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Question. Is it true that each infinite hyperbolic group has a torsion free subgroup of finite index?

Are there counterexamples, or positive results for some large subclasses of hyperbolic groups? For example, is the answer positive for orbifold fundamental groups of negatively curved orbifolds? More precisely, I am interested the most in the case of orbifolds with locally $CAT(0)$ metric. I guess it will be hard to construct a counterexample in this category.

Related question. Is it known that every hyperbolic group has a non-trivial subgroup of finite index?

Just to recall, a definition of hyperbolic group is here http://en.wikipedia.org/wiki/Hyperbolic_group .

Added. Note, that every hyperbolic group is finitely presented (thanks to Sam Nead).

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Gromov gave an example of a f.g. infinite torsion group acting on a space of nonpositive curvature in §4.5C of Mikhail Gromov, Hyperbolic groups. Essays in group theory, 75--263, Math. Sci. Res. Inst. Publ., 8, Springer, New York, 1987 –  Michele Triestino Sep 27 '10 at 10:47
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I don't think that Gromov's example is hyperbolic. Infinite hyperbolic groups contain elements of infinite order. –  Sam Nead Sep 27 '10 at 11:34
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That should be "Thanks to Gromov". :) –  Sam Nead Sep 27 '10 at 11:53
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On the CAT(0) side, let me just note that Wise constructed a fp CAT(0) group that isn't residually finite. This seems like weak evidence that, in the non-positively curved setting, your question might have a negative answer. –  HJRW Sep 27 '10 at 14:48
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To add on Henry's comment: Wise constructed a CAT(0) group with no finite quotients. Another example of such groups are due to Burger-Moses; their groups are lattices in products of trees, and are simple, hence they have no proper finite index subgroups. See Bridson's paper arxiv.org/pdf/math/9810188 and references therein. –  Igor Belegradek Sep 27 '10 at 17:06
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2 Answers 2

up vote 17 down vote accepted

This is a well known open problem. The following properties are equivalent

a) Every hyperbolic group is residualy finite

b) Every hyperbolic group has a finite index torsion-free subgroup.

The proof is either here: Olʹshanskiĭ, A. Yu. On the Bass-Lubotzky question about quotients of hyperbolic groups. J. Algebra 226 (2000), no. 2, 807--817 or here: Kapovich, Ilya; Wise, Daniel T. The equivalence of some residual properties of word-hyperbolic groups. J. Algebra 223 (2000), no. 2, 562--583 or can be given by exactly the same methods as in these two papers (I do not remember exactly which of these three possibilities hold).

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And (a) and (b) are also both equivalent to Dmitri's 'related question'. –  HJRW Sep 27 '10 at 14:38
    
Yes, that is certainly proved in both cited papers. –  Mark Sapir Sep 27 '10 at 14:52
    
Mark, thanks for the answer! –  Dmitri Sep 27 '10 at 14:53
    
I decided to ask a follow up question mathoverflow.net/questions/40170/… –  Dmitri Sep 27 '10 at 15:59
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See this page by Coornaert for an introduction and references. In particular, finitely generated hyperbolic groups are always finitely presented (so you don't need to add that condition). Also, the referenced page suggests that your first question, about finite index torsion free subgroups, is open. I looked in various references for your related question, but didn't find anything.

Edit: Aha! I finally looked in the right place. Apparently your related question is also open.

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Sam, huge thanks! –  Dmitri Sep 27 '10 at 11:48
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