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For ${\ell }_i={\lambda }_i+m-i$, the hook lengths in row $i$ are ${\ell }_i-q$, where $q$ runs over the non-beta numbers in $[0,{\ell }_i-1]$. Their product is ${\ell }_i!/{\prod }_{k>i}({\ell }_i-{\ell }_k)$. Multiplying over rows proves the identity and converts the hook-length formula into a Vandermonde determinant divided by factorials. Padding the partition with zero rows changes the beta set but not the hook product.

The conjugate partition exchanges rows and columns in a Young diagram: the cell $(i,j)$ becomes $(j,i)$. Equivalently,For $\lambda =(3,2)$, the three column lengths are $2,2,1$, so ${\lambda }^{\prime }=(2,2,1)$, as shown in Young diagram. Conjugating twice returns the original partition of an integer, and the number of cells stays five. Transposition also preserves hook lengths, with $h_{j,i}({\lambda }^{\prime })=h_{i,j}(\lambda )$; consequently the hook-length formula gives equal dimensions for the two Specht modules over the complex numbers. This row-column symmetry is already used in §1 of ^[ref].

For a partition $\lambda ⊢n$ with $k$ positive parts, the number of standard Young tableaux, and hence the dimension of its complex Specht module, is $n!\det (1/({\lambda }_i-i+j)!)$. Negative factorial arguments give zero entries. Setting $L_i={\lambda }_i+k-i$ turns the determinant into ${\prod }_{i<j}(L_i-L_j)/{\prod }_i{L}_i!$ by the Vandermonde determinant, matching the hook-length formula.

At a noncorner cell of a Young diagram, this random walk chooses uniformly one of the other cells in its hook of a Young diagram. It stops at a corner. Every step moves right or down, so it terminates. Its corner probabilities provide a probabilistic proof of the hook-length formula.

The hook graph is the Young diagram with the hook length written in each box. It displays the hook product and thus the hook-length formula for the corresponding Specht module. This use of the name is recorded in Frame, Robinson and Thrall's original hook-graph paper; it refers to a labeled array, not a directed hook-walk graph.

The hook length of cell $(i,j)$ in the Young diagram of a partition of an integer $\lambda$ is the number of cells in its Young-diagram hook:Here ${\lambda }^{\prime }$ is the conjugate partition. There is one hook length for every cell, so the multiset of hook lengths has cardinality $|\lambda |$.

For $\lambda =(3,2)$ and ${\lambda }^{\prime }=(2,2,1)$, compute $h_{1,1}=(3-1)+(2-1)+1=4$ and $h_{1,2}=(3-2)+(2-1)+1=3$. Continuing cell by cell gives the hook graph of a partitionor the row array $[[4,3,1],[2,1]]$. Its hook product is $4\cdot 3\cdot 1\cdot 2\cdot 1=24$. A cell has hook length one exactly when it is a Removable node of a Young diagram: it has no cell to its right or below. Here these are $(1,3)$ and $(2,2)$. Transposing the Young diagram gives the array $[[4,2],[3,1],[1]]$ with the same hook product; this explains geometrically why conjugate shapes have equal counts in the hook-length formula.

In the English convention of Young diagram, the hook at a cell $(i,j)$ consists of that cell itself, all cells to its right in the same row, and all cells below it in the same column:Here ${\lambda }^{\prime }$ is the conjugate partition. The cells to the right form the arm of a Young-diagram cell, and those below form the leg of a Young-diagram cell. Both extend all the way to the edge of the Young diagram, rather than containing just the immediate neighbors. The three pieces are disjoint, so their total cardinality is one plus the two lengths.

For $(3,2)$, the highlighted Young-diagram hook is $H_{1,2}=\{(1,2),(1,3),(2,2)\}$, of hook length three. The hook at $(1,1)$ is $\{(1,1),(1,2),(1,3),(2,1)\}$, of hook length four: the southeast cell $(2,2)$ is excluded because it is in neither the anchor's row nor its column. These local shapes are useful because their hook lengths give the denominator of the hook-length formula. The hook terminology and its use to calculate representation degrees appear explicitly in §1 of ^[ref].

The product of all hook lengths in a Young diagram is the denominator in the hook-length formula, and the scalar in the quasi-idempotence identity for its Young symmetrizer.

