Let $G$ be a topological abelian group and $H$ a closed subgroup of $G$. Is it true that an open subgroup of $G/H$ has the form $K/H$ where $K$ is an open subgroup of $G$ containing $H$?
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Let $\pi \colon G \to G/H$ be the canonical projection. If $M \subset G/H$ is an open subgroup, what is $\pi^{-1}(M)$? – Daniel Fischer Jul 12 '15 at 18:48
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$M=\pi^{-1}(M)/H$. Thank you – Aliakbar Jul 13 '15 at 06:39
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There is more we can say. We have the following correspondence theorem for groups:
If $G$ is a group and $N$ is a normal subgroup (which we'll express by $N\trianglelefteq G$), then there is an inclusion-preserving bijection $$\mathcal H_N(G)\to\mathcal H(G/N) \\ H\mapsto H/N$$ from the set of subgroups of $G$ containing $N$ to the set of subgroups of $G/N$.
Since $G\to G/N$ is an open map, this also restricts to a bijection between the open subgroups containing $N$ and the open subgroups of $G/N$.
Stefan Hamcke
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