$G$ is a group and $N,M$ are normal subgroups of $G$. Prove that $nm=mn$ for all $n\in N,m\in M$.

Question

My problem is the following $G$ is a group and $N,M$ are normal subgroups of $G$.

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

Prove that $nm = mn$ for every $n\in N,m\in M$.

What i did - I know that $gng^{-1}\in$ N for all g $\in G$

So also $mnm^{-1}\in N $

Now if $nm =mn$ then i can make the above be $n \in N$.

but if not .. then I can't find a contradiction.

Any help will be appreciated.

Answer 1

In knowing that $mnm^{-1} \in N$ it immediately follows that $(mnm^{-1})n^{-1} \in N$. Similarly, we have $nm^{-1}n^{-1} \in M$ so that $m(nm^{-1}n^{-1}) \in M$. Hence $mnm^{-1}n^{-1} \in M \cap N$.