Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 3

Moving a bead one step upward on an $e$-runner abacus of a partition removes an $e$-hook and decreases the size by $e$. Sliding all beads upward reaches the core of a partition after exactly $w_e(\lambda )$ moves. Therefore

The sum of the sizes of the partitions at the next level of the quotient tower of a partition isFor $e>1$, repeated passage to a quotient therefore makes every branch empty after finitely many levels. For $e=1$, the core is empty and $Q_1(\lambda )=(\lambda )$, so a nonempty partition repeats forever. Thus the tower has finite depth exactly when $e>1$ or $\lambda$ is empty.

Using the abacus of a partition in James's convention, read runner $r$ as a one-runner abacus: replace each occupied position $r+eq$ by $q$. After the usual harmless shift of its beta set, the resulting partition is ${\lambda }^{(r)}$. Doing this for $r=0,\dots ,e-1$ produces

Conjugating a partition complements beads and gaps and reflects the abacus. Reflection sends runner $r$ to runner $e-1-r$, reverses bead-gap order, and conjugates the runner partition. Hence

Removing an $e$-hook removes one cell of every residue modulo $e$. If $\lambda$ and $\mu$ have the same $e$-core and the same size, part a(i) shows that they have the same $e$-weight. Their residue multisets are therefore obtained from the common core by adjoining the same number of complete residue sets $\{0,1,\dots ,e-1\}$. Thus their $e$-contents are equal.

The beta number in row $i$ isSince $e\mid m$,

Pad both partitions to the same number $m$ of rows, with $e\mid m$. For residue $s$, subtracting the number of cells of residue $s-1$ from the number of cells of residue $s$ telescopes row by row. Since the row-start residues $1-i$ are the same for both padded diagrams, equality of $e$-contents and part ii imply that their beta sets contain the same number of elements on each abacus runner.

Sliding beads upward to obtain the $e$-core preserves the number of beads on every runner. A packed runner is uniquely determined by that number, so the two packed abaci coincide. Hence $C_e(\lambda )=C_e(\mu )$.