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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.

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.

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.

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.

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.

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 $p-1$.

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.

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.

The symmetric group acts on tabloids by permuting their entries; extending linearly gives the permutation module $M^{\mu }$. Its dimension isA tableau of type $\mu$ contains exactly ${\mu }_i$ copies of the label $i$. Fix a numbering of the $\lambda$-diagram's cells by $1,\dots ,n$. For a filling $T$, let the $i$th row set of an associated $\mu$-tabloid be the set of cell numbers whose labels in $T$ equal $i$. This is a bijection from type-$\mu$ fillings to $\mu$-tabloids.

Define $\sigma T$ by moving labels along the numbered positions: its label at cell $j$ is the old label at cell ${\sigma }^{-1}j$. The associated row set is then the image under $\sigma$ of the old row set. Thus the bijection is equivariant, and its linear extension makes $F[{\mathcal{T}}_{\lambda ,\mu }]$ and $M^{\mu }$ isomorphic representations. Extend from group elements to the group algebra by linearity.

For type $(1^n)$, all labels are distinct, so a semistandard tableau is precisely a standard Young tableau. Thus $K_{\lambda ,(1^n)}=f^{\lambda }$, and Young's rule givesThis is also the decomposition of the regular representation, since tabloids with one entry per row are naturally indexed by all permutations.

We calculate the exact multiplicity without using Young's rule, via a two-subset incidence realization of a two-row Specht module. Let $V_2$ be the permutation representation on unordered pairs from a six-element set, and $V_1$ the natural permutation representation on its singletons. The equivariant incidence mapis surjective: the transpose map sends $\sum c_i{e}_i$ to the pair coefficients $c_i+c_j$, and all such sums can be zero only when every $c_i=0$ in characteristic zero. Thus $\dim \ker D=15-6=9$.

Identify the tabloids of shape $(4,2)$ with their two-element bottom rows. A polytabloid is a signed four-edge vector $e_{cd}-e_{ad}-e_{cb}+e_{ab}$, whose incidence at every vertex is zero. Hence $S^{(4,2)}\subseteq \ker D$. The standard polytabloid basis theorem and the hook-length formula give $\dim S^{(4,2)}=\left(\genfrac{}{}{0ex}{}{6}{2}\right)-\left(\genfrac{}{}{0ex}{}{6}{1}\right)=9$, so equality holds. By Maschke's theorem,Take $H={\Sigma }_3\times {\Sigma }_2\times {\Sigma }_1$, preserving blocks of sizes three, two and one. It has three orbits on singletons and five on pairs: pairs inside the first block, inside the second, and between each of the three distinct pairs of blocks. Thus $\dim V_1^H=3$ and $\dim V_2^H=5$. Averaging over $H$ makes taking fixed vectors exact in characteristic zero, so $\dim (S^{(4,2)}{)}^H=2$.

Finally $M^{(3,2,1)}={\mathrm{Ind}}_H^{{\Sigma }_6}1$. Frobenius reciprocity, and equality of invariant and coinvariant dimensions in characteristic zero, identify the required multiplicity with this two-dimensional fixed space. Thereforewhich in particular satisfies the requested lower and upper bounds.

The simple modules are indexed by the p-regular partitions $\lambda$ of $n$, meaning no part is repeated $p$ or more times. Equip the tabloid permutation module with the bilinear form invariant under the symmetric group making its tabloid basis orthonormal. The construction isThe standard modular Specht theorem says this quotient is nonzero precisely for p-regular $\lambda$, is then simple, and these quotients give all simple modules.

To verify distinctness of labels, use the standard dominance triangularity theorem: if $D^{\nu }$ occurs as a composition factor of $S^{\lambda }$, then $\nu \ge \lambda$; for a p-regular label the diagonal multiplicity is one. If $D^{\lambda }\cong D^{\mu }$, then $D^{\lambda }$ occurs in $S^{\mu }$, so $\lambda \ge \mu$, and the symmetric argument gives $\mu \ge \lambda$. Antisymmetry of the dominance order on partitions yields $\boxed{\lambda =\mu }$. These are the general representation-theoretic results allowed in this part.

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 permutation module $M^{\lambda }$ has a basis consisting of row tabloids of shape $\lambda$: tableaux are identified when each row has the same set of entries. A permutation acts by applying it to every entry, and the action is extended linearly over $F$. Its dimension is $n!/{\prod }_i{\lambda }_i!$.

