A lemma on a compact subset of $G/H$, where $G$ is locally compact and $H$ is a closed subgroup of $G$

Question

I have this lemma from "Kazhdan’s Property (T)" book (page 343 in the link).

Here, $G$ is locally compact and $H$ is a closed subgroup of $G$.

Can't really understand why $K$ is compact. The union is obviously compact, but $p^{-1}(Q)$ not necessarily.

Also, I understand why $p(K) \subseteq Q$, but the other direction is not clear to me.

Thanks!

Answer 1

(Thank you Izaak, brain wasn't online) Quotients of Hausdorff topological groups by closed subgroups are themselves Hausdorff, so $Q$ is a compact subset of a Hausdorff space, and thus is closed. By continuity of $p$, $p^{-1}(Q)$ is hence a closed set, so the intersection of it with a compact set is thus compact. You are correct that $p^{-1}(Q)$ is not necessarily compact.

Answer 2

You can just show this by hand:

Let $q\in Q$. Since $Q\subset\bigcup_{i=1}^n p(x_iU)$ there exists $i\in\{1,\dots,n\}, a\in U$ such that $p(x_ia)=q$. Hence, $x_ia\in p^{-1}(Q)$. Since clearly $x_ia\in x_iU$, we deduce that $$x_i a\in K=p^{-1}(Q)\cap\bigcup_{i=1}^n x_iU$$ Therefore $q=p(x_ia)\in p(K)$.

This shows that $p(K)\supseteq Q$.