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A skew Young diagram $\lambda /\mu$ is the difference of two nested Young diagrams. A semistandard Young tableau is a filling of precisely those cells by positive integers, weakly increasing along each occupied row and strictly increasing down each occupied column. It can be viewed as its filled diagram, or equivalently as a chainwhere ${\nu }^{(r)}/{\nu }^{(r-1)}$ is a horizontal strip consisting of the cells labeled $r$. Indeed, the weak row and strict column inequalities ensure that the cells labeled at most $r$, together with $\mu$, form a partition diagram, and equal labels cannot share a column. Conversely such a chain gives the tableau.

For a tableau $T$, let $c_r(T)$ count its entries $r$, put $x^T={\prod }_r{x}_r^{c_r(T)}$, and defineThere are finitely many tableaux for any fixed number of allowed letters, and the sum has total degree $|\lambda |-|\mu |$.

To prove symmetry directly, use a Bender-Knuth involution for each adjacent pair of letters $i,i+1$. Pair any $i$ and $i+1$ that occur in the same column and leave them fixed; strictness makes them consecutive in that column. In a row the remaining unpaired entries have the form $i^a(i+1{)}^b$, between any fixed paired letters. Replace that block by $i^b(i+1{)}^a$. Rows stay weakly ordered. A free $i$ can be changed to $i+1$ without equaling the entry below, because such a below entry would have been paired with it; likewise changing a free $i+1$ to $i$ cannot equal the entry above. Other neighboring letters are outside $\{i,i+1\}$, so strict columns are preserved. The paired cells remain the same, and repeating the operation restores each free block. The paired letters contribute equally to the two contents, while every free block exchanges its contributions. Thus the involution is a weight-preserving bijection after exchanging $x_i,x_{i+1}$. Adjacent exchanges generate all finite variable permutations, provingThe skew Jacobi–Trudi identity gives a second proof, since its determinant entries are already symmetric.

Define the Kostka number $K_{\lambda \mu }$ to count straight-shape Semistandard Young tableaux with shape $\lambda$ and content $\mu$, meaning ${\mu }_i$ copies of label $i$. Suppose such a tableau exists. Strict columns imply that an entry in row $r$ is at least $r$. Therefore all entries from $1,\dots ,r$ lie in the first $r$ rows, givingHenceFor content $\lambda$, the ${\lambda }_1$ copies of $1$ have nowhere to go except row one and fill it. The ${\lambda }_2$ copies of $2$ then fill row two, and induction fills row $r$ with $r$. This filling exists and is unique, so $\boxed{K_{\lambda \lambda }=1}$.

Symmetry identifies equal coefficients for each rearrangement of a content, henceThe monomial symmetric functions form a basis. In any total order extending dominance, this change-of-basis matrix is triangular with diagonal one. It is invertible in each degree, so

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

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.

Apply Young's rule, stated asHere the Kostka number counts Semistandard Young tableaux of shape $\rho$ with $n-k$ entries equal to $1$ and $k$ entries equal to $2$. Their rows weakly increase and their columns strictly increase. A column can therefore have at most two cells, so only a partition of an integer $\rho =(n-j,j)$ can occur.

In such a semistandard Young tableau, all $j$ entries in the second row must be $2$ and all $j$ entries above them must be $1$. The whole first row is then fixed by the content: its first $n-k$ entries are $1$ and its remaining $k-j$ entries are $2$. This is possible exactly when $j\le k$ and $j\le n-k$. Because $k\le n/2$, the second inequality follows from the first. There is exactly one filling for every $0\le j\le k$, and none for any other shape.

Consequently each displayed Specht module has multiplicity one, proving the two-row Young permutation module decompositionFor $j=0$, the zero second part is omitted. As a dimension check, the Hook-length formula gives $\dim S^{(n-j,j)}=\left(\genfrac{}{}{0ex}{}{n}{j}\right)-\left(\genfrac{}{}{0ex}{}{n}{j-1}\right)$, with $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$; the sum telescopes to $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, the dimension of the original permutation representation.

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.

Young's rule givesthe Kostka number counting semistandard tableaux of shape $\alpha$ and content $\lambda$.

A semistandard Young tableau fills a Young diagram with positive integers, permitting repeated entries, so that entries weakly increase along rows and strictly increase down columns. Its content records the number of occurrences of each entry. The Kostka number $K_{\lambda \mu }$ counts such fillings of shape $\lambda$ and content $\mu$.

For $0\le k\le n/2$, over the complex numbers,By Young's rule, the multiplicity is the Kostka number for content $(n-k,k)$. A semistandard Young tableau on just the entries $1,2$ has at most two rows. For shape $(n-j,j)$, every lower entry is $2$ and every entry above it is $1$; the remaining top row is uniquely fixed by the content. Such a filling exists precisely for $j\le k$. Thus each displayed Specht module occurs once. A zero second part is omitted.

Over the complex numbers,where the Kostka number $K_{\alpha \lambda }$ counts semistandard tableaux of shape $\alpha$ and content $\lambda$. It is nonzero only when $\alpha$ dominates $\lambda$.