$\zeta_n^k$ is primitive if and only if $(k,n) = 1$ Show that for $k \in \mathbb{Z}$:

Is $\zeta_n$ a primitive $n$-th root of unity, then $\zeta_n^k$ is primitive if and only if $(k,n) = 1$.

I only need the backwards direction:

$\zeta_n^k$ is primitive $\Rightarrow$ $(k,n) = 1$
First of all notice that $\zeta_n^k$ is of the form
$$
e^{2\pi i\frac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(\zeta_n^k)^n=1$. If by contradiction $(k,n)=d&gt;1$ then you can write $k=ad$ and $n=bd$, with $a,b\in\Bbb Z$ and $a,b\ge1$. Thus
$$
\zeta_n^k=e^{2\pi i\frac {ad}{bd}}=e^{2\pi i\frac d{b}}.
$$
From here we deduce that $(\zeta_n^k)^b=1$ and since $b&lt;n$ (in fact $0\le k\le n-1$) our root cannot be primitive.
First of all notice that $\zeta_n^k$ is of the form
$$
e^{2\pi i\frac kn}
$$
and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(\zeta_n^k)^n=1$. If by contradiction $(k,n)=d&gt;1$ then you can write $k=ad$ and $n=bd$, with $a,b\in\Bbb Z$ and $a,b\ge1$. Thus
$$
\zeta_n^k=e^{2\pi i\frac {ad}{bd}}=e^{2\pi i\frac d{b}}.
$$
From here we deduce that $(\zeta_n^k)^b=1$ and since $b&lt;n$ (in fact $0\le k\le n-1$) our root cannot be primitive.