Suppose that finite group $G\neq 1$ and $|G : M| ∈ \mathbb P$ for every maximal subgroup $M$ of $G$. Then prove: $G$ contains a normal maximal subgroup.
(we all know that a maximal subgroup is normal then its index is prime, but the question is like a converse theorem.)
Suppose $N\leq U$, if $U/N$ is maximal subgroup of $G/N$, then $U$ is also the maximal subgroup of $G$. From $|G/N:U/N|=|G:U|\in \mathbb P$. By induction, all we need is to find normal group $1<N<G$, i.e. $G$ is not simple. If $G$ is an abelian simple group, it's so obvious. But how to find the contradiction if $G$ is non-abelian simple?
Best Answer
Suppose that $G$ is a nonabelian simple group and all of its maximal subgroups have prime index. Let $p$ be the smallest prime such that $G$ has a subgroup $H$ of index $p$. Then by considering the action of $G$ by multiplication on the left cosets of $H$ in $G$ we get an embedding of $G$ into $S_p$ (in fact into $A_p$ but we don't need that), so we can assume that $G \le S_p$.
The all primes $q$ dividing $|G|$ satisfy $q \le p$. Since $G$ is nonabelian simple it cannot be a $p$-group, so a maximal subgroup of $G$ containing a Sylow $p$-subgroup of $G$ has prime index $q<p$, contrary to the choice of $p$.