Normal subgroups and isomorphism theorem

Question

Let $G$ be a group and $N,M$ normal subgroups in $G$ such that $G=MN$.

Prove that: $G/(M\cap N)\cong G/M\times G/N$

So obviously I need to use the first Isomorphism theorem here.I thought of defining $\phi:G\rightarrow G/M\times G/N$ as : $\phi (g) =(gM,gN)$ but then I understood that it won't work since $G/M\times G/N=\{(g_1M,g_2N)|g_1,g_2\in G\}$.

Any other ideas?

Answer 1

You're very close.

Using your map $\phi$, $\mathrm{ker}(\phi)=M\cap N$ so by first isomorphism theorem $G/(M\cap N)\cong \phi(G)$ and it remains to show that $\phi$ is surjective.

So consider $(gM,hN)$. As $G=MN$, there is some $m\in M$, $n\in N$ such that $mn=g^{-1}h$. This gives $\phi(gm)=(gM,gmnN)=(gM,hN)$. So we are done.

Coming up with $mn=g^{-1}h$ may look like black magic. We come up with it by noticing $\phi(g^{-1})(gM,hN)=(M,g^{-1}hN)$ and solving $\phi(x)=(M,g^{-1}hN)$, which is conceptually a bit easier.