union of fix points of $f\in\operatorname{Gal}(K/F)\setminus\{1\}$ is a proper subset of $K$

Question

Let $F ⊆ K$ be finite fields. Show that $\operatorname{Stab}_{\operatorname{Gal}(K/F)}(z)=\{1\}$ for at least one $z\in K$.

My attempt
Firstly $z\notin F$, since any element of $F$ is fixed by $\operatorname{Gal}(K/F)$.

$\operatorname{Stab}_{\operatorname{Gal}(K/F)}(z)=\{1\}$ is equivalent to that $f(z)$ for all $f\in\operatorname{Gal}(K/F)$ are distinct.

Existence of $z$ is equivalent to that $\bigcup_{f\in\operatorname{Gal}(K/F)\setminus\{1\}}\text{Fix}(f)$ is a proper subset of $K$.

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I'm a beginner to Galois theory, could you help me proving this?

Answer 1

$K^\times$ is cyclic, let $a$ be a generator. If some $f \in \mathrm{Gal}(K/F)$ fixes $a$, then it fixes all of $K$. (Do you see why?) By Galois theory, this implies that $f=1$ is the identity.