Find the number of disjoint cycles of the map $\phi_q: \Bbb{F}_{q^p}\to \Bbb{F}_{q^p}$

Question

Let $\Bbb{F}_q$ be a finite field. I need to find the number of disjoint cycles of the map \begin{align*} \phi_q: \Bbb{F}_{q^p}&\to \Bbb{F}_{q^p}\\ \alpha &\mapsto a^q, \end{align*}

where $p$ is a prime number.

What i've done:

First I supposed $q$ was prime,

then I noticed that the constant elements $\{0,1\dots,q-1\}$, since $\Bbb{F}_{q^p}\supset \Bbb{F}_q$, are mapped to itself, because if $\alpha \in \{0,1,\dots,q-1\}$, then by the properties of the finite fields we have: $\alpha^q=\alpha$. So in this case we would have had $q$ disjoint cycles only considering the constant terms.

If $q$ is not prime, but is a prime power, then we have $\Bbb{F}_q \approx \Bbb{F}_{a^m}$, where $a$ is a prime number and $m$ is the dimension of $\Bbb{F}_a$ as an $\Bbb{F}_q$ vector space. In this case $\Bbb{F}_{q^p}= \Bbb{F}_{a^{mp}} \supset \Bbb{F}_a$, so we have $a$ constant term, which form $a$ different disjoint cycles.

Now the problem comes with the nonconstant terms of $\Bbb{F}_{q^p}$. What can be seen is that, for $\alpha \in \Bbb{F}_{a^{mp}}$,

$$\alpha \mapsto \alpha^{a^m}\mapsto \alpha^{a^{2m}}\mapsto \cdots \mapsto \alpha^{a^{(p-1)m}}\mapsto \alpha^{a^{mp}}=\alpha.$$

The question is now how many different elements satisfy this sequence.

We have tried many examples with different values for $q$ and for $p$ and it appears to be true that if $q$ is a prime number, then there are, beyond the $q$ disjoint cycles that comes from the constant terms, $q$ different cycles of length $p$. If $q$ is not prime it is even worst. Any help is greatly appreciated.

Answer 1

You need the following bits from Galois theory:

1. $\Phi_q$ is the generator of the Galois group $Gal(\Bbb{F}_{q^p}/\Bbb{F}_q)$.
2. There are no intermediate fields, because the degree of the extension is the prime $p$.
3. The size of the orbit of an element $\alpha$ is equal to the degree of its minimal polynomial, which in turn is equal to the degree $[\Bbb{F}_q(\alpha):\Bbb{F}_q]$.

In your case $\Bbb{F}_q(\alpha)=\Bbb{F}_q$, whenever $\alpha\in \Bbb{F}_q$ and $\Bbb{F}_q(\alpha)=\Bbb{F}_{q^p}$ otherwise. Therefore we can conclude that

1. There will be $q$ cycles of size one, and
2. $(q^p-q)/p$ cycles of size $p$.

The lack of intermediate fields means that any element $\alpha\in\Bbb{F}_{q^p}\setminus\Bbb{F}_q$ has $p$ conjugates.