Normalizer of a solvable maximal subgroup

Question

Let $G$ be a finite group and let $H$ be a solvable maximal subgroup of $G$, meaning that the only solvable subgroup of $G$ containing $H$ is $H$. I am trying to prove that $H=N_G(H)$.

Since $H$ is normal in $N_G(H)$, I have tried to prove that $N_G(H)/H$ is abelian by showing that $[N_G(H),N_G(H)]$ is in $H$. So far I have been unsuccessful and I am out of ideas.

Any hints would be appreciated. Thank you.

Answer 1

Suppose for contradiction that the quotient $N_G(H)/H$ is nontrivial.

Then we can find a nontrivial solvable subgroup $K/H$ of $N_G(H)/H$. For example, we could let $K/H$ be the cyclic subgroup of $N_G(H)/H$ generated by a nonidentity element of $N_G(H)/H$.

Since $K/H$ and $H$ are both solvable, so is $K$. This contradicts our assumption on $H$. Thus, $N_G(H)/H$ must be trivial, meaning that $N_G(H)=H$.

Answer 2

Hint: Let $a\in N_G(H)$ not in $H$, the elements of the group $L$ generated by $a$ and $L$ are $a^ph, h\in H$, $[L,L]\subset H$ implies $L$ is solvable.