8 Actions Group

8 L Definitions 8 Examples

how our has Up to study of largely groups

been about the groups themselves understanding

e How elements G have order K g many of

What are the subgroups of G

What are the quotients of G

Is G cyclic Abelian etc

Now we will turn our attention to how studying

a can describe the group G symmetries of

other X we see how sets Essentially will

a elements can permute the of X group

course we don't want G to Of

permute the elements of X arbitrarily

we somehow want the structure G of

in this reflected permutation

For instance how would like the you

ee to elements identity G permute the of

X If our definition is at all

reasonable it should leave everything

unchanged

be It would also reasonable to expect

that for EG X the g gz permuting by

should be the same as product g.ge

and then here permuting by gz by g

we read left like functions right to

Definition Let G be a and X be group

a set An action of G on X is a map

s h the Gx X X wi following properties

L x u e for all XEX

2 x gigs X for all g g c G g gz

and all see X

to We write GTX indicate that G

is on X acting

1 on Ex The group G Sn acts the set

X 142,3 n'T in the natural way

For TE Sm and ie X a i Oci

We'll that this is an action verify by

L f z checking properties

L e i e i i ie V for all X

2 Given a te Sn and EE X

P T.i 8 Tci

r Tc i

rot i roc E V

R acts on Ex2 The group G Gln the

I

set X R by matrix vector multiplication

c For A Glen and XER A x Ax

have We that

1 I x Ix all ne V for X

2 Given A Be Glen and KEIR

A B x A Bx

CAB x AB x V

Ex3 The G D acts on the group

set 3,4 which we can think 4 11,2 of

as the set vertices a of of square

1 2

4 3

Here we have that

Z z 1 1 v

3 3 4 4

So V 2 1

We could also think of X as the set of

the edges of square 1

4

23

In this case

I 1

V 2 7 4 2

3 43

So V 2 4 and I L

Ex 4 A G acts on itself left group by

multiplication Given a be a b ab G

an exercise is an As prove that this

action

Ex 5 A G acts on itself group by

t Given a be G a b aba conjugation

To see that this is an action note that

e a 1 eae a for all AEG

2 a b a Cbgb g 1 t a a Cbgb

ab ab ab g g

Thus this is indeed an action

It will sometimes be to view an helpful

action GTX in a different way

Suppose that G is a on group acting

a set X For G consider ge fixed

the map

X X 4g

x I se g

We can show that is a on Hg bijection

Indeed see X we have X given that

t.se a

4g g g g

x

gg

e ze R

so is To see that 4g surjective 4g

u is let X Then injective ye

Igcse a Ugly g g y

i a g g y g g

a

g g g g y

e se e Y

se

y

so is Thus is 4g injective

belongs to bijective Consequently 4g

the set

X is S f X f bijective

Exercise For set X the set any

as defined above is a under Sx group

Lcs composition Moreover X n the if

E Sx Sn

with this in mind consider once again

the action G on X of Define

E G Six

l 4g g

This OI is a between In map groups

it is a homomorphism fact group

Indeed let c We claim that g g G

o Ecg ga Ecg Ilga

That is we claim that

Ygga Vg Uga

a But of course for EX

a x Yg g gigs

x g gz

g Ygate

se as 04g 4g 4g g

we so Thus have 4g g Vge Yg

01g Eg That is I is g Iga

a homomorphism

Moral action GTX Every gives

rise to a homomorphism

G Sx

i s g 4g

where X X 4g

se I goose

this is to one knowing fact enough prove

the coolest results in the course of

Theorem 8 I Cayley's Theorem

G is isomorphic to a of Every group group

In G he permutations particular if

is a then G isomorphic to subgroup of Sh

Proof Let G act on itself by left

consider homomorphism multiplication and the

Of G described above Sq Let

the First Isomorphism Theorem By

im G I E E Koot Sq

is What turd If geker I then

01g is the of Thus 4g identity Sg

a a for all AEG a 4g g ga

cancellation e so Ker Ie By g

o is to a G isomorphic subgroup of Sq Krag

8.2 Orbits 8 Stabilizers

Given a action GAX and group

see an element G there are a few

natural one can ask questions

E Where in X can u be sent

ii elements fine u which of G

Thus we make the definitions following

a Definition Let Gnu X be group

action Given see X define

i the orbit se be the of to set

On lg.se gee

ii the stabilizer n to be the set of

stab G a se Ige g

Note that the orbit On is a subset of

X while the stabilizer Staba is a

subset G In of fact

Proposition 8.2 is an action If GTX

then for any see X Stab E G

Proof Exercise

Is a X No On subgroup of

In X is not even a general group

e C't Ex The group G feio Oer

acts on the set X c by multiplication

O e For 2 k e Z e'Oz The orbit

ZEC is of Oz Z Oz e Oz OER

WE E w Z

0 Note that stab eio eio 0 G

0 z and 2 to Stabz e ei0z 42 for

usual action Ex Consider the of Sin

on on X 11,2 in I Nci

ie we can send Given X in to any any

c r Thus X ij Oi X g using

Stab Oci i By definition lot Sn

I Sn t

n Note that Oi n Stabi t

Ex Let be the all G group of

rotations cube of the

We will consider the action

of G on the cube's vertices edges and faces g

L X vertices of the cube

Given vertex VE X we can move V

to other vertex a rotation using any

Or 8

The rotations that leave only V unchanged

e are and the rotations about the line

V and through its vertex opposite

i Stab 3

z edges the cube X of

Given E E X we can move E edge

to other a rotation edge using any

i OE 12

The rotations that leave E only unchanged

are e and rotation the about the line

and through E its opposite edge

E

or

a Stabe 2

3 faces the cube X of

Given Fe X we can move face

to other face a rotation any using

i Of 6

The only rotations that leave F unchanged

are e and the rotations about the line

F and its through opposite face

F

u

i Stab 4

Notice that in all cases Staba On 24

It turns out that this value represents

the order the G in of group and fact this occurs for every action GAX

Theorem 8.3 Orbit Stabilizer Let GAX be

For see a group action every X

G Stab On

In particular if G then

G Stab On

Proof Define 4 G stab On by

41g Staba g se We claim that 4 is bijective hence G Staba On

First we have that for all g gre G

g Stabs ga Staba gig C Stabs gig se se

g se gz K 41g Uga

Thus 4 is well defined 8 injective

Moreover given ye On we can write y g x for some gear Then

4 gstaba g x y so 9 is surjective

i 4 is bijective as claimed The final claim for finite groups follows

from Lagrange's Theorem agama

The Orbit Stabilizer Theorem can reveal lots

about a group action GAX or about the

group G itself

Ex What are the possible group actions of

G 25 on X 142,33

Well a we have given any X that

25 Staba On so 5 Staba On

Hence On L or 5 But since On E3 it must be that On L for all n That is

see g a r for all GE 25 and all X

The only action is the trivial action

Ex How does many rotational symmetries

a soccer ball have

all such Let G be the group of symmetries and let G act on the set X of all black pentagonal faces of the soccer ball Fix any face FEX This face can be

rotated to other such so any face

OF 12 Moreover the only rotations

that leave F unchanged are the 5

rotations about the axis through F 2

5 Consequently Stab By the Orbit

Stabilizer Theorem G Stab OF

5 12

GO Further Applications Burnside's Lemma

We will now see how actions can be group

to used solve some neat counting problems

Ex 1 How different can we make many ways

2 2 chess board black and white a using

squares

Here are some examples

etc wi em I 2 3 4 5

Hold on some of these boards are really the same

For instance 2 can be rotated to 3 So We should consider two boards to be the same if one can be rotated into the other That is our be restated as problem may follows

If G is the group of rotations of a

is 16 square and X the set of all 24

2 2 chess boards how orbits does many

the action GAX have

Burnside's Lemma us a to count gives way these orbits efficiently First We'll need

the following proposition

Let be a Proposition 8.4 GAX group action The orbits of the action partition

X That is

a X On

b if x yell then Ose Oy or Oan Oy 0

Proof Assignment 5

Lemma 8.5 Burnside Let G be a finite

on a set X If N group acting finite is the number of orbits then

N Fixlg g g where for geG Fix g IneX goose se Proof Let n be the number of pairs

a c such that se se g Gx X g First note that for a fixed geG the number

such is so of pairs gin Fix g

n I Fix g gEG

Also note that for fixed a X the number

of such pairs g n is staba so

n stab G l Go I k f sec On

But we have so for any rye Or Oy On

I t t t 1 yeon Oy On On On On times So I N number of orbits reX On

We conclude that I Fixg n G N gea

So as claimed N Fix g gaza g g a

We can now attempt to solve our chess board problem We wish to determine the number of orbits of all boards under the group G of rotations of a square By

