LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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.

Every element of $S_{n}$ is inverted by an involution of $S_{n - 1}$

Statement

Let $n \geq 1$ , let $S_{n - 1} \leq S_{n}$ be the stabilizer of $n$ , and let $g \in S_{n}$ . There is $h \in S_{n - 1}$ with $h^{2} = 1$ (the identity is allowed) and $h g h^{- 1} = g^{- 1}$ ; here such a self-inverse permutation is called an involution. In particular every element of $S_{n}$ is conjugate to its inverse by an element of $S_{n - 1}$ .

Facts & Assumptions

Given: An integer $n \geq 1$ and a permutation $g \in S_{n}$ , where $S_{n} = Sym ({1, \dots, n})$ (Partitions, English diagrams, and conjugation).

[F1]

Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering the factors and cyclically rotating the entries within each cycle; the identity has the empty such product (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation).

[F2]

A cycle $(a_{0} a_{1} \dots a_{k - 1})$ has support ${a_{0}, \dots, a_{k - 1}}$ , sends $a_{i}$ to $a_{i + 1}$ for $i < k - 1$ and $a_{k - 1}$ to $a_{0}$ , and fixes every point outside its support; cycles with disjoint supports are disjoint, and a cycle may be written starting at any of its entries (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[F3]

For every $g \in S_{n}$ and every cycle $c = (a_{1} a_{2} \dots a_{k})$ one has $g c g^{- 1} = (g (a_{1}) g (a_{2}) \dots g (a_{k}))$ (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

[F4]

Cycles with disjoint supports commute (Cycles with disjoint supports commute).

Proof

technique · explicit construction

1.1F1F2

By [F1] write $g = c_{1} c_{2} \dots c_{r}$ with the $c_{i}$ pairwise disjoint cycles of length at least $2$ . If $g (n) \neq n$ , exactly one factor meets ${n}$ , say $c_{s}$ ; its support is the orbit of $n$ under $g$ , and by [F2] we may write $c_{s} = (n a_{1} \dots a_{k - 1})$ with $k \geq 2$ and distinct $a_{1}, \dots, a_{k - 1} \in {1, \dots, n - 1}$ . If $g (n) = n$ , no factor meets ${n}$ ; in that case put $k = 1$ , leave the list $a_{1}, \dots, a_{k - 1}$ empty and drop the discussion of $c_{s}$ .

2.1step 1.1F1F2construct

Define $h \in Sym ({1, \dots, n})$ by the following rules on the pairwise disjoint sets listed so far: $h (n) : = n$ ; $h (a_{i}) : = a_{k - i}$ for $1 \leq i \leq k - 1$ when $c_{s}$ exists; and for each remaining factor $c_{i} = (b_{1} \dots b_{m})$ of the decomposition, $h (b_{j}) : = b_{m + 1 - j}$ for $1 \leq j \leq m$ ; every element of ${1, \dots, n}$ not yet mentioned is fixed by $h$ . The listed points are distinct, so $h$ is a well-defined bijection: each rule pairs the listed points in pairs, possibly fixing a middle point, and in every case applying the rule twice returns the point. Hence $h$ is an involution; it fixes $n$ and every point outside ${1, \dots, n - 1}$ , so $h \in S_{n - 1}$ and $h^{- 1} = h$ .

3.1step 2.1F2F3algebra

Conjugation by $h$ acts on each factor by [F3]: for $c_{s} = (n a_{1} \dots a_{k - 1})$ we get $h c_{s} h^{- 1} = (h (n) h (a_{1}) \dots h (a_{k - 1})) = (n a_{k - 1} \dots a_{1})$ , the cycle sending $n$ to $a_{k - 1}$ , sending $a_{i}$ to $a_{i - 1}$ and sending $a_{1}$ to $n$ , which is exactly $c_{s}^{- 1}$ ; for every other factor $c_{i} = (b_{1} \dots b_{m})$ we get $h c_{i} h^{- 1} = (h (b_{1}) \dots h (b_{m})) = (b_{m} \dots b_{1}) = c_{i}^{- 1}$ .

4.1step 3.1F4algebra∎

Inserting $h^{- 1} h = 1$ between consecutive factors gives $h g h^{- 1} = (h c_{1} h^{- 1}) \dots (h c_{r} h^{- 1}) = c_{1}^{- 1} \dots c_{r}^{- 1}$ by step 3.1. Reversing a product inverts it, so $c_{1}^{- 1} \dots c_{r}^{- 1} = (c_{r} \dots c_{1})^{- 1}$ ; by [F4] the pairwise disjoint factors commute, hence $c_{r} \dots c_{1} = c_{1} \dots c_{r} = g$ . Therefore $h g h^{- 1} = g^{- 1}$ with $h \in S_{n - 1}$ an involution, which is the statement.

Remarks

The source's shorter argument is incomplete as printed. The cited source proves the fact by deleting the letter $n$ from $g$ and choosing an element $h \in S_{n - 1}$ that conjugates the deletion $g^{'}$ to $g^{' - 1}$ ; it then asserts that such an $h$ realizes $g^{- 1} = h g h^{- 1}$ . That step is not correct for an arbitrary such $h$ : for $g = (1 2 3 4)$ and $h = (2 3)$ one has $h (1 2 3) h^{- 1} = (1 3 2) = (1 2 3)^{- 1}$ , while $h g h^{- 1} = (1 3 2 4) \neq g^{- 1} = (1 4 3 2)$ . The construction above chooses the explicit cycle-reversing involution, which does satisfy $h g h^{- 1} = g^{- 1}$ ; only that corrected construction is used later, in The centralizer of $C [S_{n - 1}]$ in $C [S_{n}]$ is commutative.
No choice. For each $g$ the involution $h$ is given by explicit formulas on the finitely many cycles of $g$ , so the statement is proved without any selection principle, and the argument is integral and characteristic-free: it uses only the group structure of $S_{n}$ .

Depends on

Used by

The centralizer of ℂ[Sₙ₋₁] in ℂ[Sₙ] is commutative Lemma

Dependency tree · two levels

15 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

Okounkov-Vershik, A New Approach to the Representation Theory of the Symmetric Groups, Selecta Math. (N.S.) 2 (1996) 581-605; complete arXiv repost math/0503040, Lemma 2.2, printed p. 9; its deletion argument is replaced by the explicit cycle-reversal proof below (standard reference, not scraped)
K. Conrad, Conjugacy Classes (cycle conjugation), as cited by the published cycle-conjugation lemma (standard reference, not scraped)

Every element of SnSn​ is inverted by an involution of Sn−1Sn−1​