# When is the profinite completion a pro-$p$ group?

My research area is mainly pro-$p$ groups and profinite groups. However, in the last few year I became also interested in discrete groups. Therefore, it seems to me a natural problem to look for examples of finitely generated groups such that their profinite completion is a pro-$p$ group. One trivial example is when the profinite completion is trivial or a finite $p$-group, e.g. an infinite simple group. Hence, it makes sense to require the group to be infinite and residually finite. But then you can take residually finite, $p$-groups, e.g., the Grigorchuk group. In such case tha profinite completion is pro-$p$ for an easy reason.

So here are my questions:

1. Is there an infinite finitely generated residually finite non-torsion group such that its profinite completion is pro-$p$?

2. Is there an infinite finitely generated residually finite torsion free group such that its profinite completion is pro-$p$?

(Notice that profinite completion being pro-$p$ means that all subgroups of finite index are of $p$-power index.)

• You may find examples for 1), 2) around the Grigorchuk group. There are groups acting on binary rooted tree that, like the Grigorchuk group, possess congruence subgroup property, i.e., every subgroup of finite index contains a stabilizer of certain level in the tree (therefore its index is a power of 2). Usually this property is not difficult to deduce from branching properties of the group. There are examples of non-torsion group and torsion free groups that are weakly branch or maybe even regular branch (here I am not sure). – Ievgen Bondarenko Aug 26 '14 at 13:47
• Doesn't the basilica group work? – Benjamin Steinberg Aug 26 '14 at 16:05
• The basilica group appears to have infinite abelianization. – Ian Agol Aug 26 '14 at 17:54

I have a suggested approach to problem 1. The idea is to try to construct a group with a presentation satisfying the Golod-Shafarevich inequality, but which still retains elements of infinite order.

Take $F=\mathbb{Z}/p\ast \mathbb{Z}/p$, and enumerate the homomorphisms of $F$ to finite non-cyclic simple groups $\varphi_i: F\to H_i$. We'll fix an element $f\in F$ of infinite order. Inductively construct quotient groups $F\to F_i$ such that $F_i$ does not factor through any of the first $i$ homomorphisms to finite simple groups, while requiring that $F_i$ is hyperbolic and Golod-Shafarevich with only $p$-torsion and the image of $f$ in $F_i$ is infinite order. Suppose $\varphi_{i+1}: F \to H_{i+1}$ factors through $F \to F_i \to H_i$ (if not, let $F_{i+1}=F_i$, and repeat the induction). Choose an element $g_i \in F_i$ such that $g_i$ is infinite order and $g_i$ has non-trivial image in $H_i$ of order prime to $p$. Moreover, assume that the normalizers of $g_i$ and $f$ in $F_i$ are mutually malnormal, so have trivial intersection of all conjugates (I think these conditions are not difficult to arrange via the techniques of hyperbolic groups). Then kill a high-enough $p$-power of $g_i$ to get a quotient hyperbolic group $F_{i+1}$ which does not factor through $\varphi_{i+1}$, and $F_{i+1}$ satisfies the Golod-Shafarevich presentation condition, and remains hyperbolic with $f$ of infinite order (I claim that it is possible to arrange these conditions). Now, take the infinitely presented quotient $F_\infty$ by adding all the relators defining $F_i$. This will be a Golod-Shafarevich group, and $f$ will have infinite order in the image. Clearly $F_\infty$ can only have finite $p$-quotients, since we have killed all of the finite simple non-cyclic quotients.

The issue is though whether $F_\infty$ is residually finite. One may take its maximal residually finite quotient, which is non-trivial due to the Golod-Shafarevich property. But the question is how does one show that this quotient is non-torsion? It's not clear that $f$ has infinite order in the profinite completion, which would suffice. Maybe by choosing the sequence $g_i$ more carefully, one may guarantee this property.

• Actually, I thought about this a bit more, and I don't see any way to guarantee that the RF quotient is non-torsion. – Ian Agol Aug 26 '14 at 16:13
• Thanks Ian I got some idea which I need to think about. But the point I am taking from your idea is that problem 1 might be genuinely easier than problem 2. – Yiftach Barnea Aug 27 '14 at 12:46
• @YiftachBarnea: do you have a nice example of a non-torsion finitely generated pro-p group which has finite abelianization (i.e. no map to $\mathbb{Z}_p$), and say which isn't powerful? I think Lazards result should rule out groups with powerful pro-p completion. – Ian Agol Aug 27 '14 at 18:48
• Ian every just infinite non-nilpotent pro-$p$ group is like this. A result of Zelmanov implies that infinite f.g. pro-$p$ group is non-torsion. So you can take for example the pro-$p$ completion of the Grigorchuk group or the Nottingham group or the first congruence subgroup of $SL_d(\mathbb{F}_p[[t]])$ (assuming $p$ does not divide $d$). Furthemore, Fesenko constructed an example which is torsion free, but it is not very nice. – Yiftach Barnea Aug 27 '14 at 19:12
• Here is my idea. Take $\Gamma$ to be the Grigorchuk group and let $G$ be its profinite completion (which is pro-$p$). Take $\Delta$ a subgroup of $G$ generated by $\Gamma$ and an element of infinite order. One would hope that it is possible to do this so that the profinite completion of $\Delta$ is $G$. But I don't know how to do it. Alex Lubtozky convinced me that it is not true in general for any $\Gamma$. But maybe it is possible in this particular case. – Yiftach Barnea Aug 27 '14 at 21:51

Gustavo A. Fernández-Alcober, Alejandra Garrido and Jone Uria-Albizuri gave a positive answer to question 2 and therefore also to question 1 in On the congruence subgroup property for GGS-groups.