Isomorphism between $\mathbb{R}^*$ and $\mathbb{R}\times \mathbb{Z}_2$

Question

I want to show that $\mathbb{R}^*\cong \mathbb{R}\times \mathbb{Z}_2$.

I don't know where to start, but I know that $f(1)=(0,\overline{0})$ because isomorphisms take the neutral element to the neutral element.

How can I define an isomorphism between these 2 groups? I just need a hint to lead me where to start from.

Thanks.

Answer 1

Here is my thought process. What is the most natural way to transfer addition to multiplication? The answer is the exponential, which defines an injective group homomorphism $\exp\colon\mathbb R\to\mathbb R^*$. The image of this group homomorphism is the multiplicative group $\mathbb R_{>0}$.

Thus, all that is left is to show that $\mathbb R_{>0}\times\mathbb Z_2\cong\mathbb R^*$. The natural way to do this is: $(x,\overline0)\mapsto x,(x,\overline 1)\mapsto -x$.

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Alternatively, one may observe that we have an exact sequence:

$$1\to\mathbb R\xrightarrow{\exp}\mathbb R^*\xrightarrow{sgn}\{\pm1\}\to1,$$ where $sgn$ sends postive numbers to $1$ and negative numbers to $-1$. Now observe that the inclusion $\{\pm1\}\hookrightarrow\mathbb R^*$ is a section, so the sequence splits.

Answer 2

Let $\Bbb Z_2$ be given by $(\{1,-1\}, \times)$.

Consider

$$\begin{align} \Bbb R\times \Bbb Z_2 &\to \Bbb R^*\\ (x,1) & \mapsto e^x\\ (x,-1) & \mapsto -e^{x}. \end{align}$$