CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Distinct finite automorphism groups have distinct fixed fields

Statement

For a field $K$ , the assignment $G \mapsto K^{G}$ is injective on finite groups of automorphisms of $K$ . Equivalently, distinct finite automorphism groups have distinct fixed fields.

Facts & Assumptions

Given: Finite groups $G, H$ of automorphisms of one field $K$ .

[L1]

If $G$ is a finite group of automorphisms of $K$ , then $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ (Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ ).

Proof

technique · direct

1.1

If $K^{G} = K^{H} = E$ , then [L1] applied to each group gives $G = Aut (K / E) = H$ . Thus equal fixed fields force equal groups, including when either group is trivial.

2.1

Step 1.1 is precisely injectivity of $G \mapsto K^{G}$ ; its contrapositive says that distinct finite automorphism groups have distinct fixed fields.

step 1.1∎

Depends on

Artin's fixed-field theorem: $[K:K^G]=|G|$ and $\operatorname{Aut}(K/K^G)=G$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Corollary 3.5 (standard reference, not scraped)
K. Conrad, The Galois Correspondence, Theorem 5.3 (standard reference, not scraped)