# The size of the stabilizer of the nth power of the Frobenius Map on a field

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.

$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$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 $$x^{p^{\gcd(k,n)}} - x$$.
• @SanaeKochiya, sure, the answer will be $p^{\gcd(k,n)}$. – Andreas Caranti Feb 8 at 10:00