Alphabeta MathHow 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 transposition factorisation of has parity
Statement
If is any factorisation of a finite permutation into transpositions, then
Consequently any two transposition factorisations of the same permutation have the same parity, and every transposition factorisation of the identity has even length.
Facts & Assumptions
Given: A natural and a factorisation in , where each is a transposition.
Every finite permutation has a transposition factorisation, and multiplying a permutation on either side by one transposition reverses its inversion sign (Every finite permutation is a product of transpositions, so the transpositions generate , Composing with a transposition reverses ).
Proof
The identity has no inversions, so the empty factorisation has inversion sign .
Starting with the identity and multiplying successively by the transpositions, [L1] reverses the inversion sign once at each multiplication; after multiplications the resulting sign is therefore .
The resulting permutation is , so . Applying this equality to any two factorisations proves equal parity, and applying it to the identity gives even length.
Depends on
Used by
- Two different transposition factorisations of the same permutation have the same parity Example
- Every alternating multilinear F satisfies F(A)=F(I)∑_σ∈ Sₙsgn(σ)∏ᵢ a_σ(i),i Lemma
- The Vandermonde product transforms by the sign of the root permutation Proposition
- The sign is a homomorphism Sₙ→{+1,-1}, surjective exactly when n≥ 2 Theorem
Dependency tree · two levels
10 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
- T. W. Judson, Abstract Algebra: Theory and Applications, §5.1, Lemma 5.5 and Theorem 5.6 (standard reference, not scraped)
- Stanford Math 51H, Permutations (standard reference, not scraped)