LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-05

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Normalized two-cocycles and coboundaries form groups

Statement

Under pointwise addition, $Z^{2} (G, M)$ is an abelian group and $B^{2} (G, M)$ is a subgroup of it.

Facts & Assumptions

Given: A group $G$ and an abelian $G$ -module $M$ .

[F1]

Normalized two-cocycles and two-coboundaries are defined by the displayed equations in Normalized two-cocycle and two-coboundary.

Proof

technique · direct

1.1F1givenalgebra

If $f$ and $f^{'}$ satisfy the cocycle and normalization equations of [F1], then $f + f^{'}$ does too, because each equation is linear in the values of the function. The zero function also satisfies those equations, and so does $- f$ . Hence $Z^{2} (G, M)$ is an abelian group under pointwise addition.

2.1F1step 1.1algebra

If $u$ and $v$ are normalized one-cochains, then $δ (u + v) = δ u + δ v$ by the formula in [F1], and $δ 0 = 0$ . So $B^{2} (G, M)$ is a subgroup of the abelian group from step 1.1.

3.1step 1.1step 2.1∎

Therefore $Z^{2} (G, M)$ is an abelian group and $B^{2} (G, M) \leq Z^{2} (G, M)$ .

Depends on

Normalized two-cocycle and two-coboundary

Used by

Second cohomology by factor sets Definition

Dependency tree · one level

1 result within one dependency step 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

Clara Loh, Group Cohomology, SS 2019 (standard reference, not scraped)
Caroline Lassueur, Cohomology of Groups, SS 2021 (standard reference, not scraped)