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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.

Use the pair of partitions for a generalized Specht module convention appropriate to generalized Specht modules. Put $a={\mu }^{♯}$ and $b=\mu$. Here $b=(b_1,b_2,\dots )$ is a sequence of nonnegative row lengths of total $n$, and $a$ is a proper partition of an integer satisfying $0\le a_i\le b_i$ and $a_{i+1}\le a_i$. The row-length sequence $b$ is allowed to be a composition; it need not decrease. The first $a_i$ cells of row $i$ are marked. This construction is not an arbitrary pair of unrelated partitions of an integer and is not simply a skew diagram $b/a$.

A word has type $b$ if it contains $b_i$ occurrences of $i$. Read it from left to right. Every occurrence of $1$ is good; an occurrence of $i+1$ is good exactly when there have so far been more good occurrences of $i$ than good occurrences of $i+1$. Otherwise it is bad. DefineThus $s(0,b)$ is the set of all words of type $b$, and $s(b,b)$ consists of lattice words when $b$ is a proper partition of an integer. Marking the whole first row makes no difference because every $1$ is good.

For a Young tableau $T$ with row lengths $b$, let $C_T^a$ permute labels within each column of its marked subdiagram, fixing all unmarked labels. Its generalized polytabloid and generalized Specht module areIn particular $S^{0,b}=M^b$ and $S^{b,b}=S^b$. The Young tableau shape in this definition must be the row sequence $b$.

We give the filtration argument, keeping its combinatorial and module-theoretic steps separate. Normalize $a_1=b_1$, and if $a\ne b$ choose the first row $c>1$ with $a_c<b_c$. Then $a_{c-1}=b_{c-1}$. Define $A_c(a,b)=(a+e_c,b)$ if $a_{c-1}>a_c$; otherwise the add branch is empty. Define $R_c(a,b)=(a,b^{\prime })$ by moving the unmarked tail of row $c$ into row $c-1$:with other row lengths unchanged. Re-normalize the first marked row if necessary. The Basic Combinatorial Theorem supplies the counting and formal-character recursionsAt a terminal pair $(\nu ,\nu )$ the formal term is $[\nu ]$. Thus the second recursion specifies the Specht multiplicities by the terminal leaves, with repeated leaves counted separately.

Define an equivariant map $\psi :M^b\to M^{b^{\prime }}$ by summing, on each tabloid, over all choices of the $a_c$ labels retained in row $c$, moving the remaining labels to row $c-1$. Column cancellation givesFor the first identity, terms that put two marked labels of one column in the raised row cancel in pairs; the surviving marked antisymmetrizer is the one for the raised pair, and every target generator has a preimage. For the second, the extra marked cell makes such a collision unavoidable, so the sum vanishes. Also antisymmetrizing the extra cell expresses its generator as a sum of old generators, proving containment in $S^{a,b}$.

There is a characteristic-independent lower bound $\dim S^{a,b}\ge |s(a,b)|$. Given a word in $s(a,b)$, place the label $j$ in its letter's row: good occurrences fill that row from the left and bad occurrences fill it from the right. The marked cells form increasing column chains because each good $i+1$ has an earlier unmatched good $i$. Its generalized polytabloid has the original row-assignment tabloid with coefficient one; every other term changes an earlier member of a marked chain. Ordering these row assignments gives a triangular coefficient matrix. The resulting vectors are independent over every field.

Start with $(0,v)$, where equality holds because the tabloid basis is indexed by all words of type $v$. Any pair is reached from such a pair by a sequence of add and raise operations: reverse a raise by splitting a raised tail, and reverse an add by unmarking a cell; row by row these recover an entirely unmarked composition. Suppose equality holds at a pair. The map and kernel containment above giveEvery inequality is therefore equality. In particular the containment is the whole kernel, andis exact. The process terminates because each add marks an extra cell and each raise decreases the total row index of unmarked cells. Induction from the terminal Specht modules, splicing their filtrations through these exact sequences, proves a Specht series with factors precisely $[0{]}^{[{\mu }^{♯},\mu ]}$. No splitting of these sequences in positive characteristic is asserted.

