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Abstract Algebra I

Randall R. Holmes

Auburn University

Copyright c 2012 by Randall R. HolmesLast revision: April 25, 2012

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Notation

N = {1, 2, 3 . . . }, natural numbers Z = {. . . , 2, 1, 0, 1, 2, . . . }, integers

Q =m

n| m, n Z, n = 0

, rational numbers (fractions)

R, real numbers C = {a + bi | a, b R} (i = 1), complex numbers Q, R, C, nonzero elements ofQ, R, C, respectively Q+, R+, positive elements ofQ, R, respectively Matmn(R), m n matrices over R Matn(R), n n matrices over R GLn(R), invertible n n matrices over R SLn(R), n n matrices over R having determinant 1 FR, functions from R to R

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0 Introduction

There are many familiar ways in mathematics to combine two things to getanother. Examples are

addition of numbers, multiplication of numbers, addition of matrices, multiplication of matrices,

addition of vectors, composition of functions.

There is some commonality among these operations. For instance, each isassociative ((x + y) + z = x + (y + z), (xy)z = x(yz), etc.).

We could define an abstract associative structure to be a set with an asso-ciative operation. Then we could study that abstract associative structureon its own knowing that anything we discovered would automatically applyto all of the examples above.

This is the idea behind abstract algebra.

The present course is the study of a group, which is a set with an asso-ciative operation, having an identity element, and such that each elementhas an inverse (see Section 4). With some restrictions, each of the examplesabove gives rise to a group.

Groups arise naturally in many areas of mathematics and in other areas aswell (e.g., chemistry, physics, and electronic circuit theory).

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1 Function

1.1 Notation

Let X and Y be sets. A function f from X to Y (written f : X Y)is a rule that assigns to each element x of X a unique element f(x) of Y,depicted using a Venn diagram like this:

x f(x)

fX Y

We think of x as an input to the function f and f(x) as the correspondingoutput. The set X is the domain of f and the set Y is the codomain off.

A function is sometimes described by giving a formula for the output interms of the input. For instance, one writes f(x) = x2 to refer to thefunction f : R R that takes an input x and returns the output x2. Sothe equation f(x) = x2 says that for an input x the output f(x) is x2.

1.2 Well defined function

For a function to be well defined it must assign to each element of itsdomain a unique element of its codomain. The following example showsvarious ways a function can fail to be well defined.

1.2.1 Example The following functions are not well defined:

(a) f : R R given by f(x) = 1/x. (The number 0 is in the domain R,but f(0) = 1/0 is undefined, so f does not assign an element to each

element of its domain.)

(b) g : [0, ) R given by g(x) = y where y2 = x. (We have g(4) = 2(since 22 = 4), but also g(4) = 2 (since (2)2 = 4), so g does notassign a unique element to each element of its domain.)

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(c) h : [2, ) (4, ) given by h(x) = x2. (The number 2 is in thedomain [2, ), but h(2) = 4, which is not in the codomain, so h doesnot assign to each element of its domain an element of its codomain.)

1.3 Equal functions

Two functions f and g are equal (written f = g) if they have thesame domain and the same codomain and if f(x) = g(x) for eachelement x in their common domain.

1.3.1 Example Let f : (, 0] R be given by f(x) = x and letg : (, 0] R be given by g(x) =

x2. Prove that f = g.

Solution For x (, 0], we havef(x) = x = |x| =

x2 = g(x).

Therefore, f = g.

1.4 Injective function

Let f : X Y be a function. Informally, f is injective if it never sendstwo inputs to the same output:

fX Y fX Y

f injective f not injective

Here is the formal definition:

The function f : X Y is injective if f(x) = f(x) implies x = x(x, x X).

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In words, f is injective if whenever x and x have the same outputs, it must

be the case that x and x

are just two names for the same input.An injective function is called an injection.

1.4.1 Example Prove that the function f : R R given by f(x) =2x + 3 is injective.

Solution Let x, x R and assume that f(x) = f(x). Then 2x+3 = 2x+3and we can subtract 3 from both sides and divide both sides by 2 to getx = x. Therefore, f is injective.

1.4.2 Example Prove that the function f : R

R given by f(x) = x2

is not injective.

Solution We have 1, 1 R and f(1) = 12 = 1 = (1)2 = f(1), but1 = 1. Therefore, f is not injective.

1.5 Surjective function

Let f : X Y be a function. Informally, f is surjective if every elementof the codomain Y is an actual output:

fX Y

f surjective f not surjective

fX Y

Here is the formal definition:

The function f : X Y is surjective if for each y Y there existsx X such that f(x) = y.

A surjective function is called a surjection.

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1.5.1 Example Prove that the function f : (, 0] [2, ) given byf(x) = 2 3x is surjective.Solution Let y [2, ). Put x = 13 (2 y). Since y 2, we have x 0, sothat x (, 0]. Also,

f(x) = 2 3x = 2 3 13 (2 y) = y.Therefore, f is surjective.

(Since we were seeking an x in the domain for which f(x) = y, that is,for which 2 3x = y, we just solved this equation for x. When you provesurjectivity, there is no obligation to show how you came up with an x that

works. In fact, it confuses the reader if you include this scratch work sincethe reader does not expect it.)

1.5.2 Example Prove that the function f : [1, ) (2, ) given byf(x) = x + 2 is not surjective.

Solution The number 5/2 is in the codomain (2, ). But this numberis not an output, for if f(x) = 5/2 for some x [1, ), then x + 2 = 5/2,implying x = 1/2 / [1, ), a contradiction. Therefore, there is no x [1, )for which f(x) = 5/2 and we conclude that f is not surjective.

(Since an input x is

1 and the corresponding output is x + 2, we saw thatthe output would always be 3 and this is what gave the idea to consider5/2.)

Let f : X Y be a function. The image of f is the set of all outputs.

fX Yim f

Here is the formal definition:

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The image of f : X

Y (denoted im f) is given by

im f = {f(x) | x X}.

1.5.3 Example Let f : [1, ) R be given by f(x) = 3 x. Provethat im f = (, 2].

Solution () Let y im f. Then y = f(x) = 3 x for some x [1, ).Since x 1, we have y 2 so y (, 2]. This shows that im f (, 2].() Let y (, 2]. Put x = 3 y. Since y 2, we have x 1 sox [1, ). Thus, y = 3x = f(x) im f. This shows that (, 2] im f.Therefore, im f = (, 2].(The claim was that the two sets im f and (, 2] are equal. To show twosets are equal, one shows that each is a subset of the other. We use ()to introduce the proof that im f (, 2] and () to introduce the proofthat im f (, 2].)

Let f : X Y be a function. Saying im f = Y is the same as saying everyelement of Y is in the image of f, that is, every element of Y is an output.Therefore, saying im f = Y is the same as saying that f is surjective.

1.6 Bijective function

A bijective function is a function that is both injective and surjective. Inthe Venn diagram of a bijective function, each element of the codomain hasprecisely one arrow pointing to it (it has at least one such arrow since thefunction is surjective and at most one since the function is injective).

fX Y

f bijective

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A bijective function is called a bijection. It is said to set up a one-to-one

correspondence between the elements of its domain and the elements ofits codomain.

Let X and Y be sets. We write |X| = |Y| and say that X and Y have thesame cardinality if there exists a bijection from X to Y.

IfX is finite (i.e., has finitely many elements), we denote by |X| the numberof elements in X. If X and Y are both finite sets, then they have the samenumber of elements if and only if there exists a bijection from X to Y. In thiscase, the statement |X| = |Y| causes no confusion since it is true whether itis interpreted as saying X and Y have the same number of elements or assaying there exists a bijection from X to Y.

Even if X is not finite, we think of |X| as a measure of the size of X.(Technically, |X| is the equivalence class ofX under the equivalence relation defined on sets by putting X Y if there exists a bijection from X to Y.)However, when the set is infinite, some counterintuitive things can happen.For instance, Exercise 16 shows that it is possible to have |X| = |Y| withY a proper subset of X, which certainly cannot happen if X is finite.

It has been shown that |Q| = |Z|, but |R| = |Z|.

1.7 Composition

Let f : X Y and g : Y Z be functions. The composition of f and gis the function g f : X Z given by

(g f)(x) = gf(x).

fX Y Zg

g f

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

The following theorem says that composition of functions is associative.

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1.7.1 Theorem. Iff : X Y, g : Y Z, andh : Z W are functions,then (h g) f = h (g f).Proof. Let f, g, and h be functions as indicated. For each x X we have

(h g) f(x) = (h g)f(x)= h

g(f(x))

= h

(g f)(x)

=

h (g f)(x).Therefore, (h g) f = h (g f).

Because of this theorem, we can write h g f (without parentheses) andno confusion can arise.

The following example shows that a composition of injective functions isinjective.

1.7.2 Example Let f : X Y and g : Y Z be functions. Provethat if f and g are injective, then so is g f.

Solution Assume that f and g are injective. Let x, x X and assume that(g f)(x) = (g f)(x). Then g(f(x)) = g(f(x)) and, since g is injective, weget f(x) = f(x). But since f is injective, this last equation implies x = x.Therefore g f is injective.

Exercise 17 shows that a composition of surjective functions is surjective.

1.8 Identity function

Let X be a set. The identity function on X is the function 1X : X Xgiven by 1X(x) = x.

1.8.1 Theorem. Let f : X

Y be a function from the set X to the set

Y. We have f 1X = f and 1Y f = f.

Proof. For any x X, we have

(f 1X)(x) = f

1X(x)

= f(x),

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where the last equality uses the definition of 1X. Therefore f 1X = f.The proof of the second equality is Exercise 18.

1.9 Inverse function

Let X and Y be sets and let f : X Y be a function. An inverse of f isa function f1 : Y X such that for all x X and y Y,

f1f(x) = x and ff1(y) = y.

The first equation says that if you apply f first and then apply f1, you getback the original input x. So, in a sense, f1 undoes what f does. Similarly,the second equation says that f undoes what f1 does.

In terms of the Venn diagram, f1 is just like f except with the directionof the arrows reversed:

fX1

234

Ya

bcd

f1X1

234

Ya

bcd

Another way to say that f1 : Y X is an inverse of f is to say that

f1 f = 1X and f f1 = 1Y.

