 k = k 1 = k for any k Zn; however,a multiplicative inverse for 0does not exist since 0 k = k 0 = 0for every k in Zn. Even if we consider the set Zn\ {0}, we stillmay not have a group. For instance, let 2

Z

6. Then 2 has no

multiplicative inverse since

0 2 = 0 1 2 = 22 2 = 4 3 2 = 04 2 = 2 5 2 = 4.

By Proposition 3.4, every nonzerok does have an inverse in Znifkis relatively prime ton. Denote the set of all such nonzero elementsin Zn byU(n). Then U(n)is a group called the group of unitsofZn. Table 3.12 is a Cayley table for the group U(8).

1 3 5 71 1 3 5 7

3 3 1 7 5

5 5 7 1 3

7 7 5 3 1

Table 3.12: Multiplication table for U(8)

Example 3.13. The symmetries of an equilateral triangle de-scribed in Section 3.1 form a nonabelian group. As we observed,it is not necessarily true that = for two symmetries and. Using Table 3.7, which is a Cayley table for this group, we caneasily check that the symmetries of an equilateral triangle are in-deed a group. We will denote this group by either S3 or D3, forreasons that will be explained later.

Example 3.14. We use M2(R)to denote the set of all 2 2ma-trices. Let GL2(R)be the subset ofM2(R)consisting of invertiblematrices; that is, a matrix

A=

a b

c d

is in GL2(R) if there exists a matrix A1 such that AA1 =A1A = I, where I is the 2 2 identity matrix. For A to havean inverse is equivalent to requiring that the determinant ofAbenonzero; that is, det A= adbc = 0. The set of invertible matrices


54/423


forms a group called the general linear group. The identity ofthe group is the identity matrix

I=

1 0

0 1

.

The inverse ofA

GL2(R)is

A1 = 1

ad bc

d bc a

.

The product of two invertible matrices is again invertible. Matrixmultiplication is associative, satisfying the other group axiom. Formatrices it is not true in general thatAB = B A; hence,GL2(R)isanother example of a nonabelian group.

Example 3.15. Let

1 = 1 00 1 I= 0 1

1 0J=

0 i

i 0

K=

i 0

0 i

,

where i2 =1. Then the relations I2 =J2 = K2 =1, IJ= K,JK = I, KI = J, J I =K, KJ =I, and IK =J hold.The setQ8= {1, I, J, K} is a group called the quaterniongroup. Notice that Q8 is noncommutative.

Example 3.16. Let Cbe the set of nonzero complex numbers.Under the operation of multiplication C forms a group. The iden-tity is 1. Ifz= a + bi is a nonzero complex number, then

z1 = a bia2 + b2

is the inverse ofz. It is easy to see that the remaining group axiomshold.

A group is nite, or has nite order, if it contains a nitenumber of elements; otherwise, the group is said to be inniteorto haveinnite order. Theorderof a nite group is the numberof elements that it contains. IfG is a group containing n elements,we write

|G

|= n. The group Z5 is a nite group of order 5; the

integers Z form an innite group under addition, and we sometimeswrite|Z| = .

Basic Properties of Groups

Proposition 3.17. The identity element in a group G is unique;that is, there exists only one elemente G such thateg= ge= gfor allg G.


55/423


Proof. Suppose that e and e are both identities in G. Theneg = ge = g and eg = ge = g for all g G. We need to showthat e = e. If we think of eas the identity, then ee = e; but ife is the identity, then ee =e. Combining these two equations, wehave e= ee =e.

Inverses in a group are also unique. If g and g are bothinverses of an element g in a group G, then gg = gg = e andgg = gg = e. We want to show that g = g, but g = ge =g(gg ) = (gg)g = eg = g. We summarize this fact in thefollowing proposition.

Proposition 3.18. If g is any element in a group G, then theinverse ofg, denoted byg1, is unique.

Proposition 3.19. Let G be a group. If a, b G, then (ab)1 =b1a1.

Proof. Let a, b G. Then abb1a1 = aea1 = aa1 = e.Similarly, b1a1ab= e. But by the previous proposition, inversesare unique; hence, (ab)1 =b1a1.

Proposition 3.20. LetG be a group. For anya G,(a1)1 =a.

Proof. Observe that a1(a1)1 =e. Consequently, multiplyingboth sides of this equation by a, we have

(a1)1 =e(a1)1 =aa1(a1)1 =ae = a.

It makes sense to write equations with group elements andgroup operations. Ifaand bare two elements in a group G, doesthere exist an elementx Gsuch that ax= b? If such an xdoesexist, is it unique? The following proposition answers both of thesequestions positively.

Proposition 3.21. Let G be a group and a and b be any twoelements inG. Then the equationsax= b andxa= b have uniquesolutions inG.

Proof. Suppose that ax = b. We must show that such an xexists. Multiplying both sides ofax = b by a1, we have x = ex =a1ax= a1b.

To show uniqueness, suppose thatx1and x2are both solutionsof ax = b; then ax1 = b = ax2. So x1 = a1ax1 = a1ax2 = x2.The proof for the existence and uniqueness of the solution ofxa = bis similar.


56/423


Proposition 3.22. If G is a group anda,b, c G, then ba= caimpliesb= c andab= ac impliesb= c.

This proposition tells us that the right and left cancellationlawsare true in groups. We leave the proof as an exercise.

We can use exponential notation for groups just as we do inordinary algebra. IfGis a group andg

G, then we dene g0 =e.

For n N, we denegn =g g g

n times

andgn =g1 g1 g1

n times

.

Theorem 3.23. In a group, the usual laws of exponents hold; thatis, for allg, h G,

1. gmgn =gm+n for allm, n Z;2. (gm)n =gmn for allm, n Z;3. (gh)n = (h1g1)n for all n Z. Furthermore, if G is

abelian, then(gh)n =gnhn.

We will leave the proof of this theorem as an exercise. Noticethat(gh)n=gnhn in general, since the group may not be abelian.If the group is Z or Zn, we write the group operation additivelyand the exponential operation multiplicatively; that is, we writenginstead ofgn. The laws of exponents now become

1. mg+ ng= (m + n)gfor all m, n

Z;

2. m(ng) = (mn)gfor all m, n Z;3. m(g+ h) =mg + mhfor all n Z.It is important to realize that the last statement can be made

only because Zand Zn are commutative groups.

Historical Note

Although the rst clear axiomatic denition of a group was notgiven until the late 1800s, group-theoretic methods had been em-

ployed before this time in the development of many areas of math-ematics, 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, variste Galois (18111832) succeeded in developing themathematics necessary to determine exactly which polynomial equa-tions could be solved in terms of the polynomialscoecients. Ga-lois primary tool was group theory.


57/423

3.3. SUBGROUPS 45

The study of geometry was revolutionized in 1872 when FelixKlein proposed that geometric spaces should be studied by exam-ining those properties that are invariant under a transformation ofthe space. Sophus Lie, a contemporary of Klein, used group the-ory to study solutions of partial dierential equations. One of therst modern treatments of group theory appeared in William Burn-sides The Theory of Groups of Finite Order [1], rst published in1897.

3.3 Subgroups

Denitions and Examples

Sometimes we wish to investigate smaller groups sitting inside alarger group. The set of even integers 2Z = {. . . , 2, 0, 2, 4, . . .}isa group under the operation of addition. This smaller group sitsnaturally inside of the group of integers under addition. We deneasubgroupHof a groupG to be a subsetHofG such

Documents

abstract algebra theory and practice