Closed kernel in a compact group is open

Question

The way I think it should work is that $${\rm ker} = \bigcap_{g \notin {\rm ker}} (G - g\,{\rm ker}),$$ with each $G - g\,{\rm ker}$ open. Since $G$ is compact, there should, in fact, only be finitely many distinct $G - g\,{\rm ker}$ so that the intersection is open. For some reason, I can't get all of the pieces to fit together. Any help would be greatly appreciated.

Answer 1

Compactness gives you the finiteness of covers. For any group $G$, and any proper subgroup $K$, if $gK \neq hK$ then $gK \cap hK = \emptyset$. So the cover $\{G - gK : g \in G\}$ has an obvious finite subcover:

$$(G - gK) \cup (G - hK) = G - (gK \cap hK) = G$$

But to get the intersection you wrote to be just $K$ (in particular, your kernel) you need to insure you took it over all $g \notin \ker$ or at least one representative for each coset. So the actual condition you need is that $K$ is of finite index in $G$, a purely algebraic condition.