Proving that a group is free - MathOverflow

Proving that a group is free

2010-03-16T20:56:11Z

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!

Answer by Richard Kent for Proving that a group is free

2010-03-16T21:08:33Z

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 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.")

Answer by HenrikRüping for Proving that a group is free

2010-03-16T22:15:51Z

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.

Answer by Wolffo for Proving that a group is free

2010-05-11T15:56:03Z

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.