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I'm continuing to investigate my professor's statement regarding profinite groups and Stone topological groups. See this post for more information.

A profinite group $G$ is an inverse limit of an inverse system of finite discrete groups $G_\alpha$, where $\alpha\in\Lambda$. Naturally, we can consider the automorphism group Aut$(G)$, but there is no reason to expect Aut$(G)$ to be profinite.

My question:

Is it known whether or not Inn$(G)$ is profinite?

This requires us to reason (from what I can tell) along one of two lines:

  1. That given some topology, we have Inn$(G)$ is compact, Hausdorff, and totally disconnected.
  2. Inn$(G)$ can be realized as an inverse limit of an inverse system of finite discrete groups.

For 1., we need to first determine a reasonable topology for Inn$(G)$. The first topology that comes to mind is the compact-open topology. In particular, from the fact that $G$ is Hausdorff, we obtain that Inn$(G)$ is Hausdorff as well when equipped with this topology. It would remain to prove, then, that Inn$(G)$ is compact and totally disconnected. Is this a nice topology to consider on Inn$(G)$, and how would we go about proving the remaining claims necessary for profiniteness?

For 2., a nice construction comes out of considering each of the inner automorphism groups Inn$(G_\alpha)$. It is not so hard to see that these define an inverse limit system using the same maps as the inverse limit system of the $G_\alpha$. We can take the inverse limit of this system, which I'll denote $\widehat{\text{Inn}}(G)$. It's also not hard to show that Inn$(G)$ injects into $\widehat{\text{Inn}}(G)$. To show that Inn$(G)$ and $\widehat{\text{Inn}}(G)$ are the same, it remains to show that Inn$(G)$ is compact and dense in $\widehat{\text{Inn}}(G)$. How would one go about this?

If this is not true, then we can try to show that Inn$(G)$ is a closed subspace of $\widehat{\text{Inn}}(G)$, which would demonstrate that Inn$(G)$ is profinite in the subspace topology.

Finally, if Inn$(G)$ is not profinite at all, then is it natural to consider $\widehat{\text{Inn}}(G)$ as the "true" profinite inner automorphism group?

Shaun
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Alex Byard
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1 Answers1

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$\mathrm{Inn}(G)\cong G/Z(G)$. Now in any $T_0$ topological group, $Z(G)$ is closed (see here), and the quotient of a profinite group by a closed normal subgroup is again profinite (see here). So yes, $\mathrm{Inn}(G)$ is profinite.

