Are conjugacy classes of compact groups Borel?

Question

Let $G$ be a compact Hausdorff topological group. Is it necessarily the case that every conjugacy class of $G$ is Borel? It's certainly true if $G$ is countable (in which case $G$ is actually finite and the result is trivial) or if $G$ is abelian (in which case each conjugacy class is a singleton so the result is again trivial since $G$ is Hausdorff). I haven't been able to come up with much beyond that; any thoughts would be appreciated!

Answer 1

In fact, conjugacy classes are compact (and thus closed, since $G$ is T2). The map $\rho:G\times G\times G\to G$, $\rho(x,y,z)=xzy^{-1}$ is continuous and $\operatorname{conj}(z)=\rho[\Delta\times \{z\}]$, where $\Delta=\{(x,x)\,:\, x\in G\}$ is the diagonal of $G\times G$. Since $G$ is T2, $\Delta$ is closed in $G\times G$ (and therefore compact).