# Sylow p-subgroup P of G is also Sylow p-subgroup of N[P]?

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].

It's a Sylow $p$ subgroup because of order. If a Sylow $P$ subgroup has order $p^n$ then $p^n$ is the maximal power of $p$ dividing $|G|$. Since $P \subseteq N(P)$ we get that $p^n$ divides $|N(P)|$ and since $N(P) \subseteq G$ we get that no higher power of $p$ divides $|N(P)|$. Thus $p^n$ is the maximal power of $p$ dividing $|N(P)|$ so $P$ is a Sylow $p$-subgroup of $N(P)$.
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 $g\in N(P)$.