I apologise if this has been asked before, I have searched but not found anything. Disclaimer: This is not a homework problem.
Let $\mathbb{R}^*$ be the reals but without zero, and let $\mathbb{R}_{>0}$ be the strictly positive real numbers. Consider the quotient group
$\mathbb{R}^*/\mathbb{R}_{>0}$
The subgroup is normal since $\mathbb{R}^*$ is abelian under regular multiplication. There are two cosets (?) since $r_1,r_2 $ in the same coset if we can write $r_1r_2^{-1} \in \mathbb{R}_{>0}$, so one coset will be $\mathbb{R}_{>0}$ itself, the other will be $\mathbb{R}_{<0}$. (Right?)
Now, my question: What is the geometric interpretation of this? In the case of $\mathbb{C}^*/\mathbb{R}_{>0}$ you get $S^1$. Can I make something similar here? I have tried, but not succeeded.
The similar thing here is that $\mathbb R^*/\mathbb R_{>0}$ is isomorphic to $\bigl(\{1,-1\},\times\bigr)$. To be more precise, the map from $\mathbb R^*/\mathbb R_{>0}$ into $\{1,-1\}$ which maps $\mathbb R_{>0}$ into $1$ and $\mathbb R_{<0}$ into $-1$ is a group isomorphism.
Geometrically, you can see it has follows: $\mathbb R^*$ consists of two open half-lines, of which one contains $1$ and the other one contains $-1$. The map from the previous paragrph maps the half-line containing $1$ into $1$ and the half-line containing $-1$ into $-1$. And a similar situation occurs with the standard isomorphism from $\mathbb C^*/\mathbb R_{>0}$ onto $S^1$.