Show that $G/\Phi(G)$ is a non-abelian simple group, where $\Phi(G)$ denotes the Frattini subgroup of $G$
So $G/\Phi(G)$ can't be abelian since if it were then is would be solvable and since $\Phi(G)$ is a solvable normal subgroup of $G$, it would imply that $G$ is solvable.
For the next part, I think I was able to prove that $G$ is simple which would mean $G/\Phi(G)$ is simple by the correspondence theorem, but my intuition is telling me that showing $G$ is simple is a little to over reaching so I think I made a mistake.
For the sake of contradiction suppose $G$ has a proper non trivial normal subgroup. Let $N$ be a minimal proper normal subgroup and let $P$ be a Sylow subgroup of $N$. So by the Fratini Argument $G = N_G(P)N$. Since $N$ is minimal normal $N_G(P)$ must be a proper subgroup. But $N_G(P)N/N \cong N_G(P)/N_G(P)\cap N$ which is solvable since $N_G(P)$ is solvable and the quotient group of a solvable group is solvable.
Best Answer
Let $N$ be any proper normal subgroup of $G$. If $N \not\le \Phi(G)$, then there is a maximal subgroup $M$ of $G$ with $N \not \le M$. So then $G = NM$ by maximality of $M$, and then $N$ and $M$ are both solvable and hence so is $G$.
So $N \le \Phi(G)$ and $G/\Phi(G)$ must be simple.