If $T=\lbrace -1,1 \rbrace$, show that $\mathbb{R}^{*}/\mathbb{R}^{+} \cong T$ is group under multiplication.

Question

If $T=\lbrace -1,1 \rbrace$, with the Fundamental Theorem of Homomorphism group, show that $\mathbb{R}^{*}/\mathbb{R}^{+} \cong T$ is group under multiplication, where $\mathbb{R}^*$ is set of all real number except 0. (Hint: consider $f:\mathbb{R}^* \rightarrow T$ defined by $f(x)=\frac{|x|}{x}, \forall x \in \mathbb{R}^*$.)

For attempt, first, I show that $\mathbb{R}^*/\mathbb{R}^+ \cong T$ is isomorphism, that mean $f$ must be bijective. But, from hint in the problem, its clearly that $f$ is not one-one function, so $f$ is not bijective.

Can anyone help me? Am I on the right track? Thanks for help in advanced.

Answer 1

Since

$$\begin{align} \ker(f)&:=\{k\in \Bbb R^*\mid f(k)=1\}\\ &=\Bbb R^+ \end{align}$$

and

$$\begin{align} {\rm im}(f)&:=\{y\mid \exists x\in\Bbb R^*, f(x)=y\}\\ &=\{-1,1\}, \end{align}$$

the fact that $f$ is a homomorphism gives

$$\Bbb R^*/\Bbb R^+\cong(\{-1,1\}, \times)$$

by the First Isomorphism Theorem.

Answer 2

You need to use the first isomorphism theorem:

Theorem: Let $G$ and $H$ be groups and $\varphi:G\to H$ a homomorphism. Then $G/\ker(\varphi) \cong \text{Im}(\varphi)$.

Does this help?