Talk:Profinite group

Inverse systems

Something perhaps needs to be said about the inverse systems? I mean, these are supposed to be indexed in such a way as to have a 'filtering' property? What is said about ind-finite isn't correct, anyway, unless the direct system tells you more than cyclic subgroups being finite, which is just a torsion group.

Charles Matthews 17:58, 12 Apr 2005 (UTC)

What it seems to amount to categorically is that a profinite group is a cofiltered limit of finite groups where the cofiltering category is also a partial order, and an ind-finite group is a filtered colimit where the filtering category is a partial order. The "inverse system" and "inductive system" terminology appears to be archaic from this point of view. In fact, some people even call the relevant functors "profinite groups", although this is not standard. - Gauge 05:39, 16 July 2006 (UTC)[reply]

Can we say more about the functors involved here? If limits and colimits are going to be introduced, it seems best to show how the construction of profinite groups fits that definition. Exactly which categories are involved is not even clear to somebody who isn't a specialist. Dennis Muhonen (talk) 10:35, 15 July 2013 (UTC)[reply]

natural topology

A profinite group, as defined here, doesn't have a natural topology until one selects an isomorphism with an inverse limit. It is still unclear whether the topology of a profinite group is unique, hence it's not proven that the given definitions are equivalent (i.e. that they define equivalent categories). This question is equivalent to the following:

Is every profinite group isomorphic to its profinite completion?

A group for which this holds is called strongly complete.

Zhw 00:52, 2 February 2006 (UTC)[reply]

Oh, I recently asked this and got some wonderful replies. It has been known for a long time that there are many profinite groups that are not isomorphic to their profinite completions. Indeed, the profinite completion of a profinite completion is virtually never itself (usually the cardinality bumps up). However for finitely generated profinite groups everything is now much nicer. It was recently shown (published announcement in 2006, probably somewhat announced in 2005, proof published in 2008) that the topology on finitely generated profinite group is unique and every finitely generated profinite group is strongly complete. JackSchmidt (talk) 04:02, 23 January 2009 (UTC)[reply]

Title

Why is this article called "Pro-finite group" rather than "Profinite group"? The unhyphenated form is by far the most common in the literature. I intend to move the article if there are no objections. --Zundark 15:24, 4 June 2007 (UTC)[reply]

OK, I've moved it. I should probably remove the excess hyphens in article text too. --Zundark 15:39, 5 June 2007 (UTC)[reply]

inverse system

Given an arbitrary group G, there is a related profinite group G^, the profinite completion of G. It is defined as the inverse limit of the groups G/N, where N runs through the normal subgroups in G of finite index (these normal subgroups are partially ordered by inclusion, which translates into an inverse system of natural homomorphisms between the quotients).

How can it be that the normal subgroups of finite index of an arbitrary group are partially ordered by inclusion? Take Z/10Z under addition. The subgroups <2> and <5> are both normal and of finite index but 5 belongs to <5> and 2 belongs to <2> but neither belong to the other. Thus we have neither <2> in <5> nor <5> in <2>. The set Z/pZ of subgroups of prime index of Z is also a counter example.

How can we construct the profinite completion of Z/10Z? How can we have an inverse system if (Z/10Z)/<5> is a two element group and (Z/10Z)/<2> is a 5 element group? There can never exist a surjective homomorphism from the 5 element group to the 2 element group.

--Gtg207u (talk) 01:49, 15 January 2009 (UTC)[reply]

A partial order doesn't require all elements to be comparable - that's a total order. Any set of sets is partially ordered by inclusion.

In the case of Z/10Z, there are four normal subgroups (of finite index). There's no inclusion between the subgroups of orders 2 and 5, so there's no homomorphism between the corresponding quotients in the inverse system. --Zundark (talk) 09:25, 15 January 2009 (UTC)[reply]

The profinite completion of Z/10Z is the group Z/10Z with the projections n + 10Z maps to n + kZ for k=1,2,5,10. The inverse system has two more homomorphisms in it, the projections n + kZ maps to n + 1Z for k=2,5 from Z/kZ to Z/1Z. Here I am using the natural isomorphisms of (Z/nZ)/(kZ/nZ) = Z/kZ when k divides n. JackSchmidt (talk) 14:17, 15 January 2009 (UTC)[reply]

thank you sincerely. --Gtg207u (talk) 15:41, 15 January 2009 (UTC)[reply]

Small stuff

The word 'discrete' in front of 'discrete finite groups' seems redundant. aren't all finite groups discrete? --Wishcow (talk) 18:11, 13 January 2010 (UTC)[reply]

Of course, you can give every topological space the coarse topology, where only the empty set and the whole space are open. If the space contains more than one point, this topology isn't discrete. Spaetzle (talk) 08:27, 18 July 2011 (UTC)[reply]

Krull Topology

A google for "Krull Topology" lists this page as the 2nd result, and yet this page doesn't even define the Krull topology anywhere. It simply says that "The topology we obtain on Gal(L/K) is known as the Krull topology", which is woefully inadequate in my opinion.

Can someone actually include the definition of the krull topology for a profinite group on this page? — Preceding unsigned comment added by 130.203.167.50 (talk) 21:32, 25 February 2013 (UTC)[reply]

The Krull topology is the name given to the topology on Gal(L/K) when it is viewed as a profinite group as described in that paragraph. RobHar (talk) 06:21, 26 February 2013 (UTC)[reply]

I think what you might be asking for is what is the profinite topology on an inverse limit, and quite surprisingly this doesn't seem to appear on this page (it is however discussed at Inverse limit).RobHar (talk) 06:24, 26 February 2013 (UTC)[reply]

problem with first definition

"A profinite group is a topological group that is isomorphic to the inverse limit of AN inverse system of discrete finite groups." so.. any inverse system? so for example if take an inverse system given by a single discrete group? do I get that ALL discrete groups are pro finite? this is certainly false, so I see a problem in the first definition!--78.6.83.200 (talk) 16:58, 29 January 2019 (UTC)[reply]

The definition is correct. Note that it says an inverse system of discrete finite groups. So the inverse system can consist of a single discrete group, but that group must be finite. --Zundark (talk) 17:34, 29 January 2019 (UTC)[reply]

Circle group restrict to algebraic coordinates

Could we say the Circle group, intersected with the set of algebraic points on the x-y plane, is profinite?UsingNewWikiName (talk) 15:31, 10 August 2020 (UTC)[reply]

No, because it's not compact. --Zundark (talk) 20:09, 10 August 2020 (UTC)[reply]