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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.

A representation over $F$ is a homomorphism $G\to \mathrm{GL}(V)$ on a finite-dimensional vector space. It is irreducible if it has no nonzero proper invariant subspace, and has absolute irreducibility of a group representation if it remains irreducible after every field extension. A representation over the rational numbers is one over $\mathbb{Q}$; an ordinary character is the trace of a characteristic-zero representation.

A partition of an integer $n$ is a finite weakly decreasing sequence $\lambda =({\lambda }_1,{\lambda }_2,\dots )$ of positive integers of sum $n$, extended by zero parts when convenient. Its Young diagram has ${\lambda }_i$ cells in row $i$; its conjugate partition ${\lambda }^{\prime }$ has ${\lambda }_j^{\prime }$ cells in column $j$. Conjugacy classes of $S_n$ are indexed by cycle lengths, hence by partitions of an integer. The number of ordinary irreducible characters equals the number of conjugacy classes; we now construct that many mutually distinct representations over the rational numbers.

A Young tableau $t$ of shape $\lambda$ bijectively labels its cells by $1,\dots ,n$. Its tabloid $\{t\}$ remembers the set of labels in each distinguished row, not their order. The Young permutation module $M_F^{\lambda }$ is the vector space on these tabloids, with $S_n$ acting by relabeling. Its tabloid bilinear form makes the tabloid basis orthonormal. Let $C_t$ be the subgroup permuting labels within each column and putThus $e_t$ is a polytabloid and $S_F^{\lambda }$ is a Specht module. Relabeling takes $e_t$ to $e_{gt}$, so any polytabloid generates the module. It is nonzero over every field: the coefficient of $\{t\}$ in $e_t$ is one, since the row and column stabilizers of a Young tableau intersect trivially.

Here is the elementary identity behind irreducibility. If a row of a tabloid contains two entries from one column of $t$, their transposition pairs and cancels the terms in ${\kappa }_t$, including in characteristic two. Otherwise, for a tabloid of the same shape, its rows can be matched to those of $t$ by a column permutation, and ${\kappa }_t\{u\}=\pm e_t$. Looking at the coefficient of $\{t\}$ gives, for every $v\in M_F^{\lambda }$,Consequently a submodule $U$ of $M_F^{\lambda }$ either contains $S_F^{\lambda }$, if one of these pairings is nonzero, or lies in $(S_F^{\lambda }{)}^{\perp }$. This proves the James submodule theorem used below.

Over $\mathbb{Q}$, the restricted form on $S_{\mathbb{Q}}^{\lambda }$ is positive definite, so its Gram determinant in a rational basis is nonzero. Extending scalars to any characteristic-zero field preserves that nonzero determinant and hence keeps the restricted form a nondegenerate bilinear form. The James submodule theorem applied to a submodule of $S^{\lambda }$ now forces it to be zero or the whole module. Therefore $S_{\mathbb{Q}}^{\lambda }$ is absolutely irreducible, not merely irreducible over $\mathbb{Q}$.

To distinguish shapes, the cancellation argument above works for a $\mu$-tabloid too. If ${\kappa }_t\{u\}\ne 0$, its first $r$ rows contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$, soThus $\lambda$ dominates $\mu$. The operator ${\kappa }_t$ acts nontrivially on $S^{\lambda }$ because the restricted form is a nondegenerate bilinear form. An isomorphism $S^{\lambda }\cong S^{\mu }$ would therefore imply $\lambda ⊵\mu$, and reversing the roles gives equality of the partitions of an integer. Counting conjugacy classes proves exhaustion. All matrices in a rational polytabloid basis have rational entries, giving the requested representations over the rational numbers.

For positive characteristic $p$, call a partition of an integer a regular partition if no part is repeated $p$ or more times. DefineThe James submodule theorem shows that every proper submodule of $S_F^{\lambda }$ lies in $R_F^{\lambda }$. Whenever the restricted form is nonzero, $R_F^{\lambda }$ is proper and its quotient is simple.

