120486729 Shaum Group Theory

7/29/2019 120486729 Shaum Group Theory

1/290

^^,:v-.- ,--.'.-...- ..,.-;.^. ---..- ..

iHAUM'SOUTLINESERIES

THEORY and PROBLEMSofGROUPTHEORY

by B. BAUMSLAG and B. CHANDLER

including600problems

Complefely Solved in Detail

SCHAUM'S OUTLINE SERIESMcGRAW-HlLL BOOK COMPANY


2/290

1 .'^

SCHAVM'S OVTLIISE OFTHEORY AIVD PROBLEMS

OF

GROrP THEORY

BY

BENJAMIN BAUMSLAG, Ph.D.BRUCE CHANDLER, Ph.D.

Department of MathematicsNew York University

SCHAVM'S OUTLIBTE SERIESMcGRAW-HILL BOOK COMPANY

New York, St. Louis, San Francisco, Toronto, Sydney


3/290

Copyright 1968 by McGraw-Hill, Inc. All Rights Reserved. Printed in theUnited States of America. No part of this publication may be reproduced,stored in a retrieval system, or transmitted, in any form or by any means,electronic, mechanical, photocopying, recording, or otherwise, without theprior written permission of the publisher.04124

34567890 MHUN 8 2 10 6 9

Typography by Signs and Symbols, Inc., New York, N. Y.


4/290

PrefaceThe study of groups arose early in the nineteenth century in connection with the solu-

tion of equations. Originally a group was a set of permutations with the property thatthe combination of any two permutations again belongs to the set. Subsequently thisdefinition was generalized to the concept of an abstract group, which was defined to be aset, not necessarily of permutations, together with a method of combining its elementsthat is subject to a few simple laws.

The theory of abstract groups plays an important part in present day mathematicsand science. Groups arise in a bewildering number of apparently unconnected subjects.Thus they appear in crystallography and quantum mechanics, in geometry and topology, inanalysis and algebra, in physics, chemistry and even in biology.

One of the most important intuitive ideas in mathematics and science is symmetry.Groups can describe symmetry; indeed many of the groups that arose in mathematics andscience were encountered in the study of symmetry. This explains to some extent whygroups arise so frequently.

Although groups arose in connection with other disciplines, the study of groups is initself exciting. Currently there is vigorous research in the subject, and it attracts theenergies and imagination of a great many mathematicians.

This book is designed for a first course in group theory. It is mainly intended forcollege and first year graduate students. It is complete in itself and can be used forself-study or as a text for a formal course. Moreover, it could with advantage be usedas a supplement to courses in group theory and modern algebra. Little prior knowledge isassumed. The reader should know the beginnings of elementary numbet theory, a summaryof which appears in Appendix A. An acquaintance with complex numbers is needed forsome problems. In short, a knowledge of high school mathematics should be a sufficientprerequisite, and highly motivated and bright high school students will be able to under-stand much of this book.

The aim of this book is to make the study of group theory easier. Each chapter beginswith a preview and ends with a summary, so that the reader may see the ideas as a whole.Each main idea appears in a section of its own, is motivated, is explained in great detail,and is made concrete by solved problems.

Chapter 1 presents the rudiments of set theory and the concept of binary operation,which are fundamental to the whole subject. Chapter 2, on groupoids, further exploresthe concept of binary operation. In most courses on group theory the concept of groupoidis usually treated briefly if at all. We have chosen to treat it more fully for the followingreasons: (a) A thorough understanding of binary compositions is thereby obtained, (b) Theimportant ideas of homomorphism, isomorphism and Cayley's theorem occur both in thechapter on groupoids and in the chapters on groups, and the repetition ensures familiarity.

Chapter 3 shows that the concept of group is natural by producing a large number ofexamples of groups that arise in different fields. Here are discussed groups of real andcomplex numbers, the symmetric groups, symmetry groups, dihedral groups, the group ofMobius transformations, automorphism groups of groupoids and fields, groups of matrices,and the full linear group.

Chapter 4 is concerned with the homomorphism theorems and cyclic groups. Theconcept of homomorphism is fundamental, and thus the theorems of this chapter areindispensable for further study.


5/290

Chapter 5 is on finite groups. The Sylow theorems are proved, the concept of externaldirect product is introduced, and groups up to order 15 are classified. The chapter con-cludes with the Jordan-Holder theorem and a proof that most alternating groups are simple.

Chapter 6 is on abelian groups. Two important classes of abelian groups are treated:finitely generated and divisible groups. Undergraduates will probably find their needsare met by the material through Section 6.3. Graduate students will certainly want tocontinue.

Chapter 7 is on permutational representations and extensions. Chapter 8 is on freegroups and presentations. Those who would like to study the theory of groups moredeeply will find a guide to the literature at the end of the book.

Chapters 1-4 must be read in order, although, if desired, only the first three sections ofChapter 3 need be read at first (the other sections of Chapter 3 may be studied when theyare needed). The order of reading Chapters 5-8 can be varied, although part of Chapter 7is required for the last sections of Chapter 8.

The reader need not work all the solved problems; he should decide for himself howmuch practice he needs. Some of the problems are designed to clarify the immediatelypreceding text, and the reader will find that the solutions may overcome some of hisobstacles. On the whole, however, it is advisable to attempt the problems before readingtheir solutions. The numerous supplementary problems, some of which are very difficult,serve as a review of the material of each chapter.

We thank Prof. Gilbert Baumslag for giving us access to several chapters of unpub-lished notes and for many useful suggestions. We thank Sister Weiss for reading twochapters of an early draft, Harold Brown for much helpful advice, Henry Hayden fortypographical arrangement and art work, and Louise Baggot for the typing. Finally weexpress our appreciation to Daniel Schaum and Nicola Monti for their unfailing editorialcooperation.

B. BaumslagB. Chandler

June 1968


6/290

CONTENTSPage

Chapter 1 SETS, MAPPINGS AND BINARY OPERATIONSPreview of Chapter 1 11.1 SETS 1

a. Basic notions, b. Union and intersection.1.2 CARTESIAN PRODUCTS 6

a. Definition, b. Equivalence relations, c. Partitions and equivalence relations.d. The division notation.

1.3 MAPPINGS 11a. Definition of mapping, b. Formal definition of mapping, c. Types ofmappings.

1.4 COMPOSITION OF MAPPINGS 171.5 BINARY OPERATIONS 19

a. Definition, b. The multiplication table.A look back at Chapter 1 24

Chapter 2 GROUPOIDSPreview of Chapter 2 262.1 GROUPOIDS 26

a. Definition of a groupoid. b. Equality of groupoids.2.2 COMMUTATIVE AND ASSOCIATIVE GROUPOIDS 292.3 IDENTITIES AND INVERSES IN GROUPOIDS 30

a. The identity of a groupoid. b. Inverses in a groupoid.2.4 SEMIGROUPS WITH AN IDENTITY ELEMENT 33

a. Uniqueness of inverses, b. The semigroup of mappings of a set into itself,c. Notation for a mapping, d. The order in a product.

2.5 HOMOMORPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM 40a. Definition of a homomorphism. b. Epimorphism, monomorphism, and iso-morphism, c. Properties of epimorphisms. d. Naming and isomorphisms, e. Mxand semigroups.

A look back at Chapter 2 47

Chapter B GROUPS AND SUBGROUPSPreview of Chapter 3 503.1 GROUPS 50

Definition. Examples of groups of numbers.3.2 SUBGROUPS 543.3 THE SYMMETRIC AND ALTERNATING GROUPS 56

a. The symmetric group on X. b. Even and odd permutations, c. The alter-nating groups, d. The order of A.

3.4 GROUPS OP ISOMETRIES 64a. Isometrics of the line. b. Two points determine an isometry. c. Isometriesof the plane, d. Isometries are products of reflections, translations and rota-tions, e. Symmetry groups, f. The dihedral groups.

3.5 THE GROUP OF MOBIUS TRANSFORMATIONS 77a. Defining the group, b. 2 X 2 matrices.

3.6 SYMMETRIES OF AN ALGEBRAIC STRUCTURE 83a. Automorphisms of groupoids. b. Fields of complex numbers, c. Automor-phisms of fields, d. Vector spaces, e. Linear transformations. The full lineargroup.



7/290

CONTENTSChapter 4 (SOMORPHISM THEOREMS Page

Preview of Chapter 4 944.1 FUNDAMENTALS 94

a. Preliminary remarks, b. More about subgroups, c. Exponents.4.2 CYCLIC GROUPS 101

a. Fundamentals of cyclic groups, b. Subgroups of cyclic groups.4.3 COSETS 107

a. Introduction to the idea of coset. b. Cosets form a partition. Lagrange'stheorem, c. Normal subgroups, d. Commutator subgroups, centralizers, nor-malizers. e. Factor groups.

4.4 HOMOMORPHISM THEOREMS 117a. Homomorphisms and factor groups: The homomorphism theorem, b. Corre-spondence theorem. Factor of a factor theorem, c. The subgroup isomorphismtheorem, d. Homomorphisms of cyclic groups.


Chapter 5 FINITE GROUPSPreview of Chapter 5 1305.1 THE SYLOW THEOREMS 130a. Statements of the Sylow theorems, b. Two lemmas used in the proof of the

Sylow theorems, c. Proofs of the Sylow theorems.5.2 THEORY OF p-GROUPS 139

a. The importance of p-groups in finite groups, b. The center of a p-group.c. The upper central series.

5.3 DIRECT PRODUCTS AND GROUPS OF LOW ORDER 143a. Direct products of groups, b. Groups of small order: orders p and 2p.c. Groups of small order: orders 8 and 9. d. Groups of small order: orders 12and 15.

5.4 SOLVABLE GROUPS 158a. Definition of solvable groups, b. Properties of, and alternative definitionfor, solvable groups.

5.5 COMPOSITION SERIES AND SIMPLE GROUPS 163a. The Jordan-Holder theorem, b. Proof of Jordan-Holder theorem, c. Cyclesand products of cycles, d. Transpositions, and even and odd permutations,e. The simplicity of A, n 5.

A look back at Chapter 5 174Chapter S ABELIAN GROUPS

Preview of Chapter 6 1776.1 PRELIMINARIES 178

a. Additive notation and finite direct sums. b. Infinite direct sums. c. Thehomomorphic property of direct sums and free abelian groups.

6.2 SIMPLE CLASSIFICATION OF ABELIAN GROUPS,AND STRUCTURE OF TORSION GROUPS 188a. Tentative classifications: torsion, torsion-free, and mixed, b. The torsionsubgroup, c. Structure of torsion groups. Priifer groups, d. Independenceand rank.

6.3 FINITELY GENERATED ABELIAN GROUPS 196a. Lemmas for finitely generated free abelian groups, b. Fundamental theoremof abelian groups, c. The type of a finitely generated abelian group, d. Sub-groups of finitely generated abelian groups.

6.4 DIVISIBLE GROUPS 205a. p-Prufer groups. Divisible subgroups, b. Decomposition theorem fordivisible groups.



8/290

CONTENTSChapter 7 PERMUTATION AL REPRESENTATIONS Page

Preview of Chapter 7 2147.1 CAYLEY'S THEOREM 214

a. Another proof of Cayley's theorem, b. Cayley's theorem and examples ofgroups.

7.2 PERMUTATIONAL REPRESENTATIONS 2167.3 DEGREE OF A REPRESENTATION AND FAITHFUL REPRESENTATIONS.. 217

a. Degree of a representation, h. Faithful representations.7.4 PERMUTATIONAL REPRESENTATIONS ON COSETS 2187.5 FROBENIUS' VARIATION OF CAYLEY'S THEOREM 222

a. The kernel of a coset representation, b. Frobenius' theorem.7.6 APPLICATIONS TO FINITELY GENERATED GROUPS 227

a. Subg-roups of finite index, b. Remarks about the proof of Theorem 7.4.c. Marshall Hall's theorem, d. One consequence of Theorem 7.5.

7.7 EXTENSIONS 232a. General extension, b. The splitting extension, c. An analysis of splittingextensions, d. Direct product.

7.8 THE TRANSFER 240a. Definition, b. Proof that t is a homomorphism. c. Proof that t is independentof the choice of transversal, d. A theorem of Schur.

