We say a complex number z is a primitive nth root of unity if $\displaystyle z^n=1 $ but $\displaystyle z^m\neq1 $ for 0<m<n. Show that $\displaystyle \omega^k $ is a primitive nth root of unity if and only if gcd(k,n)=1.
We say a complex number z is a primitive nth root of unity if $\displaystyle z^n=1 $ but $\displaystyle z^m\neq1 $ for 0<m<n. Show that $\displaystyle \omega^k $ is a primitive nth root of unity if and only if gcd(k,n)=1.
$\displaystyle (\Longrightarrow) $ Assume $\displaystyle g=(a,b)>1 \implies \frac ng < n$ and suppose $\displaystyle k=gd $. $\displaystyle \left(\omega^k\right)^{n/g} = \omega^{dgn/g} = \left(\omega^n\right)^d = 1 $. Thus $\displaystyle \omega^k $ is not primitive.
$\displaystyle (\Longleftarrow) $ If $\displaystyle (k,n)=1 $ and $\displaystyle i<n $, then $\displaystyle ki=dn+r $ where $\displaystyle 0\leq r<n $. Thus $\displaystyle \omega^{ki} = \omega^{dn}\omega^r = \omega^r\neq1 $. So we see $\displaystyle \omega^k $ is primitive.