Show that if $G$ is a locally compact topological group and $H$ is a subgroup, then $G/H$ is locally compact.
This seems pretty straight forward but how will I be able to prove this? I saw this property from Wikipedia that, Every closed subgroup of a locally compact group is locally compact. But if $G/H$ closed? I haven't had any lectures on subgroups and its properties yet, any help will be greatly appreciated. Thanks in advance!
Best Answer
It is not necessary for $H$ to be normal or have any other property except that it is a subgroup, but then $G/H$, in general, is not a group any more. However, it is always a set, and it becomes a topological space when equipped with the quotient topology with respect to the canonical map $\pi\colon G \to G/H$, which is then continuous, surjective, and open. If $y \in G/H$ is any element, then $y = \pi(x)$ for some $x \in G$ since $\pi$ is surjective. We find a compact neighbourhood $K$ of $x$ in $G$ because $G$ is locally compact. Then, $\pi(K)$ is a neighbourhood of $y$ in $G/H$ since $\pi$ is open, and it is also compact since $\pi$ is continuous.