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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

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 implication is false. ChooseThe inequalities $0\le 4<9$ and the relation $13=1\cdot 9+4$ hold. At cell $(1,4)$, the hook length is $12-4+1=9=a{p}^k$.

For the partition of an integer $(12,1)$, the Hook-length formula giveswhich is divisible by $3$. This also agrees with the degree of the standard representation of the symmetric group $S_{13}$.

The obstruction is visible in the core of a partition: removing the horizontal rim hook of length nine leaves $\gamma =(3,1)$, which has no hook length nine and hence is the $9$-core. Its degree is ${\chi }^{(3,1)}(1)=3$. The large hook ensures the required weight of a partition here, but the core degree is not coprime to $p$.

The equivalence is true. Write $c_r(\eta )$ for the total size at level $r$ of the core tower of a partition $\eta$, and $q_r(\eta )$ for the corresponding total size in its quotient tower of a partition. We use the size relation and the P-adic valuation of a symmetric-group character degree from the core tower:Here $d_p$ is the base-p digit sum. These imply the character degree coprime to p from the core tower criterion: if $|\eta |={\sum }_r{\alpha }_r{p}^r$, then $p\nmid {\chi }^{\eta }(1)$ exactly when $c_r(\eta )={\alpha }_r$ at every level. Indeed carrying $p$ units from one position to the next decreases the sum of coefficients by $p-1$; the valuation vanishes precisely when there are no carries.

For $k\ge 1$, we also use power-core truncation of a prime-core tower: for $\gamma =C_{p^k}(\lambda )$, its core tower agrees with that of $\lambda$ below level $k$ and is empty at every level at least $k$. Moreover $q_k(\lambda )=w_{p^k}(\lambda )$, the weight of a partition for modulus $p^k$. To see the truncation, removal of a rim hook of length $p^k$ preserves the first $p$-core and, under the abacus divisible-hook correspondence, becomes removal of a $p^{k-1}$-hook in one quotient component. Iterating preserves all the lower cores; after all such removals, quotient level $k$ is empty.

If $p\nmid {\chi }^{\lambda }(1)$, the criterion gives $c_k(\lambda )=a$, $c_r(\lambda )=0$ for $r>k$, and the lower $c_r(\lambda )$ are the base-$p$ digits of $m$. It follows that $|\gamma |=m$, $p\nmid {\chi }^{\gamma }(1)$ andConversely, suppose $w_{p^k}(\lambda )=a$ and $p\nmid {\chi }^{\gamma }(1)$. Then $|\gamma |=m$ and its lower core-tower sizes are the digits of $m$. Quotient level $k$ of $\lambda$ has total size $a<p$. Each component there has size less than $p$, hence is already a $p$-core and has empty $p$-quotient. Thus $c_k(\lambda )=a$ and every higher level is empty. Combined with truncation, this gives exactly the digits of $n$, so $p\nmid {\chi }^{\lambda }(1)$.

ThereforeIf $k=0$, necessarily $m=0$ and $n=a<p$; the $1$-core is empty, its weight is $n=a$, and the Hook-length formula shows that every degree is coprime to $p$. Thus the equivalence holds in that boundary case too, with the empty partition having degree one.

The preceding part shows that $\theta (e(t^{\prime }))\ne 0$ in characteristic zero, so $S^{{\lambda }^{\prime }}⊈\ker \theta$. Apply the James submodule theorem to the proper submodule $\ker \theta \le M^{{\lambda }^{\prime }}$ to obtainThe Hook-length formula gives $\dim S^{\lambda }=\dim S^{{\lambda }^{\prime }}$. Surjectivity of $\theta$ therefore givesand hence

Fix the original tableaux $t,u$. For a $\lambda$-tableau $w$, let $g_w$ be the unique permutation satisfying $w=g_{w}t$ and put ${\epsilon }_w=\mathrm{sgn}(g_w)$. Sincethe quotient pairing $M^{{\lambda }^{\prime }}/(S^{{\lambda }^{\prime }}{)}^{\perp }\cong (S^{{\lambda }^{\prime }}{)}^{\ast }$ gives the explicit conjugate Specht module as a sign-twisted dual isomorphismfor every $\lambda$-tableau $w$.