For $n\ge 3$, the least ordinary irreducible dimension exceeding one is $n-1$, except at $n=4$, where $S^{(2,2)}$ has dimension two. The dimension-$n-1$ shapes are $(n-1,1)$ and its conjugate partition, with the additional shapes $(3,3)$ and $(2,2,2)$ at $n=6$. For $n=2$ there is no non-linear irreducible. A proof uses the restriction branching rule for a symmetric group: a nonstandard shape with at least two corners has two non-linear predecessors, and a rectangular shape has one predecessor which in turn has two non-linear predecessors. This bounds its degree strictly above $n-1$ once $n\ge 9$; the smaller cases follow directly from the hook-length formula.

If $n=|\lambda |$ and $d_p(n)$ is its base-$p$ digit sum, thenThis combines the Hook-length formula, Legendre formula, the abacus divisible-hook correspondence, and the recurrence $q_r=c_r+p{q}_{r+1}$.

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 $(3,2,2)$.

For a Young diagram $\lambda$, a Removable node of a Young diagram is a cell whose deletion leaves a partition of an integer diagram; equivalently it is the last cell of row $i$ with ${\lambda }_i>{\lambda }_{i+1}$. Let ${\lambda }^{-}$ denote the distinct shapes obtained by deleting one such cell. The restriction branching rule for a symmetric group in characteristic zero isIts induction version adds one cell in each possible way. Over an arbitrary field, restriction instead has a Specht filtration with these successive quotients, each once; it need not split.

To prove restriction, list the removable nodes from top to bottom. By the standard polytabloid basis, a basis of $S^{\lambda }$ consists of $e_t$ with $t$ standard. Its entry $n$ is necessarily at a removable node. Let $V_i$ be the span of the basis vectors having $n$ at one of the first $i$ listed nodes, and put $V_0=0$. These are $S_{n-1}$-invariant. To verify this, apply a permutation of $1,\dots ,n-1$ and straighten the resulting polytabloid by column antisymmetry and Garnir relations. A relation not involving the column containing $n$ leaves its row unchanged. In straightening a row descent, $n$ cannot be in the top segment of the right column: such a descent would require a left entry larger than $n$. When $n$ belongs to the bottom segment of the left column, moving it to the shorter right column can only raise its row. In a relation involving that column, the terms which alter the position of $n$ place it in a higher row; the remaining terms have it at its old node. Thus straightening never moves it past the current top-to-bottom cutoff.

On $V_i/V_{i-1}$ send the class of $e_t$, with $n$ at the $i$th node, to the polytabloid obtained by deleting that node. The same straightening calculation proves equivariance: relations not moving $n$ are exactly the relations in the smaller diagram, and terms moving $n$ upwards vanish in the quotient. Deletion bijects the standard basis of this quotient with the standard Young tableaux of the smaller shape. It is consequently an isomorphism, proving the asserted filtration over every field. In characteristic zero, Maschke's theorem splits the filtration and gives the direct sum. Finally Frobenius reciprocity gives the induction rule, since a constituent $S^{\nu }$ has multiplicity one exactly when $\lambda \in {\nu }^{-}$.

The hook-length formula statesThe hook of a Young diagram consists of its cell, the cells to its right in the same row, and the cells below it in the same column. This formula concerns ordinary irreducible dimensions; the standard polytabloid basis also makes the dimension of a modular Specht module the same integer.

All transpositions are conjugate, have order two, and generate $S_n$. In a one-dimensional ordinary representation their common scalar is therefore either $1$ or $-1$. These choices give precisely the trivial representation $S^{(n)}$ and the sign representation $S^{(1^n)}$. At $n=1$ they coincide. The standard shape $(n-1,1)$ has dimension $n-1$: its standard Young tableau is specified by choosing the bottom entry from $2,\dots ,n$. Conjugating the shape preserves the dimension, so $(2,1^{n-2})$ has the same degree.

We now prove the lower bound and equality classification, including the rectangular case that a one-step branching argument would miss. Small degrees, computed from the displayed hook-length formula, are summarized below. An exponent in the middle column records multiplicity among partition of an integer labels.

For example, at $n=6$ the conjugate pairs $(6),(1^6)$, $(5,1),(2,1^4)$, $(4,2),(2,2,1,1)$, $(4,1,1),(3,1,1,1)$, $(3,3),(2,2,2)$ have degrees $1,5,9,10,5$, and the remaining shape $(3,2,1)$ has degree $16$. These checks also identify every small equality case.

