Prove that $X/A \cong (\Bbb{R^*};.)$.

Question

Let $$\begin{align} X &=\left\{r\left(\cos \dfrac{k\pi}{3}+i \sin \dfrac{k\pi}{3}\right): r \in \Bbb{R^*},k \in \Bbb{Z}\right\}, \\ A &=\{z \in \Bbb{C}|z^3=1\}. \end{align}$$

We can show that $(X;.) ,(A;.)$ is a group. Moreover, we got $(A;.)$ is a subgroup of $(X;.)$.

The problem is:

Prove that $X/A \cong (\Bbb{R^*};.)$.

I tried my best to find an isomorphism but I can't find an isomorphism which has kernel $A$ even though I used the lemma $(\Bbb{R^+};+) \cong (\Bbb{R^*};.)$, so I am stuck here.

Any help is appreciated.

Answer 1

Hint: Define $\varphi :X\to \Bbb{R^*}$ such that $re^{ik\pi/3}\mapsto (-1)^kr^3$. First show $\varphi$ is a surjective homomorphism and the kernel of $\varphi$ is $A$, then use the First isomorphism theorem.

Answer 2

A shortcut:

$A$ and $\mathbb R^*$ are both subgroups of $\mathbb C^*$ and their intersection is $\{1\}$.

Since everything is abelian, this means that $\mathbb R^*A=\{ra\mid r\in\mathbb R^*, a\in A\}$ is a direct product of $\mathbb R^*$ and $A$, and in particular $\mathbb R^*A/A\cong\mathbb R^*$.

But $\mathbb R^*A$ is exactly your $X$.