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*form singleton orbits.

^{G}
**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*such subsets. So |

^{k}**T**|=

*(n/p*. I claim that

^{k})((n-1)/1)((n-2)/2)...((n-p^{k}+1)/p^{k}-1)**|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*. Choose

^{k}*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*. It follows that

^{-1}*S*is a subset of

*Tt*and so

^{-1}*S*has at most

*p*

^{k}elements. It follows that

*S*has exactly

*p*elements and we have shown that Sylow

^{k}*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*is a subset of

^{-1}Ha.S*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.