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An ordinary Specht module of non-self-conjugate shape stays irreducible on the alternating group, and conjugate shapes give the same restriction. A self-conjugate shape splits into two irreducibles of equal degree, interchanged by an odd permutation. The index-two inner-product identity is $⟨{\chi }^{\lambda }\downarrow ,{\chi }^{\mu }\downarrow ⟩={\delta }_{\lambda \mu }+{\delta }_{{\lambda }^{\prime }\mu }$, and Frobenius reciprocity proves exhaustion. The statement applies for $n\ge 2$; $A_1$ is trivial.

The conjugate partition exchanges rows and columns in a Young diagram: the cell $(i,j)$ becomes $(j,i)$. Equivalently,For $\lambda =(3,2)$, the three column lengths are $2,2,1$, so ${\lambda }^{\prime }=(2,2,1)$, as shown in Young diagram. Conjugating twice returns the original partition of an integer, and the number of cells stays five. Transposition also preserves hook lengths, with $h_{j,i}({\lambda }^{\prime })=h_{i,j}(\lambda )$; consequently the hook-length formula gives equal dimensions for the two Specht modules over the complex numbers. This row-column symmetry is already used in §1 of ^[ref].

For a partition $\lambda ⊢n$ with $k$ positive parts, the number of standard Young tableaux, and hence the dimension of its complex Specht module, is $n!\det (1/({\lambda }_i-i+j)!)$. Negative factorial arguments give zero entries. Setting $L_i={\lambda }_i+k-i$ turns the determinant into ${\prod }_{i<j}(L_i-L_j)/{\prod }_i{L}_i!$ by the Vandermonde determinant, matching the hook-length formula.

For a partition of an integer $\lambda ⊢n$, padded to $m$ rows, let $\delta =(m-1,\dots ,0)$. If a permutation has ${\alpha }_q$ cycles of length $q$, put $p_{\alpha }={\prod }_q({\sum }_i{x}_i^q{)}^{{\alpha }_q}$. The character of the Specht module $S^{\lambda }$ is the displayed coefficient. Equivalently $p_{\alpha }{A}_{\delta }={\sum }_{\lambda }{\chi }^{\lambda }(\alpha )A_{\lambda +\delta }$, the alternant form of the Frobenius characteristic map. It gives finite coefficient computations without constructing representation matrices.

The hook graph is the Young diagram with the hook length written in each box. It displays the hook product and thus the hook-length formula for the corresponding Specht module. This use of the name is recorded in Frame, Robinson and Thrall's original hook-graph paper; it refers to a labeled array, not a directed hook-walk graph.

For any partition of an integer $\lambda ⊢n$, the number $f^{\lambda }$ of standard Young tableaux of shape $\lambda$ isThe numerator is a factorial, and the denominator is the hook product of a partition, using every cell's hook length once. The empty shape has one empty standard Young tableau, and the formula still holds with $0!=1$ and an empty product equal to one. By the standard polytabloid basis, $f^{\lambda }$ is also the dimension of the Specht module $S^{\lambda }$ over the complex numbers, or equivalently the degree ${\chi }^{\lambda }(1)$ of its irreducible character. For a teaching derivation of this dimension formula, see Etingof et al., Introduction to Representation Theory, §5.17, Theorem 5.17.1.

In the Young diagram of $(3,2)$, the hook length array $[[4,3,1],[2,1]]$ givesThis agrees with the five explicit fillings in standard Young tableau. Its conjugate partition $(2,2,1)$ has the same hook lengths after transposition, hence also gives dimension five. At the extremes, for $n\ge 1$, a single row $(n)$ and a single column $(1^n)$ each have hook product $n!$ and exactly one standard Young tableau; their Specht modules over the complex numbers are the trivial representation and sign representation, respectively.

