Let $A$ be a dense additive subgroup of $\mathbb R$ and let $B\le A$ be a subgroup such that $[A:B]=r<\infty$. I wonder if we have that $B$ is dense in $\mathbb R$ as well.
My thoughts: Write $A=\bigsqcup_{i=1}^{r} (x_i+B)$. Suppose $B$ is not dense in $\mathbb R$, say $(a,b)\cap B=\varnothing$, I want to find an interval $(c,d)\cap A=\varnothing$. I feel like some additive combinatoris argument is needed here, but I don't know how.
In fact, if $G\subset \mathbb{R}$ is a subgroup which is not dense, then G is discrete (i.e. $r\mathbb{Z}$ for some $r\in \mathbb{R}$). To see this, consider the set $D=\{a-b\mid a,b\in G, a\neq b\}$. We have $r\mathbb{Z}\subset G$ for every $r\in D$, so G being non-dense implies that $D$ has a positive infimum, i.e. $G$ is discrete.
Therefore, if $B\subset \mathbb{R}$ is a non-dense subgroup, then it cannot be of finite index in a dense subgroup.