Let $G$ be a finite group and let $H$ be a solvable subgroup of $G$. Let $G$ be a finite group and let $H$ be a solvable subgroup of $G$. Let $N=N_G(H)$ and assume that $N/H$ is a nonabelian simple group. Prove that $N=N_G(N)$.
$N=N_G(N)$. That means $N_G(H)=N_G(N_G(H))$

Let $g \in N_G(N)$. Then $H^g \lhd N$. Now  $HH^g/H \lhd N/H$. Since $N/H$ is simple, $HH^g/H=1$ or $N/H$. Since $HH^g/H \cong H^g/(H^g \cap H)$ and $H$ is solvable, we get $HH^g/H=1$, which implies $H^g=H$. Hence $N_G(N)=N$.
