# Sylow's theorem

1. Nov 8, 2009

### halvizo1031

1. The problem statement, all variables and given/known data
LET G BE A FINITE GROUP WHOSE ORDER IS DIVISIBLE BY THE PRIME P. SUPPOSE P^M IS THE HIGHEST POWER OF P WHICH IS A FACTOR OF |G|AND SET K=(|G|/P^M), THEN THE GROUP G CONTAINS AT LEAST ONE SUBGROUP OF |P^M|.

I have the proof but can someone explain it in simpler terms? Maybe even a few examples please.

2. Relevant equations

3. The attempt at a solution
Let X denote the collect of all subsets of G which have p^m elements and let G act on X by left translation so that the group element g is in G sends the subset A in X to gA. The size of X is the binomial coefficient kp^m choose p^m which is not divisible by p. Hence, there must be an orbit G(A) whose size is not a multiple of p. We have |G|=|G(A)|*|G(subA)|, consequently |G(subA)| is divisible by p^m. Now G(subA) is the stabilizer of A, so if a is in A and g is in G(subA), the ga is in A. This means that the whole right coset of G(subA)a is contained in A whenever a is in A and |G(subA)| cannot exceed p^m. Therefore, G(subA) is a subgroup of G which has order p^m.

Last edited: Nov 8, 2009
2. Nov 9, 2009

### Quantumpencil

The statement is pretty simple. It says that if you have a Group of order n, where n has factoriaztion p^k * m, where p is a prime number, then there exists a subgroup of order p^k.

The proof is a bit tricky, since it's in direct. What about it is confusing you?

3. Nov 9, 2009

### halvizo1031

could you maybe give me a few examples? It makes sense but the proof is a bit rough for me.