Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group

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

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

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

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

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

Now, verify that the conditions for Two-Step Subgroup Test are satisfied:

Closure under $\times$
Let $a, b \in \R_{>0}$.

We take on board the fact that the Real Numbers form Ordered Integral Domain.

Then $a b \in \R_{\ne 0}$ and from Positive Elements of Ordered Ring we have $a \times b > 0$, so $a b \in \R_{>0}$.

Closure under Inverse
Let $a \in \R_{>0}$.

Then $a^{-1} = \dfrac 1 a \in \R_{>0}$.

Hence, by the Two-Step Subgroup Test, $\left({\R_{>0}, \times}\right)$ is a subgroup of $\left({\R_{\ne 0}, \times}\right)$.

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

Its infinite nature follows from the nature of real numbers.