This is a continuation from the previous post. Let *G* act on set *X*, but now we assume that both *G* and *X* are finite. Since *X* is a disjoint union of transitive *G*-sets, and each transitive *G*-set is isomorphic to *G*/*H* for some subgroup *H* ≤ *G*, it follows that we have:

.

Let’s consider the case where *H _{i}* =

*G*. The corresponding

*G*-sets are merely singleton sets since |

*G*/

*H*| = 1. These are the elements

_{i}*x*of

*X*for which

*gx*=

*x*for all

*g*in

*G*. We call such elements

**fixed points**of

*X*and denote them

*X*.

_{G}Now we can modify our equation to obtain:

,

where each *H _{i}* is a proper subgroup of

*G*. Let’s specialise even further.

Definition. Let p be prime. A finite group G≠{e} is called ap-groupif its order is a power of p. A subgroup which is a p-group is also called ap-subgroup.

Plugging this into the above equation, if *G* is a *p*-group, then each |*G*|/|*H _{i}*| is a positive power of

*p*and so is divisible by

*p*. We thus obtain:

.

This will be of great use to us.

## Sylow’s First Theorem

The three Sylow theorems are extremely powerful results regarding the structure of finite groups. Let’s start with the preliminary result.

Cauchy’s Theorem. If |G| is divisible by prime p, then G has a subgroup of order p.

**Proof**. Use the following steps.

- Let .
- Let
**Z**/*p*act on it by left-rotation: , where*m*= 0, 1, …,*p*-1. [ Check that even if*G*is not abelian. ] - Easy to count
*X*: the first*p*-1 elements determine the final one uniquely, so . - The set of fixed points
*X*_{Z/p}is precisely the set of (*x*, …,*x*) (*x*in*G*) such that*x*=^{p}*e*. *X*_{Z/p}is not empty since it contains (*e*, …,*e*).- Apply the above congruence: since
**Z**/*p*is a*p*-group, |*X*_{Z/p}| ≡ |*X*| ≡ 0 (mod*p*) so there’s a non-trivial element*x*of*G*such that*x*=^{p}*e*. ♦

The first Sylow theorem is a more general version of Cauchy’s theorem, but its proof requires Cauchy’s theorem to kickstart it.

First Sylow Theorem. If |G| has order divisible by and H<G is a subgroup of order then H is contained in a subgroup K≤G of order .

In short, by Cauchy’s theorem *G* has a subgroup *H*_{1} of order *p*. First Sylow theorem says this subgroup must be contained in another subgroup *H*_{1} < *G* of order *p*^{2}. Applying this theorem repeatedly gives a sequence where each .

**Proof**.

Case 1. If *H* is a normal subgroup of *G*, then *G*/*H* is a group of order divisible by *p*. Apply Cauchy’s theorem to obtain a subgroup *G*/*H* of order *p*, which corresponds to a group *H* < *K* ≤ *G* of order *p *× |*H*|. Done.

Case 2. If not, let’s try to “make it normal”. Hence, let . This is called the **normaliser** of *H* since it is the largest subgroup of *G* for which *H* is normal in. Thus, and it remains to show [*N*(*H*):*H*] is a multiple of *p*, after which we can replace *G* by *N*(*H*) and apply case 1. To show this,

- consider the action of
*H*on the set of left-cosets*X*=*G*/*H*by left multiplication ; - |
*G*/*H*| is divisible by*p*by assumption, so |*X*| is also divisible by_{H}*p*; - but
*X*is the set of_{H}*gH*for which*hgH*=*gH*for all*h*in*H*; i.e. the set of*gH*for which*gHg*^{-1}=*H*; in short,*X*=_{H }*N*(*H*)/*H*. ♦

## Second and Third Sylow Theorems

Let *G* be of order *p ^{r}m*, where

*m*is not divisible by prime

*p*. Thanks to the first Sylow theorem, we know that there’s at least one subgroup of order

*p*. We shall call this a

^{r}**Sylow**

**.**

*p*-subgroup

Second Sylow Theorem.Any two Sylow p-subgroups of G are conjugates, i.e. if , then there is such that .More generally, if P is a Sylow p-subgroup of G, and H is a p-subgroup of G, then we can find , .

**Proof.**

Let *H* act on the set *X* = *G*/*P* of left-cosets via left multiplication. Since *P* is Sylow *p*-subgroup, |*X*| is not a multiple of *p*. Thus, neither is |*X _{H}*|. In particular,

*X*is not empty so contains some

_{H}*gP*. Since this is fixed under action by

*H*, we have

*hgP*=

*gP*for all

*h*in

*H*, and so in particular. ♦

Third Sylow Theorem. Let be the number of Sylow p-subgroups of G. Then:

- divides |G| and
- .

**Proof**.

First assertion: let *P* be any Sylow *p*-subgroup of *G*. Then all Sylow *p*-subgroups are of the form , . Repetition occurs iff

.

So the distinct Sylow *p*-subgroups correspond to the left cosets *G*/*N*(*P*), i.e. , so it divides |*G*|.

Second assertion: let *P* act on the set *X* = *G*/*P* of left cosets (by left multiplication again). By the above congruence, we have:

.

- Clearly LHS = [
*G*:*P*]. - On the other hand, the set
*X*comprises of all_{P}*gP*such that*xgP*=*gP*for all*x*in*P*. This holds iff . But this precisely means that . Hence so RHS = [*N*(*P*):*P*]. - Since LHS is not divisible by
*p*, dividing by RHS gives which completes the proof. ♦