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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!

PandaMan
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    Are you talking of the quotient space $;G/H;$ or of the quotient (topological) group $;G/H;$ ? The last one requires $;H\lhd G;$ ... – DonAntonio Feb 24 '14 at 05:26
  • I'm pretty sure $G/H$ is the quotient group. What do you mean by ⊲? – PandaMan Feb 24 '14 at 05:35
  • A subgroup $H$ of a group $G$ is said normal in $G$ if for all $x$ in $G$, $Hx = xH$. This is denoted by $H\lhd G$ – Bento Feb 24 '14 at 06:04

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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.

Lisa
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  • All this is standard but just one word: the quitent topology on $;G/;$ is precisely the one that makes the canonical projection $;\pi;$ continuous and open. From here it follows pretty smoothly what the OP wants. Nice answer +1 – DonAntonio Feb 24 '14 at 11:37
  • I guess you don't include "hausdorffness" in the definition of compactness ? – Olórin Feb 23 '15 at 23:59