display | more...
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 XG for the set of points fixed by all points of G. The points in XG form singleton orbits.

Lemma If G is a p-group then |XG| 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 XG.

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=pks 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 pk elements. There are n choose pk such subsets. So |T|= (n/pk)((n-1)/1)((n-2)/2)...((n-pk+1)/pk-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 pk 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 pk. 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 pk elements. It follows that S has exactly pk 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-1Ha.S is a subset of S, since S contains the identity element, this shows that a-1Ha.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.

Log in or register to write something here or to contact authors.