Induct on $n\ge 9$, assuming the lower bound for all smaller degrees. If a non-linear, nonstandard shape has at least two removable nodes, none of its predecessors is a single row or single column: adding one cell to a row or column could produce only a linear or standard shape. Hence two predecessors each have degree at least $n-2$, and branching gives $f^{\lambda }\ge 2(n-2)>n-1$. If it has only one removable node, it is a rectangle $(a^b)$ with $a,b\ge 2$. Its sole predecessor is $(a^{b-1},a-1)$, which has two removable nodes. Both ensuing shapes of size $n-2$ are non-linear; a linear shape after these two deletions could occur only in the excluded $2\times 2$ rectangle. Therefore two-step branching gives $f^{\lambda }\ge 2(n-3)>n-1$. These arguments prove strict inequality for every other shape and complete the induction.

Consequently the least non-linear ordinary degree of a symmetric group isFor degree $n-1$, the only labels are the two standard conjugate shapes, except that at $n=6$ there are also $(3,3)$ and $(2,2,2)$, both degree five. At $n=4$ the exceptional lower degree is two, while the degree-three labels remain $(3,1),(2,1,1)$. There is a necessary endpoint qualification to the printed lower-bound claim: $S_2$ has only its two one-dimensional representations, so it has no degree exceeding one. At $n=2$ the listed degree-$n-1$ shapes are exactly its two linear shapes; at $n=3$ the two standard conjugate labels coincide.

For $n\ge 2$, a cycle type $\alpha ⊢n$ lies in $A_n$ exactly when $n-\ell (\alpha )$ is even. Its $S_n$ conjugacy class is either one $A_n$ class or two equal-sized classes. Splitting occurs exactly when its centralizer in $S_n$ contains no odd permutation. An even-length cycle is itself odd; two equal odd-length cycles can be interchanged by an odd permutation. Conversely, for distinct odd cycle lengths the centralizer is a product of cyclic groups of odd order, all contained in $A_n$. Thus the classes that split are exactly the partitions of an integer into distinct odd parts. Repeated fixed points count as repeated parts of length one.

For $n=1$ the group is trivial, with its single trivial representation. Assume $n\ge 2$ for the index-two argument that follows. For ordinary representations work over $\mathbb{C}$. Tensoring $S^{\lambda }$ by the sign representation gives $S^{{\lambda }^{\prime }}$. The index-two restriction identity, obtained by Frobenius reciprocity, isIf $\lambda \ne {\lambda }^{\prime }$, restriction is irreducible, and the conjugate pair gives the same irreducible. If $\lambda ={\lambda }^{\prime }$, the norm is two, so restriction is the sum of two distinct irreducibles. An odd permutation interchanges them, so both have degree $f^{\lambda }/2$. Every irreducible of $A_n$ occurs in some such restriction: induce it to $S_n$ and choose an irreducible constituent, then apply reciprocity. The displayed inner product distinguishes all the listed constituents. HenceThe first family has degree $f^{\lambda }$, the second degree $f^{\lambda }/2$. A choice of labels for the two split classes and constituents fixes the otherwise interchangeable signs.

The one-dimensional representations are characters of the abelianization. For $n\ge 5$, simplicity and noncommutativity of $A_n$ make its abelianization trivial. For $n=4$, its commutator subgroup is the Klein four-group and its abelianization is $C_3$, giving the three characters obtained by sending a quotient generator to $1,\omega ,{\omega }^2$. For $n=3$, $A_3=C_3$ has the same three characters; for $n=1,2$ the group is trivial and has just one. The two additional linear characters at $n=4$ are the split constituents of shape $(2,2)$; at $n=3$ they come from $(2,1)$.

Here is a dimension argument that also proves the asserted uniqueness without assuming a classification of small character degrees. Write $f^{\lambda }$ for the number of standard tableaux. Removing the cell containing the largest label givesHook lengths, or direct Young tableau counting, give the following small cases: for $S_5,S_6$ every non-linear degree is at least $4,5$, respectively. Among non-self-conjugate shapes excluding the row, column and standard pair, the minimum degree at $n=5,6,7$ is respectively $5,5,14$. The self-conjugate shapes at these sizes have half-degrees $3,8,10$, respectively, from $(3,1^2),(3,2,1),(4,1^3)$.

Inductively, for every $n\ge 7$, a shape other than the row, column or standard pair has all its one-cell predecessors non-linear. If it has at least two removable corners, each predecessor has degree at least $n-2$, so its degree is at least $2(n-2)$. If it has only one corner, it is a nontrivial rectangle $(a^b)$. Remove that corner and then either of the two corners of $(a^{b-1},a-1)$. For $n\ge 8$ these two size-$n-2$ shapes are non-linear, so its degree is at least $2(n-3)$. At $n=7$ there is no nontrivial rectangle. Together with the size-5 and size-6 checks this proves by induction that the standard pair is the only pair of shapes of degree $n-1$ for $n\ge 7$, and that every other non-linear degree is larger.

