Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 1 / c / vii / Solution

There is a separate problem with the printed linear-span assertion, when $Z_{((n-1),1)}$ denotes the usual centralizer of a subalgebra. Let $A_{n-1}$ be the sum of all transpositions in $H$. Direct expansion in the group algebra givesThus $Y=X_n^2-2A_{n-1}$ commutes with $H$ and satisfies ${\Pi }_n(Y)=(n-1)1$. For $n\ge 3$ it contains nonzero coefficients on three-cycles involving $n$, whereas every element of $\mathrm{span}\{1,X_n\}$ is supported only on the identity and transpositions. Therefore $Y$ is an explicit counterexample to the asserted equality.

One correct description of the entire preimage is obtained by putting $C=C_{\mathbb{C}{S}_n}(\mathbb{C}H)$. Equivariance gives ${\Pi }_n(C)\subseteq Z(\mathbb{C}H)$, and ${\Pi }_n$ is the identity on that center. It follows thatA useful corrected two-dimensional statement restricts the support to the identity and the transpositions $(in)$: invariance under conjugation by $H$ then forces their coefficients to be equal, yielding exactly $\mathrm{span}\{1,X_n\}$. This additional support restriction is absent from the PDF.