Showing that z is a root of $X^n -1$ iff z is an element of the set of generators of dth roots of unity for some $d\mid n.$

Question

I would like to show that $z \in \mathbb{C}$ is the root of $X^n -1$ if and only if $z \in \mu^*_d$ for a certain $d \in \mathbb{N}$ which divides n where $\mu^*_d$ is the set of generators of the group of dth roots of unity.

My attempt for the first part of the exercise is:

(=>) Suppose $z \in \mathbb{C}$ is a root of $X^n -1$

$z^n =1$ where $n=qd$ for $q \in \mathbb{N}$

Therefore $z^{qd} =1$

z is a root of unity so:

$z^n=(\exp(\frac{2 k \pi i}{n}))^{n}=\exp(\frac{2 k \pi i}{n})^{qd}$

From here I want to argue that this will somehow show that z should be an element of $\mu^*_d$ but I can't see the missing argument.

Am I on the right path and if so what am I missing?

Answer 1

You're off a little, but playing with some of the tools.

Suppose $z^n=1$ and $z$ is not a primitive $n$th root of unity.

Then $z=e^{2\pi ik/n},$ where $(k,n)\gt1.$

But then $z$ is a primitive $n/(n,k)$-th root of unity. See here.

The converse is rather trivial.