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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 $G$.

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    @SteveD I'm not actually sure how to describe the subgroup in terms of words (I could give a list of something like 16 words which generate it, but that is far from minimal). The subgroup happens to be isomorphic to the free group on three generators (it's the fundamental group of a genus two surface with three punctures), but $z$ lies outside it. Anyway, the whole question is moot to me at this point because I found a workaround which allowed me to prove my overarching conjecture, using a different group than $G$ (specifically, $\langle a,b,c,d|a^2,b^2,c^2,d^2\rangle$).2012-12-30

2 Answers 2

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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 $z=cbaba$.

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    Indeed. Fortunately, the example turns out to be a commutator.2012-12-29
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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 $y^{-1}$.

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    Ah, OK, that is my mistake, I misread it. In that case, Seirios's answer does contain an explicit example.2012-12-29