Two disjunct normal subgroups Let $M, N$ be normal subgroups of $G$ with $M∩N=\{e\}$.
I'm trying to prove that $M\times N$ is isomorphic to $G$. I proved that $nm=mn$ for all $n$ in $N$ and $m$ in $M$. So now I'm trying to take any fixed $g$ in $G$ and represent it in terms of $m$ and $n$. I've been doing this for a while but still no answer. Any help please?
Edit: forgot to mention that $G$ is generated by $M\cup N$.
 A: We can construct an homomorphism:
$M \times N \xrightarrow{\varphi} G$
$(m,n) \mapsto mn$
Now if there are $(m,n)$ such that $mn=e \implies m=n^{-1}$  so $m \in M \cap N$ and $m=n=e$. So we know that $\varphi$ is an injection we only need to show that is surjective. That would be the case if $G=MN$ but I'm afraid we need some extra assumptions.
A: Here is a counterexample: 
Let $V_4 = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and consider $V_4 \times V_4$. 
Consider the normal subgroups $M=<((1,0),(0,0))>$, $N=<((0,0),(1,0))>$. 
It is clear that these are normal since $V_4 \times V_4$ is abelian. Furthermore, it is clear that they have trivial intersection. 
However,  $|M \times N|=4<8=|V_4 \times V_4|$.  
A: So we have that:


*

*$N,M\lhd G$ 

*$M\cap N=\{e\}$ 

*$<M\cup N>=G$ ( means subgroup generated by S. Just incase you use different notation.)


So it follows that if $x\in G$ we see that:
$x=\prod\limits_{i=1}^{k}a_i^{e_i}$ where $a_i\in M\cup N$ and $e_i=-1,0,1$. 
But since $M$ and $N$ are normal, we know that:


*

*$gm=m'g$ for all $g\in G$ and some $m'\in M$ (or some $m\in M$ depending on what direction you want to go.)

*$gn=n'g$ for all $g\in G$ and some $n'\in N$ (or some $n\in N$ depending on what direction you want to go.)


Using this information we can rearrange $\prod\limits_{i=1}^{k}a_i^{e_i}$ into $(\prod\limits_{s=1}^{k_1}m_s^{e_s})(\prod\limits_{t=1}^{k_2}n_t^{e_t})$, where $m_s\in M$, $n_t\in N$, and $k_1+k_2=k$. 
I think that is enough information for you to solve the problem.
