MS3001 Introduction To Abstract Algebra · 0 Introduction This course provides an introduction to...

MS3001

Introduction To Abstract Algebra

COURSE TEXT

Stephen Wills

Department of Mathematics

University College Cork

2005/06

Contents

0 Introduction 1Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1 Sets 4Basic definitions and constructions; Venn diagrams . . . . . . . . . . . . 4Set identities and truth tables . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Relations 20Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Properties of relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Equivalence relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Equivalence classes and partitions . . . . . . . . . . . . . . . . . . . . . 25

3 Functions 30Basic definitions and properties . . . . . . . . . . . . . . . . . . . . . . . 30Composition of functions; inverses . . . . . . . . . . . . . . . . . . . . . 36Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 Groups 54Binary operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54Cayley tables and groups . . . . . . . . . . . . . . . . . . . . . . . . . . 61Examples of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64Multiplicative and additive notation; basic group properties . . . . . . . 67Subgroups, cyclic groups and Lagrange’s Theorem . . . . . . . . . . . . 70Groups of low order and homomorphisms . . . . . . . . . . . . . . . . . 77An application: card shuffling . . . . . . . . . . . . . . . . . . . . . . . . 85

5 Answers to Selected Exercises 88

0 Introduction

This course provides an introduction to abstract algebra, a branch of mathematicsthat seeks to generalise or axiomatise familiar properties of numbers, matrices,polynomials, etc. One benefit of this process is that by stripping away extraneousdetail, the fundamental reason why many arguments or proofs work is finallyrevealed.

The process of abstraction involves deciding upon a collection of axioms orrules that we shall apply to an unknown, abstract set of objects; the axioms tellus which manipulations are permitted or defined. Once they are chosen we areno longer allowed to add or delete rules at our convenience, but must discover theconsequences of our particular choice. The choice of axioms is usually far frombeing totally arbitrary however, but normally inspired by familiar examples suchas addition and multiplication of numbers or polynomials, symmetries of regulartwo- or three-dimensional shapes and so on.

A significant impetus in the abstraction of mathematics was the development ofalgebraic geometry by Descartes, Fermat and others in the 17th century. By assign-ing numerical coordinates to points in the plane they were able to turn problemsabout plane geometry, as studied by the ancient Greeks, into algebraic problemsabout polynomial equations. Advances in the treatment of such equations thenrendered the solutions of (some of) these problems much easier, or indeed finallylead to a solution of the problem in question. On the other hand certain otherproblems remained intractable, particularly when the algebraic version involvedpolynomials of degree three or higher. Initial attempts to develop techniques tosolve these equations proved fruitless, and in the following two centuries a fun-damental shift in focus occurred. Rather than try to solve these equations, is itpossible to prove that there are such equations that do not have constructible so-

lutions? This principle was eventually established by Galois in the 19th century,by studying the properties of all polynomial equations, rather than focusing on asingle equation at any one time. This sort of study is only possible in an abstractaxiomatic framework, and hence a further justification for this approach.

In this course we will be studying groups, which are an algebraic system withone (binary) operation, and are typically thought of as an algebraic way of de-scribing symmetry. Indeed, these abstract algebraic techniques are now applied indiverse branches of mathematics including geometry, number theory and analysis.They are also fundamental in coding theory, in describing aspects of the physicalworld at the atomic scale through their use in quantum mechanics, and are usedin chemistry in studies of crystals, and in the analysis of isomers, a field that isimportant in the formulation of new drugs.

1

Logic

Logic

The twin pillars at the foundation of mathematics are set theory and logic. Theformer is the subject of our first chapter; here we recall those aspects of logic (andin particular the notation) that we shall need.

Any proof or calculation requires a careful argument which starts with clearlystated assumptions followed by a sequence of statements or propositions, each ofwhich follows from the former. In logic a proposition is any statement that iseither true or false. The proposition may depend on some variable, in which caseit is known as a conditional proposition. Examples include

(i) x2 ≥ 0 for all real numbers x.

(ii) The first month of the year is March.

(iii) x2 + x < 6.

(iv) Today is Monday.

All of these statements are either true or false. The first two are unconditionalpropositions, with (i) being true and (ii) being false. The last two, however, areconditional propositions. Proposition (iii) is true if we choose x so that −3 < x < 2;for any other value of x it is false. Similarly the proposition in (iv) depends onthe value of ‘today’.

If p and q are two propositions, and q is true whenever p is true, then we writep ⇒ q, which is read as

• p implies q, or

• q is implied by p, or

• p is a sufficient condition for q.

Note that in this case if q happens to be false then p cannot be true, so furtherequivalent ways of interpreting p ⇒ q are

• p is true only if q is true, or

• q is a necessary condition for p.

Note that p ⇒ q is again a proposition, since each of these ways of interpretingit shows that it is either true or false. The converse proposition is the statementobtained by reversing the roles of p and q, i.e. the statement q ⇒ p, which is alsowritten p ⇐ q. Further examples of propositions include

p1: You give me e500.

p2: You pass your final exam in MS3001.

p3: x is a real number greater than or equal to 0.

p4: x is a real number greater than −1.

2

INTRODUCTION

Then we have p1 ⇒ p2,∗ but, on the other hand, p2 ⇒ p1 is not true since if

you work hard then you can pass the exam without resorting to bribery. This iswritten p2 6⇒ p1. Similarly if x ≥ 0 then x > −1, and so p3 ⇒ p4, but the converseis not true since if we take x = −1

2 then p4 is true, but p3 is not true since −12 6≥ 0.

The propositions p and q are equivalent if p ⇒ q and q ⇒ p are both true.We write this in shorthand as p ⇔ q, which is read “p (is true) if and only if q(is true)”. This means that p and q are either both true, or both false. In ourexamples above we have neither p1 ⇔ p2, nor p3 ⇔ p4, because in each case theconverse statement is not true. Consider, however, the following:

p5: Today is Thursday.

p6: Tomorrow is Friday.

p7: x = 1 or x = 2.

p8: x2 − 3x + 2 = 0.

Then we have p5 ⇒ p6 and p6 ⇒ p5, hence we can write p5 ⇔ p6. Similarly, sincex2 − 3x+ 2 = (x− 1)(x− 2) for any choice of x, we see that p7 ⇔ p8. Note that inthese examples, to show equivalence we had to show two things, namely p5 ⇒ p6

(or p7 ⇒ p8) and p6 ⇒ p5 (or p8 ⇒ p7). If we were to modify p7 and insteadconsider the proposition

p7′: x = 1

Then we would have p7′ ⇒ p8, since if x = 1 then x2 − 3x+2 = 12 − 3× 1+2 = 0,

but p8 6⇒ p7′, since p8 is true if x = 2, in which case p7

′ is not true.

∗Although, in reality, it is not true!

3

1 Sets

Sets are the very building blocks of mathematics — practically everything you havedone so far in your degree could be constructed out of the content of the followingchapter, although to do this would take us on a very long and convoluted detour.Here we give a brief introduction to cover the material necessary for the course.

Basic definitions and constructions; Venn diagrams

Definition 1.1. A set is any well-defined collection of objects. The objects be-longing to the set are called elements or members of the set.

Here “well-defined” means that given an object x we can say if x belongs tothe set in question, or does not belong to the set.

Example 1.2. The following are sets:

(a) the integers between −10 and 3 (inclusive)

(b) the points on a given line in the plane

(c) the lines that pass through a given point in the plane (or 3D-space)

(d) the students at UCC.

The individual numbers, points, lines, students,. . . are the elements or members

of their respective sets.

Example 1.3. Is the collection of students present who will pass the MS3001examination this year a well-defined set? Yes, since each student in the class willeither pass or fail.

Is the collection of students who have good taste in music a well-defined set?No, since it is a matter of opinion whether or not you have a good taste in music.

Notation: Usually sets will be denoted by upper-case letters, and arbitrary ele-ments of sets will be denoted by lower case letters. Let A be a set and let x, y andz denote certain objects.

• if x is an element of A we write x ∈ A.

• if y is not an element of A we write y /∈ A.

• if both x and z are elements of A, we shall write x, z ∈ A instead of “x ∈ Aand z ∈ A”.

Since a set needs to be well-defined, there are in general there are two methodsthat we can use to specify them:

(i) By (exhaustive) listing of the elements, for example:

A = {1, 2, 3}, B = {a, b, c, d, e}

4

SETS

(ii) By a rule. The two sets above could equivalently be defined as

A = {x an integer : 1 ≤ x < 4}, and

B = {the first five letters of the alphabet}

The colon in the definition of A is read as “such that”, so A is the set consistingof “any integer x such that x is both greater than or equal to 1 and less than 4”.∗

Example 1.4. As a further example the set C = {5, 7, 11, 13, 17} could also bewritten as C = {x a prime number : 4 ≤ x < 19} since the first few prime numbersare 2, 3, 5, 7, 11, 13, 17, 19, 23, . . ., or alternatively as

C = {x an integer : x is odd, 5 ≤ x ≤ 17, x is not divisible by 3}.Exercise 1.5.

(a) Give two different rules that define the elements of the set A = {1, 4, 9}.(b) List the elements of the set B = {x ∈ Z : x is divisible by 2,−5 ≤ x < 0}.

Definition 1.6. If we can write out all of the elements of a set A and come to theend of the list then the set is finite. In this case the order of A is the number ofelements in the set, and is denoted |A|.† If we cannot complete the list then theset is infinite.

Remark. There is a more technical and precise definition of finite and infinite, butthis requires the material from Chapter 3.

Example 1.7. The sets A = {1, 2, 3} and B = {2, 3, 4} are both of order 3; theset C = {0, 1, 2, 3, . . . , 19} is of order 20.

Exercise 1.8. Determine the order of the following sets:

A = {x an integer : −7 < x ≤ 10, x divisible by 4},B = {x a prime number : 30 < x < 50}

Both methods of defining a set can be used when dealing with finite sets,although for finite sets of large order listing all of the elements can be tedious.On the other hand, if a set contains an infinite number of elements it is better, orindeed (usually) necessary, to use the second method.

The choice of letter that we use to label a given set will not generally be veryimportant. However for certain important sets of numbers there is some standardnotation used only for these particular sets. We shall need the following:

N = {1, 2, 3, 4, . . .} — the natural (or counting) numbers

Z = {. . . ,−2,−1, 0, 1, 2, . . .} — the integers

Q ={m

n: m,n ∈ Z, n 6= 0

}

— the rational numbers

R = the real numbers — the points on the line, including everything

from Q plus√

2, π, e, etc.

C = {x + iy : x, y ∈ R} — the complex numbers, where i =√−1

∗Some authors use the symbol | instead of the colon, that is they would write A ={x an integer | 1 ≤ x < 4}.

†Sometimes the order of A is denoted #A, and is also called the size or cardinality of A.

5


All of these sets are infinite, and so for the first two we are already breaking ourrule about not being able to define such a set by a list. For Q and C we are definingthe elements by a rule that gives the typical element in each case in terms of theelements of another set. So, for example, Q contains all fractions made up of aninteger divided by a nonzero integer.

In each case as we go down the list we are adding more elements — the negativeintegers and 0 to get Z from N; the fractions such as 1

2 and −53 to create Q from

Z (noting that m1 = m ∈ Q for all m ∈ Z); all square roots such as

√3 and other

numbers that are not rational to get R from Q; finally we add the square rootof −1 to R to get C. The actual definition of the reals is more involved than wehave time for in this course. The geometric model consists of all the points lyingon a line extended infinitely in either direction, with an arbitrary point chosen torepresent 0.

By adding these additional elements, we can solve more equations as we passfrom one set to the next. For example there is no x ∈ Z that satisfies 2x = 1, butx = 1

2 ∈ Q is a solution of this equation. Similarly we cannot solve x2 = 5 if weinsist on taking x ∈ Q, nor can we solve x2 +x+2 = 0 if we insist on x lying in R,but passing to R and C respectively allows us to find solutions to these equations.

Another advantage of the rule system is that it allows us to define a set withoutnecessarily “knowing” all of its elements. For example

P = {n ∈ N : n is prime}is a well-defined set since every natural number is either prime or not prime (that is,is prime or composite). However, given a large integer it may not be immediatelyobvious whether or not it is prime. Indeed the difficulty in determining whethera very large number is prime or composite, and then, if we are given a compos-ite number, determining its prime factors, lies at the heart of the cryptographicmethods used for secure internet transactions.

Two other important sets are the empty set ∅, which is the set that containsno elements, and the universal set U which contains all the elements that we aredealing with in a particular situation. A reasonable choice for U in Example 1.3would be the set of all students at UCC. For Example 1.2(a) we could take U tobe Z (or possibly Q, or R,. . . ). We do not always explicitly define our universalset U , but we are aware that it exists — it is introduced to avoid certain nastyset-theoretic paradoxes.

A useful visual aid to understanding sets are Venn diagrams. These involverepresenting a set by a closed figure in the plane, usually a circle. All points thatlie within the circle represent elements belonging to the set, and all elements thatlie outside the circle represent elements which do not belong to the set. Typicallythe circle is drawn inside a rectangle which represents the universal set. Thus thediagram for a set A lying within its universal set U would be

AU

6

SETS

Definition 1.9. Let A and B be sets.

(i) A is a subset of B (written A ⊆ B) if every element of A is also an elementof B. If in addition there is some element of B that is not in A then A iscalled a proper subset of B (written A ⊂ B). The symbols ⊆ and ⊂ can beturned round: A ⊇ B means that B is a subset of A.

(ii) A and B are equal (written A = B) if every element of A lies in B, andevery element of B lies in A. That is, A and B contain exactly the sameelements.

(iii) A and B are disjoint if they have no elements in common — that is noelement of A lies in B, and vice versa.

Symbolically we have

A ⊆ B if x ∈ A ⇒ x ∈ B

A = B if x ∈ A ⇒ x ∈ B, and y ∈ B ⇒ y ∈ A

(equivalently, if A ⊆ B and B ⊆ A)

A ⊂ B if A ⊆ B and A 6= B.

Example 1.10.

(a) Consider the three sets

A = {3, 5, 7}B = {x ∈ N : x is prime and 2 < x ≤ 9}C = {x ∈ Z : x is odd, 1 < x ≤ 9}.

Then B = {3, 5, 7} and so B = A, but C = {3, 5, 7, 9}, so that A ⊂ C sinceevery element of A is in C, but 9 ∈ C whereas 9 /∈ A.

(b) Since every element of N is also in Z we have N ⊆ Z. Similarly Z ⊆ Q,Q ⊆ R and R ⊆ C. In fact in each case we can replace the symbol ⊆ by ⊂since each set is a proper subset of the next one. For example −1 ∈ Z but−1 /∈ N so that N 6= Z, hence N ⊂ Z;

√2 ∈ R but

√2 /∈ Q, and so on.

Exercise 1.11. Consider the following four sets of integers:

A = {x ∈ Z : x2 ≤ 4} B = {x ∈ Z : x > 12}C = {−2,−1, 0, 1, 2} D = {2x : x ∈ A}

Determine which, if any, are equal, which are (proper) subsets of another, andwhich are disjoint.

Venn diagrams really become useful when we deal with two or more sets. If wehave two sets then in order to avoid making incorrect conclusions when drawing aVenn diagram we should draw the sets as follows:

7


A BU

so that we allow the possibility of there being members of A that are not membersof B and vice versa, and also the possibility of there being something in both Aand B. The following are the diagrammatic representations of what it means forA to be a subset of B, and for A and B to be disjoint respectively:

A

BU

A BU

In the first every element of A also lies inside the circle for B, and hence is anelement of B. In the second the circles do not overlap, and so the sets have nomembers in common.

Definition 1.12. The power set of a set A is the collection of subsets of A. It isdenoted P (A).

Example 1.13. If we consider the set A = {1, 2} then there are four differentsubsets of A, namely ∅, {1}, {2} and {1, 2}, the last one being A itself. Thus

P (A) ={∅, {1}, {2}, {1, 2}

}

where the elements of P (A) are now sets.

Exercise 1.14. Consider the set B = {a, b, c}. By listing all of the subsets, showthat |P (B)| = 8.

In general, if a set A is finite, and |A| = n then |P (A)| = 2n. To see this notethat when we form a subset of A we can consider each of the n elements in turnand decide whether or not it should be in our subset. Thus there are 2 choices foreach element of A, and so 2 × 2 × · · · × 2 = 2n ways to make a subset.

Exercise 1.15. What is P (∅)?

So far the only thing we have done with two sets A and B is to compare them.The following are a number of methods for constructing new sets from old ones.

8

SETS

Definition 1.16 (Operations on sets).

(i) The union of two sets A and B is A ∪ B = {x : x ∈ A or x ∈ B}. Thatis, A ∪ B consists of all those elements that are in A together with all theelements of B.

The word “or” in the definition is not used exclusive sense — elements thatlie in both A and B are included in the union.

(ii) The intersection of two sets A and B is A ∩ B = {x : x ∈ A and x ∈ B}.That is, A ∩ B consists of all those elements that are in both A and Bsimultaneously.

With this notation we see that A and B are disjoint if A∩B = ∅, since thenthey have no elements in common.

(iii) The difference of A and B is A − B = {x : x ∈ A,x /∈ B}. That is, thoseelements of A that do not also lie in B.∗

(iv) The symmetric difference of A and B is A△B = (A − B) ∪ (B − A). Thatis, those elements that lie in precisely one of A or B but not both.

(v) The complement of A is A′ = {x ∈ U : x /∈ A}. It consists of those elementsof the universal set that are not in A.†

The Venn diagrams for these operations are as follows, where in each case therelevant set is shaded:

��

��

��

��

��

��

��

��

��

��

��

��

��

��

A A

A

AA

A B B

BB

B

B

U U

UU

U

U

A ∪ B A ∩ B A − B

B − A A△B A′

Example 1.17.

(a) If we take A = {1, 3, 5} and B = {3, 5, 7} then

A ∪ B = {1, 3, 5, 7}, A ∩ B = {3, 5}, A − B = {1}, B − A = {7},A△B = {1, 7} and A′ = {2, 4, 6, 7, 8, 9, . . .} if we take U = N

∗This is sometimes written as A \ B rather than A − B.†This is sometimes written as Ac or ∁A instead of A′.

9


(b) If we take A = {2, 4} and B = {2, 4, 6} then

A ∪ B = {2, 4, 6} = B, A ∩ B = {2, 4} = A,

A − B = ∅, B − A = {6}, A△B = {6}and B′ = {. . . ,−2,−1, 0, 1, 3, 5, 7, 8, 9, . . .} if we take U = Z

Exercise 1.18. Explain the identities ∅′ = U and U ′ = ∅.

At first sight there is potentially a endless number of sets we can produce fromtwo sets A and B by using the operations ∪, ∩, −, △ and ′ repeatedly. Indeed, wecould create the sets (A∩B)∪B,

(A∪ (A ∩B)′

)∩B′, and so on. In fact it turns

out that there are at most sixteen sets that can be created this way. Moreover,this is only possible if neither A is a subset of B, nor B is a subset of A. Note thatin part (a) of Example 1.17 the sets A and B are both proper subsets of A∪B sothat this union is a distinct set, whereas in (b) we have the equality A ∪ B = B.

Definition 1.19. The Cartesian product of sets A and B is

A × B = {(a, b) : a ∈ A and b ∈ B}.

It is the set of all ordered pairs where the first element is taken from A and thesecond element from B. The order is very important; given two such pairs (a1, b1)and (a2, b2), we say (a1, b1) = (a2, b2) if and only if a1 = a2 and b1 = b2 — bothcomponents must be the same.

Example 1.20. If we take A = {1, 2} and B = {f, g} then

A × B = {(1, f), (1, g), (2, f), (2, g)}.

We create the pairs by taking each element of A in turn and then pairing it witheach element of B. We do not have (f, f) ∈ A×B, since although f ∈ B, f is not

in A. Similarly (2, 1) /∈ A × B and (g, 2) /∈ A × B. However (g, 2) ∈ B × A sinceg ∈ B and 2 ∈ A. Indeed,

B × A = {(f, 1), (f, 2), (g, 1), (g, 2)}.

The example above illustrates the fact that in general A×B 6= B ×A. Indeedit is possible to show that A × B = B × A (for nonempty sets) if and only ifA = B. For example if we take A = B = R then A × B = R × R = B × A, andis usually denoted by R2 (in general we write A2 instead of A × A). It is the setof coordinates of all points in the plane: {(x, y) : x, y ∈ R}, and is a Cartesianproduct with which you will be very familiar.

Z2 = Z×Z is a subset of R2 — it is the set of lattice points in the plane, eachvertex being distance one up, down, left or right from its nearest neighbour.

10

SETS

x

y

(1, 1)

(0, 0)

(−2,−1)

Exercise 1.21. Let A = {1, 2} and B = {0, 1, 2}. List the six elements of the setA × B. Which two pairs are not elements of B × A?

Let C = {0, 1, 2, 3}, i.e. C = B ∪{3}, the set obtained by adding the one extraelement 3 to B. Explain why |A × C| = |A × B| + |A|.

Note also in Example 1.20 we have |A| = 2, |B| = 2 and |A × B| = 4 = 2 × 2.In general if C and D are finite sets with |C| = m and |D| = n then |C×D| = mn,since when forming a general element of the Cartesian product we have m choicesfor the first component of the pair (which comes from C) and n choices for thesecond (coming from D).

Note. Many of the operations discussed above can be carried out on more thantwo sets at a time. For example if we have sets A, B, C and D then the unionA∪B ∪C ∪D is defined to be the set consisting of all those elements that belongto at least one of the four constituent sets. For example if we take

A = {1, 4}, B = {0, 1, 2}, C = {2, 5}, D = {1, 3}

then A∪B∪C∪D = {0, 1, 2, 3, 4, 5}. Similarly we can take the intersection of anynumber of sets, and the Cartesian product of any finite number of sets. Howeverwe should take care not to mix operations together, but rather use appropriatebracketing to indicate the order in which things should be done.

For example with A, B, C and D as above we have

A ∩ B = {1} ⇒ (A ∩ B) ∪ C = {1} ∪ {2, 5} = {1, 2, 5},

butB ∪ C = {0, 1, 2, 5} ⇒ A ∩ (B ∪ C) = {1, 4} ∩ {0, 1, 2, 5} = {1},

so that (A ∩ B) ∪ C 6= A ∩ (B ∪ C). This is analogous to addition, subtraction,multiplication and division of numbers, where for example we have (1 + 2) × 5 =3 × 5 = 15, but 1 + (2 × 5) = 1 + 10 = 11.

Definition 1.22. A partition of a set A is any collection P of nonempty subsets ofA such that each element of A belongs to precisely one of the subsets that makesup the collection P .

11

Set identities and truth tables

Note that if P is a partition of a set A then since each element of A is in atleast one set from P , the union of the sets making up the collection P must be allof the set A. On the other hand, since each element of A is in at most one set fromP , any pair of sets from P must be disjoint. Thus an equivalent definition to theone above is that a partition P of A is any family of pairwise disjoint nonemptysubsets of A whose union is all of A.

The following is a diagrammatic representation of partition of a set A intoseven subsets A1, . . . , A7:

A

A1

A2

A3

A4

A5

A6

A7

A jigsaw (with the pieces in position) is a good example of the idea behind apartition, the set being the completed puzzle and each piece representing a subset.

Example 1.23. Define subsets O and E of N by

E = {2, 4, 6, . . .}, O = {1, 3, 5, . . .}.

Then {E,O} is a partition of N since every natural number is either even or odd,but cannot be both. Equivalently, {E,O} is a partition since E ∩ O = ∅ andE ∪ O = N.

Example 1.24. Define subsets A, B and C of Z by

A = {1, 2, 3, . . .}, B = {−1,−2,−3, . . .}, C = {0}.

Then {A,B,C} is a partition of Z since every integer is positive, negative, or zero.Equivalently {A,B,C} is a partition since A ∪ B ∪ C = Z, A ∩ B = ∅, A ∩ C = ∅and B ∩ C = ∅.

Note that we must check each pair of subsets is disjoint in the above example.

Example 1.25. Let A = {1, 2, 3} and define subsets B,C,D by B = {1, 2},C = {2, 3} and D = {1, 3}. Then B ∪ C ∪ D = A, and B ∩ C ∩ D = ∅. However{B,C,D} is not a partition, since for example, B ∩C = {2} 6= ∅, i.e. 2 belongs tomore than one set in the collection {B,C,D}.


Suppose we are given any two (real) numbers x and y. We can then calculate thevalue of the number (x−y)(x2 +xy +y2). For example if we take x = 3 and y = 1then

(3 − 1) × (32 + 3 × 1 + 12) = 2 × (9 + 3 + 1) = 2 × 13 = 26.

12

SETS

However, we can also use the rules of arithmetic to rearrange this complicatedexpression as follows:

(x − y)(x2 + xy + y2) = x3 + x2y + xy2 − yx2 − yxy − y3 = x3 − y3,

and so we see it would have been quicker to calculate 33 − 13 = 27 − 1 = 26. Theabove is an example of an identity: it is an equation that is true independently ofwhich values of x and y we choose.

Consider instead the equation x2 + y2 = 4xy. If x, y ∈ R satisfy this equationthen

x2 + 4y2 = 4xy ⇒ x2 − 4xy + 4y2 = 0 ⇒ (x − 2y)2 = 0 ⇒ x = 2y.

That is, this equation is only true if we choose x to be twice the value of y. It isnot an identity since it is only true in certain particular circumstances.

Suppose now we are given two sets A and B. If we draw the Venn diagramsfor the sets A ∪B, A, B −A, and then A ∪ (B −A) we notice that the same areais shaded for both A ∪ B and A ∪ (B − A):

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

AA

A A A

ABB

B BB

BUU

U U U

U

B A ∪ B

A

A

B − A A ∪ (B − A)

This suggests that the sets A∪B and A ∪ (B −A) are actually the same set, andso we should be able to write A ∪ B = A ∪ (B − A), no matter what we choosefor A and B. Clearly this needs to be proved, and the fundamental technique forproving such set identities is the following idea: two sets C and D are equal ifthey have the same elements. Thus C = D if every element of C is an element ofD, hence C ⊆ D, and also every element of D is an element of C, hence D ⊆ C.So to prove C = D we first prove that C ⊆ D, and then prove that D ⊆ C. Toprove C ⊆ D we must choose an arbitrary element of C, then argue why it is alsoan element of D using the definitions above.

So to prove A ∪ B = A ∪ (B − A) we proceed in two steps as follows:

A∪B ⊆ A∪ (B −A): Let x ∈ A∪B. Then either x ∈ A or x ∈ B, by definition of∪. If x ∈ A then x ∈ A∪ (B −A), again by definition of ∪. If x /∈ A then we musthave x ∈ B, and since x /∈ A we in fact have x ∈ B − A, hence x ∈ A ∪ (B − A).

13


So in either case x ∈ A∪ (B −A), thus A∪B ⊆ A∪ (B −A), since the element xwas chosen arbitrarily.

A ∪ (B − A) ⊆ A ∪ B: Suppose now that y ∈ A ∪ (B − A). Then either y ∈ A,in which case y ∈ A ∪ B, or y ∈ B − A, which says in particular that y ∈ B, andhence y ∈ A ∪ B. Again, in either case y ∈ A ∪ B, thus A ∪ (B − A) ⊆ A ∪ B.

Putting the two inclusions together we get overall that A ∪ B = A ∪ (B − A) asrequired.

Exercise 1.26. Prove the following identities:

(a) A = (A − B) ∪ (A ∩ B)

(b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

and illustrate them on a Venn diagram.

An example of an equation involving sets that is not a set identity is A∪B = B.We saw in Example 1.17 that if A = {1, 3, 5} and B = {3, 5, 7} then A ∪ B ⊃ B,but if A = {2, 4} and B = {2, 4, 6} then A ∪ B = B. What is true in both cases,and indeed always, is that A ∪ B ⊆ B.

Proposition 1.27. Given any sets A and B, we always have B ⊆ A∪B. Moreover

we have equality, that is B = A ∪ B, if and only if A ⊆ B.

Proof. Since A ∪ B consists of everything that is in either A or B, it obviouslycontains every member of B, and so B ⊆ A ∪ B as required.

Suppose now that B = A ∪ B. We must show that A is a subset of B, sochoose any x ∈ A. Then by definition of ∪, x ∈ A ∪ B. But we supposed thatB = A ∪ B, and so in fact we must have x ∈ B. Since x was arbitrary this showsthat if B = A ∪ B then A ⊆ B.

Finally, suppose that A ⊆ B. We must show that B = A∪B. We have alreadyproved that B ⊆ A ∪ B in general, so it only remains to show that A ∪ B ⊆ Bin this instance. Pick any y ∈ A ∪ B, then y ∈ A or y ∈ B. If y ∈ A, then sinceA ⊆ B every element of A is an element of B, and so y ∈ B as well. Hence ineither case y ∈ B, and thus A∪B ⊆ B. Together these inclusions show B = A∪Bwhenever A ⊆ B.

Two important set identities are the following:

Proposition 1.28 (De Morgan’s Laws). The following identities hold for any

choice of sets A and B:

(i) (A ∩ B)′ = A′ ∪ B′

(ii) (A ∪ B)′ = A′ ∩ B′

Proof. We shall prove (i), leaving the proof of (ii) as an exercise. We have thefollowing chain of equivalent statements:

x ∈ (A ∩ B)′ ⇔ x is not in A ∩ B

⇔ x is not in A, or x is not in B

⇔ x ∈ A′ or x ∈ B′

⇔ x ∈ A′ ∪ B′

14

SETS

It is to see easy from the definitions that that each one is equivalent to the next,and so we have shown that x is in (A∩B)′ if and only if it is in A′ ∪B′, i.e. thesesets have the same elements, and so (A ∩ B)′ = A′ ∪ B′ as required.

Exercise 1.29. Prove the following set identity: A − B = A ∩ B′.

Remark. This exercise shows that the operation − is essentially redundant sincethe set difference of A and B can just as well be defined in terms of the operations∩ and ′.

Rather than arguing from the definitions, a convenient mathematical methodfor proving theorems about sets involves the use of truth tables. Consider thefollowing Venn diagrams for one, two and three sets respectively.

C

AA

A

11

U

UU

1

0

B

B

10 01

00

000

111

110

011101

010

001

100

For the case of one set, there are two distinct regions, namely inside A and outsideA. For two sets we have four regions in total: in both A and B, in one set but notthe other, and in neither of the sets. For three sets there are eight regions.

In a truth table, sets are written along the top row of the table, with the initialcomponent sets (A, B, etc.) of interest on the left, and those constructed fromthem to the right. The columns underneath are then filled in from left to right witha 1 or a 0 depending on whether or not the set contains that element, 1 indicatingmembership of the set, and 0 indicating non-membership. So, for example, if wewanted to prove an identity just involving a set A the table would begin

A · · ·1 · · ·0 · · ·

15


since there are two possibilities for where x ∈ U could live. For an identityinvolving two sets A and B the table would begin

A B · · ·1 1 · · ·1 0 · · ·0 1 · · ·0 0 · · ·

since there are four possibilities. The table for a three set identity has eight rows,and in general if there are n component sets then we need 2n rows of 0’s and 1’s.Hence this method becomes somewhat impractical (to do by hand) for n ≥ 4, since24 = 16, 25 = 32, . . .

Now, to prove an identity, we have to determine the remaining entries in thetable according to the following rules:

T1: To calculate A′ change each 0 in the column for A to a 1 and vice versa.

T2: To calculate A∪B put a 0 if there is a 0 in the column for A and the columnfor B, otherwise put a 1.

T3: To calculate A∩B put a 1 if there is a 1 in the column for A and the columnfor B, otherwise put a 0.

T4: To calculate A−B put a 1 if there is a 1 is the column for A and a 0 in thecolumn for B, otherwise put a 0.

T5: To calculate A△B put a 1 if there is precisely one 1 in the columns for Aand B, otherwise put a 0.

A simple example is the following:

A A′ (A′)′

1 0 1

0 1 0

The 1 and the 0 in the first column indicate the two possible choices of membershipor non-membership of the set A for a given object x. To calculate the column forA′ we use rule T1 to change the 1 in the column for A to a 0, and the 0 to a 1 —thus if x ∈ A then x /∈ A′, and if x /∈ A then x ∈ A′. Calculating the column for(A′)′ we look at the column for A′, and again swap 0’s and 1’s according to T1.Now note that we end up with exactly the same column for (A′)′ as we had for A.Thus this calculation proves that A = (A′)′ for any choice of set A, since a generalelement x ∈ U lies in A if and only if it lies in (A′)′.

Example 1.30. Recall from Proposition 1.28 that the second of De Morgan’sLaws state that for any sets A and B, the following identity holds:

(A ∪ B)′ = A′ ∩ B′.

16

SETS

We can prove this by considering the following truth table:

A B A ∪ B (A ∪ B)′ A′ B′ A′ ∩ B′

1 1 1 0 0 0 0

1 0 1 0 0 1 0

0 1 1 0 1 0 0

0 0 0 1 1 1 1

The third column is calculated according to T2, the fourth, fifth and sixth byusing T1, and the final column by using T3. Note that the columns for (A ∪ B)′

and A′ ∩B′ are the same, and hence the sets are equal, no matter what we choosefor A and B.

Note also that every time there is a 1 in the column for A there is also a 1 inthe same row in the column for A ∪ B. Thus every time a general element x is inA it is also an element of A ∪ B, and so we have proved that A ⊆ A ∪ B. [In factthis inclusion follows directly from Proposition 1.27 just by changing the labels —there we proved that B ⊆ A ∪ B, but this is true for any choice of sets A and B,so by changing B to A and A to B the inclusion will still be valid.]

Example 1.31. We show that A ∩ B ⊆ A ∪ B. The truth table is

A B A ∩ B A ∪ B

1 1 1 1

1 0 0 1

0 1 0 1

0 0 0 0

Thus x ∈ A ∩ B ⇒ x ∈ A ∪ B as required, since every time there is a 1 in thecolumn for A ∩ B, there is also a 1 in the column for A ∪ B.

Example 1.32. Consider the following truth table:

A B A′ B′ B − A A − B A′ − B′ A ∩ B′

1 1 0 0 0 0 0 0

1 0 0 1 0 1 0 1

0 1 1 0 1 0 1 0

0 0 1 1 0 0 0 0

Comparison of the 5th and 7th columns shows that A′ − B′ = B − A in general,and comparison of the 6th and 8th columns shows that A − B = A ∩ B′.

Example 1.33. Consider the following truth table for three sets:

A B C A△B (A△B) ∩ C A ∪ B (A ∪ B) ∩ C A ∩ B ∩ C D

1 1 1 0 0 1 1 1 0

1 1 0 0 0 1 0 0 0

1 0 1 1 1 1 1 0 1

0 1 1 1 1 1 1 0 1

1 0 0 1 0 1 0 0 0

0 1 0 1 0 1 0 0 0

0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

17


where in the last column we write D = [(A∪B)∩C]− (A∩B ∩C). The fifth andninth columns are the same, and so we have proved that the following set identityholds:

(A△B) ∩ C = [(A ∪ B) ∩ C] − (A ∩ B ∩ C).

Note that the eighth column features the intersection of three sets — this is cal-culated by a simple generalisation of the rule T3, where we now put a 1 if thereis a 1 in the columns for all three sets A, B and C, and a 0 otherwise. There isonly one row with a 1 in it, which corresponds to the single region out of the eightthat is the intersection of all three sets.

There is a similar generalisation for the rule T2, but T1 only works if we aredealing with one set, and the other two rules if we are dealing with two sets.

As a final point, one other thing that we can do with truth tables is to produceexamples of sets which do not satisfy certain equations. In the above table thecolumns for A△B and (A△B)∩C are not the same, which indicates that the setsare, in general, different. But note that the rows in which there are differences arethe fifth and sixth, which correspond, respectively, to a point x that is in A only,and to a point x that is in B only. Thus if we choose our sets A, B and C so thatthere is a point that is only in A say, then when it comes to calculating A△B and(A△B) ∩ C we will get different results. For example,

A = {1, 2}, B = {2, 3}, C = {3, 4} ⇒ A△B = {1, 3} and (A△B) ∩ C = {3}which are different; here 1 only lies in A. On the other hand, if we choose A, Band C so that there are no points that lie only in A and no points that lie only inB, then we will get equality:

A = {2, 4}, B = {2, 3}, C = {3, 4, 5} ⇒ A△B = {3, 4} = (A△B) ∩ C

Exercise 1.34. Use truth tables to establish the following set identities:

(A − B) ∪ A′ = (A ∩ B)′

(A△B)′ = (A ∩ B) ∪ (A ∪ B)′

(A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

Use your table for the second identity to first find an example of sets A and B forwhich (A△B)′ = A ∩ B, and then to find an example of sets A and B for which(A△B)′ ⊃ A ∩ B. [You will need to also specify a choice for the universal set U .]

The use of truth tables is not well suited to dealing with identities involvingthe Cartesian product ×, and so in these cases we must return to arguing fromthe definition.

Example 1.35. If A, B and C are nonempty sets such that B ⊂ C, then A×B ⊂A×C. To see this we must first show that A×B is a subset of A×C, so take anyelement (x, y) ∈ A × B. By definition of A × B, x ∈ A and y ∈ B. But B ⊂ C,and so we also know that y ∈ C. Thus (x, y) ∈ A × C as required.

Finally, we must show that A × B is a proper subset of A × C. Since A isnonempty we know that it contains at least one element, call it x say. Next, sinceB ⊂ C, there is some y ∈ C such that y /∈ B. But then (x, y) ∈ A × C, by choiceof x and y, and since y /∈ B we must have (x, y) /∈ A × B.

18

SETS

Exercise 1.36. Prove that following set identities:

(A × B) ∩ (A × C) = A × (B ∩ C)

(A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)

Find examples of sets A and B for which (A×B)∪(B×A) equals (A∪B)×(A∪B),and examples for which these two sets are unequal.

19

2 Relations

Relations

In the previous chapter we worked with sets, constructing new sets from old usingoperations such as a union and intersection, and were interested in whether theresulting sets were equal, or one was contained in another, etc. However we paidvery little attention to the nature of the elements of the given set(s), except whendefining sets by a rule. Then we wrote things like A = {x ∈ Z : 1 ≤ x < 4}, wherewe make use of the fact that there is an order on the set Z of integers which allowsus to say if one integer is larger than another. That is, given any two integersx, y ∈ Z, we can write statements such as x < y or x + y ≥ 5, which will theneither be true or false.

Consider the following subset of natural numbers: B = {1, 2, 3, 4}. In thischapter we are concerned with statements such as “2 divides 4” and “3 > 2”,which are called relations on the set B. That is a relation is a property that eitherholds or does not hold for each given ordered pair of elements from the set. In thesame way “is parallel to” and “is perpendicular to” are relations on the set L ofall lines in the plane.

In general a relation on a set A will be denoted by the letter R, and we willwrite aR b whenever a, b ∈ A are elements that satisfy the relation, that is “a isrelated to b by R”. If we need to express the fact that a and b do not satisfy therelation then we will write a 6R b. So if Rd denotes the relation “divides” on theset B considered above then we may write 1Rd 1, 1Rd 2, 1Rd 3, 1Rd 4, 2Rd 2,2Rd 4, 3Rd 3 and 4Rd 4 to denote that “1 divides 1”, 1 divides 2”, etc., notingthat m Rd n is not true for any other choice of m,n ∈ B.

Similarly, let L be the set of lines from above, R the relation “is parallel to”,and consider the lines:

l1 : 3x + y = 1 l2 : x − 3y = 5 l3 : 6x + 2y = −7.

These lines have gradients −3, 13 and −3 respectively, and so l1 R l3, but l2 6R l1,

and l2 6R l3.

In order to give a very precise set-theoretic definition of the concept of a rela-tion, note the following: given a relation R on A, and an ordered pair (a, b) ∈ A×B,either aR b or a 6R b is true, but not both. So our relation R splits the set A×A ofpairs into two parts: that for which the relation holds, and that for which it doesnot hold. We can identify the relation R with the subset of A × A for which therelation holds. For example if we take B = {1, 2, 3, 4} again, then Rd, the relation“divides”, is identified with

Rd = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)} ⊂ B × B.

Here the first number in each pair divides the second. On the other hand, if welet Rl be the relation “is less than”, then 1 < 2, 2 < 3, etc., but 3 6< 1, and so on.The subset of B × B determined by Rl is thus

Rl = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}.

20

RELATIONS

Bearing in mind this passage from an intuitive idea of relations to subsets ofCartesian products, the next definition is quite natural:

Definition 2.1. A relation R on a set A is a subset of A × A.

Example 2.2.

(a) Let A = N and let R mean “>”. Then 2R 1, 3R 1, . . . , 3R 2, 4R 2, . . . , 4R 3,5R 3, . . . are all true, and so if we take the set viewpoint of Definition 2.1 wehave

R = {(2, 1), (3, 1), . . . , (3, 2), (4, 2), . . . , (4, 3), (5, 3), . . .},since (a, b) ∈ R means that aR b, that is a > b.

(b) Let A = {1, 2, 3, 4} and let aR b mean that “a = 2b”. Then 2R 1 and 4R 2,so that R = {(2, 1), (4, 2)}.

(c) Let R = {(x, y) : x + y = 1, x, y ∈ R} ⊂ R × R = R2. This defines a relationon the set R. Geometrically each pair (x, y) ∈ R2 satisfying the relation R,i.e. each pair for which (x, y) ∈ R is a point on the graph of the equationx + y = 1

Example 2.3. Let P be a set of people and let R denote “is the father of”. TheaR b and (a, b) ∈ R are both ways of denoting that a is the father of b.

Exercise 2.4.

(a) Let R be the relation on the set N given by aR b if ab is a factor of 6. Whatis the subset of N2 determined by this relation?

(b) Consider the following subset of N × N:

{(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2)}.

Find a way to express this relation on N in words.

Properties of relations

Definition 2.1 says that any subset of A × A is a relation, which gives rise to alarge number of example. Indeed, if A is finite, with |A| = n, then |A × A| = n2,and so the number of relations on A is the number of subsets of A × A, which is2|A×A| = 2n2

. For small values of n we have

n 1 2 3 4 5 · · ·2n2

2 16 512 65536 33554432 · · ·

and so we see that the number of relations grows very quickly. However some ofthese relations are more interesting mathematically than others. The following arethree properties that are of particular concern for us.

Definition 2.5. A relation R on a set A is reflexive if (a, a) ∈ R (or aR a) forevery a ∈ A.

21

Properties of relations

Example 2.6. Let A = {1, 2, 3} and define two relations R1 and R2 on A by:

R1 = {(1, 1), (2, 2), (2, 3), (3, 3)}, R2 = {(1, 1), (2, 2), (2, 3), (3, 1)}

Then R1 is reflexive, but R2 is not reflexive. Even though (1, 1) ∈ R2 and (2, 2) ∈R2, we do not have (3, 3) ∈ R2, and so the definition of reflexivity fails for at leastone choice of a ∈ A, namely a = 3.

Example 2.7. Let L be the set of lines in the plane and let R mean “is parallelto”. Every line is parallel to itself, and so this relation R is reflexive. The relation“is perpendicular to” is not reflexive — a line cannot be at right angles to itself!

Example 2.8. The relation R on N defined by (a, b) ∈ R if b > a is not reflexive,since, for example, (1, 1) /∈ R because 1 is not greater than itself.

Note. Equality on a set is a reflexive relation.

Definition 2.9. A relation R on a set A is symmetric if for each pair (a, b) ∈ R,we also have (b, a) ∈ R. That is, if we flip the elements round in each pair in R,the resulting pairs are still in the set R. So if aR b, then we also have bR a.

Example 2.10. Let A = {1, 2, 3}, and this time define R1 and R2 by

R1 = {(1, 2), (1, 3), (2, 1), (3, 1)}, R2 = {(1, 1), (2, 2), (2, 3)}

Then R1 is symmetric, but R2 is not symmetric since we have (2, 3) ∈ R2, but(3, 2) /∈ R2.

Example 2.11. Parallelism of lines in the plane is a symmetric relation, as isperpendicularity of lines.

Note. Equality on a set is a symmetric relation.

Definition 2.12. A relation R on a set A is transitive if whenever (a, b) and (b, c)are pairs that satisfy (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R as well. This meansthat if aR b and bR c, we also have aR c.

Example 2.13. Let A = {1, 2, 3}, and now define R1 and R2 by

R1 = {(1, 2), (2, 3), (1, 3)}, R2 = {(1, 1), (1, 3), (3, 2)}.

Then R1 is transitive, since the only pairs where the second element of one is equalto the first element of the other are (1, 2) and (2, 3), and we also have (1, 3) ∈ R1.However, R2 is not transitive: we have (1, 3) ∈ R2 and (3, 2) ∈ R2, but (1, 2) /∈ R2.

Example 2.14.

(a) The relation < on N is a transitive relation: if a, b, c ∈ N such that a < band b < c, then a < c.

(b) Parallelism of lines in the plane is a transitive relation. Perpendicularity oflines in the plane is not — why not?

22

RELATIONS

Note. Equality on a set is a transitive relation.

Exercise 2.15. Let A = {1, 2, 3, 4}, and define relations R1, R2 and R3 on A by

R1 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 4), (4, 3), (4, 4)}R2 = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3), (4, 4)}R3 = {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}

In each case determine which, if any, of the three properties are satisfied.

Example 2.16. Consider the following relations on the set Z of all integers:

R1 = {(a, b) ∈ Z2 : ab is a multiple of 5}, R2 = {(a, b) ∈ Z2 : a ≥ |b|}For R1 we have the following:

• it is not reflexive: for example 1 ∈ Z such that (1, 1) /∈ R1, because 12 = 1 isnot a multiple of 5.

• it is symmetric: since ab = ba for all a, b ∈ Z, if ab is a multiple of 5 then sois ba.

• it is not transitive: for example (1, 10) ∈ R1 since 1 × 10 = 10 = 2 × 5, andso (10, 1) ∈ R1 because R1 is symmetric. But (1, 1) /∈ R1 as already shown.

For R2 we have

• it is not reflexive: for example −2 ∈ Z such that (−2,−2) /∈ R2, since−2 6≥ | − 2| = 2.

• it is not symmetric: for example 2,−1 ∈ Z such that (2,−1) ∈ R2, because2 ≥ | − 1| = 1, but the reversed pair (−1, 2) /∈ R2 since −1 6≥ |2| = 2.

• it is transitive: to see this recall that for any real number x (and hence forany integer: Z ⊂ R) we have |x| ≥ x. Thus if a, b, c ∈ Z such that (a, b) ∈ R2

and (b, c) ∈ R2, then a ≥ |b| and b ≥ |c|, hence a ≥ |b| ≥ b ≥ |c|, and so(a, c) ∈ R2.

Note. In each case of the above example we are trying to show whether a givenrelation has or does not have a certain property. These three properties are alldefined by universal statements — the definitions all feature phrases such as “foreach”, “for every”, or “whenever”. To prove that such a statement is:

TRUE one must give a general argument that holds for all possible choices ofpairs in the relation. It is not sufficient when determining whether or notR1 is reflexive in the above, for example, to check that (a, a) ∈ R1 for oneor two values of a ∈ Z. Indeed we do have (5, 5) ∈ R1 and (−25,−25) ∈ R1.But, as shown, (1, 1) /∈ R1.

FALSE it is sufficient to give a counterexample. For example, a relation R istransitive if all pairs of the form (a, b) and (b, c) in R we also have (a, c) ∈ R.So to show that R1 above is not transitive it is enough to find some valuesof a, b, c ∈ Z for which (a, b), (b, c) ∈ R1, but (a, c) /∈ R1, and a = 1, b = 10,c = 1 did the job nicely. Note that it is possible to find values of a, b and cfor which (a, b), (b, c) and (a, c) do all lie in R1 (e.g. a = 2, b = 5, c = 15),but it is not true for all choices.

23

Equivalence relations

Equivalence relations

Definition 2.17. An equivalence relation on a set A is a relation R that is (i) re-flexive, (ii) symmetric, and (iii) transitive.

Thus a R relation on a set A is an equivalence relation if

(i) (a, a) ∈ R for each a ∈ A.

(ii) whenever (a, b) ∈ R we also have (b, a) ∈ R.

(iii) whenever (a, b) ∈ R and (b, c) ∈ R, we also have (a, c) ∈ R.

When dealing with equivalence relations it is common to write ∼ rather thanR for the relation in question, particularly when writing it in the form a ∼ b.

Example 2.18.

(a) Parallelism of lines in the plane is an equivalence relation.

(b) Perpendicularity of lines is not an equivalence relation — it is not reflexive,nor is it transitive.

(c) Similarity of triangles is an equivalence relation.

(d) ≤ on N is not an equivalence relation – it is not symmetric.

(e) Equality on a set is an equivalence relation.

Note that to prove that a given relation is not an equivalence relation it isenough to show that it does not possess one of the three required properties, andto do that it is enough to provide one counterexample. On the other hand, toprove that a relation is an equivalence relation each of the properties must beshown to hold, and in each case that requires a general argument.

Example 2.19. Consider the relation R defined on the set R of all real numbersby

R = {(x, y) ∈ R2 : x − y ∈ Z}.

That is, R consists of all pairs (x, y) for which the difference x−y is integer-valued.This is an equivalence relation which can be shown as follows:

Reflexivity : for each x ∈ R we have x − x = 0 ∈ Z, and so (x, x) ∈ R.

Symmetry : if (x, y) ∈ R, then x − y = m for some m ∈ Z. But then y − x =(−1)(x − y) = −m ∈ Z, and so (y, x) ∈ R.

