Let $ G $ be a group. A element $ g\in G $ is an FC-element if it has only finitely many conjugates in $ G $. The set $ \Delta(G) $ of FC-elements of $ G $ is a characteristic subgroup of $ G $, and it is called the FC-center of $ G $.
Question: Suppose that $G$ is a topologically finitely generated profinite group. Is $ \Delta(G) $ closed in $G$?
Note that it's not true without the condition "topologically finitely generated", see Page 1281 in Profinite Groups with Restricted Centralizers. Since $G$ is topologically finitely generated, the topology on $G$ should be determined by the algebraic structure, cf. On finitely generated profinite groups, I: strong completeness and uniform bounds. Hence, it seems that the answer should be Yes in our case. Any references would be appreciated.
All I know is the following: Firstly, note that $\Delta(G)$ is just the union of all centralisers $C_G(H)$ of all (abstract) subgroup $H$ of finite index. Here each $H$ must be open since $G$ is topologically finitely generated. Also, one can show that each $C_G(H)$ is closed. For each natural number $n$, it's well known that the number of open subgroups of $G$ of index $n$ is finite. What's the next step?
No. Consider the group $\prod_{n\ge 5}\mathrm{Alt}(n)$. Then it is topologically 2-generated. Its FC-center is $\bigoplus_{n\ge 5}\mathrm{Alt}(n)$, which is dense but not closed.