# Groups whose finite index subgroups of fixed index are isomorphic

I am interested in finitely generated groups $G$ that are residually finite and have the following property: For each $d \geq 1$, $G$ has subgroups of finite index $d$, and all such subgroups are isomorphic.

I know three infinite families of such groups: (1) Free abelian groups $\mathbb Z^n$ (2) Free groups $F_n$ (3) Fundamental groups of closed orientable surfaces $\Gamma_n = \pi_1 (\Sigma_n)$.

The way to see that these are indeed examples is via topology: the classifying spaces of these groups are tori, graphs and surfaces and we understand their covering theory.

[Note: some geometric group theorist claim that these are the easiest torsion-free groups that exist. Their outer automorphism groups are given by $GL_n(\mathbb Z)$, $Out(F_n)$ and $Mod^{\pm}_n$, groups that generated a vast body of research.]

Are there more (than the above mentioned) examples of such groups?

similar, but in fact a different question

• I presume that constructing new examples from old, by taking the direct product with a group that has no proper finite index subgroups, is cheating? – Jeremy Rickard May 10 '16 at 13:19
• In a similar vein to Jeremy's comment, Baumslag gave an example of a one-relator group with every finite quotient cyclic. This group has a unique subgroup of each finite index. – HJRW May 10 '16 at 18:53
• The second two examples are mediated by the following phenomenon: you have a class of groups closed under taking finite index subgroups which have the property that they are determined up to isomorphism within the class by their Euler characteristic. This seems pretty rare, e.g. I wouldn't expect it for fundamental groups in any other dimension. – Qiaochu Yuan May 10 '16 at 20:02
• If one removes the somewhat rigid assumption that the group is torsion-free and finitely presented (and instead just assume finitely generated), we can reduce to assuming that the group is residually finite, which discards all the trivial counterexamples given so far. – YCor May 10 '16 at 20:19
• Thanks for the idea to change the conditions, it really seems more natural to ask for residual finiteness. I have edited the question as suggested. – Jens Reinhold May 10 '16 at 22:07