Why the formula is beautiful: global compatibility of all the row and column inequalities reduces to a product of local hook lengths. It is useful when enumeration is already expensive: computing one hook length per cell replaces testing up to $n!$ numbered Young tableaux. For $(3,2)$, the proportion of the $120$ bijective fillings that are standard Young tableaux is $5/120=1/24$, the reciprocal of the hook product. This product does not justify treating overlapping hook conditions as independent; Grinberg explains that tempting but invalid argument in §1.3 of his teaching slides.

Historically, the modern hook-product statement was published by J. S. Frame, G. de B. Robinson and R. M. Thrall in 1954: ^[ref]. Their manuscript was received on 1 September 1953, and their proof reorganizes an earlier irreducible character degree formula into the product of the hook lengths. The theorem unifies the counting and representation viewpoints developed earlier by Young. For proof routes within this wiki, see trace computation of Specht module dimension, determinant formula for Specht module dimension, and Greene–Nijenhuis–Wilf hook walk.

The Young–Jucys–Murphy elements commute, and symmetric polynomials in them are central. The full center theorem holds integrally as well as over fields, so it survives modular reduction. On an ordinary Specht module, the scalar of such a polynomial is its value on the cell-content multiset. The integral generation step is stronger than knowing this description only over a characteristic-zero field and is essential in the modular block argument.

For a partition of an integer $\lambda$, let ${\psi }_{i,v}:M^{\lambda }\to M^{\lambda (i,v)}$ sum all tabloids obtained by retaining $v$ elements of row $i+1$ and moving its other ${\lambda }_{i+1}-v$ elements into row $i$, leaving the other row sets unchanged. Here $0\le v<{\lambda }_{i+1}$ and $\lambda (i,v)$ replaces $({\lambda }_i,{\lambda }_{i+1})$ by $({\lambda }_i+{\lambda }_{i+1}-v,v)$; Young permutation modules are defined for compositions too. Then $S^{\lambda }={\bigcap }_{i,v}\ker {\psi }_{i,v}$ over every field. Pairwise column cancellation gives inclusion of the Specht module in each kernel. The reverse inclusion follows by straightening the tabloid coefficients using these adjacent-row relations: the remaining free leading coefficients are exactly those of the standard polytabloid basis.

At fixed degree $n$ in positive characteristic $p$, two Specht modules belong to the same block of a group algebra exactly when their labels have the same core of a partition at modulus $p$. The simple labels in that block are precisely the regular partitions with that core. The theorem follows by combining the residue-content criterion for symmetric-group blocks with the Jucys–Murphy description of the center of a symmetric-group algebra.

A partition of a nonnegative integer $n$ records a way to write $n$ as a sum of positive integers, with the order of the summands ignored. Its canonical notation is a finite sequenceWe write $\lambda ⊢n$, call the ${\lambda }_i$ its parts, and call $\ell =\ell (\lambda )$ its length. Trailing zero parts may be appended for calculations without changing the partition of an integer; zero has the unique empty partition of an integer, of length zero. Thus $(3,2)$, $(3,2,0)$ and $5=2+3$ describe the same partition of an integer of five. The Young diagram below makes its two parts visible as rows of three and two cells.

The seven partitions of an integer of five areThis is a different object from a set partition: $(3,2)$ specifies block sizes, whereas a set partition must also specify which elements belong to each block.

Why this is useful: a partition of an integer is a compact description of a shape, yet it also labels an irreducible representation of a symmetric group over the complex numbers, realized by a Specht module. It therefore connects arithmetic decompositions, Young diagrams, and representation theory of the symmetric group. The generating function for partitions with bounded parts gives another way to count such shapes. Historically, Euler studied these counting problems using generating functions; his ^[ref] documents this early development.

