LemmaStatement: 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.

A graph subgroup is a complement exactly for a crossed homomorphism

Statement

Let $G$ act on a group $M$ . A function $z : G \to M$ is a crossed homomorphism if and only if its graph subset

$Γ_{z} = {(z (g), g) : g \in G} \subseteq M ⋊ G$

is a complement to the canonical copy of $M$ in $M ⋊ G$ .

Facts & Assumptions

Given: A group action of $G$ on $M$ , and the semidirect product $M ⋊ G$ .

[L1]

A crossed homomorphism satisfies $z (g h) = z (g) (g \cdot z (h))$ (Crossed homomorphism for a G-group).

[L2]

The graph subset is $Γ_{z} = {(z (g), g) : g \in G}$ in the semidirect product (The graph subgroup attached to a map into a semidirect product).

[L3]

The semidirect-product multiplication is $(m, g) (n, h) = (m (g \cdot n), g h),$ and the canonical copy of $M$ is the kernel of the projection to $G$ ( The semidirect-product multiplication makes $N \times H$ a group, The canonical copy of $N$ is normal, the canonical copy of $H$ is a complement, and conjugation induces the action).

Proof

technique · iff

1.1

Suppose $z$ is a crossed homomorphism. Then $(z (g), g) (z (h), h) = (z (g) (g \cdot z (h)), g h) = (z (g h), g h)$ , so [L1] and [L3] show that $Γ_{z}$ is closed under products. The identity is $(z (1), 1) = (1, 1)$ , and inverses also stay in $Γ_{z}$ , so $Γ_{z}$ is a subgroup.

L1L2L3algebra

1.2

Conversely, suppose $Γ_{z}$ is a complement. Since it is a subgroup, the product of $(z (g), g)$ and $(z (h), h)$ again lies in $Γ_{z}$ . Comparing second coordinates gives $(z (g), g) (z (h), h) = (z (g h), g h)$ , and then [L3] forces $z (g h) = z (g) (g \cdot z (h))$ . So $z$ is a crossed homomorphism.

L1L2L3algebra

2.1

The projection $M ⋊ G \to G$ restricts to a bijection $Γ_{z} \to G$ by [L2], so $Γ_{z}$ intersects the kernel $M$ trivially and multiplies with that kernel to all of $M ⋊ G$ . Hence $Γ_{z}$ is a complement to the canonical copy of $M$ .

L2L3step 1.1

3.1

Steps 1.1-1.2 and 2.1 prove the equivalence.

step 2.1step 1.2∎

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)