Group of positive rationals under multiplication not isomorphic to group of rationals

Question

A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{>0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here?

Answer 1

The isomorphism would have to map some element of $(\mathbb{Q},+)$ to $2$. There is no element of $(\mathbb{Q}_{>0},\times)$ whose square is $2$, but whatever number is mapped to $2$ has a half in $(\mathbb{Q},+)$. More generally speaking, you can divide by any natural number $n$ in $(\mathbb{Q},+)$, but you can't generally draw $n$-th roots in $(\mathbb{Q}_{>0},\times)$. More abstractly speaking, you can introduce an invertible multiplication operation on $(\mathbb{Q},+)$ to turn it into a field (in fact that in a sense is the point of the construction of $\mathbb{Q}$) but you can't define a corresponding exponentiation operation within $(\mathbb{Q}_{>0},\times)$.

The isomorphism that you expected to exist exists not between $(\mathbb{Q},+)$ and $(\mathbb{Q}_{>0},\times)$ but between $(\mathbb{Q},+)$ and $(b^\mathbb{Q},\times)$ for any $b\in\mathbb{R}_{>0} \setminus\{1\}$. Since $b^\mathbb{Q}$ always contains irrational elements, this is never a subgroup of $(\mathbb{Q}_{>0},\times)$.