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

Answer 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 $n>1$.

Answer 3

Suppose, to the contrary, that there is an isomorphism $\phi:(\mathbb Q,+)\to(\mathbb Q_{>0},\times)$. Then, there must be an $a\in\mathbb Q$ such that $\phi(a)=2$. Thus, $$\phi(a)=\phi(a/2+a/2)=\phi(a/2)\times \phi(a/2)=2 \, ,$$ which is absurd, since $2$ does not have a square root in $(\mathbb Q_{>0},\times)$.

More conceptually, $(\mathbb Q,+)$ is a divisible group, whereas $(\mathbb Q_{>0},\times)$ is not. A divisible group is a group (usually assumed to be commutative) where for every positive integer $n$ and $g\in G$, there is a $h\in G$ such that $h^n=g$. In additive notation, this condition would be written $nh=g$. The above proof exploits the fact that $(\mathbb Q_{>0},\times)$ is not divisible; specificially, if $n=2$ and $g=2$, then there is no $h\in\mathbb Q_{>0}$ such that $h^n=g$.

Answer 4

Let $\pi_p$ be the projection onto prime $p$, i.e. if $r=\prod_{q~\mathrm{prime}} q^{\alpha_q}\in\mathbb Q$, then $\pi_p(r)=p^{\alpha_p}$. It is easily seen to be a endomorphism of $(\mathbb Q_{>0},\times)$. For distinct primes $p\ne p'$, $\pi_p\circ\pi_{p'}$ is trivial, which means the endomorhpism ring of $(\mathbb Q_{>0},\times)$ has zero divisors. However, $\mathrm{End}(\mathbb{Q})\cong \mathbb{Q}$ is a field, which does not have one.