The first equation here says that for all x X, (f1 f)(x) = 1X(x), thatis, f1

f(x)

= x, which is the first equation above. Similarly, the second

equation here says the same thing as the second equation above.

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An inverse function need not exist. For instance, the function f : R Rgiven by f(x) = x

2

has no inverse. Indeed, if f1

were an inverse of f, thenwe would get

f1(1)2 = ff1(1) = 1,contradicting that the square of a number is never negative.

However, the next theorem says that any bijection has an inverse (and alsothat any function that has an inverse must necessarily be bijective).

1.9.1 Theorem. A function f : X Y has an inverse if and only if itis bijective.

Proof. Let f : X

Y be a function.

() Assume that f has an inverse f1. (f injective?) Let x, x X andassume that f(x) = f(x). Then

x = f1

f(x)

= f1

f(x)

= x.

Therefore, f is injective. (f surjective?) Let y Y. Put x = f1(y). Thenx X and

f(x) = f

f1(y)

= y.

Therefore, f is surjective. Since f is both injective and surjective, it isbijective.

() Assume that f is bijective. Define f1 : Y X by letting f1(y) bethe unique x in X for which f(x) = y. (Since f is surjective, there is at leastone such x and since f is injective, there is at most one such x.) For x X,we have f1

f(x)

= x (since f1

f(x)

is defined to be the element that f

sends to f(x)). Similarly, for y Y, ff1(y) = y (since f1(y) is definedto be the element that f sends to y). Therefore, f1 is an inverse of f.

1 Exercises

11 Determine whether each function is well defined. If the function is

not well defined, explain why not.

(a) f : R R given by f(x) = ln x.(b) f : Z Z given by f(n) = n/2.

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(c) f : Z Z given by f(n) = n/2, if n is a multiple of 2,

n/3, if n is a multiple of 3,n, otherwise.

12 Prove that the function f : (, 1] R given by f(x) = (x 1)2 isinjective.

13 Let Mat2(R) be the set of all 2 2 matrices having entries in the setR of real numbers. Let f : Mat2(R) R be the determinant functiongiven by

f

a bc d

= ad bc.

(a) Prove or disprove: f is injective.

(b) Prove or disprove: f is surjective.

14 Let f : R [1, 4) be given by f(x) = 2 sin x. Prove that f is notsurjective.

15 Let f : [0, ) R be given by f(x) = 1 (x + 1)2. Prove thatim f = (, 0].

16 Let 2Z denote the set {2n | n Z} (even integers). Show that thefunction f : Z 2Z given by f(n) = 2n is a bijection. (This shows that|Z| = |2Z| even though 2Z is a proper subset of Z.)

17 Let f : X Y and g : Y Z be functions. Prove that if f and gare surjective, then so is g

f.

18 Let f : X Y be a function from the set X to the set Y. Prove that1Y f = f.

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19 It was shown in Section 1.9 that the function f : R R given byf(x) = x

2

does not have an inverse. Use this same formula, but change thedomain and codomain of f so that it does have an inverse. State what theinverse function is and prove that it satisfies the definition of an inverse.

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2 Permutation

2.1 Definition

Let X be a set. A permutation of X is a bijection from X to X. We useGreek letters, like and , to denote permutations (rather than f, g, etc.)

There are various ways to depict permutations. The figure on the left belowshows a permutation of the set X = {1, 2, 3, 4} using a Venn diagram withtwo copies of X side by side, and the figure on the right shows the samepermutation using a single copy of X.

X 1234

X1234

12

34

X

This same permutation can be represented compactly using a matrix:

=

1 2 3 42 4 1 3

.

Each number in the top row is mapped to the number directly below it.The term permutation, which ordinarily means rearrangement, is usedhere because one imagines the numbers 1, 2, 3, 4 in the usual order (top row)being rearranged in the order 2, 4, 1, 3 (bottom row).

The set of all permutations of the set X is denoted SX:

SX = { : X X| is a bijection}.

When X = {1, 2, 3, . . . , n} we write SX simply as Sn. For example, S3 hasthe following 6 elements:

1 2 31 2 3

,

1 2 32 1 3

,

1 2 33 2 1

,

1 2 31 3 2

,

1 2 32 3 1

,

1 2 33 1 2

.

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Just looking at the bottom rows of these matrices, we see that all possible

arrangements of the numbers 1, 2, and 3 are represented. We expect 3! =3 2 1 = 6 such arrangements and this is the case. In general,

|Sn| = n!.

2.2 Permutation composition

Let X be a set. For and in SX, we write the composition simplyas .

The following theorem says that the set SX is closed under composition.

2.2.1 Theorem. If , SX, then SX.

Proof. Let , SX. Then and are both bijections from X to X.Since they are both injective, so is (by 1.7.2), and since they are bothsurjective, so is (by Exercise 17). Therefore, is a bijection from Xto X, that is, SX.

2.2.2 Example Let and be the elements of S3 given by

=

1 2 32 3 1

, =

1 2 31 3 2

.

Find .

Solution We have ( )(1) =

(1)

= (2) = 3, so sends 1 to 3. Thisgives the first column in the matrix representation of . The other columnsare computed similarly and we get

=1 2 3

3 2 1

.

(Note that is a bijection and is therfore in S3 in accordance with thetheorem.)

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2.3 Identity permutation

Let X be a set. We have defined 1X : X X by 1X(x) = x. This functionis a bijection from X to X and is therefore a permutation of X, that is,1X SX. The permutation 1X is called the identity permutation of X.When X = {1, 2, 3, . . . , n} we denote the identity permutation by :

=

1 2 3 n1 2 3 n

.

The special case of Theorem 1.8.1 with Y = X says that for each permuta-tion SX we have

1X = and 1X =

so 1X acts like the number 1 if we view composition as a type of multipli-cation.

2.4 Inverse permutation

Let X be a set and let : X X be an element ofSX (so is a permutationof X). Since is bijective, its inverse 1 exists by Theorem 1.9.1.The special case of the second boxed statement in 1.9 with Y = X says that

1 = 1X and 1 = 1X.

Regarding composition as a type of multiplication, we have seen that 1Xacts like the number 1. The equations above show that 1 acts like the

multiplicative inverse of (which is the reason for the superscript 1).2.4.1 Example Let be the element of S4 given by

=

1 2 3 43 1 2 4

.

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Find 1 and verify that 1 = .

Solution Since maps each number in the top row to the number directlybelow it, the inverse function 1 maps each number in the bottom row tothe number directly above it. Therefore,

1 =

1 2 3 42 3 1 4

.

We have

1 = 1 2 3 42 3 1 4

1 2 3 43 1 2 4 =

1 2 3 41 2 3 4 =

(see Example 2.2.2 for how to compute the composition).

2 Exercises

21 Let and be the elements of S4 given by

=

1 2 3 42 3 4 1

, =

1 2 3 42 1 4 3

.

Find each of the following:

(a) (2).

(b) 1(1).

(c) .

22 Let X be a set. For each a X, define

Fa = { SX | (a) = a}.

So Fa is the set of all permutations of X that fix a (i.e., leave a unmoved).Prove that for each Fa and SX, we have 1 F(a).

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3 Binary operation

A binary operation on a set is a way of combining two elements of the setto form another element of the set.

Addition (+) is an example of a binary operation on the set Z of integers.For instance, the numbers 5 and 3 are combined to produce 5 + 3, which is8.

Multiplication () is another example of a binary operation on Z. For in-stance, the numbers 5 and 3 are combined to produce 5 3, which is 15.

3.1 Definition

Let X be a set. Informally, a binary operation on X is a rule that assignsto each pair x and y of elements of X an element x y of X.The formal definition requires the notation

X X = {(x, y) | x, y X}.

So X X denotes the set of all ordered pairs having entries coming fromthe set X.

Here is the formal definition:

A binary operation on X is a function : X X X.

Let : X X X be a binary operation on X. For an input (x, y), in-stead of denoting the corresponding output by (x, y) using usual functionnotation, it is customary to write x y instead.

The binary operation

on X is commutative if x

y = y

x for

all x, y X. It is associative if (x y) z = x (y z) for allx,y,z X.

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3.2 Examples and nonexamples

3.2.1 Example The following are binary operations:

(a) Addition (+) on Z, Q, R, and C. It is both associative and commu-tative.

(b) Multiplication () on Z, Q, R, and C. It is both associative andcommutative.

(c) Addition (+) on the set Rn of all n-tuples (x1, x2, . . . , xn) with xi Rgiven by

(x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn)

(called componentwise addition). It is both associative and com-mutative (see Exercise 33).

(d) Function addition (+) on the set FR of all functions from R to Rdefined as follows: for f, g FR, the sum f+g is given by (f+g)(x) =f(x) + g(x). It is associative (by Exercise 34) and commutative since,for f, g FR, we have

(f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x)

for all x R, implying f + g = g + f.(e) Matrix addition (+) on the set Matmn(R) (m n matrices over R).

It is both associative and commutative.

(f) Matrix multiplication () on the set Matn(R) (n n matrices over R).We show that this binary operation is associative in general but notcommutative when n 2.(Associative) Let A = [aij], B = [bij ], and C = [cij ] be n n matrices.The (i, j)-entry of the product AB is pij =

nk=1 aikbkj (= dot product

of ith row of A with jth column of B). Therefore, the (i, j)-entry ofthe product (AB)C is

nl=1

pilclj =

nl=1

nk=1

aikbkl

clj

=nl=1

nk=1

(aikbkl)clj (distributive law in R).

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It is left as an exercise to show that this is also the ( i, j)-entry of the

product A(BC) (see Exercise 35).(Not commutative when n 2.) When n = 2, we have

1 00 0

0 10 0

=

0 10 0

=

0 00 0

=

0 10 0

1 00 0

.

For general n, if A has upper left corner 1 and zeros elsewhere andB has upper right corner 1 and zeros elsewhere, then the upper rightcorner of AB is 1 and that of BA is 0, so AB = BA.

(g) Composition () on the set SX (permutations of the set X). It isassociative (see 1.7), but it is not commutative when

|X

| 3 (see

Exercise 36).

(h) Let n be a positive integer and let Zn = {0, 1, 2, . . . , n 1}. Defineaddition modulo n, denoted +, on Zn by letting a + b be the re-mainder of the usual sum a + b upon division by n. For instance, ifn = 5, then 3 + 4 = 2 (since the remainder of 7 upon division by 5is 2). Then + is a binary operation on Zn. It is both associative andcommutative. (The direct proof that it is associative is a bit tedious,but we will see an easy indirect proof later, so we postpone the proofuntil then.)

