For the statement of

Sylow's theorems see that writeup.
We will use group actions extensively so read the

group acting on set
writeup also. We begin with a lemma.

If a finite group *G* acts on a finite set *X*
then we write *X*^{G} for the set of points
fixed by all points of *G*. The points in *X*^{G}
form singleton orbits.

**Lemma** If *G* is a *p*-group then
*
|X*^{G}| is congruent to *|X|* modulo *p*

**Proof:** By the orbit-stabiliser lemma (see the
fixed point formula for a finite group acting on a finite set writeup)
we know that the size of an orbit, divides the order of the group, in
this case a power of the prime number *p*. Thus, an orbit
either has 1 element, or it is divisible by *p*. Thus modulo
*p* the only orbit sizes that show up are the singletons, which
are accounted for by *X*^{G}.

**Proof** of the theorem (this proof is not the original one that
Sylow gave but a modern version using group actions due to Wielandt).

First we show that there is a Sylow *p*-subgroup.
Write *|G|=n=p*^{k}s with *p* and *s* coprime.
If *k=0* then we are done trivially, so assume *k> 0*.

Let **T** be the set of all subsets of *G* that have
*p*^{k} elements. There are *n* choose *p*^{k}
such subsets. So
|**T**|= *(n/p*^{k})((n-1)/1)((n-2)/2)...((n-p^{k}+1)/p^{k}-1).
I claim that **|T|** is not divisible
by *p*. This is because whenever a factor in the numerator is divisible
by a power of *p* the corresponding factor in the denominator
is divisible by the same power of *p*.

Now *G* acts on **T** by left multiplication. This works because if
*g* in *G* and *T* in **T** then *gT* is also
a subset of *G* with *p*^{k} elements.
As |**T**| is not divisible by *p* it cannot be that every orbit for
this action has size a multiple of *p*. So let *X* be an
orbit with order coprime to *p*. Now choose *T* some element
of this orbit and let *S=*Stab(*T*), be the stabilser. Thus,

*S={g* in *G* such that *gT=T}*.

Note that by the orbit-stabiliser lemma again, we have that
*|S|* is divisible by *p*^{k}.
Choose *t* in *T* and suppose *g* is in *S*.
Thus, *gT=T* and so we have *gt* in *T*. Thus we have that
*g* is in *Tt*^{-1}. It follows that
*S* is a subset of *Tt*^{-1} and so *S* has at
most *p*^{k} elements. It follows that *S*
has exactly *p*^{k} elements and we have shown that
Sylow *p* subgroups exist.

Now let *H* be any *p*-subgroup we will show that *H*
is conjugate to a subgroup of *S*.
First note that *H* acts on the set of right cosets of *S* in
*G*. Here, for *h* in *H* it acts on
*aS* by *h.aS=(ha)S*. By Lagrange's theorem the number of
cosets (the index of *S* in *G*) is coprime to *p*.
Since *H* is a *p*-group the lemma above applies and we see that
the number of singleton orbits is not divisible by *p*. In particular
there exists *a* in *G* so that *HaS=aS*. Thus,
*a*^{-1}Ha.S is a subset of *S*, since *S*
contains the identity element, this shows that
*a*^{-1}Ha.S is a subset of *S*, as needed.

Note that in particular, if *H* is also a Sylow *p*-subgroup
then by counting elements it must equal *S*, so that all the
Sylow *p*-subgroups are conjugate.

Now consider the problem of how many conjugates there are.
Let *N* be the normalizer of *S* in *G*. That is

*N={g* in *G* such that *gSg*^{-1}=S}.

Note that *N* is the stabiliser of *S* for the action of *G*
on the set of conjugates of *S*. Thus, the number of
conjugates is the index *[G:N]=|G|/|N|*.
Now, obviously *N* contains *S* as a subgroup, and so it follows
that *[G:N]* is a divisor of *[G:S]*.

Let **S** be the set of all Sylow *p*-subgroups. Now *S*
acts on **S** by conjugation, if *a* in *S* and
*T* is a Sylow *p*-subgroup then *(a,T) |--> aTa*^{-1}.
Now *S* itself is in **S** and is clearly a fixed point.
It must be the only fixed point though. For suppose that *T*
is another fixed point. Thus *aT=Ta* for each *a* in *S*.
Clearly then *ST* is a subgroup
of *G* and *T* is a normal subgroup of this subgroup.
But, *S* and *T* are both Sylow *p*-subgroups of
*ST* and so are conjugate. This forces *S=T* which is impossible.
This contradiction shows that there is exactly one fixed point. By the
lemma above we have that |**S**| is congruent to the number of fixed points
modulo *p*, that is, it is congruent to 1 modulo *p*, which
finishes the proof.