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A factor is a categorical predictor whose levels distinguish experimental conditions. Two factors have a statistical interaction when the effect of changing one depends on the other's level. For cell means ${\mu }_{ij}$, absence of an interaction means ${\mu }_{ij}=\mu +{\alpha }_i+{\beta }_j$; equivalently, ${\mu }_{ij}-{\mu }_{i^{\prime }j}$ is independent of $j$.

Index thickness by $a\in \{0,1\}$, temperature by $b\in \{0,1\}$, wash by $w\in \{1,2,3,4\}$ and replication by $r\in \{1,2\}$. The full factorial design has the normal linear modelThe errors are independent with a common positive variance. Sum-to-zero constraints in each factor index, or the displayed treatment coding, identify the mean coefficients. This full model allows an arbitrary mean in each of the sixteen cells.

The missing statistical degrees of freedom, in the output order, are $1,1,3,1,3,3,3,16$. The first seven entries total fifteen; with the regression intercept there are sixteen mean coefficients among thirty-two observations, leaving sixteen residual statistical degrees of freedom. Balanced factorial orthogonality makes the term sums of squares in ANOVA independent of their ordering.

All terms involving wash have nonsignificant individual F-tests, and it is useful to test their removal jointly. The reduced model omits twelve mean coefficients. Its residual sum of squares is $57.09875$, compared with $32.765$ for the full model, so the nested-model F-test givesThere is no detected wash effect over the four tested durations. Removing its main effect and all its interaction terms respects the principle of marginality. Retaining the very large thickness-by-temperature interaction is essential: the thickness effect changes substantially with temperature.

With numeric indicators $a,b\in \{0,1\}$, treatment coding expresses the reduced fitted mean asThus the thin/low, thick/low, thin/high and thick/high fitted means are $4.225,5.6875,17.425,29.950$, respectively, at every wash level. Each is the sample mean of eight observations. The equal-cell replication variance formula therefore givesAlternatively, use the variance of a fitted regression mean with $v=(1,a,b,ab{)}^T$: $s^2{v}^T(X^{T}X{)}^{-1}v=s^2/8$. Individual coefficient standard errors cannot simply be added; their covariances enter the fitted-mean calculation.

Increasing temperature raises fitted lustre by $13.2$ for thin film and $24.2625$ for thick film. Increasing thickness raises it by $1.4625$ at low temperature and $12.525$ at high temperature. Their difference, $11.0625$, is the estimated interaction contrast; its standard error is $1.0098$, giving $t\simeq 10.956$ on twenty-eight statistical degrees of freedom. The interaction plot shows this strong departure from parallel mean lines, while the wash panel shows the much smaller variation across wash durations. Neither the nonsignificant wash test nor the plot proves that its true effect is exactly zero.

Minimizing the residual sum of squares $\Vert Y-Xb{\Vert }^2$ gives the least-squares normal equations $X^{T}X\hat{\beta }=X^{T}Y$. Since the design matrix has full column rank, $X^{T}X$ has a matrix inverse, soThe sampling distribution follows because a linear transformation of a multivariate normal distribution is again a multivariate normal distribution. Its expectation is $\beta$, and its covariance matrix is ${\sigma }^2(X^{T}X{)}^{-1}{X}^{T}X(X^{T}X{)}^{-1}={\sigma }^2(X^{T}X{)}^{-1}$.

The hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. It is the orthogonal projection matrix onto the column space of $X$: $H^T=H$, $H^2=H$ and $HX=X$. Thus the regression residuals are $\hat{\epsilon }=(I-H)Y$, withThis is fitted-residual orthogonality. The vectors are jointly normal, so independence of uncorrelated jointly normal variables strengthens both zero-covariance statements to independence.

For the quadratic linear regression, the columns of the design matrix are $1,x,x^2$. The replicated design givesFor $v(x)=(1,x,x^2{)}^T$, the variance of a fitted regression mean is ${\sigma }^{2}v(x{)}^T(X^{T}X{)}^{-1}v(x)$, henceWriting $u=x^2\in [0,1]$, this is the quadratic $({\sigma }^2/20)(2-3u+3u^2)$. Its derivative vanishes at $u=1/2$; it decreases before that point and increases afterwards. Therefore the maximum is ${\sigma }^2/10$ at $x=-1,0,1$, and the minimum is ${\sigma }^2/16$ at $x=\pm 1/\sqrt{2}$.

