4
$\begingroup$

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 $G$.

  • 2
    It depends what you mean by a "simple" description. Isn't this group (for example), the free product of an infinite dihedral group with a cyclic group of order $2$- in which case, Serre's book "Trees" may be helpful.2012-12-28
  • 0
    The relation itself can be simplified: A finite sequence of conjugations and inversions is in fact merely a single conjugation followed by an optional inversion.2012-12-28
  • 0
    @GeoffRobinson I'm not sure that $G$ is isomorphic to the group you mention, but I believe it is isomorphic to $\mathbb Z_2 *\mathbb Z_2*\mathbb Z_2$. But the group which does not admit a simple descriptions is a particular subgroup of $G$, and such subgroups can certainly be nasty.2012-12-28
  • 0
    @AlexBecker: Can you describe the subgroup here? This group is indeed the group Geoff mentioned, and the subgroups are not *that* nasty :)2012-12-28
  • 0
    @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

1

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 $z=cbaba$.

  • 0
    There is indeed a much simpler argument. As you noted, this equivalence relation just gives cosets of the commutator subgroup. So any two elements in $[G,G]$ are equivalent. So one just needs to check how many conjugacy classes there are in $[G,G]$; but this is a free group (on 5 generators), and has way more than just two conjugacy classes.2012-12-28
  • 0
    Thanks for this answer, in particular for your example (which settles the question as I gave it). However, for your first line note that I am requiring $xzy^{-1}z^{-1}=[u,v]$ for some $u,v\in G$, rather than just $xzy^{-1}z^{-1}\in [G,G]$.2012-12-28
  • 0
    Indeed. Fortunately, the example turns out to be a commutator.2012-12-29
1

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 $y^{-1}$.

  • 0
    I'm not sure I understand why the commutator condition gives cosets of $[G,G]$. For example, if we take $x\in [G,G]$ such that $x$ is not a commutator (most products of commutators should be examples) and let $y=e$, then the condition becomes $xzy^{-1}z^{-1}=x$ is a commutator, which is clearly false.2012-12-28
  • 0
    Ah, OK, that is my mistake, I misread it. In that case, Seirios's answer does contain an explicit example.2012-12-29