Transitivity : if (x, y) ∈ R and (y, z) ∈ R then x − y = m and y − z = n for someintegers m and n. But then x − z = (x − y) + (y − z) = m + n ∈ Z, and so(x, z) ∈ R.

Exercise 2.20. Prove that the following relation R on R is not an equivalencerelation:

R = {(x, y) ∈ R2 : x − y ∈ N}.

24

RELATIONS

Example 2.21. Define a relation R on the set of integers Z as follows: for a, b ∈ Z,(a, b) ∈ R if a − b is a multiple of 7. That is, (a, b) ∈ R if there is some integerm ∈ Z such that a − b = 7m, or, equivalently, a − b is (exactly) divisible by 7.

(i) R is reflexive since a − a = 0 = 7 × 0 for all a ∈ Z.

(ii) R is symmetric since if (a, b) ∈ R then a − b = 7m for some integer m ∈ Z,and so b − a = −7m = 7(−m), thus (b, a) ∈ R, since −m ∈ Z.

(iii) R is transitive: if (a, b), (b, c) ∈ R then there are integers m,n ∈ Z such thata − b = 7m and b − c = 7n. Hence

a − c = (a − b) + (b − c) = 7m + 7n = 7(m + n),

which shows that (a, c) ∈ R, since m + n ∈ Z.

This equivalence relation is called congruence modulo 7 and is denoted by “≡mod 7”. That is, we write m ≡ n mod 7 to mean that m − n is a multiple of7. The same thing can be redone with any other positive integer other than7. For example m ≡ n mod 19 means that m − n is a multiple of 19: we have3 ≡ 41 mod 19 since 41 − 3 = 38 = 2 × 19, but 3 6≡ 32 mod 19.

Equivalence classes and partitions

We saw earlier that the number of relations on a set of size n grows very quicklyas n increases. Do we have the same behaviour for equivalence relations? The firstfew values are as follows:

order of set, |A| 1 2 3 4 5 · · ·no. equiv. relations 1 2 5 15 52 · · ·

We see that the growth is very much slower — but how can we compute thesenumbers?

Example 2.22. Let C be the set of students in this class, and for any a, b ∈ C leta ∼ b mean that student a and student b are doing the same second subject. It iseasy to see that ∼ defines an equivalence relation on C. Moreover note that everystudent is doing precisely one second subject, and so the set C can be subdividedinto disjoint subsets according to these subjects.

Definition 2.23. Let A be a set and ∼ an equivalence relation on A. Let a ∈ A,then the equivalence class of a, denoted by [a], is the set of elements b ∈ A suchthat a ∼ b, i.e. a and b are related. Thus [a] = {b ∈ A : a ∼ b}.

Remark. Since any equivalence relation ∼ is reflexive, we have a ∼ a for eacha ∈ A, and so a ∈ [a]. Thus each equivalence class is always nonempty.

Example 2.24. Let L be the set of lines in the plane and ∼ denote the equivalencerelation of parallelism. If we denote the y-axis by l0, then l ∈ [l0] if it parallel tol0, that is if it is vertical. So [l0] consists of all vertical lines in the plane. Notethat if l1 denotes the line x = 5, then it is vertical, and so l1 ∈ [l0]. On the other

25


hand, we also have l0 ∈ [l1], and in fact [l1] also consists of all vertical lines, so[l0] = [l1].

The equivalence classes of L generated by ∼ are determined precisely by all thepossible directions a straight line may take. Each line has precisely one direction.

Example 2.25. Consider the set Z again, with the equivalence relation ≡ mod7,as defined in Example 2.21. Then a ≡ b mod 7 if and only if a = b + 7m for somem ∈ Z, and so the equivalence classes for this relation are

[0] = {. . . ,−14,−7, 0, 7, 14, . . .},[1] = {. . . ,−13,−6, 1, 8, 15, . . .},[2] = {. . . ,−12,−5, 2, 9, 16, . . .},[3] = {. . . ,−11,−4, 3, 10, 17, . . .},

.........

[6] = {. . . ,−8,−1, 6, 13, 20, . . .}.

and note that [0] = [7] = [−14] = · · · , [3] = [−4] = [17] = · · · , and so on. Eachinteger belongs to precisely one equivalence class, and they split up the integersas shown below:

1 2 7 8 9 140−7−6−5−14

[1][0]

[2]

Exercise 2.26. Let ∼ be the relation on N defined by a ∼ b if a = 10kb for someinteger k ∈ Z. Show that ∼ is an equivalence relation on N. Describe the sets [1],[300] and [10720].

The next result shows that some of the properties of equivalence classes thatwe have encountered in the examples above hold true for any equivalence relation.

Theorem 2.27. Let ∼ be an equivalence relation on a set A, and let a, b ∈ A.

(a) If a ∼ b then [a] = [b].

(b) If [a] ∩ [b] 6= ∅ then a ∼ b, and hence [a] = [b].

Remark. Part (a) says that the equivalence classes [a] and [b] are equal if a andb are related. A consequence of part (b) is that any two equivalence classes areeither equal or disjoint; when they are equal a and b have to be related.

Proof. (a) We are assuming a ∼ b, and must show both that [a] ⊆ [b], and that[b] ⊆ [a] in order to conclude that [a] = [b]. So let x ∈ [a], then by definition of [a]we have a ∼ x. But ∼ is symmetric, hence x ∼ a, and by our assumption we havea ∼ b. So now by transitivity we have x ∼ b. Using symmetry once more we get

26

RELATIONS

b ∼ x, and this says precisely that x ∈ [b]. But x was an arbitrary element of [a],hence we have shown [a] ⊆ [b].

An almost identical argument show that if y ∈ [b] then y ∈ [a], hence [b] ⊆ [a]as required, and the two inclusions together imply [a] = [b].

(b) Let x ∈ [a]∩ [b] — such an x exists by our assumption. Then a ∼ x and b ∼ x.Now x ∼ b by symmetry, and since a ∼ x we get a ∼ b by transitivity. Thus[a] = [b] by part (a).

This theorem gives fundamental properties about equivalence classes, andhence about equivalence relations, and is a major help in enabling us to calculatethe number of equivalence relations on a set of a given size. Indeed, recall fromDefinition 1.22 that a partition of a set A is any collection P of nonempty subsetsof A such that every element of A appears in exactly one subset from P . The dia-gram in Example 2.25 shows that the equivalence classes {[0], [1], [2], [3], [4], [5], [6]}partition the set Z of all integers. Similarly second subjects allow us to partitionthe set C of students in this class, as considered in Example 2.22. The theoremabove shows that this is always true:

Corollary 2.28. Let ∼ be an equivalence relation on a set A. Then the equivalence

classes generated by ∼ form a partition of A.

Theorem 2.27 and Corollary 2.28 show that if we have an equivalence relationon a set A then the collection of equivalence classes forms a partition of A. Nowwe show that, conversely, any partition of A defines an equivalence relation.

Theorem 2.29. Let A be a (nonempty) set. Then any partition on A defines an

equivalence relation on A.

Proof. Let P be a partition on A. We define a relation R on A as follows: fora, b ∈ A, we let (a, b) ∈ R if a and b belong to the same subset that makes up thepartition P . That is, (a, b) ∈ R if a, b ∈ B for some B ∈ P . Now note that

(i) (a, a) ∈ R for all a ∈ A, since each a ∈ A is in (exactly) one B ∈ P .

(ii) If (a, b) ∈ R then there is some B ∈ P such that a, b ∈ B. But then b, a ∈ B,and so (b, a) ∈ R.

(iii) If (a, b) ∈ R and (b, c) ∈ R then a, b belong to some subset, B say, that belongsto P , and b, c belong to some subset, C say. Thus b belongs to both B and C. Butit is only supposed to belong to one set in P , and so we must have B = C. Hencea, c ∈ B, and thus (a, c) ∈ R.

Example 2.30. For each real number r ≥ 0 let Cr = {(x, y) ∈ R2 : x2 + y2 = r2}.Thus Cr is the set of points on the circle with centre (0, 0) and radius r (andwhere C0 = {(0, 0)} consists of a single point). Each point on the plane lies onprecisely one such circle, and so the collection {Cr : r ≥ 0} is a partition of theplane R2. It is the partition generated by the equivalence relation ∼ defined by(x1, y1) ∼ (x2, y2) if the points (x1, y1) and (x2, y2) are equidistant from the origin.

27


x

y

rs

Example 2.31. For any integer n > 0 the sets

[0] = {. . . ,−2n,−n, 0, n, 2n, . . .} = {kn : k ∈ Z}[1] = {. . . ,−2n + 1, ,−n + 1, 1, n + 1, 2n + 1, . . .} = {kn + 1 : k ∈ Z}...

......

......

......

...

[n − 1] = {. . . ,−n − 1,−1, n − 1, 2n − 1, 3n − 1, . . .} = {kn − 1 : k ∈ Z}

partition the integers Z. They are the equivalence classes generated by the equiva-lence relation ≡ mod n, generalised from Example 2.21, and are known as residue

or congruence classes. The collection {[0], [1], . . . , [n − 1]} is denoted Zn. Notethat for each r ∈ Z, [r] is an infinite subset of Z, and that [n] = [0], [n + 1] = [1]etc.

Example 2.32. To find all of the equivalence relations on A = {a, b, c} it isenough to write down all of the partitions, which we can do in a systematic waysas follows:

P1 ={{a, b, c}

},

P2 ={{a}, {b, c}

},

P3 ={{b}, {a, c}

},

P4 ={{c}, {a, b}

},

P1 ={{a}, {b}, {c}

}.

We can now write down the equivalence relations corresponding to these partitions,and they are, respectively,

R1 = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)},R2 = {(a, a), (b, b), (b, c), (c, b), (c, c)},R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)},R4 = {(a, a), (a, b), (b, a), (b, b), (c, c)},R5 = {(a, a), (b, b), (c, c)}.

Example 2.33. Suppose we want to find all of the equivalence relations ∼ on theset A = {1, 2, 3, 4} for which both 1 ∼ 4 and 2 ∼ 3. Since 1 ∼ 4 they must belongto the same subset in the partition generated by ∼. Similarly 2 and 3 must belong

28

RELATIONS

to the same subset. Thus in this case there are only two possible partitions of Awhich are

P1 ={{1, 2, 3, 4}

}and P2 =

{{1, 4}, {2, 3}

}.

The corresponding equivalence relations are then

R1 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4),

(3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)},R2 = {(1, 1), (1, 4), (2, 2), (2, 3), (3, 2), (3, 3), (4, 1), (4, 4)}

29

3 Functions

Basic definitions and properties

In our story so far we have considered sets on their own, together with ways ofcomparing (internally) the elements of such a set. In this section we deal with thepossibility of moving from one set to another.

Definition 3.1. Let A and B be nonempty sets. A function or map from A to

B is a rule that associates to each element of A an element of B.

Functions will generally be denoted by lower case letters such as f or g. If f isa function from A to B we write this as f : A → B, and the unique element of Bassociated to a given element a ∈ A, known as the image of a under f , is denotedby f(a).∗ The set A is called the domain of f , and the set B the codomain of f .The range or image set of f is the set {b ∈ B : b = f(a) for some a ∈ A}. It isthe subset of B consisting of the images of all of the a ∈ A.

Two functions f : A → B and g : C → D are equal if A = C, B = D andf(a) = g(a) for all a ∈ A.

Example 3.2. As an example of a function f : N → N, consider the map thatsends a positive integer n to the integer 2n + 1. Thus, for each n we can calculatethe image f(n) by first doubling n, then adding 1. That is f(1) = 3, f(2) = 5,f(3) = 7,. . . and so on. We write the definition of this function as

f : N → N, n 7→ 2n + 1, (read: “n is mapped to 2n + 1”)

or

f : N → N, f(n) = 2n + 1. (read “f of n is equal to 2n + 1”).

It is important to note that the function is more than just the formula 2n + 1 —it is necessary to specify the domain and codomain as we shall see in examplesbelow. In this case the domain and codomain are both the natural numbers N,but the range of f is the proper subset {3, 5, 7, . . .} = {n ∈ N : n odd, n ≥ 3}.

If we were to write f : N → N, f(n) = 1 − n then this would not be a well-

defined function since for all n ≥ 1 we have 1 − n ≤ 0, and so 1 − n does not liein the specified codomain. If we change the codomain to be Z then our functionwould be well-defined.

Example 3.3. Let A = {a, b, c, d, e} and B = {1, 2, 3, 4}. Define a function g fromA to B by setting

g(a) = 1, g(b) = 2, g(c) = 3, g(d) = 3, g(e) = 1.

In this case the domain is finite, so to define g we can just list its action on eachelement of A. This can be represented graphically by

∗An alternative notation for the image of a ∈ A under f is (a)f — that is, the function iswritten to the right of the element This notation is particularly common in algebra texts, so youshould always check to see which convention is being used.

30

FUNCTIONS

a

bc

de

g

1

2

3

4

where the arrows connect the elements of the domain A to their images under gin the codomain B. That is, the domain of g is A = {a, b, c, d, e}, the codomain ofg is B = {1, 2, 3, 4}, and the range of g is {1, 2, 3}.

Exercise 3.4.

(a) Consider the following diagrams and in each case say which determine afunction from A = {1, 2, 3, 4} to B = {a, b, c}, and if so, what the range is.

4 44

1 1 1

2 2 2

3 33

a aa

b bb

c cc

(i) (ii) (iii)

(b) In each of the following cases determine if the given function is (well-)defined,and if so determine its range.

f : N → N, n 7→ 7n + 5 g : N → Q, n 7→ 1n− 1

n+1

h : Z → Q, n 7→ 1n− 1

n+1 k : R → R, x 7→√

x − 3

In Examples 3.2 and 3.3 we can use the functions f and g to specify certainsubsets of N × N and A × B respectively, namely

{(n, 2n + 1) ∈ N × N : n ∈ N} and {(a, 1), (b, 2), (c, 3), (d, 3), (e, 1)}.

These subsets contain all elements of the form (n, f(n)) for n ∈ N, and (x, g(x))for x ∈ A. That is, they pair each element of the domain with its (unique) imageunder the given function. Note that in both of these subsets each element of thedomain appears exactly once as the first component of a pair.

We can turn this observation on its head to give a set-theoretic definition offunctions, in a similar way to our definition of relations. Indeed, if A and B arenonempty sets then a function from A to B is any subset of A×B in which eachelement of A appears as the first component in one and only one of the orderedpairs in this subset. In particular, if we take this as our definition then we see thata function from a set A into A itself is a subset of A × A, and hence a particular

31


type of relation on A. In general, although each a ∈ A occurs first in one (and onlyone) ordered pair, a given element of B may occur once, several times or never inthe subset.

Example 3.5. Let A = {1, 2, 3} and consider the following subsets of A × A:

R1 = {(1, 2), (2, 1), (3, 3), (2, 3)}R2 = {(1, 2), (2, 1), (3, 3)}R3 = {(1, 2), (2, 3), (3, 3)}R4 = {(1, 2), (2, 3)}.

These are all relations on A, however only R2 and R3 are functions on A. In R1

we see that 2 appears twice as the first element of a pair, and 3 never appears inR4.

Exercise 3.6. Determine which of the following subsets determine functions, andif so specify the domain, codomain and range.

A = {(x, x2 + 3) ∈ R × R+ : x ∈ R}B = {(x + 3, x) ∈ R+ × R : x ∈ R+}C = {(x2, x) ∈ R+ × R : x ∈ R}

where R+ = {x ∈ R : x ≥ 0} = [0,∞) — the subset of the reals consisting of allnonnegative numbers.

Definition 3.7. A function f : A → B is called one-to-one or injective if theimages of distinct elements of A are distinct elements of B. That is, if a1, a2 ∈ Asuch that a1 6= a2, then f(a1) 6= f(a2) as well.

An equivalent way to say this (and indeed the formulation one would use tocheck if a given function is injective) is that a function f is one-to-one if whenevera1, a2 ∈ A such that f(a1) = f(a2), then also a1 = a2.

Note that for f to be a one-to-one function from A into B it is not requiredthat every element b ∈ B appear in the range of f . However, if b ∈ B does lie inthe range then it is equal to f(a) for only one a ∈ A.

Graphically, a function f has arrows connecting points from A to points fromB, with exactly one arrow coming from each a ∈ A. The function being one-to-one means that at most one arrow arrives at each b ∈ B. Our function f inExample 3.3 is not one-to-one since f(c) = 3 = f(d), but c 6= d. Similarly, in thetwo diagrams below the first represents an injective function, while the second isnot injective:

aa aa

bb bb

cc cc

32

FUNCTIONS

Example 3.8.

(a) Consider the function f : N → N given by f(n) = 2n+1. This is one-to-one,for if n1, n2 ∈ N such that f(n1) = f(n2) then

2n1 + 1 = 2n2 + 1 ⇒ 2n1 = 2n2 ⇒ n1 = n2.

(b) Let R+ = {x ∈ R : x ≥ 0}. The following functions are one-to-one:

(i) f : R+ → R+, f(x) = x2.

(ii) g : R+ → R, g(x) =√

x.

It is clear that both of these define functions. f is one-to-one because

f(x1) = f(x2) ⇒ x21 = x2

2 ⇒ x1 = x2,

since the domain is R+, thus by definition xi ≥ 0, and so we have excludedthe negative square root by careful choice of our domain.

(c) The following functions are not injective:

(i) f : R → R+, f(x) = x2.

(ii) g : [−1, 1] → R, x 7→ x3 − 2x2 + x.

For example, f(−1) = 1 = f(1), and similarly g(1) = 0 = g(0). Note that inthis case, to show that something is not injective we only need to find twopoints in the domain that have the same image. The fact that f(−3) = f(3)shows equally well that f is not one-to-one.

This example shows that the property of being one-to-one depends on the choiceof domain — the functions in (b i) and (c i) look alike, but one is injective whilethe other is not.

Definition 3.9. A function from A to B is called onto (or surjective) if everyelement of B is the image of some element of A. That is the range of f is equal tothe codomain of f .

An equivalent way of saying this is that a function f is onto if for every b ∈ Bthere is some a ∈ A such that f(a) = b. That is we can always solve this equation,which is the strategy one takes when determining whether or not a given functionis surjective.

The following diagrams illustrate a surjective and a nonsurjective functionrespectively:

PSfrag

aa aa

bb bb

cc cc

33


since in the first every element in the codomain has at least one arrow point to it,whereas in the second no element from the domain is taken to c.

Example 3.10.

(a) The following functions are onto:

(i) f : R → R, f(x) = 2x.

(ii) g : R → R, g(x) = x + 1.

This is so since if we choose any y ∈ R, then f(y/2) = y and g(y − 1) = y.Hence the range of both f and g is all of R.

(b) The following functions are not surjective

(i) f : N → N, f(x) = 2x, and

(ii) g : N → N, g(x) = x + 1.

This time, there is no m ∈ N such that f(m) = 1 (it would have to satisfy2m = 1 ⇒ m = 1

2 /∈ N), and there is no n ∈ N such that g(n) = 1.

Again, the property of being surjective depends delicately on the choice of domainand codomain.

Example 3.11. Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Consider the followingsubsets of A × B:

R1 = {(1, a), (2, b), (3, c), (4, d)}R2 = {(1, a), (2, d), (3, c), (4, d)}R3 = {(1, a), (2, b), (3, c)}

Then R1 is a function, and it is both one-to-one and onto. R2 is also a function,but is neither one-to-one nor onto. Finally, R3 is not a even function, since it doesnot define an image for 4.

Definition 3.12. A function from A to B is called bijective if it is both one-to-oneand onto.

Example 3.13. Let A = {1, 2, 3, 4}, and consider the following functions:

(i) f(a) = a for all a ∈ A.

(ii) g(1) = 2, g(2) = 3, g(3) = 4, g(4) = 1.

(iii) h(1) = 4, h(2) = 1, h(2) = 2, h(4) = 3.

Each of the above is a bijective function, shown in the diagrams below.

hf g

1 11 111

2 2 2 22 2

3 3 3 33 3

4 44 444

34

FUNCTIONS

Exercise 3.14. Show that the following functions from R → R are bijections:

f(x) = x, g(x) = x3 h(x) = 3x − 5.

Definition 3.15. For any set A the identity function on A is the function Idefined by setting I(a) = a for all a ∈ A. We write IA if we want to highlight theset in question. It is the only function A → A that leaves all of the elements of Aunchanged, and is clearly a bijection from A to A.

Example 3.16. There are four functions from A = {1, 2} to itself, namely

11111111

22222222

There are two choices for where we can send 1, and two choices for where we cansend 2, giving a total of 2 × 2 = 4 functions. Note, however, that there are onlytwo bijections.

Exercise 3.17. Let A = {1, 2, 3} and B = {1, 2}. List the eight functions fromA to B. How many are onto? one-to-one? bijective? Are there more or fewerfunctions from A to B than there are from B to A?

Exercise 3.18. Let A and B be finite sets with |A| = m and |B| = n. Explain whythere are nm functions from A to B. Suppose that there is an injective functionfrom A to B. What does this say about the numbers m and n? Explain why wemust have m ≥ n if there is to be a surjective map from A to B.

Exercise 3.19. Suppose that A is a finite set. Explain why every injective mapf : A → A is also necessarily onto. Find an example of an injective map g : N → N

that is not onto.

From the above exercises we see that if a set A is of finite order then everyinjective map f : A → A is also necessarily surjective. On the other hand, for theprototypical infinite set N we can find a nonsurjective one-to-one mapping. Thisallows us to formulate a more precise definition of infinite set than “it is impossibleto list all of the elements”. Namely, a set A is infinite if and only if there existsan injective map f : A → A whose range is a proper subset of A.

For a more concrete approach to injectivity and surjectivity, recall that we areused to representing a given function f : R → R by the curve in the plane that isits graph. Note that for such a curve to be the graph of a function we must havethat every vertical line cuts the curve exactly once, since this then associates tothe value of x that determines the line its unique image under the function. If thisis the case then

(i) The curve is the graph of a one-to-one function if every horizontal line cutsthe curve at most once.

(ii) The curve is the graph of a one-to-one function if every horizontal line cutsthe curve at least once.

35

Composition of functions; inverses

x

xx

y yy

not a function not 1-1

not onto

Consider the set C = {(x, y) ∈ R2 : x2 + y2 = 1}. It is a curve that consistsof the points on the circle of radius 1 centred on the origin, but does not define afunction on R for two reasons. Firstly, if x ∈ R lies outside [−1, 1] then there isno y ∈ R such that (x, y) ∈ C, and so assigns no value to such an x. Secondly, if−1 < x < 1 then (x,−

√1 − x) ∈ C and (x,

√1 − x) ∈ C, that is we would have

two choices for the value of the function.

x

y

1

1

−1

−1

C


Definition 3.20. Let f : A → B and g : B → C be functions. That is, thedomain of g is the codomain of f . The function g ◦ f : A → C with domain A andcodomain C is defined by setting (g ◦ f)(a) = g(f(a)) for all a ∈ A, and is calledthe composition of f and g.

f(a) g(f(a))

g ◦ f

f g

A B A

a

Note. Since the function is written to the left of the set element (as in f(a)),composition of functions is read from right to left. That is (g ◦ f)(a) means dof on a then do g on the result. If someone is using the alternative convention inwhich the set element is written on the left of the function, i.e. (a)f instead off(a), then composition must be written and read from left to right — the functiong ◦ f illustrated above would, in the alternative convention send an element a ∈ Ato ((a)f)g ∈ C, and so the composite function would be written f ◦ g, or, often,just fg.

36

FUNCTIONS

Example 3.21. Let A = {1, 2, 3} and define functions f : A → A and g : A → Aby

f(1) = 2, f(2) = 3, f(3) = 1, and g(1) = 1, g(2) = 3, g(3) = 2.

f ◦ gg ◦ f

ff gg1

111

11

22 22

2 2

33333

3

Note that both functions are bijections, and the composed function g ◦ f is

(g ◦ f)(1) = 3, (g ◦ f)(2) = 2, (g ◦ f)(3) = 1.

If we do the composition in the other order we have

(f ◦ g)(1) = 2, (f ◦ g)(2) = 1, (f ◦ g)(3) = 3.

So we see that (g ◦ f)(a) 6= (f ◦ g)(a) for all a ∈ A and so these functions aredifferent. Thus the order of composition is important.

Remark. Note that functions f : A → B and g : A → B are different or unequal

(written f 6= g) if f(a) 6= g(a) for at least one a ∈ A — we did not need to showabove that this was the case for all a ∈ A.

Example 3.22. Consider the functions

f : R → R, f(x) = x2, and

g : R → R, g(x) = 2x + 1.

Since in both cases the domain and codomain is R, we can compose these in eitherorder to get functions from R to R. We have

(g ◦ f)(x) = g(x2) = 2x2 + 1, and (f ◦ g)(x) = (2x + 1)2 = 4x2 + 4x + 1.

Thus once again g ◦ f 6= f ◦ g, since, for example, (g ◦ f)(1) = 3 6= 9 = (f ◦ g)(1).

Suppose now that we have functions f : A → B, g : B → C and h : C → D.Our definition of composition does not allow us to compose three or more functionsdirectly. However we can form the compositions g ◦ f : A → C and h ◦ g : B → D,and then the compositions h ◦ (g ◦ f) : A → D and (h ◦ g) ◦ f : A → D, whichare both now functions from A to D. From this it seems that we may have to becareful with bracketing. The next result, illustrated by the diagram below, showsthat in fact this is not the case.

f(a) g(f(a))

h(g(f(a)))

g ◦ f h ◦ g

h ◦ (g ◦ f) = (h ◦ g) ◦ f

f

g

h

a

37


Proposition 3.23. Let f : A → B, g : B → C and h : C → D be functions. Then

(h ◦ g) ◦ f = h ◦ (g ◦ f).

Proof. We first note that both (h ◦ g) ◦ f and h ◦ (g ◦ f) are functions from A intoD. So to see that they are equal we must show that they both map any givenelement of A to the same element of D. But this follows from the definitions, sincefor any a ∈ A

((h ◦ g) ◦ f

)(a) = (h ◦ g)(f(a)) = h(g(f(a)))

and

(h ◦ (g ◦ f)

)(a) = h((g ◦ f)(a)) = h(g(f(a))).

Thus (h ◦ g) ◦ f = h ◦ (g ◦ f) since a was arbitrary.

The next two results show how the properties of injectivity and surjectivitybehave with respect to composition.

Proposition 3.24. Let f : A → B and g : B → C be bijections. Then g ◦ f is a

bijection from A to C.

Proof. We show that g ◦ f is both one-to-one and onto. In each case the strategyis the same as used in the examples earlier.

One-to-one: Let a1, a2 ∈ A, and suppose that

(g ◦ f)(a1) = (g ◦ f)(a2).

So by definition of g ◦ f we have g(f(a1)

)= g

(f(a2)

). Now g is one-to-one, hence

we must have that f(a1) = f(a2), and now since f is also one-to-one we must havethat a1 = a2, as required.

Onto: Take any element c ∈ C — we will show that it lies in the range of g ◦ f .Since g is onto there must be some b ∈ B such that g(b) = c. But f is also onto,hence there exists some a ∈ A such that f(a) = b for this particular b. Finally,note that (g ◦ f)(a) = g(f(a)) = g(b) = c, as required.

Proposition 3.25. Let f : A → B and g : B → C be functions.

(a) If g ◦ f : A → C is onto then g is onto.

(b) If g ◦ f : A → C is one-to-one then f is one-to-one.

Proof. (a) We are assuming that g ◦ f : A → C is onto, so for any given c ∈ Cwe can find some a ∈ A such that (g ◦ f)(a) = g

(f(a)

)= c. But then f(a) is an

element of B whose image under g is our arbitrarily chosen c, and hence g is onto.

(b) Suppose that a1, a2 ∈ A such that f(a1) = f(a2). We must show thata1 = a2. But, by our choice of a1 and a2, and by the definition of g ◦ f ,

f(a1) = f(a2) ⇒ g(f(a1)

)= g

(f(a2)

)⇒ (g ◦ f)(a1) = (g ◦ f)(a2).

By hypothesis g ◦ f is one-to-one, and so we must have a1 = a2 as required.

38

FUNCTIONS

Exercise 3.26. Find examples of finite sets A and B together with functionsf : A → B and g : B → A such that

(a) g ◦ f is onto but f is not onto.

(b) g ◦ f is one-to-one but g is not one-to-one.

[Hint: from Exercise 3.19 it follows that in each case we must have |A| < |B|.Further exercise: prove why this is so!]

Recall the identity function I : A → A from Definition 3.15. One reason itis called the identity because composing it with any other function leaves thatfunction unchanged. Indeed, suppose that f is any other function A → A. Then

f ◦ I = I ◦ f = f.

To see this note first that both are functions A → A, hence so are the two compo-sitions. Moreover, for any a ∈ A,

(I ◦ f)(a) = I(f(a)

)= f(a), (f ◦ I)(a) = f

(I(a)

)= f(a).

Recall also that the identity function I is a bijection.

The identity function and composition of functions allow us to define preciselythe notion of “invertibility” for a function. Suppose f : A → B is a function. If wewere to “invert” f then we would want to be able to undo the action of f . Thatis, we should be able to find some function g : B → A that “undoes” the action off . Hence each a ∈ A is taken to B by f , and should then be mapped back to thesame a ∈ A by g, reversing the arrows in the diagram below:

a1

a2

f(a1)

f(a2)

f

f

g

g

A B

Definition 3.27. A function f : A → B is invertible if there is some functiong : B → A such that f ◦ g = IB and g ◦ f = IA.

In fact our definition is slightly stronger than the preceding discussion sincewe also require f to “undo” the action of g.

Which functions are invertible? This fundamental question is answered bythe following two results which show that any invertible function is bijective, andconversely that any bijective function is invertible. That is, the concepts of bi-jectivity and invertibility describe the same class of functions. That this is so is

39


illustrated by the following diagram of functions f1 : A → B and f2 : B → A,where A = {1, 2, 3, 4} and B = {a, b, c}.

f1 f2

AA

B B

44

1 1

22

33

a a

b b

c c

In the first the function f1 is not injective since f1(3) = f1(4) = c. When wereverse the arrows we see that we are then attempting to send c to more than onepoint in A, which is not allowed if the inverse is to be a function. Similarly, f2 isnot onto since f2(x) 6= 2 for all x ∈ B, and so when we reverse the arrows thistime we do not have anywhere to send 2 to.

Proposition 3.28. If f : A → B is an invertible function then it is a bijection.

Proof. Since f is invertible there is some map g : B → A such that f ◦ g = IB

and g ◦ f = IA. But IB and IA are both bijective, and so in particular f ◦ g issurjective and g ◦ f is injective. Thus, by Proposition 3.25, f is both surjectiveand injective, hence bijective.

To prove the converse, namely any bijection is an invertible function, we arebasically going to argue that one can “reverse the arrows”, using bijectivity toavoid the pitfalls illustrated in the diagrams above.

Proposition 3.29. Let A and B be sets and let f : A → B be a bijective function.

Then f is invertible.

Proof. Since f is bijective, it is onto, and so for each b ∈ B there is some a ∈ Asuch that f(a) = b. Also f is injective, and so if we take any other a′ ∈ A thenf(a′) 6= f(a) = b. That is, for each b ∈ B there is a unique solution to the equationf(a) = b. So we can define a function g : B → A by setting g(b) to be this uniquesolution, since this associates to each b ∈ B a unique element of A.

Now take any a ∈ A. We have f(a) ∈ B, and by construction of g we haveg(f(a)

)= a. That is, g ◦ f = IA as required. On the other hand, if we choose

any b ∈ B then b = f(a) for some a ∈ A, and by definition g(b) = a, so thatf(g(b)

)= f(a) = b. Hence f ◦ g = IB as required.

These two results show that a function is bijective if and only if it invertible.Given such a function f : A → B we would like to write f−1 for the inversefunction B → A, but if we are to do so then we need to know that there is onlyone such map.

Proposition 3.30. If f : A → B is invertible, then there is only one function

g : B → A that satisfies f ◦ g = IB and g ◦ f = IA.

40

FUNCTIONS

Proof. Since f is assumed to be invertible, there is at least one such g. Supposethat h : B → A is a function that satisfies f ◦ h = IB and h ◦ f = IA. Then, sinceg ◦ IB = g and IA ◦ h = h, we have

g = g ◦ IB = g ◦ (f ◦ h) = (g ◦ f) ◦ h = IA ◦ h = h,

where we have made use of the associativity proved in Proposition 3.23. Thusg = h, showing that there can only be one such map.

Thus if f is an invertible function then we are justified in using the symbolf−1 to denote its unique inverse function. Note also that it is immediate from thedefinitions that f−1 is also invertible, with (f−1)−1 = f .

Example 3.31. Let f : R+ → R+ and g : R+ → R+ be defined by

f(x) = x2 and g(x) =√

x.

Then clearly we have (f ◦ g)(x) = (√

x)2 = x and (g ◦ f)(y) =√

y2 = y for allx, y ∈ R+, and so f and g are invertible, with f−1 = g and g−1 = f .

However, if we change the domain and codomain of these functions then thingscan go wrong, since above we are taking the positive square root, and only squaringpositive numbers. So if we instead define

f1 : R → R+, f1(x) = x2 and g1 : R+ → R, g1(x) =√

x,

then (f1 ◦ g1)(x) = x for all x ≥ 0, that is, f1 ◦ g1 = IR+. However g1(x) ≥ 0 for all

x ≥ 0 and so we cannot have g1 ◦ f1 6= IR. Indeed, (g1 ◦ f1)(x) = |x|, the modulusof x. Note that f1 is not one-to-one and g1 is not onto, so neither function isbijective.

Example 3.32. The function f : R → R, f(x) = sinx is not one-to-one, forexample, sin 0 = sin π. Nor is f onto since −1 ≤ sin x ≤ 1 for all x ∈ R. Howeverif we restrict the domain of f to be just [−π/2, π/2] and the codomain to [−1, 1]then f is turned into a bijection, and sin−1 is then a bijection [−1, 1] → [−π/2, π/2].

xxx

yyy

1

1

π−π 2π

−2ππ

2

π

2

π

2

sin x sin x, −π

2≤ x ≤ π

2sin−1 x, −1 ≤ x ≤ 1

Permutations

For the remainder of this chapter we shall concentrate on the study of bijectionson finite sets. Such functions are called permutations. Since now the nature ofthe elements of the set is largely irrelevant we can choose any labelling we like. So

41

Permutations

if we want to consider a permutation on a set A for which |A| = n, then we willtake A = {1, 2, . . . , n}. One of our biggest challenges now is to find an efficientnotation to help us understand the nature of this class of maps.

For an example of one way to describe permutations, consider the permutationson A = {1, 2, 3}. One such permutation is the following:

1

��???

????

2

��

��

3

��

1

��

2

��

3

��or

1 2 3 2 1 3

where in the second version we have rearranged the numbers on the bottom sothat all of the arrows point straight down. Thus this permutation swaps 1 and 2,and leaves 3 fixed. Note that we could equally well rearrange the elements in thetop row in many ways, two of which are

2

��

1

��

3

��

1

��

3

��

2

��or

1 2 3 2 3 1

and all of these diagrams represent the same bijection on A. By erasing the arrowswe arrive at so-called two-line notation for permutations which involves listing theelements of A in one row, and then listing the images of each element directlyunderneath. Thus possible two-line notations for the above permutation would be

(1 2 32 1 3

)

or

(2 1 31 2 3

)

or

(1 3 22 3 1

)

.

These all describe the same map; the first representation in which the elements ofA appear in ascending order in the top row is the standard representation. Thus,in standard two-line notation, the permutations on this set A = {1, 2, 3} are:

σ1 =

(1 2 31 2 3

)

, σ2 =

(1 2 31 3 2

)

, σ3 =

(1 2 32 1 3

)

,

σ4 =

(1 2 32 3 1

)

, σ5 =

(1 2 33 1 2

)

, σ6 =

(1 2 33 2 1

)

where σ1(1) = 1, σ1(2) = 2, σ1(3) = 3, . . . , σ6(1) = 3, σ6(2) = 2, σ6(3) = 1, andthese are all of the possible permutations on this set.

The set of all permutations on the set {1, 2, . . . , n} is denoted by Sn. We have|Sn| = n!, that is, it contains n! = n × (n − 1) × · · · × 2 × 1 elements. To see thisnote that to construct a σ ∈ Sn we must first choose an image for 1, and we have npossibilities. But now, when choosing a value for σ(2), we only have n− 1 choices— σ is in particular injective and so σ(1) 6= σ(2). Similarly we only have n − 2choices for σ(3), and so on, leaving 2 choices for σ(n − 1), and then we must mapn to the one remaining free space.

Suppose we take two permutations α, β ∈ Sn. They are both functions from{1, 2, . . . , n} to itself, and so we can compose them to get another function on

42

FUNCTIONS

{1, 2, . . . , n}. Moreover α and β are bijections by definition, hence the functionsα ◦ β and β ◦ α are also bijections by Theorem 3.24. That is, these compositionsare again permutations: α ◦ β ∈ Sn and β ◦ α ∈ Sn. For example if we take α andβ to be the following two permutations from S5

α =

(1 2 3 4 52 3 4 5 1

)

, β =

(1 2 3 4 51 3 2 5 4

)

then

β ◦ α =

(1 2 3 4 53 2 5 4 1

)

.

To see this recall that the notation β ◦ α means do α then β. So we have

1

��>>>

>>>>

2

��>>>

>>>>

3

��>>>

>>>>

4

��>>>

>>>>

5

ssggggggggggggggggggggggggggg •α

��1

��

2

��>>>

>>>>

3

��

��

4

��>>>

>>>>

5

��

��

•β

��1 2 3 4 5 •

So overall 1 is taken to 3, 2 to itself, and so on. However, rather than use thenotation already developed for the composition of functions, we shall instead adoptthe algebraists convention when we are discussing permutations. In a lot of algebratext books the notation (a)f is used for the image of an element a under a functionf , and so if one has functions f : A → B and g : B → C, then we can apply fto any a ∈ A, then g to the result, to obtain the element

((a)f)g of C. In this

set up it is thus natural to write the composition as fg — where here we shallomit the ◦ symbol in order to highlight the change in convention. The advantagewhen discussing permutations is that we rarely ever need to write the element thata given permutation acts on, since the two-line notation (and the cycle notationwe shall shortly develop) shows explicitly the action of the given permutation oneach element of the set {1, . . . , n}. Thus, as we do not need to write the elementbeing acted on, we would prefer to read the functions from left to right, ratherthan right to left. So from now on we shall write αβ rather than β ◦ α for theabove permutation, and to further distinguish this, call αβ the product of thepermutations α and β, rather than talk about compositions.

Since our product of permutations is nothing but function composition in dis-guise, we can apply results from earlier to deduce results about the behaviour ofthis binary operation. In particular if we have three permutations α, β, γ ∈ Sn,then we can form the product α(βγ) by first combining β and γ, and then takingthe product of the result with α. On the other hand we could also construct thepermutation (αβ)γ. But we have shown in Proposition 3.23 that composition offunctions is associative, and so in fact we have α(βγ) = (αβ)γ. As a result ofthis we can just write αβγ, and interpret this as either one of these bracketedexpressions, since they are the same thing.

On the other hand we saw that when forming compositions the order was, ingeneral, important. This remains true in the case of permutations. Indeed with α

43

Permutations

and β as above, we find that βα =

(1 2 3 4 52 4 3 1 5

)

, as shown by

1

��

2

��>>>

>>>>

3

��

��

4

��>>>

>>>>

5

��

��

•β

��1

��>>>

>>>>

2

��>>>

>>>>

3

��>>>

>>>>

4

��>>>

>>>>

5

ssggggggggggggggggggggggggggg •α

��1 2 3 4 5 •

Thus in this case we see that αβ 6= βα.Permutations, being bijective functions, have inverses that are again permu-

tations, by Propositions 3.29 and 3.28. It is easy to calculate the inverse of anypermutation α ∈ Sn using this two-line notation, since if

α =

(1 2 · · · na1 a2 · · · an

)

then α−1 =

(a1 a2 · · · an

1 2 · · · n

)

i.e. it is got by swapping the two rows over, as demonstrated by

1

��

2

��

· · · n

��

•α

��a1

��

a2

��

· · · an

��

•α−1

��1 2 · · · n •

Note that by swapping the two rows over the resulting permutation may not be instandard form, since we need not have a1 = 1, a2 = 2 etc. For example considerthe following permutations in S4:

α =

(1 2 3 42 4 3 1

)

, β =

(1 2 3 43 1 2 4

)

.

We have

α−1 =

(2 4 3 11 2 3 4

)

=

(1 2 3 44 1 3 2

)

, β−1 =

(3 1 2 41 2 3 4

)

=

(1 2 3 42 3 1 4

)

Exercise 3.33. Consider the following elements of S5:

α =

(1 2 3 4 53 1 5 4 2

)

, β =

(1 2 3 4 52 4 3 1 5

)

, γ =

(1 2 3 4 53 4 1 5 2

)

.

Calculate the products αβ and βαγ. Calculate the inverses α−1, β−1 and (αβ)−1.Calculate the product β−1α−1, and compare this with (αβ)−1.

The trouble with this two-line notation is that it quickly becomes very cum-bersome. For instance the identity permutation in Sn involves writing out thenumbers 1, . . . , n in order twice! The key concept in introducing the more efficientalternative is the following:

44

FUNCTIONS

Definition 3.34. Let α ∈ Sn and let k ∈ {1, 2, . . . , n}. The orbit of k is the setof elements in {1, 2, . . . , n} that can be reached through repeated application of αto k.

Consider the following element of S6:

α =

(1 2 3 4 5 64 1 6 2 5 3

)

We see that α sends 1 to 4, which in turn is sent to 2, which in turn is sent to 1.Applying α again will now cycle through these values:

1α7−→ 4

α7−→ 2α7−→ 1

α7−→ 4α7−→ · · ·

Similarly, α sends 3 to 6, and then 6 back to 3, so that

3α7−→ 6

α7−→ 3α7−→ 6

α7−→ · · ·

Finally note that 5 is left fixed by α. Thus the action of α is shown by the following:

1!!4

ss2

KK 3!!6`` 5 ee

The set {1, 4, 2} is the orbit of each of the elements 1, 2 and 4; the orbit of 3 is{3, 6}, which is also the orbit of 6; and the orbit of 5 is just {5}.

Exercise 3.35. Find all the orbits of the following permutations:

α =

(1 2 3 4 53 5 1 2 4

)

, β =

(1 2 3 4 5 64 2 6 5 1 3

)

, γ =

(1 2 3 4 5 62 1 5 6 3 4

)

.

Definition 3.36. A permutation β ∈ Sn is called a cycle if there are distinctnumbers a1, a2, . . . , am ∈ {1, 2, . . . , n} such that β maps a1 to a2, a2 to a3, . . . ,am−1 to am and am to a1, and leaves all other elements in {1, 2, . . . , n} fixed(i.e. all the elements in the subset {1, 2, . . . , n} − {a1, a2, . . . , am}). It is denoted(a1a2 · · · am), and is called a cycle of length m.

Thus (142), (36) and (5) are all cycles in S6, and in two-line notation they arewritten

(142) =

(1 2 3 4 5 64 1 3 2 5 6

)

, (36) =

(1 2 3 4 5 61 2 6 4 5 3

)

,

(5) =

(1 2 3 4 5 61 2 3 4 5 6

)

Note that there is a lack of uniqueness in the way a cycle is written: in general(a1a2 · · · am) = (a2a3 · · · ama1) = · · · , so for the first of the three particular cyclesabove we have

(142) = (421) = (214).

45

Permutations

However these are distinct from the cycle (124) which maps 1 to 2, 2 to 4 and4 to 1. Also, note that any cycle of length 1 is the identity, since everything in{1, . . . , n} is left fixed. That is, (1) = (2) = · · · = (n).

Finally, note what happens if we multiply these three cycles together

(142)(36)(5) = (142)(36)

=

(1 2 3 4 5 64 1 3 2 5 6

)(1 2 3 4 5 61 2 6 4 5 3

)

=

(1 2 3 4 5 64 1 6 2 5 3

)

= α.

That is, we recover the permutation we started with. Thus our original permu-tation can be written as a product of disjoint cycles — that is, as a product ofcycles that act on (pairwise) disjoint subsets of numbers from {1, 2, . . . , 6}. Thisbehaviour is actually true of all permutations, as we shall show below. First note,however, that the order in which we wrote the disjoint cycles in the product aboveis not important. We also have

α = (36)(5)(142) = (5)(36)(142) = · · ·

This is the case since these cycles only act nontrivially on disjoint subsets of{1, 2, . . . , 6}, and so essentially act independently of each other. In general it is truethat disjoint cycles commute, which is the not the case for general permutations.

Theorem 3.37. Any permutation in Sn can be written as a product of disjoint

cycles. The product is unique up to the order of the cycles, and the inclusion or

omission of cycles of length 1.

Proof. The proof will be accomplished by induction. For S1 we only have onepermutation on {1}, namely the identity map (1), and thus the result is truein this case. So now suppose that we know the result holds for permutationsfrom S1,. . . ,Sn−1, where n ≥ 2, and consider any permutation σ ∈ Sn. Wecan apply σ to 1, then to its image, and so on, to get a sequence of numbers(1, σ(1), σ(σ(1)), . . .). Since each of these numbers lies in the finite set {1, 2, . . . , n}they cannot all be different, and so we must have

σ(σ(· · · (σ︸︷︷︸

r

(1)) · · · )) = σ(σ(· · · (σ︸︷︷︸

s

(1)) · · · ))

for some 0 ≤ r < s. But also σ is invertible, so if we apply σ−1 a total of r timesto both sides we get

1 = σ(σ(· · · (σ︸︷︷︸

s−r

(1)) · · · )).

That is, if we apply σ enough times we will get back to 1 again, and furtherapplications will cause us to cycle round the values. So let

a1 = 1, a2 = σ(1), . . . , at = σ(σ(· · · (σ︸︷︷︸

t−1

(1)) · · · ))

46

FUNCTIONS

where t is chosen so that all of the numbers a1, . . . , at are distinct. If t = nthen all n elements of {1, 2, . . . , n} are included in the above, and it follows thatσ = (a1a2 . . . at). If t < n, then we can consider the action of σ restricted tothe subset {1, 2, . . . , n} − {a1, a2, . . . , at}. Since the set {a1, a2, . . . , at} is mappedonto itself, by σ, the complement {a1, a2, . . . , at}′ is also mapped onto itself by σ.Thus σ restricted to this set can be viewed as a permutation on a set of n − telements. But note that t ≥ 1, so n − t < n. Thus, by our induction hypothesis,this restriction can be written as a product of disjoint cycles, and σ is then thisproduct multiplied by (a1a2 . . . at). The result now follows by induction.

It is perhaps easier to understand why the proof above works by seeing theprinciple behind it in action when computing some examples. For example in S5

we have(

1 2 3 4 53 2 1 5 4

)

= (13)(2)(45), and

(1 2 3 4 55 3 4 2 1

)

= (15)(234).

To find these decompositions you see where 1 is sent to, where the image is sentto, etc., until you get back to 1. This (ordered, finite) sequence of numbers givesthe first cycle. Next take the smallest element not contained in the first cycle (ifsuch an element exists) and repeat the procedure to form the second cycle, andrepeat as necessary until no elements are left. Note that any cycle of length 1is the identity permutation and so we normally omit it from the representation.Hence our first example would be more usually written as just (13)(45). Howevernote that written in full the identity permutation in Sn is equal to (1)(2) · · · (n),and if we omit all cycles of length 1 then there will be nothing left. Consequentlythe identity is usually written as (1).

Similarly it is straightforward to obtain the two-line notation form of a product

of cycles, but because we omit cycles of length 1 we must specify which set Sn weare working with. For instance in S5 we have

(14)(253) =

(1 2 3 4 54 5 2 1 3

)

, and (13)(25) =

(1 2 3 4 53 5 1 4 2

)

However, as an element of S8 the second example is

(13)(25) =

(1 2 3 4 5 6 7 83 5 1 4 2 6 7 8

)

This passage from cycles to two-line notation and back again can be used tocalculate the product of (non-disjoint) cycles, using our earlier methods. Howeverwith a little practice it is possible to calculate the product of cycles directly, andthus more efficiently. For example if we take the following cycles from S6

α1 = (1426), α2 = (153), α3 = (35), α4 = (35642)

thenα1α2α3α4 = (1426)(153)(35)(35642) = (1243)(56).

To see this recall that since we are working with products we read α1α2α3α4 as doα1 first, followed by α2 etc. Thus 1 is mapped to 4 by α1, which is left alone by

47

Permutations

α2 and then also by α3, but finally 4 is mapped to 2 by α4. So overall the effectis that 1 is mapped to 2. For all the elements we have

1α17−→ 4

α27−→ 4α37−→ 4

α47−→ 2 3α17−→ 3

α27−→ 1α37−→ 1

α47−→ 1

2α17−→ 6

α27−→ 6α37−→ 6

α47−→ 4 5α17−→ 5

α27−→ 3α37−→ 5

α47−→ 6

4α17−→ 2

α27−→ 2α37−→ 2

α47−→ 3 6α17−→ 1

α27−→ 5α37−→ 3

α47−→ 5

and from this we can now recognise the disjoint cycles making up the productα1α2α3α4.

