# Do proper dense subgroups of the real numbers have uncountable index

Just what it says on the tin. Let $G$ be a dense subgroup of $\mathbb{R}$; assume that $G \neq \mathbb{R}$. I know that the index of $G$ in $\mathbb{R}$ has to be infinite (since any subgroup of $\mathbb{C}$ of finite index in $\mathbb{C}$ has to have index 1 or 2); does it have to be uncountable, though? All the examples I can readily come up with (e.g. $\mathbb{Q}$, $\mathbb{Z}[\sqrt{2}]$, ...) have uncountable index in $\mathbb{R}$.

Thanks!

• I think this should work: Let $(\xi_i)$ be a basis for $\mathbb R$ as a vector space over $\mathbb Q$. Now what happens when you remove one element from the set of $\xi_i$? – kahen Jun 15 '13 at 18:19

No: $\mathbb{R}$ is a $\mathbb{Q}$-vector space. Choose a basis and drop finitely many basis elements to find a $\mathbb{Q}$-subspace $G$ of finite codimension. Clearly, $G$ has countable index in $\mathbb{R}$ and is dense.
Added: It can't be done without some form of the axiom of choice. The statement $\mathbb{R} \cong \mathbb{R} \oplus \mathbb{Q}$ is form number 252 in the book Consequences of the Axiom of Choice, by Howard and Rubin, as you can check on the homepage of the book.
C. J. Ash explains in A consequence of the axiom of choice, Journal of the Australian Mathematical Society (Series A), 19 (1975) 306-308, that $\mathbb{R} \cong \mathbb{R} \oplus \mathbb{Q}$ implies the existence of non-measurable sets in $\mathbb{R}$. This in turn is known not to be provable from ZF alone.