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 $\left({\Q_{\ne 0}, \times}\right)$, where $\Q_{\ne 0}$ is the set of rational numbers without zero: $\Q_{\ne 0} = \Q \setminus \left\{{0}\right\}$.

Proof
From Non-Zero Rational Numbers under Multiplication form Abelian Group we have that $\left({\Q_{\ne 0}, \times}\right)$ is a group.

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

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

Then:
 * $a b \in \Q_{\ne 0}$ and $ab > 0$

Hence:
 * $a b \in \Q_{> 0}$

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 $\left({\Q_{\ne 0}, \times}\right)$.

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

From Positive Rational Numbers are Countably Infinite, it follows that $\left({\Q_{> 0}, \times}\right)$ is an infinite group.