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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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The relevant third oriented bordism group is finite

Statement

Assume AC, as in the smooth Casson–Gordon bordism convention. For every prime-power integer d and G=Z/d⊕Z, the oriented bordism group Ω3SO(K(G,1)) is finite.

Facts & Assumptions

Given: The standing AC convention of the smooth Casson–Gordon bordism construction, a prime power d, and G=Cd×Z.

[F2]

The complete relative-source cutting, low-dimensional coefficient fillings, skeletal exact-couple and periodic cyclic-resolution calculation gives Ω3SO(K(Cd×Z,1))=Z/d, for every positive integer d, under AC. Low-degree geometric bordism over a finite cyclic group times the circle

Proof

1.1F1F2givenalgebra

Apply [F2] to the specified prime-power integer d. Its explicit cyclic resolution works for all d without assuming d prime, and its geometric skeletal comparison uses the same smooth bordism convention [F1]. It gives the isomorphism Ω3SO(K(G,1))≅Z/d.

2.1F2step 1.1algebra∎

The group Z/d has d elements and is finite, including every exponent in a prime power. This proves the assertion under the stated AC convention, rather than inferring the prime-power calculation from its prime-order special case.

Depends on

Used by

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