Let $M$ and $N$ be normal subgroups of $G$ such that $G = MN$. Show that for any $m \in M$ and $n \in N$, we can find elements $m' \in M, n' \in N$ such that $mn' = nm'$.
I've tried using properties of normal subgroups (i.e. $gNg^{-1} \subseteq N$), but I can't seem to relate m directly with n in the desired form. Any hints?
Since we need inverses, it's easier to write $m'=m_1$ and $n'=n_1$. The relation you want can be rewritten $$ m^{-1}n=n_1m_1^{-1} $$ Now $$ m^{-1}n=m^{-1}nmm^{-1} $$ so you can take $n_1=m^{-1}nm$ and $m_1=m$.