Order of the root of an irreducible polynomial over a finite field

Question

Let $p(x) \in F_p[x]$ be a monic, irreducible polynomial over $F_p$, of degree $n$. Then we know that its root satisfies $\alpha^{p^n-1}=1$ in the splitting field $F_{p^n}$ of $p(x)$ as the cyclic group $F^*_{p^n}$ has order $p^n -1$. I am struggling with finding a root of an irreducible polynomial whose order is $\textbf{not}$ $p^n-1$ in the cyclic group $F^*_{p^n}$.

Answer 1

All the roots of a given irreducible $p(x)$ have the same multiplicative order since the Frobenius automorphism $a\mapsto a^p$ permutes them transitively.

So everything depends on which irreducible you are looking at. If you take,for example, the irreducible $x^4 +x^3 +x^2 +x +1$ over $\mathbb{F}_2$, its four roots are the four elements of order $5$ in $\mathbb{F}_{16}$. The other two irreducibles of order $4$ over $\mathbb{F}_2$ give the elements of order $15$. The elements of order $3$ are roots of an irreducible of degree $2$, namely $x^2 +x +1$.And the element of order $1$ is a root of $x+1$.

I do not think there is an easier way of telling the order of the roots of an irreducible than seeing which is the first $x^{p^k}-1$ it divides.