Condition for Complex Root of Unity to be Primitive

Theorem
Let $n, k \in \N$.

Then $\alpha_k = \exp \left({\dfrac {2 \pi i k} n}\right)$ is a primitive $n$th root of unity $\gcd \left\{ {n, k}\right\} = 1$.

Proof
Let $U_n = \left\{ {\exp \left({\dfrac {2 \pi i k} n}\right): 0 \le k \le n - 1}\right\}$.

Let $V = \left\{ {1, \dotsc, \alpha_k^{n - 1} }\right\}$.

By Roots of Unity it is sufficient to show that $U_n = V$ $\gcd \left\{ {n, k}\right\} = 1$.

Let $\gcd \left\{ {n, k}\right\} = d > 1$.

Then there are $n', k' \in \N$ such that:
 * $n' = d n$

and:
 * $k' = d k$

Then we have:
 * $\alpha_k = \exp \left({\dfrac {2 \pi i k'} {n'} }\right)$

and:
 * $\alpha_k^{n'} = \exp \left({2 \pi i k'}\right) = 1$

Therefore:
 * $V = \left\{ {1, \dotsc, \alpha^{n' - 1} }\right\}$

such that $n' < n$.

So:
 * $\left\vert{V}\right\vert = n' < n = \left\vert{U_n}\right\vert$

and $U_n \ne V$.

Let $\gcd \left\{ {n, k}\right\} = 1$.

Let:


 * $\exp \left({\dfrac {2 \pi i k} n}\right)^d = \exp \left({\dfrac {2 \pi i k} n}\right) = 1$

Then it must be the case that $\dfrac {k d} n \in \Z$.

Since $\gcd \left\{ {n, k}\right\} = 1$ it follows that:
 * $n \mathrel \backslash d$

and so:
 * $d \ge n$

Therefore $\left\{ {1, \dotsc, \alpha^{n - 1} }\right\}$ are distinct

Hence $\left\vert{V}\right\vert = \left\vert{U_n}\right\vert$.

Moreover each element of $V$ can be written in the form:
 * $\exp \left({\dfrac {2 \pi i k} n}\right)$

with $0 \le k \le n - 1$.

It follows that $V = U_n$.