3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ I have added the hypothesis that $G$ is non-elementary, since otherwise the first paragraph is incorrect. $\endgroup$ Apr 21, 2012 at 14:22

2 Answers 2

5
$\begingroup$

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).

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – HJRW
    Apr 21, 2012 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.