Let $G$ be an abelian topological group and $H \subseteq G$ a dense subgroup (equipped with the subset topology). Furthermore let $V \subseteq H$ be a subgroup that is open in $H$. Does there exist a subgroup $U \subseteq G$ that is open in $G$ and such that $$ U \cap H = V $$ holds?
Of course, by definition of the subset topology there is an open set $U'$ such that this holds, but I am having trouble seeing if we can choose $U'$ to be a subgroup. One approach could be to let this set $U'$ be as stated and then let $U$ be the smallest subgroup of $G$ that contains $U'$. (This is formed by taking all the sums and inverses of elements in $U'$.) Then $U'$ is automatically open, but to show that $U' \cap H = U \cap H$ I am having difficulties. In fact I am not even sure if its true!
If someone could tell me whether this is true, either by a (sketch of a) proof or by counterexample, that would be amazing. Feel free to impose additional hypotheses on the topology, for example:
- $G$ is locally compact;
- $G$ has a "subgroup topology", i.e. there is a basis of open neighborhoods of the identity that is formed by open subgroups;
- $G$ is profinite.