# What are element with finite order at $\mathbb{C}^*/U$?

I need to find the elements with finite order at the group - $\mathbb{C}^*/U$.
$U$ - is the Circle Uint.
$\mathbb{C}^*$ - is $(\mathbb{C}/0,\cdot)$.

I need to write also the proof, and I'll be glad to get help with this...
I think that only $|z|=1,z\in\mathbb{C}$ is finite, I'm right?

I'm little bit confuse so I'd like to get help with the explanation about the finite order elements and the proof.

Thank you!

• You need to restrict even more, since you need to find $m \in \mathbb N$ such that $$z^m = z$$ Or did I oversee something? Dec 19, 2013 at 10:03
• @AlexR Everything in the unit circle has the same equivalence class. Dec 19, 2013 at 10:07
• Since the quotient map is a homomorphism of groups, it is useful to know (prove) that elements of finite order are sent by a homomorphism of groups to elements of finite order. Dec 19, 2013 at 10:18
• Someone can write how to write the proof? I'm confuse...
– CS1
Dec 19, 2013 at 10:19

Every non-zero complex number $z\in\Bbb C^\times$ can be uniquely written as $$z=re^{2\pi it}$$ where $r$ is a positive real number. Moreover $zz^\prime=(rr^\prime)e^{2\pi i(t+t^\prime)}$ so that the (norm) map $$\Bbb C^\times\longrightarrow\Bbb R^{>0},\qquad z\mapsto r$$ is a surjective homomorphism with kernel $U$. Therefore $$\frac{\Bbb C^\times}{U}\simeq\Bbb R^{>0}.$$ So the question is: what are the torsion elements (i.e. the finite order elements) in the multiplicative group $\Bbb R^{>0}$?

• I don't understand, why $z\mapsto r$? And how do you found an finite order elements?? Thank you!
– CS1
Dec 19, 2013 at 11:18
• @YoavFridman : $r$ is uniquely determined by $z$ (in fact $r=\mid z\mid^{1/2}$) so it makes perfectly sense to define a function that associates $r$ to $z$. Dec 19, 2013 at 11:22
• @YoavFridman: About the last question, which positive real numbers $r$ satisfy equations of the form $r^n=1$ for integer $n$? Dec 19, 2013 at 11:24
• @YoavFridman: If you are looking for elements of finite order, $n>0$ is to be thought fixed. What are then the $r$ such that $r^n=1$? Also $z$returns $r$, $z^\prime$ returns $r^\prime$ and $zz^\prime$ returns $rr^\prime$; this is to show that the map is a homomorphism and to apply all the theory pertaining to homomorphisms. Dec 19, 2013 at 13:17
• @YoavFridman: I don't understand what kind of example you need. Are you not familiar with complex numbers, or what? Dec 19, 2013 at 13:33

We now that every nonzero complex number $z$ can be uniquely expressed as $z=r\omega$, where $r=|z|>0$ and $\omega$ belongs to the unit circle. If $z_1=r_1\omega_1$ and $z_2=r_1\omega_2$, then $|z_1z_2|=r_1r_2$ and $\omega_1\omega_2$ belong to the unit circle.

This allows us to write the multiplicative group $\mathbb C^*$ as a product of two multiplicative groups $\mathbb R^+$ and $U$, the unit circle, i.e., $$(r,\omega)\cdot (r',\omega')=(rr',\omega\omega').$$

As $\mathbb C^*=\mathbb R^+\times U$, then $$\mathbb C^*/U=\frac{\{r\omega:r\in\mathbb R^+\,\&\,\omega\in U\}}{ \{ r\omega_1=r\omega_2:\text{for all r,\omega_1,\omega_2} \} }\cong \mathbb R^+.$$