Let M and N be normal subgroups of a group G such that $M \cap N=\{e\}$. Prove that $mn=nm$ for all $m\in M$ and all $n\in N$.

Question

Here is my progress:

$M$ and $N$ being normal in $G$, $\forall n\in N, \forall m \in M $, $nmn^{-1} \in M \implies nmn^{-1}m^{-1} \in M$.
By similar arguments, $mnm^{-1}n^{-1} \in N$ . Now, consider the product, $mnm^{-1}n^{-1}$ [ $\forall n \in N$ and $\forall m \in M$ ].

$mnm^{-1}=n_1 (say)\in N $, again $nm^{-1}n^{-1}=m_1 (say) \in M$ [by the property of normality].

But, the product $mnm^{-1}n^{-1}=n_1n^{-1}=mm_1 \in N and$ $M$ $\implies mnm^{-1}n^{-1} \in M \cap N=\{e\}$ Therefore, $mnm^{-1}n^{-1}=e \implies mn=nm$ $\forall m \in M$ and $n\in N$.