TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-11

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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.
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The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$

Statement

For every natural $n$ , the function $sgn : S_{n} \to {+ 1, - 1}$ is a group homomorphism. It is surjective exactly when $n \geq 2$ ; for $n = 0$ and $n = 1$ its image is ${1}$ .

Facts & Assumptions

Given: A natural $n$ and permutations $σ, ρ \in S_{n}$ .

[L1]

Every finite permutation has a transposition factorisation, the sign is $(- 1)^{inv (σ)}$ , and every such factorisation has that parity (Every finite permutation is a product of transpositions, so the transpositions generate $S_{n}$ , Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations, Every transposition factorisation of $σ$ has parity $(- 1)^{inv (σ)}$ ).

Proof

technique · direct

1.1givenL1

Choose transposition factorisations $σ = τ_{1} \dots τ_{r}$ and $ρ = υ_{1} \dots υ_{s}$ . Their concatenation is a transposition factorisation $σ ρ = τ_{1} \dots τ_{r} υ_{1} \dots υ_{s}$ in the library's composition order.

2.1step 1.1L1

By [L1], $sgn (σ ρ) = (- 1)^{r + s} = (- 1)^{r} (- 1)^{s} = sgn (σ) sgn (ρ)$ , and the identity has sign $1$ ; hence sign is a group homomorphism.

3.1step 2.1L1∎

If $n \geq 2$ , the transposition $(0 1)$ belongs to $S_{n}$ and has sign $- 1$ , while the identity has sign $1$ , so sign is surjective. If $n = 0$ or $n = 1$ , $S_{n}$ contains only the identity and the image is ${1}$ .

The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The sign is a homomorphism Sn→{+1,−1}Sn​→{+1,−1}, surjective exactly when n≥2n≥2

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$