How to show $U_n=\{z\in \mathbb{C } | z^n=1\}$ is cyclic

Question

I need to show that $U_n=\{z\in \mathbb{C } | z^n=1\}$ is cyclic and also work out its generator. How do I go about doing so?

Answer 1

You can verify that all $n$ roots of that polynomial are going to look like $e^{k \times 2i \pi / n}$ by an application of Euler's formula. If we choose a $k$, then the elements of $U_n$ that the corresponding element will generate look like $(e^{k \times 2i \pi/n})^m = e^{km \times 2i \pi / n}$ for $m \in \mathbb{N}$.

Once $m$ is sufficiently large, we will wrap back around to where we started: $e^{k \times 2i \pi/n}$. For example, if $n=3$ and $k=4$, we get: $(e^{2i\pi/3})^4 = e^{8i \pi/3} = e^{6i \pi/3 + 2i \pi/3} = e^{6i\pi/3}e^{2i\pi/3} = e^{2i \pi/3}$.

If a given member $x \in U_n$ is a generator, then this wrapping back around should not occur until $k=n+1$. Stated in an equivalent way, $n$ should be the smallest positive integer for which $x^n = 1$. Note that $x^m = 1$ implies that $km$ a multiple of $n$. What $k$ are such that $kn$ is the smallest multiple of both $k$ and $n$?

If it helps, try out some $k$ for small choices of $n$ and look for a relationship between the $k$ that work for a given $n$.

Answer 2

Let $\omega=e^{\frac{2 \pi i}{n}}$.

Then $\omega^k$ is a root of $z^n=1$ for all $k=0,\dots,n-1$. Therefore, $\langle \omega \rangle \subseteq U_n$.

These $n$ numbers are all the roots of $z^n=1$ because $\mathbb C$ is a field and a polynomial of degree $n$ cannot have more than $n$ roots.

Therefore, $U_n = \langle \omega \rangle$ is cyclic.