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 $\{0,1,\ldots,e-1\}$. Thus their <residue content of a partition>[$e$-contents] are equal.

Solution

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

{title2=$e\text{-content}$}

The $e$-residue of a cell $(i,j)$ is $j-i$ modulo $e$, and the $e$-content is the multiset of cell residues. Removing an $e$-hook removes exactly one cell of each residue. Among partitions of a fixed size, two partitions have the same $e$-content exactly when they have the same $e$-core.


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 Incoming links: Residue content of a partition