A look back at Chapter 7 243Chapter 8 FREE GROUPS AND PRESENTATIONS

Preview of Chapter 8 2458.1 ELEMENTARY NOTIONS 245

a. Definition of a free group, b. Length of an element. Alternative descriptionof a free group, c. Existence of free groups, d. Homomorphisms of freegroups.

8.2 PRESENTATIONS OF GROUPS 253a. Definitions, b. Illustrations of presentations.

8.3 THE SUBGROUP THEOREM FOR FREE GROUPS: AN EXAMPLE 2598.4 PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS 260

a. Plan of the proof, b. Schreier transversals, c. A look at the elements a^^.d. The proof of the subgroup theorem, e. Subgroups of finite index, f. Inter-section of finitely generated subgroups.

A look back at Chapter 8 266NUMBER THEORY ..A GUIDE TO THE LilINDEXSYMBOLS AND NOTATIONS 278

Appendix A NUMBER THEORY 269Appendix B A GUIDE TO THE LITERATURE 270

INDEX 274


9/290


10/290

chapter 1

Sets, Mappings and Binary OperationsPreview of Chapter 1

This chapter begins with a few remarks about sets. A set is a collection of objects.For example, the real numbers form a set, the objects being the numbers.The real numbers have an operation called addition. Addition essentially involves two

numbers, for the addition of a single number is meaningless, while the addition of threeor more numbers is repeated addition of two numbers. Because addition involves twonumbers it is called a binary operation.The main object of this chapter is to define precisely the notion of a binary operation.The concept of binary operation is required to define the concept of group.We introduce the important ideas of cartesian product and mapping. Welding themtogether gives rise to an explicit definition of a binary operation. Another important ideais that of equivalence relation, which is a generalization of the idea of equality. The readerwill also pick up much useful notation.1.1 SETSa. Basic notions

Set is synonymous with collection. The objects in a set are termed the elements ofthe set. Usually we denote sets by capital Latin letters, for example, B, G, T. We shalldenote

(i) the set of positive integers 1, 2, 3, ... by P(ii) the set of nonnegative integers 0, 1, 2, ... by N(iii) the set of all integers by Z(iv) the set of rational numbers by Q(v) the set of real numbers by R(vi) the set of complex numbers by C.The elements of a set will usually be denoted by small Latin letters such as s, t, u, etc.By s G S we mean "s is an element of S" or "s belongs to S". In particular, 2 GP. It s

is not an element of S, we write s ^ S and read this as "s is not an element of S" or "s doesnot belong to S" or "s is not in S". For example, 1 P.

In dealing with sets it is advantageous to abbreviate the phrase "the set whose elementsare" by using braces. Thus, for example, we write {1,2} for the set whose elements are 1and 2 and similarly we write {1, 0, 1, 2, ... } for the set whose elements are 1, 0, 1, 2, ... .A variation of this notation is useful to describe a set in terms of a property which singlesout its elements. Thus we write {x

\x has the property T} for the set of all those elements

X which have the property 'P. Here "iP stands for some "understandable" property; toillustrate: {a; | a; is a real number} is R, the set of real numbers. (T here is the property ofbeing a real number.) Notice that we read {a; | a; is a real number} as the set of all thoseelements x which have the property that a; is a real number, or the set of all those elementsX such that a; is a real number.


11/290

B

2 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1We can now introduce some useful notation.

(i) If a and b are real numbers and a


12/290

Sec. 1.1] SETS 3(iii) True, a is the only element of {a} and a S {a, b}. But 6 G {a, 6} and 6 {a}. Conse-

quently {a.} ^ {o, 5}.(iv) False. 2 {2,3} and 2S{3,4}.(v) True. Any positive integer is a rational number but not all rational numbers are positive

integers.(vi) True. All integers are rational numbers.(vii) True. All rational numbers are real numbers but B = Q.(viii) True. For if a e R, then a = a + Ot G C. But V-i ^ R implies R = C.(ix) False. Q = Q.(x) True. 3, a, b and c are the only elements of {3, a, b, 3, c, 6}. Therefore the sets are equal.

1.3. Are the following statements true?(i) Z {x\ X is real and x > 0}(ii) N = {x\ xeQ and x^ 0}(iii) Q = {x\ X = alb, where 6 # and a,b e Z}(iv) P = {x\ xGN and x^^ 1}(v) C = {x\ X = u + iv, where u,v G R and fi = 1}(vi) Z = {a;

Ia; is real and x^ S P}

Solution:(i) False. {x\ a; is real and a; > 0} has no negative elements.(ii) False, f S {x | x GQ and x ^ 0} but f N. Hence the sets are not equal.(iii) True. A rational number is defined as the set of all numbers of the form a/b where b =and a,b & Z.(iv) True. For if x & P, then xGN and x^ ^ 1. Thus xe.{x\ xGN and x^ ^ 1}. Now if

X G N and a;2 ^ i_ i^ a; G {a; | a; G A^ and a;2 ss 1}, then x = 0. Hence a; G P, as the onlyelement in A^ which is not in P is zero.(y) True. The property that x ~ u + iv where u,v e. R and ? = 1 is the defining property

for complex numbers.(vi) False, x^ G P implies x = 0. But G Z.

b. Union and intersectionLet S and T be sets. Then the union of S and T, written SUT and read "S union T", is

defined as the set whose elements are either in S or in T (or in both S and T). For example,(1,2,3) U {2,5,6} ={1,2,3,5,6} and PU{0} = N. Clearly, ScSUT and TcSUT.Indeed it follows from the definition of SuT that any set containing both S and T containsSU T, so we say SU T is the smallest set containing S and T.

Similarly it [S,T,U . . .} is any set of sets, we define SuTU C7U , the union of S andT and U and . . . , to be the set whose elements are the elements that belong to at least one ofthe sets S,T,U, ... .SuTuUU . . . is said to be the smallest set containing the sets S,T,U, . . . . To illustrate,{1,2}U{3,4}U{5,6}U = P.

If S and T are sets, we may consider the common part or intersection of S and T. Theintersection is denoted by SnT and read as "S intersection T". For example, supposeS= {1,2,3} and 7= {2,5,6}. Then SnT={2}. Repeating the definition, SnT is theset of those elements which belong simultaneously to S and to T. Here the possibility arisesthat there are no elements of S which belong also to T. We shall agree to the conventionthat there is a set, which we denote by 0, with no elements. Again we shall agree to theconvention that the empty set is a subset of every set. Two sets are termed disjoint ifthey have an empty intersection. Thus {1,2} and {3,4} are disjoint. This notion of inter-section can be generalized to any number of sets in the same way that the notion of unionwas generalized from two to any number. To be precise, the intersection of sets S,T,U, . . .,written SnTnUn -, is the set of all those elements which belong simultaneously to S,to r, to C7, . . . . Notice that SnT can be thought of as the largest subset of S which is alsoa subset of T. Similarly SnTnUn is the largest subset of S which is contained inT and in U and in ... .


13/290

4 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1If S qT we define T S, the difference between T and S, by

T - S = {x\ X GT and x^S}Thus if r= {1,2,3,4} and S- {1,2}, then T-S= {3,4}. For any sets T and S suchthat TdS, T-(T-S) = S (1.1)

We prove equation (1.1) by showing thatright side of equation {1.1) c left side of equation (1 .1)

and left side of equation (1.1) C right side of equation (1.1)To do this suppose x G S. Then clearly xGT but x^T-S. So x G r-(r-S); inother words, the right side of equation {1.1) is contained in the left side of equation {1.1).The reverse inclusion is obtained similarly. Suppose xGT {T S). Then xGT andX ^T S. Therefore x GT and x GS, i.e. xGS. So the left side is contained in theright side and we have proved equation {1.1) by virtue of Proposition 1.1.Problems1.4. Check that the following statements are correct:

(i) {1,2} U {1,2,3,4} = {1,2,3,4}(ii) {a,e} U {e,/} U {ff.fi.} = {.a,e,j,g,h)(iii) {...,-2,-1,0} U {0,1,2,...} = Z(iv) If a


14/290

Sec. 1.1] SETS 5

1.7. If S, T and U are any three sets, prove the following:(i) sur=rus (viii) SnS = s(li) SnT=Tr\S (ix) Sli(TuV) = SuTuU(iii) S cSuT and S cTuS (x) Sn(rn C7) = (Sn r)n(iv) SnTQS and TnS c S (xi) Su(rn [/) = (Sur)n(Su t/)(v) Su0=S (xii) S-S =(vi) Sn0 = (xiii) S - (S - S) = S(vii) SuS = SSolution:(i) Let xGSdT. By the definition of union of sets, x&S or x G T. Hence a; TuS andSuTcTuS. Similarly if xGTuS, it follows that TuScSuT. Consequently SuT =TuS by Proposition 1.1.(ii) If a; S SnT, then, by the definition of intersection, x G S and x G T. a; is therefore an ele-

ment of rnS and SnT C TnS. The reverse inclusion, TnS C SnT, is established in thesame manner. The equality follows from Proposition 1.1.

(iii) By the definition of union, SLIT contains all elements of S and of T. So x&S impliesX GSUT and S QSuT. Using (i) above, we also have S C TuS.(iv) xGSnT implies xGS and x e. T. In particular, a; S S. Thus SnTcS. Part (ii) allowsus to write TnS QS.

(v) ScSu0 by (iii). If a;GSu0, then x e S, for a; by definition. Therefore Su0 c S.Hence by Proposition 1.1, Su0 = S.

(vi) By (iv), Sn0 C 0. But we know that is a subset of any set and, in particular, C Sn0.Hence Sn0 = 0.

(vii) ScSuS, using (iii). Now, xGSuS implies x&S. Hence SuScS and the equalityfollows.

(viii) By (iv), SnS Q S. But x&S implies x&SnS. Thus ScSnS and the equality follows.(ix) Let xSSu{TuU). Then a; e S or x e. TuU. Thus a; G S or x . T or a; G [/. Conse-

quently xGSuTuU. Hence Su(ru 17) C SuTU [7. If x^SuTuU, then a; G S or a; G Tor a;Gf7. If a; G S, a;GSu(ruC7). If x&TorU, xGSu{TuU). Hence SuTuU QSu(TuU). The result follows.(x) X e Sn(TnU) implies xGS and xe(TnU), which in turn implies x^T and a: G t7.From a;GS and a; G T it follows that a;G(Snr) and, as x^U, xG(SnT)nU. There-fore Sn{TnU) c{SnT)nU. Similarly (Snr)n ?7 c Sn(rn t/).

(xi) Su(rnC7) C (Sur)n(Sul7). For, if a;GSu(rnC/), then x & S or xGTnU. Now, a; G Simplies xeSuT, xGSuU, and consequently x G (Su T) n (Su U). If k G Tn I/, then xGTand a; G U; hence arGSuT, x G SuU and, as before, a; G (Su T) n (Su U). (Su T) n (Su U) cSn(7'U J7) is established in a similar manner.

(xii) If S S contains an element a;, then a; G S and x ^S which is impossible. Hence S Smust be empty.

(xiii) S-S = from (xii). Thus S-{S-S) = S-0 and clearly S - = S.1.8. Prove the following statements:

(i) If S C r and U is any set, then SuU cTuU.(ii) If S c T and U is any set, then SnU cTn U.(iii) If S C r and T c t/, then S qU.(iv) S c r if and only if SnT = S.(v) T cS if and only if S = TuS.(vi) If res, then (S-r)ur = S.Solution:(i) Let a;GSuf7. Then xGS or a; G [/. If a; G S then xGT since ScT, and conse-

quently xeTuU. If x&U, then a;GruJ7. Thus SuU qTuU.


15/290

SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1(ii) xGSnU implies xGS and xGU. As S Q T, we also have x G T. Therefore xiTnUand SnV Q TnU.(iii) S Q T means that if x G S, then x G T; and since T qU, this in turn implies x G.U.Hence S C U.(iv) First, assume SnT = S. The equality implies any element a; in S belongs to SnT. But

xGSnT implies x G T. Hence xGT and S C T. Secondly, let S Q T. If x G S, thenxGT and so xGSnT. Therefore S Q SnT. The reverse inclusion SnTcS is true byProblem 1.7(iv). By Proposition 1.1, S = SnT.(v) Assume S = TuS. If x G T, then xGTuS and, since S-TuS,xGS. Hence T Q S.On the other hand if we assume T Q S, then x G TuS implies that x G S. ConsequentlyTuScS. By Problem 1.7(iii) we have SqTuS. Thus TuS = S.(vi) It follows from the definition that (S - T) c S. Using (i) above, (S-T)uT cSuT. But by

(v), TcS implies S = TuS, and -we have {S-T)UTcS. To show Sc(S-T)uT, letxGS. Either xGT or x T. If x G T, then a: (S - T) U T. If x ^ T, then a; e S - Tand we also have xG {S-T)uT. Thus Sc;(S-r)ur. The equality follows by Proposition 1.1.

1.2 CARTESIAN PRODUCTSa. Definition

The plane R^ (see Section 1.1a) consists of all pairs {x,y) of real numbers x and y. Weshall also denote R^ hy R x R; thus R X R is defined as the set of all ordered pairs (x, y) withX G R and y G R.

There is a natural extension of this notation to any two sets, S and T:SxT = {p\ p^{s,t), sGS, tGT}

For example,{1,2} X {1,2,3} = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}

In words, SxT is defined to be the set of all ordered pairs of elements (s, t), the first mem-ber of each pair always belonging to S and the second member always belonging to T. Weterm SxT the cartesian product of S and T. It is worth pointing out that, just as in R^,two elements of SxT are equal iff (if and only if) they are identical, i.e. (s, t) = (s', t') if andonly if s = s' and t = t'. If either S = or T = 0, we interpret S x T as 0.

One defines similarly the cartesian product Si x S2 x x Sn of the n sets Si, S2, . . .,Sn{n < 00) as the set of all %-tuples (si, S2, . . . , s) with Si G Si, S2GS2, . . . , s S. As withthe cartesian product of two sets, (si, S2, . . . , s) = (si , sz , . . . , Sn ) iff Si = Si , S2 = S2 , . . .,Sn = SnFor example,

{1,2} X {2,3} X {4,5} = {(1,2,4), (1,2, 5), (1,3, 4), (1,3,5),(2,2,4), (2,2,5), (2,3,4), (2,3,5)}

If any one of the sets Si, S2, . . . , S = 0, then we shall interpret Si X S2 x x Sn to be 0.If each Si = S, then Si x S2 x x Sn is denoted by S".We often are interested in certain subsets of SxT. For example, in elementary analyticgeometry one investigates lines, circles (see Section 1.1a), ellipses and other figures in theplane; these are subsets of R^.Problems1.9. Let S = (1, 2, 3}, T - {1, 5}. Verify the following statements:

(i) S X r = {(1, 1), (1, 5), (2, 1), (2, 5), (3, 1), (3, 5)}(ii) r X S = {(1, 1), (5, 1), (1, 2), (5, 2), (1, 3), (5, 3)}(iii) S X r # TXS


16/290

Sec. 1.2] CARTESIAN PRODUCTS 7(iv) S2 ^ r2 (S2 = S X S and T^ ^ T X T)(v) (SxTjXS - SX(TXS)Solution;(i) Clearly these are the only ordered pairs {x, y) with x e S and y e. T.(ii) As in (i), these are the only ordered pairs {x, y) with xG T and y & S.(iii) Looking- at (i) and (ii), we find (5, 1) S T X S and (5, 1) S X T. Hence SxT^ TxS.(iv) (3, 3) G S2 but (3, 3) fz.(v) An element (x, y) oi (S X T) X S has x G S X T and y ^ S. Thus ((1, 1), 5) e (S X T) X S. But((1,1),5)SX{TXS), since (1, 1) S.

1.10. Let S, T and 1/ be any three sets. Prove the following:(i) S X r = r X S ifF either S = T or at least one of the two is empty.(ii) If (x, y) e S2, then {y, x) e S^.(iii) Sxr^SxC/ iff either T - U or S = 0.(iv) {SXT)XU = SX(TXU) iff at least one of the sets S, T, U is empty.Solution(i) S= T implies SxT=T2=TxS; and S = or T = jZ) implies, by the definition of thecartesian product of any set and the empty set, SxT = = TxS. Therefore SxT = TxSwhenever S - T, S = 0, or T = 0. To prove the converse, assume S X T = T X S. We may

also assume S - and T -- 0. Let tGT and sGS (such elements exist since S 'and T - 0). Then {s,t)SSxT and, as SxT = TxS, (s, t) e T x S. It follows, from thedefinition of T X S, that t S S and s e. T. Therefore T qS and S Q T. We conclude, usingProposition 1.1, S = T.

(ii) {x, y) e S2 means x S S and ?/ S S. Hence {y, x) S S^.(iii) Clearly if T=U or S = 0, we have S X T = S X U. Conversely, let SxT = SxU andS - (if S = we have nothing to prove). If T # 0, let tGT. Then {s,t) e SxT forany s G S; and, as SxT ^ SXU, (s, t) G S x [/. But (s,


17/290

8 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1Solution:(i) (a) {x, x) A for every x e P, since x is not less than x.

(b) (x, v)&A implies x


18/290

Sec. 1.2] CARTESIAN PRODUCTS 9

Definition: Let X be a non-empty set and let R he a subset of X^. Then R is called anequivalence (or an equivalence relation) in X if the following conditions aresatisfied.(i) {x, x) G R for all x E Z (reflexive property),(ii) If (x, y) e R, then {y, a;) G i? (symmetric property),(iii) If {x,y)GR and {y,z):R, then {x,z)GR (transitive property).

Problem1.12. Examine Problem 1.11 for equivalence relations.

Solution:(i) A is not an equivalence relation in P, as (x, x) ^ A for all x.(ii) B is not an equivalence relation in P, as we know (x, y) & B and {y, x) G. B occurs only ifx = y. Thus though (1, 2) G B, (2, 1) g B.(iii) A is an equivalence relation in Z^ as Ac ^2 and the reflexive, symmetric and transitive

properties hold.(iv) y c i22 X i?2 and the reflexive, symmetric and transitive properties hold, so V is an equivalence

in fi2.

c. Partitions and equivalence relationsSuppose R is an equivalence relation in X. We say x is iJ-related to y, or x is related to

y by R, if (x, y) G R. If (x, y) G R we shall sometimes write xRy. To illustrate letZ= {1,2,3,4} and letR = {(1,1), (2, 2), (3, 3), (4, 4), (1,3), (3,1), (3, 4), (1,4), (4, 3), (4,1)} (1.2)

Then it is easy to check that R satisfies the three necessary conditions for it to be anequivalence relation. Now SRA since (3, 4) G R, but 2i24 is an incorrect assertion since(2,4) ^R.

Note that we have used a notation that fits in with the notation of Section 1.2b wherewe informally introduced an equivalence relation.An equivalence relation in X is intimately connected with a partition of X, i.e. a decom-

position of X into disjoint subsets of X such that every element of X belongs to some subset.Examples of partitions of {1, 2, 3, 4, 5} are

{1}, {2,3}, {4,5}and {1,3,4}, {2}, {5}On the other hand {1,2}, {2}, {3,4,5} is not a partition of {1,2,3,4,5}.

If R is the equivalence relation (1.2) in {1,2,3,4}, then all the elements of {1,3,4} arei2-related to 1, i.e., IRl, 1R3, 1R4:This suggests a means of getting a partition of a set X. In order to explain, we need someadditional notation. Let R be an equivalence relation in a set X. If x G X, we definexR= {y\ y GX and {x,y) gR}. xR is thus a certain subset of X. This subset xR iscalled the R-class of x, or the R-equivalence class of x, or the R-block of x. A subset of Xwill be called an R-class or R-block if it is the iZ-class or /2-block of some element x G X.To illustrate these terms, consider the equivalence relation R given by (1.2). Here

li2 = {1, 3, 4}, 2R = [2], and BR = 4R = IRThus the iJ-classes here are simply {1,3,4} and {2}. Notice that these iZ-classes constitutea partition of {1,2,3,4}.

More generally, we have the following


19/290

10 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1Theorem 1.2: Let Z be a non-empty set and let R be an equivalence relation in X. Then

(i) if xRnx'Rv^^, then xR = x'R,(ii) X G xR for every x e X.

Thus the J?-classes constitute a partition of X, for (i) guarantees that dis-tinct /2-classes are disjoint, while (ii) shows that every element of Xappears in at least one of the i2-classes.

Proof: First, we verify (i). Suppose xRnx'R - 0. Then there is an element y G xRwhich lies also in x'R, i.e. {x, y) G R and (x', y) G R. As R is an equivalence, it follows fromthe symmetric property that {y, x') G R. But (x, y) G R and (y, x') G R imply, by thetransitive property, {x, x') G R. Now if z e x'R, then {x', z) G R; and hence by the transi-tive property applied to {x, x') and {x', z), we find {x, z) G R. This means, by the verydefinition of xR, that z G xR. Since z was any element of x'R, we have proved x'R c xR.The reverse inequality follows by a similar argument. Hence x'R = xR as required.

The verification of (ii) is trivial since (x, x) G R means x G xR. This completes theproof of Theorem 1.2.

Problems1.13. (i) Prove that E = {(0,0), (1,1), (2,2), (3,3), (0,2), (1,3), (2,0), (3,1)} is an equivalence relation inS = {0,1,2,3}.

(ii) Find the S-equivalence blocks (a) OE, (b) IE, (c) 2E, (d) 3E.Solution:(1) The reflexive property holds, i.e. (x, x) e E for all xSS, since (0, 0), (1, 1), (2, 2), (3, 3) e E.To show that E is symmetric, let us examine all pairs (a;, y) where x = y. There are only four,

namely (0, 2), (1, 3), (2, 0), (3, 1). Clearly if (, y) is any one of the four, so is (y, x). Whenx = y, {x,y) = (y,x). Thus (x,y)&E implies (y,x)e.E. S is also transitive. Let {x,y)eEand (y,z)eE. Suppose x - y. Then (a;, j/) can be (0,2), (1,3), (2,0) or (3, 1). If (x.y) = {0,2),then {y, z) = (2, 0) or (2, 2) and {x, z) = (0, 0) or (0, 2) respectively. Hence {x, z) e E. Sim-ilarly if (x, y) = (1, 3), (2, 0) or (3, 1), it can be shown that (x, z) S E. When x = y, (y, z) & Emeans {x, z) e E. Therefore for any (x, y) G E and {y, z) E, we have {x, z) e E and^is transitive.

(ii) (a) OE = {0,2}, (6) IE = {1,3}, (c) 2E = {2,0}, (d) SE = {3,1}. Observe that OE = 2E andIE = 3E.1.14. Let A - {p\ p- (x, y) G Z^ with x-y divisible by 3}. Prove that A is an equivalence relation

in Z and find the K-equivalence classes.Solution:

It was shown in Problem l.ll(iii), page 7, that A satisfies the three conditions of anequivalence relation. The i2-equivalance classes are:(1) OA = {3? 1 g G zy, for if (0, ) G A, - a; = -x is divisible by 3. Also, (0, 3g) G A.(2) lA = {l-3g| gGZ}; for if {l,x)eA, l-x = Zq and hence x = l-Zq. If x = l-Zq,

1 a; is divisible by 3; hence (1, x) G lA.(3) 2A = {2-Zq\ q&Z}; for if (2,a;)GA, 2-x = Zq and so x = 2-Sq. If x = 2-Sq, 2-x

is divisible by 3; hence (2, x) G A.0A,1A,2A are the only fi-blocks, for any integer can be written as 3q, 1 - 3g, or 2 - 3g.

Consequently 0AulAu2A = Z.1.15. Let Z* be the set of nonzero integers and let S == Z X Z*. (Recall that Z is the set of all integers.)

Let E = {p\ p = {(r,s), {t,u)) G S^ with ru = st}. Prove that E is an equivalence relation in S.


20/290

Sec. 1.3] MAPPINGS 11Solution

:

7 is reflexive, for {r,s)BS implies ({r,s), {r,s)) S S^; and since rs = sr, {{r,s), (r,s)) G E.The symmetric property of E is established by noticing {{r,s), {t,w)) E means {r,s) and {t,w) S S^,and rw = st. But rw = st can be rewritten as ts = wr. Hence ((f , w), (r, s)) e JB7. To show E istransitive, let ((r, s), (t, m)) S 7 and {{t,u), {v,w)) G E. Then rw = sf and tw = vu. Since m ^ 0,r = and rw = = tw = vu = sv. Thus rw sv and so ((r.s), (v.w)) G E. Hence E istransitive. Therefore E is an equivalence relation.