At $100^{\circ }\mathrm{C}$, $x=0$ and the expected yield is $\alpha$. The unbiased residual variance estimate is $s^2=2.43/(30-3)=0.09$. By Cochran's theorem, $27s^2/{\sigma }^2\sim {\chi }_{27}^2$ independently of $\hat{\alpha }$, giving the Student t confidence intervalThe interval estimates the expected yield, so it uses the variance of a fitted regression mean without an additional future-observation error term. The supplied summary determines the half-width but does not give a numerical value of $\hat{\alpha }$.

Use column vectors for dollar holdings ${\theta }_t$ and put $b=\mu -r1$, $h={\sigma }^{-1}b$ and $V=\sigma {\sigma }^T$. The normalized state-price density isThe vector $h$ is the market price of risk. The density $e^{rt}{\zeta }_t$ defines an equivalent risk-neutral measure, under which $W_t^{\mathbb{Q}}=W_t+ht$ is Brownian motion and every stock has drift $r$. A terminal claim $H$ with finite state-price cost is priced at ${\zeta }_t^{-1}\mathbb{E}[{\zeta }_{T}H\mid {\mathcal{F}}_t]$. Nonsingularity of $\sigma$ and the natural Brownian market information make this a complete market; in particular the optimal nonnegative payoff below can be replicated.

Let $q=n^{-1}1$ and $v={\sigma }^{T}q$. Apply the Itô formula to each log price, whose return variance is $V_{ii}$:Summing and dividing by $n$ proves the geometric stock index formulaThus $J_t=J_0\exp (v^T{W}_t+a_{J}t)$, where $a_J=q^T\mu -\mathrm{tr}(V)/(2n)$. The factor $1/n$ applies to the whole bracket. The geometric average need not itself be a self-financing portfolio; its instantaneous drift is $a_J+|v{|}^2/2$.

If ${\theta }_t^i$ is the dollar amount invested in stock $i$, the bank account holds $w_t-1^T{\theta }_t$. The self-financing portfolio equation isEquivalently, ${\pi }_t={\theta }_t/w_t$ gives $d{w}_t/w_t=(r+{\pi }_t^{T}b)dt+{\pi }_t^T\sigma d{W}_t$. For share-holding notation replace ${\theta }_t^i$ by $S_t^i$ times the number of shares.

Static expected utility maximization uses the terminal budget $\mathbb{E}[{\zeta }_T{w}_T]\le w_0$. Since $J_T$ is an exogenous positive benchmark, differentiating $U(x/J_T)$ in $x$ gives $J_T^{-1}{U}^{\prime }(x/J_T)$. An interior Lagrange multiplier $\lambda >0$ for the budget therefore givesFor CRRA utility, write $p=1/R$. Since $U^{\prime }(y)=y^{-R}$, solving this marginal equation givesEvery positive lognormal moment is finite here, so the budget determines the constant uniquely. PutThenThis is also a sufficiency condition: concavity gives $U(X/J_T)-U(w_T^{\ast }/J_T)\le \lambda {\zeta }_T(X-w_T^{\ast })$, and taking expectations and using the budget inequality proves global optimality.

To find the whole optimal process, set $c=a_J-r-|h{|}^2/2$ andSince ${\zeta }_t{J}_t=J_0\exp ((v-h{)}^T{W}_t+ct)$, the moment-generating function of a normal distribution givesPricing the terminal payoff yields the explicit benchmark-relative power-utility portfolio wealthIts proportional diffusion coefficient is $\ell =ph+(1-p)v$. Since the wealth diffusion coefficient for dollar fractions is ${\sigma }^T\pi$, replication requiresUsing ${\sigma }^{-T}h=V^{-1}b$ gives the requested fixed proportions:The drift matches the self-financing portfolio equation as well. Indeed, $\ell =h+(1-p)(v-h)$, so the drift of $\log w^{\ast }$ from the explicit formula isConsequently an equivalent form of the solution isThe bank account fraction is $1-1^T{\pi }^{\ast }$, with negative values representing borrowing. The first stock term is ordinary power-utility risk exposure, while the second hedges the random benchmark. At the logarithmic limit $R=1$ the benchmark term disappears, consistently with $\log (w_T/J_T)=\log w_T-\log J_T$.

