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Here $[a]$ is the trivial character of $S_a$, $[0]$ is the induction-product unit, $[a]=0$ for $a<0$, and multiplication is induction of external tensor products. The Frobenius characteristic map turns the expression into the Jacobi–Trudi identity. A sign-reversing involution on intersecting lattice-path systems leaves precisely the nonintersecting systems corresponding to Semistandard Young tableaux, proving the determinant identity.

For a partition of an integer $\lambda ⊢n$, padded to $m$ rows, let $\delta =(m-1,\dots ,0)$. If a permutation has ${\alpha }_q$ cycles of length $q$, put $p_{\alpha }={\prod }_q({\sum }_i{x}_i^q{)}^{{\alpha }_q}$. The character of the Specht module $S^{\lambda }$ is the displayed coefficient. Equivalently $p_{\alpha }{A}_{\delta }={\sum }_{\lambda }{\chi }^{\lambda }(\alpha )A_{\lambda +\delta }$, the alternant form of the Frobenius characteristic map. It gives finite coefficient computations without constructing representation matrices.

Here $p_{\alpha }$ is a product of power-sum symmetric functions and $z_{\alpha }={\prod }_r{r}^{m_r}{m}_r!$ for part multiplicities $m_r$. The Frobenius characteristic map identifies this product with the ordinary character inner product. Schur functions are orthonormal, and the adjoint of multiplication by $p_k$ is $k\partial /\partial p_k$. These properties turn a power-sum multiplication formula into the Murnaghan–Nakayama rule.

The graded abelian group ${⨁}_{n\ge 0}R(S_n)$ of ordinary virtual characters has product $\chi \star \psi ={\mathrm{Ind}}_{S_m\times S_n}^{S_{m+n}}(\chi ⊠\psi )$ for degrees $m,n$. Its unit is the trivial degree-zero character. Associativity follows by transitivity of induced representations through the three-factor Young subgroup; commutativity follows by conjugating the block positions. The Frobenius characteristic map identifies this ring with the ring of symmetric functions, sending irreducible Specht characters to Schur functions. This product is different from pointwise multiplication of characters of the same symmetric group.

A skew Young diagram is connected when any two of its cells can be joined by a sequence of cells sharing edges; touching only at a corner does not connect them. A nonempty border strip, or rim hook, is a connected skew diagram containing no $2\times 2$ square. Its height is its number of occupied rows minus one.

For a weak composition $\alpha =({\alpha }_1,\dots ,{\alpha }_{\ell })$, a border-strip tableau of shape $\lambda /\mu$ and type $\alpha$ is a chainwhose $i$th difference is a border strip of size ${\alpha }_i$. Filling that strip with $i$ gives the equivalent weakly increasing row-and-column filling. A zero part corresponds to an empty difference and has height zero. Put $\mathrm{ht}(T)={\sum }_i\mathrm{ht}({\lambda }^{(i)}/{\lambda }^{(i-1)})$. For positive $r$, $p_r={\sum }_j{x}_j^r$; write $p_{\alpha }={\prod }_{{\alpha }_i>0}{p}_{{\alpha }_i}$. Thus zero type parts are ignored: one does not multiply by the finite-variable value $p_0=N$.

We first prove the one-strip multiplication formula using alternants. Choose $N$ sufficiently large and let ${\beta }_i={\mu }_i+N-i$ be the decreasing beta numbers of a partition. Expanding the determinant by permutations showsIndeed, in each determinant monomial every variable occurs in exactly one row factor; multiplying that variable by its $r$th power is equivalent to raising that row exponent.

A term vanishes if the new exponent equals another exponent. Otherwise suppose the changed exponent moves from position $i$ to position $j\le i$ when restored to decreasing order. Its determinant sign is $(-1{)}^{i-j}$. The resulting partition $\lambda$ hasand all other rows unchanged. The new diagram contains $\mu$, and its size increases by $r$. Between successive added rows, the overlap is exactly the single column ${\mu }_{t-1}+1$. Thus the added cells form a connected strip without a $2\times 2$ square, of height $i-j$. Conversely every added border strip has consecutive occupied rows and exactly this single-column overlap, forcing these formulas and the corresponding exponent shift. The correspondence is therefore bijective. Divide by the unchanged Vandermonde determinant and use the bialternant formula to get the power-sum border-strip multiplication identity

Apply this formula successively to each positive part of $\alpha$. Each sequence of choices is exactly a border-strip tableau, and its product of signs is $(-1{)}^{\mathrm{ht}(T)}$. ThereforeCommutativity of the power sums also shows that the resulting coefficient does not depend on the order of the positive parts of the type.

Take $\mu$ empty and restrict to $N\ge \ell (\lambda )$ variables. Multiply by $a_{\delta }$, with $\delta =(N-1,\dots ,0)$:In an alternant $a_{\nu +\delta }$, the exponents are distinct and their decreasing arrangement is $\nu +\delta$. Consequently the monomial $x^{\lambda +\delta }$ occurs only for $\nu =\lambda$, and in that determinant its coefficient is one. This proves the Frobenius alternant character formula

To identify these coefficients with symmetric-group characters, recall the Frobenius characteristic mapThe representation-theoretic identification $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ can be justified by Young's rule: the permutation character on ordered blocks of sizes $\nu$ is ${\sum }_{\lambda }{K}_{\lambda \nu }{\chi }^{\lambda }$. Its characteristic is $h_{\nu }$, because its fixed block decompositions allocate whole cycles to blocks, and the exponential generating series for those allocations is ${\prod }_{j}H(t_j)$. On the symmetric-function side, Hall duality and Schur orthonormality give $h_{\nu }={\sum }_{\lambda }{K}_{\lambda \nu }{s}_{\lambda }$. Inverting the unitriangular Kostka matrix establishes the identification. Thus the coefficient of $s_{\lambda }$ in $p_{\rho }$ is the irreducible character value on cycle type $\rho$. We have supplied the border-strip computation rather than assuming the character rule.

Grouping border-strip chains by their first removal of a cycle length $r$ now yields the Murnaghan–Nakayama ruleOne may choose the $r$-cycle to be last in a strip-building chain because the power sums commute. The empty-shape value on the empty type is one, providing the recursion's initial condition.

For the staircase $\delta =(m-1,m-2,\dots ,1)$, use $m$ beta numbers including its padded zero row. Its beta set isRemoving a rim hook of length $r$ lowers one beta number by $r$ into an unoccupied nonnegative position, reversing the exponent-shift description above. For a positive even $r$, this move either hits another occupied even position or goes below zero. Therefore there are no removable even-size border strips.

Under the Hall inner product of symmetric functions, the adjoint of multiplication by $p_r$ is $r\partial /\partial p_r$: this follows on the power-sum basis from $z_{\nu }/z_{\nu \setminus r}=r{m}_r(\nu )$. Taking adjoints of the one-strip multiplication rule expresses this derivative on $s_{\delta }$ as the signed sum over removable size-$r$ strips. That sum is empty for even $r$, soThe rational symmetric-function algebra is freely generated by the power sums, and its characteristic is zero. Vanishing of each of these derivatives therefore proves the staircase Schur functions use only odd power sums statement:

Finally, the cycle type of an $n$-cycle has one strip of size $n$. A straight diagram is a border strip exactly when its second row has length at most one: otherwise it contains the upper-left $2\times 2$ square; conversely the first row with a single-column tail is connected and contains none. These diagrams are exactly the hook partitions $(n-s,1^s)$, of height $s$. There is one strip filling in this case and none otherwise, giving

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 $(3,2,2)$.

Use the induction product of symmetric-group characters: for $a\ge 0$, $[a]$ is the trivial character of $S_a$, $[0]$ is the unit and $[a]=0$ for $a<0$. Products mean induction of external tensor products. The determinantal form iswhere $r\ge \ell (\lambda )$. Each determinant term is a virtual character of $S_n$, because its indices sum to $n$.

Here is a proof. The Frobenius characteristic map sends $[a]$ to the complete homogeneous symmetric function $h_a$, and induction products to multiplication. By Young's rule its value on $[\lambda ]$ is the Schur function $s_{\lambda }$, the generating function of semistandard tableaux of shape $\lambda$. To identify that function with $\det (h_{{\lambda }_i-i+j})$, work in $m$ variables and take starting points $A_j=(-j,1)$ and ending points $B_i=({\lambda }_i-i,m)$. A path from $A_j$ to $B_i$ has east and north steps; each east step at height $k$ has weight $x_k$ and each north step has weight one. Its east-step count is ${\lambda }_i-i+j$, so its weight sum is $h_{{\lambda }_i-i+j}$. Expanding the determinant sums signed systems of paths with permuted endpoints. For a system with an intersection, exchange the two tails after the first intersection. This preserves its monomial weight and reverses the endpoint permutation's sign, so all intersecting systems cancel in pairs. Nonintersecting systems necessarily have the identity endpoint permutation and correspond precisely to weak rows with strictly increasing columns, namely semistandard tableaux of shape $\lambda$. Their signs are positive. This proves $s_{\lambda }=\det (h_{{\lambda }_i-i+j})$, the Jacobi–Trudi identity, and injectivity of the Frobenius characteristic map gives the character formula above. The path sum may be taken in finitely many variables first and then stabilized, so no infinite formal sum is required in the argument.

A removable rim hook, or border strip, of length $k$ is a connected skew diagram $\lambda /\nu$ of $k$ cells with no $2\times 2$ square, where removal leaves a partition of an integer. Its leg length is the number of occupied rows minus one. The Murnaghan–Nakayama rule statesA proof sketch follows directly from the alternant form of Schur functions. Multiply an alternant by $p_k={\sum }_j{x}_j^k$. In each term one alternant exponent increases by $k$. Equal exponents cancel; otherwise reordering the exponents contributes the sign of the number crossed. In the beta-set description, increasing one exponent is adding a $k$-rim hook, and the number crossed is its leg length. Therefore $p_k{s}_{\nu }={\sum }_{\lambda }(-1{)}^{\mathrm{leg}(\lambda /\nu )}{s}_{\lambda }$. Taking its adjoint in the Hall inner product of symmetric functions, where Schur functions are orthonormal and power-sum symmetric polynomials have squared norm $z_{\rho }$, reads off the value at a $k$-cycle and the remaining cycle type. This gives the stated rule and explains its sign.

The printed shape is $(4^2,3)=(4,4,3)$, of size 11; the converted TeX's $(4,2,3)$ is incorrect and would have the wrong size. There are two removable length-5 rim hooks. They leave $(3,2,1)$ with leg length 2 and $(4,2)$ with leg length 1. ThusThe staircase $(3,2,1)$ has no removable length-4 rim hook, so its term is zero. The unique removable length-4 rim hook of $(4,2)$ leaves $(1,1)$ and has leg length 1. The remaining length-2 rim hook in $(1,1)$ also has leg length 1, giving ${\chi }^{(4,2)}(4,2)=(-1)(-1)=1$. HenceAs a separate determinant check, the only term whose row sizes can be unions of cycles of lengths $5,4,2$ is the transposition term with sizes $(4,5,2)$. Its permutation character has value one and its determinant sign is negative.

Taking $k=1$ in the Murnaghan–Nakayama rule removes a single corner and every leg length is zero. Evaluating on a permutation fixing the distinguished point therefore givesThis is the restriction branching rule for a symmetric group; ordinary semisimple representation theory converts the character equality into a direct-sum decomposition. Frobenius reciprocity gives the corresponding add-one-cell induction rule.

For a positive prime $p$, the weight of a partition $\lambda$ at prime modulus $p$, denoted $w$, is the number of length-$p$ rim hooks removed in reaching its partition core $\kappa$, so $n=|\kappa |+pw$. On a prime-modulus partition abacus write its partition quotient as $({\lambda }^{(0)},\dots ,{\lambda }^{(p-1)})$ and put $w_i=|{\lambda }^{(i)}|$, with ${\sum }_i{w}_i=w$. Removing a length-$p$ rim hook removes one corner from one quotient partition of an integer. Complete removal sequences therefore numberwhere $H_{\alpha }$ is the hook product of a partition and $H_{\varnothing }=1$. Each component's standard tableaux encode its possible removal orders, and the multinomial coefficient interleaves the components.

All these sequences have the same total sign ${\epsilon }_p(\lambda )$. To see this, label the beads on each runner in their preserved order. Each slide crosses exactly the beads counted by its rim-hook leg. Its sign is the change in the total order of bead labels, and the product telescopes to the permutation from the initial ordering to the final packed-core ordering, independently of the slides chosen. Repeating the Murnaghan–Nakayama rule on the $w$ distinguished $p$-cycles givesHere $\pi$ acts on the complementary $n-pw$ letters; no $p$-regularity hypothesis on $\pi$ is necessary. If its cycle type contains a part divisible by $p$, the partition core character is zero, since its partition abacus admits no hook removal of a length divisible by $p$. For $w=0$ the coefficient and sign are one.

Apply the Frobenius characteristic map. For even $n$, the Jacobi–Trudi identity and cancellation of consecutive terms giveThe generating function for the complete homogeneous symmetric polynomials now givesThis expansion contains only products $p_{\mu }$ for which every part of $\mu$ is even. The coefficient of $p_{\mu }$ in $\mathrm{ch}(F)$ is $F(\mu )/z_{\mu }$, so $F(\mu )=0$ whenever the cycle type $\mu$ has an odd part. Equivalently, $F(g)=0$ whenever $g$ contains an odd cycle. This is the Two-row alternating character cancellation.

For even $n$, the virtual charactervanishes on every permutation having an odd cycle. Under the Frobenius characteristic map, the Jacobi–Trudi identity identifies its characteristic with the degree-$n$ part of $H(t)H(-t)$, which contains only products of even-indexed power sums.