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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.

Rows of the symmetric-group decomposition matrix are all ordinary shapes, columns are regular partitions, and $d_{\mu \lambda }=[S^{\mu }:D^{\lambda }]$. A nonzero column antisymmetrizer of shape $\lambda$ on $M^{\mu }$ forces dominance: the first $r$ rows contain at most ${\sum }_j\min (r,{\lambda }_j^{\prime })$ column entries. For a regular shape its diagonal simple appears once, giving a rectangular triangular matrix and a unitriangular regular-row submatrix.

Fix a prime $p$. Replace each part $p^{a}m$, with $p\nmid m$, by $p^a$ copies of $m$. This is a size-preserving bijection from regular partitions to partitions of an integer whose parts are not divisible by $p$. To invert it, write each multiplicity of a part $m$ not divisible by $p$ in base $p$ and use its digit in position $a$ as the multiplicity of $p^{a}m$. Every resulting multiplicity is below $p$. The bijection explains why the number of simple symmetric-group modules equals the number of p-regular elements' conjugacy classes.

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.

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 dominance order on partitions is defined byWrite $\lambda ▹\mu$ when the shapes are distinct as well. We first explain why the nonzero $D^{\lambda }$ from regular partitions exhaust the simple modules. The argument also provides the required constraint on composition factors.

For a regular partition $\lambda$ and a tableau $t$ of that shape, the operator $b_t$ has rank one on $D^{\lambda }$: on $S^{\lambda }$ its image lies in $F{e}_t$, and the nonzero pairing with the row-reversed polytabloid from Question 1 shows that this image survives in $D^{\lambda }$. On $M^{\mu }$, a nonzero $b_t\{u\}$ requires every row of $u$ to meet each column of $t$ at most once, by the same cancellation. Consequently the first $r$ rows of $u$ contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$. Summing givesThus nonzero column action forces $\lambda ⊵\mu$. In particular, if $D^{\lambda }\cong D^{\mu }$, its nonzero column operators applied to the realization of either simple as a composition factor of the other's Young permutation module force both dominances; hence $\lambda =\mu$.

Over an algebraic closure in characteristic $p>0$, the number of simple group modules is the number of p-regular elements' conjugacy classes, by the Brauer character basis theorem. Such classes in $S_n$ have cycle lengths not divisible by $p$. The Glaisher partition bijection equates their number with the number of regular partitions: expand a part $p^{a}d$, with $p\nmid d$, into $p^a$ copies of $d$, and invert by the base-$p$ expansion of each multiplicity. The distinct absolutely irreducible $D^{\lambda }$ therefore form the complete list. Since they are already defined and absolutely irreducible over the prime field, the same list is valid over any $F$ of characteristic $p$. In characteristic zero, the analogous count uses all partitions of an integer and all conjugacy classes.

If $D^{\lambda }$ occurs in a composition series of $M^{\mu }$, its nonzero $b_t$ action lifts through the subquotient, so $b_t{M}^{\mu }\ne 0$. The inequality above proves $\lambda ⊵\mu$. When $\lambda =\mu$ is regular, $b_t$ has rank one on $M^{\mu }$ and rank one on $D^{\mu }$. An operator preserving a finite filtration has rank at least the sum of its ranks on the successive quotients: relative to an adapted basis its matrix is block triangular, and nonzero pivot minors of the diagonal blocks combine to a nonzero minor. Therefore $D^{\mu }$ can occur at most once. It occurs at least once as the quotient of $S^{\mu }\subset M^{\mu }$. We have provedFor a nonregular $\mu$, equality is impossible because $D^{\mu }=0$.

Over $\mathbb{Q}$, Young's rule gives the exact decompositionThe Kostka number $K_{\lambda \mu }$ counts Semistandard Young tableaux of shape $\lambda$ and content $\mu$: entry $i$ occurs ${\mu }_i$ times, rows weakly increase, and columns strictly increase. In particular $K_{\mu \mu }=1$ and nonzero multiplicity requires dominance.

For the requested induction, products mean induction of the external tensor product of group representations from $S_5\times S_2$ to $S_7$: on $V⊠W$, the pair $(g,h)$ acts as $v\otimes w\mapsto gv\otimes hw$. One can compute using Young's rule alone. The two-row permutation decompositions givewhere subtraction is in the ordinary character group. Reordering row blocks gives isomorphic permutation modules. Inducing after multiplication by the trivial $S^{(2)}$ gives $[M^{(3,2,2)}]-[M^{(4,2,1)}]$. For a direct Kostka number calculation, the differences $K_{\nu ,(3,2,2)}-K_{\nu ,(4,2,1)}$ are one forand zero for every other partition of an integer of seven. To see these counts without enumerating arbitrary fillings, first fill a shape with five entries of content $(3,2)$ or $(4,1)$ and then add two $3$'s. The added cells must be a horizontal strip. Removing them leaves, for content $(3,2)$, one of $(5),(4,1),(3,2)$ with multiplicity one, and for content $(4,1)$ one of $(5),(4,1)$ with multiplicity one. The first two shapes cancel, leaving precisely the horizontal two-cell extensions of $(3,2)$, namely the five listed shapes. ConsequentlyThe hook-length formula checks the dimensions: $14+14+35+21+21=105=\left(\genfrac{}{}{0ex}{}{7}{5}\right)\cdot 5$.

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.

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.

Let $\theta \in {\mathrm{End}}_{\mathbb{F}{S}_n}(S^{\lambda })$ and putSince $\lambda$ is $p$-regular, every $a_j<p$, so $q\ne 0$ in $\mathbb{F}$. Parts b and c give $b_{t}e(t^{\ast })=qe(t)$. Since $\theta$ commutes with the group-algebra action,The left side belongs to $b_t{M}^{\lambda }=\mathbb{F}e(t)$. Division by $q$ shows that $\theta (e(t))$ is a scalar multiple of $e(t)$. Part a(i) says that $e(t)$ generates $S^{\lambda }$, so $\theta$ is that scalar multiple of the identity. This proves the endomorphism theorem for a regular Specht module.

The partition is $p$-regular precisely whenthat is, when no part occurs $p$ or more times. This is the definition of a regular partition in characteristic of a field $p$.

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.