Applying the same marking, raising and straightening procedure to the two sets of column antisymmetrizers in $(S^{\mu }⊠S^{\lambda }){\uparrow }^{S_{r+s}}$, where $|\mu |=r$, $|\lambda |=s$, leaves the cells of $\mu$ fixed and records the added cells by their row labels from $\lambda$. The column relations require strict increase down columns, the row relations weak increase along rows, and the good-letter matching in a tableau word test requires a lattice word. Thus the surviving terminal multiplicity is the number $c_{\mu \lambda }^{\nu }$ of Semistandard Young tableaux of skew shape of shape $\nu /\mu$ and content $\lambda$ whose word, read right-to-left along successive rows from top to bottom, has at least as many $i$'s as $(i+1)$'s in every prefix. The same triangular leading-tabloid argument identifies each quotient with $S^{\nu }$. Consequently the Littlewood–Richardson rule isIt describes ordinary irreducible constituents in characteristic zero and Specht filtration factors over arbitrary $F$.

For induction by one letter, content $(1)$ permits exactly one added cell and each multiplicity is one. A Specht series for the requested induced module has successive quotients, ordered by addable nodes from bottom to top,Equivalently there is $0=N_0\subset N_1\subset N_2\subset N_3\subset N_4=S^{(4,2,2,1)}{\uparrow }^{S_{10}}$ with these four quotients in order. Their dimensions $567,300,768,525$ sum to $2160=10\cdot 216$, the dimension of the induced module.

Let $0\ne f:S^{\lambda }\to M^{\mu }$ be a module homomorphism. Since $S^{\lambda }$ is simple, $f$ is injective. Fix a $\lambda$-tableau $t$; its nonzero polytabloid $e_t$ then has $f(e_t)\ne 0$. The sign sum in the column group satisfiesbecause every coefficient in the square receives $|C_t|$ equal contributions. Thus ${\kappa }_t{e}_t=|C_t|e_t$, and equivariance givesThe column antisymmetrizer therefore acts nontrivially on $M^{\mu }$, so some basis tabloid $[s]$ has ${\kappa }_t[s]\ne 0$. The stated dominance from a nonzero column antisymmetrizer yieldsFor $\mu =\lambda$, the same calculation forces $f(e_t)=|C_t{|}^{-1}{\kappa }_{t}f(e_t)$ into the one-dimensional space $\mathbb{C}{e}_t$. Write $f(e_t)=c{e}_t$. Since $e_t$ generates the Specht module, equivariance gives $f(g{e}_t)=cg{e}_t$ for every $g$, so $f$ is $c$ times the inclusion $S^{\lambda }\hookrightarrow M^{\lambda }$. That inclusion is nonzero, henceAs a useful consequence, if $S^{\lambda }\cong S^{\mu }$, their inclusions into $M^{\lambda },M^{\mu }$ give dominance both ways. Antisymmetry of the dominance order on partitions then gives $\lambda =\mu$, so different partitions produce nonisomorphic simple modules.

The permutation character ${\xi }^{\mu }(g)$ is the number of $\mu$-tabloids fixed by $g$, since the trace of a permutation matrix counts fixed basis elements. It is therefore an integer. Arrange the partitions in decreasing lexicographic order. If $\lambda$ strictly dominates $\mu$, the first part at which they differ is larger for $\lambda$, so this ordering places $\lambda$ before $\mu$.

For the first partition $(n)$, there is just one tabloid, and ${\chi }^{(n)}={\xi }^{(n)}=1_{S_n}$ is integer-valued. At a general partition $\mu$, the decomposition in (iii) can be rearranged asBy induction, all characters in the sum are already integer-valued; the permutation character and all coefficients are also integral. Thus ${\chi }^{\mu }$ is integer-valued. Finite induction proves that symmetric-group characters are integer-valued:The diagonal coefficient one is essential: it lets us solve for ${\chi }^{\mu }$ without division, preserving integrality.

For integer partitions $\lambda ,\mu$ of $n$, padded by zero parts, the dominance order on partitions isA $\lambda$-tableau fills the Young diagram of $\lambda$ with $1,\dots ,n$, each once. A tabloid $[t]$ remembers the sets of entries in each row, ignoring their order within that row. The Young permutation module $M^{\lambda }=\mathbb{C}{X}_{\lambda }$ has these tabloids as basis, with $S_n$ acting by permuting the entries.

The row and column stabilizers $R_t,C_t$ independently permute entries within the rows and columns of $t$. The column antisymmetrizer is ${\kappa }_t={\sum }_{c\in C_t}\mathrm{sgn}(c)c$, and $e_t={\kappa }_t[t]$ is the associated polytabloid. Since $C_{gt}=g{C}_t{g}^{-1}$ and conjugation preserves permutation sign,Thus the Specht module is the submoduleEvery tableau is $gt$ for a fixed $t$, so any single $e_t$ generates it under $S_n$. Also $C_t\cap R_t=\{1\}$, since a permutation preserving both rows and columns must fix every cell. The tabloids $[ct]$ for $c\in C_t$ are consequently distinct, and $e_t$ has coefficient $1$ at $[t]$; in particular $e_t\ne 0$.

We use the following precise column antisymmetrizer facts. If some row of a $\mu$-tableau $s$ contains two entries from one column of $t$, the transposition interchanging them pairs equal tabloids with opposite signs in ${\kappa }_t[s]$, so ${\kappa }_t[s]=0$. Otherwise each row of $s$ meets each column of $t$ at most once. The first $r$ rows of $s$ therefore contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, whenceThis states and explains dominance from a nonzero column antisymmetrizer: ${\kappa }_t[s]\ne 0$ implies $\lambda ⊵\mu$.

When $\mu =\lambda$, the nonzero column antisymmetrizer criterion says more: either ${\kappa }_t[s]=0$, or a column permutation takes $[s]$ to $[t]$, and ${\kappa }_t[s]=\pm e_t$. Thus ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$. This follows by matching the row positions within each column when all intersections have size at most one; the equal row and column sizes force the Ferrers incidence pattern.

For the inner product, use the positive Hermitian extension of the orthonormal tabloid basis, linear in its first argument. It has the supplied self-adjointness of ${\kappa }_t$. Its orthogonal complement of $S^{\lambda }$ agrees with that obtained from the complex tabloid bilinear form, since the spanning polytabloids have real coefficients. Because the coefficient of $[t]$ in $e_t$ is $1$, self-adjointness and the one-dimensional image give the useful identityIndeed, the scalar multiplying $e_t$ is the $[t]$-coefficient of ${\kappa }_{t}u$, which is $⟨{\kappa }_{t}u,[t]⟩=⟨u,{\kappa }_t[t]⟩=⟨u,e_t⟩$.

The Young permutation module $M^{\lambda }$ has a basis of tabloids, equivalently ordered row sets of sizes ${\lambda }_1,\dots ,{\lambda }_r$. A tabloid is fixed by $\sigma$ precisely when every cycle of $\sigma$ lies entirely in one row. Assigning a length-$q$ cycle to row $j$ contributes $x_j^q$. Distinct cycles can be assigned independently, so the character isPadding to $n$ variables with zero row sizes gives exactly the same coefficient. Explicitly, the character of a Young permutation module isHere $a_{qj}$ counts the length-$q$ cycles assigned to row $j$. Rows remain distinguished even when their sizes are equal, so there is no further division by permutations of equal rows.

The Specht module construction gives a nonzero inclusion $V^{\lambda }\hookrightarrow M^{\lambda }$ over the complex numbers: a polytabloid has coefficient one at its original tabloid, because the row and column stabilizers have trivial intersection, and its span is the corresponding irreducible representation. This establishes $a=M(\lambda ,\lambda )\ge 1$ without using the multiplicity-one conclusion. Similarly $a^{\prime }=M({\lambda }^{\prime },{\lambda }^{\prime })\ge 1$.

By the corrected tensor identity for induced representations and the conjugate-partition sign twist, the multiplicity of $V^{\lambda }$ in ${\tilde{M}}^{{\lambda }^{\prime }}$ is $a^{\prime }$. The character inner product already computed is the sum of products of multiplicities of common irreducible representations. ThusBoth are positive integers, so $a=a^{\prime }=1$. In particularThis deduction avoids assuming the full Young's rule, which would already contain the answer. Alternatively, the permitted one-dimensional R-module homomorphism space gives the same character inner product immediately.

