It's known that subgroups of Gromov's hyperbolic groups are not necessarily hyperbolic. Is there any counter-example when the quotient is Abelian. More precisely, let $G$ be a Gromov's hyperbolic group and $H$ a normal subgroup of $G$ such that $G/H$ is Abelian. Is $H$ a Gromov's hyperbolic group?
$\begingroup$
$\endgroup$
2
asked Mar 14, 2014 at 17:45
-
5$\begingroup$ The answer is no. The case when H isn't f.g. is trivial, examples when H is f.g is due to Rips, and when H is f.p. this was found by Brady, see aftermath.math.ou.edu/~nbrady/papers/sub.ps and references therein. $\endgroup$– Igor BelegradekCommented Mar 14, 2014 at 18:56
-
$\begingroup$ @IgorBelegradek, why don't you post your comment as an answer? $\endgroup$– HJRWCommented Mar 17, 2014 at 9:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The answer is no. The case when $H$ isn't finitely generated is trivial (think of the free groups and its commutator subgroup), examples when $H$ is finitely generated are due to Rips (see the famous Rips construction , and when $H$ is finitely presented this was found by Brady, see here .
answered Mar 17, 2014 at 13:06