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For a type $\alpha$, each successive difference is a rim hook of size ${\alpha }_i$; zero parts give empty differences of height zero. Label its cells by $i$. Such fillings are weakly increasing along rows and columns, and the height is the sum of the occupied strips' numbers of rows minus one. Empty parts are omitted from the associated product of positive power sums; the finite-variable degree-zero power sum is not inserted.

A skew Young diagram is connected when any two of its cells can be joined by a sequence of cells sharing edges; touching only at a corner does not connect them. A nonempty border strip, or rim hook, is a connected skew diagram containing no $2\times 2$ square. Its height is its number of occupied rows minus one.

For a weak composition $\alpha =({\alpha }_1,\dots ,{\alpha }_{\ell })$, a border-strip tableau of shape $\lambda /\mu$ and type $\alpha$ is a chainwhose $i$th difference is a border strip of size ${\alpha }_i$. Filling that strip with $i$ gives the equivalent weakly increasing row-and-column filling. A zero part corresponds to an empty difference and has height zero. Put $\mathrm{ht}(T)={\sum }_i\mathrm{ht}({\lambda }^{(i)}/{\lambda }^{(i-1)})$. For positive $r$, $p_r={\sum }_j{x}_j^r$; write $p_{\alpha }={\prod }_{{\alpha }_i>0}{p}_{{\alpha }_i}$. Thus zero type parts are ignored: one does not multiply by the finite-variable value $p_0=N$.

We first prove the one-strip multiplication formula using alternants. Choose $N$ sufficiently large and let ${\beta }_i={\mu }_i+N-i$ be the decreasing beta numbers of a partition. Expanding the determinant by permutations showsIndeed, in each determinant monomial every variable occurs in exactly one row factor; multiplying that variable by its $r$th power is equivalent to raising that row exponent.

A term vanishes if the new exponent equals another exponent. Otherwise suppose the changed exponent moves from position $i$ to position $j\le i$ when restored to decreasing order. Its determinant sign is $(-1{)}^{i-j}$. The resulting partition $\lambda$ hasand all other rows unchanged. The new diagram contains $\mu$, and its size increases by $r$. Between successive added rows, the overlap is exactly the single column ${\mu }_{t-1}+1$. Thus the added cells form a connected strip without a $2\times 2$ square, of height $i-j$. Conversely every added border strip has consecutive occupied rows and exactly this single-column overlap, forcing these formulas and the corresponding exponent shift. The correspondence is therefore bijective. Divide by the unchanged Vandermonde determinant and use the bialternant formula to get the power-sum border-strip multiplication identity

Apply this formula successively to each positive part of $\alpha$. Each sequence of choices is exactly a border-strip tableau, and its product of signs is $(-1{)}^{\mathrm{ht}(T)}$. ThereforeCommutativity of the power sums also shows that the resulting coefficient does not depend on the order of the positive parts of the type.

Take $\mu$ empty and restrict to $N\ge \ell (\lambda )$ variables. Multiply by $a_{\delta }$, with $\delta =(N-1,\dots ,0)$:In an alternant $a_{\nu +\delta }$, the exponents are distinct and their decreasing arrangement is $\nu +\delta$. Consequently the monomial $x^{\lambda +\delta }$ occurs only for $\nu =\lambda$, and in that determinant its coefficient is one. This proves the Frobenius alternant character formula

To identify these coefficients with symmetric-group characters, recall the Frobenius characteristic mapThe representation-theoretic identification $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ can be justified by Young's rule: the permutation character on ordered blocks of sizes $\nu$ is ${\sum }_{\lambda }{K}_{\lambda \nu }{\chi }^{\lambda }$. Its characteristic is $h_{\nu }$, because its fixed block decompositions allocate whole cycles to blocks, and the exponential generating series for those allocations is ${\prod }_{j}H(t_j)$. On the symmetric-function side, Hall duality and Schur orthonormality give $h_{\nu }={\sum }_{\lambda }{K}_{\lambda \nu }{s}_{\lambda }$. Inverting the unitriangular Kostka matrix establishes the identification. Thus the coefficient of $s_{\lambda }$ in $p_{\rho }$ is the irreducible character value on cycle type $\rho$. We have supplied the border-strip computation rather than assuming the character rule.

Grouping border-strip chains by their first removal of a cycle length $r$ now yields the Murnaghan–Nakayama ruleOne may choose the $r$-cycle to be last in a strip-building chain because the power sums commute. The empty-shape value on the empty type is one, providing the recursion's initial condition.

For the staircase $\delta =(m-1,m-2,\dots ,1)$, use $m$ beta numbers including its padded zero row. Its beta set isRemoving a rim hook of length $r$ lowers one beta number by $r$ into an unoccupied nonnegative position, reversing the exponent-shift description above. For a positive even $r$, this move either hits another occupied even position or goes below zero. Therefore there are no removable even-size border strips.

Under the Hall inner product of symmetric functions, the adjoint of multiplication by $p_r$ is $r\partial /\partial p_r$: this follows on the power-sum basis from $z_{\nu }/z_{\nu \setminus r}=r{m}_r(\nu )$. Taking adjoints of the one-strip multiplication rule expresses this derivative on $s_{\delta }$ as the signed sum over removable size-$r$ strips. That sum is empty for even $r$, soThe rational symmetric-function algebra is freely generated by the power sums, and its characteristic is zero. Vanishing of each of these derivatives therefore proves the staircase Schur functions use only odd power sums statement:

Finally, the cycle type of an $n$-cycle has one strip of size $n$. A straight diagram is a border strip exactly when its second row has length at most one: otherwise it contains the upper-left $2\times 2$ square; conversely the first row with a single-column tail is connected and contains none. These diagrams are exactly the hook partitions $(n-s,1^s)$, of height $s$. There is one strip filling in this case and none otherwise, giving

A rim hook is a connected skew diagram containing no $2\times 2$ square, whose removal leaves a partition of an integer diagram. Its leg length is its number of occupied rows minus one. The Murnaghan–Nakayama rule states that, if a permutation has cycle type $(k,\rho )$, thenIteration removes one hook for each cycle length; the empty diagram has character one at the empty cycle type.

For shape $(3,3,3)$ and cycle type $(4,3,2)$, there are exactly two removable four-cell rim hooks. The remaining shapes are $(3,2)$ and $(2,2,1)$; the removed hooks have leg lengths one and two respectively. In shape $(3,2)$ the only removable three-cell rim hook leaves $(1,1)$ and has leg length one. In shape $(2,2,1)$ the only such hook leaves $(2)$ and again has leg length one. Since the characters of $(1,1)$ and $(2)$ at a transposition are $-1$ and $1$, we obtainand henceThe cycle lengths can be taken in any order; using $4,3,2$ makes the two branches of the computation explicit.

Now consider the ordinary induction product $[\alpha ][\beta ]$. A hook partition is $(x,1^y)$, equivalently a partition of an integer whose second row has length at most one. If a hook shape $\nu$ occurs in the Littlewood–Richardson rule, then $\alpha \subseteq \nu$, forcing $\alpha$ itself to be a hook. In the skew hook $\nu /\alpha$, the new cells consist of a segment in the first row and a segment down the first column, with no shared column between those two segments. Every first-row entry must be $1$: its rightmost entry is the first letter of the lattice word, hence is $1$, and weak row increase forces all entries to its left to be $1$ too. The column segment has strictly increasing entries. Thus only $1$ can be repeated, forcing the content $\beta$ to be a hook as well. This also proves the assertion when the top segment is empty, since then every entry lies in a strictly increasing column.

Write the two nonempty hook factors as $\alpha =(a,1^k)$ and $\beta =(b,1^l)$, with $k=n-r-a$ and $l=r-b$. Let the prospective resulting hook have top-row length $x$ and leg length $y$. Its new top segment has $H=x-a$ cells and its new column segment has $V=y-k$ cells. The content consists of $b$ ones and the single letters $2,\dots ,l+1$. All those larger letters must be in the new column; its strict increase allows either no $1$ or exactly one $1$. There are therefore only two possibilities:In the first case fill the top segment with $b$ ones and the column with $2,3,\dots ,l+1$. In the second, use $b-1$ top ones and column $1,2,\dots ,l+1$. Each filling is unique and has a lattice word, including the case $b=1$ with an empty top segment in the second filling. This provesBoth displayed multiplicities are one.