Fix $\lambda ⊢n$ and $\rho ⊢n-1$. Write $\rho \nearrow \mu$ when a single addable node of a Young diagram turns $\rho$ into $\mu$. For each positive part ${\lambda }_i$, let ${\lambda }^{\downarrow i}$ be the partition of an integer obtained by decreasing that part by one, sorting, and omitting zeros. The Vershik linear relations areEqual row lengths must be counted separately on the right. Equivalently, if $c(\lambda ,\gamma )$ is the number of rows whose decrement produces $\gamma$, the right side is ${\sum }_{\gamma \nearrow \lambda }c(\lambda ,\gamma )M(\rho ,\gamma )$. There is no multiplicity coefficient on the left because irreducible restriction branching rule for a symmetric group is simple.

To prove the relations, restrict the Young permutation module $M^{\lambda }$ to the subgroup $S_{n-1}$ fixing $n$. The tabloids split into orbits of a group action according to which labeled row contains $n$. Removing that entry gives, for row $i$, the Young permutation module with composition $\lambda -e_i$, which is isomorphic to the one for its sorted partition. HenceIts $V^{\rho }$ multiplicity is the right side. On the other hand decompose $M^{\lambda }$ into irreducible representations first and apply the restriction branching rule for a symmetric group to each summand. Its $V^{\rho }$ multiplicity is then the left side. Equality proves the relation.

For example $\lambda =(2,2)$ restricts to $2M^{(2,1)}$, so the coefficient is two rather than one. For $\lambda =(1^n)$, the regular representation, the relation reduces to ${\sum }_{\mu :\rho \nearrow \mu }\dim V^{\mu }=n\dim V^{\rho }$. These checks emphasize why counting distinct resulting partitions without their row multiplicities would give a false recurrence.

The Young permutation module $M^{(n-k,k)}$ is the symmetric-group subset permutation representation on the $k$-element subsets of $\{1,\dots ,n\}$: the second row of a tabloid specifies the subset. For $k\ge 1$, restriction of a character to $S_{n-1}$ fixing $n$ splits these subsets into two orbits of a group action, according as they contain $n$.

The corresponding permutation representations are on the $k$-subsets and the $(k-1)$-subsets of $\{1,\dots ,n-1\}$. Their Young permutation modules have two row sizes $n-1-k,k$ and $n-k,k-1$, respectively. Sort each pair into decreasing order and omit any zero part; swapping the two row labels gives an isomorphic permutation representation. This sorting matters at $n=2k$.

We use Young's rule: over the complex numbers, the multiplicity of $S^{\rho }$ in $M^{\eta }$ is the Kostka number $K_{\rho \eta }$, and it can be nonzero only if $\rho$ dominates $\eta$ in the dominance order on partitions. Since each $\eta$ above has at most two parts and size $n-1$, dominance givesThus $\rho$ has at most two nonzero parts. The character inner product measures this constituent multiplicity, soIf $k=0$, the original and restricted modules are the trivial representations, giving the same conclusion. For $n=1$, there is no partition of $n-1=0$ with three parts, so the assertion is vacuous.

Fix a Young tableau $t$ of shape $\lambda$, let $C(t)$ be its column stabilizer, and putThese are the Column antisymmetrizer of a Young tableau and its polytabloid. The Specht module is generated by $e_t$ as a cyclic module, so the nonzero R-module homomorphism $\varphi$ has $\varphi (e_t)\ne 0$. Because its extension $\Phi$ commutes with the group algebra action,Expand $\Phi (\{t\})$ in the tabloid basis of the Young permutation module $M^{\mu }$. At least one tabloid $\{u\}$ therefore satisfies $b_t\{u\}\ne 0$.

If a row of $u$ contains two entries from the same column of $t$, their transposition $\tau \in C(t)$ fixes $\{u\}$. Pairing the terms indexed by $g$ and $g\tau$ in $b_t\{u\}$ makes them cancel. This remains valid in characteristic two, since the paired coefficients sum to zero there as well. Thus each row of $u$ meets each column of $t$ in at most one entry.

The first $r$ rows of $u$ consequently contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, where ${\lambda }^{\prime }$ is the conjugate partition. HenceBy the dominance order on partitions, this provesThis is dominance from a nonzero column antisymmetrizer. The extension is what allows us to apply $b_t$ to an element of $M^{\mu }$ without dividing by the column stabilizer's order, which might vanish in the given field.