$G$ is a locally compact (may not necessarily Hausdorff) group, $H$ is a subgroup in $G$, $G/H$ is compact as a quotient space , then there exist a compact subset $K$ such that $G=KH$(or $G=HK$).
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is G/H assumed to be hausdorff? – oxeimon Jun 10 '15 at 18:25
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Use that $G/H$ is compact. – Dietrich Burde Jun 10 '15 at 18:34
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@oxeimon: $G/H$ may not necessarily be Hausdorff. – David Chan Jun 10 '15 at 18:42
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Write $G/H=K$. Then G=KH$. – Dietrich Burde Jun 10 '15 at 18:44
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Use this to write it down. – Dietrich Burde Jun 10 '15 at 19:09
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Identify ${kH}$ with ${k}\subseteq G$. – Dietrich Burde Jun 10 '15 at 19:18
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It seems to me that what you want follows from the result of the following paper:
http://www.mscand.dk/article/viewFile/12073/10089
The above paper show the existence of an interesting cross section $q$ of the fibration $\pi: G \to G/H$. Namely, a map $q: G/H \to G$ such that $\pi \circ q = id_{G/H}$, with the additional property that $q(C)$ is a relatively compact subset of $G$ for any compact subset $C \subset G/H$. So, if you assume $G/H$ to be compact then by setting $$ K := \overline{q(G/H)}$$ you get a compact subset $K \subset G$ such that $G = K H$ as the OP asked.
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