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Is it true that torsion subgroups of hyperbolic groups are finite?

I have a vague memory that this is true, perhaps due to Ol'shanskii, but have been struggling to find a reference.

(By a torsion subgroup I mean a subgroup $T\leq H$ such that every $t\in T$ has finite order. Although there are finitely many conjugacy classes of finite order elements in $H$, this does not immediately imply that $T$ is finite.)

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    $\begingroup$ See Theorem 1 in here: ams.org/journals/proc/2005-133-02/S0002-9939-04-07578-1/… $\endgroup$ Commented Nov 16, 2016 at 11:26
  • $\begingroup$ @LorenoHeer, it's reassuring to see the statement in print, but since the cited paper does not provide a specific reference for the statement, it's not very useful. $\endgroup$
    – HJRW
    Commented Nov 16, 2016 at 14:07
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    $\begingroup$ Not only they are finite but they have bounded order and form finitely many conjugacy classes. It's due to Gromov (in the 1987's original article in English) (the various authors who wrote redigests of Gromov's article reproved it... in this precise case Gromov's proof is reasonably complete). $\endgroup$
    – YCor
    Commented Nov 16, 2016 at 17:20

3 Answers 3

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This is Corollaire 36, Chapitre 8 in the (French) book "Sur les Groupes Hyperboliques d’après Mikhael Gromov", Editors: Etienne Ghys, Pierre de la Harpe, Progress in Mathematics 83. Birkhäuser Boston (1990)

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  • $\begingroup$ Great, thanks! That said, I will hold off accepting your answer in case someone comes up with an English-language proof. (Of course this book is a standard reference, but I know no French.) $\endgroup$
    – user101216
    Commented Nov 16, 2016 at 15:33
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Here is a proof that uses only the dynamics of the action of a hyperbolic group on its boundary.

Every hyperbolic group $G$ acts as a uniform convergence group on its Gromov boundary $\partial G$. This means that the action of $G$ on the space of ordered triples of distinct points of $\partial G$ is properly discontinuous and cocompact. For an elegant, simple proof of this fact, refer to Proposition 1.13 of Bowditch's article Convergence groups and configuration spaces.

Any subgroup $H$ of a convergence group whose limit set has at least three points must always contain a nonabelian free subgroup, by the Tits Alternative, see Theorem 2U of Tukia's paper Convergence groups and Gromov's metric hyperbolic spaces.

It follows from the Tits Alternative that any infinite torsion subgroup $H$ of a hyperbolic group must have a limit set with at most two points. Since subgroups with empty limit set are finite, and subgroups with two-point limit set are virtually cyclic, the only possibility is that there exists a point $p$ in $\partial G$ fixed by every element of the infinite torsion group $H$.

However, in a uniform convergence group action, no point can be fixed by infinitely many non-loxodromic elements. This is a key step in the proof, and a short argument can be found in Theorem 3A of Tukia's article Conical limit points and uniform convergence groups.

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Alternatively, one can refer to Lemma 3.4 from Osin's article Acylindrically hyperbolic groups.

Proposition: If a group acts on a hyperbolic space uniformly properly$^*$ with unbounded orbits, then it contains a loxodromic element.

This implies directly that torsion subgroups of hyperbolic groups are finite. The result is attributed to Ivanov and Olshanskii, the proof given by Osin being extracted from Lemma 17 in Hyperbolic groups and their quotients of bounded exponents.

$^*$An action $G \curvearrowright (X,d)$ is uniformly proper if, for every $R \geq 0$, there exists $N \geq 0$ such that $\# \{g \in G \mid d(x,gx) \leq R \} \leq N$ for every $x \in X$.

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  • $\begingroup$ What's the definition of "uniformly properly"? $\endgroup$
    – YCor
    Commented Jan 3, 2022 at 15:43
  • $\begingroup$ @YCor: I added a definition. $\endgroup$
    – AGenevois
    Commented Jan 3, 2022 at 17:43
  • $\begingroup$ OK, thanks. Indeed, any proper cobounded isometric action of a discrete group is then uniformly proper, and this passes to restriction to subgroups (while coboundedness is lost). $\endgroup$
    – YCor
    Commented Jan 3, 2022 at 18:33

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