Direct product of two groups.

Question

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 $G$?

Answer 1

$G/C_{G}(N)$ 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 $A$.