Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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.

Successive p-multiple layers recover the summands of C_p^2 times C_{p^3} times C_{p^4}^2

Example

For G=Cp2×Cp3×Cp42,G=C_p^2\times C_{p^3}\times C_{p^4}^2, the dimensions did_i defined by piG/pi+1G=pdi|p^iG/p^{i+1}G|=p^{d_i} are 5,3,3,2,0.5,3,3,2,0.

Facts & Assumptions

Given: The objects and hypotheses in the example.

[L1]

Suppose Gj<rCpejG\cong\prod_{j<r}C_{p^{e_j}} with ej1e_j\ge1, and in additive notation write piG={pig:gG}p^iG=\{p^ig:g\in G\}. Define did_i by piG/pi+1G=pdi|p^iG/p^{i+1}G|=p^{d_i}. Then di={j:eji+1}.d_i=|\{j:e_j\ge i+1\}|. Consequently, for every k1k\ge1, the number of summands of order pkp^k is dk1dkd_{k-1}-d_k, so the elementary divisors are intrinsic. (The successive quotients p^iG/p^{i+1}G recover the cyclic summand multiplicities of a finite abelian p-group).

[L2]

If GG and HH are finite groups, then their external direct product is finite and has order G×H=GH|G\times H|=|G|\,|H|. (For finite groups GG and HH, G×H=GH|G\times H|=|G|\,|H|).

Verification

technique · direct
1.1

At layers i=0,1,2,3,4i=0,1,2,3,4, the numbers of exponents among 1,1,3,4,41,1,3,4,4 that exceed ii are respectively 5,3,3,2,05,3,3,2,0.

givenL1L2
2.1

The differences d0d1=2d_0-d_1=2, d1d2=0d_1-d_2=0, d2d3=1d_2-d_3=1, and d3d4=2d_3-d_4=2 recover two CpC_p factors, no Cp2C_{p^2} factor, one Cp3C_{p^3} factor, and two Cp4C_{p^4} factors.

step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 55 results over 15 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