TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Conjugation $x \mapsto g x g^{- 1}$ is an automorphism

Conjugation $x \mapsto g x g^{- 1}$ is an automorphism.

For each $g \in G$ , the map $c_{g} : G \to G$ , $c_{g} (x) = g x g^{- 1}$ , is an automorphism.

Given: A group $G$ and $g \in G$ .

[L1]

[L2]

technique · direct

1.1L1L2givenalgebra

Associativity gives $c_{g} (x y) = g x y g^{- 1} = (g x g^{- 1}) (g y g^{- 1})$ , so $c_{g}$ is a homomorphism.

2.1step 1.1L1L2givenalgebra

The map $c_{g^{- 1}}$ is inverse to $c_{g}$ by cancellation.

3.1step 2.1∎

Thus $c_{g}$ is a bijective homomorphism and hence an automorphism.

Depends on