TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-13

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The adjacent transpositions $(1 2), (2 3), \dots, (n - 1 n)$ generate $S_{n}$

Statement

For $n \geq 2$ , $S_{n}$ is generated by the adjacent transpositions $s_{j} = (j j + 1)$ , $1 \leq j < n$ . For $n = 0, 1$ , the empty set generates the trivial group $S_{n}$ .

Facts & Assumptions

Given: The symmetric group $S_{n}$ and its standard ordering of symbols.

[F1]

Every element of $S_{n}$ is a product of transpositions (Every finite permutation is a product of transpositions, so the transpositions generate $S_{n}$ ).

[F2]

The subgroup generated by a set is the smallest subgroup containing it; the empty set generates the trivial subgroup (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[F3]

Products of permutations act with the right factor first (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

Proof

technique · direct

1.1

For $1 \leq a < b \leq n$ , pointwise calculation using [F3] gives $(a b) = s_{a} s_{a + 1} \dots s_{b - 2} s_{b - 1} s_{b - 2} \dots s_{a + 1} s_{a} .$ When $b = a + 1$ , this word is just $s_{a}$ .

F3algebra

2.1

Thus every transposition lies in the subgroup generated by the adjacent ones, and [F1]--[F2] show that this subgroup is $S_{n}$ .

F1F2step 1.1

3.1

If $n = 0$ or $1$ , $S_{n}$ is trivial, so [F2] says the empty displayed set generates it.

F2F3∎

Depends on

Used by

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

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

The adjacent transpositions (1 2),(2 3),…,(n−1 n)(12),(23),…,(n−1n) generate SnSn​