Write $\chi=\chi^\lambda$. By the <Hook-length formula>,
$$
\frac{|S_n|}{\chi(1)}=\prod_{(i,j)\in\lambda}h_{i,j}(\lambda).
$$
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
$$
|g|\mid\frac{|S_n|}{\chi(1)}.
$$

Solution

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

For $\chi\in\operatorname{Irr}(S_n)$, 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.

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