# Finite group extensions of lattices

I'm currently reading the proof of Geroch's conjecture in Lawson-Michelsohn's Spin Geometry book and in the proof of Proposition IV.5.8 that every Ricci-flat enlargeable manifold is flat the following situation arises:

We have the Riemannian product $$\mathbb{R}^n\times Y$$, where $$Y$$ is a simply connected, compact Riemannian manifold and a group $$\Gamma$$ acting properly discontinuously by isometries on $$\mathbb{R}^n\times Y$$ such that the quotient is a compact manifold. Since $$\mathrm{Isom}(\mathbb{R}^n\times Y) = \mathrm{Isom}(\mathbb{R}^n)\times \mathrm{Isom}(Y)$$ (also not trivial, but I was able to show that), Lawson and Michelsohn claim the following:

By passing to a subgroup of finite index we may assume that $$\Gamma$$ acts freely and properly discontinuously on $$\mathbb{R}^n$$.

The projection $$\pi : \mathrm{Isom}(\mathbb{R}^n\times Y)\rightarrow \mathrm{Isom}(\mathbb{R}^n)$$ induces a short exact sequence $$1\rightarrow \ker (\pi)\rightarrow \Gamma\rightarrow \mathrm{Im }(\Gamma)\rightarrow 1.$$ Since $$\Gamma_1:=\ker(\pi)$$ acts properly discontinuously on the compact space $$Y$$ it must be a a finite group. Further, I was able to show that $$\Gamma_2 := \mathrm{Im}(\Gamma)$$ acts properly discontinuously on $$\mathbb{R}^n$$ (this also follows from the compactness of $$Y$$). By Bieberbach's theorem on cristallographic groups there is a lattice $$\mathbb{Z}^n\subseteq \Gamma_2$$ of finite index and hence we can assume w.l.o.g. $$\Gamma_2 =\mathbb{Z}^n$$.

Hence we obtain a short exact sequence $$1\rightarrow \Gamma_1 \rightarrow \Gamma \rightarrow \mathbb{Z}^n \rightarrow 1,$$ or in other words, $$\Gamma$$ is a group extension of $$\mathbb{Z}^n$$ by a finite group $$\Gamma_1$$.

My strategy is to find a sublattice $$L\subseteq \mathbb{Z}^n$$ that lifts to a lattice $$L^\prime\subseteq\Gamma$$ of finite index but all my attempts seem to be cursed. I'm thankful for any idea on how to resolve this!

If I've understood your question (with the background removed), it is this. If a group $$\Gamma$$ has a finite normal subgroup $$\Gamma_1$$ with quotient $$\mathbb{Z}^n$$, does there exist a subgroup of $$\Gamma$$ isomorphic to $$\mathbb{Z}^n$$ whose image in the quotient has finite index? If this is your question, then the answer is yes. First, since $$\operatorname{\rm Aut}(\Gamma_1)$$ is finite, there is a finite index subgroup $$G$$ of $$\Gamma$$ mapping to the identity subgroup of $$\operatorname{\rm Aut}(\Gamma_1)$$. So $$G\cap \Gamma_1$$ is central in $$\Gamma_1$$, and $$G$$ has nilpotence class two. Commutators in $$G$$ land in $$\Gamma_1$$, and have finite order, so $$G$$ has a finite index subgroup that is commutative. This is the group you desire.

• I should mention that I'm using the fact that in a nilpotent group of class two, we have $[a,b][a,c]=[a,bc]$, so commutators are bilinear. Apr 20 at 17:02