Finding the quotient group of $\mathbb{R}^{\times}/P$, where $P = \mathbb{R}^{> 0}$.

Question

This is Artin exercise 2.10.5.

Identify the quotient group $\mathbb{R}^{\times}/P$, where $P$ denotes the subgroup of positive real numbers.

My attempt at solving it is as follows.

Given $x \in \mathbb{R} \setminus \{0\}$, the left coset of $P$ in $\mathbb{R}^{\times}$ containing $x$ is $$xP = \{xk \mid k \in P\}.$$ If $y \in xP$, then $x \sim y$, so $x^{-1} y \in P$, i.e., $\frac{1}{x} \cdot y > 0$, which is true if and only if $x,y$ have the same sign. That is, $x \sim y$ if and only if $x,y > 0$ or $x,y < 0$, so the equivalence classes are $$\mathbb{R}^{\times}/P = \{[+1], [-1]\}.$$ where $[+1]$ is the positive real numbers and $[-1]$ is the negative real numbers.

Answer 1

You're right, $\mathbb R^\times / P$ is just $\{[1],[-1]\}$ but not for the reason that you gaves: if $x>0$, then $[x] = [1]$ (since $1^{-1}x = 1x = x \in P$); and if $x<0$, then $[x] = [-1]$ (since $(-1)^{-1}x = (-1)x = -x \in P$).