For a self-conjugate diagram, compare successive row lengths along the boundary. If every difference is one, the diagram is a staircase. Otherwise the first horizontal or vertical repetition creates an even hook. Following the boundary to the last such repetition creates either a second transposed pair of the same maximal even length or a maximal even hook away from the first row and column. Both alternatives contradict part ii. Therefore every successive row length decreases by one and
$$
\lambda=(m,m-1,\ldots,2,1).
$$
Together with part b(ii), this proves the <staircase-character vanishing criterion>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 2 / c / iii / Solution

For a partition $\lambda$, the irreducible character $\chi^\lambda$ vanishes on every cycle type containing an even part exactly when $\lambda$ is a staircase $(m,m-1,\ldots,1)$. A staircase has only odd hook lengths. Conversely, vanishing first forces $\lambda$ to be self-conjugate; applying the Murnaghan–Nakayama rule at the largest even hooks then forces consecutive row lengths.


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