Tabloid

A tabloid is an equivalence class of tableaux under permutations within rows. The tabloids of shape $\lambda$ form a transitive $S_n$-set.

The Young permutation module is the permutation module on the $\lambda$-tabloids.

For $n\ge 3$, let $S_n$ permute the coordinates of $M=F^n$. Write $T$ for the constant vectors and $U=S^{(n-1,1)}$ for the augmentation submodule. These and zero and $M$ are all the submodules. Indeed a vector with two unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$; the orbit of such a difference spans $U$. A submodule without such a vector lies in $T$. If the characteristic does not divide $n$, then $M=T\oplus U$ and $U$ is simple. If it divides $n$, the lattice is the chain $0<T<U<M$, with $U/T$ simple of dimension $n-2$ and trivial end factors. The direct-sum projection in the first case is $x\mapsto x-({\sum }_i{x}_i)1/n$.

A tabloid is fixed by a permutation exactly when each of its cycles lies wholly within one row. Assigning the distinct length-$q$ cycles to rows gives the displayed coefficient. More explicitly, sum ${\prod }_q{m}_q!/{\prod }_j{a}_{qj}!$ over nonnegative integers $a_{qj}$ with ${\sum }_j{a}_{qj}=m_q$ and ${\sum }_{q}q{a}_{qj}={\lambda }_j$. Equal-length rows remain distinguished. Thus the character counts fixed ordered row sets, rather than unordered set partitions.

Over the complex numbers, the signed Young permutation module isThe isomorphism is the tensor identity for induced representations. The partition index is unchanged by this twist; conjugation instead changes the labels of its irreducible constituents. For the conjugate partition ${\lambda }^{\prime }$, the character inner product of $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ is one: Mackey restriction formula reduces it to the unique zero-one matrix with row margins $\lambda$ and column margins ${\lambda }^{\prime }$.

Over the complex numbers,where the Kostka number $K_{\alpha \lambda }$ counts semistandard tableaux of shape $\alpha$ and content $\lambda$. It is nonzero only when $\alpha$ dominates $\lambda$.

Let $M(\mu ,\lambda )$ be the multiplicity of $V^{\mu }$ in the complex Young permutation module $M^{\lambda }$. For $\lambda ⊢n$ and $\rho ⊢n-1$,Here $\nearrow$ means adding one cell, and zero parts are omitted. Equal row lengths contribute separately on the right. Restrict $M^{\lambda }$ to $S_{n-1}$ and separate the tabloid orbits by the row containing $n$ to obtain the right side. Decompose into irreducibles and use the restriction branching rule for a symmetric group to obtain the left side. Equating multiplicities proves the formula.

For $0\le k\le n/2$, over the complex numbers,By Young's rule, the multiplicity is the Kostka number for content $(n-k,k)$. A semistandard Young tableau on just the entries $1,2$ has at most two rows. For shape $(n-j,j)$, every lower entry is $2$ and every entry above it is $1$; the remaining top row is uniquely fixed by the content. Such a filling exists precisely for $j\le k$. Thus each displayed Specht module occurs once. A zero second part is omitted.

The Kostka number $K_{\alpha \lambda }$ is the number of semistandard Young tableaux of shape $\alpha$ and content $\lambda$: rows are weakly increasing and columns are strictly increasing.

Over the complex numbers, the multiplicity of $S^{\mu }$ in the Young permutation module $M^{\lambda }$ equals $K_{\mu \lambda }$. Its character is ${\mathrm{Ind}}_{S_{\lambda }}^{S_n}1$, so Frobenius reciprocity expresses this multiplicity as the average of the restricted irreducible character over the Young subgroup. For $\lambda =(2,1^{n-2})$ that subgroup contains only the identity and one transposition, giving $({\chi }_{\mu }(1)+{\chi }_{\mu }((12)))/2$. For $\lambda =(1^n)$ it is trivial, giving the character degree. This supplies a calculation from ordinary character values without enumerating semistandard tableaux. An indexing convention writing the multiplicity as $m_{\lambda ,\mu }$ reverses the conventional Kostka indices.

For $n\ge 4$, identify $M^{(n-2,1^2)}$ with the permutation module on ordered pairs $(i,j)$ with $i\ne j$. Let $p_1(i,j)=e_i$, $p_2(i,j)=e_j$, and $q(i,j)=\{i,j\}$. With $U=\ker p_1$ and $V=U\cap \ker p_2$, one has a Specht filtrationwhose successive quotients areThe same construction for $n=3$ omits the zero $S^{(n-2,2)}$ factor.