To deduce the full-cycle value from this product result, rather than quote the Murnaghan–Nakayama rule again, form the virtual characterFor every proper Young subgroup $S_m\times S_{n-m}$ and every pair of its irreducible characters, the hook-product calculation givesIf a factor is not a hook there are no hook constituents; if both are hooks, their two consecutive leg lengths cancel in this alternating inner product. By Frobenius reciprocity and the completeness of the product irreducible characters, $F_n$ restricts to zero on every such subgroup. Any permutation with more than one cycle preserves a nonempty proper subset and is conjugate into one of these subgroups. Thus $F_n$ is supported only on the class of $n$-cycles.

Its inner product with the trivial character is one, since only the one-row hook contributes. That class has size $(n-1)!$, so $F_n(\rho )/n=1$, giving $F_n(\rho )=n$. For any irreducible character, the support calculation then yieldsBy character orthogonality, the left side is $(-1{)}^{n-x}$ for $\nu =(x,1^{n-x})$ and zero otherwise. These are real integers, provingFor $n=1$ the same conclusion holds directly. This derives the requested special case from the Littlewood–Richardson rule and Frobenius reciprocity.

Choose $m\ge \ell (\lambda )$, pad the partition of an integer with zeros, and define its beta numbers byThey form a strictly decreasing set of nonnegative integers. Conversely, from its decreasing enumeration recover ${\lambda }_i={\beta }_i-m+i$. With $m=\ell (\lambda )$ these are exactly the first-column hook lengths, since $h_{i1}={\lambda }_i+{\lambda }_1^{\prime }-i={\lambda }_i+m-i$. Padding changes the numbers: increasing $m$ by one replaces $B$ by $\{0\}\cup (B+1)$, and leaves the represented partition of an integer unchanged.

Arrange positions $0,1,2,\dots$ on $p$ runners, putting position $r+pk$ on runner $r$ at height $k$. A bead occupies each position in the beta set. This is the abacus of a partition. The hook criterion in a beta set says that a bead $b$ and a gap $a<b$ correspond to a hook of length $b-a$; replacing $b$ by $a$ removes the corresponding rim hook. In particular removing a rim $p$-hook moves a bead from $b$ to the vacant position $b-p$ immediately above it on the same runner. Its leg length counts the beads in the intervening positions of the full linear numbering.

For an explicit example, take $\lambda =(4,2,1)$, $p=3$, and $m=3$. Its beta set is $\{6,3,1\}$. Runner zero has beads at $3,6$, runner one has a bead at $1$, and runner two is empty. Packing upwards gives $\{3,1,0\}$, which represents $(1)$; thus the three-core is $(1)$ and the weight is two.

Every bead slide reduces the sum of occupied positions, so repeated slides terminate. The number $n_r$ of beads on each runner is unchanged. At termination they must occupy exactlyon runner $r$, since any gap above a bead would allow a further slide. Hence the final beta set is uniquely determined by the initial runner counts, independently of the order of hook removals. Increasing the number of beads by $p$ replaces $B$ by $\{0,\dots ,p-1\}\cup (B+p)$, adds one bead to each runner, and gives exactly the same represented core after packing. General changes in padding follow by applying $B\mapsto \{0\}\cup (B+1)$; packing commutes with this operation up to the cyclic relabelling of runners. Therefore the result is independent of both removal order and padding:The weight of a partition is $w_p(\lambda )=(n-|C_p(\lambda )|)/p$.

A block of a group algebra is an indecomposable two-sided summand $F{S}_{n}e$, where $e$ is a primitive central idempotent. The identity decomposes into pairwise orthogonal such idempotents; a module belongs to a block when its idempotent acts as identity and the other block idempotents act as zero. For ordinary characters, use a splitting p-modular system $(K,\mathcal{O},F)$, or extend the residue field first: the block idempotents lift to $\mathcal{O}{S}_n$, and determine which ordinary irreducible $K{S}_n$-modules lie in each $p$-block. The symmetric group is split over its prime field, so extending $F$ introduces no change to the resulting partition of an integer labels.

The Nakayama block theorem, traditionally called the Nakayama Conjecture for symmetric groups, isThe simple modules in that block are exactly the $D^{\nu }$ with regular labels having that core. At fixed $n$, the common core also fixes the weight. Here is an essay proof route through central characters, specifying the main algebraic and combinatorial steps and why each is needed.

First introduce the commuting Young–Jucys–Murphy elementsTheir commutators vanish because contributions supported on each triple of letters cancel. Symmetric polynomials in the $J_i$ are central: for adjacent transpositions $s_i$, the relations $s_i{J}_i=J_{i+1}{s}_i-1$, $s_i{J}_{i+1}=J_i{s}_i+1$ show that $s_i$ commutes with symmetric expressions in $J_i,J_{i+1}$, and it commutes with the other $J_j$. Since these transpositions generate $S_n$, the symmetric expressions commute with the whole group algebra.

The substantial generation ingredient is the integral Jucys–Murphy description of the center of a symmetric-group algebra: every conjugacy-class sum is an integral symmetric polynomial in the $J_i$. Its proof expands symmetric monomials in these sums of transpositions and organizes the resulting class sums by the number of moved letters and cycle filtration; triangular elimination supplies integral spanning, rather than merely rational spanning. A useful first identity in this construction isIndeed, when adjoining the letter $n$, either choose $t$ and leave it as a singleton cycle, or choose a transposition $(jn)$ and insert it into the cycle containing $j$; this gives every permutation uniquely. The full generation step distinguishes the individual class sums which this first identity groups together. Its integrality is essential: a description only over $\mathbb{Q}$ could use denominators divisible by $p$ and would not prove generation of the modular center. Since class sums form the integral basis of the center, this integral theorem reduces to the same generation statement over $F$.

Next determine their scalar action. In the characteristic-zero tableau basis of $S^{\lambda }$, $J_i$ is diagonal with entry the Content of a Young-diagram cell occupied by $i$, namely column minus row. This follows recursively along the multiplicity-free branching chain: the commuting operators preserve each path line, and $J_{i+1}=s_i{J}_i{s}_i+s_i$ determines the successive eigenvalues. Consequently a symmetric polynomial $f(J_1,\dots ,J_n)$ acts on $S^{\lambda }$ byThis is independent of the ordering of the cells. The scalar identity holds on the integral Specht lattice, so after modular reduction it depends only on the multiset of contents modulo $p$. Every composition factor has this same central character of a block.

Two simple modules are in the same block exactly when their central characters agree. The center of a block is a commutative local finite-dimensional algebra; its semisimple quotient gives one scalar character, while the orthogonal block idempotents separate different blocks. By the generation theorem, equality on all symmetric polynomials in the $J_i$ is enough. Conversely, different residue multisets are detected by elementary symmetric polynomials: the monic polynomial ${\prod }_{(i,j)\in \lambda }(z-(j-i))$ over $F$ recovers the multiplicity of each residue by unique factorization, including multiplicities larger than $p$. Thus block equivalence is exactly equality of the integersThis gives the residue-content criterion for symmetric-group blocks without assuming that the sum of contents alone distinguishes blocks.

Finally identify residue content with the core. A removable rim $p$-hook has consecutive contents modulo $p$, hence removes one cell of each residue. Therefore shapes with the same core and the same size have the same residue counts. For the converse choose $m$ divisible by $p$ and at least $n$, and write $q_r=n_r-m/p$ for the runner charges. Starting from the empty partition of an integer gives $q_r=0$. Adding a cell of residue $r$ moves a beta-number bead from residue $r-1$ to residue $r$, increasing $q_r$ and decreasing $q_{r-1}$. The same changes occur in $c_s-c_{s+1}$ when $c_r$ increases by one. Induction on the cells therefore givesEqual residue counts give equal charges and hence equal packed runner configurations, so they give the same core. This completes both directions of the block criterion. The argument exposes the three essential bridges: integral center generation, its scalar content evaluation, and the abacus identification of residue data with a unique core.

Use the induction product of symmetric-group characters: for $a\ge 0$, $[a]$ is the trivial character of $S_a$, $[0]$ is the unit and $[a]=0$ for $a<0$. Products mean induction of external tensor products. The determinantal form iswhere $r\ge \ell (\lambda )$. Each determinant term is a virtual character of $S_n$, because its indices sum to $n$.

