If two quotient groups are semi-simple, then a third build from both is semi-simple too. I call a group semi-simple if it is the direct product of non-abelian simple groups. Let $G$ be a finite group and let $M, N \unlhd G$ such that $G/N$ and $G/M$ are both semi-simple. Prove that $G/(M\cap N)$ is semi-simple too.
I know that for semi-simple groups $G$, the normal subgroups have a particular simple form, i.e. they are precisely the direct products of a finite subcollection of the factors of $G$, this could be found here. This in particlar implies that for each normal subgroup, there is exactly one complemented normal subgroup. Further, for a semi-simple group the center is always trivial, otherwise it would be such an direct product of non-abelian groups, contradicting the fact that the center is always abelian.
That is all I know, but I do not know how to show that $G/(M\cap N)$ is such a direct product if $G/M$ and $G/N$ are, the closest I can think of is to consider $MN/N \unlhd G/N$ and $MN/M \unlhd G/M$, apply the above fact, and somehow build a direct product for $G/(M\cap N)$, but I have no idea how to do this concretely, so any hints?
 A: We can assume that $M \cap N = 1$.
The only normal subgroups of semisimple groups are direct products of some of the simple factors. So, since $NM/M \unlhd G/M$, we have $G/M = NM/M \times C/M$ for some $C \unlhd G$ with $M \le C$, and so $G=NC$.
Now $[C,N] \le M$, so $[[C,N],N] \le M \cap N = 1$. Then, since $N$ is perfect (i.e. $[N,N]=N$), the $3$-subgroups lemma gives $[C,N] = 1$.
So $G = N \times C$ and, since $N \cong NM/M \unlhd G/M$ and $C \cong G/N$ are semisimple, so is $G$.
A: Another way to solve the question:
Suppose $G/N= \times_{i}(A_i/N)$,$G/M= \times_{j}(B_j/M)$, where $N \leq A_i \unlhd G$, $M\leq B_j \unlhd G$ and, $A_i/N$ and $B_j/M$ both are non-Abelian simple groups, $\forall i,j$.
$(A_i \cap M) \unlhd A_i \Rightarrow (A_i \cap M)/N \unlhd A_i/N \Rightarrow (A_i\cap M)N =A_i \vee N $:
(1) $(A_i\cap M)N =N \Rightarrow A_i/(M\cap N)=A_i/(A_i\cap M) \cong A_iM/M \unlhd G/M$
so $A_i/(M\cap N)$ is semi-simple.
(2) $(A_i\cap M)N =A_i \Rightarrow A_i \leq MN$ and $MN/(M\cap N) \cong X \leq MN/N \times MN/M \unlhd G/N \times G/M$ is semi-simple.
Suppose $A_i\nleq MN, i=1,2,...,r$ without any loss of generality:
$G=\bigg(\prod_{i=1}^r A_i\bigg)(MN) \Rightarrow G/(M\cap N)=\bigg(\prod_{i=1}^r A_i/(M\cap N)\bigg)(MN/(M\cap N))$ is semi-simple.
