Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 103 / 1 / c / i

The relevant centralizer of a subalgebra isEquivalently, it is the fixed subspace of $\mathbb{C}{S}_n$ under conjugation by $S_{n-1}$. We prove the Olshanskii centralizer lemma directly, without assuming multiplicity-free restriction.

A permutation has one cycle containing $n$, of some length $r\ge 1$. Let $\nu$ record the lengths of its other nontrivial cycles, omitting fixed points. Two permutations are conjugate under $S_{n-1}$ exactly when these data $(r,\nu )$ agree: a relabelling of the other letters identifies the distinguished cycles and all remaining cycles. Hence the sums $K_{r,\nu }$ over these conjugation orbits form a basis of $Z_{(n-1,1)}$.

Filter these orbit sums by the number of moved letters other than $n$:If $r=1$, the permutation fixes $n$, and $K_{1,\nu }$ is already a conjugacy-class sum in $Z_{n-1}$. For $r\ge 2$, let $C_{\nu }\in Z_{n-1}$ be the class sum with nontrivial cycle lengths $\nu$. ExpandTerms in which the $r-1$ star transpositions use distinct letters, disjoint from the support of the chosen term of $C_{\nu }$, produce precisely $K_{r,\nu }$, each with coefficient one. Indeed, a cycle through $n$ has a unique expression as an ordered product of star transpositions using each of its other letters once; the remaining cycles uniquely determine the term of $C_{\nu }$.

Every other term moves fewer than $t$ letters outside $n$. A repeated letter uses at most $r-2$ distinct such letters in the star product, and an overlap with the support of $C_{\nu }$ likewise reduces the size of their union. Since the whole product commutes with $S_{n-1}$, these remaining terms are a linear combination of lower-filtered orbit sums. ThusInduction on $t$, starting with the identity, puts every orbit sum in $\mathrm{alg}(Z_{n-1},X_n)$. The reverse inclusion holds because both proposed generators commute with $\mathbb{C}{S}_{n-1}$. ThereforeIn particular this centralizer of a subalgebra is commutative: $Z_{n-1}$ is commutative and $X_n$ commutes with it.