Let me start this question with an example that hopefully makes clear what I am looking for:

A discrete subgroup $G$ of the group of euclidean isometries of $\mathbb{R}^d$ is called a crystallographic group if it contains a lattice $L$ of full rank. In this case, $G/L \cong P$ is a finite crystallographic subgroup of the orthogonal group. So what we get is a short exact sequence $$0 \rightarrow \mathbb{Z}^d \rightarrow G \rightarrow P \rightarrow 1,$$ which can also be expressed as stating that $G$ is isomorphic to an extension of $\mathbb{Z}^d$ by $P$. Remarkably, the converse is also true. Namely that any such extension is isomorphic to a crystallographic group. So the family of crystallographic groups is in $1$-to-$1$ correspondence with group extensions of $\mathbb{Z}^d$ by a crystallographic point group.

Now, for my question. What other interesting examples of families of groups are there that can also be described by group extensions in a similar way as crystallographic groups can?

Let me add a few sentences to maybe prevent misinterpretation of my question. I am certainly aware of the theorem of Jordan-Hölder and its consequence that one in principle could survey all finite groups by using the classification of all finite simple groups and solving the extension problem. But this also means allowing for very complicated sets of groups that can be used as a first part and last part in the above sequence. I am looking for more reasonable examples, where "reasonable" appeals to everyone's intuition.

  • $\begingroup$ Nice question, thank you. I would also add the tag of geometric topology to it. $\endgroup$ Sep 25, 2013 at 11:56
  • 1
    $\begingroup$ Would something like the class of metabelian groups qualify? en.wikipedia.org/wiki/Metabelian_group Also people study "$x$-by-$y$ groups" where $x$ and $y$ are adjectives such as cyclic, abelian, finite, nilpotent... I must confess, I think you need to make your question a little more precise. $\endgroup$
    – Mark Grant
    Sep 25, 2013 at 11:59
  • $\begingroup$ Lots of group extensions appear in Galois embedding problems. But I am not sure if they fit in your criterion for being interesting! $\endgroup$
    – user23860
    Sep 25, 2013 at 13:25
  • $\begingroup$ This example has a quasi-isometric generalization: any group QI to $\mathbb Z^d$ is ((finite)-by-($\mathbb Z^d$-by-$P$)). So this applies, for example, to any group containing a finite index $\mathbb Z^d$ subgroup. $\endgroup$
    – Lee Mosher
    Sep 25, 2013 at 13:39
  • 1
    $\begingroup$ Your statement is not quite true: Direct product of a finite group and a free abelian group is not crystallographic. A better definition is of a "c-type group", as a group which acts properly discontinuously isometrically and cocompactly on a Euclidean space. In this definition, non-effective action with finite kernel are allowed. Then, a f.g. group is virtually abelian iff it is a c-type group. $\endgroup$
    – Misha
    Sep 25, 2013 at 14:40

1 Answer 1


An example of a similar type is Gromov's Theorem that a finitely generated group has polynomial growth if and only if it is virtually nilpotent (i.e. it has a normal nilpotent subgroup of finite index).

Or Mal`cev's theorem that a polycyclic group is nilpotent by abelian by finite. So a group is polycyclic-by-finite if and only if it is (finitely generated nilpotent)-by-(finitely generated abelian)-by-finite.

  • 1
    $\begingroup$ Derek, what is your convention for "$G$-by-$H$" notation? $1 \to G \to X \to H \to 1$? or $1 \to H \to X \to G \to 1$? $\endgroup$
    – Lee Mosher
    Sep 25, 2013 at 13:43
  • $\begingroup$ I am using $G$-by-$H$ to mean normal subgroup $G$ with quotient $H$. So polycyclic groups have a normal nilpotent subgroup with virtually abelian quotient group. (I know A-by-B-by-C for classes of groups A,B,C, is not necessarily associative, but I think it makes no difference in this case.) $\endgroup$
    – Derek Holt
    Sep 25, 2013 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.