Equivalence of $x,y\in G$ given that $xzy^{-1}z^{-1}$ is a commutator for some $z$ Let $G = \langle a,b,c\:|\: a^2, b^2, c^2\rangle$. Let $\tilde{}$ by the equivalence relation on $G$ generated by conjugation and inversion (i.e., $x\tilde{} y$ if there is a finite sequence of conjugations and inversions which transform $x$ into $y$). Let $x,y\in G$ and suppose there exists a $z\in G$ such that $xzy^{-1}z^{-1}$ is a commutator. Is $x\tilde{} y$?
This question arises in the study of triangular billiards. Specifically, I am interested in a subgroup of $G$ and elements $z$ of a certain form, and there is a great deal of computational evidence that the statement holds in this case. However, the subgroup admits no simple description outside the theory of billiards, and I am hoping that I can simply hammer the statement with a combinatorial proof for the whole group $G$.
 A: First, $xzy^{-1}z^{-1} \in [G,G]$ is equivalent to $x=y$ in $G^{ab}$, so we can take $x=ac$ and $y=babc$. We can show that $x \nsim y$.
Suppose that there exists $w(a,b,c) \in G$ such that $w(a,b,c)acw(a,b,c)^{-1}=babc$ $(\ast)$. Notice that $G \simeq \mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$, so we can suppose that $w(a,b,c)$ is a reduced word over $\{a, b, c\}$ thanks to the normal form for free products. 
1) If the first letter of $w(a,b,c)$ is $b$, then $\lg(babcw(a,b,c))=4+ \lg(w(a,b,c))$ and $\lg(w(a,b,c)ac) \leq 2+ \lg(w)$, so this case is impossible.
2) Otherwise, there exists a reduced word $\tilde{w}(a,b,c)$ over $\{a, b, c\}$ such that $w(a,b,c)=cbab \tilde{w}(a,b,c)$. Now $(\ast)$ becomes $\tilde{w}(a,b,c)= cbab \tilde{w}(a,b,c)ac$. Because $w$ is reduced, $\lg(cbab\tilde{w}(a,b,c)ac)=4+ \lg(\tilde{w}(a,b,c)ac)$ so $\lg (\tilde{w}(a,b,c)ac)= \lg(\tilde{w}(a,b,c))-4$. Therefore, there exists a reduced word $r(a,b,c)$ over $\{a, b, c\}$ such that $\tilde{w}(a,b,c)=r(a,b,c)ca$. Now $(\ast)$ becomes $r(a,b,c)=cbabr(a,b,c)$. Because $w$ is reduced, $\lg(r(a,b,c))=\lg(cbabr(a,b,c))=4+\lg(r(a,b,c))$, a contradiction.
The same argument holds for $w(a,b,c)(ac)^{-1}w(a,b,c)^{-1}=babc$, and you deduce that $x \nsim y$ from the comment of Hagen von Eitzen.
It is possible that there exists a simpler argument.
EDIT: If $x=babc$ and $y=ac$ then $xzy^{-1}z^{-1}=[a,cb]$ with $z=cbaba$.
A: Here is an elaboration of my comment on Seirios's answer.
The commutator condition just gives cosets of the subgroup $[G,G]$. Thus any two elements in $[G,G]$ are equivalent. Now for $x\cong y$, we would need $x$ conjugate to either $y$ or $y^{-1}$. Consider the elements $(ab)^2$, $(ac)^2$, and $(bc)^2$; all of these are in $[G,G]$.  Let us show they are not conjugate.
First, $(ab)^2$ is normalized by the subgroup $\langle a,b\rangle$.  Similar results hold for the other two.  For them to be conjugate, then, their normalizers must be conjugate. But that would mean their normalizers all generated the same normal subgroup; this is not true, as $\langle\langle a,b\rangle\rangle$ does not contain $c$, etc. for the other two. There are thus at least $3$ conjugacy classes in $[G,G]$, so for any $y\in [G,G]$, there certainly exists an $x$ not conjugate to $y$ or $y^{-1}$.
