TITLE: Prove that $X/A \cong (\Bbb{R^*};.)$.
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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 $X$.