Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit

Theorem
Let $E$ be a field.

Let $G \leq \operatorname{Aut}(E)$ be a subgroup of its automorphism group.

Let $F = \operatorname{Fix}_E(G)$ be its fixed field.

Let $\alpha \in E$ have a finite orbit under $G$.

Then the product of polynomials
 * $p(x) = \displaystyle\prod_{\beta \in \Lambda} (x - \beta)$

is the minimal polynomial of $\alpha$ over $F$.

Proof
By Product over Finite Set with Zero Factor, we have $p(\alpha) = 0$.

By definition, $p \in E[x]$.

$p$ has coefficients in $F$
We show that $p \in F[x]$.

Let $\sigma \in G$, and denote the induced automorphism of $E[x]$ still by $\sigma$.

We show that $\sigma(p) = p$.

We have:

Because this is true for all $\sigma \in G$, indeed the coefficients of $p$ are in $\operatorname{Fix}(G) = F$.

Thus $p \in F[x]$.

$p$ is the minimal polynomial
By Product of Monic Polynomials is Monic, $p$ is monic.

In particular, $p$ is nonzero.

Thus $\alpha$ is algebraic over $F$.

Let $f$ be its minimal polynomial over $F$.

We show that $f = p$.

By definition and beacuse $p(\alpha) = 0$, $f$ divides $p$.

By Group of Automorphisms is Contained in Automorphism Group over Fixed Field, $G \leq \operatorname{Aut}(E/F)$.

By Automorphism Group of Field Extension Permutes Roots of Minimal Polynomial, each $\beta \in \Lambda$ is a root of $f$.

By Polynomial Factor Theorem, each $x-\beta$ divides $f$.

By Product of Pairwise Coprime Divisors of Polynomial over Field is Divisor, $p$ divides $f$.

By Monic Polynomials are Equal iff Divide Each Other, $p = f$.