JF Group Theory: Colm Ó Dúnlaing's Notes

7/29/2019 JF Group Theory: Colm Dnlaing's Notes

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Mathematics 1214: introduction to groups, Hilary 2011

Colm O Dunlaing

August 9, 2011

1 Sets and maps

(1.1) Mathematical symbols and . is an abbreviation for for all.

Thus

x(x + 0 = x)means for all x, x + 0 = x.

For all, for every, and for each, are pretty well equivalent in mathematics.

A more general usage:(x R) x2 = 1

means for every real number x, x2 = 1. is an abbreviation for there exists . . . such that. Thus

xy(x + y = 0)

means for all x there exists a y such that x + y = 0.

Equivalently, for some y is a better way to read

y: one can read the above formula as for all

x, for some y (x + y = 0).

A more general usage:(x C) x2 = 1

means there exists a complex number x such that x2 = 1.(1.2) Cantors intuitive definition of a set. A setis a collection of objects considered as a whole.

(1.3) Various sets of numbers. The natural numbers N are the nonnegative integers:

N =

{0, 1, 2, . . .

}The integers Z are the whole numbers, positive and negative

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Z = {. . . 3, 2, 1, 0, 1, 2, . . .}The rationalsQ consist of all fractions

Q = {pq

: p, q Z and q= 0}

The reals R consist of every number which is the limit of a Cauchy-convergent sequence of

rationals (this description must suffice).

The complex numbers C are

C = {x + iy : x, y R}where i2 = 1.

The quaternions or hypercomplex numbers H are

H = {w + ix +jy + kz: w , x , y, z R}where i2 = j2 = k2 = ijk = 1 (the Broombridge formula).

(1.4) Definition A map with domain A andcodomain B is a rule or procedure which associates witheach x A a unique element of B.

The words function, mapping, transformation are synonyms for map.

(1.5) Notation The notation f: A B means thatA andB are sets andf is a map with domain Aand codomain B.

Given x A, f(x) denotes the unique element ofB associated to x by f.Alternatively, one can write f: x y to mean thaty is the unique element ofB associated with

x by f.

(1.6) The identity map X, also written idX. IfX is any set, there is a well-defined identity map

X: X X; x xfor every x

X.

Frequently the notation idX is used instead ofX.Other examples. Squaring: R R; x x2 is the map which squares each real number x.

Parity

Z Z; x

0 ifx is even

1 ifx is odd

More examples of maps. Operations like addition and subtraction can be viewed as certain

kinds of map. They are functions of two variables, so at present they dont fit the pattern, but it will

be explained later.

Similarly, multiplication, loge

(i.e., ln), exponentiation, trigonometric functions, etcetera, can bedefined as maps with suitable domains and codomains.

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(1.7) Equality of maps.

Two maps f: A B and g : C D are equal if and only if A = C, B = D, and for all x A, f(x) = g(x).They may be given be by completely different formulae.

Example.

f : N Z; x (1)x

g : N Z; x 1 2x + 4 (x 2)

Although it is not at all obvious, f = g. (Explicitly, you can show that if x is even then f(x) =g(x) = 1 and ifx is odd then f(x) = g(x) = 1. Notice that integer division is used, so if x is eventhen 4 (x 2) = 2x and ifx is odd then 4 (x 2) = 2x 2.)(1.8) Range (or image) of a map and image of a set under a map. Given f: A B, the rangeoff is the set

{f(x) : x A}It is a subset ofB, but not necessarily all ofB. More generally, for any subset X ofA, we write

f(X) = {f(x) : x X}

and call f(X) the image ofX under f. Thus range(f) = f(A) = f(domain(f)).

(1.9) Lemma Given a map f, suppose X andY are subsets ofdomain(f).IfX Y then f(X) f(Y). (Proof: exercise.)

Example. Given

f : R R; x x2

domain(f) = codomain(f) = R range(f) = [0, ) Since N = {0, 1, 2, . . .} is (or corresponds to) a subset ofR,

f(N) = {0, 1, 4, 9 . . .}

That is, the image ofN is (or corresponds to) the set of perfect squares.

Also,Z = {. . .2, 1, 0, 1, 2, . . .} is (or corresponds to) a subset ofR, andN Z. By Lemma1.9, f(N)

f(Z). Actually, f(Z) = f(N).

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Figure 1: f, g, g

f, f

g.

(1.10) Compatible maps and composition. Let f: A B and g : C D be two maps. Ifrange(f) domain(g) (i.e., C), then we say g is compatible with f and we define the composite mapg f, g following f, as follows:

g f: A D; x g(f(x)).When g is compatible with f we may simply say that the composite map g f is defined.

(1.11) Lemma Ifdomain(g) = codomain(f) then g f is defined. (Trivial.)

(1.12) Lemma

(i) range(g f) = g(range(f)).(ii) Therefore range(g f) range(g).

Proof. Let R = range(f).

R = {f(x) : x domain(f)}g(R) = {g(r) : r R} = {g(f(x)) : x domain(f)}

range(g f) = {g(f(x)) : x domain(f)} range(g f) = g(R),

i.e., range(g f) = g(range(f)): that was (i).

R domain(g) g(R) g(domain(g)) (Lemma 1.9); i.e.,

range(g f) range(g),

so (ii) is proved.

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(1.13) Our definition differs from most (or all) textbooks, including Durbins. In Durbins book,

composition g f is defined only when f : A B and g : B D. In other words, the usualpractice is to say that g f is defined iff domain(g) = codomain(f). In our definition, g f is definediff domain(g) range(f).

Now, the codomain of a map is rather arbitrary. For example, would you write

sin : R R; x sin x, orsin : R [1, 1]; x sin x?

In the second version the range and codomain are the same, in the first, the range is smaller than the

codomain.

(1.14) Notice, however, that while the definition of g

f depends on the range (image) off rather

than its codomain, the codomain of g f is the codomain ofg:

domain(g f) = domain(f) andcodomain(g f) = codomain(g).

Example. Iff(x) = x + 1 and g(x) = x2, with A = B = C = D = R, then

g f(x) = (x + 1)2 and f g(x) = x2 + 1.Composition of maps is not commutative. This example shows that g f and f g need not

be equal, even when both are defined. That is, composition of maps is not a commutative operation.However,

(1.15) Lemma Composition of maps is associative in the following sense. Suppose that g is compat-ible with f andh is compatible with g. Then

h is compatible with g f, h g is compatible with f, and h (g f) = (h g) f

Proof. h (g f) is defined, because, since h g is defined,range(g) domain(h).

But range(g) range(g f) (Lemma 1.12), so h (g f) is defined.To show (h g) f is defined, we need to show that range(f) domain(h g). This is true,

because domain(h g) = domain(g). Next,

domain((h g) f) = domain(f)domain(h (g f)) = domain(g f) = domainf

codomain((h g) f) = codomain(h g) = codomain(h)codomain(h (g f)) = codomain(h)

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so the domains and codomains agree.

For any x domain(f),[h (g f)](x) = h((g f)(x)) = h(g(f(x))), and

[(h g) f](x) = (h g)(f(x)) = h(g(f(x)))In other words, when you evaluate both h (g f) and (h g) f at x, you are applying f, then

g, then h, and getting the same result, h(g(f(x))). Therefore h (g f) = (h g) f.For example, suppose we are considering three maps f , g , h, all with domain R and codomain R,

where

f: x x + 1, g : sin(x), h : x x2.Then g

f(x) = sin(x + 1) and h

g(x) = sin2(x).

Given x, let y = x + 1, z = sin(y), w = z2. Then h g(y) = w, g f(x) = z, and h(z) = w.Therefore (h g)(f(x)) = h(g f(x)). In other words, ((h g) f)(x) = (h (g f))(x).

It therefore makes sense to write h g f, since it evaluates to the same function that is, ofcourse

x sin2(x + 1) whichever way the parentheses are placed. This distinction is invisible to us, but there are some

simple examples of operations which are notassociative.

Example: matrix multiplication is associative. Write Rn1 for the set of column vectors of

height n. An m n matrix A represents a map (a linear map, of course)a : Rn1 Rm1; X AX.

Given matrices Bm, Ck, let b and c be the maps

b : Rm1 R1; Y BY,c : R1 Rk1; Z CZ.

Then

b a : X B(AX)c b : Y C(BY)Matrix multiplication is designed so that

X Rn1 (BA)X = B(AX).Therefore

b a : X (BA)Xc b : Y (CB)Y

c (b a) : X (C(BA))X(c b) a : X ((CB)A)X

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A A AC C

D D D

E E

F F FG G

b

c

a H H HI I

P P P

Q Q

R R RS S

a

b

c

T TU U

V V

W W

1

2

X X XY Y

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b b b

c c

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b

c

d d d

e e

f f fg g

h h h

i i

b

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a p pq q

r r

s s

2

1

(iii)(ii)i

Figure 2: injective, surjective, bijective maps.

Since composition is associative, the latter two maps are the same, so

X Rn1 (C(BA))X = ((CB)A)X.

It is easy to deduce that C(BA) = (CB)A. Thus multiplication of matrices is associative.Example. Considering addition, subtraction, multiplication, and division over R: which are

associative? Which are commutative?

(1.16) Definition Injective, surjective, and bijective maps. A map f: A B is(i) injective (or one-to-one) if it maps distinct elements to distinct elements. i.e.,

x = y = f(x) = f(y)or equivalently

f(x) = f(y) = x = y(ii) surjective (or onto) if its range equals its codomain B, and

(iii) bijective if it is both injective and surjective.

(1.17) Definition (inverse maps). Given two maps

f : A B and g : B A,if

g f = A(i.e., g(f(x)) = x for all x A, see 1.6), then

g is a left inverse forf and f a right inverse forg.

Examples. Let A = [0, ) (the nonnegative reals. i.e., A = {x R : x 0}) and B = R,f: A B taking x x and g : B A taking x x

2

. Then g is a left inverse for f and f a rightinverse for g.

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(1.18) Lemma Suppose thatf: A B has a left inverse g and a right inverse h. Then g = h andfhas a unique two-sided inverse.

Proof.

g (f h) = g B = g(g f) h = A h = h

g = h.

(1.19) Notation If f: A B is a map and Y is any set, then f1(Y) = {x A : f(x) Y}. Ify Y, f1(y) = f1({y}).

f1(Y) is called the inverse image ofY under f.

