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.

The centralizer of $C [S_{n - 1}]$ in $C [S_{n}]$ is commutative

Statement

Let $Z (C [S_{n}], C [S_{n - 1}]) = {z \in C [S_{n}] : z h = h z for every h \in S_{n - 1}}$ be the centralizer of the subalgebra $C [S_{n - 1}]$ in $C [S_{n}]$ . Then $Z (C [S_{n}], C [S_{n - 1}])$ is a commutative subalgebra of $C [S_{n}]$ .

Facts & Assumptions

Given: An integer $n \geq 1$ , the group $S_{n} = Sym ({1, \dots, n})$ with its subgroup $S_{n - 1} = Sym ({1, \dots, n - 1})$ of permutations fixing $n$ , and the group algebra $C [S_{n}]$ with basis ${[g] : g \in S_{n}}$ and multiplication $[g] [h] = [g h]$ (Partitions, English diagrams, and conjugation, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, The group ring $R [G]$ is a unital $R$ -algebra with basis $G$ , and each $g \in G$ is a unit of $R [G]$ ).

[F1]

For every $g \in S_{n}$ there is an involution $h_{g} \in S_{n - 1}$ with $h_{g} g h_{g}^{- 1} = g^{- 1}$ ; the proof of the cited lemma exhibits $h_{g}$ by explicit formulas on the cycles of $g$ (Every element of $S_{n}$ is inverted by an involution of $S_{n - 1}$ ).

[F2]

$C [S_{n}]$ has the group elements as a $C$ -basis, every element has a unique expansion $\sum_{g} c_{g} [g]$ with finitely many nonzero coefficients, and $[g] [h] = [g h]$ ; coefficients of equal basis elements are equal. The inverse in $S_{n}$ reverses products: $(g h)^{- 1} = h^{- 1} g^{- 1}$ (The group ring $R [G]$ is a unital $R$ -algebra with basis $G$ , and each $g \in G$ is a unit of $R [G]$ , The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

Proof

technique · inversion anti-automorphism

1.1F2givenalgebra

The centralizer $C : = Z (C [S_{n}], C [S_{n - 1}])$ is a $C$ -subalgebra of $C [S_{n}]$ : it contains $1$ , is closed under addition and scalar multiplication because equality with each $h \in S_{n - 1}$ is preserved by these operations, and is closed under multiplication because $y z h = y h z = h y z$ for all $h \in S_{n - 1}$ whenever $y, z \in C$ ; closure under multiplication is also checked on the basis expansions using [F2].

1.2F2givenalgebra

Define $σ : C [S_{n}] \to C [S_{n}]$ on the basis by $σ ([g]) : = [g^{- 1}]$ and extend $C$ -linearly: $σ (\sum_{g} c_{g} [g]) = \sum_{g} c_{g} [g^{- 1}]$ . Then $σ$ is an involution, since $(g^{- 1})^{- 1} = g$ , and it is an anti-automorphism: $σ ([g] [h]) = σ ([g h]) = [(g h)^{- 1}] = [h^{- 1} g^{- 1}] = [h^{- 1}] [g^{- 1}] = σ ([h]) σ ([g])$ by [F2], and the identity extends to all elements by bilinearity.

2.1step 1.2F1F2algebra

The anti-automorphism $σ$ fixes every element of $C$ . Let $z = \sum_{g} c_{g} [g] \in C$ , and fix one $g \in S_{n}$ . By [F1] there is $h_{g} \in S_{n - 1}$ with $h_{g} g h_{g}^{- 1} = g^{- 1}$ . Since $z = h_{g} z h_{g}^{- 1}$ , comparison of the coefficient of $[g^{- 1}]$ on the two sides gives $c_{g^{- 1}} = c_{g}$ ; conjugation is a bijection, so exactly the summand indexed by $g$ contributes on the right. This equality holds for every $g$ , and therefore $σ (z) = z$ by [F2]. The argument compares each coefficient separately and requires no common conjugator.

3.1step 1.1step 2.1algebra∎

For $y, z \in C$ we have $y z = σ (y z) = σ (z) σ (y) = z y$ : the first equality is step 2.1, the second holds because $σ$ is an anti-automorphism by step 1.2, and the third is step 2.1 applied to $y$ and to $z$ . Hence $C$ is a commutative subalgebra of $C [S_{n}]$ by step 1.1.

Remarks

Choice-free. The element $h_{g}$ is constructed in Every element of $S_{n}$ is inverted by an involution of $S_{n - 1}$ by reversing the cycles of $g$ , so no selection principle is used; the argument also avoids Maschke's theorem and any semisimplicity input, and works verbatim with $C$ replaced by any commutative ring.
What the centralizer contains. The centre $Z (C [S_{n - 1}])$ and the element $X_{n} = \sum_{j < n} (j n)$ both lie in $C$ : elements of the centre commute with $S_{n - 1}$ , and conjugating $(j n)$ by $h \in S_{n - 1}$ gives $(h (j) n)$ , so conjugation permutes the summands of $X_{n}$ . The opposite inclusion, that $C$ is generated by these two pieces, is The relative centralizer is generated by the previous centre and the last Jucys-Murphy element.

Depends on

Used by

The relative centralizer is generated by the previous centre and the last Jucys-Murphy element Theorem

Dependency tree · two levels

14 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, Theorem 2.1 and its complete coefficient-inversion proof, printed pp. 9-10 (standard reference, not scraped)

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