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.

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