# Why are all Frobenius elements conjugated?

Let $$K$$ be a local field, $$k$$ its residue field and $$G_K$$, $$G_k$$ be the absolute Galois groups of $$K$$ and $$k$$, respectively.

A Frobenius element is an element $$\operatorname{Frob}_K \in G_K$$ such that its image under the restriction homomorphism $$G_K \to G_k$$ is the Frobenius automorphism $$x \mapsto x^q$$ where $$q$$ denotes the cardinality of $$k$$.

Question: Why are all Frobenius elements conjugated to each other?

I have seen this statement quite often already, but I have not figured out why this is true.

I tried it to understand on the level of a finite Galois extension $$F/K$$, so a Frobenius element there would be an element $$\operatorname{Frob}_{F/K} \in \operatorname{Gal}(F/K)$$ such that under the restriction homomorphism $$\operatorname{Gal}(F/K) \to \operatorname{Gal}(f/k)$$ it gets mapped to $$x \mapsto x^q$$ as before ($$f$$ is the residue field of $$F$$). I also know that the residue field degree is exactly the cardinality of $$\operatorname{Gal}(f/k)$$. But I was not able to see any relation there.