Let $G$ be a subgroup of $(\mathbb R^2,+)$ such that $\mathbb R^2/G \cong \mathbb Q$ , then is $G$ dense in $\mathbb R^2$ ? If it was $\mathbb R$ instead of $\mathbb R^2$ then I know it would be dense because $G$ is uncountable and we know that any uncountable subgroup of real line is dense in real line . But I cannot figure out what happens in the plane . Please help . Thanks in advance
$G$ be a subgroup of $(\mathbb R^2,+)$ such that $\mathbb R^2/G \cong \mathbb Q$ , then is $G$ dense in $\mathbb R^2$?
4
$\begingroup$
real-analysis
group-theory
analysis
metric-spaces
infinite-groups
1 Answers
10
Note that such a $G$ cannot be contained in any proper $\mathbb{R}$-vector subspace of $\mathbb{R}^2$. Thus there exist $u,v\in G$ which are linearly independent over $\mathbb{R}$; we may assume that $u=(1,0)$ and $v=(0,1)$ (since a change of basis does not change the group structure or topology of $\mathbb{R}^2$). Now note that if $w\in G$ and $n$ is a nonzero integer then $w/n\in G$, since the image of $w/n$ in $\mathbb{R}^2/G$ is a torsion element and hence $0$. Thus $(1/n,0)$ and $(0,1/n)$ are in $G$ for all $n$, which implies $\mathbb{Q}\times\mathbb{Q}\subseteq G$. Thus $G$ is dense in $\mathbb{R}^2$.