TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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.

First cohomology classifies complements up to kernel conjugacy

Statement

Let $G$ act on an abelian group $A$ . Then $H^{1} (G, A)$ is in canonical bijection with the $A$ -conjugacy classes of complements to the canonical copy of $A$ in the semidirect product $A ⋊ G$ .

Facts & Assumptions

Given: An action of $G$ on an abelian group $A$ .

[L1]

First cohomology is the quotient of crossed homomorphisms by principal crossed homomorphisms (First cohomology via crossed homomorphisms).

[L2]

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

[L3]

Conjugating a graph subgroup by a kernel element changes its defining crossed homomorphism by a principal one (Kernel conjugation by an element of the coefficient group corresponds to a principal crossed homomorphism).

Proof

technique · direct

1.1

If $z : G \to A$ is a crossed homomorphism, [L2] makes $Γ_{z}$ a complement to $A$ in $A ⋊ G$ . By [L3], replacing $z$ by a cohomologous cocycle replaces $Γ_{z}$ by an $A$ -conjugate complement. Hence the rule $[z] \mapsto [Γ_{z}]$ is well defined on $H^{1} (G, A)$ .

givenL1L2L3

2.1

Every complement $H \leq A ⋊ G$ arises from some crossed homomorphism: the projection $H \to G$ is an isomorphism, so for each $g \in G$ there is a unique element of $H$ of the form $(z (g), g)$ , and [L2] says the resulting map $z$ is a crossed homomorphism. Thus the map from step 1.1 is surjective on complement classes.

L2step 1.1construct

2.2

If $Γ_{z}$ and $Γ_{w}$ are $A$ -conjugate, then [L3] says $w - z$ is principal, so $[z] = [w]$ in the quotient [L1]. Therefore the map of step 1.1 is injective.

L1L3step 1.1

3.1

Steps 1.1-2.2 give a bijection between $H^{1} (G, A)$ and the $A$ -conjugacy classes of complements to $A$ in $A ⋊ G$ .

step 1.1step 2.1step 2.2∎

Depends on

Used by

Dependency tree · two levels

10 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

David A. Craven, Finite Group Theory (standard reference, not scraped)