As a more general example of multiplying cycles together, note that the inverse

of the cycle α = (a1a2 · · · am−1am) is the cycle (amam−1 · · · a2a1), since the firstmaps ai to ai+1 for i < m, which is then mapped back to ai by the second, and am

is mapped to a1, which is then mapped back to am. All we are doing is reversingthe direction in which we move around around the following circles:

a1��

am

00

a2

��α

==

a1��

am

��

a2

nn

α−1bb

Example 3.38. Consider the following cycles:

α = (1742), β = (1345), γ = (254)

We haveαβγ = (1742)(1345)(254) = (1745)(23)

as can be seen by the following:

1α7−→ 7

β7−→ 7γ7−→ 7 5

α7−→ 5β7−→ 1

γ7−→ 1

7α7−→ 4

β7−→ 5γ7−→ 4 2

α7−→ 1β7−→ 3

γ7−→ 3

4α7−→ 2

β7−→ 2γ7−→ 5 3

α7−→ 3β7−→ 4

γ7−→ 2

noting that 6 appears in none of α, β or γ, and so will not appear in their producteither. It now follows that

(αβγ)−1 = (23)−1(1745)−1 = (32)(5471) = (1547)(23),

although the change in order when calculating the inverses of individual cycles isnot especially vital here since the cycles (and hence their inverses) are disjoint andso commute. On the other hand we have

α−1 = (2471), β−1 = (5431), γ−1 = (452)

and soγ−1β−1α−1 = (452)(5431)(2471) = (1547)(23) = (αβγ)−1

as before. Here the order is important since

α−1β−1γ−1 = (2471)(5431)(452) = (14723)(5) = (14723) = (γβα)−1.

48

FUNCTIONS

Exercise 3.39. Write the following elements of S6 as a product of disjoint cycles:

α =

(1 2 3 4 5 63 6 1 5 2 4

)

, β =

(1 2 3 4 5 62 3 1 5 6 4

)

.

Calculate the products αβ and βα, expressing the results as a product of disjointcycles. Calculate β2 = ββ and β3 = βββ. What is β7?

Since every permutation can be written as a product of cycles, these are in somesense more fundamental objects than general permutations. In fact it is possibleto generate cycles, and thus all permutations, from an even smaller subset.

Definition 3.40. A cycle of length 2 is called a transposition. That is, a trans-position is a permutation that interchanges two elements and leaves all the othersunchanged.

Example 3.41. The following are transpositions: (12), (35), (23) etc. Note thatif we consider a general transposition (ab) ∈ Sn then it is in particular a cycle soits inverse is (ba). But (ab) and (ba) are different ways of writing the same cycle,and thus (ab)(ab) = (ab)(ba) = (1). That is, every transposition is its own inverse.

Theorem 3.42. Every permutation can be written (in more than one way) as a

product of transpositions. If σ ∈ Sn has two representations as a product of trans-

positions, the first involving r transpositions and the second involving s transposi-

tions, then r and s are either both even or both odd.

Proof. First note that we can write any cycle as a product of transpositions, sincewe have

(a1a2 · · · am−1am) = (a1a2)(a1a3) · · · (a1am−1)(a1am).

But any permutation σ can be written as a product of cycles, and since each ofthese can individually be broken up into a product of transpositions, it followsthat σ can be written as a product of such maps.

The second part of the theorem requires a great deal more work. For eachn ≥ 2 define pn to be the following polynomial in n unknowns:

pn(x1, . . . , xn) =∏

1≤i<j≤n

(xi − xj).

So for low values of n we have

p2 = (x1 − x2), p3 = (x1 − x2)(x1 − x3)(x2 − x3), and

p4 = (x1 − x2)(x1 − x3)(x1 − x4)(x2 − x3)(x2 − x4)(x3 − x4),

and in general

pn = (x1 − x2)(x1 − x3) × · · · × (x1 − xn)

× (x2 − x3)(x2 − x4) × · · · × (x2 − xn)

× · · · × (xn−1 − xn).

49

Permutations

Given a permutation α ∈ Sn, define α(pn) to be the polynomial

α(pn) =∏

1≤i<j≤n

(xα−1(i) − xα−1(j)).

For example, if we take α = (24) ∈ S4, then α = α−1 = (24), and so

α(p4) = (xα−1(1) − xα−1(2))(xα−1(1) − xα−1(3))(xα−1(1) − xα−1(4))

× (xα−1(2) − xα−1(3))(xα−1(2) − xα−1(4))(xα−1(3) − xα−1(4))

= (x1 − x4)(x1 − x3)(x1 − x2)(x4 − x3)(x4 − x2)(x3 − x2) = −p4

Similarly, if we take α = (132) ∈ S3 then α−1 = (123), and so

α(p3) = (x2 − x3)(x2 − x1)(x3 − x1) = p3.

In general, since α is a bijection on {1, 2 . . . , n}, we will have the same 12n(n − 1)

factors as in pn; the only thing that can have changed is their sign, and hence thesign of pn. That is, for each α ∈ Sn we have α(pn) = ±pn. If we are given anotherpermutation β then

(αβ)(pn) =∏

i<j

(x(αβ)−1(i) − x(αβ)−1(j))

=∏

i<j

(x(α−1◦β−1)(i) − x(α−1◦β−1)(j))

=∏

i<j

(xα−1(β−1(i)) − xα−1(β−1(j)))

= α(β(pn)).

That is, acting on pn by the product αβ is the same thing as acting by β and thenby α.

Consider now the special case when the permutation is a transposition, i.e.something of the form τ = (kl) = τ−1, where we can assume that k < l. Then wehave six possibilities for what happens to each of the factors xi − xj:

neither i nor j equals k or l: in this case τ−1(i) = i and τ−1(j) = j, and so thefactor is unchanged.

i = k, j 6= l: xi − xj = xk − xj changes to xl − xj. There are sign changes whenk = i < j < l — so l − k − 1 changes.

j = k < l ⇒ i 6= l: xi − xj = xi − xk changes to xi − xl, so no sign changes.

i = l > k ⇒ j 6= k: xi − xj = xl − xj changes to xk − xj, so no sign changes.

i 6= k, j = l: xi − xj = xi − xl changes to xi − xk. There are sign changes whenk < i < j = l — so l − k − 1 changes.

i = k, j = l: xi − xj = xk − xl changes to xl − xk.

Overall there are 2(l−k−1)+1 sign changes, an odd number, and so τ(pn) = −pn.Now let us return to our σ ∈ Sn. Suppose we can write σ = α1 · · ·αr and

σ = β1 · · · βs for some transpositions α1, . . . , αr, β1, . . . , βs. Then

σ(pn) = (α1 · · ·αr)(pn) = α1(· · · (αr(pn)) · · · ) = (−1)rpn

50

FUNCTIONS

and

σ(pn) = (β1 · · · βs)(pn) = β1(· · · (βs(pn)) · · · ) = (−1)spn.

But these two expressions must be equal, and so we have (−1)r = (−1)s, andhence r and s are either both even or both odd.

What the first part of this result says is that absolutely any shuffling of theelements of the set {1, 2, . . . , n} can be accomplished by swapping two elements ata time, and repeating this procedure sufficiently often, changing the two elementsto be swapped on each occasion.

One reason that representation of a given permutation in terms of transposi-tions will not be unique is because we can insert any number of pairs of the form(ab)(ab), since this product equals the identity. Also, if we break up a given co-cycle as in the proof, then we get different results if we start with (a2a3 · · · ama1),instead of (a1a2 · · · am), etc.

Example 3.43. Consider the following permutations in S5:

(a) (23) (b) (153) (c) (123)(45).

We can write these as products of transpositions as follows:

(a) (23) = (45)(45)(23) = (12)(23)(13).

(b) (153) = (15)(13), and (153) = (315) = (31)(35).

(c) (123)(45) = (12)(13)(45) = (13)(34)(42)(34)(45).

Definition 3.44. A permutation is called even (respectively odd) if it can bewritten as the product of an even (respectively odd) number of transpositions.

In Example 3.43 above, (23) is odd, (153) is even and (123)(45) is odd.

Exercise 3.45. Determine whether the following permutations are even or odd,and write them as a product of transpositions in two different ways:

(1527)(3567)(273) (142)(243)(34)(1324).

Example 3.46. The cycle (a1a2 · · · am) can be written as the product of m − 1transpositions: (a1a2 · · · am) = (a1a2)(a1a3) · · · (a1am). Thus this cycle of lengthm can be written as a product of m− 1 transpositions, and it follows that a cycleof even length is odd, and a cycle of odd length is even!

Exercise 3.47. Explain why the product of two even or two odd permutations isan even permutation, and why the product of an even and an odd permutation is anodd permutation. Use this observation and the preceding example to (re)determinethe even- and oddness of the permutations in Exercise 3.45.

Proposition 3.48. Of the n! elements of Sn, half are even permutations and half

are odd permutations.

51

Permutations

Proof. We can write the even permutations as {α1, α2, . . . , αj} and the odd per-mutations as {β1, β2 . . . , βk}. These two sets partition Sn, so j + k = n! = |Sn|.Let τ ∈ Sn be any transposition, then τα1, τα2, . . . , ταj are all odd permutations.Moreover they are distinct, for if ταp = ταq then

αp = (1)αp = (ττ)αp = τ(ταp) = τ(ταq) = (ττ)αq = (1)αq = αq,

where we have used Proposition 3.23 and basic facts about the identity map (1).Thus the set {τα1, . . . , ταj} must be a subset of {β1, . . . , βk} and so j ≤ k. Simi-larly, the permutations τβ1, . . . , τβk are all even and distinct, and so k ≤ j. Hencej = k = n!/2 as required.

Definition 3.49. The order of a permutation σ is the smallest positive integer ksuch that σk = σσ · · · σ (k times) is equal to the identity (1), and is denoted O(σ).

An obvious question is whether or not such a k exists! That this is true canbe shown using group theory, the subject of the next chapter.

Example 3.50. A cycle of length m has order m. For example, the transposition(23) has length 2, is not equal to (1), and (23)(23) = (1). Similarly,

(123) 6= (1), (123)(123) = (132) 6= (1), (123)(123)(123) = (132)(123) = (1),

so that O((123)) = 3. To see the general case, consider (a1a2 · · · am). The effectof one application of this cycle is to rotate the circle below one time through anangle of

(360m

)◦:

a1��

am

00

a2

��α

==

It is clear that only after m such rotations will we be brought back to where westarted.

Proposition 3.51. If α and β are disjoint cycles then O(αβ) = lcm{O(α), O(β)},the lowest common multiple of their individual orders.

Proof. Since α and β are disjoint we have αβ = βα, and so for any integer m ≥ 1

(αβ)m = (αβ)(αβ) · · · (αβ)︸︷︷︸

m

= (α · · ·α︸︷︷︸

m

)(β · · · β︸︷︷︸

m

) = αmβm.

Also, again since they are disjoint, αmβm = (1) if and only if αm = (1) andβm = (1). The result now follows since αm = (1) if and only if m is a multiple ofO(α), and similarly for β.

Example 3.52. In the above result it is crucial that the cycles be disjoint. Forexample if α = (14) and β = (236), then O(α) = 2 and O(β) = 3, and so

O(αβ) = lcm{2, 3} = 6.

52

FUNCTIONS

On the other hand suppose we take α = (1254) and β = (15). Then αβ =(1254)(15) = (12)(45). But now

O(α) = 4, O(β) = 2 ⇒ lcm{O(α), O(β)} = lcm{4, 2} = 4.

However,(αβ)2 = (12)(45)(12)(45) = (12)(12)(45)(45) = (1),

so that O(αβ) = 2.

Exercise 3.53. Calculate the orders of the following permutations

(a) (132)(45)(67) (b) (132)(45) (c) (125)(354) (d) (146)(2563)(34)

The set Sn is known as the symmetric group, and arose first in the studyof polynomial equations. It nicely illustrates many geometrical symmetries aswe shall see in the next chapter. However, a more algebraic application involvesdeterminants from linear algebra. Since every permutation is even or odd we candefine a function Sn → {−1,+1} by

sgn(σ) =

{

−1 if σ is odd,

+1 if σ is even.

This allows us to give the following compact definition of the determinant of asquare matrix. If A = [aij ] is an n × n matrix then

detA =∑

σ∈Sn

sgn(σ)a1σ(1)a2σ(2) · · · anσ(n).

To see how this work, consider the case when n = 2 so that

A =

[a11 a12

a21 a22

]

.

Then we are working with S2 = {(1), (12)}. Set α = (1) and β = (12), then

detA = (sgn α)a1α(1)a2α(2) + (sgn β)a1β(1)a2β(2)

= a11a22 − a12a21.

When n = 3 we are working with S3 = {σ1, σ2, . . . , σ6} where

σ1 = (1), σ2 = (23), σ3 = (12), σ4 = (123), σ5 = (132), σ6 = (13),

and

sgn σ1 = sgn σ4 = sgn σ5 = +1, sgn σ2 = sgn σ3 = sgn σ6 = −1.

So now

detA = a11a22a33 − a11a23a32 − a12a21a33

+ a12a23a31 + a13a21a32 − a13a22a31

= a11

∣∣∣∣

a22 a23

a32 a33

∣∣∣∣− a12

∣∣∣∣

a21 a23

a31 a33

∣∣∣∣+ a13

∣∣∣∣

a21 a22

a31 a32

∣∣∣∣,

as required. However, the compact definition has many benefits over the well-known formulae for the cases n = 2 and n = 3. Indeed, the fact that we have aformula valid for all integers n ≥ 2 gives us a means to prove identities such asdet(A + B) = det A + detB, and det(AB) = detAdet B.

53

4 Groups

Binary operations

Given any two real numbers a and b (or integers, or rationals) we can add themtogether or multiply them to get new numbers, namely a + b and ab. Also we cansubtract one from another to get a − b or b − a, and if they are nonzero we canform a

band b

a. Given any 2 × 2 matrices S and T we can add them and multiply

them to get new matrices S + T and ST . Given any two functions f : A → Aand g : A → A we can form the compositions f ◦ g and g ◦ f , and if f and g arebijections then so are these compositions. In each of these examples we are takingtwo elements from some set and combining them according to some rule to forma new element of the same set. However note that in the case of subtraction ofnumbers, multiplication of matrices and composition of functions we can have thata − b 6= b − a, ST 6= TS and f ◦ g 6= g ◦ f , so the order of the pair of elements is,potentially, of great importance.

Most of the algebraic systems you have encountered so far have two centraloperations — usually addition and multiplication of numbers. In group theory werestrict our attention to just one operation at any one time.

Definition 4.1. A binary operation on a (nonempty) set A is a map from A× Ato A. When talking about a general binary operation we shall write the image ofa given pair (a, b) ∈ A × A as a ∗ b, and denote this set-up as (A, ∗).∗

Example 4.2.

(i) (Z,+) is the operation on the integers Z that takes a pair of integers (m,n) ∈Z × Z and adds them together. That is, (m,n) 7→ m + n. Similarly (R,×)is the operation on the real numbers R that takes a pair of numbers (x, y) ∈R × R and multiplies them together. That is, (x, y) 7→ xy.

(ii) (Z,−) is the operation on the integers Z that takes a pair of integers (m,n) ∈Z × Z and subtracts one from the other. In this case note that the order isimportant: 2 − 3 = −1 which is different from 3 − 2 = 1.

Note also that we can define the binary operation of subtraction on R or Q,but not on the natural numbers N, since in this case it is not closed. Thatis, for certain pairs (m,n) ∈ N×N we do not have m−n ∈ N. For example,3 − 7 = −4 /∈ N. Similarly division is not a binary operation on Z because,for example, we cannot divide by 0, and also 3

2 /∈ Z, and so on.

(iii) Let A be a set, and let F(A) denote the set of all functions from A to A.Then we have the binary operation (F(A), ◦) of composition which takes apair (f, g) of maps and returns the function f ◦ g. Note again that order isimportant (provided A has at least two elements (why?)). Note also that wehave an identity function IA which satisfies f ◦ IA = IA ◦ f , and if f is abijection is then it is invertible, that is there is a unique function f−1 such

∗Some authors do not require that a ∗ b ∈ A as part of the definition of a binary operation,only that a ∗ b is defined. In this case if a ∗ b ∈ A for all a, b ∈ A, they say that the operation isclosed.

54

GROUPS

that f ◦ f−1 = IA = f−1 ◦ f . If f is not bijective then no such map f−1

exists.

Definition 4.3. Let ∗ be a binary operation on a set A.

(i) It is associative if

a ∗ (b ∗ c) = (a ∗ b) ∗ c for all a, b, c ∈ A,

[Note: a, b and c appear in the same order on either side — it is only thebracketing that differs.]

(ii) It is commutative if

a ∗ b = b ∗ a for all a, b ∈ A,

(iii) It has an identity if there is some element e ∈ A such that

a ∗ e = e ∗ a = a for all a ∈ A.

(iv) Finally, if there is an identity, then an element a ∈ A has an inverse if thereis some a′ ∈ A that satisfies

a ∗ a′ = a′ ∗ a = e.

Note that in this case a′ also has an inverse, namely a.

Thus we see that (Z,+) is associative, is commutative, has 0 as an identity(since 0+ m = m + 0 = m for all m ∈ Z), and each m has −m as its inverse (sincem+(−m) = 0 = (−m)+m). Similarly (Z,×) is associative and commutative, and1 is an identity, but the only elements that have inverses are 1 and −1. (Exercise:convince yourself of this last fact.) In both cases associativity and commutativityare properties that are well-known and which we will take for granted. Only whengoing back to the set-theoretic foundations of mathematics would we considerbothering to prove that these properties hold for these examples.

For an easy example of a binary operation that is neither associative nor com-mutative consider (Z,−). For example we have

(1 − 2) − 3 = −1 − 3 = −4, but 1 − (2 − 3) = 1 − (−1) = 2,

and

5 − 2 = 3, but 2 − 5 = −3.

As with the properties of relations or functions, to show that a given binary op-eration is associative or commutative we must give a general argument, whereasa single counterexample will suffice to show that it does not have one or other ofthese properties.

On the other hand, to demonstrate the existence of an identity or inverse weneed merely state what it is, and show that it satisfies the relevant equation,

55

Binary operations

whereas to show that such an element does not exist requires a more generalargument. For example, our badly behaved binary operation (Z,−) has no identityelement. To see that this is the case recall that if e ∈ Z were an identity then itmust satisfy

m − e = m and e − m = m for all m ∈ Z.

Solving the first of these equations gives e = 0, no matter what we choose for m.However solving the second gives e = 2m, i.e. the answer depends on m, which isnot allowed by definition.

For a noncommutative binary operation that is associative we can take com-position on F(A), provided |A| 6= 1. Note also that this operation does have anidentity, namely the identity function, but there are many functions that do nothave inverses. If we denote the set of bijective functions on A by B(A) then ◦defines a binary operation on B(A) that is associative, has an identity, and forwhich every element has an inverse. This time it is only noncommutative if Acontains at least three elements.

Example 4.4. Let Q+ = {x ∈ Q : x > 0} = Q∩(0,∞), the set of positive rationalnumbers. So every x ∈ Q can be written as m

nfor some m,n ∈ N. Define, for each

x, y ∈ Q,

x ∗ y =1 + x

1 + y.

This is a well-defined binary operation, since we can write x = mn

and y = pq, for

some m,n, p, q ∈ N, and then

1 + x

1 + y=

1 + mn

1 + pq

× nq

nq=

(n + m)q

(q + p)n,

where (n+m)q and (q+p)n are both positive integers, so that x∗y ∈ Q+. However∗ is neither associative, nor commutative. For example

1 ∗ 2 =1 + 1

1 + 2=

2

3, but 2 ∗ 1 =

1 + 2

1 + 1=

3

2, and

(1 ∗ 2) ∗ 1 =2

3∗ 1 =

1 + 23

1 + 1=

5

6, but 1 ∗ (2 ∗ 1) =

1 + 1

1 + 32

=4

5.

[Note: if you are unsure as to whether or not the operation is associative orcommutative, then calculate the general expressions for x ∗ y and y ∗ x, or forx ∗ (y ∗ z) and (x ∗ y) ∗ z. In our case

(x ∗ y) ∗ z =2 + x + y

1 + y + z + yzand x ∗ (y ∗ z) =

1 + x + z + xz

2 + y + z.

These looks very different, so it is probable that some values of x, y and z willgive different results.]

Finally, if y were an identity for ∗ then we would have

x ∗ y =1 + x

1 + y= x ⇒ 1 + x = x(1 + y) = x + xy ⇒ 1 = xy ⇒ y =

1

x.

But y should be a particular fixed value and not depend on the choice of x, thusthere is no identity. Consequently it does not make sense to talk of inverses.

56

GROUPS

Exercise 4.5.

(a) Show that the operation x∗y = 1x+ 1

yon Q+ is well-defined and commutative,

but that it is not associative, nor does it have an identity.

(b) Show that the operation x ∗ y = x + y − 7 on Z is well-defined, associativeand commutative, that there is an identity, and that every element has aninverse.

Two very important examples of binary operations are given by matrices andmodulo arithmetic.

Example 4.6. For each n ≥ 1 let M(n, R) denote the set of n × n matrices with

real number entries. So elements of M(2, R) include[

1 −7−5 3

]and

[√2 0

3

4−2

]

but not[

i −7−i

−5+2i√

3

]

. Then we have two natural binary operations on M(n, R), namely

addition and multiplication of matrices.If we consider the operation (M(2, R),+) then this is associative, commutative,

has an identity, namely [ 0 00 0 ], and each matrix T =

[a bc d

]has an inverse −T =

[−a −b−c −d

]. All of these properties are essentially inherited from the corresponding

ones for the operation (R,+).If we consider the operation (M(2, R),×) instead, then things are not so well

behaved. It is still associative (why?), but is not commutative. For exampleconsider the matrices

S =

[1 10 0

]

, T =

[1 01 0

]

.

We have

ST =

[1 10 0

] [1 01 0

]

=

[2 00 0

]

but TS =

[1 01 0

] [1 10 0

]

=

[1 11 1

]

6= ST.

This operation does have an identity which is the matrix [ 1 00 1 ], but a given matrix

T =[

a bc d

]has an inverse if and only if its determinant (the number ad − bc) is

nonzero.The subset of M(2, R) consisting of those matrices with nonzero determinant

is denoted GL(2, R). That is,

GL(2, R) = {T ∈ M(2, R) : detT 6= 0}.Note that this set is closed under multiplication, since for any two elements T1 =[

a1 b1c1 d1

]

and T2 =[

a2 b2c2 d2

]

of M(2, R) we have

[a1 b1

c1 d1

] [a2 b2

c2 d2

]

=

[a1a2 + b1c2 a1b2 + b1d2

c1a2 + d1c2 c1b2 + d1d2

]

and so detT1T2 equals

(a1a2 + b1c2)(c1b2 + d1d2) − (a1b2 + b1d2)(c1a2 + d1c2)

= a1c1a2b2 + a1d1a2d2 + b1c1b2c2 + b1d1c2d2

− a1c1a2b2 − a1d1b2c2 − b1c1a2d2 − b1d1c2d2

= a1d1(a2d2 − b2c2) + b1c1(b2c2 − a2d2)

= (a1d1 − b1c1)(a2d2 − b2c2) = detT1 det T2.

57

Binary operations

Thus if detT1 = a1d1 − b1c1 6= 0 and detT2 = a2d2 − b2c2 6= 0 then detT1T2 6= 0.Hence (GL(2, R),×) is a well-defined binary operation that is associative, but notcommutative (exercise: show this), and which has an identity and every elementhas an inverse. Indeed, the inverse of T =

[a bc d

]is the matrix

T−1 =1

ad − bc

[d −b−c a

]

.

Example 4.7. Fix a positive integer n > 1 and consider the equivalence relationon Z given by congruence modulo n. That is, a ≡ b mod n if a − b is a multipleof n. Let Zn denote the set of equivalence classes, that is the set consisting of thefollowing subsets of Z:

[0] = {. . . ,−2n,−n, 0, n, 2n, . . .} = [n] = [−n] = [2n] = · · ·[1] = {. . . ,−2n + 1,−n + 1, 1, n + 1, 2n + 1, . . .} = [n + 1] = · · ·...

......

......

...

[n − 1] = {. . . ,−n − 1,−1, n − 1, 2n − 1, 3n − 1, . . .} = [−1] = · · ·

So Zn is a finite set with |Zn| = n, but for which each element has several differentdescriptions since for each given equivalence class we can choose infinitely manyrepresentatives. We define two binary operations on Zn as follows: for all a, b ∈ Z

set[a] + [b] = [a + b]; [a][b] = [ab].

For example consider the elements [0] = [5], [1] = [6], [2] and [4] of Z5. We have

[0] + [2] = [0 + 2] = [2], [1] + [4] = [1 + 4] = [5] = [0]

[1][4] = [1 × 4] = [4], [2][4] = [2 × 4] = [8] = [3].

Basically all we are doing is the usual addition and multiplication of integers, andthen calculating the remainder on dividing by 5. But recall that [0] = [5] and[1] = [6]. So do we have [0] + [2] = [5] + [2] and [1][4] = [6][4]? The answer is yes,since

[5] + [2] = [5 + 2] = [7] = [2], and [6][4] = [24] = [4 × 5 + 4] = [4].

But is this true in general, that is, are these binary operations well-defined? Ordoes the definition [a] + [b] = [a + b] depend on our choice of representatives aand b? What this means is if c and d are integers such that a ≡ c mod n andb ≡ d mod n (so that [a] = [c] and [b] = [d]) then do we have [a + b] = [c + d] and[ab] = [cd], so that [a] + [b] = [c] + [d] and [a][b] = [c][d]? Since a ≡ c mod n thereis some integer j ∈ Z such that a = c + jn. Similarly b = d + kn for some k ∈ Z.Hence

(a + b) − (c + d) = (a − c) + (b − d) = jn + kn = (j + k)n, and

ab − cd = (a − c)b + c(b − d) = (jb + ck)n,

hence a+b ≡ c+d mod n and ab ≡ cd mod n, and so [a+b] = [c+d] and [ab] = [cd]as required.

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GROUPS

In fact it is more common to actually work with all of Z and the equivalencerelation ≡ modn rather than the equivalence classes. The work above shows thatfor a given n we need only consider the integers 0, 1, . . . , n − 1. For example wewrite

4 × 7 = 28 ≡ 4 mod 8, 3 + 5 = 8 ≡ 2 mod 6

and so on.Since the operations (Z,+) and (Z,×) are associative, commutative and have

identities, the same is true for (Zn,+) and (Zn,×). Indeed, for (Zn,+) we have

[a] + [b] = [a + b] = [b + a] = [b] + [a] and [a] + [0] = [a + 0] = [a] = [0] + [a]

Associativity can be proved similarly, as can the same three properties for (Zn,×).In this modulo arithmetic there are other symbols used to denote the addition

or multiplication, for example we will often write +n and ×n, while other authorsmay write ⊕ and ⊗.

Some of the operations on sets considered in the first chapter provide anotherexample of binary operations, some of which are more well-behaved than others:

Example 4.8. If A is a nonempty set, and P (A) the power set of A then (P (A),∩)and (P (A),△) are both binary operations. Moreover they are both commutative,associative and have identities. However only one set has an inverse for (P (A),∩),but every set for (P (A),△) has an inverse.

The commutativity and associativity of ∩ and △ can be established by referringdirectly to the definitions of these operations, or by using truth tables. For examplegiven any subsets B and C of A, B ∩C is the subset of A consisting of everythingin both B and C, but that is the same thing as everything in C and B, and soB ∩ C = C ∩ B.

The identity for ∩ is A: for any subset B ⊆ A the set B ∩ A = A ∩ B consistsof everything that is in both B and A. But everything in B is already in A, soB ∩ A = B.

If B ∈ P (A) were to have C as an inverse then we would need B ∩ C = A.Now if B 6= A then B ⊂ A, that is it is a proper subset of A. But then B∩C ⊆ B,and so B ∩ C ⊂ A, hence cannot be equal to A. Thus such a B cannot have aninverse. However we have A ∩ A = A, and so A ∈ P (A) is its own inverse, andthus is the only set in P (A) to have an inverse.

The identity for △ is ∅, since for any B ∈ P (A) we have B − ∅ = B and∅ − B = ∅, so that

B△∅ = ∅△B = (B − ∅) ∪ (∅ − B) = B ∪ ∅ = B.

Moreover for any B ∈ P (A) we have B − B = ∅ and so

B△B = (B − B) ∪ (B − B) = ∅ ∪ ∅ = ∅.Hence each set B is its own inverse.

Note that in our list of important properties, to discuss inverses it was necessaryto assume the existence of an identity. So is it possible to have more than oneidentity? And can an element have more than one inverse? The next results givesome answers.

59

Binary operations

Proposition 4.9. Let (A, ∗) be a binary operation. Then there is at most one

identity for this operation.

Proof. Suppose that e1 and e2 are both identities for A. Then by definition

a ∗ e1 = a and e2 ∗ a = a for all a ∈ A.

So in particular, e2∗e1 = e2 by putting a = e2 in the first equation, and e2∗e1 = e1

by putting a = e1 in the second. Thus we have e2 = e1, and so there can only beone identity.

Proposition 4.10. Let (A, ∗) be a binary operation, and suppose that there is an

identity. If the operation is associative then each element a ∈ A has at most one

inverse.

Proof. Denote the identity by e, and suppose that a ∈ A has inverses b1 and b2.Then by definition b1 ∗ a = e and a ∗ b2 = e. So then

b1 = b1 ∗ e = b1 ∗ (a ∗ b2) = (b1 ∗ a) ∗ b2 = e ∗ b2 = b2,

that is b1 and b2 are the same, hence a can only have one inverse.

This result was already known in the case of function composition (cf. Propo-sition 3.30). What we have shown now is that it is true in the context of any

associative binary operation with identity.

Exercise 4.11. Find an example of a (necessarily nonassociative) binary operationwith identity in which at least one element has more than one inverse.

Thus associativity is desirable since we can talk about the inverse of an element,rather than an inverse. For another advantage suppose that (A, ∗) is a binaryoperation and that we wish to combine four elements a, b, c, d ∈ A. Since we areonly allowed to combine two elements at a time we have five different possibilities:

[(a ∗ b) ∗ c] ∗ d, [a ∗ (b ∗ c)] ∗ d, a ∗ [(b ∗ c) ∗ d], a ∗ [b ∗ (c ∗ d)], and (a ∗ b) ∗ (c ∗ d).

If (A, ∗) is associative, however, then

[(a ∗ b) ∗ c] ∗ d = [a ∗ (b ∗ c)] ∗ d (using associativity in [· · · ])= a ∗ [(b ∗ c) ∗ d] ([a ∗ x] ∗ d = a ∗ [x ∗ d] for x = b ∗ c)

= a ∗ [b ∗ (c ∗ d)] (using associativity in [· · · ])= (a ∗ b) ∗ (c ∗ d) (a ∗ [b ∗ y] = [a ∗ b] ∗ y for y = c ∗ d)

That is, all five possibilities are the actually same. More generally it can be shownby induction that all bracketing possibilities for the combination of any n elementsgives the same result.

The following table summarises a number of (possible) binary operations andthe properties that they possess:

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(Z,+) (N,+) (Z,−) (N,−) (Z,×)

Bin Op Y Y Y N Y

Assoc Y Y N Y

Comm Y Y N Y

Id 0 N N 1

Inv of a −a N

(Q,÷) (R,×) (R∗,×) (Z5,+) (Z∗3,×)

Bin Op N Y Y Y Y

Assoc Y Y Y Y

Comm Y Y Y Y

Id 1 1 0 1

Inv of a N 1a

5 − a a

(F(A), ◦) (B(A), ◦) (M(n, R),×) (GL(n, R),×)

Bin Op Y Y Y Y

Assoc Y Y Y Y

Comm N N N N

Id IA IA I I

Inv of a N a−1 N a−1

In the above R∗ = R − {0} and Z∗3 = Z3 − {0} = {1, 2}. The set A is assumed to

contain at least two elements for F(A) and at least three elements for B(A); forM(n, R) and GL(n, R) we assume n ≥ 2.

A Y or N in the first row indicates whether or not the column heading actuallydefines a binary operation. A Y or N in the second and third rows indicates ifthe binary operation is associative and/or commutative respectively. In the fourthrow an N indicates that there is no identity; on the other hand if one exists thenit is given. Finally in the fifth row there is an N if there is at least one element inthe given set that does not have an inverse; if each element does have an inversethen it is given.

Example 4.12. Convince yourself that the above table has been filled in correctly.

Cayley tables and groups

From now on we shall largely confine ourselves to groups rather than binary op-erations in general, where a group is a binary operation which has most of ouruseful properties from earlier:

Definition 4.13. A group is a nonempty set G together with a binary operation∗ on G that is associative, has an identity, and for which every element has aninverse.

Note that we do not require the operation ∗ to be commutative. This leads tothe next definition.

Definition 4.14. A group (G, ∗) is abelian or commutative if the operation ∗ onG is commutative.

61

Cayley tables and groups

A useful device for studying binary operations on finite sets is the Cayley table.In particular these can be used to help determine if the operation makes the setinto a group.

Given a binary operation (A, ∗), with A a finite set, its Cayley table has theelements of A listed in the top row and left hand column of the table, and in thebody of the table we write the outcome a ∗ b of the operation in the row labelledwith a and the column labelled with b. That is the table looks like

a b c · · ·a a ∗ a a ∗ b a ∗ c · · ·b b ∗ a b ∗ b b ∗ c · · ·c c ∗ a c ∗ b c ∗ c · · ·...

......

.... . .

The operation is commutative if our table is symmetric about the diagonal fromtop left to bottom right, since we need x ∗ y = y ∗x for all choices of x, y ∈ A. Forexample consider the tables

(a)

+3 0 1 2

0 0 1 2

1 1 2 0

2 2 0 1

and (b)

∗ 0 1 2

0 1 0 2

1 2 1 0

2 0 2 1

The first table is that for the (known) commutative operation (Z3,+3). The secondtable defines the operation ∗ on {0, 1, 2} since it lists the outcome a ∗ b for alla, b ∈ {0, 1, 2}. It is not commutative since, for instance, 0 ∗ 1 = 0 6= 2 = 1 ∗ 0.

If we have a binary operation (A, ∗) and the row and column associated tosome given element c ∈ A are copies of the labelling row and columns, then c isthe identity. That is, if the table looks like

a b c d · · ·a · · a · · · ·b · · b · · · ·c a b c d · · ·d · · d · · · ·...

......

......

. . .

since then c ∗ x = x ∗ c = x for all x ∈ A. So in table (a) above we see that 0 isthe identity element for (Z3,+3), but that the operation ∗ defined on {0, 1, 2} bytable (b) does not have an identity — although 0, 1 and 2 appear in the correctorder in the second column, they do not do so in the second row (or indeed anyrow).

Finally suppose that a given binary operation (A, ∗) has an identity and sup-

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pose it is c. If the table looks as follows

a b c d · · ·a · · a · · · ·b · · b c · · ·c a b c d · · ·d · c d · · · ·...

......

......

. . .

then b is the inverse of d and d is the inverse of b since

b ∗ d = c = d ∗ b.

Example 4.15. Consider the following tables:

(a)

× −1 1

−1 1 −1

1 −1 1

(b)

· (1) (123) (132)

(1) (1) (123) (132)

(123) (123) (132) (1)

(132) (132) (1) (123)

(c)

×8 1 3 5 7

1 1 3 5 7

3 3 1 7 5

5 5 7 1 3

7 7 5 3 1

(d)

×8 1 3 6 7

1 1 3 6 7

3 3 1 2

6 6

7 7

(e)

∗ 3 4 5

3 4 3 4

4 3 4 5

5 4 5 4

(a) is the table for ({−1, 1},×), (b) is the table for ({(1), (123), (132)}, ·), where{(1), (123), (132)} is a subset of S3 and · denotes the product of permutations, (c)is the table for the operation of multiplication modulo 8 on the set {1, 3, 5, 7} ⊆ Z8,(d) is the table for the same operation on {1, 3, 6, 7} ⊆ Z8, and (e) is a table thatdefines the operation ∗ on the set {3, 4, 5}.

Note that the first three and (e) define binary operations since the entries inthe body of the table all come from the set in question. For (d) we see that ×8

does not define a binary operation on this set, since 3 × 6 = 18 ≡ 2 mod 8, and2 /∈ {1, 3, 6, 7}.

Each of the binary operations in (a), (b), (c) and (e) is commutative, as shownby symmetry, and each has an identity. For (a) it is 1, for (b) it is (1), for (c) it is1 and for (e) it is 4. Moreover in each of these cases every element has an inverse.For (b) the inverses of (1), (123) and (132) are (1), (132) and (123) respectively.For (c) each element is its own inverse. However for (e) we see that 4 is its owninverse (this is true for any identity), but 3 has two inverses, namely 3 and 5, andsimilarly these two elements are also inverses for 5.

Note that since 3 and 5 in table (e) in the above example had two inverses,we know from Proposition 4.9 that this operation cannot be associative, and socannot define a group. Unfortunately it is not so straightforward to obtain positive

63

Examples of groups

information about associativity of a binary operation from looking directly at itsCayley table. We shall return to this problem later on.

However in this course we are taking it as given that addition and multiplicationof numbers are both associative operations, and we showed (Proposition 3.23) thatcomposition of functions is associative. Associativity of just about every otheroperation we shall encounter with can then be proved using these facts.

Examples of groups

Using our assumed knowledge of associativity for addition and multiplication ofnumbers, and the proved associativity in the case of function composition, it isnow not difficult to show that the following are all groups:

(a) (Z,+) and (R,+).

(b) (Q∗,×) and (R∗,×), where Q∗ = Q − {0} and R∗ = R − {0}.

(c) (B(A), ◦), where B(A) is the set of bijections from a set A to itself. Since Sn

is the set of bijections on {1, 2, . . . , n}, and the product of permutations isjust composition in a disguised form, (Sn, ·) is a group for all n ≥ 1.

(d) (M(n, R),+) and (GL(n, R),×).

(e) (Zn,+n), where +n denotes the operation of addition modulo n on the setZn = {0, 1, 2, . . . , n − 1}. Note that 0 is the identity, hence its own inverse,and the inverse of any k 6= 0 is n − k.

The groups in (a), (b) and (e) are all abelian as is (M(n, R),+) from (d); thosefrom (c) and (GL(n, R),×) from (d) are not abelian, except in the special casewhen |A| = 2 in (c) and n = 1 in (d). One way to show that the operation(GL(2, R),×) is associative is to calculate T1(T2T3) = (T1T2)T3 for general matricesTi =

[ai bi

ci di

]∈ GL(2, R), but this is somewhat tedious, particularly if we want to

then extend to the case of n > 2. A more subtle and slick approach is to note thateach T ∈ GL(2, R) defines a (linear) map R2 → R2 by

T

(xy

)

=

[a bc d

](xy

)

=

(ax + bycx + dy

)

.

The product T1T2 of two matrices then corresponds to the composition of thesemaps, and we know from Proposition 3.23 that composition is associative, hencetaking matrix products is associative.

Note that for any x ∈ Q or y ∈ R we have

x × 0 = 0 × x = 0 6= 1 and y × 0 = 0 × y = 0 6= 1.

But 1 is the multiplicative identity in each set, and so 0 does not have multiplicativeinverse. On the other hand, if x ∈ Q∗ or y ∈ R∗, then 1

x∈ Q∗ and 1

y∈ R∗, and

it follows that (Q∗,×) and (R∗,×) are groups as stated above. Unfortunately thesame trick of removing 0 does not work for Z: again the identity is 1, but 0 has nomultiplicative inverse. Moreover if |m| 6= 1, i.e. m is different from 1 or −1, thenits multiplicative inverse would have to be 1

m/∈ Z.

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GROUPS

The situation for (Zn,×n), multiplication modulo n, is in a certain sense bothbetter and worse than (Z,×). Its Cayley table has the form

×n 0 1 2 · · ·0 0 0 0 · · ·1 0 1 2 · · ·2 0 2 4 · · ·...

......

.... . .

and so again 1 is the identity, but 0 has no inverse. So to make a group we shouldwork with some subset A ⊂ Zn for which 0 /∈ A. If we write Z∗

n = Zn − {0} ={1, 2, . . . , n − 1}, this is sometimes, but not always, enough to create a group.

Exercise 4.16. Write out the Cayley tables for (Z∗3,×3), (Z∗

4,×4) and (Z∗5,×5).

Which of these are binary operations? Which are groups?

Proposition 4.17. The operation (Z∗n,×n) is closed if and only if n is prime.

Proof. If n is not prime then we can write n = jk for some integers j, k > 1. Thus1 < j = n

k< n and 1 < k = n

j< n, and so j, k ∈ Z∗

n. But then j×k = n ≡ 0 mod n,and so (Z∗

n,×n) is not closed.Suppose instead that n is prime, and take any j, k ∈ Z∗

n. If it were true thatjk ≡ 0 mod n then we would have jk = rn for some integer r > 1. But thenn being prime implies that it must divide into j or k, and so either j ≥ n ork ≥ n, which is not possible from the way they were chosen. Thus we must havej ×n k ∈ {1, 2, . . . , n − 1} = Z∗

n, and so the operation is closed.

Thus the situation is worse than for (Z,×) since in many cases the operationis not even closed. However, when it is closed, i.e. when n is prime, things aresignificantly better:

Proposition 4.18. For any prime number p the binary operation (Z∗p,×p) is a

group.

Proof. We already know that the operation is well-defined, associative, and has 1as its identity. The only thing left to show is the existence of inverses.

Pick any j ∈ Z∗p. We must show that there is some k ∈ Z∗

p such that jk ≡1 mod p. To do this we shall first show that the map f : Z∗

p → Z∗p given by

f(k) = jk is injective. To see this, let k1, k2 ∈ Z∗p with k1 < k2. We must show

that jk1 6≡ jk2 mod p. But recall that jk1 ≡ jk2 mod p if and only if jk2 − jk1 =j(k2 − k1) = rp for some integer r ≥ 1. Again, since p is prime, this would implythat p divides j or that p divides k2 − k1, and both of these are impossible since1 ≤ j < p and 1 ≤ k2 − k1 < p.

Thus the map f is injective as claimed. But it then must map the p−1 elementsZ∗

p to distinct elements of Z∗p, so its range must contain p− 1 elements, and hence

must be equal to all of Z∗p, i.e. f is also surjective. Thus there is some k ∈ Z∗

p suchthat f(k) = jk ≡ 1 mod p, and thus k = j−1 as required.

One final important class of examples are the groups of symmetries of shapesin the plane. Consider a rectangle:

65

Examples of groups

A B

C D

H

V

R

There are four transformations we can apply to this shape that result in the rect-angle being laid back on itself: reflection about the horizontal or vertical axes ofsymmetry, rotation through 180◦, and the identity transformation that leaves italone. Denote these H, V , R and I respectively, and let R ∗V denote the effect ofdoing R then V , that is rotating the rectangle then flipping it about the vertical.If we follow the paths of the vertices we see that: A is first mapped to the bottomright, then to the bottom left; B is first mapped to the bottom left, then to thebottom right; and similarly for C and D. That is, the combined motion R ∗ V isthe same thing as the single symmetry H. Indeed, carrying out any two of thesetransformations is equivalent to doing just one from the set {H,V,R, I}, and allpossible combinations are listed in the following Cayley table:

I V H R

I I V H R

V V I R H

H H R I V

R R H V I

Note that I is the identity, and each transformation is its own inverse. All rigidtransformations (i.e. length and angle preserving maps) in the plane can be de-scribed using a subset of matrices taken from GL(2, R) known as the orthogonal

group. From this identification and our earlier comments it follows that this com-position is associative. Thus the symmetries of a rectangle form a group.

Each regular polygon has a group of symmetries; if the polygon has n sidesthen the group of symmetries is denoted Dn, and is of order 2n. It consists of nreflections, n−1 nontrivial rotations and the identity. For the (equilateral) trianglewe have

1

2 3

f1

f2f3

r12r13

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where fj denotes the flip about the line through vertex j and the midpoint of theopposite side, and r1k denotes the rotation through 120◦ that takes 1 to k. Soafter carrying out these symmetries we end up with

11

11

1

1

2

2

22

2

2

3

3

3

3

3

3

f1 f2f3

r12r13i

Exercise 4.19.

(a) Complete the Cayley table for D3:

∗ i f1 f2 f3 r12 r13

i i f1 f2 f3 r12 r13

f1 f1 i r13 r12 f3 f2

f2 f2

f3 f3

r12 r12

r13 r13

(b) Draw a diagram illustrating the eight symmetries of a square.

Note that the symmetry group of our triangle can be described by the associ-ated permutation of the vertices 1, 2 and 3. Also note that there are six symmetriesand |S3| = 3! = 6. In a sense that we shall discuss later, the group of symmetries ofthe equilateral triangle is essentially the same as as the group S3 of permutationson {1, 2, 3}.

Multiplicative and additive notation; basic group properties

When writing down the result of applying the group law to elements a and b it isusual to use one of two different notations instead of the notation a ∗ b introducedabove. The symbol ∗ was used above in order to prevent one from thinking thatthe operation was either multiplication or addition of numbers. However it is notso efficient a notation, particularly if we want to dealing with elements of the formg ∗ g ∗ g ∗ · · · , where we take some g ∈ G and do the operation on it repeatedly.

The more general notation is multiplicative notation where we write gh insteadof g ∗ h, the identity is then often denoted by 1 instead of e, and the inverse of an

67

Multiplicative and additive notation; basic group properties

element g is denoted by g−1. Then given g ∈ G and any integer m ∈ Z we definepowers of g by

gm =

gg · · · g (m times) if m > 0,

e if m = 0,

g−1g−1 · · · g−1 (−m times) if m < 0.

The following identities, known as the laws of indices, are an immediate conse-quence of this definition:

(i) gmgn = gm+n (ii) (gm)−1 = (g−1)m (iii) (gm)n = gmn

for all choices of m,n ∈ Z. To prove these in general is quite tedious and involvesbreaking up the problem into many different cases depending on the signs of mand n. Note also that (ii) is essentially a special case of (iii) (with n = −1), butproving this first makes the proof of (iii) somewhat easier.

As an example of some of the steps involved, consider the case when m,n ≥ 1.We have

gmgn = (gg · · · g︸︷︷︸

m

)(gg · · · g︸︷︷︸

n

) = gg · · · g︸︷︷︸

m+n

= gm+n,

and

(gm)n = gmgm · · · gm

︸︷︷︸

n

= (

m︷︸︸︷gg · · · g)(

m︷︸︸︷gg · · · g)(

m︷︸︸︷gg · · · g)

︸︷︷︸

n

= gmn

as required. If m = 0 then gm = e, and so

gmgn = egn = gn = g0+n, and (gm)n = en = ee · · · e = e = emn

for any n. The more tricky cases come, for instance, when we take m and n havingdifferent sign. For instance we have

g−2g3 = g−1 g−1g︸︷︷︸

=e

gg = g−1g︸︷︷︸

=e

g = g = g1 = g−2+3.

It is very important to note that in general we do not have (gh)n = gnhn sincethe left hand side and right hand side are, respectively,

(gh)(gh) · · · (gh)︸︷︷︸

n

and gg · · · g︸︷︷︸

n

hh · · · h︸︷︷︸

n

.

If gh = hg then we can rearrange the left hand side to get the right hand side, soin particular the identity will hold if G is an abelian group.

However, if the binary operation of our group is commutative then it is perhapsmore common to use additive notation in which the symbol ∗ is replaced by +.Also, the identity in this instance is usually written as 0, and the inverse of anelement g is written as −g. Thus with this notation the following identity holdsfor all g ∈ G:

g + (−g) = (−g) + g = 0.

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When dealing with a commutative group, i.e. one for which we would use additivenotation, the exponent notation has to be changed to the following: for any g ∈ Gand any integer m ∈ Z we define

mg =

g + g + · · · + g (m times) if m > 0,

0 if m = 0,

(−g) + (−g) + · · · + (−g) (−m times) if m < 0.

The laws of indices in this notation then read

(i) mg + ng = (m + n)g (ii) − (mg) = (−m)g (ii) m(ng) = (mn)g

for all choices of m,n ∈ Z. Note that these two notations correspond to what youwould write for the groups (R∗,×) and (Z,+) respectively.

The group axioms lead immediately to some important properties as detailedin the results below. Since they are proved using only the group axioms, they areapplicable whenever we have a group G, no matter where G comes from.

Proposition 4.20. Let G be a group. Then

(i) There is only one identity element.

(ii) Each element in G has exactly one inverse.

(iii) The following cancellation laws hold for all g, x, y ∈ G:

If gx = gy then x = y, and if xg = yg then x = y.

(iv) For any g, h ∈ G, each of the equations gx = h and yg = h have unique

solutions.

Proof. Since our binary operation is associative by definition, (i) and (ii) are dealtwith by Proposition 4.9.

(iii) Suppose that gx = gy. Multiplying both sides by g−1 on the left gives

g−1(gx) = g−1(gy) ⇒ (g−1g)x = (g−1g)y ⇒ ex = ey ⇒ x = y,

where we make use of associativity. The proof of the second cancellation law isidentical and left as an exercise.

(iv) Consider first the equation. Now if we set x = g−1h, we have gx = g(g−1h) =(gg−1)h = eh = g. Thus there is a solution to the equation, namely g−1h. Ifx1, x2 ∈ G both satisfy the equation then gx1 = h = gx2, and so by (iii) we havex1 = x2. That is, the solution is unique.

Similarly, the unique solution of the second equation is y = hg−1.

Proposition 4.21. Let G be a group. Then g ∈ G is the identity if and only if

g2 = g.

69

Subgroups, cyclic groups and Lagrange’s Theorem

Proof. Let e denote the identity. That it satisfies the equation e2 = e is obviousfrom its definition. On the other hand suppose that g ∈ G satisfies g2 = g. Thengg = ge, by definition of e, and so using the cancellation laws gives g = e asrequired.

