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Let $N$ be a minimal normal subgroup of $G$. Also let $N$ and $\frac{G}{N}$ are non-abelian simple. Can we say that $G=N\times A$ where $A$ is a non-abelian simple subgroup of $G$?

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  • $\begingroup$ Presumably you mean to assume $G$ is finite, so you should add that hypothesis to the question. $\endgroup$ – Stephen Sep 9 '14 at 13:28
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$G/C_{G}(N)$ is isomorphic to a subgroup of the automorphism group $B$ of $N$. The only nonabelian simple composition factor of $B$ is $N$ itself (this requires CFSG). Since the composition factors of $G$ are $N$ and $G/N$, it follows that $G/C_{G}(N) \cong N$. Since $N \cap C_{G}(N) = \{ 1 \}$, it follows that $C_{G}(N)$ is the required $A$.

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  • $\begingroup$ I am confused---is $G$ assumed finite? $\endgroup$ – Stephen Sep 9 '14 at 12:34
  • $\begingroup$ I might be tired, but I don't really understand the answer. It could use some better formatting, at the very least. $\endgroup$ – tomasz Sep 9 '14 at 12:41
  • $\begingroup$ @Stephen: The original question has finite-groups tag, so it's kind of ambiguous. $\endgroup$ – tomasz Sep 9 '14 at 12:43
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    $\begingroup$ @tomasz, Ah, you're right. Probably the group is meant to be finite. Andreas' point is that for a finite simple non-abelian group $G$, the classification of finite simple groups implies that the only nonabelian composition factor in $\mathrm{Aut}(G)$ is $G$ itself, so every extension of one finite non-abelian simple group by another is in fact a direct product. +1 $\endgroup$ – Stephen Sep 9 '14 at 13:27
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    $\begingroup$ @tomasz, I prove that it splits, $C_{G}(N)$ being the other factor. $\endgroup$ – Andreas Caranti Sep 9 '14 at 14:26

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