For a tableau $t$, its polytabloid is

A standard polytabloid is the polytabloid associated with a standard Young tableau. The standard polytabloids of one shape form a basis of the corresponding Specht module. For linear independence, sort each row of a tabloid increasingly and compare row-reading words lexicographically. In the column antisymmetrization of a standard tableau, the identity term is its unique least row word: in the first row affected by a nontrivial column permutation, each moved-in entry is larger than the entry it replaces. Hence its coefficient is one, and all other terms are larger. Distinct standard tableaux have distinct least tabloids, so a linear relation is impossible by taking its least nonzero term. Spanning is the straightening part of the standard polytabloid basis theorem.

The column antisymmetrizer of a Young tableau $t$ is the group algebra elementOn the Young permutation module it has one-dimensional image $\mathbb{F}e(t)$.

Let $t$ be a Young tableau of shape $\lambda$, and let $u$ be a Young tableau of shape $\mu$, both on $n$ entries. If the Column antisymmetrizer of a Young tableau $b_t$ satisfies $b_t\{u\}\ne 0$ over any field, then $\lambda$ dominates $\mu$ in the dominance order on partitions.

If two entries from one column of $t$ lie in the same row of $u$, their transposition fixes the tabloid $\{u\}$. Pairing terms of $b_t$ by this transposition gives $b_t\{u\}=0$, including in characteristic two. Otherwise the first $r$ rows of $u$ contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$. Thusfor every $r$, proving dominance.

For tableaux $v,w$ of the same shape, $b_v\{w\}\ne 0$ exactly when no row of $w$ contains two entries from one column of $v$. In that case a column permutation $h\in C(v)$ satisfies $h\{v\}=\{w\}$, and $b_v\{w\}=\pm e(v)$.

The Specht module $S^{\lambda }$ is the span of the polytabloids of shape $\lambda$. It is cyclic, generated by any one polytabloid, and over $\mathbb{C}$ it is irreducible.

For the symmetric group on five letters, $S^{(4,1)}$ is the four-dimensional augmentation representation. In characteristic three it is simple, since its only possible proper nonzero submodule would be the constant line, which is excluded by $5\ne 0$ in the field. Thus it is $D^{(4,1)}$. If this module occurs as a submodule of $S^{(3,2)}$, the quotient has dimension one, since the standard polytabloid basis gives $\dim S^{(3,2)}=5$.

A transposition has trace one on $S^{(3,2)}$, by the Murnaghan–Nakayama rule, and trace two on $S^{(4,1)}$, by subtracting the trivial line from the five-point permutation representation. These integral traces reduce to the same elements in characteristic three. The quotient trace is therefore minus one, so its one-dimensional action is the sign representation. In particular an arbitrary permutation acts on it by its sign. This identifies the quotient action without asserting that the original short exact sequence splits.

Over a field of characteristic zero with $n\ge 4$, let $V_2$ be the permutation representation on unordered two-element subsets and $V_1$ the representation on singletons. The equivariant incidence map $D:V_2\to V_1$ sends $e_{ij}$ to $e_i+e_j$. It is surjective: if coefficients $c_i$ annihilate its image, then $c_i+c_j=0$ for all distinct $i,j$, forcing all coefficients to vanish. Thus its kernel has dimension $\left(\genfrac{}{}{0ex}{}{n}{2}\right)-n$.

A polytabloid of shape $(n-2,2)$, identified by its bottom-row pair, is an alternating four-edge vector $e_{cd}-e_{ad}-e_{cb}+e_{ab}$. Its incidence is zero. The hook-length formula gives the same dimension $\left(\genfrac{}{}{0ex}{}{n}{2}\right)-n$ for this Specht module, proving the identity. Maschke's theorem yields $V_2\cong S^{(n-2,2)}\oplus V_1$.

For a Young subgroup with block sizes ${\mu }_1,\dots ,{\mu }_l$, there are $l$ singleton orbits and $\left(\genfrac{}{}{0ex}{}{l}{2}\right)+\#\{i:{\mu }_i\ge 2\}$ pair orbits. Taking fixed vectors and using Frobenius reciprocity therefore gives the multiplicity of $S^{(n-2,2)}$ in $M^{\mu }$ as $\left(\genfrac{}{}{0ex}{}{l}{2}\right)+\#\{i:{\mu }_i\ge 2\}-l$. This derives the multiplicity without Young's rule; block sizes $(3,2,1)$ give exactly two.