Example. Iff(x) = x2

, A = B = R, and Y = {x R : 1 x 4}, then f1

(Y) = {x R : 2 x 2} and f1(z) = {2}.

Remark. f: A B is surjective if and only if f1(y) = for each y B, and is injective ifand only if for each y B f1(y) contains zero or 1 elements.(1.20) Lemma Except in the trivial case A = , a map f : A B has a left inverse iff it is injective.

Proof. (i) Iff is injective, we can define g : B A as follows. Fix any x0 A. For any y B,ify / range(f),

g(y) = x0

and ify

range(f),

g(y) = f(x)where x is the unique element ofA such that y = f(x). Then g f(x) = f(x) for every x A.

(ii) Iff has a left inverse g, then

f(x) = f(y) = g(f(x)) = g(f(y)), i.e.x = y

so f is injective.

(1.21) Lemma A map f : A B has a right inverse iff it is surjective.Partial proof. (i) Suppose f has a right inverse g:

(y B) f(g(y)) = y.For all y B, there exists an x A, namely, x = g(y), such that f(x) = y. Therefore f is surjective.

(ii) Conversely, if f is surjective then it has a right inverse. We omit the proof. Actually, theconverse is equivalent to the so-called Axiom of Choice in Set Theory.1

(1.22) Notation Iff: A B is bijective, then f1 denotes its unique 2-sided inverse.Thus f1 has more than one meaning. Iff is bijective, then f1(y) is a single element x ofA, or

the set {x}. Whether or not f is bijective, for any Y B, f1(Y) A, and f1(y) A under thisinterpretation.

1 In Durbins book, only 2-sided inverses are defined, so the relation of surjective maps to the Axiom of Choice is not

discussed, nor is the Axiom of Choice mentioned.

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A B

A

B

A BA B\

Figure 3: union, intersection, difference

2 Operations on sets; functions of several variables

(2.1) Definition LetA, B be any sets.

A B = {x : x A x B}( means or).

A B = {x : x A x B}( means and).

A\B = {x : x A x / B}

See Figure 3

(2.2) Definition An ordered pair of objects x and y is written (x, y). It is something like the set{x, y}, but order matters, so thatx is its first andy its second component.

The cartesian product of two sets A andB is the following set of ordered pairs.

A B = {(x, y) : x A y B}.Notation: A B is the cartesian product.

Examples.

R2 is the same as RR, hence cartesian product. B = always. {0, 1} {0, 1} = {(0, 0), (0, 1), (1, 0), (1, 1)} [0, 1] [0, 1], product of two closed unit intervals, is the unit square.Using cartesian products, we can interpret many operations as maps. For example, addition of

real numbers can now be expressed as a map:

R R R; (x, y) x + y

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3 Binary relations, equivalence relations, and partitions

A binary relation on a set A is some property connecting pairs of elements of A.This section is concerned with equivalence relations. A good example arises with fractions. The

set Q of fractions (quotients or rational numbers) is

Q = {ab

: a, b Z b = 0}To give a solid definition, we really must regard fractions as little more than ordered pairs, so

Q = Z (Z\{0})but also

1

2 =

2

4and so on, so we define equality of fractions as follows:

a

b=

c

d (def) ad = bc.

This equality relation is looser than equality as applied to ordered pairs. It is a very good example of

an equivalence relation.

Of course we also need to define sum, negative, product, and inverse, and prove, for example

(a

b=

a

b c

d=

c

d) = a

b+

c

d=

a

b+

c

d,

etcetera, but this section concentrates on the equivalence relation side.

(3.1) For example:

The relation on N, Z, R. The relation < on N, etcetera. The relation on the set of subsets of a fixed set. The relation m divides n on Z

The equality relation on any set.

The relation m and n have the same parity on Z. The relation y = x + 1 on N. Given a fixed (positive) integer d, the relation between integers x, y (i.e., x, y Z)

d divides x y.This is conventionally written

x y (mod d)or alternatively

xd yThe latter notation is not often used, but it is neater.

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Technically, a binary relation on A can be defined as a subset (any subset) ofA A. Technically,a relation R is just a set: R A Abut we can write xRy (like x y) to indicate that x and y are in relation R to one another.

(3.2) Definition A binary relation R on A is transitive if

(x,y,z A) (xRy yRz) = xRz

In the above list of examples (3.1), they are all transitive except for the relation y = x + 1.

(3.3) Definition A binary relation R on A is reflexive if

(x A) x R x

In the above list of examples, the relations < and y = x + 1 are not reflexive. The others are.

(3.4) Definition A binary relation R on A is symmetric if

(x, y A) x R y = y R x

In the above list of examples, equality, equal parity, and d

are symmetric. The others are not.

(3.5) Definition An equivalence relation is a binary relation which is reflexive, symmetric, and tran-sitive.

IfR is an equivalence relation on A andx A, then the equivalence class ofx (modulo R) is{y : x R y}.

[x]R

denotes the equivalence class of x.

Example. The equal parity relation is the same as 2

. It has 2 equivalence classes, the even

integers, and the odd integers:

[0]2

: {. . . 4, 2, 0, 2, 4, . . .}[1]

2

: {. . . 3, 1, 1, 3, 5, . . .}

(3.6) Lemma (i) Given any map f : A B, the relationxRy : f(x) = f(y)

on A is an equivalence relation.(ii) For any x A,

[x]R = f

1

(f(x))

Proof. (i)

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It is reflexive, i.e., xRx for all x, because f(x) = f(x). It is symmetric, i.e., xRy = yRx, because f(x) = f(y) = f(y) = f(x). It is transitive, i.e., (xRy yRz) = xRz because (f(x) = f(y) f(y) = f(z)) =

f(x) = f(z).

(ii)

[x]R = {y : f(x) = f(y)} = f1(f(x)).

(3.7) Lemma For any positive integerd, d

is an equivalence relation.

Proof. The recognised abbreviation for d divides x

y is

d|(x y).As a relation on Z, it means

(q Z) x y = qdReflexivity: Since 0 = 0d, it follows that every integer divides 0, hence

(x) d|(x x).In other words,

dis reflexive.

Symmetry: Suppose xd y, so for some q Zx y = qd

Then

y x = (q)d d|(y x)

so d

is symmetric.

Transitivity: Suppose x

d

y and y

d

z. This means that there exist integers q1 and q2 such that

x y = q1d, andy z = q2d

Then x z = x y + y z = q1d + q2d, so (because integer multiplication is distributive)x z = (q1 + q2)d

d|(x z)Thus

dis an equivalence relation. Q.E.D.

In the above list of examples (3.1), equal parity, equality, and d

(congruence modulo d) are

equivalence relations. The others are not.

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(3.8) Lemma LetR be an equivalence relation on A. For any x, y A, either

[x]R = [y]R

or

[x]R [y]R = .

Proof. It is enough2 to assume [x]R [y]R = and deduce that [x]R = [y]R.First, fix w [x]R. For any z [x]R,

z [x]R meansx R z

x R w (similarly)

w R x (symmetry)

w R z (transitivity), i.e.,

z [w]RThat is, for every z [x]R, z [w]R. In other words,

w [x]R = [x]R [w]R.

In particular (since x [x]R and [x]R [w]R,w [x]R = x [w]R

and also

x [w]R = [w]R [x]R.Therefore

w [x]R = [x]R = [w]R.So, if[x]R

[y]R

=

, choose some w

[x]R

[y]R. Then [x]R = [w]R and [y]R = [w]R, so

[x]R = [y]R. Q.E.D.

(3.9) Definition A partition P of a setA is a family

{Ai : i I}

of subsets ofA such that

All the sets Ai are nonempty,

their union is A, and

2No it isnt.

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they are pairwise disjoint, i.e.,(i = j I) (Ai Aj = ).

(3.10) Lemma LetP be a collection of subsets of A, all of them nonempty. Then P is a partition ofA if and only if for every x A there is a unique set S in P such thatx S.

Proof. If: to show P is a partition, we need show (a) / P: that is given; (b) their union is A,i.e., every x in A belongs to at least one set in P; that is also true; and (c) they are pairwise disjoint:otherwise some x in A would belong to more than one set in P.

Only if: from (b), every x in A belongs to at least one S in P, and from (c) no x in A belongs tomore than one S in P, hence every x belongs to a unique x P. Q.E.D.

(3.11) Corollary If R is an equivalence relation on A then its equivalence classes form a partitionofA.

Proof. (a) Every equivalence class is nonempty, (b) every x belongs to an equivalence class,namely, [x]R, and (c) the equivalence classes are pairwise disjoint (Lemma 3.8).

(3.12) Lemma For every partition P of A there is a unique equivalence relation R on A whoseequivalence classes constitute the partition P.

Proof. Write the partition P as a family of subsets ofA with an indexing set I:

P = {Ai : i I}.(Implicitly, Ai is a function with the single argument i, establishing a bijection between I and itscodomain. Does that make sense?)

Define the following map:

f : A I; x unique i in I such that x in AiFrom Lemma 3.10, f is a well-defined function. Then the relation

xRy : f(x) = f(y)

is an equivalence relation. For any x A,[x]R = f

1(f(x))

(Lemma 3.6).

Suppose x Ai. Then y [x]R iffthe unique j in I such that y Aj

=

the unique j in I such that x Aj= i, i.e.,

y Ai.Thus [x]R = Ai. The equivalence classes ofR are the sets in P.

Suppose that S is another equivalence relation whose equivalence classes are the sets in P. Forany x A, [x]S is the unique set X such that

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X P and x X.

But this coincides with [x]R; and because [x]S = [x]R for all x A, the relations R and S areidentical. Q.E.D.

Summarising. This is important.

(3.13) Theorem Equivalence relations R on A are interchangeable with partitions P of A in thesense that

the equivalence classes of any equivalence relation form a partition ofA, and every partition arises in this way from a unique equivalence relation on A.

4 Semigroups, monoids, and groups

(4.1) An associative binary operation on a set S is a map f: S S S with the property that forall x,y,z S,

f(x, f(y, z)) = f(f(x, y), z)

Recall

A map f: S T is a rule or procedure which associates with each x S a uniqueelement f(x) T.

