Let $H \leq G$, $H$ is a maximal subgroup of $G$. Is it true that if $H \lhd G$ then $G/H$ is abelian? If yes, is it also true when $H$ is not normal?
I know that if $G' \leq H$ then $G/H$ is abelian, but I don't think this is necessary.
(I'm trying to prove that $G/H$ is finite and of a prime order, and the only thing that's missing is the above)