# Proving the two groups are isomorphic by constructing an isomorphism

I was studying for the qualifying exam for graduate entrance, and I'm having hard time for a question about factor groups. Here is the question:

Let $M$ and $N$ be normal subgroups of a group $G$ such that $G=MN$. Prove that $G/(M \cap N)$ is isomorphic to $G/M \times G/N$.

I constructed a group homomorphism $\phi(a(M\cap N))=(aM,aN)$, and I already proved the well-definedness and injectivity of the homomorphism.

However, I failed to figure out the proof of surjectivity. To prove the surjectivity, for any $a,b \in G$, I should prove that there is $c\in G$ such that $aM=cM$ and $bN=cN$, i.e. $c^{-1}a \in M$ and $c^{-1}b \in N$, but I can't proceed from this any more.

I think that $G=MN$ would be a hint for proving surjectivity, but I don't know how that can be a hint.

Hint: Take $a,b \in G$ and write $a^{-1}b=mn$. Take $c=am=bn^{-1}$.

Without writing a map, it can be done by second isomoprhism theorem.

$G/(M\cap N)\cong\overline M\times \overline N$ as $\overline M \cap \overline N =1$.

$\overline M=M/M\cap N \cong MN/N=G/N$ with same manner $\overline N \cong G/M$. Hence we are done.