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How do I prove this statement?

Let $G$ be a Topological group and let $H$ be a subgroup of $G$, if both $H$ and $G/H$ are locally compact then $G$ is locally compact. (we will endow the set $G/H$ with the quotient topology)

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    quick question. how do we know that $G/H$ is locally compact for sure?2014-02-24

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This is Theorem 6.7 c) in Markus Stroppel's Locally compact groups.


Here is a sketch of the proof. Pick a neighbourhood $U$ of the identity in $G$ such that $U\cap H$ is compact (this is possible since $H$ is locally compact). Then, there exists a closed neighbourhood $T$ of the identity (in $G$) such that $TT^{-1}\subset U$. Now, look at $\pi(T)\subset G/H$, where $\pi:G\to G/H$ is the canonical projection. It is a neighbourhood of the identity (or rather the image of the identity in $G/H$, which is locally compact), and so there exists a compact neighbourhood $C$ such that $C\subset\pi(T)$. Finally, pick a closed neighbourhood $R$ of the identity (in $G$) such that $RR^{-1}\subset T$ and $\pi(R)\subset C$. The last step of the proof is to check that $R$ is compact (using, among other things, the fact that the intersection of $T$ with any right coset is compact).

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    @Sian This one I am not 100% sure. But if you pullback $C$ along $\pi$, you get a closed neighbourhood of the identity. You take $R$ to be a symmetric closed neighbourhood contained in the interior of $\pi^{-1}(C)$, and this should do the job.2012-04-20