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

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 $k=1$.

Apply the Frobenius characteristic map. For even $n$, the Jacobi–Trudi identity and cancellation of consecutive terms giveThe generating function for the complete homogeneous symmetric polynomials now givesThis expansion contains only products $p_{\mu }$ for which every part of $\mu$ is even. The coefficient of $p_{\mu }$ in $\mathrm{ch}(F)$ is $F(\mu )/z_{\mu }$, so $F(\mu )=0$ whenever the cycle type $\mu$ has an odd part. Equivalently, $F(g)=0$ whenever $g$ contains an odd cycle. This is the Two-row alternating character cancellation.

For a nonzero example, take $n=3$ and let $g$ be a transposition. The two terms are the trivial and standard characters of $S_3$, whose values at a transposition are respectively $1$ and $0$. Thus $F(g)=1$ although $g$ contains a cycle of even length.

For a zero example, take $n=5$ and again let $g$ be a transposition, of cycle type $(2,1^3)$. The values of ${\chi }^{(5)},{\chi }^{(4,1)},{\chi }^{(3,2)}$ there are $1,2,1$, respectively. Consequently