Doubt about description of group

Question

I have a small doubt. I was solving one algebra question which involved following group. $G=\{z \in \mathbb C\,|\,z^{n}=1\text{ for some }n\in\mathbb N\}$. Is this group the entire unit circle? I think yes but not sure. Any hint please.

Answer 1

No, $G$ is not the entire unit circle. If $r$ is an irrational number, then $$z=e^{r\pi i}\notin G$$

Answer 2

The unit circle is isomorphic to $\mathbb R / \mathbb Z$.

Your group is isomorphic to $\mathbb Q / \mathbb Z$, a proper subgroup of $\mathbb R / \mathbb Z$.

In both cases, the map is $t \mapsto \exp(t 2\pi i)$.

Answer 3

No, but it's a dense subset of the unit circle. For instance $\cos(1)+i\sin(1)$, which belongs to the unit circle, doesn't belong to $G$.

Answer 4

No, your set is equivalent to: $$\{e^{2i\pi q}|q\in \Bbb{Q}\cap [0,1]\}$$

if it was $\{z\in \Bbb{C}|z^x=1 \text{ for some } x \in \Bbb{R}\}$, it would be correct.

Answer 5

No, it's not.

The elements in your group clearly are in the unit circle and are of the form $z=e^{2\pi it}$ where $t\in[0,1)$. But being in $G$ is stronger than that. In fact you have:

\begin{align} z\in G&\iff \exists n\in\mathbb{N}: z^n=1\\&\iff\exists n\in\mathbb{N}: e^{2\pi int}=1\\&\iff\exists m,n\in\mathbb{N}: 2\pi int=2\pi im\\&\iff\exists m,n\in\mathbb{N}: nt=m\\&\iff t\in[0,1)\cap\mathbb{Q} \end{align}