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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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Nonabelian first cohomology classifies complements as a pointed set

Statement

Let G act on a group M. Then the pointed set Hnab1(G,M) is in canonical bijection with the M-conjugacy classes of complements to the canonical copy of M in MG, with the basepoint corresponding to the canonical complement {(1,g):gG}.

Facts & Assumptions

Given: An action of G on a group M.

[L1]

A graph subset is a complement exactly when its defining map is a crossed homomorphism (A graph subgroup is a complement exactly for a crossed homomorphism).

[L2]

Nonabelian first cohomology is the orbit set of crossed homomorphisms under the action (az)(g)=az(g)(ga)1 (First nonabelian cohomology as a pointed set).

Proof

technique · direct
1.1

For a nonabelian crossed homomorphism z, the graph Γz is a complement by [L1]. Conversely, every complement gives a unique graph map by the same lemma. So complements correspond exactly to nonabelian crossed homomorphisms.

givenL1
2.1

Conjugating (z(g),g) by (a,1) in MG gives (a,1)(z(g),g)(a,1)1=(az(g)(ga)1,g), which is the graph of the cocycle (az)(g) from [L2]. Hence M-conjugate complements correspond exactly to M-orbits of cocycles.

L2step 1.1algebra
3.1

The trivial cocycle has graph {(1,g):gG}, the canonical complement to M. Therefore the bijection of step 2.1 respects the distinguished basepoint and is an isomorphism of pointed sets.

L1L2step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources