most recent 30 from http://mathoverflow.net 2013-05-25T05:25:58Z http://mathoverflow.net/feeds/question/94737 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94737/free-groups-as-quotients-of-hyperbolic-groups Drike 2012-04-21T13:55:41Z 2012-04-22T19:41:55Z

<p>Given any infinite non-elementary hyperbolic group $G$, a theorem of Gromov asserts that there is a subgroup of $G$ isomorphic to a non-abelian free group on two generators.</p> <p>Is there a similar result for a quotient of $G$? That is, is there a normal subgroup $N$ of $G$ such that $G/N$ is isomorphic to a non-abelian free group on $r>1$ generators?</p>

http://mathoverflow.net/questions/94737/free-groups-as-quotients-of-hyperbolic-groups/94740#94740 Simon Thomas 2012-04-21T14:10:00Z 2012-04-21T14:10:00Z

<p>There exist hyperbolic groups $G$ with the Kahzdan property. Since every quotient of $G$ also has the Kahzdan property, it follows that $G$ has no nonabelian free quotients.</p>

http://mathoverflow.net/questions/94737/free-groups-as-quotients-of-hyperbolic-groups/94874#94874 Mark Sapir 2012-04-22T19:41:55Z 2012-04-22T19:41:55Z

<p>One does not need Kazhdan property (T). Take $\mathrm{PSL}_2(\mathbb{Z})$. The group is hyperbolic (it has a free subgroup of finite index), and is generated by an element of order 2 and an element of order 3 (it is the free product of two finite cyclic groups). Hence the generators die in every torsion-free homomorphic image. Thus $\mathrm{PSL}_2(\mathbb{Z})$ does not have non-trivial free homomorphic images. Of course groups with property (T) do not even contain finite index subgroups that map onto free non-trivial groups (that is a much stronger property). </p>