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 $n\in N$, $mn=nm$. My approach ; I proved the $MN$ to be a normal sub group whence $mn=nm$.
 A: $H$ being a normal subgroup of $G$ does not imply that $gh = hg$ for all $g,h\in H$. It means that for all $h\in H$, $g\in G$, $ghg^{-1}\in H$: so we might have $ghg^{-1} = h'$, or equivalently, $gh = h'g^{-1}$ for some $h'\in H$ ($h$ may not equal $h'$). For an example of a non-abelian normal subgroup, consider $A_n\subseteq S_n$ for $n> 3$, where $A_n$ is the alternating group on $n$ letters and $S_n$ is the symmetric group on $n$ letters. $A_n$ is a normal subgroup of $S_n$, but when $n > 3$, $A_n$ is not abelian, so that $ab$ may not be equal to $ba$ for $a,b\in A_n$. To prove the claim, you need to do a bit more; see the hint given by Ben.
A: 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
A: Try showing $n^{-1}m^{-1}nm\in N\cap M$.
