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

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

Then $a b \in \R^*$ and from Positive Elements of Ordered Ring we have $ab > 0$, so $a b \in \R_+^*$.


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


 * 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.