We verify exactly when this happens. Let $m_j$ be the number of rows of length $j$. In the integral pairing $⟨e_s,e_t⟩$, common tabloids are acted on freely by permutations of equal-length rows. Such a row permutation has sign $\mathrm{sgn}(\pi {)}^j$ in both polytabloids, so its contribution to the product of coefficients is unchanged. Each orbit has size ${\prod }_j{m}_j!$. Thus that integer divides every pairing. On the other hand, reverse each row of $t$ to obtain $t^{\ast }$. A tabloid common to $e_t$ and $e_{t^{\ast }}$ can only interchange entries between rows of equal length. There are $m_j!$ independent choices in each of their $j$ columns, and the two signs agree. HenceThese two products have exactly the same prime divisors. All pairings vanish modulo $p$ precisely when some $m_j\ge p$. Therefore $D_F^{\lambda }\ne 0$ exactly for $p$-regular $\lambda$.

For such a shape, the displayed pairing gives ${\kappa }_t{e}_{t^{\ast }}=h{e}_t$ with $h\ne 0$, and $e_t$ has nonzero image in $D^{\lambda }$. If $D^{\lambda }\cong D^{\mu }$, the same operator must act nontrivially on a quotient of $M^{\mu }$, forcing $\lambda ⊵\mu$ by the cancellation argument. Interchanging the shapes proves $\lambda =\mu$.

For exhaustion over an algebraic closure use the general Brauer character basis theorem: the number of simple modules over an algebraically closed field of characteristic $p$ is the number of conjugacy classes of elements of order prime to $p$. Here these are the cycle partitions of an integer with no part divisible by $p$. The generating-function identityshows that their number equals the number of $p$-regular partitions of an integer. We have already constructed that many distinct simple modules, so they exhaust all simples. The form, its radical and its quotient commute with field extension, and the same simplicity proof holds after every extension. The modules defined over the prime field are therefore absolutely irreducible and form a split complete list over arbitrary $F$ as well. In summary,For characteristic zero the list is instead all $S_F^{\lambda }$. In positive characteristic a Specht module itself need not be irreducible; the radical quotient is essential.

Use the pair of partitions for a generalized Specht module convention appropriate to generalized Specht modules. Put $a={\mu }^{♯}$ and $b=\mu$. Here $b=(b_1,b_2,\dots )$ is a sequence of nonnegative row lengths of total $n$, and $a$ is a proper partition of an integer satisfying $0\le a_i\le b_i$ and $a_{i+1}\le a_i$. The row-length sequence $b$ is allowed to be a composition; it need not decrease. The first $a_i$ cells of row $i$ are marked. This construction is not an arbitrary pair of unrelated partitions of an integer and is not simply a skew diagram $b/a$.

A word has type $b$ if it contains $b_i$ occurrences of $i$. Read it from left to right. Every occurrence of $1$ is good; an occurrence of $i+1$ is good exactly when there have so far been more good occurrences of $i$ than good occurrences of $i+1$. Otherwise it is bad. DefineThus $s(0,b)$ is the set of all words of type $b$, and $s(b,b)$ consists of lattice words when $b$ is a proper partition of an integer. Marking the whole first row makes no difference because every $1$ is good.

For a Young tableau $T$ with row lengths $b$, let $C_T^a$ permute labels within each column of its marked subdiagram, fixing all unmarked labels. Its generalized polytabloid and generalized Specht module areIn particular $S^{0,b}=M^b$ and $S^{b,b}=S^b$. The Young tableau shape in this definition must be the row sequence $b$.

We give the filtration argument, keeping its combinatorial and module-theoretic steps separate. Normalize $a_1=b_1$, and if $a\ne b$ choose the first row $c>1$ with $a_c<b_c$. Then $a_{c-1}=b_{c-1}$. Define $A_c(a,b)=(a+e_c,b)$ if $a_{c-1}>a_c$; otherwise the add branch is empty. Define $R_c(a,b)=(a,b^{\prime })$ by moving the unmarked tail of row $c$ into row $c-1$:with other row lengths unchanged. Re-normalize the first marked row if necessary. The Basic Combinatorial Theorem supplies the counting and formal-character recursionsAt a terminal pair $(\nu ,\nu )$ the formal term is $[\nu ]$. Thus the second recursion specifies the Specht multiplicities by the terminal leaves, with repeated leaves counted separately.

