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.

  • 2
    $\begingroup$ I think that needs to be $re^{ik\pi/3}\mapsto (-1)^k r^3$ -- which is just $z\mapsto z^3$ restricted to $X$. $\endgroup$ – 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 $X$.

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

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