# Nilpotent subgroups of uniform finite index

Let $G$ be a Lie group, $K\subseteq G$ be a compact group and $N\subseteq$ be a nilpotent group s.t. $N\cap K= \{e\}$. Let $H=N\rtimes K$ be the semidirect product of $N$ and $K$ and let $\Gamma$ be a discrete subgroup of $H$. Is it true that $\Gamma$ has a nilpotent subgroup of finite index. Also, can we guaranty that this index is uniform over all discrete $\Gamma\in H$. In other words, can we find a constant $C$ s.t. for all discrete subgroups $\Gamma$ of $H$, there exists $\tilde{\Gamma}$ a subgroup of $\Gamma$ s.t. $[\Gamma:\tilde{\Gamma}]\leq C$.

• You should make explicit that $N$ is normalized by $K$. – Alain Valette Mar 16 '13 at 22:27

Start with this paper: A. I. Mal′cev, On a class of homogeneous spaces, Izvestiya Akad. Nauk. SSSR. Ser. Mat. 13 (1949), 9–32 (look also here). This reduces your problem to the case when $\Gamma$ is a lattice in $N\rtimes K$. Next, observe that projection of $\Gamma$ to $K$ has to be a finite group (this should be in M.Raghunathan, "Discrete subgroups of Lie groups"), the result is due to Auslender, you can read his original paper here. Now, use Jordan-Schur Theorem: For every compact group $K$ there exists a number $j=j(K)$ so that every finite subgroup of $K$ contains an abelian subgroup of index $\le j$. This is also in Raghunathan's book, see also wikipedia article here and Tao's blog here.
• @Misha: you mean "for every compact Lie group $K$". Also I don't see why your argument is enough to conclude: assuming your argument works, you reduce to the case when the projection of $\Gamma$ on $K$ is abelian. This does not imply $\Gamma$ nilpotent, so some argument should be used, in the spirit of the fact that finite subgroups of $SL_n(\mathbf{Z})$ have a bounded order. Besides, your projection statement is suspicious: if $N\rtimes K$ is a direct product and $N=\mathbf{R}$, $K$ is the circle, you can pick $\Gamma$ to be an infinite cyclic lattice with a dense image in $K$. – YCor Mar 17 '13 at 18:11
• @Yves: You are right: I will add more details, but the proof still works. For the Auslender's result, one needs to assume that $K$ embeds in $Aut(N)$; if not, then one first splits off the compact kernel $K_2$: $H\cong (N\rtimes K_1)\times K_2$, where $K_1$ embeds in $Aut(N)$. Projection of a lattice in H to $N\rtimes K_1$ is still a lattice. – Misha Mar 17 '13 at 18:44