IB Groups, Rings and Modules

1Groups

1.7 Sylow theorems

We finally get to the big theorem of this part of the course.

Theorem (Sylow theorems). Let

be a finite group of order

· m

, with

prime and p - m. Then

(i) The set of Sylow p-subgroups of G, given by

Syl

(G) = {P ≤ G : |P | = p

is non-empty. In other words, G has a subgroup of order p

(ii) All elements of Syl

(G) are conjugate in G.

(iii)

The number of Sylow

-subgroups

|Syl

(

)

satisfies

≡

1 (

mod p

)

and n

| |G| (in fact n

| m, since p is not a factor of n

These are sometimes known as Sylow’s first/second/third theorem respec-

tively.

We will not prove this just yet. We first look at how we can apply this

theorem. We can use it without knowing how to prove it.

Lemma. If n

= 1, then the Sylow p-subgroup is normal in G.

Proof.

Let

be the unique Sylow

-subgroup, and let

g ∈ G

, and consider

−1

P g

. Since this is isomorphic to

, we must have

−1

P g|

, i.e. it is also

a Sylow

-subgroup. Since there is only one, we must have

−1

P g

. So

normal.

Corollary. Let

be a non-abelian simple group. Then

|G| |

for every prime

p such that p | |G|.

Proof.

The group

acts on

Syl

(

) by conjugation. So it gives a permutation

representation

G → Sym

(

Syl

(

))

∼

. We know

ker φ C G

. But

simple. So ker(φ) = {e} or G. We want to show it is not the whole of G.

If we had

ker

(

), then

−1

P g

for all

g ∈ G

. Hence

is a normal

subgroup. As

is simple, either

{e}

, or

. We know

cannot be

trivial since

p | |G|

. But if

, then

is a

-group, has a non-trivial center,

and hence G is not non-abelian simple. So we must have ker(φ) = {e}.

Then by the first isomorphism theorem, we know

∼

im φ ≤ S

. We have

proved the theorem without the divide-by-two part. To prove the whole result,

we need to show that in fact

(

)

≤ A

. Consider the following composition

of homomorphisms:

G S

{±1}.

φ sgn

If this is surjective, then

ker

(

sgn ◦φ

)

C G

has index 2 (since the index is the size

of the image), and is not the whole of

. This means

is not simple (the case

where |G| = C

is ruled out since it is abelian).

So the kernel must be the whole

, and

sgn ◦φ

is the trivial map. In other

words, sgn(φ(g)) = +1. So φ(g) ∈ A

. So in fact we have

∼

im(φ) ≤ A

So we get |G| |

Example. Suppose

|G|

= 1000. Then

|G|

is not simple. To show this, we need

to factorize 1000. We have

|G|

= 2

. We pick our favorite prime to be

= 5.

We know

∼

1 (

mod

5), and

= 8. The only number that satisfies this

is n

= 1. So the Sylow 5-subgroup is normal, and hence G is not normal.

Example. Let

|G|

= 132 = 2

11. We want to show this is not simple. So

for a contradiction suppose it is.

We start by looking at

= 11. We know

≡

1 (

mod

11). Also

12.

As G is simple, we must have n

= 12.

Now look at

= 3. We have

= 1 (

mod

3) and

44. The possible

values of n

are 4 and 22.

= 4, then the corollary says

|G| |

= 12, which is of course nonsense.

So n

= 22.

At this point, we count how many elements of each order there are. This is

particularly useful if

p | |G|

but

- |G|

, i.e. the Sylow

-subgroups have order

and hence are cyclic.

As all Sylow 11-subgroups are disjoint, apart from

{e}

, we know there are

(11

−

1) = 120 elements of order 11. We do the same thing with the Sylow

3-subgroups. We need 22

−

1) = 44 elements of order 3. But this is more

elements than the group has. This can’t happen. So G must be simple.

We now get to prove our big theorem. This involves some non-trivial amount

of trickery.

Proof of Sylow’s theorem. Let G be a finite group with |G| = p

m, and p - m.

(i)

We need to show that

Syl

(

)

∅

, i.e. we need to find some subgroup of

order p

. As always, we find something clever for G to act on. We let

Ω = {X subset of G : |X| = p

We let G act on Ω by

g ∗ {g

, g

, ··· , g

} = {gg

, gg

, ··· , gg

Let Σ ≤ Ω be an orbit.

We first note that if

, ··· , g

} ∈

Σ, then by the definition of an orbit,

for every g ∈ G,

−1

∗ {g

, ··· , g

} = {g, gg

−1

, ··· , gg

−1

} ∈ Σ.

The important thing is that this set contains

. So for each

, Σ contains

a set X which contains g. Since each set X has size p

, we must have

|Σ| ≥

|G|

= m.

Suppose

. Then the orbit-stabilizer theorem says the stabilizer

any

, ··· , g

} ∈

Σ has index

, hence

|H|

, and thus

H ∈ Syl

(

So we need to show that not every orbit Σ can have size

> m

. Again, by

the orbit-stabilizer, the size of any orbit divides the order of the group,

|G|

. So if

| > m

, then

p | |

. Suppose we can show that

p - |

Ω

Then not every orbit Σ can have size

> m

, since Ω is the disjoint union of

all the orbits, and thus we are done.

