Let $G$ be a locally compact second-countable Hausdorff group with a compact subgroup $H$. Assume that the modular functions of $G$ and $H$ agree on $H$. Therefore, there is a non-trivial left-invariant Radon measure $\mu$ on the quotient space of left-cosets $G/H$.
Assume now that $\mu$ is finite.
- If $H=\{1_G\}$ then $G/H=G$ has to be compact, as $\mu$ is the Haar measure on $G$.
- If $H$ were just closed and not compact I know that $G/H$ might not be compact. However, does the assumption that $H$ is compact give that $G/H$ is too, or at least that $G/H$ has finite diameter with respect to the left-invariant metric?
Thanks in advance!
Yes, if $H$ is compact, then it admits a finite Haar measure $\mu_H$. Let $\mu_{G/H}$ denote the finite measure on $G/H$, we can define a measure on $G$ by
$$\int f d\mu_G = \int_{G/H} \mathbb{E}_H(f) d\mu_{G/H}$$ where $\mathbb{E}_H(f) = \int_H f(xh) d\mu_H$ viewed as a function on $G/H$ (can define this on all compactly supported continuous functions, so everything is well defined and then extend using Riesz-Markov-Kakutani).
By construction $\mu_G$ is $G$-invariant, and we have $\int 1 d\mu_G = \mu_{G/H}(G/H)\cdot \mu_H(H)<\infty$.
It is left to prove that if $G$ admits a finite $G$-invariant measure, then $G$ is compact.
By local compactness we can find some $U\subseteq G$ open with $\overline{U}$ compact.
Since the product map is continuous, and the image of a compact set is compact, we have that $\overline{U}\cdot \overline{U}^{-1}$ is a compact set.
We have $\mu(\bigcup_{i=1}^\infty g_i \overline{U}) = \mu(G)<\infty$ and therefore, at most finitely many of these sets are disjoint. Choose a maximal set (by reordering we can assume it is $\{g_1,...,g_n\}$) so that $g_i \overline{U}$ are disjoint for all $1\leq i \leq n$. As a finite union of compact sets is compact, $\bigcup_{i=1}^n g_i \overline{U}\cdot \overline{U}^{-1}$ is compact. If by contradiction $G\not = \bigcup_{i=1}^n g_i \overline{U}\cdot \overline{U}^{-1}$, then there exists some $s\in G$ so that $s\not \in \bigcup_{i=1}^n g_i \overline{U}\cdot \overline{U}^{-1}$. But then $s\overline{U}$ does not intersect $\bigcup_{i=1}^n g_i \overline{U}$. Indeed, if it does that there are some $u,v\in \overline{U}$ and $1\leq i \leq n$ so that $s u = g_i v$, but then $s = g_i v u^{-1} \in g_i \overline{U}\cdot \overline{U}^{-1}$ by contradiction.
We deduce that $G = \bigcup_{i=1}^n g_i \overline{U}\cdot \overline{U}^{-1}$ is comapct as a finite union of compact sets.