Assume, as in the Bayesian model, that the unknown drift $\alpha$ is independent of the driving Brownian motion, with known positive prior precision ${\tau }_0$. Recover the observed signal from the positive stock price:For a fixed time $t$, the centered observed path $Y_s-(s/t)Y_t$ is a Brownian bridge independent of $(\alpha ,Y_t)$: it is independent of the prior and has zero covariance with $W_t$ in their joint Gaussian law. Thus the whole observed history has the same posterior information about $\alpha$ as its endpoint. This proves Brownian endpoint sufficiency for a constant drift rather than assuming it.

The likelihood of that endpoint, as a function of $\alpha$, is proportional to $\exp (\alpha Y_t-{\alpha }^{2}t/2)$. Multiplying by the normal distribution prior and completing the square gives the Gaussian Brownian drift filterDefine the observed innovation process ${\hat{W}}_t=Y_t-{\int }_0^t{\hat{\alpha }}_{s}ds$. It is a continuous observed-filtration martingale: for $s<u$, independence of future noise and the tower property of conditional expectation give $\mathbb{E}[\alpha -{\hat{\alpha }}_u\mid {\mathcal{F}}_s^S]=0$, so the conditional mean of its increment vanishes. Its quadratic variation is $t$, since subtracting the finite-variation drift from $Y$ does not change quadratic variation. The Lévy characterization of Brownian motion makes $\hat{W}$ an observed-filtration Brownian motion. Differentiating the posterior mean givesConversely, the last equation determines $\hat{\alpha }$ from $\hat{W}$ with deterministic coefficients and known initial value, and $Y=\hat{W}+\int \hat{\alpha }dt$. Thus the observed and innovation filtrations coincide, supplying the natural Brownian information needed for replication. Consequently the observed stock dynamics are $d{S}_t/S_t=\sigma d{\hat{W}}_t+\sigma {\hat{\alpha }}_{t}dt$. Its observed market price of risk is ${\kappa }_t={\hat{\alpha }}_t-r/\sigma$, and its normalized state-price density solvesThe posterior mean is unbounded, so bounded-coefficient Novikov condition reasoning from Question 2 is not available automatically. A direct finite-horizon argument supplies true pricing. Relative to driftless Wiener measure for $Y$, the Bayesian mixture has observation densityA law with observation drift $k=r/\sigma$ instead has density $e^{k{Y}_t-k^{2}t/2}$. ThereforeThis is the finite-horizon pricing density for Gaussian drift learning. The ratio changes the physical observed law into a genuine probability law, hence has mean one. Indeed, differentiation of the explicit likelihood, using $d{Y}_t=d{\hat{W}}_t+{\hat{\alpha }}_{t}dt$, givesThe Itô formula therefore gives $d{D}_t=-D_t{\kappa }_{t}d{\hat{W}}_t$, agreeing with the stochastic exponential above. Under this pricing law the stock drift is $\sigma k=r$.

For logarithmic utility, terminal marginal utility is $1/w_T$. The state-price budget constraint and the marginal condition imply $w_T^{\ast }=1/(y{\zeta }_T)$ and $1/y=w_0$. Since ${\zeta }_T{w}_T^{\ast }=w_0$ is constant, its conditional-expectation pricing process is constant as well. Thus the entire optimal wealth process isIndeed, for any other nonnegative admissible terminal wealth $X$, the elementary inequality $\log (X/w_T^{\ast })\le X/w_T^{\ast }-1$ givesThis directly verifies optimality. Expected candidate log wealth is finite because $\mathbb{E}{\hat{\alpha }}_t^2={\hat{\alpha }}_0^2+{\tau }_0^{-1}-({\tau }_0+t{)}^{-1}$ and its time integral is finite on every fixed horizon. Applying the Itô formula to $1/\zeta$ givesThe observed self-financing portfolio with risky dollar fraction ${\pi }_t$ has diffusion coefficient ${\pi }_t\sigma$. Matching the coefficients proves the log-optimal investment with Gaussian drift learning ruleThe remaining dollars are held in the bank account; the fraction may be negative or exceed one because the question imposes neither short-sale nor borrowing restrictions.

