Let $G = \{ z \in \mathbb{C} | z^n= 1 \ \text{for some positive integer}\ n \}$, with group structure given by multiplication of complex numbers.
Choose the correct option:
$1.$ $ G$ is a group of finite order
$2.$ $ G$ is a group of infinite order, but every element of G has finite order
$3.$ $G$ is a cyclic group
$4.$ None of these
My attempt: options $1, 3$ are true.
$G$ consists of all $n$-roots of unity for some positive integer $n$. Any element of $G$ can be uniquely written as $e^{({\frac{2i\pi k}{n}})}$ for some $k\in \{ 0,\ldots ,n-1\}$ that is prime with $n$.
$\exp\left( \frac{2k\pi i}{n}\right)^n=\exp\left( \frac{2\pi ki}{n}\cdot n\right)=\exp(2\pi k i)=1$
So $G$ is a cyclic group and $G$ is a group of finite order.
Is my solution correct?
Best Answer
No! $G$ is not cyclic( If it were, then upto isomorphism, $G \sim (\Bbb Z,+)$ .Note that $-1$ has order $2$ in $G$ but in $(\Bbb Z,+)$, there is no such element )
The group $G$ is actually union of all $n$-th root of unity. That is $$G:=\cup_{n \in \Bbb N} \{z \in \Bbb C: z^n=1\}$$
So $G$ is a group of infinite order. Also note that , for $z \in G$, order of $z$ is atmost $n$, so second option is true!