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

The deletion projection for a symmetric group removes $n$ from the cycle containing it. A cycle $(a_1\cdots a_{r}n)$ becomes $(a_1\cdots a_r)$, with a singleton understood as a fixed point; all other cycles remain unchanged. This describes a genuine permutation of $\{1,\dots ,n-1\}$ and immediately proves the two identity assertions.

Put $H=S_{n-1}$. For $\sigma ,\theta \in H$ and $k<n$, if $\pi (\theta (k))\ne n$, deletion gives $\sigma \pi (\theta (k))$. If $\pi (\theta (k))=n$, it gives $(\sigma \pi \theta )(n)=\sigma \pi (n)$. These are exactly the two cases in $\sigma {\pi }_n\theta (k)$. Thus

The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two double cosetsAn $H$-bimodule-equivariant set map is determined by $u=\Phi (1)$ and $v=\Phi (t)$. Equivariance requires $u\in Z(H)$ andbecause $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi (h)=hu$ and $\Phi (ht{h}^{\prime })=hv{h}^{\prime }$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.

For $n\ge 5$, the center of $H$ and this centralizer are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\dots ,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\ge 3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(12)\}$. The choiceis well defined and satisfies all the printed equivariance conditions, but sends $(34)$ to $(12)$ rather than to the identity. Uniqueness holds for $n\ge 5$; it is false at $n=4$ as printed. Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.

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.