Since a binary operation is technically a function, it is important, whenattempting to define a particular binary operation, to check that a well-defined function results (see Section 1.2). For instance, the function mustassign to every element (x, y) of the domain X X a unique element of thecodomain X. In other words, it must be the case that for every pair x andy in X, the expression x y is defined and is a single element of X.

3.2.2 Example Determine whether each of the following is a well-defined binary operation .

(a) On R, x y = x/y.

(b) On Z, x y = (x + y)/2.(c) On Z, x y = c, where c > x + y.(d) On Q, x y = (x + y)/2.

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Solution (a) This is not a binary operation since 1 0 = 1/0, which isundefined.

(b) This is not a binary operation since 1 2 = (1 + 2)/2 = 3/2, which isnot in Z.

(c) This is not a binary operation. Indeed, 1 2 = 4 (since 4 > 1+2), butalso 1 2 = 5 (since (5 > 1 + 2), so 1 2 is not a single element of Z.

(d) This is a binary operation since, for each pair x and y in Q, theexpression (x + y)/2 is defined and is a single element of Q.

3.3 Closed subset

Let X be a set and let be a binary operation on X. A subset Y of X isclosed under if whenever x and y are elements of Y, then x y is also anelement of Y:

x

X

y

x

y

Y

Y closed

x

X

y

x y

Y

Y not closed

Put more succinctly,

The subset Y of X is closed under if x, y Y x y Y.

If Y is closed under , then can be viewed as a binary operation on Y.3.3.1 Example Addition (+) is a binary operation on Z. Let 2Z de-note the set {2n | n Z} (even integers). Then 2Z is a subset of Z. Provethat 2Z is closed under +.

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Solution Let x, y 2Z. Then x = 2n and y = 2m for some n, m Z. Wehave x + y = 2n + 2m = 2(n + m) 2Z. Therefore, 2Z is closed under +.

3.3.2 Example Matrix multiplication is a binary operation on Matn(R).Fix n N and let GLn(R) be the set of invertible n n matrices over R.Then GLn(R) is a subset of Matn(R). Prove that GLn(R) is closed undermatrix multiplication. (GL stands for General Linear.)

Solution Let A, B GLn(R). Then the inverse matrices A1 and B1exist. We have

(AB)(B1A1) = A(BB1)A1 = AIA1 = AA1 = I

and similarly (B1A1)(AB) = I. Therefore, (AB)1 exists (it is B1A1),that is, AB GLn(R).

3.3.3 Example Let X be a set, let be a binary operation on X, andassume that is both commutative and associative. Let Y = {y X| yy =y}. Then Y is a subset ofX.

(a) Prove that Y is closed under .(b) Let X = R and let be multiplication on R. Identify the set Y in

this case and verify that it is closed under in agreement with part(a).

Solution

(a) Let y, z Y. (Must show y z is in Y, which amounts to showing(y z) (y z) = y z.) We have

(y z) (y z) = y z y z ( is associative)= y y z z ( is commutative)= y z (y, z Y)

Therefore, Y is closed under .(b) In this case, Y consists of all real numbers y such that y2 = y. Solving,

we get y(y 1) = 0 so y = 0, 1. Thus Y = {0, 1}. Since 0 0 = 0,0 1 = 0, and 1 1 = 1 (and is commutative), we have checked all

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possible products of elements of Y and have shown that they all stay

inside Y. Therefore, Y is closed under in agreement with part (a).

3.4 Table

Let X be the set {a,b,c,d}. The following table gives a binary operation on X:

a b c da a b c db b c d ac c d a bd d a b c

This table works much like a multiplication table. For instance, b c = d.In general, the element x y is located at the intersection of the row labeledx and the column labeled y.

Because the table is symmetric about the 45 line from the upper left cornerto the lower right corner, it follows that is commutative.It turns out that

is also associative (this follows from the fact that this is

the table for the group Z4, as we will see). In general, it is difficult to usethe table to tell that a binary operation is associative.

3 Exercises

31 Determine whether each of the following is a well-defined binary op-eration .

(a) On R, x y = xy.

(b) On [0, ), x y = |a + b|, where a2 = x and b2 = y (a, b R).(c) On N, x y = 2xy. (N = {1, 2, 3, . . . , }.)

(d) On (1, ), x y = 2xy 3xy 1 .

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32 Let be the binary operation on N given by n m = mn.

(a) Is commutative? Explain.(b) Is associative? Explain.

33 For fixed n, let Rn denote the set of all n-tuples (x1, x2, . . . , xn) withxi R. Componentwise addition + on Rn is given by

(x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn).

Prove that + is associative and commutative.

34 Prove that function addition on FR is associative (see Example 3.2.1).

35 Finish the proof begun in Example 3.2.1 that matrix multiplicationon Matn(R) is associative.

36 Let X be a set and assume that |X| 3. Prove that composition onSX is not commutative.

37 Matrix addition and matrix multiplication are binary operations onthe set Mat2(R) of 2 2 matrices over R. Let Y be the subset of Mat2(R)consisting of matrices of the form

a bb a

(a, b R).

Prove or disprove the following statements:

(a) Y is closed under matrix addition.

(b) Y is closed under matrix multiplication.

38 Let X be a set and let be an associative binary operation on X. LetY = {y X| y x = x y for all x X}. Then Y is a subset of X. Provethat Y is closed under .

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39 Let X = {a,b,c,d} and let be the binary operation on X given bythe following table:

a b c da d c a bb c b a cc a a d cd b d c a

(a) Find a c.(b) Is commutative? Explain.(c) Compute (a d) c and a (d c). Can you tell, based on this compu-

tation, whether is associative? Explain.(d) Compute (c a) b and c (a b). Can you tell, based on this compu-

tation, whether is associative? Explain.(e) Is the subset Y = {a,c,d} of X closed under ? Explain.

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4 Group

4.1 Definition

A group is a pair (G, ), where G is a set and is a binary operationon G satisfying the following properties:

(G1) is associative,(G2) there exists an element e ofG such that ex = x and xe = x

for all x G,

(G3) for each element x of G there exists an element x

of G suchthat x x = e and x x = e.

If (G, ) is a group, we say that G is a group under .Although the definition leaves open the possibility that there is more thanone element ofG satisfying the condition stated for e, it is a consequence ofthe assumptions that there is only one such element (see 4.3). It is calledthe identity element of the group.

Similarly, for each element x of G, it is a consequence of the assumptions

that there is only one element ofG satisfying the condition stated for x (see4.3). It is called the inverse of x.

If the binary operation is commutative, the group (G, ) is an abeliangroup (named after the Norwegian mathematician Abel, who was one ofthe first to use group theory to solve important problems of his day).

|G| is the order of G. The group G is finite if |G| is finite.

4.2 Examples, nonexamples

4.2.1 Example

(a) R is a group under addition (+).

Axiom G1 holds since + is associative. Axiom G2 holds with e = 0since 0 + x = x and x + 0 = x for all x R. Axiom G3 holds since,

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for each x R, we can let x be x noting that (x) + x = 0 andx + (x) = 0.

(b) Similarly, each set Z, Q, and C is a group under addition with 0playing the role of e and x playing the role of x.

(c) R (= the set R with 0 removed) is a group under multiplication ().Axiom G1 holds since is associative. Axiom G2 holds with e = 1since 1 x = x and x 1 = x for all x R. Axiom G3 holds since, foreach x R, we can let x be x1 (defined since x = 0) noting thatx1 x = 1 and x x1 = 1.

(d) Similarly, both sets Q and C (the symbol signifying that 0 isremoved) are groups under multiplication with 1 playing the role of eand x1 playing the role of x.

(e) For fixed n, the set Zn = {0, 1, 2, . . . , n 1} is a group under addition+ modulo n (see Section 3.2).

We are postponing the proof that + is associative. The role of e isplayed by 0. The role of 0 is played by 0 and if x Zn with x = 0,the role of x is played by n x.

(f) For fixed n, the set Rn of n-tuples (x1, x2, . . . , xn) with xi R is agroup under componentwise addition.

Componentwise addition is an associative binary operation on R

n

byExample 3.2.1. The role ofe is played by the n-tuple 0 = (0, 0, . . . , 0).Given x = (x1, x2, . . . , xn) Rn, the role of x is played by x =(x1, x2, . . . , xn).

(g) The set FR of all functions from R to R is a group under functionaddition.

Function addition is associative by Exercise 34. The role ofe is playedby the zero function 0 given by 0(x) = 0, and for f FR, the role off is played by f given by (f)(x) = f(x).

(h) For fixed m and n, Matmn(R) is a group under matrix addition.

Matrix addition is associative (proof similar to Exercise 33). Thematrix 0 having each entry 0 plays the role of e, and for A = [aij] Matmn(R), the matrix A = [aij ] plays the role of A.

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(i) For fixed n, the set GLn(R) (invertible n n matrices over R) is agroup under matrix multiplication.Matrix multiplication is a binary operation on GLn(R) by Example3.3.2. It is associative by Example 3.2.1. The role of e is played bythe identity matrix I, and for each A GLn(R), we can let A be A1(inverse matrix of A).

(j) For each set X, the set SX is a group under composition. In particular,Sn is a group under composition for all n N.Composition is an associative binary operation on SX (see Example3.2.1). The role ofe is played by the identity function 1X since 1X = and 1X = for each SX (see Section 2.3). For each SX,we can let

be the inverse function 1

(which exists because isbijective) noting that 1 = 1X and

1 = 1X (see Section 2.4).

(k) G = {e} is a group with binary operation given by ee = e. It is thetrivial group.

All of these groups are abelian except for (GLn(R), ) when n 2 (seeExercise 42) and (SX, ) when |X| 3 (see Exercise 36).

4.2.2 Example The following are not groups.

(i) Z ={

. . . ,

2,

1, 0, 1, 2, . . .}

under subtraction.

We have 0 (0 1) = 1 = 1 = (0 0) 1, so G1 fails.(ii) (0, ) under addition.

Addition is associative (G1), but G2 fails since there is no identityelement. Indeed, suppose that e (0, ) is an identity. Since 1 (0, ), we have e + 1 = 1. But this implies e = 0, contradicting thate > 0.

