Showing that $a$ is a primitive root modulo $p$

$a$ is not a primitive root mod $p \iff a^n\equiv1\bmod p$ for $0<n<p-1$.

But if $a^n\equiv1\bmod p$ then since $a^{p-1}\equiv1\bmod p$, $a^d\equiv1\bmod p$ where $d=\gcd(n,p-1)$,

and since $\dfrac{p-1}2$ is prime, the only possibilities for $d$ would be $1, 2, $ and $\dfrac {p-1}2$,

which have been ruled out.