What sort of spaces cover themselves with a finite fibre? Or, what sort of finitely generated groups contain isomorphic copies of themselves as subgroups as finite index? Is it a reasonable question to ask to classify them in any way?
-
1$\begingroup$ see mathoverflow.net/questions/15115/… $\endgroup$– Igor BelegradekMar 18, 2014 at 1:33
-
3$\begingroup$ You mean "proper subgroup of finite index". These include all direct products of finitely generated groups with $\mathbf{Z}$, so there is not much to expect from a classification. $\endgroup$– YCorMar 18, 2014 at 8:12
1 Answer
Edit. I read more into the question than it was asking. The version asked was trivial as Yves points out. The more interesting question, which my answer addresses, is about groups $G$ with a finite index subgroup isomorphic to it so that the induced sequence obtained by iteration has trivial intersection.
Besides obvious examples like finitely generated free abelian groups and some other virtually nilpotent groups, the lamplighter groups $A\wr \mathbb Z$ with $A$ finite abelian have a finite index subgroup isomorphic to the whole group, as do the solvable Baumslag groups. These can be realized as level stabilizers for the automaton group representations of these groups. This was first observed by me as a consequence of these groups being what was once called strongly fractal (I forget the current terminology) and having essentially-free action on the boundary of the tree in about 2005 or so, but never published because I didn't know people were interested in it (Benjamini had conjectured only virtually nilpotent groups could have a finite index subgroup isomorphic to themselves but I wasn't aware of his conjecture). Nekrashevych also observed it and it appears in his paper, Scale-invariant groups http://arxiv.org/abs/0811.0220, with Gabor Pete, along with lots of other results.