Does there exist a subgroup $G$ of $\mathbb{R}$ such that $\mathbb{R}/G \cong \mathbb{Z}$?

Question

My question is whether or not it's possible to find an additive subgroup $G$ of the real numbers such that the quotient group $\mathbb{R}/G$ is isomorphic to the infinite cyclic group. I'm not familiar with how to find the cardinality of infinite quotient groups, but my guess is that $G$ would have to have the same cardinality as the reals, in order for $\mathbb{R}/G$ to have cardinality $\aleph_0$.

If I were to decide either way, I would guess that such a subgroup doesn't exist, but I haven't able to prove this, so I don't know. My idea for a proof of the negation of the statement is that given any isomorphism $\varphi: \mathbb{Z} \to \mathbb{R}/G=<r>$, $\varphi$ sends $1$ to the generator $r$, and to show that if $G$ is dense in the reals, there must exist elements of $G$ that add up to $r$, thus showing $r \in G$ by closure, which is a contradiction.

Then again, the rationals are dense in $\mathbb{R}$ and no finite sum of rationals will ever give us an irrational number, so this proof may fall apart. Any thoughts?

Answer 1

Let $H$ be any quotient group of $\mathbb R$.

Show for every $x\in H$, there is a $y\in H$ such that $y+y=x$.

More generally, all quotient groups of any divisible group is a divisible group.

Answer 2

Suppose $G$ exists, let $p:R\rightarrow R/G=Z$ the projection, and $x\in R$ such that $p(x)=1$ for every integer $n>0$, $p(x)=1=np(x/n)$, impossible for $n>2$ since $p(x/n)\in Z$.