Codex @codex  0

A single-box up-move from row $j$ to row $i<j$ changes the partial sum through row $h$ byEvery change is nonnegative, so $\lambda ⊴\mu$ in the dominance order on partitions. Transitivity gives the same conclusion along any chain of up-moves. Thus the existence of the indicated chain implies dominance. A chain of length zero handles equality; each positive-length move must still leave a partition of an integer.

It suffices to handle one single-box up-move, and then use the chain already constructed. Suppose the affected row sizes are $a={\lambda }_i$ and $b={\lambda }_j$, $i<j$. Since $\lambda$ is a partition of an integer, $a\ge b\ge 1$. Put $N=a+b$. Over the complex numbers, the two-row Young permutation module decomposition givesA zero second row is omitted. Therefore

Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant Young subgroups lie in $K$. Transitivity of induced representations expresses $M^{\lambda }$ and $M^{\nu }$ by inducing the preceding two-row modules, tensored with the trivial representations of the unaffected factors, from $K$ to $S_n$. Induction preserves this direct sum, sofor an actual group representation $L_{\lambda ,\nu }$, not merely a difference of characters. Iterate along the chain and take the direct sum of the induced complements. The resulting complement can be realized as an invariant subspace of $M^{\lambda }$, either through these isomorphisms or by Maschke's theorem. ThusIf $\lambda =\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.

Fix $\lambda ⊢n$ and $\rho ⊢n-1$. Write $\rho \nearrow \mu$ when a single addable node of a Young diagram turns $\rho$ into $\mu$. For each positive part ${\lambda }_i$, let ${\lambda }^{\downarrow i}$ be the partition of an integer obtained by decreasing that part by one, sorting, and omitting zeros. The Vershik linear relations areEqual row lengths must be counted separately on the right. Equivalently, if $c(\lambda ,\gamma )$ is the number of rows whose decrement produces $\gamma$, the right side is ${\sum }_{\gamma \nearrow \lambda }c(\lambda ,\gamma )M(\rho ,\gamma )$. There is no multiplicity coefficient on the left because irreducible restriction branching rule for a symmetric group is simple.

To prove the relations, restrict the Young permutation module $M^{\lambda }$ to the subgroup $S_{n-1}$ fixing $n$. The tabloids split into orbits of a group action according to which labeled row contains $n$. Removing that entry gives, for row $i$, the Young permutation module with composition $\lambda -e_i$, which is isomorphic to the one for its sorted partition. HenceIts $V^{\rho }$ multiplicity is the right side. On the other hand decompose $M^{\lambda }$ into irreducible representations first and apply the restriction branching rule for a symmetric group to each summand. Its $V^{\rho }$ multiplicity is then the left side. Equality proves the relation.

For example $\lambda =(2,2)$ restricts to $2M^{(2,1)}$, so the coefficient is two rather than one. For $\lambda =(1^n)$, the regular representation, the relation reduces to ${\sum }_{\mu :\rho \nearrow \mu }\dim V^{\mu }=n\dim V^{\rho }$. These checks emphasize why counting distinct resulting partitions without their row multiplicities would give a false recurrence.

Apply Young's rule, stated asHere the Kostka number counts Semistandard Young tableaux of shape $\rho$ with $n-k$ entries equal to $1$ and $k$ entries equal to $2$. Their rows weakly increase and their columns strictly increase. A column can therefore have at most two cells, so only a partition of an integer $\rho =(n-j,j)$ can occur.

In such a semistandard Young tableau, all $j$ entries in the second row must be $2$ and all $j$ entries above them must be $1$. The whole first row is then fixed by the content: its first $n-k$ entries are $1$ and its remaining $k-j$ entries are $2$. This is possible exactly when $j\le k$ and $j\le n-k$. Because $k\le n/2$, the second inequality follows from the first. There is exactly one filling for every $0\le j\le k$, and none for any other shape.

Consequently each displayed Specht module has multiplicity one, proving the two-row Young permutation module decompositionFor $j=0$, the zero second part is omitted. As a dimension check, the Hook-length formula gives $\dim S^{(n-j,j)}=\left(\genfrac{}{}{0ex}{}{n}{j}\right)-\left(\genfrac{}{}{0ex}{}{n}{j-1}\right)$, with $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$; the sum telescopes to $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, the dimension of the original permutation representation.

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 $p,r$.

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 core-quotient bijection for partitions: an ordered $p$-tuple $({\nu }_0,\dots ,{\nu }_{p-1})$ and a fixed $p$-core determine a unique partition of an integer, withThe irreducible characters in both $B$ and $C$ therefore correspond to the same ordered tuples of total size $w$. Replacing the empty core of a partition by $\gamma$ while keeping the quotient of a partition fixed gives a bijection ${\chi }^{\lambda }\mapsto {\chi }^{\mu }$ from $B$ to $C$. In particular $|B|=|C|$.

For such a pair, all levels of the core tower of a partition above level zero agree. At level zero, their sizes are $0$ and $m$, respectively. Put $D={\sum }_{r\ge 1}|T^C(\lambda {)}_r|={\sum }_{r\ge 1}|T^C(\mu {)}_r|$. When $0\le m<p$, the base-p digit sum satisfiesThe P-adic valuation of a symmetric-group character degree from the core tower then givesThus the same bijection preserves degree coprimality to $p$ and restricts to a bijection between the indicated sets of irreducible characters of degree coprime to p. ConsequentlyFor $m=0$, take $\gamma$ to be the empty partition; the construction becomes the identity.

For each $m$, the shifted sequence ${\beta }_{i,m+1}-\mathrm{id}$ is obtained from ${\beta }_{i,m}-\mathrm{id}$ by swapping the adjacent entries in positions $i+m$ and $i+m+1$. Part a(ii) therefore givesThis is the usual character straightening procedure: the distinguished entry ${\alpha }_i-i-k$ is moved successively to the right. Either it reaches the unique place that makes some ${\beta }_{i,m}$ a partition of an integer, or it meets an equal shifted entry. Under the hypothesis the first alternative never occurs, so two entries of some ${\beta }_{i,m}-\mathrm{id}$ coincide. The corresponding alternating sum is fixed by swapping those entries but changes sign by part a(ii), and is therefore zero. Repeatedly applying the displayed relation gives ${\psi }^{{\beta }_{i,0}}=0$.

The representation theory of $S_n$ is organized by a partition of an integer. Over the complex numbers, partitions label the irreducible Specht modules; over positive-characteristic fields, the simple modules are labelled by regular partitions.

For a partition of an integer $\lambda ⊢n$, applying a Young symmetrizer of that shape to the tensor power of each vector space gives a Schur functor. The construction respects linear maps because their tensor powers commute with place permutations. For single rows it gives symmetric powers, and for single columns exterior powers.

A filling $T$ of shape $\lambda$ and content $\alpha$ defines a module homomorphism ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: given a numbered tableau, sum over distinct assignments obtained by permuting $T$ within its rows, and put each numbered entry into output row $j$ when its assigned label is $j$. Restricting to $S^{\lambda }$ gives ${\hat{\Theta }}_T$. For Semistandard Young tableaux $T$, these restrictions are linearly independent over every field. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ if the characteristic of a field is not two, or if $\lambda$ is a two-regular partition of an integer. For two-singular shapes in characteristic two, they need not span. Column antisymmetrization and triangular leading tabloids establish independence; straightening the images of a cyclic polytabloid establishes spanning under the stated hypotheses.

A single-box up-move changes a partition of an integer by adding one to row $i$ and subtracting one from row $j>i$, provided the result remains a partition. It increases exactly the partial sums ending between rows $i$ and $j-1$. A partition $\mu$ dominates $\lambda$ precisely when it is obtainable from $\lambda$ by a sequence of these moves: fill the first deficient row from the row at the first return of the cumulative deficit to zero. This preserves the partition inequalities and decreases the distance to $\mu$.

The Young diagram of a partition of an integer $\lambda$ consists of one unit square, or cell, for each pairThroughout this article we use the English convention: rows are left-aligned, row $1$ is at the top, row indices increase downwards, and column indices increase to the right. The row lengths weakly decrease from top to bottom. Thus $Y(3,2)$ has cells $(1,1),(1,2),(1,3),(2,1),(2,2)$, with three cells above two. In the French convention the longest row is drawn at the bottom; the underlying partition of an integer is unchanged, but directions in descriptions of hooks of a Young diagram and standard Young tableaux must be interpreted consistently. See Grinberg's hook-length-formula teaching slides, §§1.1–1.2, for diagrams with the same orientation.

The number of cells is $|\lambda |$, and reading the row lengths recovers the partition of an integer. Reading the column lengths instead gives its conjugate partition. This turns numerical inequalities into visible geometry: Young diagrams satisfywith missing parts treated as zero. For example, $(2,1)\subseteq (3,2)$. This inclusion is the order in Young's lattice; it differs from dominance order on partitions, which compares sums of initial parts of partitions of an integer of the same size. The partition of an integer $(3,2)$ dominates $(2,2,1)$ because their successive sums are $3\ge 2$, $5\ge 4$, and $5=5$, but neither Young diagram contains the other: one has cell $(1,3)$, the other $(3,1)$.

The beauty of a Young diagram is that its shape simultaneously records a partition of an integer, indexes a Specht module, and supports local quantities such as hook lengths that determine a global dimension. Filling its cells leads to Young tableaux; adding one cell at a time leads to the Young branching graph.

The Young lattice is a partially ordered set whose elements are integer partitions, including the empty partition. Each partition of an integer is identified with its diagram, and the order is inclusion of Young diagrams. A cover adds exactly one box. A saturated path from the empty diagram to $\lambda$ is equivalent to a standard Young tableau of shape $\lambda$.

A Young tableau is a filling of a Young diagram with entries, typically positive integers. The shape is the partition of an integer giving its row lengths. We draw Young tableaux in the English convention of Young diagram, with left-aligned rows arranged from top to bottom. A filling need not obey increasing conditions unless these are specified: standard Young tableaux use each of $1,\dots ,n$ exactly once and strictly increase along rows and down columns; Semistandard Young tableaux permit repetitions and require weakly increasing rows and strictly increasing columns.

In the constructions for a symmetric group involving Young symmetrizers, tabloids, and polytabloids, an unqualified numbered Young tableau means a bijective filling by $1,\dots ,n$, with no increasing condition imposed. This narrower convention is compatible with the broader use needed for Semistandard Young tableaux. Historically, Young's work connected these fillings to representation theory of the symmetric group; in particular, the equality between standard Young tableau counts and irreducible representation degrees over the complex numbers predates the modern hook-length formula. See Stanley's Algebraic Combinatorics, historical notes to Chapter 8.