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Use nonnegative integer matrix entries, including zero, and English Young diagram conventions: rows run left to right and columns run down. Replace $A$ by a two-line array containing $a_{ij}$ copies of $\left(\genfrac{}{}{0ex}{}{i}{j}\right)$, sorted lexicographically by upper letter $i$ and then lower letter $j$. Finite support makes the array finite.

In the RSK algorithm, insert each lower letter $j$ into $P$ by row insertion. In the first row replace the leftmost entry strictly greater than $j$, carrying the replaced entry into the next row; if there is no such entry, append $j$ and stop. Repeat in subsequent rows until a new cell is made. Put the upper letter $i$ in that new cell of the recording tableau $Q$.

The insertion-path facts used here are the following: row insertion preserves weak rows and strict columns; inserting $u$ followed by $v\ge u$ puts the latter new cell strictly to the right of the former; reverse insertion undoes insertion; and the reverse paths from a right-to-left horizontal strip recover its input letters in weakly decreasing order. These standard path facts ensure that $P$ is a semistandard Young tableau. The upper letters are nondecreasing, so the entries of $Q$ are weakly increasing in rows. Equal upper letters have sorted lower letters, and hence create a horizontal strip; they cannot occupy the same column. Thus $Q$ is also a semistandard Young tableau, of exactly the same shape as $P$.

Here is the inverse on any such pair. Select a largest entry $i$ in $Q$, and among the cells carrying it select the rightmost one. It is an outer corner: there can be no larger entry below it, and any entry to its right would be another maximal entry farther right. Delete that cell from $Q$ and reverse-insert the value at its cell in $P$. In each preceding row replace the rightmost entry strictly smaller than the carried value and carry that old entry upward. The value emerging from the first row is $j$. Record $\left(\genfrac{}{}{0ex}{}{i}{j}\right)$ and repeat. Equal maximal entries form a horizontal strip, so the reverse-path fact gives their $j$'s in weakly decreasing order. Reversing the recovered list therefore gives exactly a lexicographically sorted two-line array. The procedures are mutually inverse, proving the RSK correspondence bijection, including the zero matrix and the pair of empty tableaux.

Insertion only moves previous entries and adds one copy of its new letter; recording adds one copy of its upper letter. Consequently

For the trace assertion, use the insertion-path form of the RSK growth-diagram local rule. Let $\rho ,\mu ,\nu ,\lambda$ be the shapes from the northwest, northeast, southwest and southeast matrix prefixes around entry $a_{ij}$. Padding row lengths with zeros, the rule isThis local rule counts the two merging insertion paths row by row: the entry supplies the initial carry, the longer extension supplies the current row, and the overlap of the extensions beyond $\rho$ is bumped to the next row. It is a standard insertion-path fact being used in addition to the preceding monotonicity statements.

Define $o(\gamma )={\gamma }_1-{\gamma }_2+{\gamma }_3-\cdots$. Each column of the Young diagram contributes its alternating vertical sum, which is one for odd length and zero for even length. Thus $o(\gamma )$ counts odd-length columns. For symmetric $A$, transposition symmetry of RSK gives equal shapes for the two prefixes off the diagonal, so $\mu =\nu$ at entry $a_{ii}$. The rule becomesTaking alternating sums makes the two sums involving $\mu$ cancel, leavingStarting from the empty northwest corner and proceeding along the diagonal proves the trace and odd columns in symmetric RSK identityThis also accommodates arbitrary diagonal entries, rather than only permutation matrices.

For integer partitions $\mu \subseteq \lambda$, set $\lambda /\mu =\{(i,j):{\mu }_i<j\le {\lambda }_i\}$. A semistandard Young tableau on it has weakly increasing rows, strictly increasing columns and positive integer entries. Its weight is $x^T={\prod }_{(i,j)}{x}_{T(i,j)}$; their sum is the Skew Schur function $s_{\lambda /\mu }$. DefineThe Jacobi–Trudi identity, including its skew form, iswhere $\ell$ is any common padded length at least $\ell (\lambda )$. Taking $\mu$ empty gives the ordinary Schur function formula.

For a proof, first use finitely many variables $x_1,\dots ,x_N$. A directed northeast path from $A_j=({\mu }_j-j,1)$ to $B_i=({\lambda }_i-i,N)$ has one horizontal step for each unit of the horizontal displacement. Give such a step at height $r$ weight $x_r$, and vertical steps weight one. The weakly increasing sequence of step heights shows that the sum of its weights is $h_{{\lambda }_i-{\mu }_j-i+j}$.

The determinant result needed is the nonintersecting lattice-path determinant: in an acyclic weighted network, the determinant of the single-path sums is the signed sum over vertex-disjoint path families, with sign the permutation of the paired endpoints. To prove it, expand the determinant and its products of path sums. An intersecting family has a first intersection in a fixed topological order. Choose the first two indexed paths there and swap their tails. Their total weight stays the same, their endpoint permutation changes by one transposition, and the operation is its own inverse. All intersecting families cancel. Only the disjoint families remain.

In this planar network the starting and finishing points occur in the same strict left-to-right order, so disjoint paths have identity endpoint pairing. For the path in row $i$, associate its horizontal step from coordinate $j-i-1$ to $j-i$ with cell $(i,j)$. Reading its height gives the tableau entry. Weak increase follows from northward movement; nonintersection with the path for the next row is exactly the strict column condition. Conversely a tableau constructs the paths, and the strict column inequalities keep them disjoint. This weight-preserving bijection proves the determinant formula. Arbitrary $N$ gives its stable symmetric function version.

The dual determinant will also be useful below. Allow steps $(0,1)$ or $(1,1)$, assigning weight $x_r$ to a diagonal step reaching level $r$. A single path uses distinct levels, so its sum is the elementary symmetric function $e_r$, rather than $h_r$. The same cancellation proof leaves tableaux with strict rows and weak columns. Transposing them giveswith $e_0=1$ and $e_r=0$ for $r<0$.

Put $n=|\lambda |-|\mu |$. A squarefree weight $x_1\cdots x_n$ in $s_{\lambda /\mu }$ uses every label once, so it is counted by the number $f^{\lambda /\mu }$ of standard skew Young tableaux. In one determinant product ${\prod }_i{h}_{r_i}$, the signed permutation gives ${\sum }_i{r}_i=n$. If any $r_i<0$ the product is zero; otherwise the squarefree coefficient is the number of ways to distribute the distinct labels into groups of sizes $r_i$, namely $n!/{\prod }_i{r}_i!$. Coefficient extraction therefore gives the standard skew tableau determinantNegative factorial arguments mean zero entries, and $0!=1$. The empty skew shape has count one, as this convention requires.

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

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 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.

A generalised permutation is a finite multiset of pairs $(a,b)$ from two ordered alphabets, written as a two-row array, or biword. It may equivalently be specified by a finite-support matrix of nonnegative integer multiplicities $m_{a,b}$. The condition that no column is repeated means $m_{a,b}\in \{0,1\}$: each ordered pair occurs at most once. It does not forbid repeated entries in either individual row.

For this class use the dual Robinson–Schensted–Knuth correspondence. Order the columns by nondecreasing top entry $a$, and by strictly decreasing bottom entry $b$ when the top entries agree. Insert the bottom entries in that order by ordinary semistandard row insertion, which bumps the leftmost entry strictly greater than the incoming entry. Record the corresponding top entry in each newly created cell. Let $T$ be the insertion tableau and $U$ the recording tableau.

Ordinary row insertion permits repeated entries, preserving weak increase along rows and strict increase down columns; therefore $T$ is a semistandard Young tableau. The recording entries are inserted in nondecreasing order, so $U$ weakly increases along both rows and columns before considering strictness. Within one block of equal top entries, the bottom entries are strictly decreasing. The standard row-bumping comparison lemma says that a later, smaller input creates a new box strictly below the box created by the preceding input. Thus the boxes carrying one repeated top entry form a vertical strip, with at most one in each row. Consequently $U$ has strictly increasing rows and weakly increasing columns, so $U^t$ is semistandard.

Knuth's generalized insertion theorem, in this dual ordering, makes the map a bijection between finite $0$-$1$ matrices and these pairs. Its reverse description is also transparent: remove the largest recording entry, choosing its lowest box when it repeats, and reverse row insertion in $T$. Repeating recovers bottom entries in the reverse of the specified ordering. Equal top entries recover distinct bottom entries, which is exactly the no-repeated-column condition. ThusThe descending tie rule matters; the usual ascending tie rule would instead give two ordinary Semistandard Young tableaux.

Finally, the type of a tableau means its entry-multiplicity vector. Row insertion rearranges and bumps entries without changing their multiplicities. HenceThus the bottom-row multiplicities of the generalised permutation give the type of $T$, while the top-row multiplicities give the type of $U$; transposing $U$ does not change its type.

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.

Insert $x$ into the first row of a Young tableau by replacing its leftmost entry strictly greater than $x$, and carry the displaced entry into the next row. If no entry is greater, append $x$ and stop. The procedure preserves a semistandard Young tableau; when all entries are distinct it preserves a near Young tableau. It underlies the Robinson–Schensted correspondence.

Here $n=|\lambda |-|\mu |$, with zero entries for negative factorial arguments. Extract the squarefree monomial $x_1\cdots x_n$ from the skew Jacobi–Trudi identity. A product $h_{r_1}\cdots h_{r_{\ell }}$ contributes $n!/\prod r_i!$ because its distinct labels are assigned to the factors in that many ways. Summing determinant signs proves the formula, since a squarefree semistandard Young tableau is a standard skew Young tableau.

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.