Condition for Complex Root of Unity to be Primitive

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

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

Proof
Let $U_n = \set {\map \exp {\dfrac {2 \pi i k} n}: 0 \le k \le n - 1}$.

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

By Complex Roots of Unity in Exponential Form it is sufficient to show that $U_n = V$ $\gcd \set {n, k} = 1$.

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

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

and:
 * $k' = d k$

Then we have:
 * $\alpha_k = \map \exp {\dfrac {2 \pi i k'} {n'} }$

and:
 * $\alpha_k^{n'} = \map \exp {2 \pi i k'} = 1$

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

such that $n' < n$.

So:
 * $\cmod V = n' < n = \cmod {U_n}$

and $U_n \ne V$.

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

Let:


 * $\map \exp {\dfrac {2 \pi i k} n}^d = \map \exp {\dfrac {2 \pi i k} n} = 1$

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

Since $\gcd \set {n, k}\ = 1$ it follows that:
 * $n \divides d$

and so:
 * $d \ge n$

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

Hence $\card V = \card {U_n}$.

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

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

It follows that $V = U_n$.

Also see

 * Complex Roots of Unity in Exponential Form