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

The Murnaghan–Nakayama rule states that if a permutation has a $k$-cycle and remaining cycle type $\rho$, thenwhere $H$ runs over the removable rim hooks of length $k$. Iterating removes rim hooks whose lengths are the cycle lengths, and the character value is the signed sum over all complete removal sequences.

Let $s_i=(ii+1)$. Since $\mu -\mathrm{id}=s_i(\lambda -\mathrm{id})$, reindex the alternating sum defining ${\psi }^{\mu }$ by $\pi \mapsto s_i\pi$. This preserves all permutation-character terms and reverses every sign of a permutation, because $\mathrm{sgn}(s_i\pi )=-\mathrm{sgn}(\pi )$. Hence ${\psi }^{\mu }=-{\psi }^{\lambda }$.

For each $m$, the shifted sequence ${\beta }_{i,m+1}-\mathrm{id}$ is obtained from ${\beta }_{i,m}-\mathrm{id}$ by swapping the adjacent entries in positions $i+m$ and $i+m+1$. Part a(ii) therefore givesThis is the usual character straightening procedure: the distinguished entry ${\alpha }_i-i-k$ is moved successively to the right. Either it reaches the unique place that makes some ${\beta }_{i,m}$ a partition of an integer, or it meets an equal shifted entry. Under the hypothesis the first alternative never occurs, so two entries of some ${\beta }_{i,m}-\mathrm{id}$ coincide. The corresponding alternating sum is fixed by swapping those entries but changes sign by part a(ii), and is therefore zero. Repeatedly applying the displayed relation gives ${\psi }^{{\beta }_{i,0}}=0$.

Write $\chi ={\chi }^{\lambda }$. By the Hook-length formula,Suppose the order of a group element $g$ did not divide this product. Then for some prime number $p$, the highest prime power $p^a$ dividing $|g|$ would not divide the hook product. One cycle of $g$ has length divisible by $p^a$, whereas no hook length of $\lambda$ is divisible by $p^a$. In particular there is no removable rim hook having that cycle length. Applying the Murnaghan–Nakayama rule first to this cycle gives $\chi (g)=0$, a contradiction. This is the symmetric-group character co-degree vanishing criterion, and its contrapositive proves

Choose $g$ whose disjoint permutation cycles have lengths equal to the principal hook lengths of $\lambda$. Those lengths are distinct and sum to $n$, so this is a permutation in $S_n$. In the iterated Murnaghan–Nakayama rule, there is a unique complete sequence that removes the corresponding principal rim hooks. Its contribution is one sign, and hence the Principal-hook character value of a symmetric group gives ${\chi }^{\lambda }(g)\in \{1,-1\}$.

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