False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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FALSE: first cohomology with nonabelian coefficients is a group

Statement

For nonabelian coefficients, pointwise multiplication of cocycles always induces a group structure on first cohomology.

Facts & Assumptions

Given: The trivial action of $C_{2} \times C_{2} = ⟨ s, t ⟩$ on $S_{3}$ .

[L1]

Nonabelian first cohomology is defined as a pointed orbit set (First nonabelian cohomology as a pointed set).

[L2]

It classifies complements only up to coefficient-group conjugacy as a pointed set (Nonabelian first cohomology classifies complements as a pointed set).

Refutation

technique · direct

1.1

Under the trivial action, a nonabelian cocycle is exactly a homomorphism to $S_{3}$ . Define cocycles $z, w : C_{2} \times C_{2} \to S_{3}$ by $z (s) = (12), z (t) = 1, w (s) = 1, w (t) = (23) .$ Their pointwise product $u = z w$ satisfies $u (s) = (12)$ and $u (t) = (23)$ . Since $s t = t s$ , a homomorphism would require $u (s) u (t) = u (t) u (s)$ , but $(12) (23) \neq (23) (12)$ . Thus $u$ is not a cocycle.

givenL1algebra

2.1

Thus cocycles are not even closed under the proposed pointwise operation. Fact [L1] accordingly defines nonabelian $H^{1}$ only as a pointed orbit set, and [L2] identifies its natural classification target as a pointed set of complement classes. Therefore pointwise multiplication does not induce the asserted group structure.

L1L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

Chaoli Li, Class field theory: proofs (standard reference, not scraped)