(iii) R under multiplication.

Multiplication is associative (G1) and 1 is an identity (G2), but 0 hasno inverse, so G3 fails.

4.3 Uniqueness of identity, inverse

Let (G, ) be a group.

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4.3.1 Theorem.

(i) G has a unique identity element.

(ii) Each element of G has a unique inverse element.

Proof. (i) Ife and e1 are identity elements of G, then

e1 = e1 e (e is identity)= e (e1 is identity).

Therefore, G has a unique identity element.

(ii) Let x G and let x and x be inverses of x. Thenx = x e

= x (x x) (x is inverse of x)= (x x) x= e x (x is inverse of x)= x.

Therefore, x has a unique inverse.

4.4 Multiplicative notation

In the definition of group the symbol is used to emphasize the fact thatthe notation for the binary operation can be anything (e.g., +, , ).However, when referring to a group in general we always use multiplicativenotation; that is, we write x y simply as xy and call this the product ofx and y, and instead of x we write x1 (though we continue to denote theidentity element by e instead of 1).

Let G be a group and let x G. We define

x0 = e

and for a positive integer n we define

xn = xx x (n factors)

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and

xn

=

x1n

= x1

x1

x1

(n factors).This defines xn for every integer n.

In applying general results about groups to a specific group it is sometimesnecessary to translate notation. For instance, suppose that in our generalgroup G we have a statement concerning an element a = x3y2 (x, y G).If the statement is applied to the specific group (Z, +), this element wouldbe

a = x3y2

= x3 y1

2

= x x x y y= x + x + x + (y) + (y)= 3x 2y,

so 3x 2y is the additive analog of the multiplicative expression x3y2.

4.5 Generalized associativity

In the definition of group, the assumption of associativity of the binaryoperation implies that when indicating the product of three group elements

it is unnecessary to use parentheses.This is true of a product of more than three group elements as well. Forinstance, xyzuv can be computed using the grouping (x((yz)u))v, or thegrouping ((xy)(zu))v, or any of the other possible groupings, and the resultwill be the same (provided the order of the factors remains the same).

This is the generalized associativity property.

Even though parentheses are unnecessary, they are often used, nonetheless,to draw the readers attention to particular groupings.

4.6 Cancelation properties

Let G be a group and let x,y ,z G.4.6.1 Theorem.

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(i) If xy = xz, then y = z.

(ii) If xz = yz, then x = y.

Proof. (i) Ifxy = xz, then y = ey = x1xy = x1xz = ez = z.

(Second proof:

xy = xz x1xy = x1xz ey = ez y = z.)

(ii) Similar.

The statements (i) and (ii) are the left and right cancelation properties,respectively.

4.7 Properties of inverse

Let G be a group and let x G. The first theorem we have about inversessays that if an element acts as a right inverse of x, then it must be theinverse of x (and similarly for a left inverse).

4.7.1 Theorem. Let x, y G.(i) If xy = e, then y = x1.

(ii) If yx = e, then y = x1.

Proof. (i) Assume xy = e. We have xy = e = xx1, so y = x1 by the leftcancelation property (see Section 4.6).

(ii) Similar.

4.7.2 Theorem.

(i)

x11

= x,

(ii) (xy)1 = y1x1.

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Proof. (i) Since x1x = e, it follows from part (i) of the preceding theorem

that x is the inverse of x1, that is, x =

x11

.

(ii) Since

(xy)

y1x1

= x(yy1)x1 = xex1 = xx1 = e,

it follows from part (i) of the preceding theorem that y1x1 is the inverseof xy, that is y1x1 = (xy)1.

4.8 Laws of exponents

Let G be a group and let x

G. Recall that we have defined xn for every

integer n (see Section 4.4). The next theorem says that two familiar laws ofexponents are valid for groups.

4.8.1 Theorem. For x G and m, n Z,

(i) xmxn = xm+n,

(ii) (xm)n = xmn.

A careful proof of this theorem would require mathematical induction andthe consideration of cases depending on whether m and n are > 0, = 0, or< 0. Instead of giving such a proof, we consider just a few special cases to

show the main ideas:

x3x2 = (xxx)(xx) = x5 = x3+2 (see (i))

x2x3 = (x1x1)(xxx) = x1

x1x

xx = x1exx

= x1xx = ex = x = x1 = x(2)+3 (see (i))

x23

= x2x2x2 = (xx)(xx)(xx) = x6 = x23 (see (ii)).

It should be pointed out that (xy)n need not equal xnyn for x, y G andn Z. If n = 2, for instance, then we have

(xy)2 = (xy)(xy) = xyxy

and if yx = xy, then we cannot rearrange the middle factors to get xxyy,which is x2y2.

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However, if yx = xy, then we can rearrange the factors so that in this case

(xy)n

= xn

yn

.

4.9 Direct product, direct sum

Let G1 and G2 be two groups. Let G1 G2 denote the set of all orderedpairs with first component coming from the group G1 and second componentcoming from the group G2:

G1 G2 = {(x1, x2) | x1 G1, x2 G2}

Define componentwise multiplication on G1 G2 by

(x1, x2)(y1, y2) = (x1y1, x2y2).

4.9.1 Theorem. G1 G2 is a group under componentwise multiplication.

Proof. We need to check the three group axioms.

(G1) The proof of associativity of componentwise multiplication is similarto Exercise 33.

(G2) We claim that (e1, e2) is an identity element, where ei is the identityelement of Gi (i = 1, 2). For (x1, x2)

G1

G2, we have

(e1, e2)(x1, x2) = (e1x1, e2x2) = (x1, x2)

and similarly (x1, x2)(e1, e2) = (x1, x2). Therefore, (e1, e2) is an iden-tity element.

(G3) Let (x1, x2) G1 G2. We claim that (x11 , x12 ) is an inverse of(x1, x2). We have

(x11 , x12 )(x1, x2) = (x

11 x1, x

12 x2) = (e1, e2)

and similarly (x1, x2)(x11 , x

12 ) = (e1, e2). Therefore, (x

11 , x

12 ) is an

inverse of (x1, x2).

This finishes the proof that G1 G2 is a group under componentwise mul-tiplication.

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G1 G2 is the direct product of the groups G1 and G2.If the groups G1 and G2 are additive groups (i.e., the binary operationsare both +), then the direct product is called the direct sum and it isdenoted G1 G2. In this case, the operation is denoted + and it is calledcomponentwise addition:

(x1, x2) + (y1, y2) = (x1 + y1, x2 + y2).

4.10 Isomorphism

Consider the group Z = {. . . , 2, 1, 0, 1, 2, . . . } under addition. If we cre-

ate a new group Z = {. . . , 2, 1, 0, 1, 2, . . . }by putting bars over each element ofZ and by using a binary operation thatacts just like + acts on the unadorned integers (so, for instance, 2 3 = 5),then the group (Z, ) is essentially the same as the original group (Z, +). Wesay that these two groups are isomorphic (the precise definition is givenbelow).

One often encounters two groups like this that are essentially the same inthe sense that the only difference between them is the symbols used for theirelements and the symbols used for their binary operations. Usually whenthis happens it is not as obvious as it was here, so we need a methodical

way to identify such a situation. That is what this section is about.

Let G and G be two groups.

An isomorphism from G to G is a bijection : G G satisfyingthe homomorphism property

(xy) = (x)(y) for all x, y G.

Caution: On the left of the equation the product xy is computed using thebinary operation ofG, while on the right the product (x)(y) is computedusing the binary operation of G.

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4.10.1 Example The set U4 = {1, i, 1, i}, where i =1, is a

group under multiplication. Show that the function : Z4 U4 given by0 11 i2 13 i

is an isomorphism.

Solution Since never sends two elements to the same place it is injective,and since every element of the codomain has an arrow coming to it, is

surjective. Therefore, is bijective.The homomorphism property is the statement that

(x + y) = (x) (y)for each x, y Z4 (see Caution above). The easiest way to check thisproperty in this case is by looking at the binary operation tables for the twogroups:

+ 0 1 2 3

0 0 1 2 3

1 1 2 3 02 2 3 0 13 3 0 1 2

1 i 1 i1 1 i 1 i

i i 1 i 11 1 i 1 ii i 1 i 1

In words, the homomorphism property is satisfied if adding two elements ofZ4 and applying to the result yields the same thing as first applying to the elements separately and then multiplying the results. This propertyis satisfied if the second table is obtained from the first by applying toall of the entries. The reader can check that this is indeed the case here.Therefore, is an isomorphism.

Because of the existence of the isomorphism from Z4 to U4, it followsthat these two groups are essentially the same. One can think of as arenaming function: it gives to 0 the new name 1, to 1 the new name i,to 2 the new name 1, and to 3 the new name i. Moreover, because ofthe homomorphism property, this renaming is compatible with the binary

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operations of the two groups. In short, the only difference between the two

groups is the names we give their elements and the symbol we use for theirbinary operations (+ for Z4 and for U4).

The groups G and G are isomorphic, written G = G, if thereexists an isomorphism from G to G.

If : G G is an isomorphism, then 1 : G G is also an isomorphism(Exercise 410). Therefore, G = G if and only ifG = G.

In the example given at the first of the section Z is isomorphic to Z. In-deed, an isomorphism : Z Z is given by (x) = x. Note that thehomomorphism property is satisfied by the way we defined :

(x + y) = x + y = x y = (x) (y).

4.10.2 Example Let R be viewed as a group under addition and letR+ (positive reals) be viewed as a group under multiplication. Prove thatR = R+.

Solution Define : R R+ by (x) = ex. Since ex is defined and positivefor every x

R, the function is well defined.

From calculus, we know that 1 : R+ R exists (it is given by 1(y) =ln y), so is bijective by Theorem 1.9.1.

For x, y R, we have

(x + y) = ex+y = exey = (x)(y),

so satisfies the homomorphism property. Therefore, is an isomorphismand we conclude that R = R+.

4.10.3 Example Is R isomorphic to Z?

Solution It was pointed out in Section 1.6 that |R| = |Z|, that is, there isno bijection from R to Z. In particular, there cannot be an isomorphismfrom R to Z. Therefore, R is not isomorphic to Z.

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IfG = G, then G and G are indistinguishable as groups. IfG has a propertythat can be described just using its elements and its binary operation, thenG must have that same property, and vice versa.