Let $\lambda =({\lambda }_1\ge {\lambda }_2\ge \cdots )$ be a partition of an integer $n$. Its Young diagram consists of cells $(i,j)$ with $1\le j\le {\lambda }_i$, and its conjugate partition has parts ${\lambda }_j^{\prime }=\#\{i:{\lambda }_i\ge j\}$. A numbered Young tableau fills these cells bijectively with $1,\dots ,n$. Its row and column stabilizers $R_t,C_t$ permute entries within each row and each column respectively. A tabloid $\{t\}$ remembers the sets in the ordered rows, forgetting their internal order. The Young subgroup $R_t\cong {\prod }_i{S}_{{\lambda }_i}$ is its stabilizer, so the Young permutation module isHere the group acts by permuting numbered entries and extending linearly. A submodule is an invariant linear subspace. Give the tabloid basis its orthonormal symmetric bilinear form, denoted $⟨,⟩$. This form is nondegenerate and invariant under the group. Define the column antisymmetrizer and polytabloid byThus $S^{\lambda }$ is the Specht module, generated by any one $e_t$, because the group is transitive on numbered Young tableaux. The orthogonal complement $(S^{\lambda }{)}^{\perp }$ consists of vectors pairing to zero with every polytabloid.

The key calculation isTo prove it, first take $v=\{u\}$. If two entries of one column of $t$ occur in one row of $u$, their transposition $s$ fixes $\{u\}$; pairing the terms indexed by $g$ and $gs$ cancels $b_t\{u\}$. This also works in characteristic two, where the paired equal terms have sum zero. Otherwise each row of $u$ contains at most one entry from each column of $t$. The first row has ${\lambda }_1$ entries and there are ${\lambda }_1$ columns, so it uses every column once. Removing those entries, the second row must use exactly the columns of remaining positive height, and induction does the same for all rows. Consequently there is a unique $h\in C_t$ with $\{u\}=\{ht\}$. Uniqueness follows from $R_t\cap C_t=1$. In that caseLinearity proves the identity.

Now let $U\le M^{\lambda }$ be a submodule. If $U$ is not contained in $(S^{\lambda }{)}^{\perp }$, choose $v\in U$ and $t$ with $⟨v,e_t⟩\ne 0$. The displayed identity puts $e_t$ in $U$, so its group translates put all of $S^{\lambda }$ in $U$. This proves the James submodule theorem without a semisimplicity assumption.

Put $R=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. If $R\ne S^{\lambda }$, every proper submodule of $S^{\lambda }$ lies in $R$, by the same theorem. Hence $R$ is its unique maximal submodule, and $D^{\lambda }=S^{\lambda }/R$ is irreducible. If $R=S^{\lambda }$, the quotient is zero. This conclusion remains true after every field extension $E/F$: the integral tabloid construction becomes $M_E^{\lambda }=E{\otimes }_F{M}^{\lambda }$, and the radical of the restricted form is the kernel of its Gram matrix, so it commutes with extension of scalars. A nonzero Gram entry stays nonzero in $E$. Applying the theorem over $E$ makes $E{\otimes }_F{D}^{\lambda }$ irreducible. Thus a nonzero quotient is absolutely irreducible.

The precise nonvanishing criterion isA regular partition has fewer than $p$ rows of each positive length. Here is the integral pairing argument behind this condition. Write $m_j$ for the number of rows of length $j$. Expanding two polytabloids into tabloids, simultaneous permutations of equal-length rows partition of an integer all contributions to their pairing into free orbits of size ${\prod }_j{m}_j!$; the signs agree within each orbit because the same row permutation is applied in every relevant column on both sides. Therefore every pairing is divisible by ${\prod }_j{m}_j!$. If some $m_j\ge p$, the entire restricted form is zero modulo $p$. Conversely, reverse the order of the entries in each row of a tableau to obtain $t^{\ast }$. Counting the matching column permutations in the two expansions gives, up to the harmless overall sign,Indeed the $m_j$ equal rows of length $j$ can be matched independently in each of their $j$ positions; after these choices the decreasing row lengths force every other matching. For a regular partition this integer is nonzero modulo $p$, so the form and its radical quotient are nonzero. In characteristic zero it is nonzero for every shape. This supplies the requested justification of the criterion.

Since $S^{\mu }\subset M^{\mu }$, Question 2 shows that every composition factor of $S^{\mu }$ is $D^{\lambda }$ for a regular partition $\lambda ⊵\mu$. If $\mu$ is regular, its radical quotient $D^{\mu }$ occurs once: it occurs as a quotient, and its multiplicity is bounded above by the multiplicity one in $M^{\mu }$. All remaining factors have strictly dominating labels. A nonregular row has only strictly dominating regular labels.

Define $d_{\mu \lambda }=[S^{\mu }:D^{\lambda }]$. Order the row labels, all partitions of an integer of $n$, and the column labels, the regular partitions of an integer, by the same total order refining decreasing dominance. ThenThe matrix is generally rectangular. Its submatrix on regular rows is lower unitriangular in this decreasing convention; placing least dominant shapes first would transpose the visual triangular convention, not the indices.

This is the decomposition matrix, not just an abstract composition factor table. The standard polytabloid basis makes $S_{\mathbb{Z}}^{\mu }$ a free invariant integral lattice whose extension to characteristic zero is the ordinary irreducible $S^{\mu }$. Reducing this same lattice modulo $p$ gives exactly the modular $S^{\mu }$. On p-regular elements, reduction of its ordinary character is the sum of the Brauer characters of its composition factors, with coefficients $d_{\mu \lambda }$. These are the defining decomposition numbers; the resulting class in the modular representation ring is independent of the chosen invariant lattice.

For $S_3$, take rows $(3),(2,1),(1,1,1)$ in that order. In characteristic zero, or $p>3$, all three Specht modules are irreducible by Maschke's theorem, andFor $p=2$ the regular labels are $(3),(2,1)$ andIndeed the sign representation equals the trivial representation. The two-dimensional $S^{(2,1)}$ is the sum-zero subspace of the three-point permutation module. Over an algebraic closure, a $3$-cycle has two distinct nontrivial third-root eigenvalues there, and a transposition exchanges their lines. Neither line is invariant under $S_3$, proving absolute irreducibility.

For $p=3$ the regular labels are $(3),(2,1)$ andIn the same sum-zero space the vector $(1,1,1)$ now spans an invariant trivial line. Its one-dimensional quotient is sign: the determinant of the sum-zero representation is sign, and the determinant of the invariant line is trivial. Therefore $S^{(2,1)}$ has one trivial and one sign factor, with sign its simple head $D^{(2,1)}$. The bottom Specht module is itself sign. This proves every entry of the three possible matrices.

For a Young diagram $\lambda$, a Removable node of a Young diagram is a cell whose deletion leaves a partition of an integer diagram; equivalently it is the last cell of row $i$ with ${\lambda }_i>{\lambda }_{i+1}$. Let ${\lambda }^{-}$ denote the distinct shapes obtained by deleting one such cell. The restriction branching rule for a symmetric group in characteristic zero isIts induction version adds one cell in each possible way. Over an arbitrary field, restriction instead has a Specht filtration with these successive quotients, each once; it need not split.

To prove restriction, list the removable nodes from top to bottom. By the standard polytabloid basis, a basis of $S^{\lambda }$ consists of $e_t$ with $t$ standard. Its entry $n$ is necessarily at a removable node. Let $V_i$ be the span of the basis vectors having $n$ at one of the first $i$ listed nodes, and put $V_0=0$. These are $S_{n-1}$-invariant. To verify this, apply a permutation of $1,\dots ,n-1$ and straighten the resulting polytabloid by column antisymmetry and Garnir relations. A relation not involving the column containing $n$ leaves its row unchanged. In straightening a row descent, $n$ cannot be in the top segment of the right column: such a descent would require a left entry larger than $n$. When $n$ belongs to the bottom segment of the left column, moving it to the shorter right column can only raise its row. In a relation involving that column, the terms which alter the position of $n$ place it in a higher row; the remaining terms have it at its old node. Thus straightening never moves it past the current top-to-bottom cutoff.