1.16. Prove that S X S is an equivalence relation in S.Solution:

S X S is reflexive since (x, x) S S X S for all x & S. If {x, y) G S X S, then x and y G Sand, by definition of S X S, {y, x) G SxS. Hence SxS is symmetric. Now (x, y) G S and{y, z) GS imply x, y and z G S. But then (x, z) S S X S and S X S is transitive. Thus S X S isan equivalence relation on S.

1.17. Prove that a set X is infinite if and only if there are infinitely many equivalences in X. (Hard.)Solution:

Assume there are an infinite number of equivalences in X. If X is finite, then there are at mosta finite number of distinct subsets of X^. Therefore when X is finite there are at most a finitenumber of equivalences in X, which contradicts our hypothesis. Hence X must be infinite. Con-versely, assume X is infinite. We exhibit an infinite number of equivalences in X as follows: Foreach pair a,b G B, a = b, we define

^(a,b) = {p| P = (. 2/) e X^, where either x = y, (x,y) = (a,b) or (x,y) = (b,a)}Now R(a,b) = ^(c,d) if and only if {a,b} = {c,d}. Therefore since X is infinite, we can find aninfinite number of different pairs a,bGX each of which gives a distinct set iZ(a,b)- Furthermore,each R^a.b") is an equivalence. To prove that i2(o,b) satisfies the three conditions of an equivalencerelation, we first notice {x,x) G R^a.b^ for all xGX, by the very definition of fi(a,b)- Secondly,R(o,b) is symmetric since (x, ?/) G i2(o,b) means (x,y) = (a,b) or (b, a), in which case {y,x) =(b,a) or (a, 6) respectively, or x = y and then (x,y) = (y,x). Thirdly, if (x,y) and (y,z) G R(a,b),then {x,z) G R^^.b)- To see this, notice that (x,y) can only be (a, 6), (6, a) or ix,x). (x,y) = (a, b)implies (y,z) = (b, a) or (b,b); hence (x, z) = (a, a) or (a, 6), which are both elements of fi(a,b)-Similarly, (x,j/) = (6, a) implies (x,z) G Rca.bi- Finally, {x,y) = (x,x) means {x,z) {y,z) efi( ,,)

d. The division notationWe find it useful to introduce a notation for the i?-classes of an equivalence relation R

in a set X, namely X/R.Problems1.18. What is S/E in Problem 1.13?

Solution: S/E = {0E,1E}.1.19. What is Z/A in Problem 1.14?Solution: Z/A = {OA, lA, 2A}.1.3 MAPPINGSa. Definition of mapping

Assign to each even integer the value 1, and to each odd integer the value 1. Let usgive the name a to this assignment; thus a assigns to each even element in Z, the set of allintegers, the unique element +1 in the set {1, 1} and to each odd element in Z the uniqueelement 1 in {1,-1}. In less detailed terms a assigns to each element in Z a unique ele-ment in (1,-1}. Such an assignment is termed a mapping from Z into {1,-1} or a map


21/290

12 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1from Z into {1,-1}. More generally, if S and T are any two non-empty sets, a mapping ora map from S into T is an assignment of a unique element of T to each element of S. Forthe most part we shall denote mappings by lower case Greek letters such as a, p, y. If a is amapping from -S into T, we shall express this fact more briefly by writing a:S^T; this isread "a is a mapping from S into T". We call S the domain and T the codomain of a.

We find it useful to provide further notation and definitions. Suppose that a: S-* T.If a assigns to s in S the element t in T, we write a: s^t and read this as "a sends s intot". We call t the image of s (under ) if a : s -* i. It is convenient to have a number ofnotations for the image of an element s in S under a mapping a: S^T; thus we shall writeSff or s", or even a{s) for the image of s under a. For the most part we use the first notation.If t e r and sa = t, we call s a preimage of t. By Sa we mean {sa j s S S). We call Sathe range of a.Problems1.20. Suppose a : P - P is defined by

(i) a: n-* n^ for all re e P (iv) a : Ji -^ 1 for all n e P(ii) a : re -* w + 1 for all w e P (v) a : 1 -^ 2, re -^ 1 for all re e P, re > 1(iii) a : re -> 2re for all re S P

Afofe: In (i)-(v) above, the mapping a is defined by describing its "action" on every element ofP. For example, in (i), la = 12 = 1, 2a = 4, Sa = 9, .... Note that each element of P has a uniqueelement assigned to it.In each case determine: (a) 2a, 5a, 6a; (b) a preimage (under a) of 2, 5, 6, 27. (c) Is every elementof P an image of some element of P in (i)-(v)? How many preimages (under a) does 2 have in (v)?How many preimages does 1 have in (iv)? in (v)?Solution:(1) (a) 2a = 4; 5a = 25; 6a = 36.

(6) There is no preimage of 2, 5, 6 or 27.(c) Not every element of P is an image, e.g. 2 is not an image.

(ii) (a) 2a = 3; 5a = 6; 6a = 7.(6) la = 2; 4a = 5; 5a = 6; 26a = 27. Hence 1, 4, 5, 26 are the required preimages.(c) The only element of P which has no preimage is 1.

(iii) (a) 2a = 4; 5a = 10; 6a = 12.(6) 1 is a preimage of 2; 5 has no preimage; 3 is the preimage of 6; 27 has no preimage.(c) 1 has no preimage, so not every element of P is an image.

(iv) (a) 2a = 1; 5a = 1; 6a = 1.(6) 2, 5, 6 and 27 have no preimages.(c) 1 has every element of P as a preimage and no other element of P has a preimage.

(v) (a) 2a = 1; 5a = 1; 6a = 1.(6) 1 is a preimage of 2; 5, 6 and 7 have no preimages.(c) 2 has one preimage, namely 1. 1 has an infinite number of preimages. In fact any ele-

ment of P not equal to 1 is mapped onto 1. 1 and 2 are the only elements of P which havepreimages.

1.21. Let S = {1, 2, 3}, T = {1, 4, 5}.(i) Write down a mapping of S into T.(ii) Let a:S^T be defined by la = 4, 2a = 5, 3a = 4. Is every element of T an image of some

element in S under a? Is every element of T the image of more than one element in S?Give preimages of 1 and 4.

Solution:(i) For example, a ; S -> T defined by la = 1, 2a = 4, 3a = 5.(ii) Not every element of T is an image, for 1 has no preimage. Only 4 is an image of more than

one element in S. 1 has no preimage, and 4 has preimages 1 and 3.


22/290

Sec. 1.3] MAPPINGS 131.22. Suppose a: P -* C is defined by

(i) a : a; - a;2 {v) a: x -> ix + 1(ii) a: X-* 2x + 27 , .^ ix + 1(vi) a: X -^ . _^(iii) a: x^ z where z S C is such that z^ x. ix i(iv) a: x^z where z 6 C is such that z^ = a; - 1. ("^"^ a: x^ logjo cc(a) Do all of these descriptions really define mappings of P into C?(6) Is every element in C a preimage of some element of P in (i), (ii), (v), (vi), (vii)?Solution:(a) (i) and (ii) define mappings since every x G P has a unique image, (iii) does not define a map-

ping; for example, either 2 or 2 could be taken as 4a. Similarly (iv) is not a mapping, sincethere are three complex cube roots of x 1, so that each x has three different images, (v), (vi)and (vii) define mappings.

(6) In (i), (ii), (v) and (vii), i has no preimage: for (i) i = x^ implies x = yfi^ P; (ii) 2a; + 27 Ij 27implies x = P; (v) ix + X-i gives x = 1 + t g P; (vii) if logjo x - i, then 10* = x

and thus x S P. In (vi) 1 has no preimage because -: = 1 implies 1 = 1.IX 1.1.23. What is Pa in each of the cases in Problem 1.20?

Solution:(i) Pa is the set of squares {1, 4, 9, 16, . . .}. (iv) Pa = {1}(ii) Pa = {2, 3, 4, 5, ... (v) Pa = {1, 2}(iii) Pa {2,4,6,8, . . .}, i.e. all the even integers.

b. Formal definition of mappingThe reader may ask whether our definition of mapping is precise. After all, it dependsupon an English word, assignment, a word that is used in many different ways.A comparison with Section 1.2b is valuable. In Section 1.2b we introduced the concept

of equivalence relation in X, but as we felt uneasy about it, we redefined it in terms of asubset of X^. Here too we feel uneasy about our definition of mapping and so we shallredefine it in terms of sets.A subset a of S X r is called a mapping of S into T if (s, ti) and (s, ^2) e a occurs only ifU = ti, and for each s GS there exists an element (s, t) G a. S is called the domain and Tthe codomain of a. If a is a mapping of S into T (written briefly as a: S^T) and (s, t) G a,we call t the image of s under a and write : s -* t. We also write t = sa.

It is easy to see the relationship between the old definition and the new. In the olddefinition the elements of S were assigned unique elements of T.

Consider the subset of SxT consisting of the pairs (s, t) where t is assigned to s. Thetwo conditions of the new definition are satisfied by this subset.

In the sequel we will use the definition of Section 1.3a, being confident that if necessarywe could justify our arguments using the definition of a mapping in terms of a subset.c. Types of mappings

We have talked of mappings without defining what is meant by the equality of twomappings. We will now remedy this. Suppose a: S ^ T and (S: S' -^ T'. Then we definea = ;8 if and only if S = S', T T', and, for every element s & S, sa = sa'. In other words,two mappings are equal if and only if they have the same domain, the same codomain, andthe same "action" on each element of S. For example, let S {1,2,3,4}, r = {4,5,6}.Let a:S^T be defined by 1 = 4, 2a = 5, 3a = 6, 4a = 4; let /8 : S - T be defined by1/3 = 4, 213 = 5, 3j8 = 6, 4^8 = 5. Then a' 13 since 4a ^ 4:/3.


23/290

14 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1It is important to distinguish certain types of mappings. Thus suppose a: S^ T. Thenwe say a is a mapping from S onto T (notice that into has now been replaced by onto) if every

element in T has at least one preimage in S, i.e. if for every t GT there is at least oneelement s G S for which Sa = t; in this case we call a an onto mapping.

On the other hand we say a is one-to-one if sa = s'a implies s = s', i.e. distinct elementsof -S have distinct images in T (under a). Finally, we say a is a matching of S and T or thata matches S with T or is a bijeetion if a is both onto and one-to-one. Two sets are termedequipotent or of the same cardinality if there exists a matching of the one with the other.If S is finite and a matches T, then we say S and T have the same number of elements. Wedenote the number of elements in a set S by |S|. If S is infinite this definition no longermakes sense unless one takes from all the sets which match S a single fixed set which wethen term \S\, the cardinality of S+. A set which matches P is called enumerable, countable,or countably infinite. Important results are that the set of rational numbers is enumerableand the set of real numbers is not+.

We require one further definition. Suppose a: S^T and suppose S' c


24/290

Sec. 1.3] MAPPINGS 151.27. Suppose m is a fixed positive integer. Every integer n can be written uniquely in the form

n qm + r where the remainder r is an element of the set {0, 1, . . .,m 1}. For example, ifm = 4, then every integer can be expressed in precisely one of the following forms:4k, 4k + 1, 4k + 2, 4k + 3. Let a : Z - {0, 1, . . .,m 1} be defined by na = r, if n = qm, + r wherer e {0, 1, . . . ,m 1}. Prove:(a) a is onto {0, 1, . . ., m 1}.(6) If TOj, ^2 ^re any two integers, then (n-^nâ (n-ân2a)a.Solution:(a) a is onto because 0, 1, . . ., 7n 1 have preimages 0, 1, . . ., tn 1 respectively.(h) Let Ml =: QiWi + ri and n2 = 92 + '"2' Then

iW2 = (qim + n){q2Vi + rg) = 9192^ + 'i92 + ^29ii + *'i'2 = (9i92 + -i92 + ^'agO'w + ^^2Now letting r^rj = qstn + rg, we obtain (wiW2) = ''s = ('i''2)'>' (wia'K2)-

1.28. Check which of the following mappings are onto, one-to-one, bijections:(i) a: C -^ R defined by a: a + ib-â^+b^(ii) a.Z-^P defined by a: n -^ n^ + 1(iii) a: P ^ Q defined by a: n-^ r

.

Solution :(i) a is neither onto nor one-to-one, because a^+ b^ 0, for any o, 6 e i?, and la = la = 1.

