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I am looking for a reference for the following:

Proposition. Suppose that $n \in \mathbb{N}$ and $G$ is a group with a normal subgroup $N \cong \mathbb{Z}^n$ and with a finite quotient $Q=G/N$. Then $G$ embeds into a semidirect product $\mathbb{R}^n \rtimes Q$, where $Q$ acts on $\mathbb{R}^n$ by Euclidean isometries.

If one assumes that the natural action of $Q$ on $N$ by conjugation is faithful, then the statement follows from Zassenhaus' theorem about crystallographic groups (see, for example, Theorem 2.2 in Section 2.2 of Szczepański, Andrzej, Geometry of crystallographic groups., Algebra and Discrete Mathematics (Hackensack) 4. Hackensack, NJ: World Scientific (ISBN 978-981-4412-25-4/hbk; 978-981-4412-26-1/ebook). xi, 195 p. (2012). ZBL1260.20070.).

I know two proofs of the proposition: one that reduces it to Zassenhaus' theorem and another that simply repeats the proof of the latter. But I suspect that the statement must have already been recorded somewhere in the existing literature.

Ak3.14
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Ashot Minasyan
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  • Did you check Wolf's book "Spaces of constant curvature"? But the proof is a straightforward application of the trivial fact that every finite linear group over $\mathbb R$ preserves a positive definite bilinear form. – Moishe Kohan Feb 11 '25 at 15:46
  • @MoisheKohan Wolf's book has a verion of Zassenhaus' theorem (Theorem 3.2.9) with a more direct but less detailed argument. It does not seem to contain the proposition as stated above. – Ashot Minasyan Feb 11 '25 at 15:55
  • Here is how I would prove it (sorry, do not know a reference): You have a virtually abelian group $G$ (containing a finite index subgroup $\mathbb Z^n$). Start with a faithful affine isometric action of $\mathbb Z^n$ on $\mathbb R^n$. Then use the induced representation construction to induce an action of $G$ on $\mathbb R^N$ where $N$ is some multiple of $n$: Then check (it is straightforward) that the induced action is again affine and isometric. qed (Or, just channel your inner Gromov and say "well-known and easy to prove.") – Moishe Kohan Feb 13 '25 at 03:36
  • @MoisheKohan The argument using the induced representation works but it increases the dimesion of the abelian subgroup. Zassenhaus' argument allows to keep the same dimension. In any case, I have written the proof down for future reference. – Ashot Minasyan Feb 13 '25 at 10:39
  • Of course, but if you want to get a minimal dimension, the action will be ineffective; you get minimal dimension by applying one of the Bieberbach's theorems (3rd?) to the action I constructed. – Moishe Kohan Feb 13 '25 at 11:34

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