Consider the finite field $\mathbf{F}_{p^n}$.
If $3$ divides $p^n - 1$, can it be shown that $\mathbf{F}_{p^n}$ contains a primitive third root of unity?
I am interested in the answer to this question because it arose while trying to prove that every element of $\mathbf{F}_{p^n}$ has a cube root in $\mathbf{F}_{p^{3n}}$. Specifically, I want to use that $a^{p^{3n}} = a$ and $a^{p^n} = a$. If $p^{3n} \equiv 0 \;(\text{mod 3})$ then we're done. Otherwise, $a^{p^{3n}} = a^{p^{3n} - (p^n - 1)}$. If $p^n - 1 \equiv 1 \text{ or } 2 \;(\text{mod}3)$, then we can do this a couple of times and be done. This doesn't work if $3$ divides $p^n - 1$. However, if $F_{p^n}$ contains a primitive third root of unity, then I can be done by using the lemma in the answer here: Is it true that every element of $\mathbb{F}_p$ has an $n$-th root in $\mathbb{F}_{p^n}$?
Since $p^n - 1 = (p-1)(p^{n-1} + p^{n-2} + \dots + 1)$ we can reduce to the case that $3$ divides either factor.
If $3$ divides $p-1$, then the answer is yes because the group of units $(\mathbf{F}_p)^{\times}$ is cyclic of order $p-1$.
What about the case that $3$ divides $p^{n-1} + p^{n-2} + \dots + 1$?
I also know the group of units $(F_{p^n})^{\times}$ is cyclic of order $p^{n-1}(p-1)$.
Thanks for your help.
As said in the comments, $|\Bbb F_{p^n}^{\times}|=p^n-1$. Since 3 divides it, Cauchy's Theorem in group theory gives the existence of an element $a$ of order 3 in this group, i.e. such that $a^3=1$ and $a\neq 1$.
Edit: If we know that $\Bbb F_{p^n}^{\times}$ is cyclic, then we know that it has a cyclic subgroup of order 3, which gives the result without using Cauchy's Theorem.