TITLE: If $N$ and $M$ are normal subgroups and $N$ and $M$ have no common element other than $e$ then prove that for all $m \in M$ and $n\in N$, $mn=nm$.
QUESTION [5 upvotes]: My approach ; I proved the $MN$ to be a normal sub group whence $mn=nm$.

REPLY [4 votes]: Consider $nmn^{−1}m^{−1}$ in two ways. Firstly $(nmn^{−1})m^{−1} \in M$ as $nmn^{−1} \in M$ and $m^{−1} \in M$, secondly $n(mn^{−1}m^{−1}
) \in N$ as $n \in N$ and $mn^{−1}m^{−1} \in N$. Hence $nmn^{−1}m^{−1} \in N \cap M = {e}$. We obtain $nmn^{−1}m^{−1} = e$ i.e. $nm = mn$ for all $n \in N$ and $m \in M$.
Source: http://www.metu.edu.tr/~matmah/Graduate-Algebra-Solutions/Undergraduate-Algebra-Problems%20and%20Solutions.pdf