TITLE: Group of positive rationals under multiplication not isomorphic to group of rationals
QUESTION [29 upvotes]: 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?

REPLY [4 votes]: 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 $n>1$.