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A multiset records a nonnegative integer multiplicity for each element of a set. A finite multiset has finite support and finite total multiplicity. Its cardinality is the sum of its multiplicities; its disjoint union with another multiset adds multiplicities. The hook lengths of a Young diagram form a multiset, since different cells can have equal lengths.

Put $L=h_{i,j}=pr$. The Hook-interval decomposition at a Young-diagram cell is the disjoint unionHere is a proof of the identity using the hook criterion in a beta set. Keep only the cells weakly southeast of $(i,j)$, translating that cell to $(1,1)$; this gives the partition of an integerLet $\ell ={\lambda }_j^{\prime }-i+1$ and use its beta set of a partition $b_s={\nu }_s+\ell -s$. These numbers are precisely the hook lengths down the original column from $(i,j)$, with $b_1=L$. The hooks along the first row of $\nu$ are the differences $L-c$ for the gaps $0\le c<L$. The remaining positions in that interval are the beads $b_2,\dots ,b_{\ell }$, giving exactly the displayed disjoint decomposition. The first-row and first-column hook lengths of $\nu$ agree with those in the original Young-diagram hook.

Because $p\mid L$, a leg cell satisfies $p\mid h_{x,j}$ exactly when $p\mid L-h_{x,j}$. Thus divisibility by $p$ identifies the relevant cells of the Young-diagram hook with the multiples of $p$ in $\{1,\dots ,pr\}$. There are exactly $r$ of those, provingThe count includes the corner cell $(i,j)$ itself.

The converse is false. Take $p=2$, $r=3$, $n=6$ and the partition of an integer $\lambda =(3,3)$. Its hook lengths, arranged in its Young diagram, areBoth $2$ and $3$ occur, but $6$ does not, even though $pr=6=n$. Henceis a counterexample with distinct $p,r$.

Use the hook criterion in a beta set: for a finite beta set of a partition $X$, each hook length is a difference $b-c$ with $b\in X$, $0\le c<b$ and $c\notin X$, and every such difference is a hook length.

A hook length $pr$ therefore gives a bead $b$ and a gap $c=b-pr$. Along the progressionthe first position is a bead and the last is a gap. Some adjacent positions are consequently a bead followed by a gap. Their difference is $p$, so $p$ is a hook length. Similarly the progression from $b$ to $c$ in steps of $r$ gives a hook length $r$. Every position used is nonnegative because it lies between $c$ and $b$.

Thus the divisor closure of hook lengths yieldsNeither $p$ nor $r$ needs to be a prime number.

The implication is false. ChooseThe hook lengths areThus $3$ does not divide this irreducible character degree, but no hook has length $a{p}^k=6$.

There is no contradiction with part (i): removing two horizontal rim hooks of length three gives the empty core of a partition, so $w_3(\lambda )=2=a$ and the core degree is one. Having total weight of a partition two does not force a single hook of length twice the modulus.

The implication is false. ChooseThe inequalities $0\le 4<9$ and the relation $13=1\cdot 9+4$ hold. At cell $(1,4)$, the hook length is $12-4+1=9=a{p}^k$.

For the partition of an integer $(12,1)$, the Hook-length formula giveswhich is divisible by $3$. This also agrees with the degree of the standard representation of the symmetric group $S_{13}$.

The obstruction is visible in the core of a partition: removing the horizontal rim hook of length nine leaves $\gamma =(3,1)$, which has no hook length nine and hence is the $9$-core. Its degree is ${\chi }^{(3,1)}(1)=3$. The large hook ensures the required weight of a partition here, but the core degree is not coprime to $p$.

We use the abacus divisible-hook correspondence in its multiset form: the hook lengths of $\lambda$ that are divisible by $p$, divided by $p$, form exactly the disjoint union of the multisets of hook lengths of the components of its quotient of a partition for modulus $p$.

Iterating this correspondence gives, for every $r\ge 1$,as multisets. Each Young diagram has one hook for every cell, so the cardinality of the right-hand side is $|T^Q(\lambda {)}_r|$, the sum of the component sizes.

For any positive integer $h$, its P-adic valuation is the number of positive $r$ for which $p^r\mid h$. Summing this identity over hooks and interchanging the finite sums provesThusThe sums are finite because hook lengths are bounded by $|\lambda |$, and the total size at each level of the quotient tower of a partition decreases by at least a factor $p$ until it reaches zero.