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Inside a longer proof I saw the following statement and Im not sure why this is true (but I think this is easy), maybe someone can help me. Look at the subgroup:

$H = \{a^n \mid a \in F_{p}^*\} \subset F_p^*$. The claim is that the index is smaller or equal to $n$. I remember that $F_p^*$ is cyclic of order p-1, but I dont know the order of $H$ in general (do I?) If I would I could just divde this by $p-1$ and hope that this is smaller than $n$.

user1072285
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    This is the set of $n$th powers, not $n$th roots! To show the index is at most $n$, you could consider the map $\phi: \Bbb F_p^\times \to H$ given by $\phi(a) = a^n$. This is a surjective group homomorphism, so the index of $H$ in $\Bbb F_p^\times$ will be at most the size of $\ker \phi$. What can you say about the number of $b$ such that $b^n = 1$? – Izaak van Dongen Oct 18 '23 at 17:03
  • such bs are all roots of x^n- ^1 i.e they are less then n, right? Thank you! – user1072285 Oct 18 '23 at 17:10
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    Indeed! Feel free to write this in a 1-or-2-sentence answer to this post and then accept it :) – Izaak van Dongen Oct 18 '23 at 17:11
  • I will do, I would wait a bit because now Im wondering if one can give the index explictly and not just a bound – user1072285 Oct 18 '23 at 17:13
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    The exact index, $\gcd(n,p-1)$, follows from the properties of cyclic groups, but is not obvious without cyclicity of $\Bbb{F}_p^*$. – Jyrki Lahtonen Oct 18 '23 at 17:17
  • The "dual" problem settles that one also. – Jyrki Lahtonen Oct 18 '23 at 17:25
  • I know that $F_p*$ is cyclic of order $p-1$ but how does it follow that the index is gcd(n,p-1)? – user1072285 Oct 18 '23 at 17:38
  • You know that $\Bbb{F}_p^*$ is generated by an element $g$ of order $p-1$. The subgroup $H$ is thus generated by $g^n$. The linked result tells that the order of $g^n$ is equal to $(p-1)/\gcd(n,p-1)$. Therefore the index of $H$ is ____? – Jyrki Lahtonen Oct 19 '23 at 11:27

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