Burnside we must find FixLg for all rotations gear Note that

G e Rao Riso 12270 g e Rao Riso 12220 2 22 2 Fix g 24

Thus the number of distinct boards orbits

is N Fix Ig Egfr g

L 16 2 4 2 6 4

Boat atheroma

How necklaces Ex many 6 bead can be

and 3 made using 3 black beads white beads Solution We can choose the location of

3 beads Es 20 the black in ways

and the beads must be white remaining

So let X be the set of these 20

possible necklaces and let G D6 be

the a We symmetry group of hexagon

consider 2 necklaces in X to be the

same to the same orbit if they belong

under the action Gnr X What are the symmetries in G

Rotations V

C R R2 Rs R Rs

F Fy Fr Flips

Fsf Fl Fz Fz Fy E5 Foo

F6

We now Fix each compute g for gEG

g e R R R R Rs F Fa F F Fo Foo

20 O 2 O 2 O O O O I 22 22 Fix g By Burnside's Lemma the number of orbits i.e the number of necklaces is

Zoot 2 2 22 2722 1g Fix g plz g

L 36 3 12

They are

1

can one label the Ex How many ways

sides a G sided die the of using

each of the numbers 1 G exactly once Solution There are 6 ways to put the numbers on but some of the dice

be the rotation may same after

Let X be the set of all 6 720

dice and possible let G be the group of rotations of the cube We consider two dice to be the same if they

are in the same orbit of the action

GAX Note that the only group element that

the cube is e fixes every face of

Since all faces are marked differently

we have that Fix g o Vgte and

Fix e X 720

By Burnside's Lemma the number of orbits

i.e the number of dice is

Fix 720 30 1g Egg g 214

Ex How can one many ways paint the edges of a tetrahedron red blue or green 6 Solution There are 3 different ways but to paint the edges some of these

be the same rotation colorings may after

Let G be the group of all rotational symmetries of the tetrahedron and X be

6 the set of all 3 possible colourings

Since we must compute Fix g for each geG we should first try to understand what the

rotations in G look like Note that G acts on g the faces of the

tetrahedron For a fixed

face F there are 3 rotations that fix

F F so Stab 3 We can send to any other face so OF 4 By Orbit Stabilizer

G 3 4 12 The 12 possible rotations are

as follows 1 identity e Fixle 3

8 rotations T about vertex and opposite face

Fix IT 3 J y 3 rotations 6 about opposite edges

Fixlov 3

1 8 We have 3

V r V g e T d

Fix 36 32 34 g

So N I 1.36 8.32 3.34 87 12

The Class Equation

a and Let G be finite group let G act

b on itself by conjugation a aba Let Og Og Og denote the disjoint orbits of the action that are not contained in 2 G Using Proposition 8.4 one can prove that

G 2 G t G stabg F 1 where Stabg at G agia g Iaea agi gia Cgi centralizer of gi

The equation

G 2 Ca t Z G Clg F 1

is called the class equation and it has many remarkable consequences

corollary 8.6 Let p be a prime

1 is a order If G group of p for

some K 1 then 2 G t let

2 G is a order then If group of p2

G is Abelian

The details are left to the assignment

Here is another amazing application

Theorem 8.7 Cauchy's Theorem

is If G a finite group and p is a prime that divides G then G contains an element of order p

Proof By induction assume that the

C result holds for groups of order G

i Case I p 2CG

By Cauchy's theorem in the Abelian case

2 G and hence G has an element of

order p

Case I pl 2 G

Let G act on itself by conjugation and

let Og Og Ogr be the distinct

orbits not contained in 2 G By the Class equation 2Ca G I G Ccgi

be and since p 2CG there must an integer k such that pl G Ugk

Since divides does not p G yet p divide G Ugk G it must be Ugk that p divides Ign Note that Ugh is a group and Ugn FG otherwise agk gka Fae G so get 2 G contradiction By induction CCgu and hence G contains an element of order p Bombard