If $G$ is a topological group and $H$ a normal subgroup, then if $H$ and $G/H$ are both totally disconnected then so is $G$.

Question

I encountered a statement in a proof that said if $G$ is a topological group and $H$ a normal subgroup, then if $H$ and $G/H$ are both totally disconnected then so is $G$. I have no idea how it is obvious or why it is left without any proof or hints but to me it seems really tough to see a proof. Has someone here encountered such theorem or can help with its proof?

Answer 1

Assume that $C\subseteq G$ is a connected subset. We will show that $C$ is a point which proves that $G$ is totally disconnected.

Consider the projection $\pi:G\to G/H$. It follows that $\pi(C)$ is connected and since $G/H$ is totally disconnected then $\pi(C)$ is a point. Therefore $C\subseteq gH$ for some $g\in G$. And since $H$ is totally disconnected then so is $gH$ and therefore $C$ has to be a point, which completes the proof. $\Box$

Note that "being a group" is not really important here. This can be generalized as follows:

Assume that $X$ is a topological space and $\sim$ an equivalence relation such that $X/\sim$ is totally disconnected and all equivalence classes $[x]_\sim$ are totally disconnected. Then $X$ is totally disconnected as well.

The only property of topological groups that we used is that $x\mapsto gx$ is a homeomorphism and so all cosets are totally disconnected. In particular $H$ doesn't have to be normal.