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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.

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?

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I have added the hypothesis that $G$ is non-elementary, since otherwise the first paragraph is incorrect. –  Simon Thomas Apr 21 '12 at 14:22
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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).

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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.

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For instance, Zuk proved that many random groups have property T. See Żuk, A, Property (T) and Kazhdan constants for discrete groups. Geom. Funct. Anal. 13 (2003), no. 3, 643–670. –  HJRW Apr 21 '12 at 16:35
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