Is $(\mathbb Q^+, . )$ isomorphic with a subgroup of $(\mathbb R,+)$ ?

Question

Does there exist any injective group homomorphism from $(\mathbb Q^+,.)$ ( the multiplicative group of positive rational numbers ) to $(\mathbb R,+)$ ? I know that $(\mathbb Q^+,.) \cong (\mathbb Z[x] , +)$ but I am still getting no idea ... I have tried like checking divisibility , essential subgroups , elements of finite order ... Please help . Thanks in advance

Answer 1

You may want to try $$ x\mapsto \ln x$$

Answer 2

The subgroup $\mathbb{Z}[\pi]$ of $(\mathbb{R},+)$ (with respect to addition) is a free abelian group over a countable basis. So it is isomorphic to $(\mathbb{Q}^+,\cdot)$. Any transcendental number instead of $\pi$ is good as well.

Alternatively, $(\mathbb{R},+)$ is isomorphic to $\mathbb{Q}^{(\mathfrak{c})}$ (direct sum of $\mathfrak{c}=2^{\aleph_0}$ copies of $\mathbb{Q}$) and it's easy to embed in it a free abelian group over a countable basis.

_{Of course, using the logarithm is way easier.}