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Let $α$ be a primitive element of $F_q$ . Show that $α^k$ is also a primitive element if and only if gcd$(k, q − 1) = 1$.

$1=ak+b(q-1) \implies α^1=α^{(ak+b(q-1))}=α^{(ak)}α^{(q-1)}a=α^ka=α$ i cant forward the proof and i need someone to confirm me that $F_9,F_{19},F_{25},F_{32}$ has $4,6,8,2$ primitive roots, respectively.

If i were to open up for $F_9,$ $\phi(9-1)=4 \{1,2,4,8\}$

thank you for your help.

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For the "forward" direction, suppose that $d\gt 1$ divides both $k$ and $q-1$. Then $(\alpha^{k})^{(q-1)/d}=1$, contradicting the fact that $\alpha$ has order $q-1$.

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