Use the standard polytabloids for tableaux with rows $12\mid 34$ and $13\mid 24$, as an ordered basis $e_1,e_2$. Adjacent transpositions act byIn characteristic three their common fixed line is $F(e_1+e_2)$. On its quotient every transposition acts by minus one, giving the sign representation. Thus the displayed sequence has the trivial representation and sign as its composition factors, each once. It is nonsplit: the minus-one eigenspace of $\rho (12)$ is $F{e}_2$, which is not preserved by $\rho (23)$.

The natural tabloid form restricts to Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$. Its radical in characteristic three is exactly that fixed line. Therefore $D^{(2,2)}$ is the sign quotient, while the trivial factor is $D^{(4)}$. This gives an explicit small failure of semisimplicity in modular symmetric-group representation theory.

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.

Let $n\ge 4$ and let the coefficient field have characteristic two. The subgroup fixing one permutes the standard hook basis $e_{1ij}$ indexed by unordered pairs from $\{2,\dots ,n\}$. Its fixed space is the line generated by $v={\sum }_{2\le i<j\le n}{e}_{1ij}$. Record a hook tabloid by its second- and third-row entries. In $v$, every recorded ordered pair avoiding one occurs once, and every pair involving one occurs $n-2$ times: a polytabloid is the sum of the six ordered pairs from its three column entries. For odd $n$, $v$ is therefore the sum of all tabloids and is fixed by the full symmetric group. For even $n$, it is the sum of the tabloids avoiding one and is moved by $(1,2)$. No other full-group fixed vector is possible because the point-stabilizer fixed space is one-dimensional. In characteristic different from two, swapping $i,j$ negates the $e_{1ij}$ coefficient and forces it to vanish, so even the point-stabilizer fixed space is zero.

The polytabloids $e_t$ indexed by standard Young tableaux of shape $\lambda$ form a basis of $S^{\lambda }$ over every field, and a basis of its integral lattice over $\mathbb{Z}$. Column antisymmetry and Garnir relations straighten any polytabloid into this set. Ordering leading tabloids by the rows containing successive entries makes the corresponding coefficient matrix triangular with diagonal entries one, proving independence. In particular the dimension is the standard-tableau count in every characteristic.

Take two adjacent columns of a numbered Young tableau, with a bottom segment $A$ of the left column and a top segment $B$ of the right column such that $|A|+|B|$ exceeds the height of the left column. Alternating over permutations of $A\cup B$, modulo permutations preserving $A$ and $B$ separately, gives a signed relation among polytabloids. The full alternation vanishes because each tabloid has two of these entries in one row. Solving for a nonstandard polytabloid replaces its offending row inversion by earlier column-standard tableaux. The relation is proved over the free integral tabloid lattice, where cancellation and division by nonzero integer factors are valid; reduction then makes it valid in every characteristic. Together with column antisymmetry these relations give the standard polytabloid basis.

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

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.

In positive characteristic $p$, $(S^{\lambda }{)}^{S_n}$ is nonzero exactly when ${\lambda }_i\equiv -1(\mathrm{mod}{p}^{t_i})$ for every adjacent pair, where $t_i$ is least with ${\lambda }_{i+1}<p^{t_i}$. The only invariant line in $M^{\lambda }$ is generated by the sum $z_{\lambda }$ of all tabloids. The map ${\psi }_{i,v}$ sends it to $\left(\genfrac{}{}{0ex}{}{{\lambda }_i+{\lambda }_{i+1}-v}{{\lambda }_i}\right)z_{\lambda (i,v)}$. Thus the kernel intersection theorem for Specht modules gives vanishing of $\left(\genfrac{}{}{0ex}{}{{\lambda }_i+j}{j}\right)$ modulo $p$ for $1\le j\le {\lambda }_{i+1}$. By Lucas's theorem, all these binomial coefficients vanish exactly when the last $t_i$ base-$p$ digits of ${\lambda }_i$ are $p-1$.

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.