Use ${\theta }_t$ for the dollar amount invested in the stock; the number of shares is ${\theta }_t/S_t$. The bank account contains $w_t-{\theta }_t$ dollars. With no intermediate consumption, the self-financing portfolio equation isIf instead $\theta$ denotes share holdings, replace each dollar ${\theta }_t$ here by $S_t{\theta }_t$. The equation says that trading only reallocates existing portfolio wealth; gains come from cash interest and stock returns.

For the standard unconstrained complete-diffusion formulation, assume the completed natural Brownian filtration, initial wealth $w_0>0$, and nonnegative admissible wealth. Define the market price of risk ${\kappa }_t=({\mu }_t-r_t)/{\sigma }_t$. The boundedness assumptions make $\kappa$ bounded. The state-price density normalized at zero isIts money-market-adjusted process $Z_t=e^{{\int }_0^t{r}_{s}ds}{\zeta }_t$ is a true martingale by the Novikov condition. Under $d\mathbb{Q}=Z_{T}d\mathbb{P}$, the Girsanov theorem gives $W_t^{\mathbb{Q}}=W_t+{\int }_0^t{\kappa }_{s}ds$ as a Brownian motion, and the discounted stock has zero drift.

The Itô product rule applied to the wealth equation givesFor nonnegative admissible wealth this is a nonnegative local martingale, hence a supermartingale. Therefore every affordable terminal payoff $X=w_T$ satisfies the state-price budget constraint $\mathbb{E}[{\zeta }_{T}X]\le w_0$. Nonzero volatility and the Brownian martingale representation theorem allow replication of nonnegative claims with finite state-price cost.

For a differentiable strictly concave utility function with the usual conditions ensuring an interior integrable optimizer, put $I=(U^{\prime }{)}^{-1}$, the inverse marginal utility. Pointwise maximization of $U(x)-y{\zeta }_{T}x$ givesThe second equation determines the positive constant $y$. To prove optimality, use the concave supporting-tangent inequality:Taking expectations and using the budget bound proves that the candidate dominates every admissible affordable payoff, whenever the utility expectations are well defined. This is the complete-market terminal utility optimizer.

The actual wealth and dollar portfolio are as explicit as general adapted coefficients permit. LetThe Brownian martingale representation theorem, with localization if only $L^1$ integrability is initially known, gives $h$. The dollar portfolio from a deflated wealth martingale isThe bank holds $w_t^{\ast }-{\theta }_t^{\ast }$ dollars. Equivalently, with $B_t=e^{{\int }_0^t{r}_{s}ds}$, $w_t^{\ast }=B_t{\mathbb{E}}^{\mathbb{Q}}[w_T^{\ast }/B_T\mid {\mathcal{F}}_t]$. These formulas need no unjustified differentiability of a value function with random coefficients.

The printed assumption that $U$ is only increasing and concave does not guarantee that the interior formula exists, or even that the optimal value is finite. In general replace the inverse formula by an affordable selectionwhere $\mathcal{D}$ is the wealth domain; boundary choices or supporting slopes handle nondifferentiability. Existence, finite cost and replication still have to hold. A concrete obstruction is the unbounded linear terminal-wealth utility: take $U(x)=x$, $r=0$, $\mu =\sigma =1$, and the Brownian filtration. Then ${\zeta }_T=e^{-W_T-T/2}$. For $A_n=\{{\zeta }_T<1/n\}$, the bounded nonnegative claimis replicable, costs $w_0$, and has $\mathbb{E}{X}_n>n{w}_0$. Thus there is no finite optimal expected utility under the literal utility hypotheses alone. Also, bounded coefficients do not ensure market completeness in an arbitrarily enlarged filtration containing untraded randomness; the full replication formula uses the Brownian market information assumption stated above.

