Fix a finite field $\mathbb{F}_{p^k}$. Since its multiplicative group $\mathbb{F}_{p^k}^\times$ is cyclic,the primitive $(p^k-1)^{th}$ roots of unity are precisely the generators of $\mathbb{F}_{p^k}^\times$ of which there are $\varphi(p^k-1)$, where $\varphi$ is the Euler totient function. This makes sense so far.

From this mathoverflow answer, the number of $n$-th roots equals $\gcd{(n,p^k-1)}$. I am struggling to understand why. Letting $g$ be a generator for $\mathbb{F}_{p^k}^\times$, the first step in justifying the claim is

For an element $x$ of the group $x^n=1$ holds iff $x=g^m$ with $nm$ divisble by $p^k-1$

Why is this the case? I'm probably missing something super basic.

  • 2
    $\begingroup$ In a cyclic group of order $\ell$ the equation $x^r=1$ has $d=\gcd(r,\ell)$ solutions. This is because $x$ can be written as a power of the generator $g$, $x=g^i$. And $x^r=1$ if and only if $\ell\mid ri$. This happens if and only if $i$ is divisible by $\ell/d$. $\endgroup$ Nov 3, 2022 at 5:15
  • $\begingroup$ Start from here. $\endgroup$ Nov 3, 2022 at 5:45

1 Answer 1


As you said, the group $\mathbb{F}_q^*$ ($q=p^k$) is cyclic of order $q-1$, so every element is of the form $g^m$ for some generator $g$. Of course, $g$ has order $q-1$. If $x=g^m$ is an $n$th root of unity, that is the same as saying $x^n=g^{mn}=1$ which is the same as $q-1\mid mn$.


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