Self normalizing maximal subgroup of a non solvable group

Question

Let $G$ be a finite non-solvable group and $H$ its maximal subgroup. Prove that if $H$ is solvable then $H=N_G(H)$ I think I found different ways to prove it but I don't know how to begin:
-if $N_G(H)=G$ then $H$ is not solvable;
-if $G/H$ is simple then $H$ is not solvable;
-prove that $G/H$ is not simple;
-prove that $G/H$ is abelian;

Answer 1

Here is an outline proof. Suppose that $N_G(H) \ne H$. Then $N_G(H)=G$, so $H \lhd G$. But $H$ is maximal in $G$, so $G/H$ has no proper nontrivial subgroups. Hence $G/H$ is cyclic of prime order, and so is solvable. Also $H$ is solvable, so $G$ is solvable, contradiction.