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 nth root of unity and let d|n. Prove that \zeta^d is a primitive n/dth 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/dth 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 mth and nth roots of unity, respectively. Show that \zeta_m\zeta_n is a primitive mnth 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 mnth 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 mnth root of unity.