Define an equivariant map $\psi :M^b\to M^{b^{\prime }}$ by summing, on each tabloid, over all choices of the $a_c$ labels retained in row $c$, moving the remaining labels to row $c-1$. Column cancellation givesFor the first identity, terms that put two marked labels of one column in the raised row cancel in pairs; the surviving marked antisymmetrizer is the one for the raised pair, and every target generator has a preimage. For the second, the extra marked cell makes such a collision unavoidable, so the sum vanishes. Also antisymmetrizing the extra cell expresses its generator as a sum of old generators, proving containment in $S^{a,b}$.

There is a characteristic-independent lower bound $\dim S^{a,b}\ge |s(a,b)|$. Given a word in $s(a,b)$, place the label $j$ in its letter's row: good occurrences fill that row from the left and bad occurrences fill it from the right. The marked cells form increasing column chains because each good $i+1$ has an earlier unmatched good $i$. Its generalized polytabloid has the original row-assignment tabloid with coefficient one; every other term changes an earlier member of a marked chain. Ordering these row assignments gives a triangular coefficient matrix. The resulting vectors are independent over every field.

Start with $(0,v)$, where equality holds because the tabloid basis is indexed by all words of type $v$. Any pair is reached from such a pair by a sequence of add and raise operations: reverse a raise by splitting a raised tail, and reverse an add by unmarking a cell; row by row these recover an entirely unmarked composition. Suppose equality holds at a pair. The map and kernel containment above giveEvery inequality is therefore equality. In particular the containment is the whole kernel, andis exact. The process terminates because each add marks an extra cell and each raise decreases the total row index of unmarked cells. Induction from the terminal Specht modules, splicing their filtrations through these exact sequences, proves a Specht series with factors precisely $[0{]}^{[{\mu }^{♯},\mu ]}$. No splitting of these sequences in positive characteristic is asserted.

Applying the same marking, raising and straightening procedure to the two sets of column antisymmetrizers in $(S^{\mu }⊠S^{\lambda }){\uparrow }^{S_{r+s}}$, where $|\mu |=r$, $|\lambda |=s$, leaves the cells of $\mu$ fixed and records the added cells by their row labels from $\lambda$. The column relations require strict increase down columns, the row relations weak increase along rows, and the good-letter matching in a tableau word test requires a lattice word. Thus the surviving terminal multiplicity is the number $c_{\mu \lambda }^{\nu }$ of Semistandard Young tableaux of skew shape of shape $\nu /\mu$ and content $\lambda$ whose word, read right-to-left along successive rows from top to bottom, has at least as many $i$'s as $(i+1)$'s in every prefix. The same triangular leading-tabloid argument identifies each quotient with $S^{\nu }$. Consequently the Littlewood–Richardson rule isIt describes ordinary irreducible constituents in characteristic zero and Specht filtration factors over arbitrary $F$.

For induction by one letter, content $(1)$ permits exactly one added cell and each multiplicity is one. A Specht series for the requested induced module has successive quotients, ordered by addable nodes from bottom to top,Equivalently there is $0=N_0\subset N_1\subset N_2\subset N_3\subset N_4=S^{(4,2,2,1)}{\uparrow }^{S_{10}}$ with these four quotients in order. Their dimensions $567,300,768,525$ sum to $2160=10\cdot 216$, the dimension of the induced module.

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.

A partition of an integer $n$ is a finite sequence ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }>0$ with sum $n$; append zeros when comparing lengths. Its Young diagram has cells $(i,j)$ with $1\le j\le {\lambda }_i$. A Young tableau is a bijective filling of these cells by $1,\dots ,n$. Write $R_t,C_t$ for its row and column stabilizers, and fix the conventionPermutations act on entries on the left, with the rightmost factor acting first. This defines the Young symmetrizer used throughout. In the dictionary order on integer partitions, $\lambda >\mu$ means that at the first differing part ${\lambda }_i>{\mu }_i$; equality is allowed in $\ge$.