The definition of the inverse h of an element g ∈ G of a group is that it shouldsatisfy gh = e and hg = e. That is, we appear to have two equations to check.The next result shows that, if we know we are dealing with a group, then we onlyneed to check that gh = e holds.

Proposition 4.22. Let G be a group with identity e, and suppose that g, h ∈ Gsuch that gh = e. Then hg = e, and so h = g−1.

Proof. Consider the element (hg)(hg) ∈ G. By associativity and the definingproperties of the identity we have

(hg)(hg) = ((hg)h)g = (h(gh))g = (he)g = hg.

So now, by Proposition 4.21, we have that hg = e as required.

The next result shows how products and inverses are related.

Proposition 4.23. Let G be a group and let g1, g2, . . . , gn ∈ G. Then

(g1g2 · · · gn)−1 = g−1n · · · g−1

2 g−11 .

Proof. It is easy to verify that

(g1g2 · · · gn)(g−1n · · · g−1

2 g−11 ) = e

as each pair in the middle gives e and so disappears from the product. Thus theresult follows by the previous proposition.


Consider the group S3. Given any subset H ⊂ S3 we can take the product of anytwo elements α, β ∈ H, and will get something from S3. Is it possible to chooseα and β so that αβ ∈ H? Moreover, is it possible to choose the subset H so thatαβ ∈ H for all choices of α, β ∈ H?

Indeed, since we are now dealing with sets on which we have defined a binaryoperation with good properties, we should not be concerned with all subsets ofthat group, but rather with subsets that behave well with respect to the groupoperation.

Example 4.24. Consider the subsets {−1, 1} ⊂ R∗, {0, 2, 4} ⊂ Z6 and the per-mutations {(12), (13), (23)} ⊂ S3. The Cayley tables are

× −1 1

−1 1 −1

1 −1 1

+6 0 2 4

0 0 2 4

2 2 4 0

4 4 0 2

· (12) (13) (23)

(12) (1) (123) (132)

(13) (132) (1) (123)

(23) (123) (132) (1)

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GROUPS

We see that the first two sets are closed under the operation of the larger set, buttaking the products of any two permutations in {(12), (13), (23)} takes us outsideof this subset.

Definition 4.25. A subset H of a group G is a subgroup if, with respect to thebinary operation on G, H is itself a group.

This definition, while precise, does not give a totally clear impression of whatis involved in the concept of being a subgroup. The next result clarifies this bygiving an easy method for checking if a given subset is indeed a subgroup.

Proposition 4.26. A H subset of a group G is a subgroup if and only if

(i) gh ∈ H for all g, h ∈ H, and

(ii) g−1 ∈ H for all g ∈ H.

Proof. Suppose that H is a subgroup of G. Then the operation from G mustdefine a (closed) binary operation on H, and so (i) must hold. Also H must havean identity for this operation. Call it f since we do not know a priori that f needbe the same as e, the identity of G. But since f is an identity in H it must satisfyf2 = f , and so, by Proposition 4.21, we must have f = e after all. So now take anyg ∈ H, and let g′ denote its inverse in H, and g−1 its inverse in G. We want toshow that these coincide so that g−1 ∈ H. But this follows from the cancellationlaws of Proposition 4.20 since gg′ = f = e = gg−1.

Assume instead that H is some subset of G for which (i) and (ii) hold. Wehave a binary operation on H by (i), so it remains to show that H has an identityfor which there are inverses, and that this operation is associative. Associativityis inherited from G, and taking any g ∈ H we can combine (i) and (ii) to get

g−1g = e ∈ H,

so that H has an identity, and hence contains the inverses of all of its elementsby (ii).

Example 4.27.

(i) ((0,∞),×) is a subgroup of (R∗,×) (where (0,∞) = {x ∈ R : x > 0}), sincethe product of any two positive numbers is positive, and if x > 0 then 1

x> 0.

(ii) (2Z,+) is a subgroup of (Z,+), where 2Z = {. . . ,−4,−2, 0, 2, 4, . . .}.

(iii) For any group G the subsets {e} and G are both subgroups of G and so areknown as trivial subgroups. Any other subgroup is called nontrivial — butthere are many examples of groups that have no nontrivial subgroup.

Definition 4.28. Let G be a group. For each g ∈ G define the following subset:

〈g〉 = {gk : k ∈ Z}.

It is not hard to see that 〈g〉 is a subgroup of G for each choice of g — justapply Proposition 4.26 and the laws of indices. But what form can this subgrouptake?

71


Proposition 4.29. Let G be a group and let g ∈ G. If gr = gs for some integers

r, s ∈ Z with r > s then there is a positive integer m ∈ N such that

(i) gm = e.

(ii) 〈g〉 = {e, g, g2, . . . , gm−1} and all of these elements are distinct.

(iii) gj = gk for j, k ∈ Z if and only if j ≡ k mod m.

Proof. Since r > s and gr = gs we have by the law of indices that

grg−s = gsg−s = gs−s = g0 = e.

But now r − s > 0 is a positive integer. Hence the subset P of Z defined by

P = {k ∈ Z : k > 0 and gk = e}

is nonempty. Let m be the smallest element of P , so in particular we must havegm = e.

Now suppose that 0 ≤ i < j < m and that gi = gj . Then the same computationshows that gj−i = e where now j − i > 0 so that j− i ∈ P and j− i < m. But thiscontradicts our choice of m, since m is the smallest positive integer in P . Thuswe must in fact have that gi 6= gj as required.

To see that 〈g〉 = {e, g, g2 , . . . , gm−1} note that gm = e, gm+1 = gmg = eg = g,and so on, thus gk for any k > m must lie in this set. Also we have gm−1g = gm = e,and so g−1 = gm−1. Then for any n < 0 we have gn = (g−1)−n = (gm−1)−n =g(1−m)n, where (1 − m)n > 0, and so gn also lies in the given set.

That gj = gk if and only if j ≡ k mod m is now clear from the structure of thesubgroup 〈g〉.

Definition 4.30. Let G be a group and let g ∈ G. The order of g is defined tobe the smallest positive integer k such that gk = e, if such a k exists. If no suchk exists then g is said to be of infinite order.

Remark. Note that the identity e of any group has order 1, and is the only elementof this order.

Corollary 4.31. If G is a finite group then every element has finite order.

Remarks. In particular we see that every element of Sn must have finite order.

Proof. Let g ∈ G. Since G is finite the elements g, g2, g3, . . . cannot all be distinct,and so we must have gr = gs for some integers r and s satisfying 0 < s < r. Thuswe can apply Proposition 4.29.

Example 4.32.

(i) Consider the group (Z9,+9). If we take 3 ∈ Z9, then 3+3 = 6, and 3+3+3 =9 ≡ 0 mod 9. Thus 〈3〉 = {0, 3, 6} is a subgroup of Z9, and 3 ∈ Z9 is of order3.

Note also that 6 is the inverse of 3 in this group, and we have 6 + 6 = 12 ≡3 mod 9, and 6 + 6 + 6 = 18 ≡ 0 mod 9. Hence 〈6〉 = 〈3〉 = {0, 3, 6}.

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GROUPS

(ii) Consider the group (R∗,×), and the element 2 ∈ R∗. We have 21 = 2, 22 = 4,23 = 8, 24 = 16, and so on. But there is no integer k > 0 such that 2k = 1.(Thus there is also no integer j < 0 such that 2j = 1.) That is, 2k only everequals the multiplicative identity when k = 0, and so 2 is of infinite order.

Exercise 4.33. Determine the order of the given elements of the following groups:

(i) 2 ∈ Z7 (under +7) (ii) 2 ∈ Z8 (under +8) (iii) (154)(23) ∈ S5

(iv)7

2∈ Q (under +) (v)

[0 1−1 0

]

∈ GL(2, R).

A particular class of groups of interest are those that are equal to 〈g〉 for someelement g ∈ G.

Definition 4.34. A group G is called cyclic if there is some g ∈ G such thatG = 〈g〉. That is, for every h ∈ G, there is some n ∈ Z such that h = gn (wherethe number n will depend on h).

If G is cyclic with G = 〈g〉, then g is called a generator for G.

Example 4.35. Examples of cyclic groups include

(i) 〈2〉 = {21, 23, 22, 24} = {1, 2, 3, 4} = (Z∗5,×)

(ii) 〈i〉 = ({1,−1, i,−i},×) ⊂ (C∗,×)

(iii) 〈(1234)〉 = {(1), (1234), (13)(24), (1432)} ⊂ S4

The above are all cyclic groups for which we use multiplicative notation, andnote that the last two are cyclic by construction since we deliberately chose asubgroup of the form 〈g〉 of some larger group.

If we use additive notation then a (commutative) group G is cyclic if there issome g ∈ G such that for each h ∈ G there is some n ∈ Z satisfying h = ng.

Example 4.36. Further examples of (additive) cyclic groups are

(i) 〈1〉 = (Z,+), where 1 is of infinite order.

(ii) 〈1〉 = (Z4,+), where 1 is of order 4.

In both sets of examples the generator is not unique. For instance:

(Z∗5,×) = 〈2〉 = 〈3〉 and (Z,+) = 〈1〉 = 〈−1〉.

More generally, if G is a cyclic group with generator g, i.e. if G = 〈g〉, thenG = 〈g−1〉. That is, G is also generated by the inverse of g.

On the other hand in our examples (and in general) not every element is agenerator. Indeed in any group the identity can only ever generate a (sub)groupof order 1.

In general it is not true that a group is cyclic, nor that every element in a givencyclic group is a generator.

73


Example 4.37. We have, for G = S3,

〈(1)〉 = {(1)}, 〈(12)〉 = {(1), (12)}, 〈(13)〉 = {(1), (13)},〈(23)〉 = {(1), (23)}, and 〈(123)〉 = {(1), (123), (132)} = 〈(132)〉.

In all cases the subset 〈g〉 is a proper subset of the group S3, and so S3 is notcyclic.

Similarly if we take g = i ∈ C∗ then 〈i〉 is the subgroup {±1,±i} of (C∗,×),so that |〈i〉| = 4, even though the group (C∗,×) has infinitely many elements.

However there is a cyclic group of every finite order:

Example 4.38. For each n ≥ 1 the group (Zn,+n) is cyclic, with Zn = 〈1〉 ={0, 1, 2, . . . , n − 1}.

For the remainder of this subsection we want to try and find out what sort ofsubsets of a group can be subgroup. Recall that if G is a finite group then thereare 2|G| subsets of G. We have already seen that any subgroup of G must containthe identity e, and to form a subset of G that contains e involves picking a subsetof G − {e}. Thus there are potentially 2|G|−1 subgroups.

If the group in question is cyclic then we are already quite restricted, as thefollowing shows:

Proposition 4.39. Every subgroup of a cyclic group is cyclic. Furthermore, if Gis an infinite group then every nontrivial subgroup is also infinite.

Proof. Let G be a cyclic group with generator g, and let H be a subgroup ofG = 〈g〉. If H 6= {e} then there must be some positive integer k such thatgk ∈ H, since we must have gj ∈ H for some j 6= 0 and if j < 0 then note that(gj)−1 = g−j ∈ H. So let m be the smallest positive integer such that gm ∈ H.We claim that H = 〈gm〉.

Clearly we must have 〈gm〉 ⊆ H since gm ∈ H. Now let gn ∈ H for somen ∈ Z. The Euclidean algorithm supplies us with some j ∈ Z and 0 ≤ r < m suchthat n = mj + r. But then

gn = gmj+r = (gm)jgr = ejgr = gr,

and so by our choice of m we must have r = 0, that is, gn = (gm)i ∈ 〈gm〉 asrequired.

Finally, suppose that G is infinite, and let H be a subgroup such that H 6= {e}.Then we have H = 〈gm〉 for some m ≥ 1. If H were to be finite then we wouldhave (gm)k = e for some k ≥ 1, which would imply that |G| ≤ mk, contradictingour assumption that G be infinite.

Exercise 4.40. Use the result above, together with Example 4.35(ii) to show thatthe group (C∗,×) is not cyclic.

When G is not necessarily cyclic, then the main tool for the analysis of whichsubsets are subgroups is an equivalence relation on G defined in terms of a given(hypothetical) subgroup.

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Definition 4.41. Let G be a group and H a subgroup of G. For each a ∈ G theright coset of H determined by a is the subset

Ha = {ha : h ∈ H}.

Similarly, the left coset of H determined by a is aH = {ah : h ∈ H}.

Note that in particular we have He = {he : h ∈ H} = {h : h ∈ H} = H, thatis, the subgroup H is itself a right coset.

Example 4.42. Let G = S3, and let H = {(1), (123), (132)}. Then we have

H(12) = {(1)(12), (123)(12), (132)(12)} = {(12), (23), (13)}.

In fact we have H(12) = H(13) = H(23). Also,

H(123) = {(1)(123), (123)(123), (132)(123)} = {(123), (133), (1)}= H(1) = H(132).

Example 4.43. The set of even integers 2Z = {2n : n ∈ Z} is a subgroup of(Z,+). There are two (right) cosets, namely 2Z and

2Z + 1 = {2n + 1 : n ∈ Z} = 2Z + 3 = 2Z − 7 = · · · ,

the set of odd integers.

Proposition 4.44. Let H be a subgroup of a group G. Then the right cosets of Hform a partition of G. Hence any two right cosets are equal or disjoint, and their

union is all of G.

Proof. Define a relation R on H by (a, b) ∈ R if a ∈ Hb. We will show that it isan equivalence relation.

Reflexive: for any a ∈ G, a = ea ∈ Ha, since we know that e ∈ H.

Symmetric: suppose that (a, b) ∈ R, so then a = hb for some h ∈ H. But thenb = h−1a, and h−1 ∈ H, hence b ∈ Ha, and so (b, a) ∈ R.

Transitive: suppose that (a, b) ∈ R and (b, c) ∈ R, then there are elements h1, h2 ∈H such that a = h1b and b = h2c. Substituting for b we get a = (h1h2)c, and so(a, c) ∈ H, since we know that h1h2 ∈ H.

Thus R is an equivalence relation, and its equivalence classes form a partitionof G by Corollary 2.28. So take a ∈ G, and consider the subsets [a] and Ha. Ifx ∈ [a] then (x, a) ∈ R and so x = hxa for some hx ∈ H. Thus x ∈ Ha, and hence[a] ⊆ Ha. Conversely, if y ∈ Ha, then y = hya for some hy ∈ H, and so (y, a) ∈ R.Hence y ∈ [a] and so Ha ⊆ [a]. Putting these two together we get that [a] = Ha,and so the right cosets are the equivalence classes of R, and form a partition of Gas required.

Proposition 4.45. Let H be a subgroup of a group G and suppose that H is finite.

Then |Ha| = |H| for all a ∈ G.

75


Proof. Let H = {h1, h2, . . . , hn}, then for the given a ∈ G we have that Ha ={h1a, h2a, . . . , hna}. But now if hia = hja then we must have hi = hj by thecancellation law of groups. Hence the elements making up Ha are distinct, and so|H| = |Ha| as required.

Theorem 4.46 (Lagrange’s Theorem). Let H be a subgroup of a finite group

G. Then the order of H divides the order of G.

Proof. Note that H must be finite, since it is a subset of a finite set. For the samereason there can only be a finite number of distinct right cosets of H,

Ha1,Ha2, . . . ,Hak

for some a1, a2, . . . , ak ∈ G. These form a partition of G by Proposition 4.44, soin particular we have

G = Ha1 ∪ Ha2 ∪ · · · ∪ Hak

with Hai ∩ Haj = ∅ if i 6= j. Thus we must have

|G| = |Ha1| + |Ha2| + · · · + |Hak|,

but by Proposition 4.45 we have |Hai| = |H| for all i and so

|G| = k|H|.

Thus |H| divides |G| as required.

Corollary 4.47. The order of an element g in a finite group G divides the order

of the group.

Proof. This follows since the order of g is equal to the size of 〈g〉, the cyclicsubgroup of G generated by g.

Corollary 4.48. Let G be a group of order n and let g ∈ G. Then gn = e.

Proof. Let g have order m, then, by the previous corollary, this is a divisor of nand so n = mk for some k ≥ 1. But then

gn = gmk = (gm)k = ek = e.

Corollary 4.49. A group of prime order is cyclic.

Proof. Let G be a group of order p for some prime p, and let g ∈ G with g 6= e.Since g is not the identity its order is larger than 1, and so its order must be p,since the only divisors of p are 1 and p. Thus |〈g〉| = p = |G|, which forces G = 〈g〉,so that G is cyclic.

Remark. As an extension of this result, it follows from the proof that if G is agroup of prime order then G = 〈g〉 for any g 6= e, i.e. all nonidentity elements aregenerators of G.

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Example 4.50. We will use Lagrange’s Theorem to help us find all the subgroupsof (Z6,+6). As with all groups we know that Z6 is a subgroup of Z6, and since 0is the identity of Z6 we have that {0} is a subgroup. If H is any other subgroupthen |H| has to divide |Z6| = 6 by Lagrange’s Theorem. Thus |H| must equal 1, 2,3 or 6. The first and last possibilities have already been taken care of, so the onlyother possibilities are |H| = 2 or |H| = 3. Since these are both prime numberssuch a subgroup H can only occur if it is cyclic and generated by an element ofZ6 of that order.

So now note that

1 + 1 = 2, 1 + 1 + 1 = 3, 1 + 1 + 1 + 1 = 4, 1 + 1 + 1 + 1 + 1 = 5,

1 + 1 + 1 + 1 + 1 + 1 = 0 ⇒ 〈1〉 = Z6,

2 + 2 = 4, 2 + 2 + 2 = 0 ⇒ 〈2〉 = {2, 4, 0},3 + 3 = 0 ⇒ 〈3〉 = {3, 0},4 + 4 = 2, 4 + 4 + 4 = 0 ⇒ 〈4〉 = {4, 2, 0},5 + 5 = 4, 5 + 5 + 5 = 3, 5 + 5 + 5 + 5 = 2, 5 + 5 + 5 + 5 + 5 = 1,

5 + 5 + 5 + 5 + 5 + 5 = 0 ⇒ 〈5〉 = Z6.

Hence the nontrivial subgroups of (Z6,+6) are

〈2〉 = 〈4〉 = {0, 2, 4} and 〈3〉 = {0, 3},

and so there are four subgroups overall. Note that since |Z6| = 6 it has 26 = 64subsets, 25 = 32 of which contain the identity 0.

Remark. It is possible to show that for any group G and any g ∈ G that 〈g〉 =〈g−1〉, which would help to avoid some work above since this automatically shows〈1〉 = 〈5〉 and 〈2〉 = 〈4〉, because 5 is the inverse of 1 and 4 is the inverse of 2.

Exercise 4.51. Show that there are six subgroups of the group S3.

Groups of low order and homomorphisms

We have seen that groups of prime order are necessarily cyclic. This means that ifG is a finite group with |G| = p and p prime, then G essentially looks like (Zp,+p).To see this, pick any g ∈ G − {e}, then g must have order p, hence G = 〈g〉, andso the group tables for these two groups are

∗ e g g2 · · · gp−1

e e g g2 · · · gp−1

g g g2 g3 · · · gp = e

g2 g2 g3 g4 · · · g...

......

.... . .

...

gp−1 gp−1 e g · · · gp−2

and

+p 0 1 2 · · · p − 1

0 0 1 2 · · · p − 1

1 1 2 3 · · · 0

2 2 3 4 · · · 1...

......

.... . .

...

p − 1 p − 1 0 1 · · · p − 2

These tables have the same basic pattern: everywhere there is a gk in the tablefor G, we have k in the corresponding spot in the table for Zp. If we define a

77


map f : Zp → G by f(k) = gk, then it is easy to see that f is a bijection, andmoreover f(j + k) = gj+k = gjgk = f(j)f(k). That is, the map f respects thebinary operation on each of the sets.

Definition 4.52. Let G and H be groups. A homomorphism from G into H is afunction f : G → H that satisfies

f(ab) = f(a)f(b) for all a, b ∈ G.

That is, a homomorphism is a function that is compatible with the group structuresof G and H. If f is a bijection then it is called an isomorphism, and the groupsG and H are said to be isomorphic.

To see that this definition corresponds with our intuitive idea of “the same basicpattern”, suppose that G and H are finite groups, and that there is a bijectionf : G → H. We can use f to transform the Cayley table for G into two tablesinvolving elements of H as follows:

(i) Apply f to all the entries in the table.

(ii) Apply f only to the labelling row and column, then recalculate the entriesin the body of the table according to the group law of H.

These are illustrated by

G · · · y · · ·...

...

x · · · xy · · ·...

...(i)

}}||||

|||| (ii)

!!DDDD

DDDD

· · · f(y) · · ·...

...

f(x) · · · f(xy) · · ·...

...

H · · · f(y) · · ·...

...

f(x) · · · f(x)f(y) · · ·...

...

If f is an isomorphism then these two processes will yield the same result, and soG and H will have the same structure. In particular we see that that the mapf : Zp → G considered earlier is an isomorphism, and so G is isomorphic to Zp.But this is true for all groups of order p, so there is essentially only one groupstructure on a set of prime order.

Example 4.53. The groups (Z3,+3) and 〈(123)〉 ⊂ S3 are isomorphic. To seethis, consider their Cayley tables

+3 0 1 2

0 0 1 2

1 1 2 0

2 2 0 1

· (1) (123) (132)

(1) (1) (123) (132)

(123) (123) (132) (1)

(132) (132) (1) (123)

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GROUPS

It is obvious that the map f : Z3 → 〈(123)〉 given by

f(0) = (1), f(1) = (123), f(2) = (132)

is a bijection. Moreover it is an isomorphism since the Cayley tables follow thesame pattern.

Note that if we swap the rows and columns for (123) and (132) then we get

· (1) (132) (123)

(1) (1) (132) (123)

(132) (132) (123) (1)

(123) (123) (1) (132)

It is now obvious that the map g : Z3 → 〈(123)〉 given by

g(0) = (1), g(1) = (132), g(2) = (123)

is also an isomorphism.Thus there can be more than one isomorphism between isomorphic groups.

However in our example here any isomorphism from Z3 to 〈(123)〉 must map 0 to (1)(cf. Proposition 4.57 below), so these maps f and g are the only two isomorphisms.

Example 4.54. The subset

R =

{[x −x−x x

]

: x ∈ R∗}

⊂ M(2, R)

is a group since

(i)

[x −x−x x

] [y −y−y y

]

=

[2xy −2xy−2xy 2xy

]

∈ R for all x, y ∈ R∗, since 2xy 6= 0

if x 6= 0 and y 6= 0.

(ii) Matrix multiplication is associative.

(iii)

[12 −1

2−1

212

]

is the identity since

[x −x−x x

] [12 −1

2−1

212

]

=

[12 −1

2−1

212

] [x −x−x x

]

=

[x −x−x x

]

.

(iv)

[x −x−x x

]

has inverse

[14x

− 14x

− 14x

14x

]

:

[x −x−x x

] [14x

− 14x

− 14x

14x

]

=

[14x

− 14x

− 14x

14x

] [x −x−x x

]

=

[12 −1

2−1

212

]

.

Also, the map f : R∗ → R given by

f(x) =

[x2 −x

2−x

2x2

]

79


is clearly injective and onto, hence bijective. But note also that

f(x)f(y) =

[x2 −x

2−x

2x2

] [y2 −y

2−y

2y2

]

=

[xy2 −xy

2−xy

2xy2

]

= f(xy)

and so is a homomorphism. Thus the group R is isomorphic to (R∗,×).

[Note: both groups are infinite, and so we cannot make use of Cayley tables inthis example. Also, R ∩ GL(2, R) = ∅, which is why the identity and inverses aredifferent from usual.]

Exercise 4.55.

(a) Let ∗ be the binary operation on Z defined by a ∗ b = a + b + 1. Show that(Z, ∗) is a group, and that it is isomorphic to (Z,+).

(b) Write out the Cayley tables for the two operations ({1, 2, 3, 4},×5) and({1, 3, 7, 9},×10). Hence show that they are both groups, and that theyare isomorphic to each other.

We showed above that, up to isomorphism, there is only one group structure ona set of prime order. Contrast this with the number of distinct binary operationson a set A of order n: we can map each pair (a, b) ∈ A × A to any of the nelements of A. There are n2 pairs whose destinations we must decide, and thereare n destinations. Thus there are n × n × · · · × n = nn2

binary operations on A,a function of n that grows very rapidly:

n 1 2 3 4 5 · · ·n2 1 4 9 16 25 · · ·nn2

1 16 19683 4294967296 3 × 1018 · · ·

But 4 is not prime, so how many of the approximately 4 billion binary oper-ations actually define group structure? We can start to analyse this question byconstructing candidate Cayley tables, since the next result brings into play a greatrestriction in what form they can take if we are working with a group.

