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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!

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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 at 5:26
    
I'm pretty sure $G/H$ is the quotient group. What do you mean by ⊲? –  PandaMan Feb 24 at 5:35
    
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 at 6:04

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up vote 5 down vote accepted

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.

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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 at 11:37

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