Do the positive rationals under multiplication contain a subgroup of finite index? 
Do the positive rationals under multiplication contain a subgroup of finite index?

Similar questions usually rely the fact a subgroup is divisible, however, this is not the case in this question. I have a feeling that the answer should be "no", but have been unable to show this.

The group $(\mathbb{Q}^{>0},\cdot )$ is a free abelian group with a countably infinite basis. Hence there exists a surjective homomorphism $h: \mathbb{Q}^{>0}\rightarrow G$ to every countably generated abelian group $G$. In particular $G$ can be any finite abelian group, so that subgroups of any finite index exist.
