Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02
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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.
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Conjugation x↦gxg−1 is an automorphism

Statement

Conjugation x↦gxg−1 is an automorphism.

For each g∈G, the map cg:G→G, cg(x)=gxg−1, is an automorphism.

Facts & Assumptions

Given: A group G and g∈G.

[L1]

An automorphism is a bijective group homomorphism (Group isomorphisms, automorphisms and the set Aut⁡(G)).

Proof

technique · direct
1.1

Associativity gives cg(xy)=gxyg−1=(gxg−1)(gyg−1), so cg is a homomorphism.

L1L2givenalgebra
2.1

The map cg−1 is inverse to cg by cancellation.

step 1.1L1L2givenalgebra
3.1

Thus cg is a bijective homomorphism and hence an automorphism.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

8 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