prove that if $N\lhd G$, $ M\lhd G$, $M\bigcap N=\{e\}$ so: $mn=nm , \forall n \in N,\forall m\in M$

Question

prove that if $N\lhd G$, $ M\lhd G$,

$M\bigcap N=\{e\}$

so: $mn=nm , \forall n \in N,\forall m\in M$

Answer 1

Let $m \in M$ and $ n \in N$, consider $g=n^{-1}mnm^{-1}$. Now by normality of $M$, the element $n^{-1}mn \in M$ and by closure $g=(n^{-1}mn)m^{-1} \in M$. Likewise by normality of $N$, the element $mnm^{-1} \in N$ and by closure $g=n^{-1}(mnm^{-1}) \in M$. But $M \cap N = \{e\}$. Thus $n^{-1}mnm^{-1}=e$.