= Solution
Use the induction product of symmetric-group <characters>: for $a\ge0$, $[a]$ 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 is
$$
\boxed{[\lambda]=\det\bigl([\lambda_i-i+j]\bigr)_{1\le i,j\le r},}
$$
where $r\ge\ell(\lambda)$. 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 $2\times2$ square, where removal leaves a <partition of an integer>. Its leg length is the number of occupied rows minus one. The <Murnaghan–Nakayama rule> states
$$
\boxed{\chi^\lambda(k,\rho)=\sum_{\nu:\lambda/\nu\text{ is a }k\text{-rim hook}}
(-1)^{\operatorname{leg}(\lambda/\nu)}\chi^\nu(\rho).}
$$
A proof sketch follows directly from the alternant form of <Schur functions>. Multiply an alternant by $p_k=\sum_jx_j^k$. 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 $p_ks_\nu=\sum_\lambda(-1)^{\operatorname{leg}(\lambda/\nu)}s_\lambda$. 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. Thus
$$
\chi^{(4,4,3)}(5,4,2)=\chi^{(3,2,1)}(4,2)-\chi^{(4,2)}(4,2).
$$
The 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 $\chi^{(4,2)}(4,2)=(-1)(-1)=1$. Hence
$$
\boxed{\chi^{(4,4,3)}(5,4,2)=-1.}
$$
As 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 gives
$$
\operatorname{Res}^{S_n}_{S_{n-1}}\chi^\lambda=\sum_{\nu\in\lambda^-}\chi^\nu.
$$
This 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 $(\lambda^{(0)},\ldots,\lambda^{(p-1)})$ and put $w_i=|\lambda^{(i)}|$, with $\sum_iw_i=w$. Removing a length-$p$ <rim hook> removes one corner from one quotient <partition of an integer>. Complete removal sequences therefore number
$$
N_p(\lambda)=\binom{w}{w_0,\ldots,w_{p-1}}\prod_{i=0}^{p-1}f^{\lambda^{(i)}}
=\frac{w!}{\prod_iH_{\lambda^{(i)}}},
$$
where $H_\alpha$ 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 $\varepsilon_p(\lambda)$. 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 gives
$$
\boxed{\chi^\lambda(\pi\rho)=\varepsilon_p(\lambda)N_p(\lambda)\chi^\kappa(\pi).}
$$
Here $\pi$ 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.
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