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

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.

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.

• I think that needs to be $re^{ik\pi/3}\mapsto (-1)^k r^3$ -- which is just $z\mapsto z^3$ restricted to $X$. – Henning Makholm Dec 30 '18 at 4:49

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 $$X$$.

• I like this one best of the two so far :) – Shaun Dec 30 '18 at 10:40
• Thank you for your guidance ^^ – 129492 Dec 30 '18 at 10:40