We must additionally exclude a self-conjugate shape of degree $2(n-1)$, since its restriction splits. For $n\ge 8$, if such a shape has at least three corners, the preceding lower bound gives $f^{\lambda }\ge 3(n-2)>2(n-1)$. With two corners, its predecessors are a conjugate pair and are not standard shapes; the bounds just proved give $f^{\lambda }\ge 4(n-4)>2(n-1)$. With one corner it is a square $(a^a)$. For $a\ge 4$, the two size-$n-2$ predecessors obtained after two removals are nonstandard, giving $f^{\lambda }\ge 4(n-5)>2(n-1)$. The remaining square $(3^3)$ has degree $42>16$ by the hook-length formula. The size-7 self-conjugate case has degree $20>12$. This excludes both smaller split degrees and equality with the standard degree.

It follows that the lowest non-linear ordinary degree is $3$ for $A_4$ and $n-1$ for every $n\ge 6$. For $A_1,A_2,A_3$ no non-linear irreducible exists. The excluded case $A_5$ has lowest degree $3$, although its standard degree-$4$ representation is unique. At $A_6$ there are two degree-$5$ irreducibles, from the standard conjugate pair and the pair $(3,3),(2,2,2)$.

Finally, for every $n\ge 7$ the preceding proof gives a unique degree-$n-1$ irreducible, namely the restriction of $S^{(n-1,1)}$. The small cases $n=4,5$ have unique degree $3,4$, respectively, and $A_2$ has its unique trivial degree-$1$ representation. These prove exactly the requested uniqueness range; the omitted cases $n=3,6$ genuinely fail.

A standard Young tableau has its entries increasing from left to right in each row and from top to bottom in each column. For the convention $h_t=c_t{r}_t$, use column-reading order of standard Young tableaux: read columns from left to right, each from top to bottom, and compare the resulting words lexicographically. This makes the requested product direction explicit. The order and the symmetrizer multiplication convention must be chosen together.

We prove the vanishing claim. Suppose the rows of $t$ and the columns of $u$ have no collision. The argument in Question 1 shows that every column of $u$ contains one entry from each eligible row of $t$. Its first column thus selects one entry from every row. Each selected entry is at least the first entry of that row in $t$. Sorting the selected entries, as the standard tableau $u$ does, gives a column componentwise at least the first column of $t$: increasing individual entries cannot decrease any order statistic. If the columns are equal, equality of their sums forces every selected entry to be that row's first entry. Delete this common column and repeat. At the first differing column, its first differing entry in $u$ is therefore larger; otherwise $u=t$. Thus absence of a collision implies $u\ge t$ in column-reading order of standard Young tableaux.

If $t>u$, there must instead be a transposition in $R_t\cap C_u$. It fixes $r_t$ and negates $c_u$, giving $r_t{c}_u=0$. HenceThis is triangular vanishing of Young-symmetrizer products; it does not assert vanishing in the opposite order.

Normalize to $e_t=h_t/H_{\lambda }$. List all standard tableaux in increasing shape dictionary order on integer partitions, and within each shape in increasing column-reading order of standard Young tableaux. Then $e_i{e}_j=0$ for $i>j$, using Question 1 between different shapes and the result above within a shape. The left ideals $A{e}_i$ have an internal direct sum: if ${\sum }_i{z}_i=0$ with $z_i\in A{e}_i$, multiply on the right by $e_1$ to get $z_1=0$, since $z_1{e}_1=z_1$ and every later $z_i{e}_1=0$. Repeat with $e_2,e_3,\dots$. This proves directness without incorrectly treating all the idempotents as mutually orthogonal.

Let $f_{\lambda }$ count the standard tableaux of shape $\lambda$ and let $d_{\lambda }=\dim S^{\lambda }$. In the regular representation a simple module of dimension $d_{\lambda }$ occurs $d_{\lambda }$ times, by the Artin–Wedderburn theorem. The direct sum just constructed contains $f_{\lambda }$ copies of $S^{\lambda }$, so $f_{\lambda }\le d_{\lambda }$ for every shape. We supply the needed counting identity independently of the dimension conclusion.

