## Normalizers of Sylow subgroups are self-normalizing

Let $H$ and $K$ be a finite groups, $P$ a Sylow $p$-subgroup of $H$, and $H \leq K$. If $P$ is normal in $H$ and $H$ is normal in $K$, prove that $P$ is normal in $K$. Deduce that if $P \in \mathsf{Syl}_p(G)$ then $N_G(N_G(P)) = N_G(P)$. (That is, normalizers of Sylow subgroups are self-normalizing.)

Let $k \in K$. Now $kHk^{-1} = H$ since $H$ is normal, so that conjugation by $k$ is an automorphism of $H$. Since $P$ is the unique subgroup of its order in $H$, $P$ is characteristic in $H$, and we have $kPk^{-1} = P$. Thus $P$ is normal in $K$.

Now let $P$ be a Sylow subgroup of $G$. Note that $N_G(P)$ is the $\leq$-largest subgroup of $G$ in which $P$ is normal. Now if $N_G(P) \leq K$ is normal, we have $P \leq K$ normal by the above argument, so that $K \leq N_G(P)$. In particular, $N_G(N_G(P)) = N_G(P)$.