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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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6 results · all verified · 6 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 6 also cleared it.

Schur Multipliers and Universal Central Extensions — Examples

1 · Prerequisites

2 · Summary

The examples distinguish a universal central extension, which requires a perfect base, from finite Schur covers, which need not be unique.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Multiplier of a cyclic group

Example

M(C)=0 for every cyclic group C.

Facts & Assumptions

Given: Let C be cyclic.

Verification

technique · direct
1.1

Multiplier of a cyclic group applies to every finite or infinite cyclic group and gives M(C)=0.

given
2.1

This is exactly the claimed cyclic-group calculation.

step 1.1
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Multiplier of a finite abelian group

Example

For ACn1××Cnr, M(A)1i<jrCgcd(ni,nj).

Facts & Assumptions

Given: Use the invariant-factor decomposition of A.

Verification

technique · direct
1.1

Exterior squares take finite direct sums to the sum of the exterior squares of the summands and the pairwise tensor products. Each cyclic summand has zero exterior square, while CmCnCgcd(m,n).

givenalgebra
2.1

Applying Multiplier of an abelian group gives the displayed formula, including the empty sum 0 when r1.

step 1.1algebra
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Binary icosahedral cover of A5

Example

SL(2,5)PSL(2,5)A5 is the sourced binary-icosahedral universal-cover example.

Facts & Assumptions

Given: Use Weibel, §6.9, Example 6.9.1, cited above.

[L1]

Weibel's Example 6.9.1 identifies SL(2,5)PSL(2,5) as a universal central extension with kernel {±I}.

Verification

technique · direct
1.1

The central quotient map SL(2,5)PSL(2,5) has kernel {±I}C2, and the cited example identifies it as the universal central extension.

L1given
2.1

With the standard isomorphism PSL(2,5)A5, this is the binary-icosahedral universal cover asserted in the statement.

step 1.1algebra
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Hopf formula from a one-relator presentation

Example

For ⟨x | x^n⟩, Hopf’s quotient is trivial.

Verification

Given: Take F=x and R=xn.

1.1

As F is cyclic, [F,F]=1 and R[F,F]=1.

given
2.1

Thus Hopf’s quotient is trivial.

step 1.1algebra
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

A stem extension that is not universal

Example

The central extension C2→D8→C2×C2 is stem but not universal because its base is not perfect.

Verification

Given: The center and commutator subgroup of D8 are both r2.

1.1

Thus C2D8C2×C2 is stem.

given
2.1

Its nontrivial abelian base is not perfect, so it cannot be universal.

step 1.1algebra
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Nonuniqueness of Schur covers

Statement refuted

D8 and Q8 are nonisomorphic Schur covers of C2×C2.

Counterexample

Given: Both D8 and Q8 have central commutator subgroup of order two and quotient C2×C2.

1.1

The exterior-square calculation gives M(C2×C2)C2, so both are Schur covers.

given
2.1

D8 has five involutions while Q8 has one, hence they are not isomorphic.

step 1.1algebra

Sources