Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Conjugation xgxg1x\mapsto gxg^{-1} is an automorphism

Statement

Conjugation xgxg1x\mapsto gxg^{-1} is an automorphism.

For each gGg\in G, the map cg:GGc_g:G\to G, cg(x)=gxg1c_g(x)=gxg^{-1}, is an automorphism.

Facts & Assumptions

Proof

technique · direct
1.1

Associativity gives cg(xy)=gxyg1=(gxg1)(gyg1)c_g(xy)=gx yg^{-1}=(gxg^{-1})(gyg^{-1}), so cgc_g is a homomorphism.

L1L2givenalgebra
2.1

The map cg1c_{g^{-1}} is inverse to cgc_g by cancellation.

step 1.1L1L2givenalgebra
3.1

Thus cgc_g is a bijective homomorphism and hence an automorphism.

step 2.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 12 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources