Groups, normal subgroups and isomorphism

Question

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

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

Any ideas/hints how to begin?

Answer 1

Hint: Define $\varphi: G \to G/M \times G/N$ by $\varphi(g) = (gM,gN)$. Identify the kernel of $\varphi$, show $\varphi$ is surjective, and then apply the First Isomorphism Theorem

Answer 2

Let $f:G\rightarrow G/M\times G/N$ such that $f(x)=(g(x),h(x))$ where $g:G\rightarrow G/M$ and $h:G\rightarrow G/N$ are the canonical projection map. $f(x)=(e,e)$ if and only if $g(x)=e, h(x)=e$, this is equivalent to saying that $x\in N\cap M$. Let $u,u'\in G$, we can write $u=mn, u'=m'n', m,m'\in M, n,n'\in N$. $f(nm')=(g(nm'),h(nm'))=(g(n),h(m'))=(g(u),g(u'))$. This implies that $f$ is surjective, since the kernel of $f$ is $M\cap N$, we deduce that $f$ factors by an isomorphism $G/M\cap N\rightarrow G/M\times G/N$.