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A finitely generated virtually abelian group $G$ must have $\mathbb Z^n$ as a normal subgroup with $G/\mathbb Z^n$ finite. For each $n$ there are a finite number of examples of such groups arising as space or crystallographic groups, represented by isometries on $\mathbb R^n$, where the action of the finite group on $\mathbb Z^n$ is faithful, but of course there are many more finitely generated virtually abelian groups than there are space groups, most obviously by taking a direct product of a space group and a finite group.

It would be very convenient for me if any finitely generated virtually abelian group with $n=2$ was of this form, i.e. a direct product of one of the 17 wallpaper groups and some finite group, but I think perhaps that's a bit too much to hope for. Could anybody point me to a proof or a counterexample? I only need $n=2$ but I'm curious about general $n$.

I found the claim

If $G$ is finitely generated virtually abelian then there is a projection with finite kernel to a crystallographic group.

in https://core.ac.uk/download/pdf/82725144.pdf but I can't work out what projection means there. I don't see how it helps to just mod out by the kernel of the map from the finite group to $\operatorname{Aut}(\mathbb Z^n)$, and I think perhaps that's what is considered there.

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I think projection just means surjective homomorphism.

A counterexample to your conjecture is the group $\langle x,y \mid y^4=1, y^{-1}xy=x^{-1} \rangle$. It has the infinite cyclic group $\langle x \rangle$ as a subgroup of index $4$. It projects onto the infinite dihedral group (which is a space group) with kernel of order $2$, but it is not a direct product.

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  • $\begingroup$ Thank you; I was sort of hopeful that things would be easier at low $n$, but your example sets that straight. I do see how there must be an epimorphism with finite kernel but I doubt that helps me. Back to the drawing board! $\endgroup$ – Søren Eilers Jul 20 '18 at 11:12

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