Hence a is not a bisection.(ii) As 3 has no preimage, a is neither onto nor a bijection. Also a is not one-to-one, since la = 2

and la = 2.71

(iii) 1 has no preimage, since ^x-r = 1 implies n. = 1 P. Therefore a is not onto and hencenot a bijection. na n'a means -rr = ., , . ^ or 2n'n + n = 2n'n + n'; hence n = n'.2n + l 2n' + 1Thus each image has a unique preimage and a is one-to-one.

1.29. Let S be the set of open intervals (a, b) on the real line and let T be the set of closed intervals[a,b]. Define a.S^T by {a,b) a = [a,b]. Is a one-to-one? Onto?Solution:

The mapping is one-to-one and onto. For, if {a,b} a = {a',b') a then [a,b] = [a',b']. But thisequality holds iff a = a' and b = b'. a is therefore one-to-one. a is also onto since a closed interval[a, b] has a preimage {a, b).

1.30. How many mappings are there from {1,2} into itself? From {1,2,3} into itself? In each case,how many of these mappings are one-to-one? Onto?Solution:

There are four mappings of {1, 2} into itself, namely: aj defined by laj = 1 and 2q!i = 2; a2defined by la2 = 1 and 2a2 = 1; a^ defined by las 2 and 2a3 = 1; a^ defined by la4 = 2 and2a4 = 2. Only a^ and ag are one-to-one and onto.

To find the number of mappings of {1,2,3} into itself, we proceed as follows: 1 may haveany of three images under such a mapping, i.e. 1 -> 1 or 1 -> 2 or 1 -> 3. Also, 2 may have any of 3images, either 1, 2 or 3. So we have in all 3 X 3 possibilities for the actions of mappings on 1 and2. Then 3 can be sent into 1, 2 or 3, giving 3 X 3 X 3 = 27 possible mappings of {1, 2, 3} into itself.There are 3X2X1 = 6 possible one-to-one and onto mappings; for when we once choose an imagefor 1 there are only two possible images for 2, and then the image of 3 is uniquely determined.

1.31. Let S = {1, 2, 3}, T = {3, 4, 5}, U = {4, 5, 6}, and let a:S^T be defined by a : 1 -^ 3, 2 -^ 3, 3 ^ 5.Let p and y be the mappings from T into U given by /8 : 3 -^ 4, 4 -> 6, 5 - 4, y : 3 -* 4, 4 -> 4, 5 -* 4.Compute (la)/3, (2a)j8, (3a)y8, (la)y, (2a)y, (3a)y.Solution:

(la)y8 = 3y8 = 4; (2a);8 = 3/3 = 4; (3a);8 = 5,8 = 4; (la)y = 3y = 4; (2a)y = 3y = 4; (3a)y = 5y = 4.


25/290

16 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 11.32. Let a, p, y be the mappings of Q into Q defined by a: x^-^ , p : x ^ x + 1, y:x^x-l. Prove that

x + 1if X is any element of Q, then ((xy)a)li = = . What is a^p ? a^^. ' -^^ "ip ^ "iz ^Solution: _ i _i_ o _i_ i

({xy)a)li = {(x - l)a)p = ^ (3 = ^ + 1 = ^ = ^~ |P '^ ^ mapping of Pinto Q. a,^ is a mapping of Z into Q. a^p = a^^ since the domain of a^p is not the same as thedomain of a^^,

1.33. If S is a non-empty set, prove that S is infinite if and only if there are infinitely many mappingsof S into S.Solution:

First we show that if there are infinitely many mappings of S into S, then S is infinite. Assume,on the contrary, that S is finite and \S\ = n. Now each of the n elements can be mapped onto atmost n images. Hence there are at most m" different mappings of S into S. This contradicts ourassumption that there is an infinite number of such mappings. Hence S is infinite. Conversely, letS be infinite. We define, for each se.S, the mapping a^'- S -^ S by a^: x -^ s for all x S.{qis I s e S} is an infinite set since as = s' if and only if s = s', and we assumed S to be infinite.Therefore we have found an infinite number of mappings of S into S.

1.34. If S is any non-empty set, verify that S matches S.Solution:

Define the mapping a: S -^ S by a: s -* s for each s S S. a is clearly one-to-one and onto.Hence a is a matching.

1.35. If S matches T, prove that T matches S.Solution:

If S matches T, then there is a mapping a: S ^ T which is one-to-one and onto.Define a: T -^S as follows. Let t G T. Then there is an sG S such that sa = t. The image

of t under a is defined to be s.We now show a is a matching. In the first place, 5 is a mapping. For if ta = s and ts = s',for some t e T, then by definition of a, sa = t and s'a = t. But a is one-to-one, so that sa = s'a

implies s = s'. Thus the image of an element under a is unique. Secondly, 5 is one-to-one, becauseta = t'a = s implies Sa = t and sa = t', which in turn implies, since a is a mapping, t = t'.Thirdly, if s G S, then sa = t for some t G T. By definition of a, ta = s. Hence every elementof S has a preimage under a and a is onto.

1.36. If S matches T and if T matches U, prove that S matches U. (Hard.)Solution:

Let a:S-*T and p-.TÛ be matchings. Then a:SÛ, defined by sa = (sa)p, is amatching, a is a mapping of S into U; for sa S U, and if sa = u^ and sa = M2 then {sa)P uând {sa)p = M2, which implies u^ = 1*2 since sa has a unique image under p. a is one-to-one, forsa = s'a implies (sa)p = is'a)p. But a and /3 are one-to-one, so that sa = s'a and s s'. a is alsoonto, for if M e C7 then there is a t 6 T such that tp = u. Now t has a preimage s e S undera. Thus sa = {sa)P =^ tp = u.

1.37. Let a be the mapping of S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} into itself given by a : 1 ^ 3, 2 ^ 4, 3 -* 5,4 -^ 7, 5 -* 9, 6 -^ 10, 7 - 1, 8 -> 3, 9 ^ 4, 10 -^ 5. Then there is a useful alternative way of describinga: we list the elements of S on one line (in any order) and on the following line we place undereach element of S its image under a, enclosing the entire description in parentheses as follows.123466789 1034579 10 1345Describe the following mappings of S into S, using this notation.


26/290

1 2 3 4 5 6 7 8 9 102 2 2 2 2 3 4 4 4 51 2 3 4 5 6 7 8 9 106 7 8 9 10 1 1 1 1 1

Sec. 1.4] COMPOSITION OF MAPPINGS 17(i) /3 : 1 - 2, 2 ^ 2, 3 ^ 2, 4 - 2, 5 - 2, 6 -^ 3, 7 -> 4, 8 ^ 4, 9 -^ 4, 10 - 5

.

(li) y:1-*6, 2-^7, 3^8, 4-^9, 5^10,6-*1,7^1, 8->l, 9^1, 10^1.Use these descriptions to decide whether (iii) p = y, (iv) /? is one-to-one, (v) y is onto.Solution

:

(i)

(ii)

(iii) j8 7^ Y since 1/3 = 2 and ly = 6. It is only necessary to compare the bottom rows.(iv) /3 is not one-to-one, e.g. 1/8 = 2(3 = 2. It is only necessary to find a repetition on the bottom row.(v) y is not onto, e.g. 2 has no preimage. It is only necessary to check whether all the integers

1, 2, . . . , 10 appear in the bottom row.

1.4 COMPOSITION OF MAPPINGSDefinition

Let a: S-^T, p-.T^lJ. Because Sa T, we can compute {sa)p. This suggests"composing the mappings a and ^", i.e. defining a mapping of S into U by performing aand p in succession on each of the elements in S. More precisely we define a op, the com-position of a with p (in this order) as a mapping of S into JJ defined by

s{a P) = {sa)p, for all s in S(Some authors use exactly the opposite order, so that their ap \& our p o a.) For example,let

S = {1,2}, T - {3,4,5}, U = {6,7}and let a:S^T, p : T -* U be defined by

a: 1^3, 2-* 5, /3:3^6, 4-7, 5^6Then l(aoj8) = (la)/3 = 3/3 = 6

2(ao/3) = (2a);8 = 5/3 = 6Hence ao/3: {1,2} ^ {6,7} is defined by

aop: l-6, 2^6This notion of the composition of two mappings is of tremendous importance; hence we

give the following drill problems.

Problems1.38. Let a: P-* C be defined by na = in + 1 and let p : C -* P be defined by p: a + ib ^b^, where

i2 = 1. What do aOyS, (a/3) a, and Q:(/3a) map n G P to? Why is a p = 13 a"!Solution

:

Let n&P. Then n(a fi) = {na)p {in + l)/8 = n^. Now n{{a p) a) (n{a p))a = nâ =in^ + 1 and n{a (/3 a)) = (na){p a) = {(in + V)P)a = nâ in^ + 1. Hence {a p)a = a {jS a).a p: P -^P while pa: C -* C. Hence a p = pa.

1.39. Let a:Q-*Q be defined by a: aâ^ + 2 and let P Q -^ Q be defined by p: a^ â-2. Com-pute a p, /3a. Are these mappings equal? Compute {a p)a, a{p a).


27/290

18 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1Solution:

Let aGQ. a{a p) = (aa)/3 = {a^ + 2)/3 = A(a2 + 2) - 2 = la2 - 1. a(/3 a) = (a^)a = (|a - 2)a =(Aa - 2)2 + 2. Clearly a 13^13 a. Furthermore, a([ap)a) - (a(a p))a - (â^ - l)a -(lo2-l)2 + 2 and a{a{lia)) = (aa)(/3a) = (a^ + Z)l3a = ((a2 + 2)/3)a = ((|(a2 + 2) - 2)a =(i(i2 - 1)2 + 2. Note then that {a li)a = a o (/3 a).

1.40. Employing the notation of Problem 1,37, compute the following:^. /I 2 3 4 5\ /I 2 3 4 5\ ,...^ /I 2 3 4 5\ /I 2 3 4 5

2 3 1 5 4/ Vl 2 4 5 3/ ' ' ^2 4 5 3 1/ ^1 2 4 3 5,..^ /I 2 3 4 5\ /I 2 3 4 5\ ,. , /I 2 3 4 5\ / 1 2 3 4 5

2 14 3 5/123451/ 112345/125341(i) ('i ! ? ! I) (ii) r! ! ! ! f^ (Hi) (lll'l^ (iv) /I 2 ^ ^ ^Solution

,24135/ '""' VS 2 5 4 1/ '"*' ^2 3 5 4 l/ '"'' V2 5 3 4 11.41. Let a: S-* T, p-.T-Û, y.U^V. Prove that (a /3) y = a (/3 o y). (Hard.)

Solution

:

If s e S, then s{{ap)y) = (s{a j3))y = ((sa)/3)y and s(ao(/3y)) = (sa)(/3oy) = ((sa)/?)y.Consequently (a o j8) o y = a o (/3 y).

1.42. Prove that if a: S -^ T and p : T ^ U and a = /3 is onto, then /3 is onto. Is a onto? (Hard.)Solution

:

Let u & U. As a j8 is onto, we can find a preimage of u under a /3. Let s & S be a preimageof M under a fS, i.e. s(a! o /3) = u. Thus s(a! ;3) = (sa)y8 = tt and so: is a preimage of u under p.Hence /3 is onto, a need not be onto, e.g. let S = {!}, T = {1,2}, [7 = {1}. Define a: S-*T byla = 1, and 13: T ^ U by 1/8 = 2,8 = 1. o /? is onto but a is not.

1.43. Prove that if a: S^ T, p : T ^ U and a;8 is one-to-one, then a is one-to-one. Is /3 one-to-one?(Hard.)Solution

:

Let Si, S2 G S and sâ = S2a. Sjo: = 820: implies {sia)p = {S2a)p and, by definition of a y8,! y8 = S2a ,8. But a p is one-to-one, so that Sj = 82- Hence a is one-to-one. p is not necessarilyone-to-one, e.g. let S = {1}, T = {1, 2} and U = {1}. Define a: S->T by la = 1, andP : T ^ V by 1/3 = 1 and 2/3 = 1. a o y8 is one-to-one but p is not one-to-one.

1.44. Prove that p : T -* U {T = Uand p': T ->U such that ap - ap', it follows that p = p'. (Hard.)Solution:

Let us assume that for every set U and every pair of mappings p and p' of T into U such thata p = a p', it follows that p = p'. Say a is not onto, and t^ is an element of T which has no


28/290

Sec. 1.5] BINARY OPERATIONS 19preimage under a. Let U {1, 2} and define the mappings /3 and p' of T into U as follows: tji = Xfor all t e r, t/3' = 1 for all t ?^ tj in T and tiyS = 2. Now if s&S, sap = (sa)p = 1 andsa p' = {sa)p' = 1, since sa # ij. Hence a 13 a p' and /3 t^ y8'. This contradicts the as-sumption that a p = tt p' implies p = /?'. Thus a must be onto.

Conversely, assume a is onto and we can find a set U and two mappings, p and /?', of T" into Usuch that ap = ap' and P - P'. P - p' means there is a tj S T such that *iy3 # ti,8'. Further-more, since a is onto, we can find s S S such that sa = i,. But a y8 = a p' implies (sa)p = (sa)/3'or tiP = tiP'. Here we have a contradiction because we chose t^p ' t^p' . We therefore concludeP = P'.

1.5 BINARY OPERATIONSa. Definition

The idea of a binary operation is illustrated by the usual operation of addition in Z,which may be analyzed in the following way. For every pair of integers (m, n) there isassociated a unique integer m + n. We may therefore think of addition as being a briefdescription of a mapping ot Z x Z into Z where the image of {m,n) G Z xZ is denoted bym + n. Any mapping /3 ot SxS into S, where S is any non-empty set, is called a binaryoperation in S. We shall sometimes write instead of (s, f)/3 (the image of (s, t) under (3) oneof sot, s-t, st, s + t or sxt. We stress that in all these cases the meaning of the variousexpressions sot, S't, st, s + t and s x Ms simply the image of (s, t) under the given mapping^ of Sx S into S. These notations suppress the binary operation /3, so there is danger ofconfusion. However, we will work with binary operations and the various notations sofrequently that the reader will become familiar with the pitfalls. Incidentally, we read

Sot as "ess circle tee"s i as "ess dot tee" or "ess times tee"st as "ess tee" or "ess times tee"s + t as "ess plus tee"sxt as "ess times tee".

The notation s o Ms called the circle notation, the notations s t and st are termed multi-plicative, and the notation s + 1 is termed additive. We sometimes refer to s-t or st as theproduct of s and t, and s + t as the sum of s and t. The following problems will help tomake the various notations clear.Problems1.46. Convince yourself that the following mappings are binary operations in P.

(i) a: PXP ^ P defined by a : (i, j) -* i^, where {i, j) S P.(ii) a: P XP -* P defined by a : {i, j) -* i + j, where {i, j) e P.(iii) a: P X P -* P defined by a : (i, j) -^ ix j (regular multiplication of integers), where {i, j) G P.(iv) a: PXP-*P defined by a : (i, j) -^2i+ Sj, (i, j) e P X P.(v) a: PXP ^ P defined by a : {i, j)^i + j + 1, (i, j) e P X P.

1.47. Which of the following are binary operations in P (throughout {i,j) e P^) ?(i) a : (i, j) - i + i (iii) a : {i, j) ^i-^ j (v) a : (i, j) -^ j(ii) a : {i, j) -* i j (iv) a : {i, j) ^ i + j + i^Solution :

In (i), a is clearly a mapping from P X P into P. So a is a binary operation in P. (ii) and(iii) do not define binary operations in P because not every element in P X P has an image in P:e.g. in (ii), a: (1,2) -^ 1 -2 = -1 gP; and in (iii), a: (1,2) ^ 1 h- 2 = | g P. (iv) and (v) definemappings from P X P into P. Hence they define binary operations in P.


29/290

20 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 11.48. Interpret the following (abbreviated) definitions of binary operations in Z, where i and j denote

arbitrary elements of Z.(i) i j = {i + i)^ (iii) ij = i-k-j (v) 1X3= V(ii) i + j = i j {ix j) (iv) ij = i ix j (vi) i' j = t + 27jSolution:

Throughout, (i, j) e Z^.(i) a: (i,j)^(t + ;)2 (iii) a : (i, j) î + j (v) a: (i,j)îi(ii) a : {i, j)î- j - (ix j) (iv) a: {i, j) -* i i X j (vi) a : (i, j) ^ i+ 27}

1.49. Check that the following are binary operations in the plane R^.(i) (^, V) (*' y') = midpoint of the line joining the point (x, y) e R^ to the point (x', y') G B^

if {x,y) = (x',y'). If {.x,y) = (x',y'), define ix,y){x',y') = {x,y).(ii) (x, y) + [x', y') = {x + x',y + y'), where (x, y), {x', y') e jR2.(iii) (x,y)'(x',y') = {xx',yy'), where (x,y),(x',y') B R^.(iv) (a;, 2/) (a;'. 2/') = (x x', 2/ j/'), where (x,y),(x',y') e. R^.(v) (x, 2/) (x', 3/') = (x + x', x'2/ + 2/'), where (x, 2/), (x', 2/') S jR^.(Notice here how we have abbreviated the definitions of the binary operations considered.)Solution:

(i) o is a binary operation since two points determine a unique line and each line has a uniquemidpoint.

(ii)-(v) are binary operations because each has a unique image by virtue of the fact that addition,multiplication and subtraction are binary operations in R.

1.50. Let S = {1, 2, 3} and let a and p be the following binary operations in S:a : (1, 1) -* 1, (1, 2) -* 1, (1, 3) -* 2, (2, 1) -* 2, (2, 2) -> 3, (2, 3) ^ 3, (3, 1) ^ 8, (3, 2) -* 2, (3, 3) -* 1

;

p: (1,1) -*1, (1,2)^1, (1,3)^2, (2,1)^3, (2,2)->3, (2, 3) -^3, (3, 1) -^ 3, (3,2)->l, (3,3)^2.(i) Is a = pi(ii) Compute ((l,l)a,l)/3, ((i, l)/3, l)a.(iii) Compute ((l,2)a,3)a, (1, (2, 3)a)a, ((1, 2)^, 3)^3, (1,(2,3)/?)^.Solution :(i) a # ;8 for (2, l)a = 2 and (2, l)y8 = 3.(ii) ((l,l)a,l)/3 = (l,l);8 = l; ((l,l)/3,l)a = (l,l)a-l.(iii) ((1, 2)a, 3)a = (1, 3)a = 2; ((1, 2)^8, 3)/3 = (1, 3)^8 = 2;

(1, (2, 3)a)a = (1, 3)a = 2; (1, (2, 3)/3);8 = (1, 3)/3 = 2.1.51. Let S = {1,2,3} and let X be the set of all mappings of S into S.

(i) Compute \X\.(ii) Verify that the composition o of mappings is a binary operation in X.(iii) Is a p = p a for all elements a, p S XlSolution :(i) |Z| = 27 (see Problem 1.30, page 15).(ii) The composition of mappings a: S -* S and p : S -> S is again a mapping of S into S.

Therefore ir:X^^X, defined by (a, P)t!- = a p, {a, p) & X^, is a binary operation.(iii) No. For example, if a: S^ S defined by sa = 1 for all s.S, and p : S ^ S defined by

sp - 2 for all s e S, are two mappings of S into S, then s{a p) = {sa)p = ip 2 andspo a = (sp)a = 2a = 1. Hence a P ^ P a.1.52. Let Q* be the set of nonzero rational numbers. Make sense of the remark that division (denoted as

usual by -^) is a binary operation in Q*. Check whether the following statements hold for alla,b,ce Q*.(i) a H- 6 = 6 -=- a. (iv) If a -^ b a ^ c, then b = c.(ii) (a-.-6)- c a -rib-i-c).(iii) {(a-^b)-^c)-.- d a -(6

(v) If 6 -^ a = c -i- a, then 6(c^d)).


30/290

Sec. 1.5] BINARY OPERATIONS 21Solution:

Division is a binary operation in Q*; for if w/x G Q* and y/z&Q* where w,x,y,z areintegers, then w/x -^ y/z = wz/xy S Q*. Hence -^ is a mapping of Q* X Q* -* Q*.(i) False; e.g. 2 -^ 3 # 3 -^ 2.(ii) False; e.g. (2 h- 1) -=- 2 = 1 and 2 -h (1 ^ 2) = 2 ^ | = 4.(iii) False; e.g. ((1 ^ 2) -^ 2) -^ 2 = 1 and 1 h- (2 -r- (2 ^ 2)) = |.(iv) True, a -^ & = a ^ e implies ac = ab and, as a = 0, c = b.(v) True, b -^ a c ^ a implies ab ca. Hence b = c, since a = 0.

1.53. Let o be the binary operation in J? defined by ab = a+ b + ab. Verify that:(i) For all a,b,cGR, {ab) o c = a {b c).(ii) For all a,b e. R, ab = ba.(iii) Prove that if a # 1, then ah = ac iff b = c.Solution:(i) (oofe)oc = (a + b + ab)c = a + b + ab + c + {a+ b + ab)c= a + b + c + bc + ab + ac + abc

a{bc) = ao{b + c+bc) a + 6 + c + 6c + a(6 + c4- 6c)= a-\-b + c-\-bc-\-ab-\-ac + abc(ii) ab = a + b + ab = b + a + ba = b a(iii) If 6 = c, then a o 6 = a o c for any a. If ab = ac and a = 1, then a + 6 + a6 =

a + e + ac. Therefore 6 + a6 = c + ac, 6(1 + a) = c(l + a) and, since a = 1, 6 = c.1.54. Let o be the binary operation in R^ defined by (x, y) (x', y') = (xx' yy', yx' + xy'). Verify that for

all (x,y),(x',y')Ax",y")&Rh(i) (x,y)(x',y') = (x',y')o(x,y)(ii) {(x, y) o (x', y')) (x" , y") = {x, y) {(x', y') (x", y"))Solution:(i) (x, y) o (x', y') = (xx' - yy'

,

yx' + xy') = (x'x - y'y, y'x + x'y) = (x', y') o (x, y)(ii) ((, y) o (x', y')) (x", y") = {xx' ~ yy', yx' + xy') o (", y")= {{xx' yy')x" {yx' + xy')y", {yx' + xy')x" + {xx' yy')y")

{xx'x" yy'x" yx'y" xy'y", yx'x" + xy'x" + xx'y" yy'y"){x,y)o{{x',y')o{x",y")) = {x,y)o(x'x" - y'y" , y'x" + y"x')= {x{x'x" - y'y") - y{y'x" + y"x'), y{x'x" - y'y") + x{y'x" + y"x')) {xx'x" xy'y" yy'x" yy"x', yx'x" yy'y" + xy'x" + xy"x')= {{x,y)o{x',y'))o{x",y")1.55. Let be the binary operation in Q defined by

(a) aob = a-b + ab, (b) 6 = + ^-"'""^ , (c) a6=^!-t-^^ 3Determine which of the above binary operations satisfy(i) (a o 6) o c = o o (6 o c) for all a, 6, c S Q(ii) a 6 = 6 o a for all a, 6 S QSolution:(i) (a) {aob)c' ao(boc) for some a, 6, c G Q; e.g. (2o0)o2 = (2 + 0)o2 = 2o2 =2-2 + 4 = 4 and 2 (0 2) = 2 (0-2 + 0) = 2 - = 2- (-2) -4 = 0.

(6) (oo6)oc ^ ao(6oc) for some a, 6, cGQ; e.g. (loO)oO = ^oO = ^ and 1(00) =l0 = f(c) (ao6)oc # ao(6c) for some a, 6,ceQ; e.g. (loO)oO = ^oO = A and lo(OoO) =loO = J.

(ii) (a) aob=ba since 10 = 1 and 01 = -1./. V . a + 6 + 0.6 6 + a + 6a .(6) a6 = = = 6oa/\ I. a + b b + a(c) o6 = - = I' = bao 3


31/290

22 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1b. The multiplication table

So far we have introduced a number of definitions and notations and familiarized our-selves with them. The object of this section is to introduce a "table" as a convenient wayof either defining a binary operation in a finite set S or tabulating the effect of a binaryoperation in a set S. To explain this procedure, suppose S = {1, 2, 3} and let /i be the binaryoperation in S defined by

^: (1,1)^1, (1,2)^1, (1,3)^2, (2,1)^2, (2,2) - 3,(2, 3) -3, (3,1)-1, (3, 2) -3, (3, 3) -2

Then a table which sums up this description of /x is12 31122 3 313 223

We put the number (2, 3)/^ = 3 in the square that is the intersection of the row facing2 (on the left) and the column below 3 (on the top). More generally, in the {i,j)th square,i.e. the intersection of the ith row (the row labelled or faced by i) and the ith column (thecolumn labelled by j), we put {i, y)/i.

