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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.

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.

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.

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.