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Let $G$ be a topological group such that for each $x \in G$ the mapping $x\mapsto xy$ is a homeomorphism. If $H$ is a open subgroup of $G$, prove that $H$ is also closed

Could anyone just give hint for this one?

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2 Answers 2

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We just need to show that the complement $G \setminus H$ is open, so take $y \in G \setminus H$. Note that $Hy = \{ xy : x \in H \}$ is an open neighbourhood of $y$ (since $1 \in H$, and the mapping $x \mapsto xy$ sends open sets to open sets). This set is also disjoint from $H$, since given $x \in H$, if $xy \in H$ then $y = x^{-1}(xy) \in H$, which is false by assumption).

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Every coset is open and $G\setminus H$ is a union of cosets. –  egreg Nov 23 at 14:26

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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