Let $H\leq G$ where $H$ is max simple. Prove either $G$ is simple or there exists a minimal normal subgroup $N$ of $G$ such that $G/N$ is simple.

Question

Question: Let $G$ be a finite group and $H$ a maximal, simple subgroup of $G$. Prove that either $G$ is simple or there exists a minimal normal subgroup $N$ of $G$ such that $G/N$ is simple.

Thoughts: If $H=G$ or if $H$ is trivial, then $G$ must be simple. Otherwise, assume that there is a subgroup $N$ of $G$ such that $H\subset N\subseteq G$. I know that any group that isn't simple can always be split up between a normal subgroup and its corresponding quotient group, but the "maximal" and "minimal" stuff here is throwing me off.

Any help is greatly appreciated! Thank you!

Answer 1

If $H$ is normal, take $N=H$. Otherwise choose any minimal normal subgroup $N$. Then $N\cap H=1$ since $H$ is simple and $G=NH$ since $H$ is maximal. Now $G/N\cong H/H\cap N\cong H$ is simple.