A table of this kind is termed a multiplication table because it looks like the usual multi-plication tables. One often calls fi a multiplication in S. Thus when we talk about amultiplication /j, in a set S, we mean that ft is a binary operation in S.

There is a reverse procedure to the one described above. For example, suppose we startout with a table 112

2 3 313 223Then there is a natural way of associating with this table a binary operation /x in {1,2,3}.We simply define {i, j)ti to be the entry in the {i, i)th place in the table. For example,

(1,1)^1 = 1, (2,3)^0 = 3, (3,2)ft = 3We shall usually define multiplications in a finite set by means of such tables.Problems1.56.' Write down the multiplication tables for the following binary operations in S = {1,2,3}.

(i) a : (1, 1) -* 2, (1, 2) -> 3, (1, 3) ^ 1, (2, 1) ^ 3, (2, 2) -^ 1, (2, 3) ^ 2, (3, 1) ^ 1, (3, 2) -> 2, (3, 3) ^ 3.(ii) /3 : S2 -^ S defined by (i, j)p = 1 for all (i, j) e S^.(iii) y : (1, 1) -* 1, (2, 2) -> 1, (3, 3) -^1, (1,3)^ 2, (3, 1) -^ 2, (2, 3) ^ 1, (3, 2) -^ 1, (2, 1) -* 3, (1, 2) - 3.Solution: 12 3 12 3 12 3

1

(i)

2 3 1

3 1212 3 (ii)1 1 1

1 1 1

1 1 1

(iii)

13 23 1 1

2 11


32/290

Sec. 1.5] BINARY OPERATIONS 231.57. Does the following table define a binary operation in {1,2,3}? In {1,2,3,4}?12 3

1

23

Solution:The table does not give a binary operation in {1, 2, 3} since (1, 3) -^ 4 {1, 2, 3}. The table

also does not define a binary operation in {1,2,3,4}, because (1,4), (2,4), (3,4), etc., have no images.

1.58. Write down explicitly the binary operations in {1,2,3,4} defined by the following tables.

1 3 4

2 1 31 3 2

1 2 3 41

2

3

4

1 2 3 41 2 3 4 1 2 3 42 3 4 1 2 4 1 33 4 1 2 3 1 4 24 1 2 3 4 3 2 1

11112 2 2 23 3 3 3

4 4 4 4(i) (ii) (iii)

Solution:(i) (I,:;) ^i (;' = 1,2,3,4); (;, 1) ^ j (j = 2,3,4); (2,2)^3; (2, 3) -^ 4; (3, 2) -^ 4; (3,3)^1;

(3, 4) -^ 2; (4, 3) ^ 2; (4, 4) ^ 3; (4, 2) -* 1; (2, 4) ^ 1.(ii) (1, i) ^ ; (i = 1, 2, 3, 4); (j, 1) ^ ; (j = 2, 3, 4); (2, 2) -^ 4; (2, 3) ^ 1; (3, 2) ^ 1; (3, 3) -* 4;(3,4)-2; (4,3)^2; (2,4)^3; (4,2) -> 3; (4,4)->l.(iii) (t, j) ^i {i= 1, 2, 3, 4 and i = 1, 2, 3, 4).

1.59. Rewrite the three binary operations in Problem 1.58, using(a) circle notation, i.e. write {i, j)/3 as t /,(6) additive notation, i.e. write (i, i)/3 as i + j,(c) multiplicative notation, i.e. write (i, j)p as i j.Solution:(a) Problem 1.58(i): 1 i = i (i = 1, 2, 3, 4); ;" o 1 = i (; = 1, 2, 3, 4); 2 o 2 = 3; 2 o 3 = 3 o 2 = 4;3o3 = l; 3o4 = 4o3 = 2; 4o4 = 3; 4o2 = 2o4 = l.

Problem 1.58(ii): 1 i = i (; = 1, 2, 3, 4); jo\ = j (j = 2,ZA); 2 o 2 = 4; 2 = 3 = 3 2 = 1;303 = 4; 304 = 403 = 2; 204 = 402 = 3; 404 = 1.Problem 1.58(iii): ioj = i (i = 1, 2, 3, 4 and j = 1, 2, 3, 4).

(6) Problem 1.58(i): 1 + i = i (i = 1, 2, 3, 4); ; + 1 = j (; = 1, 2, 3, 4); 2 + 2 = 3; 2 + 3 = 3 + 2 = 4;3 + 3 = 1; 3 + 4 = 4 + 3 = 2; 4 + 4 = 3; 4 + 2 = 2 + 4 = 1.Problem 1.58(ii): 1 + i = i and i + 1 = i (i = 1, 2, 3, 4); 2 + 2 = 4; 2 + 3 = 3 + 2 = 1;

3 + 3 = 4; 3 + 4 = 4 + 3 = 2; 2 + 4 = 4 + 2 = 3; 4 + 4 = 1.Problem 1.58(iii): i + j = i (t = 1, 2, 3, 4 and j = 1, 2, 3, 4).

(c) Problem 1.58(i): !} - j and / 1 = y (/ = 1, 2, 3, 4); 2-2 = 3; 2 3 = 3 2 = 4; 3 3 = 1;3.4 = 4.3 = 2; 4.4 = 3; 4. 2 = 2-4 = 1.Problem 1.58(ii): !} = j and i 1 = i (j = 1, 2, 3, 4); 2 2 = 4; 2 3 = 3 2 = 1; 3-3 = 4;3-4 = 4-3 = 2; 2-4 = 4-2 = 3; 4-4 = 1.Problem 1.58(iii): i'j - i (i = 1, 2, 3, 4 and j = 1, 2, 3, 4).


33/290

24 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1A look back at Chapter 1

We began with a few remarks about sets. We then introduced the idea of cartesianproducts. This led to the idea of an equivalence relation on a set. Then the notion of amapping was defined, followed by the definition of a binary operation.

In this book we are mainly concerned with binary operations in sets. At this stagethe reader may wonder what one could possibly say about binary operations in a set.Without some specialization we can say very little. In Chapter 2 we begin to place re-strictions on binary operations.

Supplementary ProblemsSETS1.60. The sets Sj (i = 1, 2, ..., n) are such that Sj C Sj+i (i = 1,2, . . ., w-1). Find SinSaO nS andSiUS'2U---uS.1.61. Let E = {n\ nG Z and n even}, O = { | nG Z and n odd}, T = {n\ nG Z and n divisible by

3}, and F={m| nGZ and w divisible by 4}. Find (i)EnT, (ii) EuT, (iii) TnFnEnO, (iv) TuF,(v) OnF, (vi) OnT.

1.62. Given A = {-3, -2, -1, 0, {1,2, 3,}} and B = {-3, -1, {1,3}}. Find AuB and AnB.1.63. Prove Sn{TuU) = {SnT)D{SnU).1.64. Let A = {-5,-4,-3, ...,3,4,5}, B = {-4,-2,0,2,4}, C = {-5,-3,-1,1,3,5}, D = {-4,4},E = {3, 2, 1, 0}, F = 0. Which, if any, of these sets take the place of X if (i) X C = 0,

(ii) XnB = C, (iii) X C C but X is not a subset of A, (iv) XqB and X is not a subset of E,(v)ZnCcA, (vi) Zu(BnD) = A?1.65. Prove ScT if and only if (TnC)uS = Tn(CuS) for every set C.CARTESIAN PRODUCTS1.66. Prove S X (TuW) = (SxT)u(SxW) for any sets S, T and W.1.67. Let P be the set of positive integers and S = P^. Show that E = {p\ p = {{r,s), (t,w)) G S^ and

r + w = s + t} is an equivalence relation on S. Find the E-class determined by (4, 7).1.68. Find the equivalence relation, E, on Z: (i) if the equivalence classes of F are {n\ n = Aq forqeZ}, {n\ n = l + 4q for q e Z}, {n\ n = 2 + 4q for g S Z}, and {n | n = 3 + 4g for q G Z};

(ii) if every equivalence class consists of a single integer; (iii) if the equivalence classes are{q, q} for each q G Z.

1.69. If E and F are equivalence relations on S, is (i) EnF, (ii) EuF a.n equivalence relation on S?1.70. What is wrong with the following argument: B is a non-empty subset of S^ which has the symmetric

and transitive properties. If (a, b) S E, then by the symmetric property, (6, a) G E. But by thetransitive property, {a,b)GE and (b,a)GE implies (a,a)GE. Therefore 7 is also reflexive.

1.71. (a) If P is the set of positive integers, show that E = {(a, 6) i (a, 6) G pa and a divides 6} isreflexive and transitive but not symmetric.

(6) Find an example of a subset E of P^ which is both symmetric and transitive but not reflexive,(c) Find an example of a subset E of P^ which is reflexive and symmetric but not transitive.


34/290

CHAP. 1] SUPPLEMENTARY PROBLEMS 25MAPPINGS1.72. Show that the following define mappings of Z into Z.

(i) a: X ^ \ \x\ is the absolute value of x, i.e. Ixj = \yxl\x\ if cc y^ ' ' ' ' \-x if a; sin2 x + cos^ x

1.73. Which of the mappings a,/3,y,S of the preceding problem are equal?1.74. Find subsets S(^0) and T(^0) of the real numbers i2 such that the mappings (i) a : x -* cos x,

(ii) /3 : X ^ sin X, and (iii) y: x -^ tan x are one-to-one mappings of S onto T. Do a, /?, y definemappings of R into Rt1.75. Let E = {n\ nG P and n even}. Define a: E -* E by na n for all n&E, and p : E -^ Eby M;8 = 2w for all w S . Find an infinite number of mappings a', p' of P into P such that

[e = and Z?;^ = ^.1.76. Suppose a is a mapping of a set S into a set T and, for any subset W of S, Wa - {t\ t^T and

= sa for some s&W). If A and B are any subsets of S, show: (i) (AuP)a = AaUBo:;(ii) (AnB)acAanPa; and (iii) A CB implies AaQBa.

1.77. Let S be a subset of a set T, and a: T -^W. Prove: (i) a one-to-one implies a is one-to-one;(ii) a onto implies a is onto.

COMPOSITION OF MAPPINGS1.78. Suppose a: S ^ T is onto and p : T -> U is onto. Show a p is an onto mapping.1.79. Given a:S-*T is one-to-one and p : T ^ U is one-to-one. Prove a /3 is one-to-one.1.80. (i) a : X - sin (x^), (ii) /? : x -^ sin (sin (x)), and (iii) y : x -> ^/l-x^ define mappings of non-empty

subsets of the real numbers R into R. First, find an appropriate subset in each case. Secondly,write a, p and y as the composition of two mappings, giving in each case the domain and codomainof each mapping defined.

BINARY OPERATIONS1.81. Let S be a set and gf the set of all subsets of S, i.e. gf = {A | A C S}. Show that intersection andunion define binary operations on gf.1.82. How many different binary operations can be defined on a set of 3 elements?1.83. Consider the set, F, of mappings /j (i = 1, 2, . . . , 6) of jB - {0, 1} into R defined for each x G ft - {0, 1}

by: fi-.x-^x; f^-.x^--; f^-.x-^^^^; f^:x-^^; f^-.x-^^; /g : x ^ 1 - x. Show thatcomposition of mappings is a binary operation on F and write a multiplication table for the operation.


35/290

chapter 2

GroupoidsPreview of Chapter 2

In this chapter we define a set G together with a fixed binary operation o to be a groupoid.As we remarked at the end of Chapter 1, there is little one can say about binary operationswithout making restrictions.

The first restriction is that of associativity. A groupoid with set G and operation o issaid to be associative if gi (,92 gs) = {gi fl'2) gs for all gi, gi, ga in G. Such a groupoid iscalled a semigroup. In order of increasing specialization we have the concepts of groupoid,semigroup, and group.

To define a group we need the concepts of identity and inverse. Hence we discuss theseideas.

We introduce the semigroup Mx of mappings of X into X. The importance of Mx isthat, but for the names of the elements, each semigroup is contained in some Mx.

