Is a compact subset of a topological group G/N closed if G is hausdorff?

Question

I just used this hypothesis when I was proving a theorem.But I was not sure if this hypothesis is correct.

Answer 1

Edit: This is a response to the original question:

If $G$ is a Hausdorff topological group and $N$ is a normal subgroup is the quotient map $G\to G/N$ always a closed map?

Assuming $G$ is Hausdorff, one necessary condition for the quotient map $q\colon G \to G/N$ to be closed is that $N$ is closed in $G$. Otherwise $q^{-1}(q(\{1\})) = N$ isn't a closed set, which means that $q(\{1\})$ isn't closed in $G/N$.

For example, the quotient map $q\colon \mathbb{R}\to \mathbb{R}/\mathbb{Q}$ isn't closed since $q^{-1}(q(\{0\})) = \mathbb{Q}$ isn't a closed set in $\mathbb{R}$.

However, there are lots of examples where $N$ is closed in $G$ but the quotient map is not closed. For example, if $G = \mathbb{R}^2$ and $N=\{0\}\times\mathbb{R}$, then the quotient map $G\to G/N$ is essentially the projection of $\mathbb{R}^2$ onto its first factor. This is not a closed map since, for example, the set $$ \{(x,\tan x) \mid -\pi/2<x<\pi/2\} $$ is closed in $\mathbb{R}^2$, but its projection onto the first factor is not.

The quotient map is always closed if $G$ is compact and $G/N$ is Hausdorff, since every map from a compact space to a Hausdorff space is a closed map. It follows that the quotient map is closed whenever $G$ is compact Hausdorff and $N$ is closed, since then it follows that $G/N$ is Hausdorff (see here).