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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.

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 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: $14+14+35+21+21=105=\left(\genfrac{}{}{0ex}{}{7}{5}\right)\cdot 5$.

Use $[\lambda ]$ for the ordinary irreducible character of $S_{|\lambda |}$. Multiplication of different degrees is the product in the induction ring of symmetric-group characters: it induces an external tensor product of group representations. For $\alpha \subseteq \nu$, the skew Young diagram $\nu /\alpha$ is the set of cells of $\nu$ outside $\alpha$. A filling of it is a semistandard Young tableau when entries weakly increase along rows and strictly increase down columns. Its content $\beta$ means that $i$ occurs ${\beta }_i$ times. Its reading word takes each row right to left and the rows top to bottom. A lattice word has, in every initial segment, at least as many $i$'s as $(i+1)$'s for every $i$.

The Littlewood–Richardson rule sayswhere the Littlewood–Richardson coefficient $c_{\alpha \beta }^{\nu }$ counts such skew semistandard fillings with content $\beta$ and lattice reading word. If $\alpha$ is not contained in $\nu$, the coefficient is zero. Over characteristic zero this is an actual direct-sum multiplicity statement; over other fields the analogous statement describes a Specht filtration, not necessarily a decomposition into simple modules.

For the required coefficient, the five skew cells areTheir columns are all distinct. Content $(3,2)$ uses three $1$'s and two $2$'s. The first reading letter must be $1$ by the lattice word condition, and row monotonicity forces both top cells to be $1$. The remaining three cells, in reading order, can then be filled by $122$, $212$, or $221$. The resulting words $11122$, $11212$, $11221$ all satisfy the lattice inequalities, and exhaust the possibilities. Thus

For the determinantal form of a symmetric-group character, set $[0]=1$ in degree zero and $[r]=0$ for $r<0$. The one-row character $[r]$ is trivial. The one-row specialization of the Littlewood–Richardson rule is the Pieri rule: multiplication by $[r]$ adds a horizontal strip of $r$ cells, since all entries are $1$ and no column can contain two of them.

Here is an explicit determinant argument, so that the desired form is genuinely deduced. Under the Frobenius characteristic map, $[r]$ corresponds to the complete homogeneous symmetric polynomial $h_r$, whose monomials enumerate weakly increasing sequences of $r$ entries. Fix $N$ variables and $l$ rows. Let a directed lattice path run right or up from $A_i=(-i,1)$ to $B_j=({\lambda }_j-j,N)$, and assign weight $x_k$ to a horizontal step at height $k$ and weight one to a vertical step. The sum of its weights is $h_{{\lambda }_j-j+i}$. Expanding the determinant of this path matrix sums signed families connecting the sources to a permutation of the destinations.

Cancel intersecting families as follows. Choose the first intersection in a fixed order of vertices, and then the first pair of paths meeting there. Exchange their tails after that vertex. The total weight is unchanged, while the endpoint permutation changes by a transposition, reversing the sign. Choosing the intersection in this fixed order makes the operation an involution. The surviving families are nonintersecting; the ordered sources and ordered destinations force the identity endpoint pairing. Record the heights of the horizontal steps of path $i$ as row $i$ of a tableau. Rows weakly increase, and nonintersection of successive paths is precisely strict increase down the shared columns. Conversely every semistandard Young tableau gives one surviving path family. Thus the determinant has the semistandard-tableau generating function $s_{\lambda }$.

Young's rule identifies this generating function with the characteristic of $[\lambda ]$: its coefficient at a monomial of weight $\mu$ is $K_{\lambda \mu }$, the multiplicity obtained by pairing with the Young permutation module of type $\mu$. The Frobenius characteristic map is injective and respects induction products. Transposing the path matrix does not change its determinant, soThis is the character version of the Jacobi–Trudi identity.

For the example the matrix isExpanding gives the concise answerAt the identity the six induction-product dimensions are $210,140,105,35,42,21$, respectively; their alternating sum is $21$, agreeing with the hook-length formula for $(3,2,2)$.

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.

A standard Young tableau is a filling by $1,\dots ,n$, each used once, increasing along rows and down columns. A semistandard Young tableau is a filling by positive integers, weakly increasing along rows and strictly increasing down columns. Its content $\mu$ means that the entry $i$ occurs ${\mu }_i$ times. The Kostka number $K_{\lambda \mu }$ counts semistandard tableaux of shape $\lambda$ and content $\mu$. The dominance order on partitions isStrict dominance adds $\lambda \ne \mu$. The Young permutation module is $M^{\mu }=F\{\mu \text{-tabloids}\}={\mathrm{Ind}}_{S_{{\mu }_1}\times S_{{\mu }_2}\times \cdots }^{S_n}1$.