For $A_n$, the least non-linear ordinary degree is $3$ at $n=4,5$ and $n-1$ for $n\ge 6$. For $n\le 3$ no non-linear irreducible exists. The degree-$n-1$ representation is unique for $n\ge 2$, except $n=3$ where it does not exist and $n=6$ where two distinct degree-five irreducibles occur. The proof uses tableau branching to bound the degrees of nonstandard shapes and of the half-degree constituents of self-conjugate shapes, with the small cases checked by the hook-length formula.

For a row composition $b$ and a weakly decreasing marking $a$ with $0\le a_i\le b_i$, mark the first $a_i$ cells of each row. Antisymmetrize within the marked columns and span the resulting polytabloids in the Young permutation module $M^b$. This gives $S^{a,b}$; $S^{0,b}=M^b$ and $S^{b,b}=S^b$ for a proper partition $b$. Its characteristic-independent Specht filtration is obtained by successively marking a cell or raising an unmarked row tail, with column-cancellation maps and the good-letter counting recursion.

The first component is a proper partition marking an initial segment of each row of the second component. The second component may be a composition rather than a decreasing partition. This convention specifies a generalized Specht module and differs from specifying a skew shape by two nested proper partitions.

Read a word of type $b$ from the left. Every $1$ is good; an $i+1$ is good when the number of previous good $i$'s exceeds the number of previous good $(i+1)$'s. The set $s(a,b)$ contains words with at least $a_i$ good $i$'s for each $i$. Their matched chains yield independent marked-column polytabloids; column antisymmetrization and the add-or-raise recursion give $\dim S^{a,b}=|s(a,b)|$ over every field.

A Specht filtration is a finite chain of submodules whose successive quotients are Specht modules. In positive characteristic it need not split and is not a composition series, because its Specht factors may themselves be reducible. Dimensions and ordinary lifted characters can still be added along the filtration.

For induction of $S^{\mu }⊠S^{\lambda }$ to the symmetric group on $|\mu |+|\lambda |$ letters, the Specht filtration multiplicity of $S^{\nu }$ is the number of Semistandard Young tableaux of skew shape $\nu /\mu$, content $\lambda$, and lattice word read from right to left in each row, taking rows from top to bottom. In characteristic zero these are irreducible constituents; in positive characteristic they remain Specht factors rather than simple constituents. Content $(1)$ gives the add-one-cell induction rule.

The nonnegative integer $c_{\alpha \beta }^{\nu }$ is the multiplicity of $S^{\nu }$ in ${\mathrm{Ind}}_{S_{|\alpha |}\times S_{|\beta |}}^{S_{|\nu |}}(S^{\alpha }⊠S^{\beta })$ in characteristic zero. The Littlewood–Richardson rule counts it by Semistandard Young tableaux of skew shape $\nu /\alpha$, content $\beta$, and lattice word in right-to-left, top-to-bottom reading order. For example $c_{(3,2,1),(3,2)}^{(5,3,2,1)}=3$: the two new top cells must be ones, and the remaining reading letters can be $122,212,221$. This coefficient also multiplies Schur functions in the induction ring of symmetric-group characters.

In the induction ring of ordinary symmetric-group characters, $[\lambda ][r]$ is the sum, with multiplicity one, of $[\nu ]$ for which $\nu /\lambda$ is a horizontal strip of $r$ cells. Here $[r]$ is the trivial representation of $S_r$. In the Littlewood–Richardson rule, content $(r)$ means every skew entry is $1$; strict column increase is possible exactly when there is at most one new cell in every column. This proves the rule.

In the complex symmetric-group algebra, normalize a Young symmetrizer by $e_t=h_t/H_{\lambda }$. The row-column collision lemma proves $e_{t}A{e}_t=\mathbb{C}{e}_t$, so $e_t$ is a primitive idempotent and $A{e}_t$ is an irreducible left module. Different shapes are separated by the vanishing corner $h_{\lambda }A{h}_{\mu }=0$ for $\lambda >\mu$. Counting conjugacy classes then proves that these modules exhaust the simple modules.

Let $U$ be a submodule of the Young permutation module $M^{\lambda }$ over any field. Then eitherfor the tabloid bilinear form. The key identity is $b_{t}u=⟨u,e(t)⟩e(t)$: if one pairing is nonzero, the cyclic generator $e(t)$ and hence the whole Specht module lies in $U$.

Over a field of characteristic zero,The map from $M^{{\lambda }^{\prime }}$ sending the tabloid of $g{t}^{\prime }$ to $ge(t)\otimes ge(u)$ is a surjection whose kernel is $(S^{{\lambda }^{\prime }}{)}^{\perp }$.

