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I'd appreciate it if you consider this question and together with its hint:

Let $G=AB$ be a finite group which is the internal direct product of $A$ and $B$ which are non-abelian simple groups. Show that the only proper, nontrivial normal subgroups of $G$ are $A$ and $B$.

Hint: If $N$ is another such subgroup, we would have to consider the commutator $[N,A]$.

In fact I am not sure how to use this hint!

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    $\begingroup$ $[N,A]$ is a normal subgroup, and is contained in both $A$ and $N$. $\endgroup$ – user641 Aug 12 '13 at 0:02
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Suppose $N \leq G$ is normal, and $A$ is not contained in $N$, nor is $B$. By normality of $N$ and $A$, $[N,A] \leq A \cap N$, and $A \cap N$ is normal in $A$. By simplicity, $[N,A]=1,$ so for all $n \in N, a \in A,$ $nan^{-1}a^{-1}=1,$ and therefore $(n^{-1})^{a}=n^{-1}$, which implies that $A \subset C_G(N)$. Similarly, $B \subset C_G(N)$. It follows that $N \leq Z(G).$

Assuming that $A,B$ are nonabelian, we get that $Z(G)=Z(A)\times Z(B)=1.$ Therefore, $N$=1. The case of $A \lneq N$ should contradict the simplicity of $B$.

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    $\begingroup$ The normal subgroups of $G$ above $A$ corresponds to the normal subgroups of $G/A \cong B$. $\endgroup$ – Dune Aug 12 '13 at 0:37
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    $\begingroup$ It's worth noting that nowhere in this argument is the assumption that $G$ is finite used. $\endgroup$ – B. Mackey Aug 12 '13 at 1:08

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