So we have to show p - |Ω|. This is just some basic counting. We have

|Ω| =



|G|







−1

j=0

m − j

− j

Now note that the largest power of

dividing

m −j

is the largest power

dividing

. Similarly, the largest power of

dividing

−j

is also the

largest power of

dividing

. So we have the same power of

on top and

bottom for each item in the product, and they cancel. So the result is not

divisible by p.

This proof is not straightforward. We first needed the clever idea of letting

act on Ω. But then if we are given this set, the obvious thing to do

would be to find something in Ω that is also a group. This is not what we

do. Instead, we find an orbit whose stabilizer is a Sylow p-subgroup.

(ii)

We instead prove something stronger: if

Q ≤ G

is a

-subgroup (i.e.

|Q|

, for

not necessarily

), and

P ≤ G

is a Sylow

-subgroup, then

there is a

g ∈ G

such that

−1

Qg ≤ P

. Applying this to the case where

is another Sylow

-subgroup says there is a

such that

−1

Qg ≤ P

, but

since g

−1

Qg has the same size as P , they must be equal.

We let Q act on the set of cosets of G/P via

q ∗ gP = qgP.

We know the orbits of this action have size dividing

|Q|

, so is either 1 or

divisible by

. But they can’t all be divisible by

, since

|G/P |

is coprime

. So at least one of them have size 1, say

{gP }

. In other words, for

every

q ∈ Q

, we have

qgP

. This means

−1

qg ∈ P

. This holds for

every element q ∈ Q. So we have found a g such that g

−1

Qg ≤ P .

(iii)

Finally, we need to show that

∼

1 (

mod p

) and

| |G|

, where

|Syl

(G)|.

The second part is easier — by Sylow’s second theorem, the action of

Syl

(

) by conjugation has one orbit. By the orbit-stabilizer theorem,

the size of the orbit, which is

|Syl

(

)

, divides

|G|

. This proves the

second part.

For the first part, let

P ∈ Syl

(

). Consider the action by conjugation

Syl

(

). Again by the orbit-stabilizer theorem, the orbits each

have size 1 or size divisible by

. But we know there is one orbit of size 1,

namely

{P }

itself. To show

|Syl

(

)

∼

1 (

mod p

), it is enough to

show there are no other orbits of size 1.

Suppose {Q} is an orbit of size 1. This means for every p ∈ P , we get

−1

Qp = Q.

In other words,

P ≤ N

(

). Now

(

) is itself a group, and we

can look at its Sylow

-subgroups. We know

Q ≤ N

(

)

≤ G

. So

| |N

(

)

| | p

. So

is the biggest power of

that divides

(

)

So Q is a Sylow p-subgroup of N

(Q).

Now we know

P ≤ N

(

) is also a Sylow

-subgroup of

(

). By Sylow’s

second theorem, they must be conjugate in

(

). But conjugating

anything in

by something in

(

) does nothing, by definition of

(

). So we must have

. So the only orbit of size 1 is

{P }

itself.

So done.

This is all the theories of groups we’ve got. In the remaining time, we will

look at some interesting examples of groups.

Example. Let

(

Z/p

), i.e. the set of invertible

n × n

matrices with

entries in

Z/p

, the integers modulo

. Here

is obviously a prime. When we do

rings later, we will study this properly.

First of all, we would like to know the size of this group. A matrix

A ∈

(

Z/p

) is the same as

linearly independent vectors in the vector space

(

Z/p

)

. We can just work out how many there are. This is not too difficult,

when you know how.

We can pick the first vector, which can be anything except zero. So there

are

−

1 ways of choosing the first vector. Next, we need to pick the second

vector. This can be anything that is not in the span of the first vector, and this

rules out

possibilities. So there are

− p

ways of choosing the second vector.

Continuing this chain of thought, we have

|GL

(Z/p)| = (p

− 1)(p

− p)(p

− p

) ···(p

− p

n−1

What is a Sylow

-subgroup of

(

Z/p

)? We first work out what the order of

this is. We can factorize that as

|GL

(Z/p)| = (1 · p · p

· ··· · p

n−1

)((p

− 1)(p

n−1

− 1) ···(p − 1)).

So the largest power of

that divides

|GL

(

Z/p

)

(

)

. Let’s find a subgroup

of size p

(

)

. We consider matrices of the form

U =

















1 ∗ ∗ ··· ∗

0 1 ∗ ··· ∗

0 0 1 ··· ∗

0 0 0 ··· 1







∈ GL

(Z/p)











Then we know

|U|

(

)

as each

∗

can be chosen to be anything in

Z/p

, and

there are





∗s.

Is the Sylow

-subgroup unique? No. We can take the lower triangular

matrices and get another Sylow p-subgroup.

Example. Let’s be less ambitious and consider GL

(Z/p). So

|G| = p(p

− 1)(p − 1) = p(p − 1)

(p + 1).

Let

be another prime number such that

` | p −

1. Suppose the largest power of

` that divides |G| is `

. Can we (explicitly) find a Sylow `-subgroup?

First, we want to find an element of order

. How is

p −

1 related to

(apart

from the obvious way)? We know that

(Z/p)

= {x ∈ Z/p : (∃y) xy ≡ 1 (mod p)}

∼

p−1

So as

` | p −

1, there is a subgroup

≤ C

p−1

∼

(

Z/p

)

. Then we immediately

know where to find a subgroup of order `

: we have

× C

≤ (Z/p)

∗

× (Z/p)

≤ GL

(Z/p),

where the final inclusion is the diagonal matrices, identifying

(a, b) ↔



a 0

0 b



So this is the Sylow `-subgroup.