Galois group of a finite field extension

Remarks

The map $\varphi$ is a power of the Frobenius automorphism, so it is a field automorphism of $\F_{q^n}$. It fixes $\F_q$, because every element of $\F_q$ satisfies $a^q = a$. If $\varphi^k = \id$ with $k \ge 1$, then every element of $\F_{q^n}$ is a root of $x^{q^k} - x$, so $q^n \le q^k$ and $k \ge n$; and $\varphi^n = \id$. So $\varphi$ has order $n$, and since the Galois group has order at most the degree $n$, the group is $\langle \varphi \rangle$ and the extension is Galois. For example, $\Gal(\F_4/\F_2) = \set{\id, \varphi}$ with $\varphi(x) = x^2$, and $\varphi$ swaps the two roots $\omega$ and $\omega + 1$ of $x^2 + x + 1$.

Going deeper

For a student

With the fundamental theorem of Galois theory, the subgroups of a cyclic group of order $n$ match the fields between $\F_q$ and $\F_{q^n}$. This gives again that these fields are the $\F_{q^m}$ with $m$ dividing $n$, as in subfields of a finite field. The absolute Galois group of $\F_q$ is the profinite completion $\widehat{\Z}$ of $\Z$, topologically generated by $\varphi$. The Frobenius element of a prime in algebraic number theory comes from this theorem applied to residue fields.