Two other important concepts we deal with are homomorphism and isomorphism.Homomorphism is a more general concept than isomorphism. There is an isomorphismbetween two groupoids if they are essentially the same but for the names of their elements.

2.1 GROUPOIDSa. Definition of a groupoid

Consider the set Z of integers. Z has two binary operations, addition (+) and multi-plication (X). The set Z is one thing, a binary operation in Z is another; the two togetherconstitute a groupoid. Repeating this definition in general terms.Definition: A groupoid is a pair (G,/x,) consisting of a non-empty set G, called the carrier,

and a binary operation [j. in G.We shall mostly use a multiplicative notation when dealing with groupoids. Thus we

write g-k or simply gh for {g, h)ti, g,hGG. This notation has been used in Chapter 1in our consideration of binary operation. As an example, let G= {1,2,3} and let /x be thebinary operation in G defined by the following table.

13 22113 2 3

Then the pair (G, f^) is a groupoid. Suppose we use multiplicative notation ; then1-1 = 1, 1-2 = 3, 2-2 = 1, 3-2 = 2, etc.

26


36/290

Sec. 2.1] GROUPOIDS 27These products look bizarre unless we recall that the notation employed is a shorthandversion of

{l,l)ix = l, (1,2)m = 3, (2,2)f.= l, (3,2)m = 2, etc.If we use the expression "the groupoid G," where G is a set, it is understood that wehave already been given a binary operation

i^.in G, and that we have been talking aboutthe groupoid (G, /*).

Suppose now that (G, ix) is a groupoid. If we use the circle notation for /j., i.e. we writegoh instead of {g,h)ix, we shall sometimes write (G,o) to refer to the groupoid (G,/x). Sim-ilarly we write (G, ), (G, +), (G, x) if we employ g'h,g + h,gxh respectively for {g, h)ix.

Example 1: Let G = {1,2} and let n be the binary operation in G defined as follows:(1, 1)m = 1, (1, 2)/i = 2, (2, 1)m = 1, (2, 2)ix = 2

If we use the circle notation, we have11 = 1, lo2 = 2, 2ol = i, 22 = 2

The pair (G, /j.) or (G, ) is then a groupoid.Example 2:

Example

Example 4:

Let S be the set of all mappings of {1, 2, 3} into {1, 2, 3}. Then (S, o) is a groupoid,where is interpreted as the usual composition of mappings. The compositionmakes sense, for if a, j3 G S, thena: {1,2, 3} ^{1,2, 3} and /? : {1, 2, 3} ^ {1, 2, 3}

Therefore a p, the composition of a and /?, is defined by aa /3 = (aa)p, a G {1, 2, 3},and is once again a mapping of {1, 2, 3} into itself. (S, ) is indeed a groupoid.Let o be the binary operation in Q, the rational numbers, defined by ab a +b + ah. Then (Q, ) is a groupoid, because for every pair of rational numbersa and h, ab defines a unique rational number a + h + ab.Let B2 be the plane. Further, let there be a cartesian coordinate system in R^ andlet C be the disc of radius 1 with center at the point (2, 0) of the coordinate system.Consider the region R^ - C, the unshaded area in the diagram. We term any pathbeginning and ending at 0, which does not meet any point of C (i.e. it is entirely inR^ C), a loop in R'^ C. By a path we mean any line which can be traced out bya pencil without raising the point from the paper. For example, I and m areloops in R^ - C. Let L be the set of all such loops in R'^ - C. Then there is anatural binary operation in L which we denote by ; thus if l^, l^ e L, then l^ I2is the loop obtained by first tracing out l^, followed by tracing 4- This type ofgroupoid {L, ) is of considerable importance in modern topology.


37/290

28 GROUPOIDS [CHAP. 2

Example 5: Let F be the set of all mappings of R, the real numbers, into R. Consider twoelements a,p & F. We define the mapping a + p : R -^ R by a{a + p) aa + ap,aG R. a + /3 is clearly a unique mapping of R into R and hence is an element in F.Therefore + is a binary operation in F and (F, +) is a groupoid. Notice that {F, +)is not the groupoid with F as the carrier and the composition of mappings as thebinary operation.

Problems2.1. Are the following groupoids?

(i) (S,o) where S = {1,2, 3, 4} and ij = l for i and J elements of S.(ii) {Z, ), the set of integers with the usual subtraction of integers as binary operation,(iii) (F, ), the set of positive integers with the usual subtraction as binary operation.(iv) {Q, -^), the set of rational numbers with the usual binary operation of division,(v) {Z, +), the set of integers with the usual binary operation of division.Solution:(i) (S, o) is a groupoid since " is clearly a binary operation in S.(ii) {Z, ) is a groupoid; for if a, 6 Z, then ab is a unique element of Z.(iii) (F, ) is not a groupoid since a-b^P for all a, 6 e F. Therefore - is not a binary

operation in P.(iv) (Q, -^) is not a groupoid because a -^ is not defined for any aS Q and hence -^ is not a binary

operation in Q.(v) (Z, -V) is not a groupoid since a^bZ for all a, b e Z, e.g. 2 -^ 3 Z. Therefore division

is not a binary operation in Z.

2.2. Is {Z, ) a groupoid if o is defined as (i) a. 6 = ^/a+b , (ii) ab = {a+ 6)2, (iii) aob-a-b-ab,(iv) a o 6 = 0, (v) a o 6 = a?Solution:

All but (i) define a binary operation in Z. Therefore (Z, ) is a groupoid in (ii) through (v).The multiplication in (i) does not define a binary operation in Z since y/a + b is not always aninteger.

2.3. Let S be any non-empty set and T the set of all subsets of S. Are (T, n) and (T, U) groupoids?Solution

:

Both intersection n and union U are binary operations on T, for the intersection or union oftwo subsets of S is again a unique subset of S. Thus (T, n) and {T, U) are groupoids.

b. Equality of groupoidsTwo groupoids are equal if and only if they have the same carriers and the same binary

operation. Remember, a binary operation was defined as a mapping and two mappingsare equal if and only if they have the same domain and codomain, and the image of eachelement is the same under both mappings. Thus the groupoids described in Examples 1-5are all different.Problems2.4. Are any two of the groupoids in Problems 2.1-2.3 equal?

Solution: No.

2.5. Which of the following pairs define equal groupoids?(i) (Z, +) and (Z, m), where (a, b)ii = a + b.(ii) (Z, ) and (Z, ), where a^b = ab.(iii) (Z, o) where ab - a for all a and 6 in Z, and (Z, X) where a X 6 = & for all a and 6 in Z.Solution:

The groupoids in (i) are clearly the same. So too are the groupoids in (ii). In (iii) (Z, o) is notthe same as (Z, X); for if a # 6, ab - aX b.


38/290

Sec. 2.2] COMMUTATIVE AND ASSOCIATIVE GROUPOIDS 292.2 COMMUTATIVE AND ASSOCIATIVE GROUPOIDSDefinition of commutative and associative groupoids

Let (Z, +) be the groupoid of integers under the usual operation of addition. Thena + 5 = b + a {2.1)

and {a + h)-\- c = a-^{h + c) {2.2)for all a,b,c G Z.

Similarly if (Z, ) is the groupoid of integers under multiplication,a 6 = b ' a {2.3)

and {a-b)- c = a- {h-c) {2.A)for all a,b,c G Z.

The analog of {2.1) and {2.3) in an arbitrary groupoid {G,o) isaob boa {2.5)

for all a,b GG. Similarly the analog of {2.2) and {2.4) in (G, o) is{aob)oc = ao{boc) {2.6)

for all a,b,c G G. We term a groupoid satisfying {2.5) commutative or abelian, and agroupoid satisfying {2.6) associative or a semigroup. Thus a semigroup is an associativegroupoid. Of course it is not clear that there are non-commutative groupoids, i.e. groupoidswhich are definitely not commutative, and similarly it is not clear that there are non-associative groupoids. We settle the issue now. Let G = {1,2} and let o be the followingbinary operation in G:

1 21

2

Then (G,o) is a groupoid. Observe that lo2 = 1 but 2! = 2, so G is not commutative.Furthermore, (2 o 1) o 2 = 2 o 2 = 1 but 2 o (1 o 2) = (2 o 1) = 2, so G is also non-associative,i.e. G is not a semigroup.

For the most part we shall use the multiplicative notation for a groupoid {G,^) andsimply talk about the groupoid G. If the groupoid is commutative we will use the additivenotation instead of the multiplicative notation, since we are accustomed to addition as acommutative binary operation, e.g. in the integers.The order of a groupoid (G, /x) is the number of elements in G and is denoted by |G[;

(G,/x) is infinite if \G\ is infinite, and fi,nite if \G\ is finite.Problems2.6. Which of the groupoids in Examples 1, 2, 3 and 5 are commutative and which are associative?

SolutionThe groupoid of Example 1 is not commutative, since 1 o 2 = 2 and 2 o 1 = i, but is associative.To shovir (G, o) is associative we must examine the following 8 cases:

(a) lo(loi) = ioi = i^ (loi)oi = ioi = i (e) 2o(2oi) = 2oi = i, (2o2)ol = 2ol = l(6) 2(11) = 21 = 1, (2ol)l = loi = i (/) 2o(lo2) = 2o2 = 2, (2 ol) 02 = 102 = 2(c) l(2ol) = ioi = i_ (lo2)ol = 2oi = i {g) lo(2o2) = io2 = 2, (lo2)2 = 22 = 2(d) lo(lo2) = lo2 = 2, (lol)o2 = lo2 = 2 {h) 2o(2o2) = 2o2 = 2, (2o2)o2 = 2o2 = 2

1 1

2 1


39/290

30 GROUPOIDS [CHAP. 2

The groupoid of Example 2 is not commutative; for if a is defined by la = 1, 2a = 3 and 3a = 2,and p is defined by 1^ = 2, 2/3 = 1 and 3/3 = 1, then la ^ = (la),8 = 1/3 = 2, 1/3 a = (l/3)a =2a = 3. Hence a p = 13 a. Since the binary composition of mappings is an associative binaryoperation (Problem 1.41, page 18), (S, ) is an associative groupoid.

The groupoid in Example 3 is both commutative and associative. ah = a + h + ah = h + a +ba ba, since addition and multiplication are commutative binary operations in Q. Also,a o (6 o c) = a o (6 + c + 6c) = (a + (6 + c + 6c)) + a(6 + c + 6c) = a + 6 + c + 6c + a6 + ac + a6c and(ao6)oc = (a+6 + a6)c = a+6 + a6 + c + (a + 6 + a6)c = a+6 + a6 + c + ac+6c + a6c.Using the associative and commutative properties of addition and multiplication in the rationals,we see a o (6 o c) = (a o 6) o c.

In Example 5 the groupoid (F, +) is commutative because a(a + p) = aa + ap = ap + aa =a(p + a), a,pe.F and aG R (here we use the fact that aa, ap G R and addition is a commutativebinary operation in R). {F, +) is also a semigroup, for a{{a + P) + y) = a{a + p) + ay = aa +(a/3 + ay) = aa + a(P + y) = a{a + (/3 + y)) (here we use the associativity of addition in R).

2.7. Construct an example of a commutative groupoid of order 3.Solution:

Let S = {a, 6, c} and the binary operation o be defined by the multiplication tablea 6 c

a a 6 cb a cc c a

(S, ) is clearly a commutative groupoid.

2.8. Show that the set Q* of nonzero rational numbers with binary operation the usual division ofrational numbers, is a groupoid. Is it commutative? Is it associative?Solution:

If I and I are any two elements of Q*, then f "^ | = |^ is ^ unique element in Q*/^T^O since a, 6,c,d#oy Therefore division is a binary operation in Q*. However, divisionis neither commutative (e.g. |--^i = 2#| = i-|) nor associative (e.g. ^ "^ (i - i) =i - f = I ^ 6 = (i ^ i) ^ i).

2.9. Which of the groupoids in Examples 1-3, 5 and in Problems 2.7 and 2.8 are finite?Solution:

The groupoid of Example 1 is clearly finite of order 3.In Example 2 the set S of all mappings of {1,2,3} into itself contains 27 elements, and so

is finite.In Example 3, (Q, ) is not finite as there are an infinite number of rational numbers.In Example 5 the set F is infinite. To show that F is not finite we construct an infinite number

of mappings of R into R as follows. Let pj : iJ -* fi (

Documents

120486729 Shaum Group Theory