Suppose there is no row-column collision between $t$ of shape $\lambda$ and $u$ of shape $\mu$. Each column of $u$ contains at most one entry from each row of $t$. The first $k$ rows of $t$ therefore contain at mostentries, where ${\mu }_j^{\prime }$ is a column length. Thus ${\sum }_{i\le k}{\lambda }_i\le {\sum }_{i\le k}{\mu }_i$ for every $k$: $\mu$ dominates $\lambda$ in dominance order on partitions. If $\lambda \ge \mu$ in dictionary order on integer partitions, a first strictly larger part would contradict the corresponding prefix inequality. Hence $\lambda =\mu$.

All the bounds must now be equalities. Every column $j$ of $u$ contains exactly one entry from row $i$ of $t$ whenever $i\le {\lambda }_j^{\prime }$. Choose $r_0\in R_t$ sending that entry to the entry of $t$ in cell $(i,j)$. These prescriptions are bijections within the rows. The columns of $r_{0}u$ then have exactly the same sets of entries as the columns of $t$, so $r_{0}u=c_{0}t$ for some $c_0\in C_t$. ConsequentlyIf a collision was present instead, its two entries supply the first alternative. This proves the row-column collision lemma, including the prescribed order of the two stabilizer factors.

We next prove the Specht module classification. By Maschke's theorem, the group algebra $A=\mathbb{C}{S}_n$ is a semisimple algebra. Use the permitted basic quasi-idempotence of a Young symmetrizer,and put $e_t=h_t/H_{\lambda }$. The coefficient of $1$ in $h_t$ is one, because $R_t\cap C_t=\{1\}$, so $e_t\ne 0$.

For $\pi \in S_n$, consider $r_t\pi c_t$. A collision between the rows of $t$ and the columns of $\pi t$ gives a transposition $\tau \in R_t\cap \pi C_t{\pi }^{-1}$. Row symmetrization fixes $\tau$, whereas column antisymmetrization changes its sign, so $r_t\pi c_t=0$. If there is no collision, the proved lemma gives $\pi =rc$ with $r\in R_t,c\in C_t$, andIt follows that $h_t\pi h_t$ is zero or $\mathrm{sgn}(c)H_{\lambda }{h}_t$. Since permutations span $A$, $e_{t}A{e}_t=\mathbb{C}{e}_t$. In a semisimple algebra this means that $e_t$ is a primitive idempotent, so $A{e}_t=A{h}_t$ is an irreducible left module.

If $\lambda >\mu$ in dictionary order on integer partitions, the collision lemma applied to every $\pi u$ gives $r_{t}A{c}_u=0$, and therefore $h_{t}A{h}_u=0$. Sincethe two simple modules cannot be isomorphic. Conversely, Young tableaux of the same shape are related by a permutation, which conjugates their Young symmetrizers and gives isomorphic left ideals. Finally, the center of $\mathbb{C}{S}_n$ has the conjugacy-class sums as a basis. Conjugacy classes are indexed by cycle-type partitions, so its dimension is the number of partitions of $n$. The Artin–Wedderburn theorem gives exactly that many simple-module isomorphism classes. We have already produced one for each partition. Thus the $A{h}_{\lambda }$ form a complete set of pairwise nonisomorphic irreducible modules.

The annihilating polynomial excludes any free $\mathbb{C}[X]$-summand. In a cyclic primary decomposition summand $\mathbb{C}[X]/(P^a)$, the same condition requires $P^a$ to divide $(X-2{)}^4$. Thus $P=X-2$, up to a scalar unit, and $1\le a\le 4$. The complex dimension of this summand is $a$.

Write $A_j=\mathbb{C}[X]/((X-2{)}^j)$. The exponents must form a partition of an integer $4$, giving exactlyThey are distinct module isomorphism types by uniqueness in the structure theorem for finitely generated modules over a principal ideal domain; equivalently they are the five possible Jordan block patterns for eigenvalue $2$ in dimension four.

The Young lattice, or Young poset, has one vertex for each partition of an integer, including the empty partition, ordered by inclusion of their Young diagrams. A diagram covers another exactly when it adds one box. The Young branching graph is its graded graph: vertices at level $i$ are the partitions of $i$, and edges join diagrams differing by one box.

A descending path from $\lambda ⊢n$ to $(1)$ determines a standard Young tableau: label the successive removed boxes $n,n-1,\dots ,2$, then label the remaining box $1$. Every removed box is a Removable node of a Young diagram, so the resulting labels increase along rows and columns. Conversely, deleting boxes in decreasing label order from a standard Young tableau gives the path. Extending by the unique edge from $(1)$ to the empty diagram yields the usual path-to-tableau correspondence.

For the operator identity, compare the coefficients of each $\mu ⊢i$ in $DU(\lambda )$ and $UD(\lambda )$. If $\mu \ne \lambda$, a common upper cover exists exactly when a common lower cover exists, and each is then unique: the two diagrams differ by exchanging one box, their union is the upper cover and their intersection the lower cover. Since unions and intersections of partition diagrams are again partition diagrams, these off-diagonal coefficients agree.

The coefficient of $\lambda$ in $DU(\lambda )$ is the number of addable nodes of a Young diagram; its coefficient in $UD(\lambda )$ counts the choices of a Removable node of a Young diagram. Along the diagram's boundary the two types of corners alternate, beginning and ending with addable corners. Hence there is exactly one more addable than removable corner. For the empty partition the counts are $1$ and $0$.

Therefore the up and down operators satisfyor $D_{i+1}{U}_i-U_{i-1}{D}_i=I_i$ at level $i$. At level zero, interpret $D_0$ and the lowering contribution as zero. This makes the Young lattice a differential poset.

Use the Young–Jucys–Murphy elements in the group algebra $\mathbb{C}{S}_n$. Their joint spectrum isEquivalently one can use the regular representation, which contains every irreducible representation. The commuting elements are self-adjoint in a unitary representation, so there is a simultaneous eigenbasis.

A standard Young tableau has entries increasing along each row and down each column. If entry $r$ occupies cell $(i_r,j_r)$, its Content of a Young-diagram cell is $c_T(r)=j_r-i_r$. Define the set of content vectors of standard Young tableaux byThe shape here is a partition of an integer; it is not itself the vector of eigenvalues. For $n=1$ both sets consist of $(0)$.

Necessity is a property of standard Young tableaux alone. The first entry occupies $(1,1)$, with Content of a Young-diagram cell zero. Every subsequent cell has a cell immediately above or to its left, of content one larger or one smaller, already present. If two entries occupy the same diagonal, the later cell lies strictly southeast of the earlier. The cells immediately right of and immediately below the earlier one exist and have contents $a+1,a-1$; their entries lie strictly between the two given entries. This proves all three conditions for content vectors of standard Young tableaux.

For sufficiency, build the Young diagram one cell at a time. All coordinates are integers, because each new coordinate differs by one from an earlier coordinate, starting at zero. Cells of any fixed content $a$ are ordered northwest to southeast. For $a\ge 0$ their positions are $(r,r+a)$, $r=1,2,\dots$; for $a<0$ they are $(r-a,r)$.

At the first occurrence of a positive $a$, the candidate is $(1,a+1)$. The occurrence of $a+1$ earlier would already force a cell of content $a$ in its row, so the neighbor condition must instead supply the content $a-1$. This supplies the left predecessor. The negative case is symmetric, supplying the upper predecessor, and the first zero gives $(1,1)$.

At any later occurrence of $a$, take the next position on its diagonal. The preceding occurrence already supplied all but the next necessary predecessor on each neighboring diagonal. The repeated-entry condition supplies both $a-1$ and $a+1$ after that preceding occurrence, so those next predecessors are present. More explicitly, for $a>0$ and the $r$th occurrence, the left predecessor is the $r$th cell of content $a-1$, and the upper predecessor is the $(r-1)$st cell of content $a+1$. For $a=0$ both are the $(r-1)$st cells on their respective diagonals; for $a<0$ the roles are reversed. These are exactly the predecessors supplied by the two new neighboring occurrences.

