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It suffices to handle one single-box up-move, and then use the chain already constructed. Suppose the affected row sizes are $a={\lambda }_i$ and $b={\lambda }_j$, $i<j$. Since $\lambda$ is a partition of an integer, $a\ge b\ge 1$. Put $N=a+b$. Over the complex numbers, the two-row Young permutation module decomposition givesA zero second row is omitted. Therefore

Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant Young subgroups lie in $K$. Transitivity of induced representations expresses $M^{\lambda }$ and $M^{\nu }$ by inducing the preceding two-row modules, tensored with the trivial representations of the unaffected factors, from $K$ to $S_n$. Induction preserves this direct sum, sofor an actual group representation $L_{\lambda ,\nu }$, not merely a difference of characters. Iterate along the chain and take the direct sum of the induced complements. The resulting complement can be realized as an invariant subspace of $M^{\lambda }$, either through these isomorphisms or by Maschke's theorem. ThusIf $\lambda =\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.

There is an index error in the displayed identity of the original PDF. With its definition of the signed Young permutation module, the tensor identity for induced representations givesBoth superscripts must be $\alpha$. In particular ${\tilde{M}}^{{\lambda }^{\prime }}\cong \mathrm{sgn}\otimes M^{{\lambda }^{\prime }}$, rather than the printed sign twist of $M^{\lambda }$. For $\lambda =(n)$, $n\ge 2$, the printed left side has dimension $n!$, while the printed right side has dimension one. Thus that identity cannot hold as written.

The requested character inner product between $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ nevertheless has a well-defined answer. Let $R$ be the row Young subgroup of a tableau of shape $\lambda$, and let $C$ be its column subgroup, conjugate to $S_{{\lambda }^{\prime }}$. By Frobenius reciprocity and the Mackey restriction formula, the inner product is a sum over $R\backslash S_n/C$ ofThese double cosets are encoded by nonnegative integer matrices $A=(a_{ij})$ with row sums ${\lambda }_i$ and column sums ${\lambda }_j^{\prime }$, recording intersection sizes of row and column blocks. The intersection subgroup is a product of $S_{a_{ij}}$. Its sign representation is trivial exactly when every $a_{ij}\le 1$. Each zero-one matrix contributes one, and every other matrix contributes zero.

There is exactly one zero-one matrix with these margins. Its first row has length ${\lambda }_1$, equal to the number of nonzero column sums, so that row must consist entirely of ones. Delete it, subtract one from every column margin and discard zero columns. The remaining margins are those of $({\lambda }_2,{\lambda }_3,\dots )$ and its conjugate partition; induction forces the rest. The unique matrix is $a_{ij}=1$ precisely for $j\le {\lambda }_i$. ThereforeThis calculation uses the actual two modules requested in the PDF and does not depend on its erroneous displayed isomorphism.