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!



Proving that a group is free
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.")