The cartesian productS T is the set of all ordered pairs (x, y) where x S and y T:

S T = {(x, y) : x S, y T}.The formal f(x, y) notation for maps is seldom used: rather infix form like x + y, x y, x y. Thebinary refers to the fact that f is a function with two variables.

(4.2) Definition A semigroup consists of a nonempty setS together with an associative binary oper-ation on S,

The operation is often called multiplication and sometimes addition.

As in ordinary algebra, the can be omitted so x y becomes justxy. (As in ordinary algebra, ifthe operator symbol is + then it is not omitted.)

(4.3) Examples of semigroups

N,Z,Q,R,C and H under addition

N,Z,Q,R,C and H under multiplication

Any nonempty set S with , a strange operation defined as follows: x y = y, for all x, y S.

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a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

fg p q

=

Figure 4: All maps from {a, b} to itself; qp f = q.

As above, except now x

y is defined as x.

The set of maps f: X X where X is any set and the operation is , composition of maps. The set of2 2 matrices with coefficients in N,Z, ,Q,R, or C under addition (see below) the set of2 2 matrices . . . under multiplicationFor an example involving maps S S, let S = {a, b}. There are four maps, illustrated in Figure

4.

f: a a, b a, g : a a, b b, p : a b, b a, q: a b, b b(4.4) Identity and inverses. A left identity a in a semigroup S, is an element such that for allx S,

a x = x.Similarly, a right identity b satisfies x b = x for all x A.A two-sided identity, or identity for short, is an element e which is both a left- and a right-identity

for S.

(4.5) Lemma IfSpossesses a left identity and a right identity they they are equal and S possesses aunique (two-sided) identity. (Easy proof omitted.)

(4.6) Notational conventions. Often one leaves out the operation sign, so xy might mean x y.However, one usually uses + to represent the operation when the semigroup is commutative, and theaddition sign is not omitted. Just like in ordinary algebra.

Generally, a two-sided identity will be represented either as 1 or as e. In a commutative group,where + is the symbol used, the identity is written as 0.

(4.7) Definition A monoid is a semigroup containing a (two-sided) identity.

As already remarked, the identity is unique. Since it is unique, we may as well give it a special name:

1, 0, or e, depending on context.

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(4.8) Definition LetS be a monoid, e its identity. If a andb are elements of S with a b = e, then ais a left inverse forb andb is a right inverse fora.Ifb is both a left- and a right-inverse for a the it is called a two-sided inverse fora orinverse forshort.

(4.9) Lemma Letb be an element of a monoid S. If b has a left inverse a and a right inverse c, thena = c andb has a two-sided inverse which is unique. (Proof exercise.)

(4.10) Definition A group G is a monoid in which every element has a (two-sided) inverse.

As noted, the inverse of x depends uniquely on x, so we can show it as a function of x: usuallyx1, or if the group is commutative we usually write + for the group operation, 0 for the indentity,

and x for the inverse ofx: thus x + x = 0.The inverse ofx is usually written x1, or x if the group operation is denoted +.

5 Remainder modulo n and integer division

Let d be a fixed positive integer.

(5.1) Proposition The relation xd

y on Z, meaning x y is an (integer) multiple ofd, is an equiva-lence relation on Z.

(5.2) The principle of well-ordering forN. In the definition below, we invoke a very useful property

of the integers, the principle of well-ordering, or least integer principle according to Durbin.

If P is a property of integers, and at least one nonnegative integer has property P, thenthere exists a smallest nonnegative integer with property P, and it is unique.

This principle is equivalent to the principle of Mathematical Induction.

(5.3) Lemma Given a positive integerd, and an integern, the equivalence class

[x]d = {r : nd r}contains a nonnegative integer, and hence a (unique) smallest nonnegative integer.

Proof. This class contains n, so ifn 0 then there is nothing more to prove. If n < 0, look atn nd. It, too, is in the class, and

n nd = n(1 d) = (n)(d 1).

It is the product of two nonnegative integers, therefore nonnegative. Q.E.D.

We write n mod d for the smallest nonnegative integer in the class [n]d

. The proof below applies

certain facts about integers without always justifying them: they could ultimately be deduced from a

suitable axiom system, such as Peano Arithmetic.

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(5.4) Theorem (division algorithm.) Given a positive integerd and an integern, there exist unique

integers q andr such that n = qd + r and 0 r < d .

Proof. (i) Existence. Let r = (n mod d). 0 r: that is given.Also, r < d, because otherwise r d would be a smaller nonnegative integer in the class ofn.Also, n

dr, i.e., d divides n r, i.e., n r = qd for some integer q. Therefore

n = qd + r

where 0 r < d, as required.(ii) Uniqueness. Suppose

n = q1d + r1 = q2d + r2

where 0 r1 < d and 0 r2 < d. Without loss of generality, r1 r2. Therefore

(q1 q2)d = r2 r1But 0 r1 r2 < d, so

0 r2 r1 < d,

and alsoq1 q2 0.

But the only nonnegative multiple ofd which is less than d is 0. Therefore r1 = r2. Finally, q1q2 = 0since otherwise (q1 q2)d would be nonzero. Q.E.D.

(5.5) Definition Given a positive integerd and an integern, the unique qandr such thatn = qd + rwhere q is an integer and r an integer between 0 andn 1, are called the quotient andremainder,respectively, on dividing n by d. The remainder is, of course, n mod d.

We write x n for the quotient.

(5.6) Integers mod n under addition and multiplication.Congruence modulo d is an equivalence relation. It is more than that (hence the word congru-

ence is used, and congruence class rather than equivalence class), because for any m1, m2, n2, n2 Z,

(5.7) Lemma

(m1d

m2 n1d

n2) = m1 + n1 d

m2 + n2

(Proof: exercise.)

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We can then define a semigroup whose elements are the congruence classes (mod d) with the fol-

lowing operation

[m]d

+ [n]d

= (def) [m + n]d

Any congruence class [m]d

has infinitely many elements, so for any two congruence classes there are

infinitely many ways to define [m + n]d

. Bowever, no matter which way is chosen, the answer

which is another congruence class is the same.

Moreover, it is a commutative group traditionally called an abelian group for the following

reasons. For any integers ,m,n,

(5.8) Lemma

To make it more readable, write [m] instead of[m]d

.

[] + ([m] + [n]) = ([] + ([m]) + [n]. [m] + [n] = [n] + [m]. [m] + [0] = [0] + [m] = [m] [m] + [m] = [m] + [m] = [0].

Similarly, for any positive integer d, the set {0, . . . , d 1} is a commutative monoid under mul-tiplication (mod d). It is not a group, except in the trivial case d = 1.

What is the order of this group? By the division algorithm, every congruence class [n] contains aunique integer r where 0 r < d. Therefore there are at most d congruence classes

[0], [1], . . . , [d 1].

Moreover, they are all different, since otherwise there would exist a class containing more than

one r such that 0 r d 1.

(5.9) Corollary For any positive integer d, the relation d

has d congruence classes [0], . . . [d 1],and these congruence classes form an abelian group oforder d under the operation

[m] + [n] = [m + n].

We can do with those square brackets by using the numbers 0, . . . , d 1 as representatives fortheir congruence classes. Since

[m] + [n] = [m + n] = [(m + n) mod d],

we can make the following definition.

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(5.10) Definition Given a positive integerd, the additive group Zd of integers modd is the set

{0, . . . , d 1}together with the operation

r + s = (def) ((r + s) mod d).

It is an abelian group of order (i.e., cardinality) d. It is also called the cyclic group of order d.

Similarly,

(5.11) Definition The multiplicative monoid Zd of integers modd is the set{0, . . . , d 1} togetherwith the product

rs = (def) (rs mod d).

+ 0 1 2 30 0 1 2 31 1 2 3 02 2 3 0 13 3 0 1 2

0 1 2 30 0 0 0 01 0 1 2 32 0 2 0 23 0 3 2 1

6 Groups

Recall a semigroup is a nonempty set together with an associative binary operation on it, a monoid

is a semigroup with a 2-sided identity, necessarily unique, and a group is a monoid in which everyelement has a two-sided inverse, necessarily unique.

In a monoid or group, e or 1 usually denote the identity. In a group, x1 usually denotes theinverse ofx.

(6.1) Definition The order |S| of a semigroup, group, or monoid S is the number of elements itcontains..

There is, essentially, just one semigroup of order 1, and it also happens to be a group. For the

only possible operation on {a} is aa = a.There are two monoids of order 2, in the sense that given two objects {e, a}, where e is to be the

identity, so ee = e and ea = ae = e, we have two choices for aa: either a or e.e ae e aa a a

e ae e aa a e

Two see that these tables define associative multiplication, you need to consider all 8 possible

values of x , y, z and show in each case that (xy)z = x(yz). In the first table, x(yz) and (xy)z bothevaluate to e if x = y = z = e, otherwise they both evaluate to a. In the second table, (xy)z andx(yz) both evaluate to a if an odd number of the objects x,y,zis a, otherwise they both evaluate toe.

A quicker way of showing that the operation is associative is to note that the multiplication table

for{

1, 0}

has the same form, and that we know to be associative. In the second case, the multiplica-

tion table for {1, 1} has the same form, and that also is associative. The second one gives a groupshince a and e both have inverses. The first is not a group since a does not have an inverse.

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(7.1) Lemma Suppose f : A B is a map. IfA is infinite andf is injective then B is infinite IfB is infinite andf is surjective then A is infinite.

(Proof omitted, because although this is all quite plausible, it needs some set theory.)

When infinite sets are involved, even if domain and codomain are the same, maps can be injective

without being surjective and vice-versa. For example let

f : N N; x 2xg : N

N; x

x

2

Integer division is used, so 3 2 = 1 etcetera.Clearly f is injective but not surjective and g is surjective but not injective.

(7.2) Lemma Letf : A B be a map where A andB are finite sets, i.e., |A| < and|B| < .

(i) Iff is injective then |A| |B|(ii) Iff is surjective then |A| |B|

(iii) If|A| = |B| andf is injective then f is bijective(iv) If|A| = |B| andf is surjective then f is bijective .

(Proof omitted; some of this has appeared in problem sets.)

Finally, a useful fact:

(7.3) Lemma (Pigeonhole principle). IfA andB are finite sets and|A| > |B| then no map from Ato B can be injective (see above lemma, part (i)).

