# Let $G$ be a finite non-solvable group, each of whose proper subgroups is solvable.

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.

• You have just proved that $P=N$. Jul 21, 2020 at 18:23
• You will not be able to prove that $G$ is simple, because there are counterexamples. For example, $\mathrm{SL}_2(5)$ has order $120$, is not soluble (it has a quotient $A_5$) and has no subgroup $A_5$ as it has a single element of order $2$). Jul 21, 2020 at 18:28

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.