If the intersection of two normal subgroups is trivial, then their elements commute

Question

How to show that if $N \ \& \ M$ are 2 normal subgroups of group $G$ and $N\cap M=\{e\}$ (identity element), then for any $n\in N \ \&\ m\in M $, $nm=mn$?

Answer 1

If we can show that $m^{-1}nmn^{-1}=e$, then multiplying by $m$ and $n$ from the respective sides you'll get the desired result. Now, regarding the element $m^{-1}nmn^{-1}$: use the normality of $M$ and $N$ to show that it lies in both subgroups.