Alphabeta Math
PropositionStatement: 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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Blocks in a finite transitive action have a common size

Statement

Let G act transitively on a finite set Ω, and let B be a block system. Then any two blocks in B have the same finite cardinality. In particular, if B∈B, then Ω is a disjoint union of finitely many copies of B, so ∣B∣ divides ∣Ω∣.

Facts & Assumptions

Given: A transitive action of G on a finite set Ω and a block system B.

[L1]

A transitive action sends any chosen point to any other point by some element of G (Left group actions, transitive actions, and faithful actions).

[L2]

A block system is a partition of Ω into blocks, and if B is a block then every translate g⋅B is again a block (Blocks and block systems for a group action).

Proof

technique · direct
1.1L1choose

Let B,C∈B. Choose b∈B and c∈C. By transitivity there is g∈G with g⋅b=c.

2.1step 1.1L2

Since g⋅b∈(g⋅B)∩C, the two blocks g⋅B and C meet. Because B is a partition into blocks, they are equal.

3.1step 2.1algebra

The map B→C, x↦g⋅x, is a bijection because every group element acts bijectively on Ω. Hence ∣B∣=∣C∣.

4.1step 3.1given∎

As B and C were arbitrary, all blocks in B have the same size. Since the blocks are pairwise disjoint and cover the finite set Ω, the cardinality of any block divides ∣Ω∣.

Depends on

Used by

Dependency tree · two levels

3 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