Let $G$ be a f.g. abelian-by-finite group, i.e. there exists a f.g. abelian group $N$ which is normal in $G$ and such that the quotient $G/N = Q$ is finite.
The problem of classifying all such $G$, given $N$ and $Q$, is equivalent to classifying all the group extensions of $Q$ by $N$.

Does there exist such a classification, perhaps for $Q$ of small order?

EDIT: I'm briefly presenting a possible way to tackle the problem.
One can note that $G$ is virtually $\mathbb{Z}^d$, so it is $\mathbb{Z}^d$-by-finite. Let us call $N$ a normal subgroup, of finite index in $G$, isomorphic to $\mathbb{Z}^d$.
We know that $Q = G/N$ is an arbitrary finite group and we want to solve the extension problem for $N$ and $Q$. What we have to do is to find, for all fixed $Q$, the homomorphisms $$\varphi\ :\ Q\ \rightarrow\ Aut(N)$$ in order to reconstruct the action of $Q$ on $N$. Since $Aut(\mathbb{Z}^d) \cong GL(d,\mathbb{Z})$, one just has to consider the $k$-involutory elements of $GL(d,\mathbb{Z})$.
I found some references with a classification of these up to at least $d \le 7$ (and their number is finite $\forall\ d$). Once we have such integer $k$-involutory matrices, we can hope to conclude something about $\varphi$ and then about $G$.
Is that a good way to deal with the problem? Any other suggestion or insight into the problem would be greatly appreciated.

  • $\begingroup$ There is a very similar question -equivalent, I would say- here. $\endgroup$
    – recobrama
    May 6 '15 at 8:48

Such a group has a maximal finite normal subgroup $W(G)$, and $G/W(G)$ is called a crystallographic group. Crystallographic groups are the lattices in the isometry groups of Euclidean spaces, and are classified in small dimension (at least $\le 3$).

  • $\begingroup$ What you're saying is that, in order to classify abelian-by-finite groups, one takes the crystallographic groups and then try to solve the extension problem for them considering all the possible finite W, right? $\endgroup$
    – recobrama
    Apr 28 '15 at 16:47
  • $\begingroup$ We can do this in principle, but then the bottom group is an arbitrary finite group and I don't have a great hope to get something very interesting. In a sense, I would say that the more interesting part of the picture is this quotient group (the crystallographic group). $\endgroup$
    – YCor
    Apr 28 '15 at 17:02
  • $\begingroup$ Thank you very much for your answer. I edited my question in order to briefly explain an alternative way to tackle this problem. $\endgroup$
    – recobrama
    May 4 '15 at 9:56

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