Here is a proof. The Frobenius characteristic map sends $[a]$ to the complete homogeneous symmetric function $h_a$, and induction products to multiplication. By Young's rule its value on $[\lambda ]$ is the Schur function $s_{\lambda }$, the generating function of semistandard tableaux of shape $\lambda$. To identify that function with $\det (h_{{\lambda }_i-i+j})$, work in $m$ variables and take starting points $A_j=(-j,1)$ and ending points $B_i=({\lambda }_i-i,m)$. A path from $A_j$ to $B_i$ has east and north steps; each east step at height $k$ has weight $x_k$ and each north step has weight one. Its east-step count is ${\lambda }_i-i+j$, so its weight sum is $h_{{\lambda }_i-i+j}$. Expanding the determinant sums signed systems of paths with permuted endpoints. For a system with an intersection, exchange the two tails after the first intersection. This preserves its monomial weight and reverses the endpoint permutation's sign, so all intersecting systems cancel in pairs. Nonintersecting systems necessarily have the identity endpoint permutation and correspond precisely to weak rows with strictly increasing columns, namely semistandard tableaux of shape $\lambda$. Their signs are positive. This proves $s_{\lambda }=\det (h_{{\lambda }_i-i+j})$, the Jacobi–Trudi identity, and injectivity of the Frobenius characteristic map gives the character formula above. The path sum may be taken in finitely many variables first and then stabilized, so no infinite formal sum is required in the argument.

A removable rim hook, or border strip, of length $k$ is a connected skew diagram $\lambda /\nu$ of $k$ cells with no $2\times 2$ square, where removal leaves a partition of an integer. Its leg length is the number of occupied rows minus one. The Murnaghan–Nakayama rule statesA proof sketch follows directly from the alternant form of Schur functions. Multiply an alternant by $p_k={\sum }_j{x}_j^k$. In each term one alternant exponent increases by $k$. Equal exponents cancel; otherwise reordering the exponents contributes the sign of the number crossed. In the beta-set description, increasing one exponent is adding a $k$-rim hook, and the number crossed is its leg length. Therefore $p_k{s}_{\nu }={\sum }_{\lambda }(-1{)}^{\mathrm{leg}(\lambda /\nu )}{s}_{\lambda }$. Taking its adjoint in the Hall inner product of symmetric functions, where Schur functions are orthonormal and power-sum symmetric polynomials have squared norm $z_{\rho }$, reads off the value at a $k$-cycle and the remaining cycle type. This gives the stated rule and explains its sign.

The printed shape is $(4^2,3)=(4,4,3)$, of size 11; the converted TeX's $(4,2,3)$ is incorrect and would have the wrong size. There are two removable length-5 rim hooks. They leave $(3,2,1)$ with leg length 2 and $(4,2)$ with leg length 1. ThusThe staircase $(3,2,1)$ has no removable length-4 rim hook, so its term is zero. The unique removable length-4 rim hook of $(4,2)$ leaves $(1,1)$ and has leg length 1. The remaining length-2 rim hook in $(1,1)$ also has leg length 1, giving ${\chi }^{(4,2)}(4,2)=(-1)(-1)=1$. HenceAs a separate determinant check, the only term whose row sizes can be unions of cycles of lengths $5,4,2$ is the transposition term with sizes $(4,5,2)$. Its permutation character has value one and its determinant sign is negative.

Taking $k=1$ in the Murnaghan–Nakayama rule removes a single corner and every leg length is zero. Evaluating on a permutation fixing the distinguished point therefore givesThis is the restriction branching rule for a symmetric group; ordinary semisimple representation theory converts the character equality into a direct-sum decomposition. Frobenius reciprocity gives the corresponding add-one-cell induction rule.

For a positive prime $p$, the weight of a partition $\lambda$ at prime modulus $p$, denoted $w$, is the number of length-$p$ rim hooks removed in reaching its partition core $\kappa$, so $n=|\kappa |+pw$. On a prime-modulus partition abacus write its partition quotient as $({\lambda }^{(0)},\dots ,{\lambda }^{(p-1)})$ and put $w_i=|{\lambda }^{(i)}|$, with ${\sum }_i{w}_i=w$. Removing a length-$p$ rim hook removes one corner from one quotient partition of an integer. Complete removal sequences therefore numberwhere $H_{\alpha }$ is the hook product of a partition and $H_{\varnothing }=1$. Each component's standard tableaux encode its possible removal orders, and the multinomial coefficient interleaves the components.

All these sequences have the same total sign ${\epsilon }_p(\lambda )$. To see this, label the beads on each runner in their preserved order. Each slide crosses exactly the beads counted by its rim-hook leg. Its sign is the change in the total order of bead labels, and the product telescopes to the permutation from the initial ordering to the final packed-core ordering, independently of the slides chosen. Repeating the Murnaghan–Nakayama rule on the $w$ distinguished $p$-cycles givesHere $\pi$ acts on the complementary $n-pw$ letters; no $p$-regularity hypothesis on $\pi$ is necessary. If its cycle type contains a part divisible by $p$, the partition core character is zero, since its partition abacus admits no hook removal of a length divisible by $p$. For $w=0$ the coefficient and sign are one.

