I've been struggling with this proof lately. It says
Let $M$ and $N$ be normal subgroups of a group $G$ such that $M\cap N=\langle e\rangle $. If $a \in M$ and $b \in N$ then $ab=ba$.
The proof starts with
[...] consider $a^{-1}b^{-1}ab$. Since $M$ is normal, $b^{-1}ab \in M$; similarly normality in $N$ implies that $a^{-1}b^{-1}a \in N$.
What leaves me confused is the above statement : how can I know that $b^{-1}ab \in M$ and how can I know that $a^{-1}b^{-1}a \in N$, and why is it not the other way around?
The subgroup $M$ is normal in $G$ if and only if, for every $x\in M$ and every $g\in G$, $g^{-1}xg\in M$.
In your case, you can consider, for $a\in M$ and $b\in N$ (so also $b^{-1}\in N$), $$ a^{-1}b^{-1}ab=a^{-1}\underbrace{(b^{-1}ab)}_{\in M}\in M $$ and $$ a^{-1}b^{-1}ab=\underbrace{(a^{-1}b^{-1}a)}_{\in N}\,b\in N $$ because $M$ and $N$ are subgroups, so closed under products.
Therefore $a^{-1}b^{-1}ab\in M\cap N$ and so, by assumption, $a^{-1}b^{-1}ab=e$. Hence $$ ba(a^{-1}b^{-1}ab)=bae $$ which means $ab=ba$.