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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 $(3,2,2)$.