EDIT Just realized that this question
Conjugacy classes of a compact matrix group
is related, but I think that the answer use specific properties of matrix groups, so it doesn't apply.
QUESTION
Let $G$ be a Hausdorff topological group.
Is it true that every conjugacy class is closed in $G$?
I found this assertion in a particular case in the context of profinite groups and I tried and tried but just can't justify it.
I guess this is true for any topological group $G$, but I'm not sure.
I think that we need the Hausdorff condition, because it is possible that each element is its own conjugacy class, so we couldn't get a general result for topological groups, since $\{1\}$ closed is equivalent to Hausdorff.
Since this was found in the context of profinite groups, maybe we could try arguing with compactness too, if the Hausdorff condition alone isn't enough.
Any hints on how to solve this problem? I appreciate it.
If neither of these conditions is enough, go ahead and suppose $G$ is profinite or even procyclic.
As observed in the comments, conjugacy classes need not be closed in noncompact topological groups.
As for compact Hausdorff topological groups, it is true that all conjugacy classes are closed. Indeed, let $G$ be a compact Hausdorff topological group. Let $x\in G$. The map $$ f\colon G\rightarrow G $$ defined by $f(g)=gxg^{-1}$ is continuous. Since $G$ is compact, the image $f(G)$ of $f$ is compact. Since $G$ is Hausdorff, $f(G)$ is closed in $G$. Since $f(G)$ is nothing but the conjugacy class of $x$, this conjugacy class is closed.