TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck pass

How 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.

The standard $n$ -cycle and a neighbour transposition generate $S_{n}$

Statement

For $n \geq 2$ , use the one-based model $Sym ({1, \dots, n})$ , identified with the library's $S_{n} = Sym ({0, \dots, n - 1})$ by conjugation with $κ_{n} (k) = k + 1$ (The finite symmetric group $S_{n}$ , one-line notation, and cycle notation, Adjacent transpositions generate the finite symmetric group $S_{n}$ ). In this model, $c = (1 2 \dots n)$ and $t = (1 2)$ satisfy $⟨ c, t ⟩ = S_{n}$ . Equivalently, in the library's zero-based model $c_{0} = (0 1 \dots n - 1)$ and $t_{0} = (0 1)$ satisfy $⟨ c_{0}, t_{0} ⟩ = S_{n}$ .

Facts & Assumptions

Given: $n \geq 2$ , the one-based model fixed in the Statement, $c = (1 2 \dots n)$ , and $t = (1 2)$ .

[F1]

Conjugating a cycle relabels its entries (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F2]

The standard adjacent transpositions generate $S_{n}$ (Adjacent transpositions generate the finite symmetric group $S_{n}$ ).

[F3]

A generated subgroup contains the generators and is closed under products and inverses (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

Proof

technique · direct

1.1F1algebra

For $0 \leq j \leq n - 2$ , [F1] gives $c^{j} t c^{- j} = (j + 1 j + 2)$ .

2.1F3step 1.1

Every element in step 1.1 belongs to $⟨ c, t ⟩$ by [F3], so that subgroup contains all standard adjacent transpositions.

3.1F2step 2.1∎

By the transported one-based generation clause of [F2], $⟨ c, t ⟩ = S_{n}$ . Conjugating by $κ_{n}^{- 1}$ preserves products and generated subgroups and carries $c, t$ to $c_{0}, t_{0}$ , so it gives the equivalent zero-based assertion. At $n = 2$ , $c = t = (1 2)$ in the one-based model and $c_{0} = t_{0} = (0 1)$ in the zero-based model; the argument in either case has just its single adjacent transposition.

Depends on

Used by

FALSE: any transposition together with any n-cycle generates Sₙ False statement

Dependency tree · two levels

11 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

M. Macauley, Math 4120 lecture notes: Generating sets for $S_n$ (standard reference, not scraped)

The standard nn-cycle and a neighbour transposition generate SnSn​