PutThe defining relation between the quotient tower of a partition and the core tower of a partition isSumming the resulting telescoping identities gives

By the Hook-length formula,The abacus divisible-hook correspondence says that the number of hooks divisible by $p^r$ is $q_r$, soIf $d_p(n)={\sum }_r{\alpha }_r$, the digit-sum form of the Legendre formula isCombining the three displayed identities proves the P-adic valuation of a symmetric-group character degree from the core tower formula

Write $\chi ={\chi }^{\lambda }$. By the Hook-length formula,Suppose the order of a group element $g$ did not divide this product. Then for some prime number $p$, the highest prime power $p^a$ dividing $|g|$ would not divide the hook product. One cycle of $g$ has length divisible by $p^a$, whereas no hook length of $\lambda$ is divisible by $p^a$. In particular there is no removable rim hook having that cycle length. Applying the Murnaghan–Nakayama rule first to this cycle gives $\chi (g)=0$, a contradiction. This is the symmetric-group character co-degree vanishing criterion, and its contrapositive proves

For $A_n$, the least non-linear ordinary degree is $3$ at $n=4,5$ and $n-1$ for $n\ge 6$. For $n\le 3$ no non-linear irreducible exists. The degree-$n-1$ representation is unique for $n\ge 2$, except $n=3$ where it does not exist and $n=6$ where two distinct degree-five irreducibles occur. The proof uses tableau branching to bound the degrees of nonstandard shapes and of the half-degree constituents of self-conjugate shapes, with the small cases checked by the hook-length formula.

A standard Young tableau of shape $\lambda ⊢n$ is a Young tableau that uses $1,\dots ,n$ exactly once and strictly increases left to right along every row and top to bottom down every column. Formally, a bijection $T:Y(\lambda )\to \{1,\dots ,n\}$ is standard whenwhenever the indicated cells belong to the Young diagram. We write $\mathrm{SYT}(\lambda )$ for the set of such fillings and $f^{\lambda }=|\mathrm{SYT}(\lambda )|$ for its cardinality.

For $(3,2)$, the five standard Young tableaux are listed below, with each pair giving the top row followed by the bottom row:There are no others: the upper-left entry must be $1$, and choosing the two bottom-row entries fixes both increasing row orders. Of the six choices of two entries from $\{2,3,4,5\}$, only bottom row $(2,3)$ fails, because its $3$ would lie below $4$. Hence $f^{(3,2)}=5$ by direct enumeration, independently of the hook-length formula.

The cells occupied by $1,\dots ,r$ form a Young diagram: every cell above or to the left of an occupied cell has a smaller label. Consequently a standard Young tableau is equivalently a path from the empty shape to $\lambda$ in the Young branching graph, adjoining one addable node of a Young diagram at each step. For $T_2$, the successive shapes areConversely, number the cell added at step $r$ by $r$ to recover the standard Young tableau. This path interpretation also makes a standard Young tableau a linear extension of a partially ordered set defined by the row and column requirements; see Stanley's Algebraic Combinatorics, Chapter 8, on tableaux and paths in Young's lattice.

The usefulness goes beyond counting: the standard polytabloid basis of a Specht module $S^{\lambda }$ over the complex numbers is indexed by $\mathrm{SYT}(\lambda )$, so ${\dim }_{\mathbb{C}}{S}^{\lambda }=f^{\lambda }$. Each of the five pictured standard Young tableaux therefore labels one basis element of the irreducible representation $S^{(3,2)}$ of the symmetric group $S_5$, whose dimension over the complex numbers is five. The filled diagrams index the basis; the actual vectors are polytabloids. The hook-length formula computes this dimension without listing all the standard Young tableaux.

For $\chi \in \mathrm{Irr}(S_n)$, if the order of $g\in S_n$ does not divide $|S_n|/\chi (1)$, then $\chi (g)=0$. The Hook-length formula turns the co-degree into a product of hook lengths, while the Murnaghan–Nakayama rule detects a cycle whose required prime-power hook cannot be removed.

The left side is an alternating polynomial of total degree one more than $\Delta (x)$. Dividing by the Vandermonde determinant gives a symmetric homogeneous polynomial of degree one, necessarily $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$; differentiation in $t$ and the Euler theorem for homogeneous functions give $b=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Thus the identity holds as a polynomial identity, even at repeated coordinates. At shifted partition coordinates with $t=-1$, it proves the removable-corner recurrence in the hook-length formula.

A Young tableau is a filling of a Young diagram with entries, typically positive integers. The shape is the partition of an integer giving its row lengths. We draw Young tableaux in the English convention of Young diagram, with left-aligned rows arranged from top to bottom. A filling need not obey increasing conditions unless these are specified: standard Young tableaux use each of $1,\dots ,n$ exactly once and strictly increase along rows and down columns; Semistandard Young tableaux permit repetitions and require weakly increasing rows and strictly increasing columns.

In the constructions for a symmetric group involving Young symmetrizers, tabloids, and polytabloids, an unqualified numbered Young tableau means a bijective filling by $1,\dots ,n$, with no increasing condition imposed. This narrower convention is compatible with the broader use needed for Semistandard Young tableaux. Historically, Young's work connected these fillings to representation theory of the symmetric group; in particular, the equality between standard Young tableau counts and irreducible representation degrees over the complex numbers predates the modern hook-length formula. See Stanley's Algebraic Combinatorics, historical notes to Chapter 8.