R^x/<-1> is Isomorphic to R^+

Proposition 1: $\mathbb{R}^{\times}/\langle-1 \rangle$ is isomorphic to $\mathbb{R}^+$.

Recall that $\mathbb{R}^{\times} = \mathbb{R} \ \{ 0 \}$ and $\mathbb{R}^+ = \{ x \in \mathbb{R} : x > 0 \}$. Both are multiplicative groups with identity $1$.

• Proof: Let $\phi : \mathbb{R}^{\times} \to \mathbb{R}^+$ be defined for all $x \in \mathbb{R}^{\times}$ by:

(1)

\begin{align} \quad \phi(x) = |x| \end{align}

• Observe that $\phi$ is a homomorphism from $\mathbb{R}^{\times}$ to $\mathbb{R}^+$ since for all $x \in \mathbb{R}^{\times}$ we have that $\phi(x) = |x| > 0$ (since $x \neq 0$), and for all $x_1, x_2 \in \mathbb{R}^{\times}$ we have that:

(2)

\begin{align} \quad \phi(x_1x_2) = |x_1x_2| = |x_1||x_2| = \phi(x_1)\phi(x_2) \end{align}

• Now observe that:

(3)

\begin{align} \quad \ker(\phi) &= \{ x \in \mathbb{R}^{\times} : \phi(x) = 1 \} \\ &= \{ x \in \mathbb{R}^{\times} : |x| = 1 \} \\ &= \{ 1, -1 \} \\ &= \langle -1 \rangle \end{align}

• So by The First Group Isomorphism Theorem we have that:

(4)

\begin{align} \quad \mathbb{R}^{\times}/\langle -1 \rangle = \mathbb{R}^{\times}/\ker(\phi) \cong \phi(\mathbb{R}^{+}) \end{align}

• But observe also that $\phi$ is surjective since for all $y \in \mathbb{R}^{+}$ we have that $y > 0$, and so $y \in \mathbb{R}^{\times}$ is such that $\phi(y) = |y| = y$. Therefore $\phi(\mathbb{R}^{\times}) = \mathbb{R}^+$, and so from above:

(5)

\begin{align} \quad \mathbb{R}^{\times}/\langle -1 \rangle \cong \mathbb{R}^+ \quad \blacksquare \end{align}