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I've got a group $G$ that I'm trying to prove is free. I already know that $G$ is torsion-free. Moreover, I can "almost" prove what I want : I can find a finite index subgroup $G'$ of $G$ that is definitely free.

This leads me to the following question. Can anyone give me an example of a torsion-free group $G$ that is not free but contains a free subgroup of finite index? I've tried pretty hard to find groups like this, but i can't seem to avoid introducing torsion. Thanks!

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It's a theorem of Stallings and Swan that a group of cohomological dimension one is free.

By a theorem of Serre, torsion-free groups and their finite index subgroups have the same cohomological dimension.

So, a torsion-free group is free if and only if one of its finite index subgroups are free.

(Here are the references. For Stallings-Swan, see

John R. Stallings, "On torsion-free groups with infinitely many ends", Annals of Mathematics 88 (1968), 312–334.

and

Richard G. Swan, "Groups of cohomological dimension one", Journal of Algebra 12 (1969), 585–610.

Serre's theorem is in Brown's book "Cohomology of Groups.")

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    $\begingroup$ Thank you very much! Given the people whose work you had to use to prove this, I don't feel so bad for not being able to do it myself =) $\endgroup$
    – Nick Hildebrand
    Commented Mar 16, 2010 at 22:31
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If you don't like cohomological dimension:

Given a group that acts properly (and cocompactly) on a tree. Then any finite extension of this group also acts properly and cocompactly on a tree. The idea of the construction is contained in the article Dunwoody, "Accessibility and Groups of Cohmological Dimension One".

It is shown there, that any such action determines a system of "almost invariant subsets" and the other way round. The existence of such a system passes directly to a finite extension. So your finite extension also acts properly and cocompactly on a tree and (as it is torsionfree) is free.

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If a torsion free group is quasi-isometric to a (nontrival) free product, then it is free product.(Gromov). And we know that a finite index subgroup of G is quasi-isometry to G. So G is also free.

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    $\begingroup$ The meat of the theorem you state is due to Stallings, not Gromov. Gromov is presumably responsible for phrasing it in terms of quasiisometries, but Stallings certainly knew that a group and its finite-index subgroups have the same number of ends. So your answer adds nothing to the answer that Richard Kent gave above. -1. $\endgroup$
    – HJRW
    Commented May 11, 2010 at 17:44
  • $\begingroup$ Yes, I know it is essentially Stallings theorem, but sometimes quasiisometry is easy to see. $\endgroup$
    – Wolffo
    Commented May 12, 2010 at 2:29
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    $\begingroup$ And sometimes Gromov gets credit for the whole of geometric group theory (whereas he only deserves credit for most of it). $\endgroup$
    – HJRW
    Commented May 12, 2010 at 21:15
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    $\begingroup$ Gromov deserves credit to none of geometric group theory done before, say, 1978. This includes a lot of things, including all this Stallings stuff, and work of many people (among others: Dehn, Siegel, Freudenthal, Specker, Milnor, Dunwoody, Abels). Gromov's great contributions, did not only consist in proving strong results and introduce and study fundamental notions (esp. hyperbolicity), but also to phrase concepts in a language that make previously known results more intuitive. $\endgroup$
    – YCor
    Commented Apr 18, 2018 at 22:47

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