Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 2 / a

The Murnaghan–Nakayama rule states that if a permutation has a $k$-cycle and remaining cycle type $\rho$, thenwhere $H$ runs over the removable rim hooks of length $k$. Iterating removes rim hooks whose lengths are the cycle lengths, and the character value is the signed sum over all complete removal sequences.

Let $s_i=(ii+1)$. Since $\mu -\mathrm{id}=s_i(\lambda -\mathrm{id})$, reindex the alternating sum defining ${\psi }^{\mu }$ by $\pi \mapsto s_i\pi$. This preserves all permutation-character terms and reverses every sign of a permutation, because $\mathrm{sgn}(s_i\pi )=-\mathrm{sgn}(\pi )$. Hence ${\psi }^{\mu }=-{\psi }^{\lambda }$.

For each $m$, the shifted sequence ${\beta }_{i,m+1}-\mathrm{id}$ is obtained from ${\beta }_{i,m}-\mathrm{id}$ by swapping the adjacent entries in positions $i+m$ and $i+m+1$. Part a(ii) therefore givesThis is the usual character straightening procedure: the distinguished entry ${\alpha }_i-i-k$ is moved successively to the right. Either it reaches the unique place that makes some ${\beta }_{i,m}$ a partition of an integer, or it meets an equal shifted entry. Under the hypothesis the first alternative never occurs, so two entries of some ${\beta }_{i,m}-\mathrm{id}$ coincide. The corresponding alternating sum is fixed by swapping those entries but changes sign by part a(ii), and is therefore zero. Repeatedly applying the displayed relation gives ${\psi }^{{\beta }_{i,0}}=0$.