Also, if A andB are finite sets and |A| < |B| then no map from A to B can be surjective. (see(ii)).

8 The symmetric group Sn

Let A = {1, . . . , n}. The set of all maps from A to A is a monoid (of size nn), and the set of allbijective maps from A to itself is a group of order n!.

This is almost obvious, since the identity is bijective and bijective maps are invertible. But the

following facts need to be shown. The proofs are easy and omitted.

(8.1) Lemma Suppose f andg are maps andg f is defined.

Iff andg are injective, so is g f. Iff andg are surjective, so is g f.

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t t

t t

v v

v v

w w

w w

x x

x x

y y

y y

Figure 5: Inverse of a bijective map

Iff is bijective, so is f1. A is bijective.

(8.2) Definition A bijective map f :

{1, . . . , n

} {1, . . . , n

}is called a permutation of n letters.

The group of permutations ofn letters isSn,

called the symmetric group on n letters.

When maps are given diagrammatically, it is easy to see if they are bijective and to give their

inverse. The inverse is computed just by taking the mirror image of the diagram. See Figure 5

The word permutation ofn letters used to mean an arrangement ofn different letters, like acbd.So there should be 2 permutations of2 letters

12, 21

and 6 of 3

123, 132, 213, 231, 312, 321

Ambiguity of this representation. How are we to interpret these as bijective maps? Certainly

123 should represent the identity permutation. Also 132 is not a problem: 2 and 3 are swapped.Similarly, 213: 1 and 2 are swapped. But 231 is not so easy.

Im inclined to think of this as a map taking

1 2, 2 3, 3 1.This is a map taking positions to labels.

Under such an interpretation, let us look at the composition

213

231 : 1

2

1, 2

3

3, 3

1

2

Result: 132.On the other hand, you could also interpret the arrangements as mapping labels to positions,

so 231 maps2 1, 3 2, 1 3

and the composite map

213 231 : 2 1 2, 3 2 1, 1 3 3is 321.

So, there are two natural ways of interpreting 231, position value and value position, andI find the first more natural. Sometimes the second is more natural. For example, when it comes

to permuting rows of a matrix, or the symmetries of a triangle, the second is more natural. The

ambiguity is not in Sn, it is in how we interpret arrangments and figures.

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(8.3) Notation (nonstandard). Every permutation describes an arrangement, and the permutation

1 r, 2 s, 3 t , . . . is represented by the arrangement rst. . .. In this notation 123 . . . representsthe identity permutation.Example. S0 contains one element, the empty map. S1 = {1}, S2 = {12, 21},

S3 = {123, 132, 213, 231, 312, 321}, andS4 = {1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413,2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321}.(8.4) Lemma |Sn| = n! (from school).(8.5) Function composition operator omitted. Since Sn is a group, we write rather than for the composition of two permutations.

The multiplication table (Cayley table) for S1 is completely trivial and for S2 is trivial.12 21

12 12 2121 21 12

For S3 it is not (this is the smallest noncommutative group). In the table, we use the position value interpretation.

123 132 213 231 312 321123 123 132 213 231 312 321132 132 123 312 321 213 231213 213 231 123 132 321 312231 231 213 321 312 123 132312 312 321 132 123 231 213321 321 312 231 213 132 123

9 Parity and the alternating group

In linear algebra you have seen the parity and signature of a permutation, so most of the work has

been done.

An elementary row operation on an m n matrix is scaling, subtracting, or swapping. Here weare only interested in swapping.

For any permutation in Sm (one uses small greek letters for permutations) there is a permutationof the rows ofA producing a new matrix A:

if : i jthen the i-th row ofA will be the j-th row ofA.

There is a matrix P representing the permutation in the sense that for any matrix A of height m,P A is the result of the permutation on A.

To study permutations, we can restrict ourselves to the standard unit column vectors of heightm i.e., the column-vectors appearing in the m m identity matrix. If X is one of these column-vectors then P X is another.

(9.1) Lemma For any permutation , let P be the corresponding permutation matrix. The identitymatrix can be converted to P by a series of adjacent swaps. Hence is a product of adjacent swaps.

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Also, det P = 1.Note: this uses the value position interpretation. An adjacent swap is equivalent to swappingtwo adjacent rows of the matrix. In other words, it is a permutation

1 1, . . . , i 1 i 1, i i + 1, i + 1 i, i + 2 i + 2, . . . , m m

(9.2) Definition The signature (or sign) of a permutation is the determinant of the correspondingpermutation matrix.3

A permutation is even if its sign is +1.

From known properties of the determinant, we have the following facts.

(9.3) Lemma

A permutation is even/odd iff it can be expressed as a product of an even/odd

number of swaps, adjacent or otherwise.

=

( means ). 1 = 1 (the identity permutation is even). 1 = .

Hence the set of even permutations is a group in its own right. It is written An and called thealternating group.

10 Generators for Sn

(10.1) Generators of a group. A set of generators for a group G is a set a1, . . . , ap of elementsofG with the property that every element of G can be expressed as a product of powers (positive ornegative) of these elements.

(10.2) Cycle notation. A cycle is a permutation which fixes some letters and permutes the others in

a cyclic order.In other words, there is a subset i1, . . . , ik of letters such that

i1 i2, i2 i3, . . . , ik1 ik, ik i1and the other letters are left fixed.

(10.3) Definition The above cycle is represented

(i1i2 . . . ik)

its length is k and it is called a k-cycle; k

1; a 1-cycle is just the identity map.

3This looks like a circular definition, but it can be put in the proper sequence. Alternatively, you already know a lot

about the sign of a permutation.

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1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6 7 98

9

98

7 8

7

Figure 6: disjoint cycle decomposition of a permutation.

Thus (153) maps 1

5

3

1. The cycle can begin at any place: (153) = (531) = (315). A

2-cycle is a transposition or swap.

(10.4) Lemma Disjoint cycles commute.

Proof. When and are cycles, they are disjoint when

(i) = i = (i) = i(i) = i = (i) = i

So

Either (i) = i (i) = i, in which case (i) = (i) = i, or (i) = i. Let j = (i), so j = i.

Since i = j and is injective, (i) = (j), so (j) = j.Therefore (i) = i and (j) = j.

(i) = (j) = j (i) = (i) = j, or (i) = i: same as previous case by symmetry.

(10.5) Lemma Every permutation can be written as a product of (commuting) disjoint cycles.

Sketch proof. Figure 6 illustrates the idea. The permutation is (under position value)154382967

Begin at 1:

(1)

Next at 2:(2586)

Next at 3:

(34)

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And last at 7:

(79)Thus we get

(79)(34)(2586)(1)

Cycles of length 1 are redundant. This is (79)(34)(2586).Even-length cycles have odd parity and odd-length cycles have even parity: hence

(10.6) Lemma The parity of a permutation is the parity of the number of even-length cycles.

In the example illustrating the previous lemma, there are 3 even-length cycles: the permutation is

odd.

From the previous section,

(10.7) Proposition Ifn 2 then Sn is generated by its 2-cycles.Moreover, Sn is generated by adjacent transpositions, i.e., 2-cycles of the form (i i + 1).

Our main result is

(10.8) Theorem Ifn 2 then (12), (12 . . . n) together generate Sn.Proof. Abbreviation: let = (12 . . . n). From the above proposition, it is enough to show that

every adjacent transposition

(i i + 1)

can be expressed in terms of (12) and . The trick is conjugation, meaning a product of the kindxyx1.

1 : 3 2 1 n . . .(12)1 : 4 3 3; 3 2 1; 2 1 2; 1 n n

(12)1 : 4 3 3 4; 3 2 1 2; 2 2 2 3; 1 n n 1Thus (12)1 = (23). In the same way, (23)1 = (34), and so on.

(10.9) Proposition Forn 3

, The 3-cycles generateAn

.

Proof. Let be an even permutation: ignore the case = 1. It can be expressed as a product ofan even number of swaps: write this as

1122 kk.Each product jj is a product of two 2-cycles; either they are the same and cancel, or they have

one letter in common, or they are disjoint.

For the second case we are given (pq)(pr), or without loss of generality (12)(13). But this equals(132), a 3-cycle.

The third case is, without loss of generality, (12)(34). This can be written as (12)(13)(13)(34), aproduct of two 3-cycles.

Thus is a product of 3-cycles.

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11 Subgroups (and submonoids)

(11.1) Definition LetS be a semigroup (or monoid, or group) with operation . A subsetT ofS is

a sub-semigroup ifT is closed under the operation, i.e., x,y, T = xy T, a submonoid if it is a subsemigroup and S is a monoid and1 T, or a subgroup ifS is a group andT is a submonoid closed under inverse, i.e., x T = x1

T.

(11.2) Lemma A sub-semigroup, submonoid, or subgroup is itself a semigroup, monoid, or group

under the operation/operations given for the semigroup, monoid, or group containing it. (Trivial.)

Example. Because of the properties given for , for any , Sn,

1 is even If and are even then so is If is even then so is 1.

Hence the even permutations form a subgroup An ofSn, the alternating group An.

11.1 Subgroup generated by a set of elements

Let G be a group, S a subset of G. It can be proved that there exists a smallest subgroup of Gcontaining S. (Smallest with respect to the set inclusion relation .) This is written

S

or, ifS = {a1, . . . , ak},a1, . . . , ak.

=

{1

}(which is a subgroup of G). For the purposes of this course S is always finite and

nonempty.

(11.3) Lemma Given a1, . . . , ak G. We call them generators,4

a1, . . . , ak

is the set (including 1) of all products of the form

a1i1 a1i2

a1i (11.4)

of generators and/or their inverses.

4 for lack of a better word. It does not mean that they generate the group. Rather, they generate the subgroup we are

looking at.

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Proof. Let T denote the given set of products. We need to show (a) T is closed under products,

(b) 1 T, (c) T is closed under inverse, (d) aj T (1 j k), and (e) for any subgroup H ofG,ifH contains all the generators ai, then T H.(a) Obvious. (b) Obvious.

(c) T is closed under inverse, because

(a1i1 a1i2

a1i )1 =a1i a1i2 a1i1 .

(d) Obvious.

(e) If H is a subgroup containing the given generators, then it contains all products of the form11.4, i.e., T

H.

(There are similar ways of describing sub-semigroups and submonoids, not covered in these

notes.)

