# 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?

• This is Exercise II.4.6 in Aluffi Chapter 0. I was also struggling with it. – user125763 Sep 6 '14 at 13:50

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 $(\mathbb{Q}_{>0},\times)$.
• More-or-less, joriki's answer can be rephrased like this. If we find a property, which is preserved by isomorphisms and which is fulfilled only for one of the groups, then the groups are not isomorphic. In this case, the property of the group $(G,\circ)$ is $(\forall x\in G)(\exists y\in G) y\circ y=x$. – Martin Sleziak Apr 18 '11 at 9:11
• To clarify that slightly (since "being able to divide" is a property of humans and computers, not of groups :-): A group is called divisible if for each element $x$ and each natural number $n$ there is an element $y$ such that $ny$, defined as the $n$-fold sum of $y$, is $x$. Then $(\mathbb{Q},+)$ is divisible and $(\mathbb{Q}_{>0},\times)$ isn't. – joriki Apr 18 '11 at 10:19
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 $n>1$.