Over the complex numbers, restriction from $S_n$ to $S_{n-1}$ is multiplicity-free:Equivalently, one removes one Removable node of a Young diagram in every possible distinct way.

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.

For $\lambda ⊢n$ with $n\ge 2$,Indeed, adjoining the trivial character to the standard character gives the point-permutation character, and Frobenius reciprocity turns its multiplicity in the self-product into the norm of the multiplicity-free restriction.

Order tabloids lexicographically by the row containing $1$, then the row containing $2$, and so on. For a standard tableau $t$, the tabloid $\{t\}$ occurs with coefficient one in $e(t)$ and precedes every other tabloid occurring in it. Distinct standard tableaux have distinct leading tabloids, so their polytabloids are linearly independent.

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.

For $n\ge 3$, the least ordinary irreducible dimension exceeding one is $n-1$, except at $n=4$, where $S^{(2,2)}$ has dimension two. The dimension-$n-1$ shapes are $(n-1,1)$ and its conjugate partition, with the additional shapes $(3,3)$ and $(2,2,2)$ at $n=6$. For $n=2$ there is no non-linear irreducible. A proof uses the restriction branching rule for a symmetric group: a nonstandard shape with at least two corners has two non-linear predecessors, and a rectangular shape has one predecessor which in turn has two non-linear predecessors. This bounds its degree strictly above $n-1$ once $n\ge 9$; the smaller cases follow directly from the hook-length formula.

Right multiplication by $e_t=h_t/H_{\lambda }$ on the group algebra is an idempotent with image $A{e}_t$. Every diagonal entry in the permutation basis is the coefficient of the identity in $e_t$, namely $1/H_{\lambda }$. Thus its rank equals its trace, giving $\dim S^{\lambda }=n!/H_{\lambda }$. Triangular standard-tableau ideals and the Robinson–Schensted correspondence identify this dimension with the standard-tableau count.

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.

At a noncorner cell of a Young diagram, this random walk chooses uniformly one of the other cells in its hook of a Young diagram. It stops at a corner. Every step moves right or down, so it terminates. Its corner probabilities provide a probabilistic proof of the hook-length formula.

For a target corner $(\alpha ,\beta )$ in the Greene–Nijenhuis–Wilf hook walk, put $A_i=h(i,\beta )-1$ and $B_j=h(\alpha ,j)-1$. The probability of terminating there from $(a,b)$ factors as $F_a{G}_b$, where $F_{\alpha }=G_{\beta }=1$, $F_a=A_a^{-1}{\prod }_{i=a+1}^{\alpha -1}(1+A_i^{-1})$ for $a<\alpha$, and $G_b=B_b^{-1}{\prod }_{j=b+1}^{\beta -1}(1+B_j^{-1})$ for $b<\beta$. It is zero when the start is outside the target's northwest rectangle. The factorization uses $h(i,j)-1=A_i+B_j$ and the first-step recurrence.

If a permutation has a $k$-cycle and remaining cycle type $\rho$, thenwhere the sum is over removable rim hooks of length $k$.

Multiplication of an alternant by $p_r$ is the sum of alternants obtained by raising one row exponent by $r$. Equal exponents give zero. Otherwise sort the shifted beta number of a partition back into decreasing order. The resulting partition adds a size-$r$ rim hook; the number of crossed exponents is its height and supplies the determinant sign. This proves the formula through the bialternant formula. Iteration counts signed border-strip tableaux.

With $m$ beta numbers, the staircase has set $\{0,2,\dots ,2m-2\}$. Lowering a bead by a positive even number either hits an occupied bead or a negative position, so no even-size rim hook can be removed. The adjoint of multiplication by $p_r$ under the Hall inner product of symmetric functions is $r\partial /\partial p_r$. The power-sum border-strip multiplication formula therefore gives zero derivative with respect to every even power sum. Since the rational symmetric-function algebra is freely generated by the power sums, the stated conclusion follows.

Let $\kappa$ be the core of a partition $\lambda$ at prime modulus $p$, with quotient component sizes $w_i$ and total weight $w$. Complete $p$-hook removal sequences number $N_p(\lambda )=\left(\genfrac{}{}{0ex}{}{w}{w_0,\dots ,w_{p-1}}\right){\prod }_i{f}^{{\lambda }^{(i)}}$. Their signs agree because abacus bead-crossing parities telescope to the permutation taking the initial labeled beads to the packed core order. Repeated Murnaghan–Nakayama rule gives the displayed formula for any permutation $\pi$ on the remaining letters.