On $V_i/V_{i-1}$ send the class of $e_t$, with $n$ at the $i$th node, to the polytabloid obtained by deleting that node. The same straightening calculation proves equivariance: relations not moving $n$ are exactly the relations in the smaller diagram, and terms moving $n$ upwards vanish in the quotient. Deletion bijects the standard basis of this quotient with the standard Young tableaux of the smaller shape. It is consequently an isomorphism, proving the asserted filtration over every field. In characteristic zero, Maschke's theorem splits the filtration and gives the direct sum. Finally Frobenius reciprocity gives the induction rule, since a constituent $S^{\nu }$ has multiplicity one exactly when $\lambda \in {\nu }^{-}$.

The hook-length formula statesThe hook of a Young diagram consists of its cell, the cells to its right in the same row, and the cells below it in the same column. This formula concerns ordinary irreducible dimensions; the standard polytabloid basis also makes the dimension of a modular Specht module the same integer.

All transpositions are conjugate, have order two, and generate $S_n$. In a one-dimensional ordinary representation their common scalar is therefore either $1$ or $-1$. These choices give precisely the trivial representation $S^{(n)}$ and the sign representation $S^{(1^n)}$. At $n=1$ they coincide. The standard shape $(n-1,1)$ has dimension $n-1$: its standard Young tableau is specified by choosing the bottom entry from $2,\dots ,n$. Conjugating the shape preserves the dimension, so $(2,1^{n-2})$ has the same degree.

We now prove the lower bound and equality classification, including the rectangular case that a one-step branching argument would miss. Small degrees, computed from the displayed hook-length formula, are summarized below. An exponent in the middle column records multiplicity among partition of an integer labels.

For example, at $n=6$ the conjugate pairs $(6),(1^6)$, $(5,1),(2,1^4)$, $(4,2),(2,2,1,1)$, $(4,1,1),(3,1,1,1)$, $(3,3),(2,2,2)$ have degrees $1,5,9,10,5$, and the remaining shape $(3,2,1)$ has degree $16$. These checks also identify every small equality case.

Induct on $n\ge 9$, assuming the lower bound for all smaller degrees. If a non-linear, nonstandard shape has at least two removable nodes, none of its predecessors is a single row or single column: adding one cell to a row or column could produce only a linear or standard shape. Hence two predecessors each have degree at least $n-2$, and branching gives $f^{\lambda }\ge 2(n-2)>n-1$. If it has only one removable node, it is a rectangle $(a^b)$ with $a,b\ge 2$. Its sole predecessor is $(a^{b-1},a-1)$, which has two removable nodes. Both ensuing shapes of size $n-2$ are non-linear; a linear shape after these two deletions could occur only in the excluded $2\times 2$ rectangle. Therefore two-step branching gives $f^{\lambda }\ge 2(n-3)>n-1$. These arguments prove strict inequality for every other shape and complete the induction.

Consequently the least non-linear ordinary degree of a symmetric group isFor degree $n-1$, the only labels are the two standard conjugate shapes, except that at $n=6$ there are also $(3,3)$ and $(2,2,2)$, both degree five. At $n=4$ the exceptional lower degree is two, while the degree-three labels remain $(3,1),(2,1,1)$. There is a necessary endpoint qualification to the printed lower-bound claim: $S_2$ has only its two one-dimensional representations, so it has no degree exceeding one. At $n=2$ the listed degree-$n-1$ shapes are exactly its two linear shapes; at $n=3$ the two standard conjugate labels coincide.

The kernel intersection theorem for Specht modules uses maps that move entries between adjacent rows. For $1\le i<\ell (\mu )$ and $0\le v<{\mu }_{i+1}$, let $\mu (i,v)$ replace the row lengths $({\mu }_i,{\mu }_{i+1})$ by $({\mu }_i+{\mu }_{i+1}-v,v)$, keeping every other length unchanged. This is allowed to be a composition rather than a decreasing partition of an integer, and a zero row can be omitted. For a tabloid with row sets $R_j$, defineby summing over all $v$-subsets $B\subset R_{i+1}$: the new row $i+1$ is $B$, the new row $i$ is $R_i\cup (R_{i+1}\setminus B)$, and all other rows are unchanged. These are module homomorphisms, since relabelling commutes with choosing subsets. The theorem isIt holds over every field. Column cancellation shows the inclusion of $S^{\mu }$ in each kernel; the full theorem identifies the common kernel with the Specht module.