For a cell $(i,j)$ in the Young diagram of $\lambda$, its hook of a Young diagram consists of that cell, the cells to its right in row $i$ and the cells below it in column $j$. Its length isA removable rim hook is a connected border strip $\lambda /\mu$ containing no $2\times 2$ square, whose removal leaves a Young diagram. Its leg length is the number of rows it meets minus one. The hook graph is the Young diagram labeled at each cell by $h_{ij}$; here “graph” means this labeled array, not a graph of possible hook-walk moves.

Choose $l\ge \ell (\lambda )$, pad with zeros and form the beta numbersThey are distinct nonnegative integers in decreasing order and recover ${\lambda }_i={\beta }_i-l+i$. For $l=\ell (\lambda )$ they equal the first-column hook lengths, since $h_{i1}={\lambda }_i+l-i$. If $l$ is increased, old beta numbers shift upward by one and a new bead at zero is inserted.

For a prime-modulus partition abacus arrange the nonnegative positions in $p$ runners: position $pq+r$ lies at level $q$ on runner $r$, $0\le r<p$. Put beads at the beta numbers and leave the other positions empty. Replacing a bead at $b$ by a bead at a vacant $b-k\ge 0$ removes a rim hook of length $k$; its leg length is the number of beads strictly between these positions. This follows by tracing the boundary of the diagram: bead positions record vertical steps and gaps record horizontal steps, and exchanging one bead and gap removes the intervening border strip. In particular a removable $p$-hook is exactly a bead that can slide up one place on its own runner.

The partition core at modulus $p$ is the result of sliding beads up until there are no gaps above any bead on a runner. If runner $r$ initially has $b_r$ beads, its final occupied levels must be $0,1,\dots ,b_r-1$, independently of slide order. Thus its packed beta set, and hence its core partition of an integer, is unique. Increasing the number of beta numbers merely re-expresses the same boundary by shifting positions and adding the new initial bead; the packed configuration represents the same partition of an integer. This proves well-definedness both with respect to removal order and to the chosen padding. The number of slides isThe partition quotient components record how far each runner's beads lie above their packed positions, giving the hook-removal interpretation used in Question 5.

A block of a group algebra in characteristic $p$ for $F{S}_n$ is a two-sided ideal $B=F{S}_{n}e$ for a primitive central idempotent $e$, meaning a nonzero central idempotent that is not a sum of two nonzero orthogonal central idempotents. The identity is the sum of these block idempotents and the group algebra is their direct product. An indecomposable module $V$ belongs to $B$ when $eV=V$, equivalently every other block idempotent acts as zero. Indeed $V={⨁}_{e}eV$; indecomposability makes exactly one summand nonzero.

The Nakayama block theorem, historically called the Nakayama Conjecture, statesEquivalently the ordinary characters assigned to one modular block have a common partition core at modulus $p$. Its simple modules are the $D^{\lambda }$ with regular $\lambda$ having that core. Only cores $\kappa$ for which $n-|\kappa |$ is a nonnegative multiple of $p$ occur. The block weight is $(n-|\kappa |)/p$; a core alone determines the block at this fixed degree.

One route to the proof separates the combinatorial core invariant from the algebraic center invariant. Let $c_r(\lambda )$ be the number of cells of residue $r=j-i(\mathrm{mod}p)$. A length-$p$ rim hook has one cell of every residue: along its border strip the contents are consecutive integers. Therefore removal subtracts one from every $c_r$. More precisely, choose the number $l$ of beta numbers divisible by $p$ and write the runner charges as $q_r=b_r-l/p$. Reading the boundary, or adding a cell and checking its two changed beta positions, givesThe charges determine the packed runners and hence the core. For fixed $n$, two equal cores have the same weight, so their full residue multisets are equal. Conversely equal residue multisets give equal charges and the same core. This establishesfor partitions of an integer of the same integer. The fixed-degree condition matters.

For the center invariant, use the commuting Jucys–Murphy elements $J_k={\sum }_{i<k}(ik)$, with $J_1=0$. In the Young seminormal form basis, $J_k$ has eigenvalue equal to the content of the cell containing $k$. A symmetric polynomial in them therefore acts on $S^{\lambda }$ as the same scalar for every Young tableau: its value on the multiset of all contents of $\lambda$. The integral Specht lattice makes this a scalar identity before modular reduction, so modulo $p$ these central scalars depend only on the residue multiset, and every composition factor inherits them.

