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

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?

Solution 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 $(\mathbb{Q}_{>0},\times)$.

Solution 2:

The fundamental theorem of arithmetic exactly says that $(\mathbb{Q}_{>0}, \times)$ is an abelian free group with the set of primes as a basis. Therefore, if $(\mathbb{Q}_{>0}, \times)$ and $(\mathbb{Q},+)$ were isomorphic, $(\mathbb{Q},+)$ would be an abelian free group.

Suppose by contradiction that $X$ is a free basis of $(\mathbb{Q},+)$ and let $x \in X$. Then there exist $x_1,\dots,x_m \in X$ and $a_1, \dots,a_m \in \mathbb{Z}$ (uniquely determined) so that $$ \frac{x}{n} = a_1x_1+ \dots + a_mx_m,$$ hence $$x= na_1x_1+ \dots+ na_mx_m.$$ Consequently, there exists $1 \leq i \leq m$ such that $x=na_ix_i=na_ix$ ie. $na_i=1$; a contradiction with $a_i \in \mathbb{Z}$ when $n>1$.