Let $p$ be a prime and let $F$ be a field with $p^k$ elements and $K$ a subfield. Show that $Gal(F : K)$ is a cyclic group of order $[F : K]$.

Question

My attempt:

Consider the map $σ : F→ F$ such that $∀a ∈ F, σ(a) = a^p$. Then $σ$ is a field automorphism of $F$.

$ker(σ) = 0$ and since $F$ is a finite field, $σ ∈ Gal(F : K)$, the galois group of order $k$. Since the order of the galois group is $k$ it is generated by $σ$ so we can deduce that $Gal(F : K)$ is a cyclic group $Z_n$.

By the Fundamental Theorem of Galois Theory, $|G| = |Gal(F : K)| = [F : K]$, for the field extension $F : K$ of finite degree.

$Gal(F : K)$ is a cyclic group of order $[F : K]$.

Answer 1

consider $\sigma$ from F to F such that $\sigma$(a)=$a^p$ then you can easily show that this an element of Galois group of order k And order of Galois group is k so it generated by $\sigma$.Therefore Galois group is cyclic.