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

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.