Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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.

Central factors are equivalent to adjacent commutator containments

Statement

Let N⊴G and N≤H≤G. Then H/N≤Z(G/N)⟺[G,H]≤N. Consequently a normal series 1=H0≤H1≤⋯≤Hc=G is central, meaning Hi+1/Hi≤Z(G/Hi), exactly when [G,Hi+1]≤Hi for every 0≤i<c.

Facts & Assumptions

Given: A normal subgroup N⊴G and a subgroup H containing N.

[F1]

[G,H] is generated by all [g,h] with g∈G and h∈H (Subgroup commutators and the lower central series).

[F2]

An element is central exactly when it commutes with every element of the group (The center Z(G) of a group).

[F3]

Multiplication in G/N is (gN)(hN)=ghN (The quotient group G/N and coset product (gN)(hN)=ghN).

Proof

technique · direct
1.1

Suppose H/N≤Z(G/N). For every g∈G and h∈H, the cosets gN and hN commute by [F2]. Expanding their products with [F3] gives ghN=hgN, equivalently ghg−1h−1∈N; hence [F1] gives [G,H]≤N.

assume-hypF1F2F3
2.1

Conversely, suppose [G,H]≤N. Then [F1] gives ghg−1h−1∈N for all g∈G,h∈H. Reversing the coset calculation in step 1.1 shows gN and hN commute, so H/N≤Z(G/N) by [F2].

assume-hypstep 1.1F1F2F3
3.1

Applying the equivalence of steps 1.1 and 2.1 with (N,H)=(Hi,Hi+1) for each adjacent pair proves the series criterion, including H0=1 and Hc=G.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

9 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