Let $G$ be a group and $N$ and $M$ normal subgroups of $G$ such that $M\cap N = \{1_G\}$. Show that $\forall m\in M, n\in N$ we have $mn=nm$
My attempt: In the given conditions we have $mn\in mN = Nm$, the last equality because $N$ is normal. Then $nm = am$ for some $a\in N$. But then by the cancellation law we have $a=n$ and then $nm=mn$.
But this seems strange to me because I have not used the condition that $M$ is normal and that $M, N$ have only the identity element in common. Or did I and I don't see where? Can someone help me clarify this question?
(Note that this is not the same question as, e.g., this, because I'm asking for verification on my attempt).
Let $m \in M$ and $n \in N$. Then $mnm^{-1} \in N$ and $n^{-1} \in N$ so that $mnm^{-1}n^{-1} \in N$.
Also, $nm^{-1}n^{-1} \in M$ so that $m(nm^{-1}n^{-1}) \in M$.
Hence, $mnm^{-1}n^{-1} \in M\cap N = \{1_{G}\} \implies mnm^{-1}n^{-1} = 1_{G}$ and the result follows.