The hook-divisibility vanishing criterion

Let $m$ be the order of the $p$-singular permutation $\sigma$, so $p\mid m$, and let $d_1,\dots ,d_r$ be its cycle lengths, each dividing $m$ by [F1]. Then some $d_i$ is divisible by $p$: otherwise $p\nmid d_i$ for every $i$, and writing $m=p{m}^{\mathrm{\prime }}$ with $m^{\mathrm{\prime }}=m\mathrm{/}p\ge 1$ and $m^{\mathrm{\prime }}<m$, fix $i$; from $d_i\mid m=p{m}^{\mathrm{\prime }}$ write $p{m}^{\mathrm{\prime }}=d_{i}k$; by [F2] the divisibility $p\mid d_{i}k$ together with $p\nmid d_i$ gives $p\mid k$, say $k=p{k}^{\mathrm{\prime }}$, and cancelling $p$ by [F3] gives $m^{\mathrm{\prime }}=d_i{k}^{\mathrm{\prime }}$, so $d_i\mid m^{\mathrm{\prime }}$. Thus $m^{\mathrm{\prime }}$ would be a positive integer, smaller than the order $m$, divisible by every cycle length $d_i$, contradicting the minimality in [F1]. Hence $\sigma$ contains a cycle of some length $r=d_i$ with $p\mid r$.

By the hook length formula [F5], $\mathrm{\mid }{S}_n\mathrm{\mid }\mathrm{/}{f}^{\lambda }={\prod }_{x\in [\lambda ]}h(x)$; since $p$ is prime, $p\mid {\prod }_{x}h(x)$ holds if and only if $p\mid h(x)$ for at least one node $x$, by the finite-product form of Euclid's property [F2], while the converse divisibility is immediate from a product being a multiple of each factor. Hence $p\nmid \mathrm{\mid }{S}_n\mathrm{\mid }\mathrm{/}{f}^{\lambda }$ if and only if no hook length of $\lambda$ is divisible by $p$.

Assume no hook length of $\lambda$ is divisible by $p$, and let $r$ be the cycle length of $\sigma$ with $p\mid r$ found in step 1.1. Then no hook length of $\lambda$ equals $r$, since such a hook length would be divisible by $p$; by [F4] the character vanishes, ${\chi }^{\lambda }(\sigma )=0$. Step 1.2 identifies this hypothesis with $p\nmid \mathrm{\mid }{S}_n\mathrm{\mid }\mathrm{/}{f}^{\lambda }$, so the criterion holds in both equivalent forms.