TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-05

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The factor-set model agrees with the inhomogeneous cochain model in degree two

Statement

Let $C^{n} (G, M)$ be the inhomogeneous cochain groups with differentials

(d^{1} u) (g, h) = g \cdot u (h) - u (g h) + u (g)

and

(d^{2} f) (g, h, k) = g \cdot f (h, k) - f (g h, k) + f (g, h k) - f (g, h) .

Then the normalized factor-set quotient from Second cohomology by factor sets agrees with the degree-two cohomology of this inhomogeneous cochain complex:

$H^{2} (G, M) = \ker d^{2} / im d^{1} .$

Facts & Assumptions

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

[F1]

Normalized two-cocycles are exactly the functions satisfying the displayed cocycle equation, and normalized two-coboundaries are exactly the functions of the form $(g, h) \mapsto g \cdot u (h) - u (g h) + u (g)$ (Normalized two-cocycle and two-coboundary).

[F2]

The factor-set model defines $H^{2} (G, M)$ as $Z^{2} (G, M) / B^{2} (G, M)$ (Second cohomology by factor sets).

Proof

technique · direct

1.1

Let $f \in C^{2} (G, M)$ satisfy $d^{2} f = 0$ , and put $a : = f (1, 1)$ . Substituting $g = 1$ into the cocycle equation gives $f (1, h k) = f (1, h)$ for all $h, k$ , so $f (1, g) = a$ for every $g$ . Substituting $k = 1$ gives $g \cdot f (h, 1) = f (g h, 1)$ for all $g, h$ , and then taking $h = 1$ yields $f (g, 1) = g \cdot a$ .

givenalgebra

1.2

By [F1], a normalized two-cocycle is exactly a normalized function satisfying the displayed cocycle equation, so every normalized two-cocycle lies in $\ker d^{2}$ .

F1given

1.3

For an arbitrary one-cochain $u : G \to M$ , one has $(d^{1} u) (1, g) = u (1), (d^{1} u) (g, 1) = g \cdot u (1) .$ Hence $d^{1} u$ is normalized if and only if $u (1) = 0$ , that is, if and only if $u$ is a normalized one-cochain. So the normalized elements of $im d^{1}$ are exactly $B^{2} (G, M)$ from [F1].

F1givenalgebra

2.1

Let $u : G \to M$ be the constant one-cochain $u (g) = a$ . Then $(d^{1} u) (1, g) = a, (d^{1} u) (g, 1) = g \cdot a .$ So the cohomologous two-cochain $f_{0} : = f - d^{1} u$ satisfies $f_{0} (1, g) = f_{0} (g, 1) = 0$ for every $g$ . A direct cancellation shows $d^{2} (d^{1} u) = 0$ , hence $d^{2} f_{0} = 0$ . Therefore every class in $\ker d^{2} / im d^{1}$ has a normalized representative.

step 1.1algebra

3.1

Steps 2.1, 1.2, and 1.3 show that every class in $\ker d^{2} / im d^{1}$ has a normalized representative, and two normalized cocycles represent the same class there exactly when they differ by a normalized two-coboundary. Thus the quotient $\ker d^{2} / im d^{1}$ is naturally the same as $Z^{2} (G, M) / B^{2} (G, M)$ .

step 2.1step 1.2step 1.3

4.1

Combining step 3.1 with [F2] gives $H^{2} (G, M) = Z^{2} (G, M) / B^{2} (G, M) = \ker d^{2} / im d^{1} .$

F2step 3.1∎

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Sources

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