Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group

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

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

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

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

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


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

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


 * Let $$a \in \R_+^*$$. Then $$a^{-1} = \frac 1 a \in \R_+^*$$.


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

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

Its infinite nature follows from the nature of real numbers.