primitive roots $g^a \mod{p}$

Question

$p$ prime, $g$ primitive root $\mod{p}$, $0 \leq a \leq p-2$

Show:

$g^a \mod{p}$ is a primitive root $\mod{p}$ $\Leftrightarrow$ gcd($a,p-1) = 1$

Ideas:

$g^a \mod{p}$ is a primitive root if $ord(g^a)=p-1$, so thats the left side. But how can i show the equivalence?

Answer 1

Hint: you can show that for any cyclic group $G$ of order $n$ and any generator $g$ of $G$, the order of $g^a$ is equal to $\dfrac n{n\wedge a}$.