# Inner automorphism group of a profinite group is profinite

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?

$$\mathrm{Inn}(G)\cong G/Z(G)$$. 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.
• @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. Commented Jun 30, 2023 at 13:57
• @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 $G$. Commented Jun 30, 2023 at 19:28