The parity of the sum of the leg lengths in any sequence that removes all $e$-hooks from $\lambda$ is independent of the sequence. Its sign ${\epsilon }_e(\lambda )$ is therefore well defined and supplies the common sign in repeated applications of the Murnaghan–Nakayama rule.

For even $n$, the virtual charactervanishes on every permutation having an odd cycle. Under the Frobenius characteristic map, the Jacobi–Trudi identity identifies its characteristic with the degree-$n$ part of $H(t)H(-t)$, which contains only products of even-indexed power sums.

At a permutation whose cycle lengths are the principal hook lengths of $\lambda$, the Murnaghan–Nakayama rule has a unique complete removal sequence. Consequently ${\chi }^{\lambda }$ has value $1$ or $-1$ there.

For a partition $\lambda$, the irreducible character ${\chi }^{\lambda }$ vanishes on every cycle type containing an even part exactly when $\lambda$ is a staircase $(m,m-1,\dots ,1)$. A staircase has only odd hook lengths. Conversely, vanishing first forces $\lambda$ to be self-conjugate; applying the Murnaghan–Nakayama rule at the largest even hooks then forces consecutive row lengths.

Conjugating a Young diagram twists its complex Specht module by the sign representation:

For the alternating expression ${\psi }^{\lambda }$ indexed by an integer composition, swapping two adjacent entries of $\lambda -\mathrm{id}$ negates ${\psi }^{\lambda }$. If two entries coincide the expression vanishes; otherwise sorting produces, up to sign, the irreducible character indexed by the resulting partition.

Over a field of characteristic $p$, the invariant tabloid form on $S^{\lambda }$ may be degenerate. Its radical is $S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$.

The Specht module $S^{(n-1,1)}$ is the augmentation subspace of the natural permutation module $F^n$, with the standard tabloid bilinear form. Its orthogonal complement in $F^n$ is the constant-vector line. That line lies in the augmentation subspace exactly when the field characteristic divides $n$, giving the displayed radical. Equivalently, the basis $v_i-v_n$ has Gram matrix $I_{n-1}+J_{n-1}$. The James submodule theorem makes every nonzero radical quotient simple, with dimension $n-2$ when $p\mid n$ and $n-1$ otherwise. The case $n=2,p=2$ gives the zero quotient rather than a nonzero simple module.

The tabloid bilinear form makes the tabloid basis orthonormal and restricts to an invariant bilinear form on each Specht module. In positive characteristic of a field, its radical controls the corresponding simple quotient.

A partition is $p$-regular when no part occurs $p$ or more times.

Write $n=q(p-1)+r$, $0\le r<p-1$. The regular partition labelling the sign representation in characteristic $p>0$ is $\lambda =((q+1{)}^r,q^{p-1-r})$, with zero parts removed. Its conjugate partition is $((p-1{)}^q,r)$, which satisfies the invariant vector criterion for a Specht module. The sign-twisted duality of Specht modules turns this invariant line into a sign quotient of $S^{\lambda }$, necessarily its unique simple head. Conversely, regularity says consecutive column heights differ by less than $p$; combining this with the invariant congruences forces these heights to be $p-1$ except possibly the last. For $p=2$, the label is $(n)$ because sign equals the trivial representation.

For a $p$-regular partition,is nonzero and absolutely irreducible. Distinct $p$-regular partitions label nonisomorphic simple modules.

The James submodule theorem follows from ${\kappa }_{t}v=⟨v,e_t⟩e_t$. A nonzero restricted form therefore gives a simple radical quotient. If $m_j$ counts equal-length rows, all integral polytabloid pairings are divisible by ${\prod }_j{m}_j!$, while pairing with row reversal gives ${\prod }_j(m_j!{)}^j$. Thus the form is nonzero modulo $p$ exactly for regular partitions. The construction and the proof commute with every field extension, so the quotient is absolutely irreducible, including over the prime field.

If $\lambda =(n^{a_n},\dots ,1^{a_1})$ is $p$-regular, thenFor a tableau $t$ and its row reversal $t^{\ast }$,which is nonzero in characteristic $p$. Applying a column antisymmetrizer to an endomorphism at $e(t^{\ast })$ therefore forces its value on the cyclic generator $e(t)$ to be scalar.