Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group

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

The structure $\left({\Q_+^*, \times}\right)$ is an infinite abelian group.

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

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

We know that $\Q_+^* \ne \varnothing$, as (for example) $1 \in \Q_+^*$.


 * Let $a, b \in \Q_+^*$.

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


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


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

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

Its infinite nature follows from the nature of rational numbers.