3
$\begingroup$

I am in a situation where a discrete, finitely generated group $H$ satisfies property (T), and was wondering if I was able to conclude anything about the pair $(G,H^G)$, where $G$ is a finitely generated discrete group containing $H$, and $H^G$ is the normal closure of $H$ in $G$. In particular, does this pair necessarily also satisfy (T)? If not, are there conditions which would guarantee that it did?

Thanks for your time!

Improve this question
$\endgroup$
3
  • 5
    $\begingroup$ Clearly no: for instance when $H$ is finite $H^G$ can be large. For instance in a free product $G=H\ast K$ with $H,K\neq 1$ and $K$ having Property T the answer is always no. It's even false when $H$ is a commensurated subgroup, using slightly more refined counterexamples, such as $\mathbf{Z}[1/2]^2\rtimes (\langle 2\rangle \mathrm{SL}_2(\mathbf{Z}))$. $\endgroup$
    – YCor
    Commented Jan 22, 2020 at 21:53
  • 3
    $\begingroup$ Also in my paper Relative Kazhdan Property, Proposition 1.13: I showed that there is a group $\Gamma$ (lattice in some semisimple group) with some infinite subgroup $\Lambda$ such that $(\Gamma,\Lambda)$ has relative Property T, but such that for every subgroup $\Gamma'$ of $\Gamma$ and normal infinite subgroup $\Lambda'$ of $\Gamma$, the pair $(\Gamma',\Lambda')$ does not have relative Property T. $\endgroup$
    – YCor
    Commented Jan 22, 2020 at 21:58
  • $\begingroup$ Meant normal subgroup of $\Gamma'$, sorry. $\endgroup$
    – YCor
    Commented Jan 23, 2020 at 4:57

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .