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

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For a finite group, uniquely divisible coefficients have trivial first cohomology

Statement

Let $G$ be finite and let $A$ be an abelian $G$ -group such that, for every positive integer $m$ , multiplication by $m$ on $A$ is bijective. Then

$H^{1} (G, A) = 0.$

Facts & Assumptions

Given: A finite group $G$ of order $m$ , an abelian $G$ -group $A$ , and a crossed homomorphism $z : G \to A$ .

[L1]

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

Proof

technique · direct

1.1

Put $s = \sum_{x \in G} z (x) \in A$ . For any fixed $g \in G$ , $\sum_{x \in G} z (g x) = \sum_{x \in G} (z (g) + g \cdot z (x)) = m z (g) + g \cdot s$ , and the left-hand side is just $s$ because $x \mapsto g x$ permutes $G$ . Hence $m z (g) = s - g \cdot s$ .

givenL1algebra

2.1

Because multiplication by $m$ is bijective on $A$ , choose $a \in A$ with $m a = - s$ . Then $m (z (g) - (g \cdot a - a)) = m z (g) - g \cdot (m a) + m a = (s - g \cdot s) - (- g \cdot s) + (- s) = 0.$ Bijectivity of multiplication by $m$ forces $z (g) = g \cdot a - a$ for every $g$ . So $z$ is principal.

step 1.1choosealgebra

3.1

Every crossed homomorphism is principal, so the quotient in [L1] is zero. Therefore $H^{1} (G, A) = 0$ .

L1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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)