# Topological examples of profinite groups

I am preparing a course on profinite groups, to be delievered to early graduate students. The first part of the course will discuss the equivalent characterizations of profinite groups. I will first define a profinite group as a Hausdorff, compact topological group such that the open subgroups form a base for the neighbourhoods of the identity.

I would like to give my students a few ‘natural’ examples. I am looking for examples of profinite groups that give the students something they can wrap their heads around without knowing about products of finite groups and inverse limits. Ideally, I am looking for examples which have more of a topological emphasis.

As an example of what I am looking for: The $p$-adic integers $\mathbb{Z}_p$ can be constructed as the completion of $\mathbb{Z}$ with the metric space structure induced from the $p$-adic absolute value.

I would particularly like to see an example of a group acting on some object which induces some topology on the group. I would like to exclude Galois groups: these will be covered in another part of the course. I would also like to exclude the fundamental groups encountered in algebraic geometry: I do not expect my students be familiar with this material.

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Why do you want to avoid products of finite groups? It's the simplest way to give an example without introducing any new ideas. By the way, I think it's a bad idea to start off with a definition of a profinite group using its topological characterization. That will be largely meaningless to the students, hence setting a bad tone for the course. I think instead you should start by explaining the inverse product construction (alongside the $p$-adic integers or perhaps formal power series over a field). After they wrap their heads around that they'll appreciate the topological viewpoint. –  KConrad Jan 3 '12 at 2:38
"I would like to exclude Galois groups". Actually, all profinite groups are Galois groups, so you may be in trouble. More seriously, I agree with KConrad: for most us of a profinite group is a projective limit of finite groups, so it's better to start with that as the definition. –  anon Jan 3 '12 at 3:17
You could of course give the profinite or pro-p completion of any discrete group. If the group is f.g. it can be given by completing with respect to an obvious metric generalizing the p-adic metric. –  Benjamin Steinberg Jan 3 '12 at 3:25
To be a devil's advocate, I would like to point out that the Krull topology is nothing more than the topology of pointwise convergence (equals compact-open) for the action of the Galois group on the field (with the discrete topology on K). This trivially gives the Galois group is totally disconnected. The fact that an element has only finitely many Galois conjugates implies immediately compactness so one trivially gets profiniteness of the Galois group from this viewpoint. –  Benjamin Steinberg Jan 3 '12 at 3:31
Profinite groups can be viewed in various ways and that is, in some sense, their strength. How to present them slightly depends on how you plan to use them. If you would like to study the asymptotic behaviour of finite groups, then inverse limit is the way. If on the other way you are interested in Lie groups, then compact totally disconnected is the way. Intuitively, I like to think about them as residually-finite groups which are “determined” by their finite quotients. This is not completely accurate (more accurate if they are finitely generated) but it gives the right feeling about them. –  Yiftach Barnea Jan 3 '12 at 16:48

If you view the Cantor space as a sequence space, then the isometry group is profinite (using the usual sort of metric that words are close if they have a long common prefix). The automorphism group of a locally finite rooted tree is profinite or more generally the stabilizer of a vertex is profinite in the automorphism group of a locally finite graph. Here use the compact-open topology for the path metric.

Added. More generally the isometry group of any compact totally disconnected metric space is profinite.

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@Benjamin: I'm a bit confused (as usual) - what if anything is the difference between the automorphism group of such a tree and the isometry group of such a tree? I seem to recall that the latter is one of the standard examples of a unimodular, totally disconnected, non-compact group. –  Yemon Choi Jan 2 '12 at 23:42
@Yemon, there is no difference really between automorphisms and isometries. The automorphism group of a tree is non-compact in general, but a vertex stabilizer is compact and do the automorphism group of a locally finite ROOTED tree is profinite. The point is it stabilizes the ball of radius n around the root, which is finite, and this separated points. –  Benjamin Steinberg Jan 3 '12 at 0:44
@Benjamin: my mistake, I somehow forgot to parse "automorphism of rooted tree" as preserving the root. –  Yemon Choi Jan 3 '12 at 0:46
Thank you Benjamin, this is exactly the sorts of examples I was looking for! –  candl Jan 3 '12 at 10:00
You are welcome. –  Benjamin Steinberg Jan 3 '12 at 14:03

One natural source of examples is matrix groups over either $\mathbb{Z}_p$ or $\mathbb{F}_q[[t]]$ (formal power series over a finie field). So for instance $SL_n(\mathbb{Z}_p)$ or $SL_n(\mathbb{F}_q[[t]])$.

Another good example is the Nottingham group: one way to describe it is as the normalized automorphisms (you can require continuous, but it is the same) of $\mathbb{F}_q[[t]]$. Another way is as power series in $\mathbb{F}_q[[t]]$ of the form $t+a_1t^2+a_2t^3+\cdots$ where the product is composition.

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Thank you for these examples Yiftach. I will certainly be using the matrix groups at some point. The Nottingham group is an excellent example, but, at the moment, I see it more as an algebraic example. As you mention in your answer: you do not need to require continuity in the automorphisms. Perhaps there is more geometric interpretation that I am unfamiliar with. –  candl Jan 3 '12 at 10:20
One reason I like these examples is that you can undersatnd them without knowing anything about profinite groups. But then you can demonstarte all the defining properties of profinite groups directly on the examples. –  Yiftach Barnea Jan 3 '12 at 16:39

I may be way off on what you're looking for (learned from Serre's Galois Cohomology):
If $M$ is a torsion abelian group, then its dual $Hom(M,\mathbb{Q}/\mathbb{Z})$ with the topology of pointwise convergence, is a commutative profinite group.
This gives a "Pontryagin duality": $\lbrace$torsion abelian groups$\rbrace\Longleftrightarrow \lbrace$commutative profinite groups$\rbrace$.

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How does this relate to Pontryagin Duality? –  Sean Tilson Jan 3 '12 at 0:25
I guess a better way to phrase it, is as an equivalence via Pontryagin duality. –  Chris Gerig Jan 3 '12 at 0:30
This is a great example Chris, thank you. In a sense, this is a way off from what I had in mind. However, I think it actually fits my requirements rather well. It has the added bonus of planting the seed for when I later cover Pontryagin duality! –  candl Jan 3 '12 at 10:23

One idea would be to introduce the profinite completion. If the students have already met the connection of the fundamental group with covering spaces, why not look at finite covering spaces. (This is essentially doing the SGA1 algebraic fundamental group idea without mentioning the links with alg. geom nor with Galois theory.) You are looking at a group $G$ and the category of finite $G$-sets. As all the permutation groups of the objects concerned are finite, the group acts via its profinite completion. There is a nice master's thesis by Robalo at the IST in Lisbon which looked at various aspects of this situation and may provide a useful reference document for the students as it is written at only slightly above the level at which they will be working.

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