A lie group $G$ is compact iff $G/H$ is.

Question

Suppose that $G$ is compact Lie group and $H$ is closed subgroup. Than $G/H$ is compact since canonical projecton is continuous. Is the inverse true, namely can one prove that:

G is compact iff $G/H$ is compact where $H$ is a compact subgroup of $G$ (It won't work for H closed only).

I will be gratefoul for any kind of helpe.

Answer 1

You have a little more general result. You only need to be in a topological group not, a Lie group.

Proposition : Let $G$ be a topological group, and $H$ a subgroup of $G$. If $H$ and $G/H$ are compact then $G$ is compact.

Sketch of a proof : Consider a set $(F_i)_{i \in I}$ of closed sets of $G$. Suppose that for all finite $J \subseteq I$, $\cap_{j \in J} F_j \neq \emptyset$, we will show that $\cap_{i \in I} F_i \neq \emptyset$.

1. Show that we can suppose $(F_i)$ stable by intersection. Let $p$ the canonical projection, show that all finite intersection of $p(F_i)$ is not empty. Let $G_i$ be the closure of $p(F_i)$ in $G/H$.
2. Show that $\cap_{i \in I} G_i \neq \emptyset$. So $\exists a \in G$ such as for all neighbourhood $U$ of $e$ in $G$, we have $F_i \cap UaH \neq \emptyset$, id est $U^{-1}F_i \cap aH \neq \emptyset , \forall i \in I$.
3. Let $H_{i,U}=U^{-1}F_i \cap aH,i \in I, U$ neighbourhood of $e$ in $G$. Show that all the finite intersections of these items are not empty.
4. Show that it implies that it exists $b \in aH$ which is adherent to all the $H_{i,U}$. Finally show that $b$ is adherent to all the $F_i$, and you can conclude.

This is and exercice in the book (Chap 2, ex 4) : Introduction à la théorie des groupes classiques by R.Mneimné & F.Testard. (This book is in french).