Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group

Theorem
Let $\Q_{> 0}$ be the set of strictly positive rational numbers, i.e. $\Q_{> 0} = \left\{{ x \in \Q: x > 0}\right\}$.

The structure $\left({\Q_{> 0}, \times}\right)$ is an infinite abelian group.

In fact, $\left({\Q_{> 0}, \times}\right)$ is a subgroup of $\left({\Q^*, \times}\right)$, where $\Q^*$ is the set of rational numbers without zero, i.e. $\Q^* = \Q \setminus \left\{{0}\right\}$.

Proof
From Multiplicative Group of Rational Numbers we have that $\left({\Q^*, \times}\right)$ is a group.

We know that $\Q_{> 0} \ne \varnothing$, as (for example) $1 \in \Q_{> 0}$.


 * Let $a, b \in \Q_{> 0}$.

Then $a b \in \Q^*$ and $ab > 0$, so $a b \in \Q_{> 0}$.


 * Let $a \in \Q_{> 0}$. Then $a^{-1} = \dfrac 1 a \in \Q_{> 0}$.


 * So, by the Two-Step Subgroup Test, $\left({\Q_{> 0}, \times}\right)$ is a subgroup of $\left({\Q^*, \times}\right)$

From Subgroup of Abelian Group is Abelian it also follows that $\left({\Q_{> 0}, \times}\right)$ is an abelian group.

Its infinite nature follows from the nature of rational numbers.