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The real numbers with addition, $\left( \mathbb{R}, + \right)$, and the positive-real numbers with multiplication, $\left( \mathbb{R}_{>0}, \cdot \right)$, both are Abelian groups. For reasons of symmetry, I think that it should be also possible to construct the group of negative-real numbers.

How does the operation $$ \ast : \mathbb{R}_{<0} \to \mathbb{R}_{<0} $$ have to look like to make $$ \left( \mathbb{R}_{<0}, \ast \right) $$ an Abelian group?

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The simplest I have in my mind is this: $$x * y = -xy$$

Actually, if you have any abelian group $(G,+)$ and any bijection $f: \mathbb{R}_{<0} \to G$, then $$x*y = f^{-1}(f(x)+f(y))$$ endows $\mathbb{R}_{<0}$ with a group structure isomorphic to $G$, and $f$ becomes an isomorphism of groups. In my first case, if you take the map $x \mapsto -x$, you can recognize that $(\Bbb{R}_{< 0}, *)$ is just isomorphic to $(\Bbb{R}_{> 0}, \cdot)$

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If you must, define $a*b = -(ab)$.

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