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Let $\Gamma$ be a hyperbolic group. Let $g$, $\gamma\in \Gamma$ freely generate a non-abelian semigroup (in particular, they don't commute and have infinite order). Does the equation $g\gamma^n=h^m$ only have finitely many solutions of triples $(n,h,m)\in \mathbb N\times\Gamma\times \mathbb N$ with $m\neq \pm 1$?

More generally, let $(X,d)$ be a Gromov hyperbolic space, and $\Gamma$ be a subgroup of the group of isometries of $X$: $\mathrm{Isom}(X)$. What is the answer to the same question when $\gamma$ is a hyperbolic isometry of $X$?

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    $\begingroup$ You need to add that $\gamma$ has infinite order. $\endgroup$
    – IJL
    Commented Nov 7, 2023 at 14:40
  • $\begingroup$ For honest hyperbolic groups the answer is "essentially yes"; translation lengths of elements are rational, and this means that if there are infinitely many solutions, then $g \gamma^k = h^l$, $\gamma^m = h^n$ for some $k, l, m, n$. This can be found in Bridson-Haefliger's book somewhere. $\endgroup$
    – Denis T
    Commented Nov 7, 2023 at 14:44
  • $\begingroup$ @DenisT T Sorry, I am confused. The $h$ could be different. How do you have the conclusion here? Do you mean $g\gamma^m=h^n$? I believe the answer is yes, but have no good reasoning. $\endgroup$
    – Yanlong Hao
    Commented Nov 7, 2023 at 14:59
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    $\begingroup$ You have to add extra conditions, as the answer is negative in general. For example, let $\Gamma$ be the infinite dihedral group and $g,f \in \Gamma$--two reflections. Then $\gamma=gf$ has infinite order and doesn't commute with $g$, $h=g\gamma^n$ is a reflection for every $n$, so $h^m=h$ for every odd $m$. Hence $g\gamma^n=h^m$ for infinitely many $m$. $\endgroup$
    – Ashot Minasyan
    Commented Nov 8, 2023 at 21:54
  • $\begingroup$ For each $n$, the assumption (that the semigroup is free) forces $g\gamma^n$ to have infinite order and $\Gamma$ hyperbolic ensures it has finitely many roots. Hence the question amounts to: is it true that for $n$ large enough, $g\gamma^n$ is not a proper power? $\endgroup$
    – YCor
    Commented Nov 9, 2023 at 16:43

1 Answer 1

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The statement about the hyperbolic group $\Gamma$ is true if $\Gamma$ is torsion-free. This follows from work on special elements in hyperbolic (and relatively) hyperbolic groups.

In the torsion-free case, an element $\gamma \in \Gamma\setminus\{1\}$ is special if $\langle \gamma \rangle$ is a maximal cyclic subgroup of $\Gamma$ (equivalently, if $\gamma$ is not a proper power of another element). Since every non-trivial element of $\Gamma$ (as $\Gamma$ is torsion-free hyperbolic) is contained in a maximal infinite cyclic subgroup, we can assume that the element $\gamma$ from the question is special (if not, replace it with its maximal root). Now, since $g$ does not commute with $\gamma$, we know that $g \notin \langle \gamma \rangle $, whence we can apply Lemma 3.6 from [Minasyan, A.; Osin, D., Normal automorphisms of relatively hyperbolic groups., Trans. Am. Math. Soc. 362, No. 11, 6079-6103 (2010). ZBL1227.20041.] to conclude that for some $N \in \mathbb N$ the element $g\gamma^n$ is special as long as $|n| \ge N$. Thus $g\gamma^n$ is not a proper power in $\Gamma$ when $|n| \ge N$.

This shows that there can only be finitely many integers $n$ from the question. From here it's easy to conclude that for a given $n$ there can only be finitely many $h$ and $m$ because every element has only finitely many roots in a torsion-free hyperbolic group.

If $\Gamma$ is not torsion-free then one has to require, at least, that $g \notin E(\gamma)$, where $E(\gamma)$ is the maximal virtually cyclic subgroup of $\Gamma$ containing $\gamma$ (otherwise we have examples like the dihedral group in the comments). Similar arguments may work in this case, but it would certainly be more technical.

The answer is negative for $Isom(X)$, where $X$ is a Gromov hyperbolic metric space in general. Indeed, consider the free product $\Gamma=A*B$, where $A$ has a non-trivial element $a \in A$ such that $a$ admits infinitely many roots in $A$ (e.g., $a$ belongs to a copy of $\mathbb{Q}$ inside $A$).

Then $\Gamma$ acts faithfully on the Bass-Serre tree $X$ associated to the free product decomposition, thus $\Gamma \hookrightarrow Isom(X)$. Choose any non-trivial element $b \in B$ and set $\gamma=ba \in \Gamma$ and $g=b^{-1} \in \Gamma$. Note that $\gamma$ is a hyperbolic isometry of $X$ and $\gamma$ does not commute with $g$. By construction, $g\gamma=a$ has infinitely many roots in $\Gamma$, so the property from the original question fails.

I think that to have any chance of a positive answer in this case, one has to require that the action of $\Gamma$ on $X$ is at least proper (but probably more is needed to avoid problems with torsion).

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  • $\begingroup$ Thanks for your answer and reference. I want to consider the solution for $n$ big enough. But you example is very interesting. $\endgroup$
    – Yanlong Hao
    Commented Nov 9, 2023 at 13:03
  • $\begingroup$ "If $\Gamma$ is not torsion-free one has to assume at least $g\notin E(\gamma)$": this is of course not the case if $g,\gamma$ freely generate a semigroup. But this assumption was added after this answer was posted. $\endgroup$
    – YCor
    Commented Nov 9, 2023 at 23:47
  • $\begingroup$ @YanlongHao If you do not require properness of the action then, I think, even for large $n$ it is still possible to construct counter-examples among groups acting on trees. $\endgroup$
    – Ashot Minasyan
    Commented Nov 15, 2023 at 17:46

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