# Homework Help: Abstract->p-sylow groups

1. Dec 21, 2009

### TheForumLord

1. The problem statement, all variables and given/known data
Let P be a p-sylow sbgrp of a finite group G.
N(P) will be the normalizer of P in G. The quotient group N(P)/P is cyclic from order n.

PROVE that there is an element a in N(P) from order n and that every element such as a represnts a generator of the quotient group N(P)/P

2. Relevant equations
3. The attempt at a solution

Welll.... there is mP in N(P)/P such as (mP)^n = P -> m^n*P=P -> m^n = 1 ...
If m has order that is less the n, we'll get a contradiction to the fact that mP is from order n.
It's pretty obvious that every element of this kind is a generator of this group...But I realy feel I'm missing something... It's a 20 points question and the answer will take me 2 lines...

Where is my mistake?

2. Dec 21, 2009

### ystael

You cannot deduce that $$m^n = 1$$ from $$m^n P = P$$; that tells you only that $$m^n \in P$$. However, there is a simple way to produce the element of order $$n$$ you need, using $$m^n$$.

3. Dec 21, 2009

### TheForumLord

Hmmmm...Yep, you're right...
So we have m^n is in P... We need to produce an element k of order n that is in N(P)... Hmmmm we have m^n*P=P, and m^n is in P, m is in N(P). We know something more about this m? We know nothing about its order but we know that mP=Pm, which means m^n*P=P*m^n, and it's still gives us nothing...

Can you please give more detailed directions?
I'll appreciate any kind of further help...

TNX a lot

4. Dec 21, 2009

### TheForumLord

Wait a sec! We know that m^n is in P and that P has order p^r for some r in N...So m^n must have an order that is a power of p, say p^k... so we know that m^(p^k) has order n in N(P)...
But how will we discover what k is? And why it's a generator of N(P)/P?