Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 19H / b / Solution

For permutation characters, Burnside lemma on a product givesTwo pairs $(A,B)$ and $(A^{\prime },B^{\prime })$ lie in the same $S_n$-orbit exactly whenWhen $0\le l\le k\le n/2$, every value $0,1,\dots ,l$ is possible because $k+l\le n$. Therefore the symmetric-group subset permutation representation satisfies

For $1\le r\le n/2$, the inclusion map between subset permutation modules embeds $\mathbb{C}[X_{r-1}]$ into $\mathbb{C}[X_r]$, sois a character. Its norm isBy character orthogonality, ${\chi }_r$ is the character of an irreducible representation.

For $r>n/2$, complementation is an $S_n$-equivariant bijection $X_r\cong X_{n-r}$, soConsequentlythe negative of one of the irreducible characters just found, and is not itself the character of a representation.