Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Every transposition factorisation of σ\sigma has parity (−1)inv⁡(σ)(-1)^{\operatorname{inv}(\sigma)}

Statement

If σ=τ1⋯τr\sigma=\tau_1\cdots\tau_r is any factorisation of a finite permutation into transpositions, then

(−1)r=(−1)inv⁡(σ).(-1)^r=(-1)^{\operatorname{inv}(\sigma)}.

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 nn and a factorisation σ=τ1⋯τr\sigma=\tau_1\cdots\tau_r in SnS_n, where each τi\tau_i is a transposition.

[L1]

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 SnS_n, Composing with a transposition reverses (−1)inv⁡(σ)(-1)^{\operatorname{inv}(\sigma)}).

Proof

technique · direct
1.1

The identity has no inversions, so the empty factorisation has inversion sign 1=(−1)01=(-1)^0.

givenL1
2.1

Starting with the identity and multiplying successively by the rr transpositions, [L1] reverses the inversion sign once at each multiplication; after rr multiplications the resulting sign is therefore (−1)r(-1)^r.

step 1.1L1
3.1

The resulting permutation is σ\sigma, so (−1)r=(−1)inv⁡(σ)(-1)^r=(-1)^{\operatorname{inv}(\sigma)}. Applying this equality to any two factorisations proves equal parity, and applying it to the identity gives even length.

step 2.1L1∎

Depends on

Used by

Dependency tree · next 3 levels

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