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Fix a prime $p$. Let $F$ be a field of order $p^k$ and with characteristic $p$. Let $\phi$ be the Frobenius map, which maps $a$ to $a^p$; one can check that, for a field of characteristic $p$, it is a field homomorphism.

Pick a positive integer $n$ and define $G = \{a \in F| \phi^n(a) = a\}$. It is obvious that $G$ is a subfield of $F$. Now I am asked to calculate the size of $G$.

This question appeared in an Algebraic Graph Theory textbook, so I wonder if anyone could figure out a way to construct a Cayley Graph and work this out in that graph. I believe someone might have a solution using only knowledge from Algebra. If you have such a solution and are willing to share with me, I will be glad to see your ideas.

Any responses will be appreciated.

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$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Once you make the definition of $G$ explicit, you find $$ G = \Set{ a \in F : a^{p^{n}} - a = 0} = \Set{ a \in F : \text{$a$ is a root of $x^{p^{n}} - x$}}. $$

Now you should know that $F$ is the set of the roots of $x^{p^{k}} - x$ (in a suitable extension that contains them).

Therefore $G$ is the set of the common roots of $x^{p^{k}} - x$ and $x^{p^{n}} - x$, and thus is the set of the roots of $$\gcd(x^{p^{k}} - x, x^{p^{n}} - x),$$ which one can see is $$x^{p^{\gcd(k,n)}} - x$$.

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  • $\begingroup$ Could you help clarify how your answer tell me the size of G? I do not quite see that ..... $\endgroup$ – Sanae Kochiya Feb 8 at 5:32
  • $\begingroup$ @SanaeKochiya, sure, the answer will be $p^{\gcd(k,n)}$. $\endgroup$ – Andreas Caranti Feb 8 at 10:00

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