Let $D^{\lambda }$ be a composition factor of $M^{\mu }$. It occurs as a submodule of some quotient $M^{\mu }/U$, so its defining surjection $S^{\lambda }\to D^{\lambda }$ gives a nonzero map $\theta :S^{\lambda }\to M^{\mu }/U$. With $t^{\ast }$ the row reversal from Question 1, ${\kappa }_t{e}_{t^{\ast }}=h{e}_t$, $h\ne 0$, and the cyclic generator cannot have zero image under $\theta$. Therefore ${\kappa }_t$ acts nontrivially on $M^{\mu }/U$, and hence on $M^{\mu }$. The column-cancellation argument proves $\lambda ⊵\mu$.

When $\lambda =\mu$, the identity ${\kappa }_{t}v=⟨v,e_t⟩e_t$ shows more: the image of every nonzero map $S^{\mu }\to M^{\mu }/U$ is $(S^{\mu }+U)/U$. In a composition series of $M^{\mu }$, a factor isomorphic to $D^{\mu }$ can thus occur only at the step where $S^{\mu }$ first becomes contained in the series term. Once contained, every later such map would have zero image. Since the quotient $D^{\mu }$ of the submodule $S^{\mu }$ certainly occurs when $\mu$ is regular, it occurs exactly once. If $\mu$ is not regular, no simple is indexed by $\mu$. Consequently every other factor has a strictly dominating regular label. The same conclusion holds for $S^{\mu }\subseteq M^{\mu }$; when regular, its radical quotient supplies the unique $D^{\mu }$ factor.

The integral standard-basis theorem says that the standard polytabloids form a basis over every field. Thus $\dim S_F^{\mu }=f^{\mu }$ is independent of characteristic. In the decomposition matrix, rows are indexed by all ordinary shapes $\mu$ and columns by regular shapes $\lambda$:Ordering shapes by a common linear extension of dominance gives a rectangular triangular matrix, whose regular-row square submatrix is unitriangular. The standard basis also gives $f^{\mu }={\sum }_{\lambda }{d}_{\mu \lambda }\dim D^{\lambda }$.

Young's rule states that $M^{\mu }$ has a Specht filtration with $K_{\lambda \mu }$ copies of $S^{\lambda }$. In characteristic zero this is a direct-sum decomposition into irreducibles. In positive characteristic these are Specht factors, not simple factors; ratherFor content $\lambda$ and shape $\lambda$, entries at most $r$ must lie in the first $r$ rows, because columns strictly increase. Their number is exactly the size of those rows, so successive rows must be filled by their row number. Hence $K_{\lambda \lambda }=1$. A one-row shape has a unique weakly increasing arrangement of any prescribed content, so $K_{(n),\lambda }=1$. Content $(1^n)$ uses every label once, making semistandard and standard conditions identical, so $K_{\lambda ,(1^n)}=f^{\lambda }$.

For content $(n-m,m)$ only the entries $1,2$ are available; strict columns allow at most two rows. In shape $(n-j,j)$ every bottom entry is $2$ and every entry above it is $1$. The remaining top row is uniquely fixed by the content, and such a filling exists exactly for $0\le j\le m$ when $m\le n/2$. Young's rule therefore giveswhere multiplication denotes induction from the indicated Young subgroup, not pointwise multiplication of characters of the same group. The dimension of this permutation module is $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$. Subtracting the analogous identity with $m-1$ yieldswith $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$ for $m=0$. This dimension holds over every field.

The preceding parts give a simple Specht module for each partition and show that different partitions give nonisomorphic simples. The conjugacy classes of $S_n$ are indexed by partitions of $n$, namely cycle types. We use the standard finite-group character theorem: the number of irreducible complex characters equals the number of conjugacy classes, and they form an orthonormal basis of complex class functions. Consequently the ${\chi }^{\lambda }$ are the complete list of irreducible characters of $S_n$.

By Maschke's theorem, the Young permutation module $M^{\mu }$ decomposes as a direct sum of these simple modules. For a semisimple complex module, Schur lemma makes the multiplicity of a simple module equal to the dimension of the homomorphism space from that simple; character orthogonality identifies the same number with the character inner product. HencePart (ii) makes $a_{\lambda \mu }=0$ unless $\lambda ⊵\mu$, and gives $a_{\mu \mu }=1$. Taking characters provesThus the permutation-character decomposition is triangular in the dominance order on partitions, with diagonal multiplicity one. These are the dominance and diagonal-multiplicity consequences of Young's rule, obtained here without assuming its full multiplicity formula.

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.

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.

The character of a Young permutation module counts fixed tabloids and is integer-valued. Dominance order on partitions makes its decomposition into Specht modules triangular with diagonal multiplicity one. Descending induction in any order refining dominance therefore expresses each irreducible Specht character as an integer combination of permutation characters, proving integrality.

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 a composition $\lambda$ of $n$, a Young subgroup permutes independently the blocks of a set partition of $\{1,\dots ,n\}$ with sizes ${\lambda }_i$, and is isomorphic to $S_{{\lambda }_1}\times \cdots \times S_{{\lambda }_r}$. Reordering the block sizes gives a conjugate subgroup. Inducing its trivial representation gives a Young permutation module.