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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.

Over ${\mathbb{F}}_2$, the Specht module $S^{(1,1)}$ for $S_2$ is generated by the nonzero polytabloid $\{1\mid 2\}+\{2\mid 1\}$, which the transposition fixes. It is therefore a one-dimensional trivial module, as is the Young permutation module $M^{(2)}$. There is a nonzero equivariant map $S^{(1,1)}\to M^{(2)}$, although $(1,1)$ does not dominate $(2)$. The familiar dominance implication for a map from an entire permutation module cannot be transferred automatically to a map defined only on a Specht submodule: in modular characteristic an equivariant extension to the larger module need not exist.

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.

In characteristic zero, inducing a Specht module from ${\Sigma }_{n-1}$ to ${\Sigma }_n$ gives one copy of each Specht module obtained by adding one box in a valid Young-diagram position. By Frobenius reciprocity, its multiplicity in the induced module equals the multiplicity of $S^{\lambda }$ in the restriction of that larger Specht module. The restriction branching rule for a symmetric group says this multiplicity is one exactly when removing one corner gives $\lambda$, and zero otherwise. Semisimplicity in characteristic zero turns these multiplicities into the displayed direct-sum decomposition.

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.

A tabloid records the sets of entries in each row of a tableau. Let $C_t$ be the subgroup permuting entries independently within each column. The polytabloid isand the Specht module $S^{\lambda }$ is spanned by these vectors. A standard polytabloid is one whose tableau increases along rows and down columns.

Order tabloids by first sorting each row increasingly, then comparing the resulting row-reading words lexicographically. For a standard tableau $t$, every nonidentity column permutation produces a strictly larger word. In the first row affected by that permutation, each replacement comes from farther down the same column and is larger than the entry it replaces; all earlier rows remain unchanged. Sorting the affected row therefore increases its word. Also no nontrivial column permutation fixes its row sets, since a cell is the unique intersection of its row and column. Hence the least tabloid in $e_t$ is $\{t\}$ with coefficient one.

Distinct standard tableaux have distinct row tabloids. In a nonzero linear relation among their polytabloids, choose the least leading tabloid with a nonzero coefficient. None of the vectors with larger leading tabloids can contribute to it, so its coefficient cannot cancel. This contradiction proves linear independence of the standard polytabloids.

Young's rule states that in characteristic zero,where the Kostka number $K_{\lambda \mu }$ counts Semistandard Young tableaux of shape $\lambda$ and type $\mu$: rows are weakly increasing, columns strictly increasing, and label $i$ occurs ${\mu }_i$ times. The summands are irreducible Specht modules, and the coefficients are nonnegative integer multiplicities.

Realize $S^{(n-1,1)}$ as the augmentation submoduleof the natural permutation module: its polytabloids are the differences $e_i-e_j$, which span exactly $W$. Let $U$ be a submodule. If it contains a vector $v$ with two unequal coordinates $v_i,v_j$, thenThe coefficient is nonzero, so $e_i-e_j\in U$; its symmetric-group orbit spans $W$, forcing $U=W$. Therefore every proper nonzero submodule consists only of constant-coordinate vectors. Those form at most a one-dimensional trivial submodule. If such a trivial submodule is absent, no proper nonzero submodule exists, so the Specht module is simple.

Let $f^{\lambda }$ be the number of standard tableaux of shape $\lambda$. First the associated polytabloids are linearly independent. Order tabloids lexicographically by reading each row after sorting its entries. For a column-increasing tableau $t$, the term $\{t\}$ in $e_t$ has coefficient one; every other tabloid term has larger row word. To see this, take the first row changed by a nontrivial column permutation. Earlier rows are fixed entry by entry, since each entry belongs to one original column. The changed row can only receive entries from lower positions of those increasing columns, so after sorting it is lexicographically larger. In a linear relation between standard polytabloids, choose the smallest row word with a nonzero coefficient. Its own tabloid coefficient cannot be cancelled by a larger standard polytabloid, a contradiction. Thus $\dim S^{\lambda }\ge f^{\lambda }$.

Work first over $\mathbb{C}$. The RSK bijection between permutations and pairs of standard tableaux of the same shape gives ${\sum }_{\lambda ⊢n}(f^{\lambda }{)}^2=n!$. The assumed pairwise nonisomorphic irreducible Specht modules, and the decomposition of the regular representation with irreducible multiplicities equal to their dimensions, give ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2\le n!$. Combining the two inequalities forces equality term by term: $\dim S^{\lambda }=f^{\lambda }$. Hence the independent standard polytabloids are a basis.

The same conclusion holds over every characteristic-zero field. All polytabloid coordinates in the tabloid basis are integers, so the ranks of their coordinate matrices are the same over any characteristic-zero field as over $\mathbb{C}$. In particular the dimension equality and independence transfer without requiring an embedding of the whole field into $\mathbb{C}$.

The equality ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2=n!$ exhausts the dimension of the regular representation. There is no room for another irreducible constituent, since it would contribute its positive squared dimension. ThusEach has the standard-polytabloid basis just proved. The usual split characteristic-zero statement follows over any characteristic-zero field because these integral Specht models remain irreducible after extension to an algebraic closure and already exhaust all dimensions there. In the literal wording of the source, polytabloids themselves are vectors; it is the Specht modules with those bases that form the family of representations.

The two partition diagrams are transposes, so their characteristic-zero Specht modules differ by tensoring with the sign representation. A transposition has sign minus one. ConsequentlyThe middle character is $n-3$, since the standard representation is the point-permutation representation minus its trivial line, and a transposition fixes $n-2$ points.

Choose the two standard tableauxand use their polytabloids $e_1,e_2$, with column-vector coordinates. Writing $[ab\mid cd]$ for the tabloid with those two row sets givesThe coefficients at $[12\mid 34]$ and $[13\mid 24]$ prove independence over every field. Applying the generators directly givesand $(34)$ acts in the same way as $(12)$. Their span is invariant under the generators, contains $e_1$, and contains its whole orbit, hence is the Specht module. Thus it is a basis without needing a characteristic-zero argument. The matrices areThese integral matrices may be reduced in the chosen characteristic.

For a tableau $t$, let $C_t$ be its column stabilizer and $e_t={\sum }_{\sigma \in C_t}\mathrm{sgn}(\sigma )\{\sigma t\}$ its polytabloid. The Specht module is the span of all such polytabloids inside the tabloid permutation module.

A tabloid with second row of size $k$ is determined by that subset, so $\dim M^{(n-k,k)}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The two-row hook formula givesAt $k=3$ this provesBy Maschke's theorem, characteristic-zero permutation modules are completely reducible, so their Specht submodules are summands. The nonzero sum of all three-subset tabloids is fixed by every permutation; its span is a copy of the trivial module $S^{(n)}$.