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The factor set of a section is a normalized two-cocycle

Statement

If $s$ is a normalized section of an abelian-kernel extension, then its factor set $f_{s}$ is a normalized two-cocycle.

Facts & Assumptions

Given: An extension $1 \to M \to E \to G \to 1$ with abelian kernel, and a normalized section $s : G \to E$ .

[F1]

Normalized two-cocycles are characterized by the cocycle and normalization equations (Normalized two-cocycle and two-coboundary).

[F2]

The factor set of a normalized section is defined by $s (g) s (h) s (g h)^{- 1} = i (f_{s} (g, h))$ (Normalized set-theoretic section and factor set).

Proof

technique · direct

1.1

Because $s (1) = 1$ , [F2] gives $i (f_{s} (1, g)) = s (1) s (g) s (g)^{- 1} = 1$ and likewise $i (f_{s} (g, 1)) = 1$ . The kernel map $i$ is injective, so $f_{s} (1, g) = f_{s} (g, 1) = 0$ .

F2givenalgebra

1.2

Compute $(s (g) s (h)) s (k)$ and $s (g) (s (h) s (k))$ using [F2]. The left-associated expansion is $i (f_{s} (g, h)) i (f_{s} (g h, k)) s (g h k),$ while the right-associated expansion is $s (g) i (f_{s} (h, k)) s (g)^{- 1} i (f_{s} (g, h k)) s (g h k) .$ Translating the conjugation term by the given $G$ -action yields $g \cdot f_{s} (h, k) - f_{s} (g h, k) + f_{s} (g, h k) - f_{s} (g, h) = 0.$

F2algebra

2.1

Steps 1.1 and 1.2 are exactly the conditions of [F1], so $f_{s} \in Z^{2} (G, M)$ .

F1step 1.1step 1.2∎

Depends on

Used by

Changing the section changes the factor set by a coboundary Lemma

Dependency tree · two levels

3 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

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