Alex Kruckman
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  • Thank you. Is there any more-general reason why the inner automorphism group of a pro-object should be profinite? – Alex Byard Jun 28 '23 at 23:53
  • @AlexByard What do you mean by an inner automorphism of a pro-object? – Alex Kruckman Jun 28 '23 at 23:54
  • I mean "pro-object" to represent a profinite thing not necessarily belonging to the category of groups. Sorry if I was/am still being too vague. The approach I proposed with the inverse limit of inner automorphism groups would generalize beyond groups to other algebraic structures for which generating a group by symmetries makes sense. Is there a general reason beyond groups by which Inn(X) might be profinite? – Alex Byard Jun 29 '23 at 00:56
  • @AlexByard I understand what you mean by a pro-object. What I was asking about was the "inner automorphism" part - as far as I know, this is a purely group-theoretic concept. Do you have other examples of algebraic structures where it makes sense to talk about inner automorphisms? – Alex Kruckman Jun 29 '23 at 00:59
  • Say, quandles, for which Inn(Q) is generated by the symmetries $y\mapsto x\triangleright y$. – Alex Byard Jun 29 '23 at 01:12
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    @AlexByard I'm afraid I don't have any particular insight on the case of quandles. In any situation where you associate an "inner" automorphism to an element of your algebraic structure $X$, you get a map from the structure to $\mathrm{Aut}(X)$, whose image is the inner automorphism group. So the inner automorphism group is a quotient of $X$ by some equivalence relation. If the induced topology on the quotient is a profinite topology, and the group is a topological group with respect to this topology, then it is a profinite group. These things should be easy to check in most cases of interest. – Alex Kruckman Jun 30 '23 at 13:57
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    (If they're true.) So in this sense the argument from my answer will generalize. – Alex Kruckman Jun 30 '23 at 13:58
  • The argument regarding the quotient of a profinite group by a closed normal subgroup uses the Stone space result, right, i.e., shows that Inn$(G)$ is compact, Hausdorff, and totally disconnected. Do you know how to construct an inverse limit system whose inverse limit is equal to Inn$(Q)$? – Alex Byard Jun 30 '23 at 19:16
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    @AlexByard Of course: take the system of all finite quotients of $\mathrm{Inn}(G)$ (this works for any profinite group). This is unlikely to satisfy you: you want a more concrete description. But if you look a bit closer, this provides one: Since $\mathrm{Inn}(G)$ is already a quotient of $G$ (by $Z(G)$), a finite quotient of $\mathrm{Inn}(G)$ is the same as a finite quotient of $G$ whose kernel contains $Z(G)$. So for an inverse limit system presenting $\mathrm{Inn}(G)$, we can take a subsystem of the canonical inverse limit system presenting $G$. – Alex Kruckman Jun 30 '23 at 19:28
  • Thank you that makes sense. Do you know of a construction similar to my $\widehat{\text{Inn}}(G)$? I'm at the point of proving that Inn$(G)$ is dense inside of the construction. If I can do that, then they are the same, which provides another concrete description for the innet automorphism group of a profinite group. – Alex Byard Jun 30 '23 at 19:55
  • @AlexByard I believe they are the same if you start with the canonical presentation of $G$ as an inverse limit. If you start with a presentation of $G$ where the morphisms are not all surjective then your construction doesn't make sense, since $\mathrm{Inn}$ is only functorial on maps $f\colon H\to H'$ such that $f(Z(H))\subseteq Z(H')$. This is always true when $f$ is surjective, but not in general. If you start with some inverse system of groups where all maps are surjective, but which isn't the canonical presentation of $G$, I'm not sure whether $\widehat{\mathrm{Inn}}(G)=\mathrm{Inn}(G)$. – Alex Kruckman Jun 30 '23 at 20:34
  • What do you mean by the canonical presentation? – Alex Byard Jul 01 '23 at 21:09
  • @AlexByard the system of all finite quotients of $G$. – Alex Kruckman Jul 01 '23 at 21:26
  • I see. Does this mean it is the case that if I build a profinite group out of an inverse limit of any finite discrete groups, then the inverse limit can be realized as an inverse limit of a system of finite quotients for some group $G$? Is there a way to find the group $G$? – Alex Byard Jul 02 '23 at 18:30
  • Correction: in all of the above, when I talked about finite quotients of a profinite group, I should have specified that these should be quotients by clopen normal subgroups, not arbitrary normal subgroups. And the connecting maps in the system are all those commuting with the quotient maps. – Alex Kruckman Jul 02 '23 at 18:49
  • But yes, this is essentially the proof that every Stone topological group is an inverse limit of finite groups: you look at the inverse system consisting of all finite quotients (by clopen normal subgroups) and check that you recover the original group as the inverse limit. I don't understand your last question: the group $G$ is the inverse limit of your system. – Alex Kruckman Jul 02 '23 at 18:51
  • Sorry, I was abusing notation and was not clear. Is your statement that a profinite group $\widehat{G}$ can be realized as the profinite completion of a group $G$? If we arrive at a profinite group $\widehat{G}$ by taking the inverse limit of an inverse system of finite discrete groups $G_i$, is there a way to tell what the group $G$ is that has been completed? – Alex Byard Jul 02 '23 at 18:54
  • The second question being: Can you retrieve a canonical presentation from only knowledge of the profinite group and a noncanonical presentation? – Alex Byard Jul 02 '23 at 18:57
  • Let us continue this discussion in chat. – Alex Kruckman Jul 02 '23 at 19:00