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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.

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  • $\begingroup$ You have just proved that $P=N$. $\endgroup$
    – Derek Holt
    Jul 21 '20 at 18:23
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    $\begingroup$ 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 $2$). $\endgroup$ Jul 21 '20 at 18:28
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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.

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