The new cell is therefore an addable node of a Young diagram. Induction produces a partition of an integer at every stage, and filling the added cell by its insertion time gives a standard Young tableau. No choice was possible because addable nodes of a Young diagram have distinct contents. Thusand the construction also proves uniqueness of the tableau with a specified vector.

The local spectral interchanges preserve an irreducible representation, and standard Young tableaux of a fixed shape are connected by admissible adjacent interchanges. Different shapes have different multisets of Young-diagram cell contents. Their elementary symmetric functions in the Young–Jucys–Murphy elements are central: the identityproves this coefficient by coefficient. By the Schur lemma, one irreducible representation cannot therefore contain two different shape classes. The multiset determines a diagram because the counts on its positive and negative diagonals determine its arm and leg lengths at the diagonal cells.

Different irreducible representations cannot share a joint eigenvalue vector: the Gelfand–Tsetlin algebra is generated by the Young–Jucys–Murphy elements, so all its elements would act identically on that vector in both representations, whereas a central primitive idempotent distinguishes their two blocks. Thus distinct irreducibles give distinct shape classes.

The number of irreducible representations is the number $p(n)$ of conjugacy classes, each indexed by a partition of an integer. Every spectral vector gives a tableau by the preceding construction. Closure under admissible interchanges makes every shape class that occurs occur in full. The two counts $p(n)$ then force every shape to occur, with exactly one irreducible representation per shape. This supplies $\mathrm{Spec}(n)=\mathrm{Cont}(n)$.

On restricting to $S_{n-1}$, delete the last coordinate. For tableaux of shape $\lambda$, the entry $n$ occupies one Removable node of a Young diagram. Grouping by that node gives one copy of the corresponding module; the path decomposition has simple multiplicities. Hence the restriction branching rule for a symmetric group isThe different removable nodes produce distinct partitions. The multiplicity-free structural input defining the Gelfand–Tsetlin basis is distinct from identifying this graph with the graph of Young diagrams.

Let $\lambda ,\mu ⊢n$ each be a partition of an integer. Pad the lists of parts with zeros and use the dominance order on partitionsThe total sums agree, so only finitely many inequalities matter. This is a partial order: reflexivity and transitivity hold term by term, and equality of all partial sums forces equality of every part, proving antisymmetry. For example $(2,2)⊴(3,1)$. It compares concentration in the top rows, whereas diagram inclusion compares individual cells and is a different relation.

Suppose $\lambda ⊴\mu$ in the dominance order on partitions and $\lambda \ne \mu$. Let $i$ be the first unequal row. Then ${\lambda }_i<{\mu }_i$. PutWe have $D_i>0$, and eventually $D_h=0$. Let $j>i$ be its first return to zero. Then $D_h\ge 1$ for $i\le h<j$ and ${\lambda }_j>{\mu }_j$.

Move one box from row $j$ to row $i$. Increasing row $i$ is legal: if $i>1$, then ${\lambda }_{i-1}={\mu }_{i-1}\ge {\mu }_i>{\lambda }_i$. Decreasing row $j$ is also legal. If ${\lambda }_j={\lambda }_{j+1}$, then ${\mu }_{j+1}\le {\mu }_j<{\lambda }_j={\lambda }_{j+1}$, forcing $D_{j+1}<0$, a contradiction. Also ${\lambda }_j>0$. Thus the new list $\nu$ is a partition of an integer.

The new deficits equal $D_h-1$ on $i\le h<j$ and $D_h$ elsewhere, so $\lambda ⊴\nu ⊴\mu$. The quantity ${\sum }_r|{\lambda }_r-{\mu }_r|$ decreases by two, since the changed rows were a deficit and a surplus. Repeating must terminate at $\mu$. HenceThe construction proves both termination and preservation of the intermediate dominance inequalities.