The Robinson–Schensted correspondence bijects permutations with pairs of standard tableaux of the same shape. Here is its row insertion construction and inverse. Insert the successive permutation entries into an increasing row by replacing its first entry larger than the incoming entry, bumping that replaced entry into the next row; if no entry is larger, append at the row end. Continue until a new cell is created. Record the insertion time in that cell of a second tableau. For completeness, the successive bumped entries strictly increase, and their column indices weakly decrease: an entry below a bumped entry was originally larger, so the next replacement occurs no farther right. The entry newly placed in each row is smaller than the entry removed and larger than the entry above it. At a strictly earlier column, that last inequality follows from row increase in the preceding row; at the same column, it follows from the preceding bump. Hence the insertion tableau keeps increasing rows and columns. If a new cell is appended below the first row, the preceding bump guarantees that the row above reaches that column, so the shape stays a Young diagram. Recording times also increase in rows and columns, since every new cell is an outer corner of the current diagram. Thus both tableaux are standard at the end. Conversely, remove the cell with the largest recording label. Reverse its bumping path upward, replacing in each preceding row the rightmost entry smaller than the moving entry and moving the displaced entry upward. This recovers the last inserted letter; iterating recovers the entire permutation. The two rules undo one another at each row. ThusSemisimplicity also gives $n!={\sum }_{\lambda }{d}_{\lambda }^2$. Since $0\le f_{\lambda }\le d_{\lambda }$ termwise, equality of these sums forces $f_{\lambda }=d_{\lambda }$ for every shape. The internal direct sum has dimension $n!$ and hence fills $A$:There is also a useful numerical form. Right multiplication by $e_t$ is an idempotent with image $A{e}_t$. In the permutation basis of $A$, each diagonal coefficient is the coefficient of $1$ in $e_t$, namely $1/H_{\lambda }$. Its trace equals its rank, so $\dim S^{\lambda }=n!/H_{\lambda }$. Together with the count just obtained this recovers the hook-length formula.

For a box $(i,j)$ of a Young diagram, its hook of a Young diagram contains that box, all boxes to its right in its row, and all boxes below it in its column. Its hook length is $h_{ij}={\lambda }_i-j+{\lambda }_j^{\prime }-i+1$. The hook graph of a partition is the diagram with each box labeled by its hook length. Write $H_{\lambda }={\prod }_{(i,j)\in \lambda }{h}_{ij}$ for the hook product of a partition. The hook-length formula is

Pad $\lambda$ to $m$ rows and use the beta set of a partition ${\ell }_i={\lambda }_i+m-i$, so ${\ell }_1>\cdots >{\ell }_m\ge 0$ and ${\sum }_i{\ell }_i=n+d$, where $d=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. We prove the beta-set hook-product identity row by row. For $1\le j\le {\lambda }_i$, put $q_j=m+j-{\lambda }_j^{\prime }-1$; then $h_{ij}={\ell }_i-q_j$. The $q_j$ are distinct integers in $[0,{\ell }_i-1]$, and none equals a beta number. Indeed, if $k\le {\lambda }_j^{\prime }$ then ${\ell }_k\ge q_j+1$, whereas if $k>{\lambda }_j^{\prime }$ then ${\ell }_k\le q_j-1$. Exactly $m-i$ beta numbers lie below ${\ell }_i$, so the remaining ${\lambda }_i$ integers in that interval are precisely these $q_j$. ThereforeThis gives the equivalent Specht module dimension expression $f^{\lambda }=n!\Delta (\ell )/{\prod }_i{\ell }_i!$.

The standard Young tableaux of shape $\lambda$ form the dimension count for the Specht module. The largest entry $n$ must occupy a removable corner. Deleting it bijects tableaux with the disjoint union of standard tableaux of the shapes obtained by deleting a corner, provingSet a term to zero whenever $\lambda -e_i$ is not a partition: this includes equal adjacent rows and an attempted deletion from a zero row. The empty diagram has dimension $1$.

For an algebraic proof that the proposed formula has the same recurrence, establish the Vandermonde shift identityIts left side is an alternating polynomial in the $x_i$, because permuting the variables permutes the summands and changes the sign of every Vandermonde factor. It is therefore divisible by $\Delta (x)$. The quotient is symmetric in $x$ and homogeneous of total degree one in $(x,t)$, so has the form $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$. Differentiating in $t$ at zero and using the Euler theorem for homogeneous functions gives ${\sum }_i{x}_i{\partial }_i\Delta =d\Delta$, so $b=d$. This proves the identity as a polynomial identity, including repeated coordinates.

