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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 $\mathbb{R}^n$.

I've been thinking about this sentence quite a lot and this is how far I got:

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!

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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.

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  • $\begingroup$ 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. $\endgroup$ Apr 20 at 17:02

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