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$G$ 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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  • $\begingroup$ is G/H assumed to be hausdorff? $\endgroup$ – oxeimon Jun 10 '15 at 18:25
  • $\begingroup$ Use that $G/H$ is compact. $\endgroup$ – Dietrich Burde Jun 10 '15 at 18:34
  • $\begingroup$ @oxeimon: $G/H$ may not necessarily be Hausdorff. $\endgroup$ – David Chan Jun 10 '15 at 18:42
  • $\begingroup$ Write $G/H=K$. Then G=KH$. $\endgroup$ – Dietrich Burde Jun 10 '15 at 18:44
  • $\begingroup$ Use this to write it down. $\endgroup$ – Dietrich Burde Jun 10 '15 at 19:09
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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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