Proving that a group is free - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:30:47Z http://mathoverflow.net/feeds/question/18423 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18423/proving-that-a-group-is-free Proving that a group is free Nick Hildebrand 2010-03-16T20:56:11Z 2010-05-11T15:56:03Z <p>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.</p> <p>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!</p> http://mathoverflow.net/questions/18423/proving-that-a-group-is-free/18425#18425 Answer by Richard Kent for Proving that a group is free Richard Kent 2010-03-16T21:08:33Z 2010-03-16T21:16:18Z <p>It's a theorem of Stallings and Swan that a group of cohomological dimension one is free.</p> <p>By a theorem of Serre, torsion-free groups and their finite index subgroups have the same cohomological dimension.</p> <p>So, a torsion-free group is free if and only if its finite index subgroups are free.</p> <p>(Here are the references. For Stallings-Swan, see</p> <p>John R. Stallings, "On torsion-free groups with infinitely many ends", Annals of Mathematics 88 (1968), 312–334. </p> <p>and</p> <p>Richard G. Swan, "Groups of cohomological dimension one", Journal of Algebra 12 (1969), 585–610.</p> <p>Serre's theorem is in Brown's book "Cohomology of Groups.")</p> http://mathoverflow.net/questions/18423/proving-that-a-group-is-free/18432#18432 Answer by HenrikRüping for Proving that a group is free HenrikRüping 2010-03-16T22:15:51Z 2010-03-16T22:15:51Z <p>If you don't like cohomological dimension:</p> <p>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 <a href="http://plms.oxfordjournals.org/cgi/pdf_extract/s3-38/2/193" rel="nofollow">Dunwoody, "Accessibility and Groups of Cohmological Dimension One"</a>.</p> <p>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.</p> http://mathoverflow.net/questions/18423/proving-that-a-group-is-free/24253#24253 Answer by Wolffo for Proving that a group is free Wolffo 2010-05-11T15:56:03Z 2010-05-11T15:56:03Z <p>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.</p>