I know how to prove the following result. However, my proof is a little bit long and complicated and only uses fairly low tech results in group cohomology. It would be nice if I could find a citation in the literature that directly implies this claim but I am not experienced in group theory or homological algebra and to be honest I'm not sure where to start looking for such a result.
$\textbf{Lemma}:$Lemma: Suppose $\mathbb{T}^n = \mathbb{R}^n/ \mathbb{Z}^n $ is the standard $n$-dimensional torus and $\Gamma$ is a finite group of order $k$. Consider a short exact sequence: $$ 0 \to \mathbb{T}^n \to E \to \Gamma \to 1 $$$$ 0 \to \mathbb{T}^n \to E \to \Gamma \to 1. $$ Suppose $ R := \{ t \in \mathbb{T}^n \ | \ kt = 0 \} $$ R := \{ t \in \mathbb{T}^n \mathrel| kt = 0 \} $. Then the following short exact sequence splits: $$ 0 \to \mathbb{T}^n/R \to E/R \to \Gamma \to 1 $$$$ 0 \to \mathbb{T}^n/R \to E/R \to \Gamma \to 1. $$
The proof basically boils down to the fact that $ H^2( \Gamma, \mathbb{T}^n) $ is annihilated by $k$ and using that to show that the induced map $ H^2(\Gamma, \mathbb{T}^n) \to H^2(\Gamma, \mathbb{T}^n/R) $ is the zero map and therefore the curvature class of the top short exact sequence projects to zero.
I guess the things I would like to know are:
- Is this 'obvious' to those experienced with finite group extensions or Lie theory?
- Is there a textbook/paper where I can find a result or exercise that directly implies this lemma?
