The center of compact groups

Question

Let $G$ be a compact Hausdorff topological group. Let $H$ be a closed subgroup of $G$ and $Z$ be the center of $G$. Is the product $H\cdot Z$ closed? What about $H\cdot Z^0$? Where $Z^0$ is the connected component of the unit.

Answer 1

Yes: $Z$ is closed. Thus $H\cdot Z$ is compact and hence closed, being the continuous image of the compact space $H\times Z$ under the multiplication map. The same argument works for $Z^0$.

Answer 2

In a Hausdorff topological group, the product of a closed set and a compact set is closed. Thus, all you need to do is note that $Z$ and $Z^0$ are closed. To show the latter, you just need to show that the closure of a connected set is connected.

To show the former, it suffices to show that, given a particular element $g \in G$, the set $$Z_g = \lbrace h \in G : gh = hg \rbrace$$ is closed, as the centre is the intersection of all such sets. These sets are kernels (i.e. inverse images of the identity) of the functions $$f_g : G \to G : h \mapsto ghg^{-1}h^{-1},$$ which are evidently continuous. Since $G$ is Hausdorff, $\lbrace 0 \rbrace$ is closed, hence $Z_g = f_g^{-1} \lbrace 0 \rbrace$ is closed, hence $Z$ is also closed. Thus, $H \cdot Z$ is closed too (and indeed compact).