Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26
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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.
  • 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.

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The translates of a block partition its orbit

Statement

Let G act on Ω, and let B⊆Ω be a block. Then the family TB:={ g⋅B:g∈G } has pairwise equal-or-disjoint members, its union is G⋅B:={ g⋅b:g∈G, b∈B }, and it is preserved by the action of G. Hence TB is a G-invariant partition of G⋅B.

Facts & Assumptions

Given: A left action of G on Ω and a block B⊆Ω.

[L1]

A block is a nonempty subset B such that for every g∈G one has either g⋅B=B or (g⋅B)∩B=∅ (Blocks and block systems for a group action).

Proof

technique · direct
1.1L1choose

If g⋅B meets h⋅B, choose x∈(g⋅B)∩(h⋅B). Then h−1⋅x∈(h−1g)⋅B∩B, so [L1] gives (h−1g)⋅B=B.

1.2givenalgebra

By definition every point of ⋃TB has the form g⋅b with g∈G and b∈B, and every such point lies in the translate g⋅B. So ⋃TB=G⋅B.

2.1step 1.1

From step 1.1, g⋅B=h⋅B whenever the two translates meet. Thus distinct translates are disjoint.

3.1step 2.1step 1.2algebra∎

For k∈G one has k⋅(g⋅B)=(kg)⋅B, which is again in TB. Hence G permutes the members of TB, and steps 2.1 and 1.2 make it a G-invariant partition of G⋅B.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources