Prove $G/(M\cap N) \cong M/(M\cap N) \times N/(M\cap N)$ where $G=MN$ and $M,N\triangleleft G$

Question

If we consider the map $\phi: M\times N \rightarrow M/(M\cap N) \times N/(M\cap N)$, I was able to show that this is onto and the kernel of the map is $(M\cap N) \times (M\cap N)$ and hence by first isom. theorem,
$$(M\times N)/\bigl((M\cap N)\times (M\cap N)\bigr) \cong \bigl(M/(M\cap N)\bigr) \times \bigl(N/(M\cap N)\bigr).$$ But how would I go about showing
$$(M\times N)/\bigl((M\cap N)\times (M\cap N)\bigr) \cong MN/(M\cap N)$$ to prove the final result? Appreciate your help.

Answer 1

Use first isomorphism theorem again. Can you show that $M\times N\to G/(M\cap N)$ given by $(m,n)\mapsto mn(M\cap N)$ is a group homomorphism and has kernel $(M\cap N)\times (M\cap N)$?

Second edit after reading comments on Servaes' answer on the first question you asked related to this Servaes appears to have suggested this homomorphism in a prior version of his answer according to the comments below it and a skim of revision history. Since you couldn't see why it is a homomorphism, I'll add the following hint: Using that $M$ and $N$ are normal, can you show that the element $mnm^{-1}n^{-1}$ is in $M\cap N$ when $m\in M$, $n\in N$. Can you see that this implies that $M/M\cap N$ and $N/M\cap N$ commute with each other in $G/M\cap N$?

Edit Or alternatively you can follow Arturo's hint.