33

Is there a profinite group $G$ which is not its own profinite completion?

Surely not, I thought. But upon looking into it, I found that there is a special name given to a $G$ which is its own profinite completion, namely "strongly complete". And a recent (2003) hard theorem (which according to Wikipedia uses the classification of finite simple groups) due to Nikolov and Segal asserts that, if $G$ is finitely generated (as a topological group), then it is "strongly complete".

So the $G$ I'm looking for cannot be topologically finitely generated. An equivalent question to the above is:

Is there a profinite group $G$ which admits a non-open subgroup of finite index?

Now here's my problem; the only exposure to profinite groups I've had has been in the context of number theory, absolute Galois groups, local fields, etc. In particular, the only non-topologically-finitely-generated profinite group I'm aware of is the absolute Galois group of a number field, say $\mathbb{Q}$. But I reckon the Krull topology demands that the finite index subgroups of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ be open.

Maybe there is a more 'exotic' example of such a $G$...

Giuseppe
  • 801
  • "... which admits a non-open subgroup of finite index" <- don't we also have to require that it is a normal one? – Martin Brandenburg Nov 30 '11 at 08:31
  • @Martin: No, I don't think so. The paper of Nikolov and Segal states that strongly equivalent is equivalent to "every subgroup of finite index is open"; no mention of normality. In fact, I have a feeling that a normal subgroup of finite index in a profinite group is forced to be open. (But I may be wrong, since this is way outside my own area.) – Giuseppe Nov 30 '11 at 15:43
  • 12
    The statements are equivalent. Clearly, if every finite index subgroup is open then every finite index normal subgroup is open. Conversely, every finite index subgroup contains a finite index normal subgroup (the intersection of its conjugates, for example) so if every finite index normal subgroup is open then so is every finite index subgroup. – candl Jan 05 '12 at 13:39
  • This is related to my question http://mathoverflow.net/questions/106216/without-choice-can-every-homomorphism-from-a-profinite-group-to-a-finite-group – Will Sawin Mar 02 '14 at 17:16

4 Answers4

35

The absolute Galois group $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ has lots of subgroups of finite index that aren't open. See the section "Nonopen subgroups of finite index" in Chapter 7 of Milne's notes on field theory for a construction (it uses the axiom of choice).

Keenan Kidwell
  • 3,340
  • 8
    I now wonder: What is the profinite completion of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$? I've never seen anybody refer to this object. – Giuseppe Nov 29 '11 at 16:52
  • 3
    This means that $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ is not finitely generated as a topological group. What a monster. – Spice the Bird Nov 30 '11 at 00:53
  • @Spice the Bird: On the bright side, Shaferevich conjectured that the the Galois group ramified at only finitely many primes is topologically finitely generated. I don't know if it's proved or not. – Moosbrugger Nov 30 '11 at 02:09
  • 8
    @Spice the Bird, Moosbrugger: You're probably aware of this, but here's an analogy which might explain why both of these observations are to be expected. In the dictionary between number fields and function fields, one thinks as Galois groups as follows--take the etale fundamental group of your curve minus some points, and take the limit as one removes all the points. Removing a point gives a factor of \hat Z; the topological generator is analogous to a Frobenius lift on the number field side. On the other hand, if you only allow finite amounts of ramification, this is like removing... – Daniel Litt Sep 07 '12 at 00:21
  • 7
    (cont.) ... finitely many points, whence the (at least prime to p) part of the Galois group of the curve still remains manifestly finitely generated. – Daniel Litt Sep 07 '12 at 00:22
28

Example taken from Ribes and Zalesskii's book "Profinite groups". Take an infinite set $I$ and a finite group $T$. You can let $G$ be the profinite group $\prod_I T$. Denote its elements by $(g_i)_{i\in I}$. Let $\mathcal F$ be an ultrafilter which contains the filter of all cofinite subsets of $I$. If you denote $H$ to be the subgroup of elements with $\lbrace i\in I \mid g_i=1\rbrace\in \mathcal F$, it is clear that $H$ is proper normal and that it is not open because it is dense and has finite index $|T|$.

