Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 2 / d / i / Solution

Take $g$ to be an $n$-cycle. For $n\ge 2$, the Murnaghan–Nakayama rule gives ${\chi }^{(n)}(g)=1$ and ${\chi }^{(n-1,1)}(g)=-1$; every remaining two-row diagram contains a $2\times 2$ square and is not a rim hook, so its value at $g$ is zero. ThereforeThe exhibited cycle has length $k=n$, so $k\equiv n(\mathrm{mod}2)$. For $n=1$, the sole character has value one at the identity and the same conclusion holds with $k=1$.