Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 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.

Every finite p-group is nilpotent

Statement

Every finite p-group is nilpotent. The trivial group is included and has nilpotency class zero.

Facts & Assumptions

Given: A prime p and a finite group G of order pn for some nN.

[F1]

Zr+1(G) is the inverse image of Z(G/Zr(G)) under the quotient map (The upper central series).

[F2]

G is nilpotent if Zc(G)=G for some c; the trivial group has class zero (Nilpotent groups and nilpotency class).

Proof

technique · induction
1.1

If n=0, then G=1, so G=1 is nilpotent of class zero by [F2].

baseF2
1.2

Assume n>0 and that every p-group of order smaller than pn is nilpotent.

ih
2.1

By [L1], Z(G) is nontrivial. Its order is a positive power pk with 1kn, and [L2] gives G/Z(G)=pnk<pn.

step 1.2L1L2
3.1

By induction, G/Z(G) is nilpotent, so its upper central series reaches the whole quotient at some term c.

step 2.1ihF2
4.1

Starting with Z1(G)=Z(G), [F1] shows inductively that the inverse image in G of Zr(G/Z(G)) is Zr+1(G). Since the quotient series reaches G/Z(G) at r=c, one has Zc+1(G)=G.

step 3.1F1L3
5.1

Thus G is nilpotent by [F2], completing the induction.

step 4.1F2discharge-induction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 89 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources