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:
- That given some topology, we have Inn$(G)$ is compact, Hausdorff, and totally disconnected.
- 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?
Best Answer
$\mathrm{Inn}(G)\cong G/Z(G)$. 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.