Galois group of a finite field extension is cyclic

Let $p$ be prime and let $\mathbb{F}_{p^n}$ be the finite field with $p^n$ elements.

Define the Frobenius map

It is a field automorphism of $\mathbb{F}_{p^n}$ fixing $\mathbb{F}_p$.

Theorem.

1. The extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is a finite Galois extension .
2. Its Galois group is cyclic of order $n$: $\mathrm{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)=\langle \mathrm{Frob}_p\rangle \cong C_n.$ In particular, $\mathrm{Frob}_p^n=\mathrm{id}$ on $\mathbb{F}_{p^n}$, and no smaller positive power is the identity.
3. More generally, if $m\mid n$, then $\mathbb{F}_{p^n}/\mathbb{F}_{p^m}$ is Galois with $\mathrm{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p^m})=\langle x\mapsto x^{p^m}\rangle \quad\text{and}\quad \big|\mathrm{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p^m})\big|=n/m.$

Combined with the fundamental theorem of Galois theory , this implies that intermediate fields of $\mathbb{F}_{p^n}/\mathbb{F}_p$ correspond exactly to divisors of $n$.

Examples

1. $\mathbb{F}_{p^2}/\mathbb{F}_p$.
   The Galois group has order $2$, generated by $x\mapsto x^p$. This is the unique nontrivial $\mathbb{F}_p$-automorphism of $\mathbb{F}_{p^2}$.

2. $\mathbb{F}_{2^3}/\mathbb{F}_2$.
   The Galois group is cyclic of order $3$, generated by $x\mapsto x^2$. The intermediate fields correspond to divisors of $3$, so there are no proper intermediate fields besides $\mathbb{F}_2$.

3. $\mathbb{F}_{p^6}/\mathbb{F}_{p^2}$.
   Since $2\mid 6$, this is Galois of degree $3$ and

   $$\mathrm{Gal}(\mathbb{F}_{p^6}/\mathbb{F}_{p^2})=\langle x\mapsto x^{p^2}\rangle\cong C_3.$$