Here the two singleton rows form an ordered pair of distinct entries, soA standard tableau of the same shape must put one in its top-left corner. Choose two entries from $2,\dots ,n$ for the two remaining boxes of the first column, in increasing order; all other entries then fill the top row in increasing order. Thus the standard-polytabloid basis gives

Write $v_{ij}$ for the tabloid with second-row entry $i$ and third-row entry $j$. Equivariance gives $\varphi (v_{ij})=w_i+w_j$ for every distinct pair. For a tableau $t$, the only nontrivial column antisymmetrizer is on the three entries in its first column. In its alternating sum, pair each column permutation with the permutation obtained by interchanging the two singleton-row positions. Their images under $\varphi$ are the same unordered sum of two $w$'s, while their signs are opposite. Hence $\varphi (e_t)=0$ for every polytabloid, proving $\boxed{S^{(n-2,1,1)}\subseteq \ker \varphi }$.

For distinct $i,j,k$,Each summand is an image and two is invertible in characteristic zero. Thus every basis vector of $M^{(n-1,1)}$ is in the image, proving surjectivity. rank-nullity theorem then gives

Give $M^{\lambda }$ the invariant symmetric bilinear form for which its tabloid basis is orthonormal. DefineIn characteristic $p>0$, this quotient is nonzero and simple when $\lambda$ is $p$-regular, meaning that no part occurs $p$ or more times. For other partitions it is zero, so the label “simple module” includes the nonzero/$p$-regular qualification.

The James submodule theorem states that for every $F{\Sigma }_n$-submodule $U\subseteq M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Applied to submodules of $S^{\lambda }$, it says that every proper submodule lies in the radical of the restricted form, giving the simplicity of the displayed nonzero quotient.

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 the characteristic-three Specht module for the two-by-two partition, reduce the matrices modulo three. The vector $e_1+e_2$ is fixed by every displayed transposition: $(12)$ and $(34)$ send it to $e_1-2e_2=e_1+e_2$, and $(23)$ interchanges its two terms. Thus its span is a proper nonzero trivial submodule.

On the quotient impose $e_2=-e_1$. Each generating transposition then acts as minus one, so the quotient is the sign representation. Consequentlyand the two composition factors are trivial and sign, each once. In the $D$ notation they are $D^{(4)}$ and $D^{(2,2)}$ respectively. Indeed the restricted tabloid form has Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$, whose characteristic-three radical is precisely the trivial line, so the simple quotient is $D^{(2,2)}$.

The extension does not split: the minus-one eigenspace of $\rho (12)$ is the line through $e_2$, but $\rho (23)$ sends that line to the line through $e_1$. Thus there is no sign submodule complementing the trivial line, consistent with the failure of semisimplicity.

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 $S^{(n)}$.

Use the standard invariant tabloid bilinear form, with the tabloid basis orthonormal. Each permutation acts by an orthogonal matrix, so $⟨gx,y⟩=⟨x,g^{-1}y⟩$. Inversion permutes $C_t$ and preserves its signs. ThereforeThe James submodule theorem states that for any submodule $U\le M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Use the allowed fact ${\kappa }_t{M}^{\lambda }=F{e}_t$. The coefficient of $\{t\}$ in $e_t$ is one, so the displayed adjoint identity gives ${\kappa }_{t}x=⟨x,e_t⟩e_t$. If $U$ is not orthogonal to $S^{\lambda }$, some $x\in U$ and some tableau $t$ have $⟨x,e_t⟩\ne 0$. Hence $e_t\in U$, and its permutation translates span $S^{\lambda }$. This proves the theorem.

Let $R=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. Every proper submodule of $S^{\lambda }$ lies in $R$ by the theorem. If the restricted form is nonzero, $R$ is proper, hence is the unique maximal submodule andIn the stronger nondegenerate case specified in the question, $R=0$, so $S^{\lambda }$ itself is simple. The nonzero-form version also covers the possibly degenerate case needed in the final calculation.

For shape $(n-1,1)$, identify $M$ with $F^n$ with basis $v_1,\dots ,v_n$. Its Specht module is the augmentation subspace $S=\{\sum a_i{v}_i:\sum a_i=0\}$, spanned freely by $v_i-v_n$, $1\le i<n$. Its Gram matrix is $G=I_{n-1}+J_{n-1}$, with diagonal entries two and off-diagonal entries one. Equivalently, $S^{\perp }$ in $M$ is the span of $v_1+\cdots +v_n$. This vector belongs to $S$ exactly when $p\mid n$. Thus the restricted radical has dimension one if $p\mid n$ and zero otherwise, givingFor $n\ge 3$ this quotient is nonzero and simple by the theorem. In the small exceptional case $n=2,p=2$, the canonical radical quotient is zero, as the same dimension formula states; it is then not a nonzero simple module. This qualification distinguishes the numerical formula from the nonzero-form hypothesis.