The next example illustrates this principle. (Recall that G is abelian ifxy = yx for all x, y G.)

4.10.4 Example Assume that G = G. Prove that G is abelian if andonly if G is abelian.

Solution Since G = G, there exists an isomorphism : G G.() Assume that G is abelian. Let x, y G. (Must show xy = yx.)Since is surjective, we have x

= (x) and y

= (y) for some x, y G.Thenxy = (x)(y)

= (xy) (homomorphism property)

= (yx) (G is abelian)

= (y)(x) (homomorphism property)

= yx.

Therefore, G is abelian.

(

) Assume that G is abelian. Let x, y

G. (Must show xy = yx.) We

have

(xy) = (x)(y) (homomorphism property)

= (y)(x) (G is abelian)

= (yx) (homomorphism property),

and since is injective, we conclude that xy = yx. Therefore, G is abelian.

Recall that the symmetric group S3 has order 3! = 6. Since Z6 also hasorder 6, there exists a bijection from S3 to Z6, so one might wonder whether

these two groups are actually isomorphic.

4.10.5 Example Is S3 isomorphic to Z6?

Solution According to Exercise 36 the group S3 is not abelian. Since Z6 is

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abelian, we conclude from the preceding example that S3 is not isomorphic

to Z6.

4 Exercises

41 In each case, the set with the stated binary operation is not a group.State the first of the three group axioms (G1, G2, G3) that fails to hold andjustify your claim.

(a) N = {1, 2, 3, . . . } under multiplication.(b) R (nonzero reals) under

given by x

y = x/y.

(c) R under given by x y = |xy|.

42 Prove that GLn(R) is nonabelian.

Hint: First handle the case n = 2. (Recall that a matrix is invertible if andonly if its determinant is nonzero, so in particular the matrices of Example3.2.1(f) cannot be used.) For the general case, use the fact that if A and Bare 2 2 matrices, then

A 00 I

B 00 I

=AB 0

0 I

,

where the first matrix on the left denotes the n n matrix with A in theupper left corner and the (n 2) (n 2) identity matrix I in the lowerright corner and zeros elsewhere, and similarly for the other matrices.

43 Let G = R\{1} (reals without 1). Define on G byx y = x + y + xy.

Prove that (G,

) is a group. (Be sure to check first that

is a well-defined

binary operation on G.)

44 Let G be a group and let a, b G. Prove that there exist elements xand y of G such that ax = b and ya = b.

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45 Let G be a finite group. Prove that each element of G appears pre-

cisely once in each row and each column of the binary table of G.

Hint: Use a cancelation property (4.6) to show that an element of G canappear at most once in a given row of the table. Then use the fact that eachrow has as many entries as there are elements in the group to argue thatevery element of the group appears in each row.

46 Let G be a group and assume that x2 = e for all x G. Prove thatG is abelian.

47 Let G1 and G2 be groups. Prove that G1 G2 = G2 G1.

48 Let G be a finite group and let x be an element of G. Prove thatxn = e for some positive integer n.

Hint: Use the fact that the elements x1, x2, x3, . . . cannot be distinct.

49 Let G be a group of even order. Prove that there exists a nonidentityelement x of G such that x2 = e.

410 Let : G G be an isomorphism from the group G to the groupG. In particular, is bijective so that its inverse 1 : G G exists (see1.9.1). Prove that 1 is an isomorphism.

411 Prove that Q is not isomorphic to Z (both groups under addition).

Hint: Suppose that : Q Z is an isomorphism. Then 1 = (x) for somex Q. Write x = x2 + x2 and derive a contradiction.

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5 Subgroup

Sometimes it happens that a subset of a group is a group in its own right (us-ing the same binary operation as in the larger group). When this happens,the smaller group is called a subgroup of the larger group.

For instance, Z is a subset of the additive group R and it is also a groupunder addition, so Z is a subgroup of R.

5.1 Definition

Let G be a group.

A subset H of G is a subgroup (written H G) if it is a groupunder the binary operation of G.

5.1.1 Example We have Z Q R C, meaning each is a subgroupof the next (under addition).

It is possible for one group to be a subset of another group without theformer being a subgroup of the latter. For instance, R+ (positive reals) is a

subset of R, and we have that R+ is a group under multiplication and R isa group under addition. But R+ is not a subgroup of R since R+ does notuse the same binary operation as R.

The following theorem provides the most common way to check whether asubset of a group is a subgroup.

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5.1.2 Subgroup Theorem. A subset H of the group G is a subgroup if

and only if the following hold:

(i) e H,(ii) x, y H xy H,

(iii) x H x1 H.x

G

y

xy

H

e

x1

Note: (i) says that the identity element e of G is in H, (ii) says that H isclosed under the binary operation of G, and (iii) says that H is closed underinversion.

Proof. Let H be a subset of G.

() Assume that H is a subgroup of G. Then H has an identity elemente1. Since e1e = e1 = e1e1 it follows from the left cancelation property that

e = e1 H, so (i) holds. By the definition of subgroup, H is closed underthe binary operation of G, so (ii) holds. Let x H. Then x has an inversex in H. Since xx1 = e = e1 = xx

it follows from the left cancelationproperty that x1 = x H, so (iii) holds.() By property (ii), H is closed under the binary operation on G, so we getan induced binary operation on H. Since (xy)z = x(yz) for every x,y ,z G,this equation holds for every x,y ,z H, so (G1) is satisfied. The identitye of G is in H by property (i) and this same element acts as an identityelement of H, so (G2) is satisfied. Finally, if x H, then its inverse x1(which exists in G) is actually in H by property (iii) and since x1x = eand xx1 = e, (G3) is satisfied. Therefore, H is a group.

The group G is a subgroup of itself. If H is a subgroup of G and H = G,then H is a proper subgroup of G.

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The subset {e} of G consisting of the identity alone is a subgroup of G. Itis the trivial subgroup.

5.2 Examples

5.2.1 Example Put 3Z = {3m | m Z} (multiples of 3). Prove that3Z is a subgroup of the additive group Z.

Solution We use the Subgroup Theorem (5.1.2) with G = Z and H = 3Z.The role of e in the group Z is played by the integer 0. Since the binaryoperation of Z is +, we need to translate the multiplicative notation of thetheorem into additive notation (see Section 4.4).

(i) (0 3Z?) We have 0 = 3 0 3Z as desired.(ii) (x, y 3Z x + y 3Z?) Let x, y 3Z. We have x = 3m and

y = 3m for some m, m Z. Sox + y = 3m + 3m = 3(m + m) 3Z.

(iii) (x 3Z x 3Z?) Let x 3Z. We have x = 3m for some m Z.So

x = (3m) = 3(m) 3Z.

By the Subgroup Theorem, 3Z is a subgroup of Z.

More generally, nZ = {nm | m Z} is a subgroup of Z for every integer n(same proof with n replacing 3).

5.2.2 Example Let X be a set and let x0 be a fixed element of X.Define

H = { SX | (x0) = x0}.Prove that H is a subgroup of SX.

Solution We use the Subgroup Theorem (5.1.2). The binary operation ofSX is composition (but we have decided to write simply as ) andthe role of the identity e is played by the identity function 1X.

(i) (1X H?) We have 1X(x0) = x0, so 1X H.

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(ii) (, H H?) Let , H. We have()(x0) = ((x0))

= (x0) ( H)= x0 ( H).

Therefore, H.(iii) ( H 1 H?) Let H. We have

1(x0) = 1

(x0)

( H)= x0 (definition of inverse function).

Therefore, 1 H.

By the Subgroup Theorem, H is a subgroup of SX.

5.2.3 Example Let G be an abelian group and let H = {x G | x2 =e}. Prove that H is a subgroup of G.

Solution We use the Subgroup Theorem (5.1.2).

(i) (e H?) We have e2 = ee = e, so e H.(ii) (x, y H xy H?) Let x, y H. We have

(xy)2 = xyxy

= xxyy (G is abelian)= x2y2

= ee (x, y H)= e.

Therefore, xy H.(iii) (x H x1 H?) Let x H. We have

x12

= x2 (law of exponents)

=

x2

1

(law of exponents)

= e1 (x

H)

= e.

Therefore, x1 H.

By the Subgroup Theorem, H is a subgroup of G.

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5.3 Cyclic subgroup

Let G be a group and let a be a fixed element of G. Put

a = {am | m Z}.5.3.1 Theorem. a is a subgroup of G.

Proof. We use the Subgroup Theorem (5.1.2).

(i) (e a?) We have e = a0 a.(ii) (x, y

a

xy

a

?) Let x, y

a

. Then x = am and y = an for

some m, n Z. So xy = aman = am+n a.(iii) (x a x1 a?) Let x a. Then x = am for some m Z.

So x1 = (am)1 = am a.

By the Subgroup Theorem, a is a subgroup of G.

We call a the subgroup of G generated by a.

5.3.2 Example

(a) Find 3 in Q

(under multiplication).

(b) Find 3 in Z (under addition).

Solution

(a) In Q we have

3 = {3m | m Z} = {. . . , 32, 31, 30, 31, 32, . . . }= {. . . , 19 , 13 , 1, 3, 9, . . . }.

(b) In Z we have

3 = {m3 | m Z} (m3 is the additive notation of 3m)= {. . . , (2)3, (1)3, (0)3, (1)3, (2)3, . . . }= {. . . , 6, 3, 0, 3, 6, . . . } = 3Z.

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The cyclic subgroup a is the smallest subgroup of G containing a in thefollowing sense:

5.3.3 Theorem. If H is any subgroup of G with a H, then a H.

Proof. Let H G with a H. We have, using closure, a1 = a H,a2 = aa H, a3 = a2a H and in general am H for every positiveinteger m. Also, for each positive integer m, we have, using closure underinversion, am = (am)1 H (since am H as was just shown). Finally,a0 = e H. Therefore, am H for all m Z, whence a H.

5.4 Order of element

Let G be a group and let x G.Ifxn = e for some nonzero integer n, then the least such integer is the orderof x, written ord(x). If no such positive integer exists, then x has infiniteorder, written ord(x) = .

5.4.1 Example In the multiplicative group C, find

(a) ord(i) (i = 1),(b) ord(3).

Solution In this group, 1 plays the role of e.

(a) Since

i1 = i

i2 = 1i3 = i2i = i

i4

= i3

i = i2

= 1 = e,

it follows that ord(i) = 4.

(b) We have 31 = 3, 32 = 9, 33 = 27, and in general, 3n = 1 for everypositive integer n. Therefore, ord(3) = .

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5.4.2 Example Find the order of the element 3 in Z12.

Solution Put x = 3. Since Z12 uses additive notation, the expression xn in

the definition really means nx. We have

1x = 3

2x = 3 + 3 = 6

3x = 3 + 3 + 3 = 9

4x = 3 + 3 + 3 + 3 = 0 = e (addition modulo 12)

Therefore, 3 has order 4.

If we continue computing multiples of x = 3 in the preceding example, weget

1x = 3 5x = 3 9x = 3 . . .

2x = 6 6x = 6 10x = 6 . . .

3x = 9 7x = 9 11x = 9 . . .

4x = 0 8x = 0 12x = 0 . . . .

Evidently mx = 0 if and only if 4 divides m. Noting that 4 is the order ofxin the example, we see that this illustrates the general property of the orderof an element stated in the following theorem (expressed in multiplicativenotation as usual).

5.4.3 Theorem. Let G be a group and let x be an element of G of finiteorder n. If m is an integer, then xm = e if and only if n divides m.

The proof of this theorem requires the following standard fact about integers.

5.4.4 Theorem (Division algorithm). Let m and n be integers withn > 0. There exist unique integers q and r with 0 r < n such that

m = qn + r.

We omit the proof and instead appeal to the readers knowledge of longdivision, which is used to divide m by n as follows:

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qn ) m

. . .

r

mn = q+ rn

m = qn + r.

So the q in the theorem corresponds to the whole part of the quotient andthe r corresponds to the remainder. In long division, one continues thealgorithm until the remainder is less than the divisor, that is, 0 r < n (cf.theorem).

Proof of Theorem 5.4.3. Let m be an integer.

() Assume that xm = e. By the Division Algorithm (5.4.4), there exist(unique) integers q and r with 0 r < n such that m = qn + r. So

xr = xmqn = xm(xn)q = eeq = e,

where we have used laws of exponents (see 4.8). Since n is the least positiveinteger for which xn = e and 0 r < n, we conclude from this equationthat r = 0. Therefore, m = qn, implying that n divides m.

() Assume that n divides m. Then m = kn for some integer k. Hence,

xm = xkn = (xn)k = ek = e,

as desired.

The following theorem relates the order of an element to the cyclic subgroupthe element generates.

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5.4.5 Theorem. Let G be a group and let x

G.

(i) If ord(x) = , then

x = {. . . , x2, x1,e ,x,x2, x3, . . . }

and the elements xi, i Z, are all distinct.(ii) If ord(x) = n, then

x = {e,x,x2, x3, . . . , xn1}

and the elements xi, 0

i < n, are all distinct.

(iii) |x| = ord(x).

Proof.

(i) Assume that ord(x) = . By definition, x consists of all powers xiwith i Z, so the equation follows once we observe that x0 = e andx1 = x. (Distinct?) Suppose that xi = xj with i j. Then

xji = xjxi = xj(xi)1 = xj(xj)1 = e.

Now j i is a nonnegative integer and, since ord(x) = , this numbercannot be positive. Therefore, j i = 0, that is, j = i. This provesthat the elements xi, i Z, are all distinct.

(ii) Assume that ord(x) = n. Put S = {e,x,x2, x3, . . . , xn1}. We provethat x = S by showing that each set is contained in the other.() Let y x. Then y = xm for some m Z. By the DivisionAlgorithm (5.4.4), there exist (unique) integers qand r with 0 r < nsuch that m = qn + r. Therefore,

y = xm = xqn+r = (xn)qxr = eqxr = xr S.

() This follows immediately from the definition of x.(Distinct?) Assume that xi = xj with 0 i j < n. Then 0 j i < n and, arguing just as in the proof of (i), we get xji = e.Since ord(x) = n, it follows that j i = 0, that is, j = i. This provesthat the elements xi, 0 i < n, are distinct.

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(iii) This follows from (i) and (ii). (In the case ord(x) = , we interpretx = as simply saying that the set x is not finite.)

5.4.6 Example Find the cyclic subgroup of Z12 generated by 3.

Solution In Example 5.4.2 it was shown that ord(3) = 4. Put x = 3.According to part (ii) of the preceding theorem, we have

3 = x = {e,x, 2x, 3x} (additive notation)= {0, 3, 6, 9}.

5.4.7 Example Find , where is the element of S3 given by

=

1 2 32 3 1

.

Solution We have

=

1 2 32 3 1

,

2

= =1 2 3

2 3 11 2 3

2 3 1

=1 2 3

3 1 2

,

3 = 2 =

1 2 33 1 2

1 2 32 3 1

=

1 2 31 2 3

= .

Therefore, has order 3 and, according to Theorem 5.4.5,

= {,,2}

=

1 2 31 2 3

,

1 2 32 3 1

,

1 2 33 1 2

.

5.5 Subgroup diagram

We sometimes draw a subgroup diagram to show the inclusion relation-ships between subgroups of a given group. For instance, if a group G has

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subgroups H and K, then H K is also a subgroup ofG (Exercise 56) andwe draw

G

H K

H KThe line segment joining H to G indicates that H is contained in G (thesmaller set is lower than the larger one) and so forth.

Here is the subgroup diagram for the group Z12 = {0, 1, 2, . . . , 11}:Z12

{0, 2, 4, 6, 8, 10} {0, 3, 6, 9}

{0, 4, 8} {0, 6}

{0}

An easy way to tell that the indicated sets are actually subgroups is to notethat each is a cyclic subgroup. For instance, {0, 2, 4, 6, 8, 10} = 2. Here isthe diagram again using just the cyclic subgroup notation:

Z12

2 3

4 6

{0}It turns out that this is the full subgroup diagram of Z12, meaning that allof the subgroups of Z12 are represented (see 6.3.1).

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5 Exercises

51 Let G be a group. The center ofG, denoted Z(G), is the set of thoseelements of G that commute with every element of G:

Z(G) = {z G | zx = xz for all x G}.Prove that Z(G) is a subgroup of G.

52 Fix n N. Let SLn(R) be the set of all n n matrices over R havingdeterminant 1:

SLn(R) =

{A

Matn(R)

|det(A) = 1

}.

Prove that SLn(R) is a subgroup of GLn(R) (= invertible n n matricesover R). (SL stands for Special Linear.)

Hint: From linear algebra, we know that a square matrix is invertibleif and only if its determinant is nonzero. Use the fact that det(AB) =det(A)det(B) for A, B Matn(R).

53 Compute 6 in the group Z15.

54 Compute , where is the element of S4 given by =

1 2 3 43 1 4 2

.

55 Draw a subgroup diagram for the group Z18 including the subgroups{0}, 2, 3, 6, 9, and Z18.

56 Let G be a group and let H and K be subgroups of G.

(a) Prove that the intersection H K is a subgroup of G.(b) Give an example to show that the union HK need not be a subgroup

of G.

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6 Cyclic group

6.1 Definition and examples

Let G be a group. Recall that if g is an element of G, then g denotes theset of all powers of g:

g = {gm | m Z}.

G is cyclic if G = g for some g G.

In other words, G is cyclic if there exists some element g in G having theproperty that every element of G equals gm for some integer m.

IfG is cyclic, then any g G for which G = g is called a generator ofG.

6.1.1 Example Prove that the group Z (under addition) is cyclic andfind all generators of Z.

Solution The set 1 consists of all multiples of 1, so Z = 1. Therefore,Z is cyclic and 1 is a generator.

For any integer n, the set

n

consists of all multiples of n, so

n

= Z if and

only if n = 1. Therefore, 1 are the only generators of Z.

6.1.2 Example Let n N. Prove that the group Zn (under additionmodulo n) is cyclic.

Solution We claim that Zn = 1. The element 1 of Zn has order n, so byTheorem 5.4.5, we have

1 = {0, 1, (2)1, (3)1, . . . , (n 1)1}= {0, 1, 2, 3, . . . , n 1}= Zn

as claimed. Therefore, Zn is cyclic.

In Section 6.4 we will see that Z and Zn (n N) are essentially the onlycyclic groups in the sense that any cyclic group is isomorphic to one of these.

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6.1.3 Example Prove that the group Q (under addition) is not cyclic.

Solution We give a proof by contradiction. Suppose that Q is cyclic. ThenQ = g for some g Q. We can write g = m/n for some integers m and n.Let x = 1/(2n). Then x Q = g which says that x = kg for some k Z.This gives

1

2n= k

mn

,

implying km = 12 . But this last equation contradicts that km is an integer(since both k and m are integers).

We conclude that Q is not cyclic.

6.2 Cyclic group is abelian

6.2.1 Theorem. Every cyclic group is abelian.

Proof. Exercise 62.

6.3 Subgroup of cyclic group is cyclic

The group Z is cyclic since Z = 1. The set 3Z = {3m | m Z} (multiplesof 3) is a subgroup of Z by Example 5.2.1. Note that

3Z = {. . . , 6, 3, 0, 3, 6, . . . }= {. . . , (2)3, (1)3, (0)3, (1)3, (2)3, . . . }= 3,

so 3Z itself is cyclic.

The following theorem shows that this is no coincidence.

6.3.1 Theorem. Every subgroup of a cyclic group is cyclic.

Proof. Let G be a cyclic group and let H be a subgroup of G. Since G iscyclic, there exists g G such that G = g.If H = {e}, then H = e and H is cyclic. Now suppose H = {e}. Sinceevery element of G, and hence H, is a power of g, it follows that gn Hfor some nonzero integer n. If n < 0, then gn = (gn)1 H, so we may

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(and do) assume that n is positive and that it is the least positive integer

for which gn

H.We claim that H = gn. Let h H. We have h = gm for some integer m.By the division algorithm, there exist integers q and r with 0 r < n suchthat m = nq+ r. Therefore,

gr = gmnq = gm(gn)q H,

which implies, due to the minimality ofn, that r = 0. Thus, h = gm = gnq =(gn)q gn. This proves that H gn and, since the other inclusion isimmediate, the theorem follows.

Since Z is cyclic, every subgroup of Z is cyclic and hence of the form n =nZ for some n Z.

6.4 Classification of cyclic groups

The following theorem says that every cyclic group is isomorphic to eitherZ or Zn, some n.

6.4.1 Theorem. Let G be a cyclic group.

(i) If |G| is infinite, then G = Z.(ii) If |G| = n, then G = Zn.

Proof. Since G is cyclic, we have G = g for some g G.

(i) Assume that |G| is infinite. By Theorem 5.4.5, g has infinite order,

G = g = {. . . , g2, g1, g0, g1, g2, . . . },

and the powers gk with k Z are distinct.Define : Z G by (m) = gm. We claim that is an isomorphism.

( injective?) Let m, n Z and assume that (m) = (n). Thengm = gn and, since the powers gk with k Z are distinct, it followsthat m = n. Hence, is injective.

( surjective?) Let x G. Since G = g, we have x = gm for somem Z. Then (m) = gm = x. Hence, is surjective.

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(Homomorphism property?) For m, m Z, we have

(m + m) = gm+m

= gmgm

= (m)(m),

so satisfies the homomorphism property.

Therefore, : Z G is an isomorphism. We conclude that Z = Gand hence G = Z (see Section 4.10).

(ii) Assume that |G| = n. By Theorem 5.4.5, g has order n,

G = g = {g0, g1, g2, . . . , gn1},

and the powers gk with 0 k < n are distinct.

Define : Zn G by (m) = gm. An argument very similar to thatin part (i) shows that is bijective.

(Homomorphism property?) Let m, m Zn. By the division algo-rithm (Theorem 5.4.4), there exist integers q and r with 0 r < nsuch that m + m = qn + r. By the definition of addition modulo n,the sum m + m computed in Zn is r. Therefore,

(m + m) = (r) = gr = gm+mqn

= gmgm

(gn)q = gmgm

= (m)(m),

using the fact that g has order n so that gn = e. Hence, satisfies the

homomorphism property.We have shown that is an isomorphism, so we conclude that G = Zn.

6.5 Direct sum of cyclic groups

Let m and n be positive integers. We denote by gcd(m, n) the greatestcommon divisor of m and n. For instance gcd(12, 18) = 6.

6.5.1 Theorem. Zm

Zn= Zmn if and only if gcd(m, n) = 1.

Proof.

() Assume that Zm Zn = Zmn. Then there exists an isomorphism : Zmn Zm Zn. We have (1) = (a, b) for some a Zm and b Zn.

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Since the element 1 of Zmn has order mn, the element (a, b) of Zm Znmust also have order mn.Let k = gcd(m, n). Then m = km and n = kn for some integers m andn. We have

kmn(a, b) = (kmna,kmnb) = (nma,mnb) = (0, 0)

so by Theorem 5.4.3, the order of (a, b), which is mn, divides kmn. Inparticular, mn kmn. So

mn kmn kmkn = mn.

Since the ends of this string are the same, the intermediate

signs must

actually be equalities. Thus kmn = kmkn which implies k = 1 as desired.

() Assume that gcd(m, n) = 1. Let k be the order of the element (1, 1) ofZm Zn. Then

(k1, k1) = k(1, 1) = (0, 0).

so by Theorem 5.4.3, m divides k and n divides k. Since gcd(m, n) = 1,it follows that mn divides k. In particular, mn k. On the other hand kis the order of the cyclic subgroup of Zm Zn generated by (1, 1), so k isless than or equal to the order of Zm Zn, which is mn. We conclude thatk = mn, so in fact (1, 1) = Zm Zn. Therefore Zm Zn is cyclic. ByTheorem 6.4.1, Zm

Zn = Zmn.

6 Exercises

61 Let G be a cyclic group of order n. Prove that ifd is a positive integerdividing n, then G has a subgroup of order d.

Hint: The desired subgroup is necessarily cyclic by Theorem 6.3.1. Firstwork with the special case G = Z12 to get an idea for the proof.

62 Prove Theorem 6.2.1, which says the following: Every cyclic group isabelian.

63 Prove or disprove: The group S12 is cyclic.

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64 Draw the full subgroup diagram for Z10.

Hint: Theorem 6.3.1.

65 Let G be a cyclic group and let g be a generator ofG. Prove that g1

is also a generator of G.

66 Let m and n be positive integers.

(a) Prove that H := {am + bn | a, b Z} is a subgroup of Z.(b) By part (a) and Theorem 6.3.1, H = d for some integer d. By

Exercise 65, we may assume that d is positive. Prove that d is thegreatest common divisor of m and n in the following sense:

(i) d|m and d|n,(ii) ifk is a positive integer such that k|m and k|n, then k|d.

(The notation d|m stands for d divides m, which means that m = difor some integer i.)

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7 Symmetric group

Recall that if X is a set, then SX denotes the group of all permutations ofX with binary operation being function composition. It is the symmetricgroup on X.

In the special case X = {1, 2, . . . , n} (n N) this group is denoted Sn andis called the symmetric group of degree n.

We study the symmetric group in this section, focusing primarily on Sn.

7.1 Cycle

A cycle is an element ofSn represented by an ordered list of numbers fromthe set {1, 2, . . . , n}. For instance, in S5 the cycle (1, 5, 2, 4) is the functionthat does this:

1 5 2 4 1 3 3.So the cycle (1, 5, 2, 4) sends each number in the list to the next numberto the right, except for the last number, which gets sent to the first. Anynumber not appearing in the list (3 in this case) is fixed, that is, sent toitself.

The matrix representation of this cycle is

=

1 2 3 4 55 4 3 1 2

.

More generally, let n be a positive integer and let i1, i2, . . . , ir be distinctintegers with 1 ij n. Let be the element of Sn that satisfies

(ij) =

ij+1 j < r,

i1 j = r,

and (k) = k for all k / {i1, i2, . . . , ir}. This permutation is written =(i1, i2, . . . , ir). It is called an r-cycle (or a cycle of length r) and we write

length() = r.

A cycle is unchanged if the last number is moved to the first. For instance:

(1, 5, 2, 4) = (4, 1, 5, 2) = (2, 4, 1, 5) = (5, 2, 4, 1).

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These arrangements of the numbers 1, 5, 2, 4 are called cyclic permuta-

tions. If the numbers are arranged in order around a circle, then a cyclicpermutation corresponds to a rotation of the circle.

The inverse of a cycle is obtained by writing the entries in reverse order.For example,

(1, 5, 2, 4)1 = (4, 2, 5, 1).

A transposition is a 2-cycle. The transposition (i1, i2) transposes (inter-changes) the two numbers i1 and i2 and fixes every other number.

A 1-cycle (i1) is the identity since it fixes i1 as well as every other number.

7.1.1 Example Let and be the cycles in S7 given by

= (1, 3, 7, 2, 5) and = (1, 7, 4, 5).

(a) (3) = 7.

Reason: sends 3 to the number in the list just to its right, which is7.

(b) ()(1) = 2.

Reason: The notation really means , so recalling how to com-pose functions, we get ()(1) = ((1)) = (7) = 2. One can avoid

writing so much by just thinking moves 1 to 7 and then moves 7to 2, so moves 1 to 2. Using symbols,

1 7 2.

In computing the product it is important to remember to workfrom right to left.

(c) ( )(4) = 7.

Reason:4

5 1 7.

(d) ()(3) = 7.Reason:

3 3 7.

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(e) 1(2) = 7.

Reason: Since moves numbers to the right, 1 moves numbers tothe left, so 1 moves 2 to 7.

7.2 Permutation is product of disjoint cycles

The cycles (1, 4, 2, 6) and (3, 8, 7) in S8 have no numbers in common; theyare said to be disjoint.

Generally, two cycles (i1, i2, . . . , ir) and (k1, k2, . . . , ks) are disjoint if

{i1, i2, . . . , ir} {k1, k2, . . . , ks} = .

The following theorem says that disjoint cycles commute.

7.2.1 Theorem. If and are disjoint cycles in Sn, then = .

Proof. Let and be disjoint cycles in Sn. We can write = (i1, i2, . . . , ir)and = (k1, k2, . . . , ks). Put I = {i1, i2, . . . , ir} and K = {k1, k2, . . . , ks}.To show that the two functions and are equal, we need to show that()(m) = ( )(m) for all m in their common domain, which is {1, 2, . . . , n}.Let m {1, 2, . . . , n}. First assume that m / I K (in other words, assumethat m does not appear in either cycle). Then and both fix m giving

()(m) = ((m)) = (m) = m = (m) = ((m)) = ( )(m).

Now assume that m I. Then (m) I, as well. Since I and K are disjointby assumption, m and (m) are not in K, so they are fixed by . Therefore,

()(m) = ((m)) = (m) = ((m)) = ( )(m).

Similarly, if m K, then ()(m) = ( )(m).We have shown that in all cases ()(m) = ( )(m). Since m was an

arbitrary element of{1, 2, . . . , n}, we conclude that = as desired.

Let

=

1 2 3 4 5 6 7 8 93 7 8 9 5 4 2 1 6

S9

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We can write as a product (meaning composition) of pairwise disjoint

cycles by following an algorithm as illustrated here:

Start with 1. We have : 1 3 8 1 (back to where we started).This completes the cycle

(1, 3, 8).

Pick the smallest number not yet appearing, namely, 2. We have :2 7 2. This completes the cycle (2, 7), which we compose withthe cycle above to get

(1, 3, 8)(2, 7).

Again, pick the smallest number not yet appearing, namely, 4. Wehave : 4 9 6 4 giving the cycle (4, 9, 6), so we now have

(1, 3, 8)(2, 7)(4, 9, 6).

Only 5 remains, and since : 5 5 we get the cycle (5). Since thiscycle is the identity permutation (as is every cycle of length one), ithas no effect on the composition and we omit it.

We have accounted for all of the numbers 1, 2, . . . , 9, and we are leftwith

(1, 3, 8)(2, 7)(4, 9, 6).

It is a fact that actually equals this product:

= (1, 3, 8)(2, 7)(4, 9, 6) (*)

The product on the right of this equation is the composition of the threeindicated cycles.

Each side of (*) represents a function from the set {1, 2, . . . , 9} to itself, sothe equation asserts that both functions send each of these numbers to thesame output.

Lets use 7 as a test. The effect of the right-hand side of (*) on 7 is

7(4,9,6)

7

(2,7)

2

(1,3,8)

2,

so 7 is sent to 2. Since also sends 7 to 2, both sides of (*) send 7 to thesame output as expected.

The process we used to write as a product of disjoint cycles generalizes toan arbitrary element of Sn:

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7.2.2 Theorem. Any element of Sn can be written as a product of disjoint

cycles. Moreover, any two such factorizations are the same except possiblyfor the order of the factors (provided all cycles of length one are included).

Proof. Let Sn. The algorithm illustrated above produces a product12 m of disjoint cycles. The proof that actually equals this productis Exercise 76. The proof of the uniqueness statement in the theorem isomitted.

In the statement of the theorem the word product has a broad meaningincluding any number of factors, with the case of a product of one factormeaning that element itself.

7.3 Permutation is product of transpositions

Recall that a cycle of length two, like (3, 5), is called a transposition. Thetheorem of this section states that any element of Sn can be written as aproduct (meaning composition) of transpositions. For instance,

1 2 3 4 5 64 1 3 2 6 5

= (1, 4)(4, 2)(5, 6)

as the reader can check.

7.3.1 Theorem. Any element of Sn can be written as a product of trans-positions.

Note: The case n = 1 is included by allowing the possibility of a productwith no factors, which is interpreted to be the identity. (This is necessarysince S1 = {} has no transpositions at all.)

Proof. By Section 7.2 it suffices to show that every cycle can be written asa product of transpositions. Let = (i1, i2, . . . , ir) be a cycle. We proceedby induction on r. If r = 1, then is the identity, which, according to our

convention, is a product of transpositions with no factors.Assume that r > 1. We claim that (i1, i2) =

, where = (i2, i3, . . . , ir).

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We have,

(i1, i2)(i1) = i1 = (i1),

(i1, i2)(ir) = i2 = (ir),

(i1, i2)(ij) = ij+1 = (ij), 1 < j < r,

(i1, i2)(k) = k = (k), k = i1, i2, . . . , ir,so the claim is established. Now the cycle has length r 1, so the in-duction hypothesis says that it is a product of transpositions. Therefore, = (i1, i2)

is a product of transpositions as well.

The proof of the theorem provides an algorithm for writing a given element of

Sn as a product of transpositions: write the element as a product of disjointcycles and then write each cycle as a product of transpositions using theformula

(i1, i2, . . . , ir) = (i1, i2)(i2, i3)(i3, i4) (ir1, ir).

7.3.2 Example Write the permutation

=

1 2 3 4 5 6 7 8 94 1 3 6 9 2 5 8 7

as a product of transpositions.

Solution We first use the method of Section 7.2 to write the permutationas a product of disjoint cycles, and then we use the formula above to writeeach of the resulting cycles as a product of transpositions:

= (1, 4, 6, 2)(5, 9, 7)

= (1, 4)(4, 6)(6, 2)(5, 9)(9, 7).

It is important to note that an element of Sn can be written as a productof transpositions in more than one way. For instance, if = (1, 2, 3), then

= (1, 2)(2, 3), = (2, 3)(1, 3),

= (1, 3)(2, 3)(1, 2)(1, 3),

= (1, 3)(2, 3)(1, 2)(1, 3)(1, 2)(1, 2)

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as the reader can check.

However, we will see in the next section that if one such factorization hasan even number of factors, then every other such factorization will alsohave an even number of factors (as the example suggests). Likewise, if onefactorization has an odd number of factors, then every other factorizationwill have an odd number of factors as well.

7.4 Even permutation, odd permutation

Let n be a positive integer. An element of Sn is even if it can be writtenas a product of an even number of transpositions. An element of Sn is odd

if it can be written as a product of an odd number of transpositions.By Theorem 7.3.1 an element of Sn is either even or odd, but possibly bothfor all we know at this point. Choosing to apply these terms to permutationswould be a bad idea if a permutation could be both even and odd. However,this is not the case:

7.4.1 Theorem. An element of Sn is not both even and odd.

Proof. Let Sn. Let = 12 t be a factorization of as a productof disjoint cycles. Define N() =

i(length(i) 1) (well defined by the

uniqueness statement of 7.2.2). Let 1 a, b n with a = b. Any cyclein which a and b appear can be written (a, c1, . . . , ch, b , d1, . . . , dk) (afterapplying a cyclic permutation to the elements, if necessary). For such acycle, a routine check verifies the equation

(a, b)(a, c1, . . . , ch, b , d1, . . . , dk) = (b, d1, . . . , dk)(a, c1, . . . , ch).

Multiplying both sides of this equation by (a, b)1 = (a, b) gives

(a, b)(b, d1, . . . , dk)(a, c1, . . . , ch) = (a, c1, . . . , ch, b , d1, . . . , dk).

It follows from these equations that

N((a, b)) =

N() 1, if a and b appear in the same i,N() + 1, otherwise.

Indeed, assuming that a and b both appear in i for some i, we may(and do) assume that i = 1 (since disjoint cycles commute) and 1 =

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(a, c1, . . . , ch, b , d1, . . . , dk), so that, writing = 2 t,

N((a, b)) = N((a, b)1) + N()

= k + h + N()

= N(1) 1 + N()= N() 1,

and similarly for the other case. In particular, N((a, b)) and N() alwayshave opposite parities (i.e., if one is even, then the other is odd).

Let = 12 s be a factorization of with each i a transposition.Then ss1 1 = , so N(ss1 1) = N() = 0, which is an evennumber. By repeated application of the observation above, we find thatN() is even or odd according as s is even or odd. This is to say that theparity of the number s of factors in our factorization = 12 s is thesame as the parity of the number N(). Since this latter depends only on the proof is complete.

7.4.2 Example Determine whether the permutation

=

1 2 3 4 5 6 7 8 94 1 3 6 9 2 5 8 7

is even or odd.

Solution In Example 7.3.2 we found that

= (1, 4)(4, 6)(6, 2)(5, 9)(9, 7).

Therefore, is odd (since it can be expressed as a product of an odd numberof transpositions, namely, five).

7.5 Alternating group

Let n N. Let An be the set of all even permutations in Sn:

An = { Sn | is even}.

7.5.1 Theorem. An is a subgroup of Sn.

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Proof. We use the Subgroup Theorem (5.1.2).

( An?) By convention, is a product of zero transpositions, and sincezero is even, we have An.(, An An?) Let , An. We can write

= 12 s and = 12 tfor some transpositions i and i with s and t both even. Then

= 12 s12 t.

This expresses as a product of transpositions. The number of factors is

s + t, which is even since both s and t are even. We conclude that An.( An 1 An?) Let An. We can write

= 12 sfor some transpositions i with s even. Using Theorem 4.7.2 and then thefact that each transposition is its own inverse, we get

1 = (12 s)1= 1s

1s1 11

= ss1

1.

This expresses 1 as a product of transpositions. The number of factors iss, which is even. Therefore, 1 An.By the Subgroup Theorem, An is a subgroup of Sn.

The group An is the alternating group of degree n.

7.5.2 Theorem. If n 2, then |An| = n!/2.

Proof. See Exercise 78.

We note that A1 = {}, so that |A1| = 1.

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7.6 Dihedral group

Let and be the elements of the symmetric group S4 given by

=

1 2 3 42 3 4 1

= (1, 2, 3, 4)

and

=

1 2 3 41 4 3 2

= (24).

The subset D4 of S4 given by

D4 =

{,,2, 3, , , 2, 3

}is a subgroup of S4 called the dihedral group (of degree 4).

To verify that D4 is indeed a subgroup, one can use the Subgroup Theoremand the following multiplication (i.e., composition) table:

2 3 2 3 2 3 2 3

2 3 2 3

2 2 3 2 3

3

3

2

3

2

3 2 3 2

3 2 3 2

2 2 3 2 3

3 3 2 3 2

The following formulas aid in the verification of the table:

4 = , 2 = , i = i

(see Exercise 710).

There is a geometrical interpretation of the group D4. Consider a squareoriented in the plane like this:

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1

2

3

4 x

y

With the corners numbered as indicated, the permutation = (1, 2, 3, 4)corresponds to 90 clockwise rotation of the square. The corner labeled 1 issent to the 2-position (1 2), the corner labeled 2 is sent to the 3-position(2 3), the corner labeled 3 is sent to the 4-position (3 4), and thecorner labeled 4 is sent to the 1-position (4 1).Similarly, the permutation = (2, 4) corresponds to a flip of the squareabout the y-axis. The corner labeled 2 is sent to the 4-position (2 4), thecorner labeled 4 is sent to the 2-position (4 2) and the other two cornersare fixed (1

1, 3

3).

The reader can check that the group elements have these geometrical inter-pretations:

= (1) no movement

= (1, 2, 3, 4) 90 clockwise rotation

2 = (1, 3)(2, 4) 180 clockwise rotation

3 = (1, 4, 3, 2) 270 clockwise rotation

= (2, 4) flip about y-axis

= (1, 2)(3, 4) flip about line y = x

2

= (1, 3) flip about x-axis3 = (1, 4)(2, 3) flip about line y = x.

These movements are called symmetries of the square. They represent allpossible ways of reorienting the square.

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The definition ofD4 generalizes. Let n be any integer greater than two. The

set Dn of all symmetries of a regular n-gon is a subgroup of the symmetricgroup Sn called the dihedral group of degree n.

So D3 is the group of symmetries of an equilateral triangle; D4, a square;D5, a regular pentagon; etc.

We have Dn = {ij | 0 i < n, 0 j < 2}, where is a rotation and isa flip. In particular |Dn| = 2n.

7.7 Equivalence relation

Let S be a set. A relation on S is a subset R of S

S =

{(x, y)

|x, y

S

}.

If (x, y) R, we write xRy and if (x, y) / R, we write xRy. Informally, Ris a rule that some pairs (x, y) of elements satisfy and some dont.

For example, < is a relation on Z. We have 3 < 5, but 4 1.

A relation on S is an equivalence relation if it satisfies these threeproperties:

(Reflexive) x x for all x S; (Symmetric) ifx y, then y x (x, y S);

(Transitive) ifx

y and y

z, then x

z (x,y ,z

S).

If is an equivalence relation on S and x y, we say that x is equivalentto y.

Equality (=) is an equivalence relation on the set Z.

7.7.1 Example Let be the relation on Z given by x y ifxy 3Z,where 3Z = {3m | m Z}. Prove