# If both $H$ and $G/H$ are locally compact then $G$ is locally compact (topological Group)

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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5.25 in Hewitt-Ross' Abstract Harmonic Analysis I. I don't know of any "instant" proof... –  Xabier Domínguez Apr 18 '12 at 14:15
Is my answer good enough, or do you need clarifications? –  M Turgeon May 22 '12 at 15:28

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).
@Sian Using just the continuity of the multiplication and inverse maps, you can show that there exists a symmetric neighbourhood $T$ such $TT=TT^{-1}\subset U$. Regularity allows you take it closed. –  M Turgeon Apr 19 '12 at 11:23
@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. –  M Turgeon Apr 20 '12 at 12:09