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 $$a^{-1} = \frac 1 a \in \Q_+^*$$.


 * 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.