Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 103 / 2

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 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 $p,r$.

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.