Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
How statement and proof provenance work

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.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Integral Stallings theorem for finite lower-central quotients

Statement

Assume AC. Let ϕ:A→B be a group homomorphism inducing an isomorphism H1(A;Z)→H1(B;Z) and an epimorphism H2(A;Z)→H2(B;Z). For a group X, define Γ1X=X and Γn+1X=[X,ΓnX]. Then for every finite n≥1, the induced homomorphism A/ΓnA⟶B/ΓnB is an isomorphism.

Facts & Assumptions

Given: AC and a group homomorphism ϕ:A→B satisfying the stated H1 and H2 hypotheses.

[A1]

Full AC, written AC, is The Axiom of Choice; it is inherited from the group-extension homology sequence supplier.

[F1]

For every group extension 1→N→G→Q→1, the exact sequence H2(G)→H2(Q)→N/[G,N]→H1(G)→H1(Q)→0 is natural in maps of extensions, including its transgression (the supplier proves this naturality using maps of classifying-space fibrations). Five-term homology sequence of a group extension

[L1]
[F2]

Under DC with supplied projective-resolution data, the bar complex computes group homology. In the bar complex, degree one has generators [x] and degree-two boundaries [y]−[xy]+[x], so H1(X;Z)≅X/[X,X]. Diagonal bar coinvariants compute group homology

Proof

Given: The group map and hypotheses above.

1.1A1L1F2givenconstruct

Supply the resolution convention for [F2] by taking, for every module over each relevant group ring, the free module on its underlying set, augmented [m]↦m, and repeating this construction on successive kernels. These surjections give an exact free resolution. By [A1] and [L1], DC also holds. The case n=1 is the map between two trivial quotients. For n=2, the bar degree-one calculation in [F2] identifies A/Γ2A→B/Γ2B with the assumed isomorphism on H1. These are the base cases.

1.2A1F1F2givenalgebra

Fix n≥2 and assume A/ΓnA→B/ΓnB is an isomorphism. Apply [F1] to the extensions 1→ΓnA→A→A/ΓnA→1 and 1→ΓnB→B→B/ΓnB→1. Since ΓnX⊆[X,X] for n≥2, the maps H1(X)→H1(X/ΓnX) are isomorphisms. Exactness therefore makes the transgression H2(X/ΓnX)→ΓnX/[X,ΓnX] onto, with kernel the image of H2(X). By definition [X,ΓnX]=Γn+1X.

2.1F1step 1.2algebra

Naturality of [F1] makes the map H2(A/ΓnA)→H2(B/ΓnB) compatible with the two transgressions. It is an isomorphism because the quotient-group map is an isomorphism by the induction hypothesis. The epimorphism H2(A)→H2(B) makes the image of H2(A) in H2(A/ΓnA) map onto the image of H2(B) in H2(B/ΓnB). The induced map on the two cokernels is consequently an isomorphism. By step 1.2 and exactness, those cokernels are ΓnA/Γn+1A and ΓnB/Γn+1B.

3.1step 2.1algebrachoose

The inclusions and quotient maps give exact central extensions 1→ΓnX/Γn+1X→X/Γn+1X→X/ΓnX→1,X=A,B, because [X,ΓnX]=Γn+1X. The maps on the kernel and quotient terms are isomorphisms by step 2.1 and the induction hypothesis. The middle map is injective: an element in its kernel projects to the identity in the quotient, hence lies in the kernel term, where injectivity holds. It is surjective: for a target element, lift its quotient class through the quotient isomorphism and then correct the chosen middle lift by a kernel element using the kernel isomorphism. Thus A/Γn+1A→B/Γn+1B is an isomorphism.

4.1A1step 1.1step 3.1∎

Steps 1.1 and 3.1 prove by induction the isomorphism for every finite n. Apart from the inherited AC used to obtain the homology sequence [F1], the induction uses only exactness and a single element lift at a time; no additional choice principle is used.

Depends on

Used by

Dependency tree · two levels

22 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