For a cell $(i,j)$ in the Young diagram of $\lambda$, its hook of a Young diagram consists of that cell, the cells to its right in row $i$ and the cells below it in column $j$. Its length isA removable rim hook is a connected border strip $\lambda /\mu$ containing no $2\times 2$ square, whose removal leaves a Young diagram. Its leg length is the number of rows it meets minus one. The hook graph is the Young diagram labeled at each cell by $h_{ij}$; here “graph” means this labeled array, not a graph of possible hook-walk moves.

Choose $l\ge \ell (\lambda )$, pad with zeros and form the beta numbersThey are distinct nonnegative integers in decreasing order and recover ${\lambda }_i={\beta }_i-l+i$. For $l=\ell (\lambda )$ they equal the first-column hook lengths, since $h_{i1}={\lambda }_i+l-i$. If $l$ is increased, old beta numbers shift upward by one and a new bead at zero is inserted.

For a prime-modulus partition abacus arrange the nonnegative positions in $p$ runners: position $pq+r$ lies at level $q$ on runner $r$, $0\le r<p$. Put beads at the beta numbers and leave the other positions empty. Replacing a bead at $b$ by a bead at a vacant $b-k\ge 0$ removes a rim hook of length $k$; its leg length is the number of beads strictly between these positions. This follows by tracing the boundary of the diagram: bead positions record vertical steps and gaps record horizontal steps, and exchanging one bead and gap removes the intervening border strip. In particular a removable $p$-hook is exactly a bead that can slide up one place on its own runner.

The partition core at modulus $p$ is the result of sliding beads up until there are no gaps above any bead on a runner. If runner $r$ initially has $b_r$ beads, its final occupied levels must be $0,1,\dots ,b_r-1$, independently of slide order. Thus its packed beta set, and hence its core partition of an integer, is unique. Increasing the number of beta numbers merely re-expresses the same boundary by shifting positions and adding the new initial bead; the packed configuration represents the same partition of an integer. This proves well-definedness both with respect to removal order and to the chosen padding. The number of slides isThe partition quotient components record how far each runner's beads lie above their packed positions, giving the hook-removal interpretation used in Question 5.

A block of a group algebra in characteristic $p$ for $F{S}_n$ is a two-sided ideal $B=F{S}_{n}e$ for a primitive central idempotent $e$, meaning a nonzero central idempotent that is not a sum of two nonzero orthogonal central idempotents. The identity is the sum of these block idempotents and the group algebra is their direct product. An indecomposable module $V$ belongs to $B$ when $eV=V$, equivalently every other block idempotent acts as zero. Indeed $V={⨁}_{e}eV$; indecomposability makes exactly one summand nonzero.

The Nakayama block theorem, historically called the Nakayama Conjecture, statesEquivalently the ordinary characters assigned to one modular block have a common partition core at modulus $p$. Its simple modules are the $D^{\lambda }$ with regular $\lambda$ having that core. Only cores $\kappa$ for which $n-|\kappa |$ is a nonnegative multiple of $p$ occur. The block weight is $(n-|\kappa |)/p$; a core alone determines the block at this fixed degree.

One route to the proof separates the combinatorial core invariant from the algebraic center invariant. Let $c_r(\lambda )$ be the number of cells of residue $r=j-i(\mathrm{mod}p)$. A length-$p$ rim hook has one cell of every residue: along its border strip the contents are consecutive integers. Therefore removal subtracts one from every $c_r$. More precisely, choose the number $l$ of beta numbers divisible by $p$ and write the runner charges as $q_r=b_r-l/p$. Reading the boundary, or adding a cell and checking its two changed beta positions, givesThe charges determine the packed runners and hence the core. For fixed $n$, two equal cores have the same weight, so their full residue multisets are equal. Conversely equal residue multisets give equal charges and the same core. This establishesfor partitions of an integer of the same integer. The fixed-degree condition matters.

For the center invariant, use the commuting Jucys–Murphy elements $J_k={\sum }_{i<k}(ik)$, with $J_1=0$. In the Young seminormal form basis, $J_k$ has eigenvalue equal to the content of the cell containing $k$. A symmetric polynomial in them therefore acts on $S^{\lambda }$ as the same scalar for every Young tableau: its value on the multiset of all contents of $\lambda$. The integral Specht lattice makes this a scalar identity before modular reduction, so modulo $p$ these central scalars depend only on the residue multiset, and every composition factor inherits them.

The substantive center theorem used in this proof is that the center of the symmetric-group algebra, integrally and over a field, consists of symmetric polynomials in the Jucys–Murphy elements. Centrality of elementary symmetric polynomials follows fromthe coefficients are central class-sum combinations. The reverse inclusion requires the integral triangular class-sum argument: expand symmetric monomials in the $J_k$ as products of transpositions, group the leading terms by reduced cycle type, and eliminate these leading class sums in the resulting unitriangular order. The integral coefficients, with no division by multiples of $p$, are crucial; knowing the center only over $\mathbb{Q}$ would not prove the modular assertion. This is the central algebraic ingredient of this proof route.

Thus equal residue multisets yield equal characters of the entire modular center. Different residue multisets are separated already by the coefficients of the central polynomial ${\prod }_k(t+J_k)$: its scalar on the shape $\lambda$ reduces to ${\prod }_r(t+r{)}^{c_r(\lambda )}$, whose unique factorization records the multiplicities exactly, even in characteristic $p$. In a finite-dimensional split algebra, two simple modules are in the same block exactly when their central characters agree: the center of each block is local, while different block idempotents separate different blocks. It follows that equal cores give one block and different cores give distinct blocks. This proves both directions rather than just observing that core is constant on known blocks.

Finally, each ordinary Specht lattice has a single reduced central character, so all constituents of its reduction lie in that one block. Question 1 supplies every simple module and its label; the regular labels therefore identify all the simples in each core block. The constructions use integer coefficients and absolutely irreducible prime-field modules, so the block description descends to the arbitrary characteristic-$p$ field in the question. This completes the main steps of the block proof without presupposing semisimplicity in positive characteristic.

The Gelfand–Tsetlin basis spans $V^{\lambda }$, and the scalar computed on its vectors depends only on the shape. Thusfor every $w\in V^{\lambda }$. The product is the sum of the $(n-1)!$ permutations in the conjugacy class of an $n$-cycle. Taking traces therefore gives $(n-1)!{\chi }^{\lambda }((n))$ equal to the displayed scalar times $\dim V^{\lambda }$.

For a hook partition, a standard Young tableau is uniquely determined by the choice of its $k$ entries below the top cell, selected from $\{2,\dots ,n\}$. The column and the remaining row are then forced to increase. Hence $\dim V^{(n-k,1^k)}=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$, and cancellation of the factorials yieldsThis uses the central character value of a conjugacy-class sum and tableau counting, without a character rule for removing rim hooks.

The inclusion here is inclusion of Young diagrams, not the dominance order on partitions. The skew Young diagram $\lambda /\mu$ is the set difference of their cells. Two cells are adjacent when they share an edge. It is connected when any two cells can be joined by such steps, and is a rim hook when it is connected and contains no $2\times 2$ square. Under the usual edge-adjacency convention a totally disconnected skew Young diagram has only singleton components, equivalently no two cells share an edge. A horizontal strip instead means at most one cell per column; this distinction matters for the last part of this question.

A standard skew Young tableau is a linear extension of a partially ordered set: the cells are ordered by the row and column inequalities, and the tableau lists them in increasing label order. Write the current list as $x_1,\dots ,x_k$ and let $r_j$ be the desired label of $x_j$ in $R$. Then $(r_1,\dots ,r_k)$ is the one-line notation of the unique permutation $\pi$ with $\pi T=R$.

Whenever this list of desired labels is not increasing, there is an adjacent descent $r_j>r_{j+1}$. The cells $x_j,x_{j+1}$ are incomparable in the cell partial order: if they were comparable, both $T$ and $R$ would have to put them in the same order. Interchanging their consecutive current labels is therefore admissible. It removes exactly one inversion of a permutation from the list of desired labels. Repeating ends at $R$ after exactly the original inversion count, which is the Coxeter length $\ell (\pi )$. ThusThis proves the reduced adjacent-swap path between linear extensions and applies equally to ordinary tableaux.

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.