To show that $H$ has index $|T|$ in $G$ consider all elements $a_t=(t,t,\dots)$ for $t\in T$. For any $g\in G$, consider $I_t=\lbrace i\in I \mid g_i=t\rbrace$. Since we have $\bigcup_{t\in T} I_t=I$ then $I_t\in \mathcal F$ for some $t$, and therefore $ga_t^{-1}\in H$.

Johannes Hahn
  • 9,539
Gjergji Zaimi
  • 85,056
18

Yes, this is possible. Take as $G$ the product of countable many copies of $\mathbb Z_p$. It has a countable basis of open subgroups, hence only countably many open subgroups. But it has many more subgroups of finite index!

Ricky
  • 3,674
  • Thank you for your answer. But why does $G$ have many more subgroups of finite index than open subgroups? Does 'many more' here mean uncountably many? – Giuseppe Nov 29 '11 at 15:03
  • It does mean uncountably many.

    The easiest way to see this is to note that it maps onto the product of countably many copies of $C_{p}$, which is of course isomorphic to a $2^{\aleph_{0}}$-dimensional $F_{p}$ space. (This is itself a pro-$p$ group with the properties you want to see.)

    I have a paper on the arxiv all about these sorts of groups. The new version of it should be up tomorrow morning. I will provide the link then.

     – Jonathan Kiehlmann Nov 29 '11 at 18:13
  • 1
    More generally, every pro-$p$ group $G$ that is not topologically finitely generated maps onto the one Jonny describes, so $G$ has a non-open subgroup of index $p$. – Colin Reid Nov 29 '11 at 18:22
  • This is exactly what I mean. – Ricky Nov 29 '11 at 20:52
  • Also a free profinite group of countable rank is an example since it maps onto this one and so has uncountably many finite image but it is still countably based. – Benjamin Steinberg Nov 29 '11 at 23:24
  • Benjamin - you need to make sure that your generators tend to the trivial element, otherwise a free profinite group on countable generators isn't free or countably-based. One of those breaks. – Jonathan Kiehlmann Nov 30 '11 at 03:22
  • @Jonathan, I thought it was standard that a free profinite group of countable rank means a countable set of generators converging to 1. At least this is the convention used by Ribes and Zalesskii. – Benjamin Steinberg Nov 30 '11 at 03:33
  • @Benjamin, it is standard when talking about profinite groups. However this is not obvious, and so it is worth mentioning for the benefit of the readers who have not read a textbook on them. – Jonathan Kiehlmann Nov 30 '11 at 12:26
6

It's hard to pick many non-finitely-generated pro-$p$ groups which are isomorphic their own profinite completions. Partially as non-finitely-generated pro-$p$ groups aren't generally that easy to construct.

My paper here http://arxiv.org/abs/1101.3005 outlines the construction of infinitely many topologically non-isomorphic pro-$p$ groups isomorphic to $\prod_i C_{p^{i}}$. These will thus all have the same profinite completion, but are non-isomorphic.

Jonathan Kiehlmann
  • 1,016
  • out of curiousity, what exactly is the profinite completion of $\prod_i C_{p^i}$ – candl Jan 05 '12 at 13:54
  • 1
    Very good question.

    The easiest response is "something very big". It is $\aleph_{1}$-based and has a direct summand isomorphic to $\prod_{\aleph_{0}}\mathbb{Z}{p}$. (The latter as $\prod{i\in\mathbb{N}} C_{p^{i}}$ has a direct summand isomorphic to $\prod_{\aleph_{0}}\mathbb{Z}_{p}$.)

    Not sure of an exact answer, but will think more of it. Feel free to e-mail me with any more questions.

     – Jonathan Kiehlmann Jan 05 '12 at 20:56