Exercise 1.6.41(a) of Bourbaki's Algebra goes like this:
Let $x,y$ be two elements of a group $G$. For there to exist $a,b$ in $G$ such that $bay=xab$, it is necessary and sufficient that $xy^{-1}$ be a commutator.
The necessity part is nothing else than the statement that for arbitrary elements $x,a,b\in G$, $bayb^{-1}a^{-1}y^{-1}$ is a commutator. But this is not obvious to me at all. I have the feeling I am missing something very simple. Can someone help?