The substantive center theorem used in this proof is that the center of the symmetric-group algebra, integrally and over a field, consists of symmetric polynomials in the Jucys–Murphy elements. Centrality of elementary symmetric polynomials follows fromthe coefficients are central class-sum combinations. The reverse inclusion requires the integral triangular class-sum argument: expand symmetric monomials in the $J_k$ as products of transpositions, group the leading terms by reduced cycle type, and eliminate these leading class sums in the resulting unitriangular order. The integral coefficients, with no division by multiples of $p$, are crucial; knowing the center only over $\mathbb{Q}$ would not prove the modular assertion. This is the central algebraic ingredient of this proof route.

Thus equal residue multisets yield equal characters of the entire modular center. Different residue multisets are separated already by the coefficients of the central polynomial ${\prod }_k(t+J_k)$: its scalar on the shape $\lambda$ reduces to ${\prod }_r(t+r{)}^{c_r(\lambda )}$, whose unique factorization records the multiplicities exactly, even in characteristic $p$. In a finite-dimensional split algebra, two simple modules are in the same block exactly when their central characters agree: the center of each block is local, while different block idempotents separate different blocks. It follows that equal cores give one block and different cores give distinct blocks. This proves both directions rather than just observing that core is constant on known blocks.

Finally, each ordinary Specht lattice has a single reduced central character, so all constituents of its reduction lie in that one block. Question 1 supplies every simple module and its label; the regular labels therefore identify all the simples in each core block. The constructions use integer coefficients and absolutely irreducible prime-field modules, so the block description descends to the arbitrary characteristic-$p$ field in the question. This completes the main steps of the block proof without presupposing semisimplicity in positive characteristic.

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.

Use the pair of partitions for a generalized Specht module convention appropriate to generalized Specht modules. Put $a={\mu }^{♯}$ and $b=\mu$. Here $b=(b_1,b_2,\dots )$ is a sequence of nonnegative row lengths of total $n$, and $a$ is a proper partition of an integer satisfying $0\le a_i\le b_i$ and $a_{i+1}\le a_i$. The row-length sequence $b$ is allowed to be a composition; it need not decrease. The first $a_i$ cells of row $i$ are marked. This construction is not an arbitrary pair of unrelated partitions of an integer and is not simply a skew diagram $b/a$.

A word has type $b$ if it contains $b_i$ occurrences of $i$. Read it from left to right. Every occurrence of $1$ is good; an occurrence of $i+1$ is good exactly when there have so far been more good occurrences of $i$ than good occurrences of $i+1$. Otherwise it is bad. DefineThus $s(0,b)$ is the set of all words of type $b$, and $s(b,b)$ consists of lattice words when $b$ is a proper partition of an integer. Marking the whole first row makes no difference because every $1$ is good.

For a Young tableau $T$ with row lengths $b$, let $C_T^a$ permute labels within each column of its marked subdiagram, fixing all unmarked labels. Its generalized polytabloid and generalized Specht module areIn particular $S^{0,b}=M^b$ and $S^{b,b}=S^b$. The Young tableau shape in this definition must be the row sequence $b$.

We give the filtration argument, keeping its combinatorial and module-theoretic steps separate. Normalize $a_1=b_1$, and if $a\ne b$ choose the first row $c>1$ with $a_c<b_c$. Then $a_{c-1}=b_{c-1}$. Define $A_c(a,b)=(a+e_c,b)$ if $a_{c-1}>a_c$; otherwise the add branch is empty. Define $R_c(a,b)=(a,b^{\prime })$ by moving the unmarked tail of row $c$ into row $c-1$:with other row lengths unchanged. Re-normalize the first marked row if necessary. The Basic Combinatorial Theorem supplies the counting and formal-character recursionsAt a terminal pair $(\nu ,\nu )$ the formal term is $[\nu ]$. Thus the second recursion specifies the Specht multiplicities by the terminal leaves, with repeated leaves counted separately.