For completeness, the semistandard homomorphism theorem concerns a partition of an integer $\lambda$, a composition $\alpha$, and fillings $T$ of shape $\lambda$ with ${\alpha }_j$ occurrences of $j$. Such a filling gives ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: replace its labels by the numbered entries of an input tableau, and sum the output tabloids over distinct permutations of the labels within each input row. Label $j$ assigns an entry to output row $j$. The use of distinct assignments avoids dividing by any factorial. Let ${\hat{\Theta }}_T$ be its restriction to $S^{\lambda }$. The restrictions indexed by Semistandard Young tableaux are linearly independent. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ when $p\ne 2$, or when $\lambda$ is two-regular. In characteristic two for a two-singular shape they need not span; omitting this exception would make the statement incorrect.

We now prove the invariant vector criterion for a Specht module directly from the first theorem. The group acts transitively on the tabloid basis of $M^{\mu }$, so any invariant vector has all coefficients equal. Thus $(M^{\mu }{)}^{S_n}=F{z}_{\mu }$, where $z_{\mu }$ is the sum of all its tabloids and is nonzero over every field. Fix $i,v$, write $a={\mu }_i$, $b={\mu }_{i+1}$, and count preimages of one output tabloid. The $v$ retained entries are prescribed; among the $a+b-v$ entries of the enlarged row, exactly $a$ must have come from its old row. HenceThe common-kernel criterion becomesfor every adjacent pair, where $j=b-v$.

Suppose $p>0$, and let $t$ be least with $b<p^t$. If $a\equiv -1(\mathrm{mod}{p}^t)$, its lowest $t$ base-$p$ digits are all $p-1$. For any $0<j<p^t$, adding $j$ to $a$ creates a carry at the lowest nonzero digit of $j$. At that position the digit of $a+j$ is smaller than the corresponding digit of $j$. Lucas's theorem therefore gives $\left(\genfrac{}{}{0ex}{}{a+j}{j}\right)\equiv 0$. Conversely, if one of the first $t$ digits $a_d$ is not $p-1$, take $j=p^d$. Minimality of $t$ gives $p^d\le b$, while addition at that digit creates no carry. Lucas then yieldsThis proves the claimed equivalence, and thereforeThe invariant space, when present, is one-dimensional. At the last row put ${\mu }_{i+1}=0$; then $t_i=0$ and the condition is vacuous. This congruence formulation is for positive characteristic. In characteristic zero the positive binomial coefficients cannot vanish, so an invariant vector exists precisely for the one-row shape $(n)$.

Finally suppose $\mu$ is regular. It has a unique simple head $D^{\mu }$, by Question 1. The given sign-twisted duality of Specht modules impliesThus $D^{\mu }$ is sign exactly when the conjugate shape ${\mu }^{\prime }$ satisfies the invariant congruences.

Here is an explicit solution of those congruences under regularity. Write the positive column heights of $\mu$ as $a_1\ge \cdots \ge a_l>0$, so ${\mu }^{\prime }=(a_1,\dots ,a_l)$. Regularity says $a_i-a_{i+1}<p$, including $a_l-a_{l+1}=a_l<p$, because this difference counts rows of length $i$. Starting with $0<a_l<p$, the congruence for its preceding row forces $a_{l-1}\equiv -1(\mathrm{mod}p)$. Since $a_l\le a_{l-1}<a_l+p\le 2p-1$, the only possibility is $a_{l-1}=p-1$. Repeating upwards forces every earlier height to be $p-1$. Consequentlywhere a zero final part is omitted. Conversely this shape satisfies every invariant congruence, so its conjugate is the desired regular label of the modular sign representation:Zero parts are omitted. For $p=2$ this is $(n)$, as it should be because sign is trivial. In characteristic zero the answer is $(1^n)$.

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.