Doubt about description of group 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.
 A: 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 $t \mapsto \exp(t 2\pi i)$.
A: 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 $G$.
A: No, $G$ is not the entire unit circle. If $r$ is an irrational number, then
$$z=e^{r\pi i}\notin G$$
A: 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.
A: 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}
