## Tag Archives: primitive root

### If ζ is a primitive nth root of unity and d divides n, then ζᵈ is a primitive (n/d)th root of unity

Let $\zeta$ be a primitive $n$th root of unity and let $d|n$. Prove that $\zeta^d$ is a primitive $n/d$th root of unity.

Certainly $(\zeta^d)^{n/d} = 1$. Now if $(\zeta^d)^t = 1$, we have $n|dt$, and so $n/d$ divides $t$. So $n/d$ is the order of $\zeta^d$, and thus $\zeta^d$ is a primitive $n/d$th root of unity.

### If m and n are relatively prime, the product of a primitive mth root of unity and a primitive nth root of unity is a primitive mnth root of unity

Suppose $m$ and $n$ are relatively prime, and let $\zeta_m$ and $\zeta_n$ be primitive $m$th and $n$th roots of unity, respectively. Show that $\zeta_m\zeta_n$ is a primitive $mn$th root of unity.

Note first that $(\zeta_m\zeta_n)^{mn} = (\zeta_m^m)^n (\zeta_n^n)^m = 1$, so that $\zeta_m\zeta_n$ is an $mn$th root of unity.

Now let $t$ be the order of $\zeta_m\zeta_n$; we have $(\zeta_m\zeta_n)^t = 1$, so that $\zeta_m^t = \zeta_n^{-t}$. In particular, $\zeta_m^t$ and $\zeta_n^{-t}$ have the same order, which must be a divisor of both $m$ and $n$. Since $m$ and $n$ are relatively prime, the order of $\zeta_m^t$ is 1, so $\zeta_m^t = 1$. Likewise $\zeta_n^t = 1$. So $m|t$ and $n|t$, and again since $m$ and $n$ are relatively prime, $mn|t$. So $|\zeta_m\zeta_n| = mn$, and $\zeta_m\zeta_n$ is a primitive $mn$th root of unity.