Why Does G Equal N_G(P)N in Group Theory?

[bham10246]

This shouldn't be too hard but I'm having a hard time putting a few pieces together.

Let $G$ be a finite group with $N$ a normal subgroup. Let $P$ be a Sylow $p$-subgroup of $N$. Prove that $G=N_G(P)N$, where $N_G(P)$ denotes the normalizer of $P$ in $G$.

My attempts: I know that the number of conjugates of P in G equals the index of the normalizer of P in G. What I don't understand is: why does the number of conjugates of P in G equal the normal subgroup N?

 

[mathwonk]

well it looks plausible ebcause NP(G) is th stuff in G that leaves P inside P, and N moves P around to all its conjugates inside N. Since N is normal also all elments of G leave P inside of N, so if you have the stuff that leaves P fixed when acting by conjugation, and also have a transitive subgroup acting by conjugation, you should have everyone.

i.e. let G act on P by conjugation and prove NP(G)N contains a a transitive subgroup for this action, plus the full isotropy subgroup. then it is done isn't it? by the usual trick for finding automorphism groups. i.e. any group that acts transitively and contains an isotropy group of on point is the full automorphism group. (see cartan's wonderful little book on complex variables, in the section on complex automorphisms.)