Let $\omega$ be a primitive $n$-th root of unity. Prove that $\omega^k$ is a primitive $\tfrac{n}{\text{gcd}(n,k)}$-th root of unity.
To prove this, I have to show that if$\quad \omega^n = 1 \quad$ then $\quad (\omega^k)^{\tfrac{n}{\text{gcd}(n,k)}} = 1 \quad$,
as well as to show that $ \quad \forall m < \tfrac{n}{\text{gcd}(n,k)} : (\omega^k)^m \neq 1 \quad$.
The first one is easy:
$(\omega^k)^{\tfrac{n}{\text{gcd}(n,k)}} = (\omega^n)^{\tfrac{k}{\text{gcd}(n,k)}} = 1^{\tfrac{k}{\text{gcd}(n,k)}} = 1$
But i am having a hard time proving the second identity. Does anyone know how to prove this and can help me with that?
Greetings, Octavius
