Finding isomorphism between two groups

Question

I'm reading Vern Paulsen's notes.

I don't know how can I show $\Bbb R^*/\Bbb R^+$ and $\Bbb Z_2$ are isomorphic because I'm not sure about operation of $\Bbb R^*/\Bbb R^+$. Should I consider it as multiplicative group while $\Bbb Z_2$ is additive group?

In addition I appreciate any help or hint for constructing a bijective homomorphism between them.

Thanks!

Answer 1

Consider the map $\operatorname{sgn}\colon \mathbb{R}^*\to \{-1,1\}^\times$; note that $\{-1,1\}^\times$ is a group as it is finite and closed under multiplication; by the rule of signs, $\operatorname{sgn}(xy)=\operatorname{sgn}(x)\operatorname{sgn}(y)$, and thence $\operatorname{sgn}$ is a group homomorphism; moreover, $\operatorname{sgn}$ is surjective; then (first homomorphism theorem), $\mathbb{R}^*/\operatorname{ker}(\operatorname{sgn})\cong \{-1,1\}^\times$. Now, $\operatorname{ker}(\operatorname{sgn})=\{x\in \mathbb{R}^*\mid\operatorname{sgn}(x)=1\}=\mathbb{R}_{>0}$ and $\{-1,1\}^\times \cong \mathbb{Z}_2$ (*), and finally $\mathbb{R}^*/\mathbb{R}_{>0}\cong \mathbb{Z}_2$.

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(*) $\varphi\colon \mathbb{Z}_2\to \{-1,1\}^\times$, defined by $0\mapsto 1$ and $1\mapsto -1$ is an isomorphism.

Answer 2

Yes, the operation that you should consider on $\Bbb R^*$ is the multiplication. And the operation on $\Bbb Z_2$ is induced by the addition on $\Bbb Z$. Now, consider the map$$\begin{array}{ccc}\Bbb R^*/\Bbb R^+&\longrightarrow&\Bbb Z_2\\\lambda\Bbb R^+&\mapsto&\begin{cases}0&\text{ if }\lambda>0\\1&\text{ otherwise.}\end{cases}\end{array}$$and prove that it is a group isomorphism.