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If $n=|\lambda |$ and $d_p(n)$ is its base-$p$ digit sum, thenThis combines the Hook-length formula, Legendre formula, the abacus divisible-hook correspondence, and the recurrence $q_r=c_r+p{q}_{r+1}$.

The preceding part shows that $\theta (e(t^{\prime }))\ne 0$ in characteristic zero, so $S^{{\lambda }^{\prime }}⊈\ker \theta$. Apply the James submodule theorem to the proper submodule $\ker \theta \le M^{{\lambda }^{\prime }}$ to obtainThe Hook-length formula gives $\dim S^{\lambda }=\dim S^{{\lambda }^{\prime }}$. Surjectivity of $\theta$ therefore givesand hence

Fix the original tableaux $t,u$. For a $\lambda$-tableau $w$, let $g_w$ be the unique permutation satisfying $w=g_{w}t$ and put ${\epsilon }_w=\mathrm{sgn}(g_w)$. Sincethe quotient pairing $M^{{\lambda }^{\prime }}/(S^{{\lambda }^{\prime }}{)}^{\perp }\cong (S^{{\lambda }^{\prime }}{)}^{\ast }$ gives the explicit conjugate Specht module as a sign-twisted dual isomorphismfor every $\lambda$-tableau $w$.

PutThe defining relation between the quotient tower of a partition and the core tower of a partition isSumming the resulting telescoping identities gives

By the Hook-length formula,The abacus divisible-hook correspondence says that the number of hooks divisible by $p^r$ is $q_r$, soIf $d_p(n)={\sum }_r{\alpha }_r$, the digit-sum form of the Legendre formula isCombining the three displayed identities proves the P-adic valuation of a symmetric-group character degree from the core tower formula

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

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.