similar to (applying) group isomorphism theorems

Question

$G$ is a topological group, $H$ is a subgroup of $G$, $K$ is a compact subset of $G$, if $G=HK$, then $G/H$ is compact?

Is this right?

I said this conclusion is similar to group isomorphism theorems, that is, if $K$ is a group, then $(HK)/H$ "is" $ K/(H\cap K)$, so $(HK)/H$ is compact if $K$ is compact.

I hope it is a simple question.

Thanks a lot.

Obviously, $H$ is considered a normal subgroup (but it is not stated). — Crostul, Jun 10 '15 at 12:07
@Crostul: It means taking the quotient of $G$ under the right action of the subgroup $H$ by multiplication, then equipping it with the quotient topology. — Cihan, Jun 10 '15 at 13:49

score 1 · Accepted Answer · answered Jun 10 '15 at 13:44

If we assume $G = KH$, the function \begin{align*} K &\rightarrow G/H \\ k &\mapsto kH \end{align*} is continuous, and surjective because $G = KH$. Hence as $K$ is compact, so is its image $G/H$. I am not sure when you change the order of $H$ and $K$, yet still insist to use left cosets.

similar to (applying) group isomorphism theorems

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