Proposition 4.56. Suppose that G is a finite group. Then each element of Goccurs exactly once in each row and exactly once in each column of its Cayley

table.

Remark. A Cayley table of this form is known as a Latin square. Thus the tableof a group is always a Latin square, but the converse does not hold. There arebinary operations whose tables are Latin squares but which do not define a groupstructure on the set in question.

Proof. Consider the row in table labelled by some g ∈ G and pick any h ∈ G. Nowby part (iv) of Proposition 4.20 there is a solution to the equation gx = h, namelyx = g−1h, and moreover that solution is unique. But this says precisely that hdoes appear in the row for g, and that it appears exactly once. Moreover it appearsin the column labelled by g−1h. A similar argument works for columns.

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GROUPS

With this result we can now give the tables for groups of small order as follows,in which we will always denote the identity by e. If |G| = 1 then we must havethat G = {e}, and so

e

e e

If |G| = 2 then we have G = {e, a} where a is an element distinct from theidentity. The only possible table is

e a

e e a

a a e

So we have a 6= e but a2 = e. Note that 2 is prime, and from our earlier workwe only expected to be able to find one table, indeed, a table that looked like thetable for (Z2,+2).

If |G| = 3 then we have G = {e, a, b} and the table must start off as:

e a b

e e a b

a a

b b

But now we must place both e and b in the two remaining places in the secondrow, bearing in mind that they can only appear once in each column. Similarlywe must place e and a in the third row according to these rules. We end up beingforced to take:

e a b

e e a b

a a b e

b b e a

Again, we only found one structure, which agrees with our earlier work since 3 isprime.

Suppose now that G is of order 4, so that G = {e, a, b, c}. We now end up withtwo possibilities. In the first case let us see if we can find a group G such thatg2 = e for all g ∈ G. Thus the table starts as

e a b c

e e a b c

a a e

b b e

c c e

and we see that we can complete it as

e a b c

e e a b c

a a e c b

b b c e a

c c b a e

81


If we do not have g2 = e for all g then we can always label our elements so thata is an element that satisfies a 6= e and a2 6= e. Note also that we cannot havea2 = a since this would imply a = e by Proposition 4.21. Thus a2 must be equalto b or c, and we can always adjust our labels once again so that a2 = b. Hencethe table begins:

e a b c

e e a b c

a a b

b b

c c

There is now only one way to complete this table, namely

e a b c

e e a b c

a a b c e

b b c e a

c c e a b

One very important point to note is that it is difficult to tell directly from theseCayley tables whether or not the binary operations defined by them are associative,and hence whether or not we really do have a group. These calculations shouldinstead be thought of as a first step in finding possible group structures, giving aguide to what might be. Then, rather than actually check associativity by hand(if |G| = n then there are 2n3 computations to do since we must calculate a(bc)and (ab)c for all choices of a, b and c) it is preferable to find a natural example ofa group having the particular table, and deduce associativity by other means.

For example, for a group of order 2 we could consider S2 = {(1), (12)}. Thishas Cayley table

(1) (12)

(1) (1) (12)

(12) (12) (1)

Thus if we make the identifications e ↔ (1) and a ↔ (12) then we see thatabstract Cayley table that we wrote down for a group of order 2 does indeeddefine an associative operation, since we know that the product of permutationsis associative.

The other three (nontrivial) tables can be obtained as follows: for |G| = 3 take

G = (Z3,+), where e ↔ 0, a ↔ 1, b ↔ 2

For the first table with |G| = 4 take

G = ({1, 3, 5, 7},×8), where e ↔ 1, a ↔ 3, b ↔ 5, c ↔ 7.

and for the second take

G = (Z4,+), where e ↔ 0, a ↔ 1, b ↔ 2, c ↔ 3

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GROUPS

Thus both of our possible tables for groups of order 4 really do occur. Moreoverevery group of order 4 must be of one form or the other. That is, we can partitionthe set of all groups of order 4 into those that are isomorphic to (Z4,+4) and thosethat are isomorphic to ({1, 3, 5, 7},×8). For instance the group of symmetries of arectangle has Cayley table

I V H R

I I V H R

V V I R H

H H R I V

R R H V I

Each nonidentity element is self-inverse, i.e. of order 2, and so it follows that thisgroup is isomorphic to ({1, 3, 5, 7},×8), whereas the subgroup 〈(1234)〉 of S4 oforder 4 is isomorphic to (Z4,+4).

The following are some consequences of the definition of homomorphism andisomorphism, which lead to useful techniques for spotting whether or not certaingroups are isomorphic.

Proposition 4.57. Let G and H be groups and suppose that f : G → H is a

homomorphism. Then f maps the identity of G to the identity of H and f(g−1) =f(g)−1 for all g ∈ G. Moreover the sets

ker f := {g ∈ G : f(g) = e} and f(G) := {f(g) : g ∈ G}

are subgroups of G and H respectively.

Proof. Let eG denote the identity in G and eH denote the identity in H. TheneGeG = eG, so by the homomorphism property of f we have

f(eG) = f(eGeG) = f(eG)f(eG),

and hence f(eG) = eH by Proposition 4.21. So now for each g ∈ G we have

f(g)f(g−1) = f(gg−1) = f(eG) = eH

and thus f(g)−1 = f(g−1) by Proposition 4.22.Now take any g1, g2 ∈ ker f . Then

f(g1g2) = f(g1)f(g2) = eHeH = eH and f(g−11 ) = f(g1)

−1 = e−1H = eH

so that g1g2 ∈ ker f and g−11 ∈ ker f . Hence ker f is indeed a subgroup by Propo-

sition 4.26. That f(G) is a subgroup is proved similarly.

That ker f and f(G) are always subgroups of G and H can be combined withLagrange’s Theorem to prove that there are not too many homomorphisms betweencertain pairs of groups. Note that for any choice of G and H we always have onehomomorphism given by

f : G → H, f(g) = eH for all g ∈ G.

In some cases this trivial homomorphism is the only one.

83


Example 4.58. Suppose that G and H are groups of order 4 and 25 respectively,and let f : G → H be a homomorphism. Now f(G) is a subgroup of H, and|H| = 25. So, by Lagrange’s Theorem, |f(G)| must be one of 1, 5 or 25. However|G| = 4, which means that the range of f can contain at most 4 elements, i.e.|f(G)| ≤ 4. Thus we must have |f(G)| = 1, and so f(G) = {eH}. Hence there isonly one possible homomorphism, namely the trivial one.

Proposition 4.59. Let G and H be groups and suppose that there is an isomor-

phism f : G → H. Then for each g ∈ G, g and f(g) are of the same order.

Proof. Let g ∈ G be of finite order, say it is of order k. Then gk = eG, the identityof G. But then

f(g)k = f(g)f(g) · · · f(g)︸︷︷︸

k

= f(gg · · · g︸︷︷︸

k

) = f(eG) = eH .

So f(g) is of finite order, with order no greater than k. Suppose it has order lessthan k, that is f(g)j = eH for some j < k. Then we have

eH = f(g)j = f(gj) ⇒ f−1(eH) = f−1(f(gj)

)⇒ eG = gj ,

and this cannot happen by definition of k. Thus f(g) must have order k. Thisargument can be tweaked to show that if g is of infinite order then so is f(g).

Thus when deciding whether or not a given group of order 4 is isomorphicto (Z4,+4) or ({1, 3, 5, 7},×8) it is enough to check if it contains an element oforder 4. If it does then it is isomorphic to (Z4,+4), otherwise it is isomorphic to({1, 3, 5, 7},×8).

Exercise 4.60. Show that the following are all groups, and in each case determinewhether or not they are isomorphic to (Z4,+4) or ({1, 3, 5, 7},×8).

(i) 〈i〉 = {1, i,−1,−i} ⊂ (C∗,×)

(ii)

{[1 00 1

]

,

[−1 00 1

]

,

[1 00 −1

]

,

[−1 00 −1

]}

⊂ GL(2, R)

(iii)

{[1 00 1

]

,

[0 −11 0

]

,

[−1 00 −1

]

,

[0 1−1 0

]}

⊂ GL(2, R)

A modification of the argument in Proposition 4.59 gives

Corollary 4.61. Let G and H be groups, and suppose that f : G → H is a

homomorphism. Then for each g ∈ G of finite order, the order of f(g) divides the

order of g.

Exercise 4.62. Suppose that G and H are groups, with |G| = p and |H| = q,where p and q are distinct primes. Prove that the only homomorphism between Gand H is the trivial one.

The permutations groups Sn are perhaps the most important examples of finitegroups. Indeed, every finite group can be found “inside” some Sn as follows:

84

GROUPS

Theorem 4.63 (Cayley’s Theorem). Let G be any finite group, then there is

an injective homomorphism F : G → Sn for some n ≥ 1.

Proof. Let B(G) denote the group of bijective maps from G to G, and for eachg ∈ G define fg : G → G by fg(h) = gh. Then fg ∈ B(G) since it is

one-to-one: if fg(h1) = fg(h2) then gh1 = gh2, which implies h1 = h2 by thecancellation laws, and

onto: for any h ∈ G, g−1h ∈ G satisfies fg(g−1h) = gg−1h = eh = h.

We can define a map F : G → B(G) by F (g) = fg. Since

(fg1◦ fg2

)(h) = fg1

(fg2

(h))

= fg1(g2h) = g1(g2h) = (g1g2)h = fg1g2

(h)

we have fg1◦ fg2

= fg1g2, that is, F (g1) ◦ F (g2) = F (g1g2), and so F is a homo-

morphism. Moreover, if F (g1) = F (g2) then fg1= fg2

, and so

g1 = g1e = fg1(e) = fg2

(e) = g2e = g2.

Thus F is an injective homomorphism. Finally, since G is finite, we can (re)labelits elements by 1, 2, . . . , n, and then view B(G) as the permutation group S|G|.

This result shows that the permutation groups are indeed important, but theusefulness of this result is somewhat mitigated by the inefficiency of the method ofembedding G into some Sn. In particular recall that |Sn| = n!, hence |S|G|| = |G|!,and so we are embedding our group G into a much larger group. For the group(Z4,+4) we have |Z4| = 4, and the smallest value of n such that n! is a multiple of4 is n = 4. Thus, by Lagrange’s Theorem, S4 is the smallest permutation groupinto which we can embed (Z4,+4). On the other hand, the following exercise givesa group of order 6 that is isomorphic to S3 — but the method of Cayley’s Theoremembeds it into S6, where |S6| = 6! = 720.

Exercise 4.64. Let X = R − {0, 1}, and define maps fi : X → X, i = 1, 2, . . . , 6,by

f1(x) = x, f2(x) =1

x, f3(x) = 1 − x,

f4(x) =1

1 − x, f5(x) =

x

x − 1, f6(x) = 1 − 1

x.

Show that f2 ◦ f5 = f6, f3 ◦ f5 = f4 and f6 ◦ f2 = f3. Construct the Cayleytable for the set {f1, f2, . . . , f6} with respect to the operation of function compo-sition. Hence show that this is a binary operation that is moreover a group thatis isomorphic to S3.

An application: card shuffling

As a final application of some of these results on group theory, consider shufflinga pack of cards. There are 52 cards in a standard pack, and shuffling them justrearranges their order in the pack. That is, we are performing a permutation of52 objects — so any shuffle corresponds to some σ ∈ S52, and there are 52! = |S52|

85

An application: card shuffling

possible shuffles. Moreover by Corollary 4.48 we know that σ52! = (1), the identitypermutation. That is, if we repeat the same shuffle 52! times then the pack willreturn to its original state. However this will take a very long time since

52! = 80, 658, 175, 170, 943, 878, 571, 660, 636, 856, 403, 766, 975,

289, 505, 440, 883, 277, 824, 000, 000, 000, 000

≈ 8.07 × 1067.

One particular shuffle is the perfect riffle shuffle which involves splitting thepack exactly in half, and then interleaving the two halves with the top card re-maining on top. This is illustrated for a pack of 6 cards by:

11

2

2

3

34

4 55

66

σ

Note that since the top and bottom cards remain in the same position we are onlyreally moving 4 cards, and for a pack of n cards we would only move n−2 of them.So after (n − 2)! riffle shuffles we would get back to the original state.

So when n = 6 we know that 6! = 720 repeats of any shuffle bring the packback to its original state, but that (6 − 2)! = 4! = 24 riffle shuffles will suffice.In fact we can do even better than this — what we are really after for any givenshuffle σ is its order, that is the minimum positive number of repeats required toproduce the identity. When n = 6 the riffle shuffle in two-line notation is

σ6 =

(1 2 3 4 5 61 3 5 2 4 6

)

since the top half of the pack end up in the odd numbered positions and the bottomhalf of the pack end up in the even numbered spots. Writing this as a product ofdisjoint cycles we have

σ6 = (2354),

a cycle of length 4. So O(σ6) = 4, that is only 4 riffle shuffles after all are requiredto restore the pack.

For a pack of 8 cards any shuffle repeated 8! = 40320 times will restore thepack, but naively we know that (8 − 2)! = 6! = 720 repeats is sufficient for theriffle shuffle. However in two-line and cycle notation we have

σ8 =

(1 2 3 4 5 6 7 81 3 5 7 2 4 6 8

)

= (235)(476)

Thus O(σ8) = lcm{3, 3} = 3, and so only 3 shuffles are needed — fewer than for apack of 6 cards!

Exercise 4.65. Calculate the orders for the perfect riffle shuffle of packs of 10, 12and 14 cards.

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GROUPS

Even more remarkably, if we return to our standard pack of 52 cards then theperfect riffle shuffle is

σ52 =

(1 2 3 · · · 26 27 28 · · · 51 521 3 5 · · · 51 2 4 · · · 50 52

)

= (2 3 5 9 17 33 14 27) (4 7 13 25 49 46 40 28)

× (6 11 21 41 30 8 15 29) (10 19 37 22 43 34 16 31)

× (12 23 45 38 24 47 42 32) (20 39 26 51 50 48 44 36) (18 35).

Since this is a product of six disjoint cycles of length 8 and one cycle of length 2,

O(σ52) = lcm{8, 8, 8, 8, 8, 8, 2} = 8.

So the pack will return to its original state after only 8 such shuffles!Clearly there are many possibilities for creating variations on such shuffles.

One possibility would be to interleave the cards in a similar way as above, butput the top half in the even number positions rather than the odd numbered ones,so that the top and bottom cards will change position. Further, we could firstperform some new shuffle on the top half, and then riffle shuffle the two halvestogether (see question 6(b) of the Summer 2005 exam).

As another example suppose instead that we have 3n cards in the pack, take offthe top third, and interleave these into the bottom two thirds so that the originalcard number 1 stays in position 1, the original card number 2 now goes to position4, the original card number 3 goes to position 7, and so on. For packs of size 6, 9and 12 the shuffles are

σ6 =

(1 2 3 4 5 61 4 2 3 5 6

)

= (243)

σ9 =

(1 2 3 4 5 6 7 8 91 4 7 2 3 5 6 8 9

)

= (24)(3765)

σ12 =

(1 2 3 4 5 6 7 8 9 10 11 121 4 7 10 2 3 5 6 8 9 11 12

)

= (2 4 10 9 8 6 3 7 5)

with orders 3, 4 and 9 respectively.

87

5 Answers to Selected Exercises

1.5 (a) A ={x2 : x ∈ {1, 2, 3}

}or A = {x ∈ R : x3 − 6x2 + 11x − 6 = 0} would do.

1.8 A = {−4, 0, 4, 8}, so |A| = 4. B = {31, 37, 41, 43, 47}, so |B| = 5.

1.15 The empty set is a subset of every set, and so ∅ ⊆ ∅. But ∅ cannot have a nonemptyset as a subset, since this would force ∅ to contain something. Thus P (∅) = {∅},and so |P (∅)| = 1 = 20 = 2|∅|.

1.21 A×B = {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}. (1, 0) /∈ B×A and (2, 0) /∈ B×A,since 0 /∈ A, even though 1 ∈ B and 2 ∈ B.

1.26 We have the equivalences

x ∈ A ∩ (B ∪ C) ⇔ x ∈ A and x ∈ B ∪ C

⇔ x ∈ A and (x ∈ B or x ∈ C)

⇔ (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C)

⇔ x ∈ A ∩ B or x ∈ A ∩ C

⇔ x ∈ (A ∩ B) ∪ (A ∩ C),

and so A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

1.34 Consider the truth table

A B A − B A′ (A − B) ∪ A′ A ∩ B (A ∩ B)′

1 1 0 0 0 1 0

1 0 1 0 1 0 1

0 1 0 1 1 0 1

0 0 0 1 1 0 1

The fifth and seventh columns are the same, hence (A − B) ∪ A′ = (A ∩ B)′.

1.36 (A × B) ∩ (A × C) ⊆ A × (B ∩ C): Suppose (x, y) ∈ (A × B) ∩ (A × C), then(x, y) ∈ A × B, so that x ∈ A and y ∈ B, and also (x, y) ∈ A × C, so that x ∈ Aand y ∈ C. Hence we have x ∈ A, and y is in both B and C, i.e. y ∈ B∩C. Hence(x, y) ∈ A × (B ∩ C).

(A × B) ∩ (A × C) ⊇ A × (B ∩ C): Suppose (x, y) ∈ A × (B ∩ C). Then x ∈ A,and y ∈ B ∩ C. Thus y ∈ B, so that (x, y) ∈ A × B, and also y ∈ C, so that(x, y) ∈ A × C. Hence (x, y) ∈ (A × B) ∩ (A × C).

These two inclusions together prove that (A × B) ∩ (A × C) = A × (B ∩ C).

2.4 (a) The factors of 6 are 1, 2, 3 and 6, and so aR b if ab = 1, 2, 3 or 6. Hence

R = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (1, 6), (6, 1)}

(b) This is the relation R on N given by (a, b) ∈ R if a + b ≤ 4.

88

ANSWERS TO SELECTED EXERCISES

2.15 R1 is not reflexive since (3, 3) /∈ R1. It is symmetric. It is not transitive since(2, 3) ∈ R1 and (3, 4) ∈ R1, but (2, 4) /∈ R1.

R2 is reflexive, is not symmetric, and is transitive.

R3 is not reflexive, is symmetric, and is transitive.

2.26 Reflexive: a = 1 × a = 100 × a, and so a ∼ a for all a ∈ N.

Symmetric: if a ∼ b then a = 10kb for some k ∈ Z, hence b = 10−ka, where−k ∈ Z. Thus b ∼ a.

Transitive: if a ∼ b and b ∼ c then a = 10kb and b = 10jc for some j, k ∈ Z. Thusa = 10k10jc = 10k+jc, for k + j ∈ Z, and so a ∼ c.

Now a ∈ [1] ⊆ N if and only if a = 10k for some k ∈ Z. We need k ≥ 1 to ensurea ∈ N, and so [1] = {1, 10, 100, 1000, . . .}. Similarly [300] = {3, 30, 300, 3000, . . .}.

3.4 (b) If n is an integer, so is 7n + 5. If n ≥ 1 then 7n + 5 ≥ 12, hence 7n + 5 ∈ N forall n ∈ N, and so f is well-defined. We have f(1) = 12, f(2) = 19, , f(3) = 26,. . . ,so the range is {12, 19, 26, 33, 40, . . .}.If n ≥ 1 is an integer then n and n + 1 are positive integers, and so 1

n, 1

n+1 ∈ Q,

which implies 1n− 1

n+1 ∈ Q, and so g is well-defined. Note that 1n− 1

n+1 = 1n(n+1) ,

and so the range of g is

{ 1n(n+1) : n ≥ 1} = {1

2 , 16 , 1

12 , 120 , 1

30 , . . .}.

1n

is not defined for 0 ∈ Z, and 1n+1 is not defined for n = −1 ∈ Z. Thus h is not

well-defined.

If x − 3 < 0 (⇔ x < 3), then√

x − 3 /∈ R, and so k is not well-defined.

3.6 A defines a function f : R → R+, f(x) = x2 + 3, with domain R, codomain R+,and range [3,∞).

B does not define a function since there is no y ∈ R for which (0, y) ∈ B ⊆ R+×R.So it does not define an image for 0.

(12, 1) = (1, 1) ∈ C, and ((−1)2,−1) = (1,−1) ∈ C, so C has two pairs with 1 asthe first component, hence C does not define a function.

3.18 To construct a function A → B we must decide on an image for each of the melements of A. But for each image we have n elements of B to choose from, so thenumbers of functions is

n × n × · · · × n︸︷︷︸

m

= nm.

If the function f : A → B is injective then distinct elements of A are mapped todistinct elements of B, and so the range of f has the same number of elements asA, namely m. But the range of f is a subset of B, and so we must have m ≤ n.

If g : A → B is surjective, then each element of B is reached by applying g tosome element of A. Thus there must be at least as many elements in A as thereare in B, and so |A| = m ≥ n = |B|.

89

3.26 Set A = {1}, B = {a, b}, and define f : A → B and g : B → A by f(1) = a,g(a) = 1 and g(b) = 1. These satisfy the requirements of both (a) and (b).

3.33 αβ =

(1 2 3 4 53 2 5 1 4

)

α−1 =

(1 2 3 4 52 5 1 4 3

)

β−1 =

(1 2 3 4 54 1 3 2 5

)

β−1α−1 =

(1 2 3 4 54 2 1 5 3

)

= (αβ)−1

3.35 β: The orbit of 1, 4 and 5 is {1, 4, 5}. The orbit of 2 is {2}. The orbit of 3 and 6is {3, 6}.

3.39 α = (13)(2645) and β = (123)(456). Hence αβ = (24653) and βα = (16542). Alsoβ2 = (132)(465) and β3 = (1), the identity. Thus β7 = β3β3β = (1)(1)β = β.

3.45 (1527)(3567)(273) = (15)(12)(17)(35)(36)(37)(27)(23) = (57)(51)(56)(53), and sois even.

3.53 (a) The cycles are disjoint, and hence the order is lcm{3, 2, 2} = 6.

(b) 6.

(c) (125)(354) = (12435), so the order is 5.

(d) 5.

4.5 (b) For any integers x and y, x + y ∈ Z, and x+ y− 7 ∈ Z, so x ∗ y is well-defined.

(x ∗ y) ∗ z = (x + y − 7) ∗ z = (x + y − 7) + z − 7 = x + y + z − 14, andx ∗ (y ∗ z) = x ∗ (y + z − 7) = x + (y + z − 7) − 7 = x + y + z − 14, thus ∗ isassociative. (Here we have used associativity of (Z,+).)

x ∗ y = x + y − 7 = y + x − 7 = y ∗ x, and so ∗ is commutative.

x ∗ 7 = 7 ∗ x = x + 7 − 7 = x, and so 7 is the identity.

x ∗ (14−x) = (14−x) ∗x = x+ (14−x)− 7 = 7, and so 14−x is the inverse of x.

4.16 The Cayley tables are

×3 1 2

1 1 2

2 2 1

×4 1 2 3

1 1 2 3

2 2 0 2

3 3 2 1

×5 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

The first and third are both binary operations, and indeed groups. The table for(Z∗

4,×4) shows that the operation is not closed.

4.19 (a)∗ i f1 f2 f3 r12 r13

i i f1 f2 f3 r12 r13

f1 f1 i r13 r12 f3 f2

f2 f2 r12 i r13 f1 f3

f3 f3 r13 r12 i f2 f1

r12 r12 f2 f3 f1 r13 i

r13 r13 f3 f1 f2 i r12

90

ANSWERS TO SELECTED EXERCISES

4.33 (i) 7.

(ii) 2 + 2 = 4, 2 + 2 + 2 = 6 and 2 + 2 + 2 + 2 = 8 ≡ 0 mod 8, so 2 is of order 4.

(iii) 6.

(iv) Infinite order, since n72 6= 0 for all n ≥ 1.

(v) 4.

4.51 These are {(1)}, S3 and the four cyclic subgroups listed in Example 4.37.

4.55 That (Z, ∗) is a group can be shown in the same way as Exercise 4.5(b), with theidentity now being −1, and the inverse of x being −x − 2.

Consider the map f : Z → Z given by f(n) = n + 1. Then f is obviously bijective,and

f(x ∗ y) = f(x + y + 1) = (x + y + 1) + 1 = (x + 1) + (y + 1) = f(x) + f(y).

Thus f : (Z, ∗) → (Z,+) is an isomorphism.

4.60 (i) and (iii) are isomorphic to (Z4,+4), (ii) is isomorphic to ({1, 3, 5, 7},×8).

4.65 For 10 cards

σ10 =

(1 2 3 4 5 6 7 8 9 101 3 5 7 9 2 4 6 8 10

)

= (2 3 5 9 8 6) (4 7)

and so O(σ10) = lcm{6, 2} = 6.

For 12 cards

σ12 =

(1 2 3 4 5 6 7 8 9 10 11 121 3 5 7 9 11 2 4 6 8 10 12

)

= (2 3 5 9 6 11 10 8 4 7)

and so O(σ10) = 10.

For 14 cards O(σ14) = 12.

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