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For the staircase $\delta =(m,m-1,\dots ,1)$,whenever $(i,j)$ is a cell, so every hook length is odd. A removable rim hook of length $k$ corresponds to a hook of length $k$ in the original diagram. If a cycle type $\mu$ contains an even part $k$, apply the Murnaghan–Nakayama rule to that part first. There are no terms in the sum, and therefore ${\chi }^{\delta }(\mu )=0$.

The Murnaghan–Nakayama rule states that if a permutation has a $k$-cycle and remaining cycle type $\rho$, thenwhere $H$ ranges over removable rim hooks of length $k$ and $\mathrm{leg}(H)$ is one less than the number of rows occupied by $H$.

Because $\lambda$ is self-conjugate, every off-diagonal hook of even length is paired with its transpose, while every diagonal hook has odd length. Suppose even hooks exist and let $L$ be their maximum length. Apply the Murnaghan–Nakayama rule to cycle types beginning with $L$ and complete the remaining cycle type with the principal hooks of the residual diagram. The principal-hook character value of a symmetric group makes each surviving residual character equal to $1$ or $-1$.

The assumed vanishing forces cancellation among the removable $L$-hooks. The standard maximal-hook comparison shows that the only possible cancellation is one transposed pair: the hooks must be $H_{1,j}$ and $H_{j,1}$ for a single $j>1$. Any further hook of length $L$, or a maximal hook with both indices greater than one, can be isolated by the residual principal-hook cycle type and would give a nonzero value. Thus either there are no even hooks or the maximum even length occurs exactly at that pair.

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.

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 Murnaghan–Nakayama rule successively to the $w$ disjoint $e$-cycles. A complete term requires a sequence of $w$ removable $e$-hooks. If $w>w_e(\lambda )$, no such sequence exists after the Weight of a partition is exhausted, so the character value is zero.

Suppose $w=w_e(\lambda )$. Every complete sequence ends at the Core of a partition $C_e(\lambda )$. Under the abacus divisible-hook correspondence, a removal chooses one cell from one component ${\lambda }^{(i)}$ of the quotient of a partition. The choices of which runner is used occur inorders. Within runner $i$, the signed complete removal sum is the degree ${\chi }^{{\lambda }^{(i)}}(1)$, and all inter-runner removal orders have the common Sign of an abacus hook-removal sequence $\epsilon \in \{1,-1\}$. The remaining permutation $\gamma$ acts on the core, giving

The parity of the sum of the leg lengths in any sequence that removes all $e$-hooks from $\lambda$ is independent of the sequence. Its sign ${\epsilon }_e(\lambda )$ is therefore well defined and supplies the common sign in repeated applications of the Murnaghan–Nakayama rule.

For $\chi \in \mathrm{Irr}(S_n)$, if the order of $g\in S_n$ does not divide $|S_n|/\chi (1)$, then $\chi (g)=0$. The Hook-length formula turns the co-degree into a product of hook lengths, while the Murnaghan–Nakayama rule detects a cycle whose required prime-power hook cannot be removed.