Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 2 / a

Write $a=\lambda -\mathrm{id}$. The two modified entries satisfyThus $\mu -\mathrm{id}$ is obtained from $\lambda -\mathrm{id}$ by the adjacent transposition $\tau =(i,i+1)$. In the alternating definition of $\psi$, reindexing $\pi$ by $\tau \pi$ preserves every induced permutation character and reverses every sign. Therefore the straightening of a symmetric-group character indexed by a composition gives

The restriction form of the restriction branching rule for a symmetric group iswhere ${\lambda }^{-}$ contains the distinct partitions obtained by deleting one Removable node of a Young diagram. In particular, the restriction is multiplicity-free.

Restrict the alternating expressionThe supplied restriction formula for a Young permutation character, with $k=1$, says that each term restricts by subtracting one from each possible component. After collecting the alternating sums, this givesIf row $i$ has no removable node, part i straightens ${\psi }^{\lambda -{ϵ}_i}$ against the adjacent term with the opposite sign, or makes it zero when two shifted entries coincide. The surviving terms are exactly ${\psi }^{\mu }$ for $\mu \in {\lambda }^{-}$. Since $\lambda$ and each surviving $\mu$ are partitions, ${\psi }^{\lambda }={\chi }^{\lambda }$ and ${\psi }^{\mu }={\chi }^{\mu }$. We obtainComplex representations of a finite group are semisimple, so equality of characters proves the asserted module decomposition.