Let G be a finite group and let p be a prime dividing the order of G. Let P be a Sylow p-subgroup of G. The problem is to show that N[N[P]]=N[P], where N[P] is a normalizer of P in G.
Solution says conjugation of P by an element of N[N[P]] yields a subgroup of N[P] that is a Sylow p-subgroup of G, and also of N[P].
I understand that conjugation is also Sylow p-subgp of G(by 2nd Sylow thm), but I don't know why it is also subgroup and Sylow p-subgroup of N[P].
Please help me!
Best Answer
If $g\in N(N(P))$, then $gN(P)g^{-1}\subseteq N(P)$. As $P\subseteq N(P)$ then $gPg^{-1}\subseteq N(P)$. By reasons of its order, $gPg^{-1}$ is a Sylow subgroup of $N(P)$. But $N(P)$ only has one Sylow $p$-subgroup (why?) namely $P$ then $gPg^{-1}=P$. Therefore $g\in N(P)$.