Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 3

Use a beta set $X$ on an $e$-runner partition abacus. A hook of length divisible by $e$ corresponds to a bead $x$ and a gap $x-eh$ on the same runner. Divide both runner positions by $e$. They become a bead and a gap at distance $h$ in the runner partition ${\lambda }^{(s)}$, and hence determine a hook there. This gives the required bijection, with $|H|=e|f(H)|$.

Removing $H$ replaces the bead $x$ by the gap $x-eh$. On its runner this is exactly the bead move that removes $f(H)$, while every other runner is unchanged. Thus hook removal commutes with the construction andThis is the abacus divisible-hook correspondence.

For $\lambda =(6,5,3)$ use the three-bead beta set $\{8,6,3\}$. Its runners give the 3-quotient of a partitionThe hook lengths divisible by three and their images areThere are no others, as the complete hook-length rows are $(8,7,6,4,3,1)$, $(6,5,4,2,1)$, and $(3,2,1)$.

True. A hook length divisible by $ef$ is divisible by $e$. If $\lambda$ is its own Core of a partition $C_e(\lambda )$, it has no such hook, so it is also its own $ef$-core.

False. Let $\lambda =(3)$, $e=2$, and $f=3$. ThenThus core operations for different moduli need not commute.

True. On an $f$-runner partition abacus, taking the $e$-quotient of each runner refines its positions by their residue modulo $e$. The resulting $ef$ runners are indexed by pairs of residues. Ordering those pairs may differ from the conventional residue order modulo $ef$, but only by a permutation. Hence $Q_e(Q_f(\lambda ))$ is a permutation of $Q_{ef}(\lambda )$.

True. Slide beads upward separately in each of the refined residue classes modulo $ef$. Grouping the refined runners first by residue modulo $f$ and then packing each component modulo $e$ produces $C_e(Q_f(\lambda ))$. Packing modulo $ef$ before forgetting the refinement produces $Q_f(C_{ef}(\lambda ))$. These are the same runner configurations, so

Apply the Murnaghan–Nakayama rule successively to the $w$ disjoint $e$-cycles. A complete term requires a sequence of $w$ removable $e$-hooks. If $w>w_e(\lambda )$, no such sequence exists after the Weight of a partition is exhausted, so the character value is zero.

Suppose $w=w_e(\lambda )$. Every complete sequence ends at the Core of a partition $C_e(\lambda )$. Under the abacus divisible-hook correspondence, a removal chooses one cell from one component ${\lambda }^{(i)}$ of the quotient of a partition. The choices of which runner is used occur inorders. Within runner $i$, the signed complete removal sum is the degree ${\chi }^{{\lambda }^{(i)}}(1)$, and all inter-runner removal orders have the common Sign of an abacus hook-removal sequence $\epsilon \in \{1,-1\}$. The remaining permutation $\gamma$ acts on the core, giving