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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?

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    Or rather, under Bourbaki's convention, of $a^{-1}b^{-1}$ with $a^{-1}y^{-1}$.2012-02-07

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$\begin{align*} x^{-1}y^{-1}z^{-1}xyz &= y(y^{-1}x^{-1}y^{-1}z^{-1}xyzy)y^{-1}\\ &= y\left((xy)^{-1}(zy)^{-1}(xy)(zy)\right)y^{-1}\\ &= y[xy,zy]y^{-1}\\ &= [xy,zy]^{y^{-1}}\\ &= \left[(xy)^{y^{-1}},(zy)^{y^{-1}}\right]\\ &= \left[yx,yz\right] \end{align*}$ for all $x,y,z\in G$.