Take $x=\ell$ and $t=-1$. Since ${\sum }_i{\ell }_i=n+d$, it gives ${\sum }_i{\ell }_i\Delta (\ell -e_i)=n\Delta (\ell )$. For $n\ge 1$, this is exactlyThe summand is the proposed dimension for $\lambda -e_i$; if two beta numbers collide its Vandermonde is zero, and if ${\ell }_i=0$ its coefficient is zero, so no negative factorial is needed. For the empty partition, $\ell =(m-1,\dots ,0)$ and $\Delta (\ell )={\prod }_{k=0}^{m-1}k!$, giving initial value $1$. Induction now proves the hook-length formula from the tableau recurrence.

For the final sum, tuples with repeated coordinates contribute zero. Sorting each distinct nonnegative tuple with sum $\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)$ gives one beta set of a partition of size $m$, since subtracting the staircase removes $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ from the sum. Conversely every partition of $m$, padded to $m$ rows, supplies exactly $m!$ ordered tuples, with the same squared summand. Hence the square-sum identity for shifted partition coordinates isThe middle equality uses the Artin–Wedderburn theorem for the group algebra and the complete classification of its simple modules by Specht modules. It explains why the last identity is a representation-dimension count rather than an accidental cancellation.

For a cell $x=(i,j)$ in a Young diagram, its hook of a Young diagram consists of $x$, all cells to its right in row $i$, and all cells below it in column $j$. If ${\lambda }^{\prime }$ is the conjugate partition, its hook length isThe hook-length formula is

In the Greene–Nijenhuis–Wilf hook walk, a walk at a noncorner cell chooses uniformly one of the other $h(i,j)-1$ cells in its hook, and stops at a corner. Moves are strictly rightwards or downwards, so termination is certain. A terminal cell must be a Removable node of a Young diagram. For a chosen corner $(\alpha ,\beta )$, its probability is zero unless the start satisfies $a\le \alpha$ and $b\le \beta$.

For a start in that northwest rectangle, defineThe corner conditions are ${\lambda }_{\alpha }=\beta$ and ${\lambda }_{\beta }^{\prime }=\alpha$, so $A_{\alpha }=B_{\beta }=0$, and all earlier $A_i,B_j$ are positive. DefineandThen the closed conditional probability isIn particular, if both inequalities are strict, this isThe separate definitions of $F_{\alpha },G_{\beta }$ handle starts in the terminal row, in the terminal column, or at the corner itself without division by zero.

To prove the formula, the product definitions give the identitieswith the terminal zero identities interpreted as empty sums. For example, summing the $F$ recurrence backwards gives ${\sum }_{r=i+1}^{\alpha }{F}_r={\prod }_{r=i+1}^{\alpha -1}(1+A_r^{-1})$. AlsoThe first-step equation for the hook walk terminal probability isJumps beyond the target rectangle contribute zero. Substituting $p_{i,j}=F_i{G}_j$ proves this equation, and $p_{\alpha ,\beta }=1$ gives its boundary value. Since every step increases a coordinate, backward recursion determines the probability uniquely. This proves the claimed expression.

Set $L_i={\lambda }_i+k-i$, for $1\le i\le k$. These are distinct nonnegative integers with $L_1>\cdots >L_k$. Using the falling factorial, the matrix entry can be writtenWhen $k-j>L_i$, the falling factorial vanishes, agreeing with the stipulated convention for a negative factorial argument.

The column polynomials $(X{)}_{k-1},(X{)}_{k-2},\dots ,(X{)}_0$ are monic of successive descending degrees. Subtracting lower-degree combinations of columns turns them into $X^{k-1},X^{k-2},\dots ,1$ without changing the determinant. The Vandermonde determinant therefore givesThe positive sign comes from reversing the usual increasing-degree Vandermonde columns at the same time as writing the differences as $L_i-L_j$.

We now match this denominator with the hook-length formula. The hook lengths in row $i$ are preciselyTo verify the claim, the first hook in the row has length $L_i$. As the column index passes from $j$ to $j+1$, its hook length drops by $1$ plus the number of lower rows ending at column $j$. The skipped lengths are exactly $L_i-L_r$ for those rows: if ${\lambda }_r=j$, then $L_i-L_r={\lambda }_i-j+r-i$. At the right-hand end, lower rows having the same length as row $i$ account for the missing smallest lengths by the same formula. This describes all omitted values, and the remaining count is $L_i-(k-i)={\lambda }_i$.

ConsequentlyCombining the two identities with the hook-length formula proves the determinantal dimension formula