TITLE: Proving that a group is free
QUESTION [31 upvotes]: 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!

REPLY [3 votes]: 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.