Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13
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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

All 3-cycles are conjugate in An for n≥5

Statement

For n≥5, all 3-cycles in An lie in one An-conjugacy class.

Facts & Assumptions

Given: n≥5.

[F1]

Two permutations in Sn are conjugate exactly when their cycle types agree (Two elements of Sn are conjugate if and only if they have the same cycle type).

[F2]

For n≥2 and σ∈An, the Sn-class of σ splits into two An-classes of equal size exactly when all cycle lengths in its decomposition, including 1-cycles for fixed points, are odd and no two are equal (For n≥2, an Sn-class of an even permutation splits in An exactly when all cycle lengths, including 1-cycles, are odd and distinct).

Proof

technique · direct
1.1

Every 3-cycle has cycle type consisting of one 3-cycle and n−3 fixed points, so [F1] puts all of them in one Sn-class.

F1
2.1

Since n−3≥2, the length 1 is repeated. Thus [F2] says that this class does not split in An.

F2step 1.1
3.1

Therefore all 3-cycles form one An-conjugacy class.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources