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!
Best Answer
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 $B$.