TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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Every transposition factorisation of $σ$ has parity $(- 1)^{inv (σ)}$

Statement

If $σ = τ_{1} \dots τ_{r}$ is any factorisation of a finite permutation into transpositions, then

$(- 1)^{r} = (- 1)^{inv (σ)} .$

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 $n$ and a factorisation $σ = τ_{1} \dots τ_{r}$ in $S_{n}$ , where each $τ_{i}$ 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 $S_{n}$ , Composing with a transposition reverses $(- 1)^{inv (σ)}$ ).

Proof

technique · direct

1.1givenL1

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

2.1step 1.1L1

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

3.1step 2.1L1∎

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

Every transposition factorisation of $σ$ has parity $(- 1)^{inv (σ)}$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Every transposition factorisation of σσ has parity (−1)inv⁡(σ)(−1)inv(σ)

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Every transposition factorisation of $σ$ has parity $(- 1)^{inv (σ)}$