Examples.

In S3, (12) = {1, (12)}. In S3, (12), (23) = S3. In Sn, 3-cycles = An.

In S3,

(123)

= A3.

In Z (under addition), 2 is the subgroup of even integers.

Notation for subgroups. IfG is a group, then we write

H G (def) H is a subgroup ofG.

12 Matrix groups GL(n), SL(n), O(n), SO(n)

GL(n,R) is the general linear group . . . n. It is the set of invertible n n real matrices under matrixmultiplication with identity I and matrix inverse.

It is equivalent to the group of invertible linear maps Rn Rn.

GL(n,C) is defined similarly.

SL(n,R) is the special linear group . . . n, the matrices of determinant 1.

SL(n,R) = {A Rnn : det A = 1}

Similarly, SL(n,C).

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(12.1) Lemma

SL(n,R) GL(n,R) and SL(n,C) GL(n,C).

O(n) is the orthogonal group . . . n:

O(n) = {A Rnn : ATA = I}

consisting of all n n orthogonal matrices.

O(n) is equivalent to the group of distance-preserving (linear) maps from Rn to Rn.SO(n) is the special orthogonal group . . . n, orthogonal matrices of determinant 1.

SO(n) = {A SO(n) : det A = 1}.

SO(1) = {1}. O(1) = {1, 1}. SO(2) is the group of rotation matricescos sin sin cos

.

O(2) = {A , F A : A SO(2)}, where

F = 1 0

0 1 One could say (loosely) that O(2) can be formed by including one orientation-reversing orthogo-

nal matrix to SO(2).SO(3) is the group of orientation-preserving orthogonal 3 3 matrices, which can be shown to

be the set of rotation matrices. O(3) can be formed by including one orientation-reversing orthogonalmatrix, such as I (this is orientation-reversing in odd dimensions).

Then there is U(n), the unitary group . . . n: the group ofn n complex unitary matrices.

U(n) = {A Cnn : A A = I}.

A is the (complex) adjoint ofA, 5

A is the componentwise complex conjugateof the transpose AT ofA.

U(n) is equivalent to the group of distance-preserving (linear) maps Cn Cn.Then there is SU(n),

SU(n) = {A U(n) : det(A) = 1}.Later, we shall connect SU(2) with SO(3) and with quaternions.

5not to be confused with the other adjoint matrix, the matrix of cofactors.

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13 Dihedral group Dn and symmetries of the cube

(13.1) Symmetries of plane figures. We are interested in a certain group of bijections from the plane

to itself, namely, the rigid transformations: rotations around a point, reflections in a line, translations,

etcetera. (This group includes O(2), but in O(2) all rotations are around O, all reflections are inlines through O, and there are no notrivial translations.) Given a nonempty set T in the plane, Thesymmetries of T are those rigid transformations of the plane which map T onto T. The symmetriesform a group. We are mostly interested in the case where T is a regular n-sided polygon, such asan equilateral triangle, centred at the origin. The group of symmetries of a regular n-gon (n 3) iscalled the n-th dihedral group D2n.

Its order is 2n. To see this, label the corners from 1 to n in anticlockwise order. A symmetrymust take the adjacent pair 12 to another adjacent pair i, (i mod n) + 1 or i, ((i

2)mod n) + 1.

Different symmetries correspond to different pairs, and each such pair is the image of 12 under aunique symmetry: hence, 2n symmetries.6

1 2

3 2

3

1

2 31

2 1

3

1

2

3

1

23

312

132 321

231

213

If we label the corners of the n-gon as described, then each symmetry corresponds to a uniquepermutation of the corners. This way, Dn can be viewed as part of Sn. However, notice that the waysymmetries are defined, it makes no sense for the permutations to act on values. They should act

on positions, as illustrated. Notice that D6 corresponds exactly to S3, and this helps us to understandS3 better.

In general there is the identity map, n 1 rotations, and n reflections, one in each of the n axesof symmetry of the regular n-gon.

The figure below illustrates the ten symmetries of a regular pentagon:

35

4

1 2 1

1

1

12

2

2

2

3

3

3 3

4

4 4

4

5 5

5

5

e

1

1

42

5

12

3 5

43

4

5

2 5 1

3

1

5 4

3

2

3

25

F F F F F

R R R

1 4 5

R432

2 3

Given a letter 1 j 5, j + 1, j + 2, and so on are taken with wraparound at 5, so j + s shouldbe interpreted as 1 + ((j + s 1)mod 5).

6This can be made plausible in the following way. By polygon we mean the edges and corners, not the 2-dimensional

interior. The edge joining 1 and 2 is a line-segment of a certain length, and all rigid transformations take it to a line-segment of the same length. The only subsets of the polygon (not its interior) which are line-segments of that length are

other edges.

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7(1,1,1)

1(1,1,1)

(1,1,1)

4 (1,1,1)

3(1,1,1)2

(1,1,1)

5 (1,1,1)

6(1,1,1)

8

Figure 8: Symmetries of the cube. Three of the 13 axes of rotation are shown.

Subgroups ofD10. Apart from

{e

}and D10 themselves, there is also the subgroup

{e,R,R2, R3, R4

},

which has the same structure as Z5 (under addition).There are also groups of order 2: {e, F1}, {e, F2}, {e, F3}, {e, F4}, {e, F5}.These are all the subgroups, for the following reasons.

A subgroup H either is trivial, or it contains a nontrivial power of R, or it contains some Fi, or itcontains both a nontrivial power ofR and some Fi, or it contains some Fi and Fj, i = j.

If a subgroup H contains any nontrivial power of R, it contains R, because (R2)3 = R6 = R,(R3)2 = R6 = R, and (R4)4 = R16 = R. Hence it contains all powers ofR.

Any nontrivial power of R, and any Fi, together generate D10, because you can generate allpowers ofR, and every other Fj which can be expressed in the form FiR

s (refer to the Cayley table).

Two Fi, Fj , i = j, together generate D10, because then Fj can be written as FiRs for somenontrivial power R

s

ofR, so the subgroup contains FiFiRs

= Rs

, a nontrivial power ofR, as well asFi.

13.1 Symmetries of the cube

The cube is centred at (0, 0, 0) R3 and has corners 1 : (1, 1, 1) etcetera. See Figure 8.The group consists of those A O(3) which permute the corners.There is one obvious orientation-reversing symmetry: I, or as a permutation (15)(26)(37)(48).The orientation-preserving symmetries are rotations around various axes.

The identity map is the trivial rotation.

There are 4 axes passing through opposite corners, and there are two nontrivial rotations aroundeach. This contributes 8 rotations.

There are 6 axes bisecting diagonally opposite edges, with one nontrivial rotation around each.This gives 6 rotations.

There are 3 axes passing through opposite faces, with three nontrivial rotations around each.This gives 9 rotations.

This gives 24 rotations. Multiplication by

I (which commutes with everything) gives another

24 orientation-reversing symmetries. This is everything: there are 48 symmetries of the cube.

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14 Cosets and Lagranges Theorem

(14.1) Lemma LetH be a subgroup of a group G. (i) The relation

x1y H

is an equivalence relation on G.(ii) The relation

xy1 His an equivalence relation on G.

Proof. Exercise.

(14.2) Definition Given x G the set

{xh : h H}

is called a left coset ofH and written xH.

Remember the quiz questions about products of subsets of a semigroup: xH = {x}H in thatnotattion.

Similarly there is a right coset Hx.

(14.3) Lemma LetH be a subgroup ofG, x G.(i) xH is the equivalence class of x modulo the relation x1y H.

(ii) Hx is the equivalence class ofx modulo the relation xy1 H.

Proof. (i)

xH = {xh : h H}y xH (h H) (y = xh)

y xH (h H) (x1y = h)y

xH

x1y

H

(ii) is similar.

For a simple example, let G = Z and let H be the even integers. The relation is just y x H,which is just the relation x y mod 2. There are two distinct (left) cosets, [0], the even integers,and [1], the odd integers.

Here is a result about finite sets which possibly belongs somewhere else.

(14.4) Lemma Iff : A B is injective andX A is finite, then

|f(X)| = |X|.

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Proof. The collection

f

1

(y) : y f(X)is a partition ofX.

|X| =

yf(X)

|f1(y)|

But f is injective, so each for each y f(X), |f1(y)| = 1, so|X| =

yf(X)

1 = |f(X)|.

(14.5) Lemma For any a G, the maps x ax andx xa are both bijective. (Exercise.)(14.6) Lemma Suppose H is finite. All left- (and right-) cosets ofH in G have the same cardinality,namely, |H|.

Proof. We shall only consider left-cosets. The proof for right-cosets is much the same.

aHis the image ofHunder the injective map x ax, so |aH| = |H|. Similarly |Ha| = |H|.(14.7) Corollary (Lagranges Theorem). IfG is finite andH G, then |H| divides |G|.

Proof. Since G is finite, G is a finite union of disjoint left cosets

a1H . . . akHwhich partition G into sets all of the same size, i.e., |H|. Therefore

|G| = k|H|.(14.8) Subgroup generated by x. Given any member x of any group G, the set of all powers ofx,

{xn : n Z}is a subgroup ofG, called the subgroup generated by x, and denoted

x.For example, x = {e} x = e;

ifG = S3,(12) = {e, (12)}, and (123) = {e, (123), (132)}.

IfG =Z

under addition then 10is the subgroup {. . . 30, 20, 10, 0, 10, 20 . . .}.(14.9) Definition A group G is cyclic if there exists x G such thatG = x.

For example, for any positive integer n, the additive group Zn is cyclic.Recall (not yet mentioned in this course) that a prime number p is an integer 2 whose only

divisors are 1 and p.

(14.10) Corollary Every group of prime order is cyclic.

Proof. If

|G

|is prime, choose any x

= e in G. The subgroup

x

generated by x is nontrivial, so

|x| > 1,and |x| divides |G|, so |x| = |G| and x = G. Q.E.D.

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15 Additive subgroups ofZ

(15.1) Definition For groups, the adjectives abelian and commutative are synonymous, that is,

an abelian group is a commutative group.

Usually one uses + to denote the group operation in an abelian group, the identity is called 0,andx is the inverse of x.(15.2) Definition In a group, xn means (x1)n.

Recall that in a group G, the subgroup generated by the elements a , b , c . . . is denoted

a , b , c, . . ..It consists of all elements which can be formed as products of powers (positive and negative) of

a , b , c, . . .If the group is abelian then it consists of elements which can be formed as sums of multiples ofa , b , c, . . ., i.e.,

a , b , c, . . . = {ra + sb + tc + . . . : r, s, t , . . . Z}(15.3) Theorem Every subgroup H ofZ is generated by a unique nonnegative integer n; thus theonly subgroups ofZ are of the form n for some unique nonnegative integern.

Explicitly, n = 0 ifH = {0}, otherwise n is the smallest positive integer in H.Proof. The trivial subgroup {0} contains only 0 and is generated by 0 and no other integer.Let H be any nontrivial subgroup. H contains some nonzero element x; if x < 0 then x H

also; so H contains a positive integer. By the least integer principle H contains a smallest positive

integer n (as opposed to a smallest nonnegative integer, which would be 0).Let m be any other element ofH. H contains all linear combinations

rm + sn, r, s Zand in particular

n Hand H contains

m mod n

which has the form m qn.Since 0

m mod n < n and n is the smallest positive integer in H, m mod n is zero. There-

foreH n

so H = n.It remains to show that if k and n are distinct positive integers then k = n. Equivalently, ifk

and n are positive integers and k = n then k = n. This is easy, becausek n

n|k, andn k k

|n.

But it can be shown that if two positive integers divide each other, then they are equal.

Example. A little experimentation will show that 12, 8 = 4.

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16 Greatest common divisor

Recall that when a, b belong to an Abelian group, in particular, a, b Z,

a, b = {ra + sb : r, s Z}.

(16.1) Definition Given a, b Z, not both zero, the greatest common divisor (or highest commonfactor) gcd(a, b) is the unique positive generator of the subgroup a, b ofZ.

(16.2) Lemma When a and b are integers, not both zero, gcd(a, b) is a positive integer and thereexist integers r ands such thatgcd(a, b) = ra + sb.

Proof. Since they are not both zero, a, b is nontrivial, so it is generated by a unique positiveinteger, i.e., gcd(a, b).

By definition, gcd(a, b) = a, b, so gcd(a, b) is of the form ra + sb for some integers r and s.Q.E.D.

(16.3) Corollary gcd(a, b) is the largest positive integer which divides both a andb.

Proof. Let g = gcd(a, b). If d is a positive integer dividing both a and b, then it divides allexpressions ra + sb, r, s Z, so it divides g.

But a g and b g, so g divides both a and b: it is the greatest common divisor. Q.E.D.The definition of gcd gives no way of computing it. There is an efficient way, based on the

following lemma which will not be proved.

(16.4) Lemma Ifn = 0 then gcd(m, n) = gcd(n, m mod n).

(16.5) Euclids gcd algorithm. To calculate gcd(m, n) (assuming m n > 0), generate a sequence

d1, d2, d3, . . .

where

d1 = m, d2 = n, and ifdj = 0,dj+1 = dr1 mod dj.

For example, to compute gcd(91, 35):

91, 35, 91 mod 35 = 21, 35 mod 21 = 14, 21 mod 14 = 7, 14 mod 7 = 0

When dr+1 reaches 0, dj is the gcd.This can be extended to compute constants r and s such that 91r + 35s = 7.We develop three other sequences, rj, sj, and qj.

r0 = 1, s0 = 0, r1 = 0, s0 = 1

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Proof. Let

gcd(a, b) = ra + sb = 1, gcd(a, c) = ta + uc = 1.

Multiplying,

(ra + sb)(ta + uc) = (rat + sbt + ruc)a + (su)bc = 1.

Therefore the gcd ofa and b, which divides both, divides 1, and must equal 1. Q.E.D.

(17.3) Corollary For any n 2, letZn be the set of integers in {0, . . . , n 1} which are relativelyprime to n. Then Zn is a commutative group under multiplication (mod n).

Proof. Since multiplication is associative and commutative, and 1 is a multiplicative identity(modulo n), and 1 is relatively prime to n, Zn is a commutative monoid.If a and b are relatively prime to n then so is ab. Let c = (ab mod n). As we have seen before

(Lemma 16.4) gcd(n,ab) = gcd(n, c). Thus c Zn, so Zn is closed under multiplication.Again, for any a Zn choose r, s so ra + sn = 1. Then ra 1(mod n), so ramod n = 1 and

r is the multiplicative inverse of a in Zn. Therefore Zn is a group. Q.E.D.

For example, Z9 = {1, 2, 4, 5, 7, 8}. Its Cayley table is1 2 4 5 7 8

1 1 2 4 5 7 82 2 4 8 1 5 7

4 4 8 7 2 1 55 5 1 2 7 8 47 7 5 1 8 4 28 8 7 5 4 2 1

(17.4) Definition A prime is an integerp > 1 whose only factors are 1 andp.

(17.5) Lemma Ifp is prime then Zp = {1, . . . , p 1}.Proof. In other words, if 1 a p 1, then gcd(a, p) = 1. But gcd(a, p) is a positive integer

dividing p, and gcd(a, p) = p, so gcd(a, p) = 1. Q.E.D.

(17.6) Corollary (Fermats Little Theorem.) If p is prime and x is an integer not divisible by p,then xp1 1 is divisible by p.

Proof. Given x, let y = (x mod p): y = 0, so y Zp. Now the (multiplicative) subgroup ygenerated by y is cyclic of order k, say: yk = 1. But k = |y| divides |Zp| by Lagranges theorem:i.e., k divides p 1, so yp1 = 1 in Zp, i.e., (yp1 mod p) = 1. But xp1

pyp1, so xp1

p1.

Remark. For a positive integer n, Eulers totient function

(n) = |{r : 1 r < n gcd(r, n) = 1}.Therefore (n) =

|Zn

|. Fermats little theorem can be generalised effortlessly:

gcd(n, x) = 1 = x(n)n

1.

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18 Direct products of groups

IfS1, S2 are semigroups (with operation symbols implicit) then there is an easy way to make S1 S2into a semigroup:

(a, b)(c, d) = (ac, bd)

If they are monoids, so is the product. If they are groups, so is the product. If additive notation is used (implicitly, they are abelian) then it is used in the product group as

well.

Example. Here is an addition table for Z2 Z2.

+ (0, 0) (0, 1) (1, 0) (1, 1)(0, 0) (0, 0) (0, 1) (1, 0) (1, 1)(0, 1) (0, 0) (0, 1) (1, 0) (1, 1)(1, 0) (1, 0) (1, 1) (0, 0) (0, 1)(1, 1) (1, 1) (1, 0) (0, 1) (0, 0)

19 Group isomorphism

Consider

Kleins 4-group, which has four elements 1, a , b , c, where a2 = b2 = c2 = 1 (so it is abelian),,and ab = c.

The group{1, (12)(34), (13)(24), (14)(23)} S3

Z2 Z2

Z4

i 00 1

C22

0 11 0

C22

{eni/2 : n

Z

}This leads to definition

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(19.1) Definition An isomorphism between two groups7 G, G is a bijective map f such that

(x, y G) (f(xy) = f(x)f(y)).This corresponds to an invertible linear map between vector spaces.

20 Normal subgroups and quotient groups

Zd can be defined in three ways (of course, the groups are isomorphic).

One can define the elements ofZd to be congruence classes [x]. There are d different congru-ence classes. One can define [x] + [y] as [x + y]. We know this is unambiguous.

One can define the elements to be {0, . . . , d1} with an operation : r s = (r + s) mod d. One can define the elements to be Z, with the usual operation, but a different equality relation,

namely, d

.

The question is: given an equivalence relation on a group G, when can it be used to define a newgroup, like Zd was defined from a congruence d on Z? In order to do this, we need the idea ofcongruence on a group (or semigroup or monoid).

(20.1) Definition A congruence on a group G is an equivalence relation such thatfor all x1, y1, x2, y2 G,

(x1 x2 y1 y2) = x1y1 x2y2.(20.2) Lemma If is a congruence on G, write H for[1]. Then H G.Proof.

1 1, 1 H Suppose x, y H. Thus x 1 and y 1. Therefore xy 1, so xy H. Suppose x H. Then x 1. Then x1x x11, so x1 H.

(20.3) Lemma If is a congruence on G andH = [1], then the equivalence classes (congruenceclasses) of are the left cosets of H, and also the right cosets left and right cosets are the same.Proof.

x y = x1y 1 = x1y Hx1y H = x1y 1 = x y

so coincides with the relation x1y H, whose equivalence classes are the left cosets ofH:[x] = xH

Similarly, coincides with the relation xy1 H, whose equivalence classes are the right cosetsofH:

[x] = Hx7or, for that matter, semigroups or monoids

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(20.4) Lemma Given groups H G, suppose that left and right cosets of H coincide. Then therelations x1y H and xy1 Hcoincide and are a congruence on G.

Proof. Certainly the relations coincide, since they produce the same partition.

Write for the relation.Suppose x1 y1 and x2 y2. Then y1 Hx1 and y2 x2H, so

y1y2 (Hx1)(x2H)

(refer to the quiz about products of subsets in a semigroup). It follows almost immediately from the

associative nature of products of subsets that

(Hx1)(x2H) = (Hx1x2)H = x1x2HH = x1x2H.

Therefore x1x2 y1y2 and is a congruence on G.

(20.5) Lemma Suppose H G. Then its left and right cosets coincide if and only if for every x G

xHx1 = H

Proof.

xH = Hx = xHx1 = Hxx1 = HxHx1 = H = xH = xHx1x = Hx.

(20.6) Definition A subgroup H ofG is a normal subgroup if

(x G) (xHx1 = H).

There is a special notation for normal subgroups:

HG.

(20.7) Lemma IfG is abelian, then every subgroup is a normal subgroup. (Trivial.)

(20.8) Lemma An Sn.

Proof. Given an even permutation and any permutation , 1 is even by the parity theorem.Q.E.D.

Sn/An is (isomorphic to) what?

Z/d is isomorphic to Zd.

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21 Group homomorphisms and the first isomorphism theorem

(21.1) Definition A homomorphism of groups G, G (the group operations are left implicit) is a maph : G G such that for all x, y G,

h(xy) = h(x)h(y).

Thus an isomorphism is a bijective homomorphism.

(21.2) Lemma Ifh : G G is a homomorphism then h(G) G.

Proof. (i) h(1) = h(12) = (h(1))2, and it follows that h(1) is the identity in G.

(ii) h(x)h(y) = h(xy) h(G), so h(G) is closed under product.(iii) h(x)(h(x))1 = h(1), so h(G) is closed under inverse.

(21.3) Examples of homomorphisms.

Z Zn; x x mod n. Sn {1, 1}; 1 if is an even permutation, 1 if is odd. Let G be a group, x an element ofG. Then the map Z G; k xk is a homomorphism.

IfHG, then h : G

G/H; x

[x] (i.e., x

xH), is a homomorphism.

From Z2 (under addition mod 2) to {1, 1} (under multiplication): 0 1, 1 1.

(21.4) Definition Leth : G G be a group homomorphism. The kernel ofh is the set of elementsofG mapped to the identity e ofG:

kernel(h) = h1(e) = {x G : h(x) = e}

(21.5) Lemma Leth : G G be a homomorphism. Then

ker(h)G

Proof. Let K = ker(h).(i) h(1) = 1, so 1 K.

(ii) Ifh(x) = h(y) = 1 then h(xy), so K is closed under product.(iii) Ifh(x) = 1, then (h(x))1 = h(x1) = 1, so K is closed under inverse.

(iv) Given x G, y K,

h(xyx1) = h(x)1h(x))1 = 1,

and y

K is arbitrary, so xKx1

K, for all x

G.

Since x1Kx K, xx1Kxx1 xKx1.Thus xKx1 = K for all x G: KG.

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In this case k = |Zk| = |x| = |x|.First, x

k

= h(k) = h(0) = 1, so x

k

= 1.Second, if xn = 1, then n ker h, so n k, so k|n. Therefore the order of x is the smallestpositive integer k such that xk = 1.

Application. Chinese Remainder Theorem. Recall

gcd(a, b) = 1 gcd(a, c) = 1 = gcd(a,bc) = 1Again

gcd(a, b) = 1 a|n b|n = ab|nProof. Let n = ax = by. Choose r, s so that

ra + sb = 1

Then

rax + sbx = x = rby + sbx

so b divides x and ab divides ax = n.

(21.9) Corollary Let k > 0 and n1 > 1, . . . , nk > 1,8 be pairwise prime integers, and let N =

n1n2 . . . nk. Then for any integern,

ni|n(1 i k) = N|n

Proof. Obvious ifk = 1. Otherwise let a = n1 and b = n2n3 nk. By an inductive argument,since gcd(a, nj ) = 1, 1 j k, gcd(a, b) = 1. Then

ni|n(2 i k) = b|nso

ni|n(1 i k) = (a|n b|n) = ab = N|n.(21.10) Theorem (Chinese Remainder Theorem). If n1 > 1, . . . , nk > 1 are pairwise primeintegers, then for any vector

i1, . . . , ik

where 0 ij nj 1, 1 j k, there exists an integern, 0 n < N, such thatn mod nj = ij , 1 j k

Proof. Let

P = Zn1 . . . Znkh : Z P

n (n mod n1, . . . , n mod nk)The map h is a homomorphism, and its kernel is

N

, so there is an isomorphism : ZN

h(Z).

But |ZN| = N = |Zn1| |Znk| = |P|, so h(Z) = P and ZN = P.8Occurrences of1 could be accommodated, but are of no interest.

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22 Group of the cube, again

7(1,1,1)

1(1,1,1)

(1,1,1)

4 (1,1,1)

3(1,1,1)2

(1,1,1)

5(1,1,1)

6(1,1,1)

8

There are 8 vertices on the cube; a vertex must go to another and the 3 edges incident must go to

edges incident to the other. That gives a bound of 8

6 = 48 on the number of symmetries.

There are 4 axes passing through diagonally opposite pairs of vertices; each accounts for 2proper rotations, or 8.

There are 6 axes passing through the centres of diagonally opposite pairs of edges: each gives1 more proper rotation, or 6.

There are 3 axes passing through the centres of opposite faces, each giving 3 proper rotations,or 9.

Together with the identity we get 24 rotations. Reflection in the centre, I, is orientation-reversing and commutes with everything: it gives

another 24 symmetries.

Therefore the group has order 48.Let K denote the group of symmetries of the cube, so K is a subgroup of O(3), and can be

identified with a group of matrices.

There is a map

K SO(3) : S (det S) SSince ((det S)S)(det T)T = (det ST)ST, this map is a homomorphism. Let

R = K SO(3)the subgroup of 24 rotations; we have a homomorphism K R. Also, a homomorphism

K R {1} : S ((det, S)S, det, S)The kernel consists of all S such that (det S)S = I and det S = 1. That is, S = I and

det S = 1: S = I. The kernel is trivial, which implies (by the isomorphism theorem) that the rangeof the map is isomorphic to K. But the codomain, like K itself, has order 48, so K is isomorphic toR {1}.

Whenever v goes to v under a symmetry, the vertex diagonally opposite v goes to that opposite

v

. That is, every symmetry S induces a bijective map from the set

{1, 5}, {2, 6}, {3, 7}, {4, 8}.

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to itself.

If we label these diagonal pairs as 1, 2, 3, 4, respectively, every symmetry Sinduces a permutationS in S4. S(i) is the unique a such that S(i) {a, a + 4}.

(22.1) Lemma This is a homomorphism from K to S4.

Proof. Suppose S(i) = j and T(j) = k. S(i) {j, j + 4} and T(j) {k, k + 4}.We need to show

T S(i) = k.

It is easiest to break into cases.

Case: S(i) = j and T(j) = k. Then T S(i) = k and T S(i) = k.

Case: S(i) = j and T(j) = k + 4. Then T S(i) = k + 4 and T S(i) = k. Case: S(i) = j + 4 and T(j) = k, so T(j + 4) = k + 4. Then T S(i) = k + 4, so T S(i) = k Case: S(i) = j + 4 and T(j) = k + 4, so T(j + 4) = k. Then T S(i) = k, so T S(i) = k.Since R K, we also get a homomorphism from R into S4. Its kernel consists of rotations S

which fix the four diagonal axes: only I does that. Thus we have an injective homomorphism from Rwhose range has the same order, 24: R = S4. Obviously, {1} (under multiplication) is isomorphicto Z2 (under addition). Therefore

K = S4 Z2.

23 SO(3), SU(2), quaternions, and S3

In this section we show that conjugation by unit quaternions on real vectors has the effect of rotating

them through twice the angle.

23.1 SO(2) and SO(3).

O(3) is the group of distance-preserving (linear) maps on R3. SO(3), its special subgroup consistsof those elements with determinant 1.

Equivalently, O(3) is the group of orthogonal 3 3 matrices and SO(3) is the subgroup withdeterminant 1.

(23.1) Lemma SO(2) consists of all rotations ofR2 aroundO.

Proof. A matrix a cb d

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is in SO(2) if and only if

a2 + b2 = c2 + d2 = 1

ac + bd = 0

ad bc = 1Since a2 + b2 = 1, a = cos and b = sin for a unique angle . Likewise, c = sin and d = cos for a unique . Since ac + bd = 0,

sin( + ) = 0

so = or = , i.e., c = sin and d = cos , or c = sin and d = cos . Sincead bc = 1, the first and not the second case applies and the matrix is

cos sin sin cos

a rotation matrix.

(23.2) Lemma SO(3) consists of all rotations ofR3 (where the axis of rotation passes through O).

Proof. Fix A SO(3). Since the dimension is odd, there exists a real eigenvalue, so we canwrite the eigenvalues as

1, 2, 3

where 3 is real and the others are either real or complex conjugate. Since A is distance-preserving,all eigenvalues have absolute value 1. If 1 is complex, then |2|23 = 1, so 3 = 1. Suppose theyare all real: then they are all 1, and the product is 1, so 1 or 3 equal 1.

In any case, A has an eigenvector with eigenvalue 1: without loss of generality, 3 = 1. Choosea unit eigenvector X3 with eigenvalue 1. A preserves distances, and also preserves scalar products:(AX)T(AY) = XTY. Therefore A maps the plane

{X : XTX3 = 0}to itself. The restriction to this plane gives a length-preserving map with determinant 12 = 1,

9 so

it is a rotation within the plane through angle , say, and A is a rotation ofR3 through the angle

around the axis OX3.

23.2 Quaternions and SU(2)

U(2) is the group of distance-preserving (linear) maps on C2, and SU(2) is the subgroup of mapswith determinant 1. Distance in C2 is

(z1, z2) =

|z1|2 + |z2|2.It is associated with a scalar product, but the scalar product is linear in one variable and conjugate

linear in the other.

9This can be justified, and there are better ways to show that the determinant of the restricted map is 1. We shall leave

this as a gap in the proof.

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(23.3) Definition LetA be a complex matrix. Its complex adjoint A is the complex conjugate of its

transpose. That is, ifA = [aij ], then the (i, j) entry in A

is

aji .

The scalar product of two column vectors X andY (with complex coefficients) is

XY i.e., (X)Y

A square matrix A is unitary ifA1 = A

(i.e., AA = I).

This corresponds to orthogonal real matrices.

The upshot is that the matrix of such that

A1 = A

where A is the complex adjointofA, meaning the complex conjugate ofAT.Recall that quaternions are formal linear combinations

a + bi + cj + dk

where i2 = j2 = k2 = ijk =

1. They can be encoded as complex matrices under the mapping

1

1 00 1

, i

i 00 i

, j

0 1

1 0

, k

0 ii 0

(i has been used ambiguously to denote a quaternion and an imaginary number). Then

a + bi + cj + dk

a + ib c + idc + id a ib

(23.3)

(23.4) Lemma Unit quaternions (under this encoding) coincide with SU(2).

Partial proof. (Proof omits some cases). The matrix (23.3) has determinant

a2 + b2 + c2 + d2

so for unit quaternions, the determinant is 1. Inverting the matrix in the usual way (the adjointformula; the determinant is 1):

A =

a + ib c + id

c + id a ib

, A1 =

a ib c idc id a + ib

.

