What is the significance of p-sylow groups in finite groups?

[TheForumLord]

Homework Statement

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

Homework Equations

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 really feel I'm missing something... It's a 20 points question and the answer will take me 2 lines...

Where is my mistake?

TNX in advance

 

[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.

 

[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

 

[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?


Will you help me please?
:(


TNX

 

[TheForumLord]

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