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A word in positive integers is a lattice word if every initial segment has at least as many $i$'s as $(i+1)$'s for every $i$. This is the reading-word condition in the Littlewood–Richardson rule. Under good-letter matching in a tableau word, every letter of a lattice word is good.

The nonnegative integer $c_{\alpha \beta }^{\nu }$ is the multiplicity of $S^{\nu }$ in ${\mathrm{Ind}}_{S_{|\alpha |}\times S_{|\beta |}}^{S_{|\nu |}}(S^{\alpha }⊠S^{\beta })$ in characteristic zero. The Littlewood–Richardson rule counts it by Semistandard Young tableaux of skew shape $\nu /\alpha$, content $\beta$, and lattice word in right-to-left, top-to-bottom reading order. For example $c_{(3,2,1),(3,2)}^{(5,3,2,1)}=3$: the two new top cells must be ones, and the remaining reading letters can be $122,212,221$. This coefficient also multiplies Schur functions in the induction ring of symmetric-group characters.

For induction of $S^{\mu }⊠S^{\lambda }$ to the symmetric group on $|\mu |+|\lambda |$ letters, the Specht filtration multiplicity of $S^{\nu }$ is the number of Semistandard Young tableaux of skew shape $\nu /\mu$, content $\lambda$, and lattice word read from right to left in each row, taking rows from top to bottom. In characteristic zero these are irreducible constituents; in positive characteristic they remain Specht factors rather than simple constituents. Content $(1)$ gives the add-one-cell induction rule.

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 rim hook is a connected skew diagram containing no $2\times 2$ square, whose removal leaves a partition of an integer diagram. Its leg length is its number of occupied rows minus one. The Murnaghan–Nakayama rule states that, if a permutation has cycle type $(k,\rho )$, thenIteration removes one hook for each cycle length; the empty diagram has character one at the empty cycle type.

For shape $(3,3,3)$ and cycle type $(4,3,2)$, there are exactly two removable four-cell rim hooks. The remaining shapes are $(3,2)$ and $(2,2,1)$; the removed hooks have leg lengths one and two respectively. In shape $(3,2)$ the only removable three-cell rim hook leaves $(1,1)$ and has leg length one. In shape $(2,2,1)$ the only such hook leaves $(2)$ and again has leg length one. Since the characters of $(1,1)$ and $(2)$ at a transposition are $-1$ and $1$, we obtainand henceThe cycle lengths can be taken in any order; using $4,3,2$ makes the two branches of the computation explicit.

Now consider the ordinary induction product $[\alpha ][\beta ]$. A hook partition is $(x,1^y)$, equivalently a partition of an integer whose second row has length at most one. If a hook shape $\nu$ occurs in the Littlewood–Richardson rule, then $\alpha \subseteq \nu$, forcing $\alpha$ itself to be a hook. In the skew hook $\nu /\alpha$, the new cells consist of a segment in the first row and a segment down the first column, with no shared column between those two segments. Every first-row entry must be $1$: its rightmost entry is the first letter of the lattice word, hence is $1$, and weak row increase forces all entries to its left to be $1$ too. The column segment has strictly increasing entries. Thus only $1$ can be repeated, forcing the content $\beta$ to be a hook as well. This also proves the assertion when the top segment is empty, since then every entry lies in a strictly increasing column.

Write the two nonempty hook factors as $\alpha =(a,1^k)$ and $\beta =(b,1^l)$, with $k=n-r-a$ and $l=r-b$. Let the prospective resulting hook have top-row length $x$ and leg length $y$. Its new top segment has $H=x-a$ cells and its new column segment has $V=y-k$ cells. The content consists of $b$ ones and the single letters $2,\dots ,l+1$. All those larger letters must be in the new column; its strict increase allows either no $1$ or exactly one $1$. There are therefore only two possibilities:In the first case fill the top segment with $b$ ones and the column with $2,3,\dots ,l+1$. In the second, use $b-1$ top ones and column $1,2,\dots ,l+1$. Each filling is unique and has a lattice word, including the case $b=1$ with an empty top segment in the second filling. This provesBoth displayed multiplicities are one.

To deduce the full-cycle value from this product result, rather than quote the Murnaghan–Nakayama rule again, form the virtual characterFor every proper Young subgroup $S_m\times S_{n-m}$ and every pair of its irreducible characters, the hook-product calculation givesIf a factor is not a hook there are no hook constituents; if both are hooks, their two consecutive leg lengths cancel in this alternating inner product. By Frobenius reciprocity and the completeness of the product irreducible characters, $F_n$ restricts to zero on every such subgroup. Any permutation with more than one cycle preserves a nonempty proper subset and is conjugate into one of these subgroups. Thus $F_n$ is supported only on the class of $n$-cycles.

Its inner product with the trivial character is one, since only the one-row hook contributes. That class has size $(n-1)!$, so $F_n(\rho )/n=1$, giving $F_n(\rho )=n$. For any irreducible character, the support calculation then yieldsBy character orthogonality, the left side is $(-1{)}^{n-x}$ for $\nu =(x,1^{n-x})$ and zero otherwise. These are real integers, provingFor $n=1$ the same conclusion holds directly. This derives the requested special case from the Littlewood–Richardson rule and Frobenius reciprocity.

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.