# Finitely generated, infinite, residually finite groups whose finite quotients are $p$-groups.

This question grew out of this post.

Question: Is there a finitely generated, infinite, residually finite group such that every finite index subgroup has $p$-power index for a fixed prime $p$?

The $p$-adic integers $\mathbb Z_p$ give an example of a non-finitely generated such group. This is not entirely obvious, but follows from a result of Jean-Pierre Serre, which states that every finite index subgroup of $\mathbb Z_p$ is closed.

EDIT: André Henriques has pointed out that Rostislav Grigorchuk constucted a finitely generated, infinite, residually finite $2$-group. All finite quotients of this group have to be $2$-groups. Colin Reid asked in a comment whether there is a torsionfree example. So let me take the freedom to extend my question:

Question: Is there a torsionfree example?

• I suppose you want to exclude p-groups from this; if not, any finitely generated infinite residually finite p-group is an example. I wonder if there are any torsion-free examples? – Colin Reid Nov 20 '10 at 13:11
• @Andreas: I will remove my comment which you refer to together with my answer there. So you may want to modify your question accordingly. You did not copy my comment precisely. I asked for a non-torsion f.g. example. Of course I know about torsion ones. There are also misprints in the title of your question. – user6976 Nov 20 '10 at 15:41
• @Mark: Sorry if I misread your comment. – Andreas Thom Nov 20 '10 at 20:56

Yes, there are torsion free examples. I do not know who constructed them first, but some examples can be found in papers by Grigorchuk and his co-authors. For example Bartoldi and Grigorchuk proved that a certain Fabrykowski-Gupta group $\Gamma$ has the following properties (see Propositions 6.4 and 6.5 in arXiv:math/9911206):

(a) It is a subgroup of the automorphism group of a rooted tree.

(b) It is virtually torsion free.

(c) It satisfies the following 'congruence property': every finite index subgroup of $\Gamma$ contains a level stabilizer (i.e., the stabilizer of a level of the tree) and the index of every level stabilizer is a power of 3.

By (a) $\Gamma$ is residually finite. By (b) there is a torsion free subgroup $K$ of finite index in $\Gamma$. Since every finite index normal subgroup of $K$ contains a finite index normal subgroup of $\Gamma$, (c) implies that every finite quotient of $K$ is a 3-group.

Grigorchuk's group $G$ is an example of what you're looking for: http://en.wikipedia.org/wiki/Grigorchuk_group.

Every element of $G$ has finite 2-power order, and so every finite quotient of $G$ is a 2-group.

• For the moment, let me not count this as an answer, since the question for a torsionfree example (raised by Colin Reid in a comment) is really a bit more interesting. – Andreas Thom Nov 20 '10 at 13:31