But this last is the (complex) adjoint A ofA, so the matrix is in SO(2).We need to prove the converse: a matrix in SU(2) represents a unit quaternion. Such a matrix is

z1 z2z3 z4

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(the notation has been chosen to fit in with the quaternion encoding). The matrix is unitary, meaningz1 z2

z3 z4

z1 z3z2 z4

=

1 00 1

,

and the determinant is 1:z1z4 + z2z3 = 1.

Suppose z1 = 0 and z2 = 0 (there are a few other cases). Taking the (1, 2) entry in the product,

z1z3 + z2z4 = 0z3 + z2z4/z1 = 0

Substitute z2z4/z1 for z3 in

z1z4 + z2z3 = 1

z1z4 + z2z2z4/z1 = 1

Now, the (1, 1) entry in the product gives

|z1|2 + |z2|2 = 1

so we reach the conclusion below

|z1|2z4 + |z2|2z4 = z1z4 = z1

z4 = z1.

Again

|z1|2 + z2z3 = 1so z2z3 is real, which means that z3 is a real multiple ofz2, say z3 = tz2. Then

z1 z2

tz2 z1

has determinant |z1|2 + t|z2|2 = 1. We have assumed z2 = 0, and |z1|2 + |z2|2 = 1, so t = 1, and thematrix has the form

z1 z2z2 z1

with |z1|2 + |z2|2 = 1. This encodes a unit quaternion.

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23.4 Conjugation by unit quaternions

Quaternions are composed of a 1-dimensional real part and a 3-dimensional imaginary or spatialpart. Hence

a + x

is a natural way to write it. For multiplication,

(a + x)(b + y) = ab x y + ay + bx + x y

The (quaternionic) conjugate ofa + x isa x.

Multiply a quaternion by its conjugate:

(a + x)(a x) = a2 (x2) + xa ax + x x = a2 + x2.

In particular, ifa + x is a unit quaternion, i.e,

a + x = 1,

then its conjugate is its (multiplicative) inverse.

A unit quaternion can be represented as

cos + v sin

where v is a unit vector in R3. Let x be a vector in R3, treated as a quaternion with zero real part.Finally we will show that conjugation of real vectors by a unit quaternion

cos + v sin

is rotation around Ov through an angle 2.To reduce clutter we write c for cos and s for sin .First,

(c + sv)x =

sv x + cx + sv x

Then multiply by c sv.

(sv x + cx + sv x)(c sv) =scv x + csx v s2(v x) v

+s2(v x)v + c2x + cs(v x)cs(x v) s2(v x) v

The real part vanishes because (v

x)v = 0. Referring to the calculations leading to Theorem 23.5,

x = w + y

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where w = v x)v, y is perpendicular to v, z = v y, and and y = z v. We can rewrite the aboveexpansion

s2 w + c2x + csz + csz s2y.

Substituting w + y for x

(s2 + c2) w + c2y + csz + csz s2y =w + (c2 s2)y + 2csz =

w + cos y + sin z.

But this is just the formula 23.3 preceding Theorem 23.5. Therefore

(23.6) Theorem Conjugation of a real vectorx by a unit quaternion cos + sin v is the same asrotating x through the angle

= 2

around the axis Ov.

(23.7) Corollary There is a homomorphism h from SO(2) onto SO(3) where h(U)X is defined asfollows.

Interpeting X as a quaternion with zero real part, convert X to a complex matrix MX underthe encoding of quaternions.

Compute U MX U. This is MY for a real vectorY. Y = h(U)X, an image ofX under a rotation h(U).The homomorphism is surjective, and its kernel is {I}.

24 Group actions

(24.1) Definition Given a setS, Aut(S) denotes the group of bijections from S onto S.

(24.2) Definition A left action of a group G on a set S is an injective homomorphism of G intoAut(S), the group of bijective maps from S to S.

Variant: there is a map

m : G S SWe use gx as an abbreviation for m(g, x)

such that

gx = gy x = y (g1g2)x = g1(g2x)

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1x = x

((x S) gx = x) = g = 1.(24.3) Definition For any x S, the fixing group Gx is

{g G : gx = x}.(24.4) Lemma Gx G.

Proof.

1 Gxg1, g2 Gx = g2x = x, so g1g2x = g2x = x, so g1g2 Gx.g Gx = gx = x = g1gx = g1x = x = g1x, i.e., g1 Gx.(24.5) Definition The Orbit Ox of any x S is

{gx : g G}Examples.

The map (, i) (i) is a left action ofSn on {1, . . . , n}. The orbit of any i is all of{1, . . . , n}.(One says in this case the Sn acts transitively on {1, . . . , n}). For any i, the group fixing i is trivial.

For any Sn, the same map as above is a left action of Sn on {1, . . . , n}. The orbitsmatch the disjoint cycle decomposition of . For any i, let k be the size of the orbit Oi. Then thegroup fixing i is

{r : k|r}.(24.6) Definition S2 is the unit sphere in R3. S3 corresponds to the unit quaternions.

O(3) and SO(3) act on S2 since every distance-preserving linear map sends S2 bijectively ontoitself. The orbit of any point is all ofS2: the actions are transitive. The fixing subgroups are trivial.

Let S1 (parametrised by angle) act on S2 so that m(, X) rotates X through angle anticlock-

wise around the z-axis. Orbits are lines of longitude. The fixing subgroups are trivial except for thenorth and south poles where they coincide with S1.

(24.7) Lemma Let G be a group acting (on the left) on a set S and let x S. Then there is abijection between Ox and the left cosets of Gx.

Proof. There is a map

f : G Ox; g gx.The relation f(g) = f(g) is an equivalence relation on G. Note

f(g) = f(g

) gx = gx g1gx = x g1g Gx g gGx

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(24.8) Corollary IfG is finite then

|Ox| = |G

||Gx|Let G be a finite group, k a positive integer, k |G|. Conventionally,

G(k)

is the family of all k-element subsets ofG.G acts on G(k) as follows:

gS = {gx : x S}.

(24.9) Lemma Suppose G is finite. Under this action, for any S G(k)

, |GS| k.Proof. Let S = {a1, . . . , ak}. If g GS then ga1 = ai for some i aand g = a11 ai. Thus

|GS| k.

25 A Sylow theorem

This section is about the existence of p-subgroups of a finite group G. It will be shown that if pk isthe highest power ofp dividing the group order |G|, then G has a subgroup of order pk.

In order to prove the Sylow Theorem below, we consider a set S consisting of all pk-element

subsets ofG:

S = {P G : |P| = pk}and the following left action on S:

g(P) = {gx : x P}.First,

(25.1) Lemma Ifn is an integer andp is prime andpk is the highest power ofp dividing n, then

n

pk

is not divisible by p.

Proof. Write n = mpk where gcd(p, m) = 1. Use induction on k. In the case k = 0,m

1

= m

is not divisible by p. Assume the result for k.mpk+1

pk+1

=

mpk+1(mpk+1 1)( )((m 1)pk+1 + 1)pk+1(pk+1 1)( )(2)(1)

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We are only interested in those terms which are divisible by p, so we separate the terms

mpk+1(mpk+1 p)( )((m 1)pk+1 +p)pk+1(pk+1 p)( )(p)

But we can divide each term by p:

mpk(mpk 1)( )((m 1)pk + 1)pk(pk 1)( )(2)(1)

This is mpk

pk

and, by induction, is not divisible by p.

(25.2) Theorem (a Sylow theorem). Let G be a finite group, let p be prime dividing |G|, pk thehighest power ofp dividing |G|. Then G has a subgroup of orderpk.

Proof. We consider the set S and left action g introduced above. The cardinality ofS is

|S| =

n

pk

and by the above lemma, |S| is not divisible by p. The orbits form a partition of S, so there exists aP such that |orbit(P)| is not divisible by p.

This implies that p does not divide

|G

|/

|GP

|, so

|GP

|is divisible by pk.

But according to Lemma 24.9, |GP| pk. Therefore |GP| = pk, and GP is the required subgroup.Q.E.D.

Application. Show that there are just two groups of order 6.Solution. Let G be a group of order 6. We must show that G is isomorphic to Z3 Z2 or to S3.

By the Sylow theorem, G contains subgroups of prime orders 3 and 2: these are cyclic, hence thereare elements a and b of orders 3 and 2 respectively.

a b = {1}since a and a2 both have order 3 and b has order 2.

Therefore

a

b

=

a

and

a abhas cardinality 6 and coincides with G. Thus

G = {1, a , a2,b,ab,ab}We can complete part of the Cayley table for G.

1 a a2 b ab a2b1 1 a a2 b ab a2ba a a2 1 ab a2b ba2 a2 1 a a2b b ab

b b 1ab ab aa2b a2b a2

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Exercise. For any finite group F and any subgroup H ofF, if |F| = 2|H| then H F.

Therefore aG, so left and right cosets are the same.

{b,ab,a2b} = {b,ba,ba2}Therefore either

ab = ba

in which case G is commutative and isomorphic to Z3 Z2, or

ab = ba2.

In this case,

a2b = aba2 = ba4 = ba, bab = a2b2 = a2, ba2b = ab2 = a,

1 a a2 b ab a2b1 1 a a2 b ab a2ba a a2 1 ab a2b ba2 a2 1 a a2b b abb b a2b ab 1 a2 a

ab aa2b a2

Now the fifth and sixth rows are easily completed by multiplying the fourth row on the left by a anda2.

1 a a2 b ab a2b1 1 a a2 b ab a2ba a a2 1 ab a2b b

a2 a2 1 a a2b b abb b a2b ab 1 a2 a

ab ab b a2b a 1 a2

a2

b a2

b ab b a2

a 1

This is the Cayley table for S3. (Put another way, the map a (123) and b (12) extends to anisomorphism with S3).

26 Classification of finite abelian groups

(26.1) Theorem Letn = pe11 pekk where pj are distinct primes and ej are positive integers. Thenevery abelian group of ordern is isomorphic to a direct product of cyclic groups of the form Zprj, andthe product is unique (up to re-ordering).

Let A be an abelian group of order n.

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26.1 Product of maximalp-groups

By Sylows Theorem, for 1 j k, A has a subgroup of order pejj : call it Aj .There is a map

A1 A2 . . . Ak A(x1, x2, . . . , xk) x1 + x2 + . . . + xk

which (since A is abelian) is easily seen to be a homomorphism (exercise).

(26.2) Lemma If Z is an abelian group andX, Y are finite subgro

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