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could any one just give hint for this one?

$G$ be a topological group such that $\forall x\in G, x\mapsto xy$ Homeomorphism, $H$ is a open subgroup of $G$, we need to prove $H$ is also closed

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Hint: If $x \in G \setminus H$, then $xH \cap H = \emptyset$.

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so when $x\in H^c$, there exist a nbd $V$ of $x$ such that $V\cap H=\phi$ hence $H^c$ is open so $H$ is closed? well,if $x\in H^c$, and $xH\cap H\neq\phi$, then $y=xh$ for some $y\in H\cap xH$, what is the contradiction? –  Une Femme Douce Nov 1 '12 at 16:16
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@Flute: Yep. (Though it is preferred to use either \emptyset ($\emptyset$) or \varnothing ($\varnothing$) to denote the empty set.) –  Arthur Fischer Nov 1 '12 at 16:19

In a topological group the group multiplication is by definition continuous (and thus translations are homeomorphisms). You're probably trying to say that if $G$ is a group with topology such that right translations are homeomorphisms, then any open subgroup is also closed.

To show that, notice that $H$ is closed iff its complement is open, which you can write out explicitly using the group operations.

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