In the group $\left( \mathbb{C} \setminus\{0\}, \times \right)$ find all elements of order $12$.

Question

In the group $\left( \mathbb{C} \setminus\{0\}, \times \right)$ find all elements of order $12$.

The order of an element $g$ of a group $G$ is the smallest positive integer $m$, such that $g^{m}=e$, the identity element. So we need to find all $12$th roots of $1$. There are $12$ of them but some of them will equal $1$ in other degree that is less than $12$ and therefore will not suit our definition. So how to find the elements I need?

Answer 1

The $12^{th}$ roots of unity are $\exp(2\pi ki/12)$ where $k=0, 1, ..., 11$.

If $\gcd(k,12)=d,$ then $\exp(2\pi ki/12)^{12/d}=\exp(2\pi ki/d)=1,$

so $\exp(2\pi ki/12)$ is a $12/d^{th}$ root of unity.

So $\exp(2\pi ki/d)$ is a primitive $12^{th}$ root of unity when $d=1$,

i.e., $\gcd(k,12)=1,$ i.e., $k=1, 5, 7,$ or $11.$

Answer 2

Every finite subgroup of $\Bbb C^{\times}$ is cyclic. The cyclic group $C_{12}$ has $\phi(12)=4$ different generators.