An outer-induction multiplicity greater than one

Listing the partitions of $7$ and retaining those with ${\lambda }_1\ge 2$ and ${\lambda }_2\ge 1$ from [F1] gives $(6,1),(5,2),(5,1,1),(4,3),(4,2,1),(4,1,1,1),(3,3,1),(3,2,2),(3,2,1,1),(3,1,1,1,1),(2,2,2,1),(2,2,1,1,1),(2,1,1,1,1,1)$. The two remaining partitions of $7$, $(7)$ and $(1^7)$, do not contain $\mu$ and have coefficient zero by [F4].

Let $f^{\lambda }$ count standard tableaux. In every nonempty standard tableau the largest entry is at a removable corner, that is, a node with no box to its right or below, by [F11]; deleting it leaves a partition and gives a bijection with the standard tableaux of the predecessor shapes, while appending the largest entry at any removable corner reverses the deletion. Thus $f^{\mathrm{\varnothing }}=1$ and $f^{\lambda }={\sum }_{x\in \mathrm{Rem}(\lambda )}{f}^{\lambda -x}$. The one-row and one-column shapes each have count $1$. Repeated application gives $f^{(2,1)}=1+1=2$, $f^{(3,1)}=2+1=3$, $f^{(4,1)}=3+1=4$, $f^{(5,1)}=4+1=5$, $f^{(2,2)}=2$, $f^{(3,2)}=2+3=5$, $f^{(4,2)}=5+4=9$, $f^{(5,2)}=9+5=14$, $f^{(3,3)}=5$, $f^{(4,3)}=5+9=14$, $f^{(1,1,1)}=1$, $f^{(2,1,1)}=1+2=3$, $f^{(3,1,1)}=3+3=6$, $f^{(4,1,1)}=6+4=10$, $f^{(2,2,1)}=3+2=5$, $f^{(3,2,1)}=5+6+5=16$, $f^{(2,2,2)}=5$, $f^{(2,1,1,1)}=1+3=4$, $f^{(2,2,1,1)}=4+5=9$, $f^{(3,1,1,1)}=4+6=10$, $f^{(4,1,1,1)}=10+10=20$, $f^{(5,1,1)}=10+5=15$, $f^{(4,2,1)}=16+10+9=35$, and $f^{(3,2,1,1)}=9+10+16=35$. By [F7], the input dimensions are $2,3$ and the eight summand dimensions in statement order are $14,15,14,35,20,21,21,35$, where $f^{(3,3,1)}=16+5=21$ and $f^{(3,2,2)}=16+5=21$.

Each retained skew diagram has four boxes, so by [F2] a filling is determined by the position of its unique $2$. Checking the row and column inequalities [F3], the semistandard positions for that $2$, in the same order as step 1.1, are $\{(1,6)\};\{(1,5),(2,2)\};\{(1,5),(3,1)\};\{(2,3)\};\{(1,4),(2,2),(3,1)\};\{(4,1)\};\{(2,3)\};\{(3,2)\};\{(4,1)\}$; the last four shapes admit no semistandard filling. In $(4,1,1,1)\mathrm{/}(2,1)$, $(3,2,2)\mathrm{/}(2,1)$ and $(3,2,1,1)\mathrm{/}(2,1)$, the column with two skew boxes forces the $2$ into its lower box, respectively $(4,1)$, $(3,2)$ and $(4,1)$. For $(3,1,1,1,1)$, $(2,2,1,1,1)$ and $(2,1^5)$ a column has at least three boxes, which cannot be strictly filled with only $1$'s and $2$'s; for $(2,2,2,1)$ each of columns $1$ and $2$ forces a separate $2$.

For a word with one $2$ and three $1$'s, the lattice condition in [F4] holds exactly when at least one $1$ is read before the $2$: after that first $1$, every prefix has at least as many $1$'s as $2$'s, and no letters exceed $2$. Removing the first-read top-right position from the semistandard-position lists in step 2.1 when it occurs, the LR-valid $2$-positions are $(5,2):(2,2)$, $(5,1,1):(3,1)$, $(4,3):(2,3)$, $(4,2,1):(2,2),(3,1)$, $(4,1,1,1):(4,1)$, $(3,3,1):(2,3)$, $(3,2,2):(3,2)$, and $(3,2,1,1):(4,1)$. Thus the coefficients in the order of step 1.1 are $0,1,1,1,2,1,1,1,1,0,0,0,0$, with zero also for $(7)$ and $(1^7)$.

Applying the outer Littlewood–Richardson theorem [F6] to the coefficient list in step 3.1 gives exactly the direct sum in the Statement; all other partitions of $7$ have coefficient zero. In particular, the two LR-valid positions $(2,2)$ and $(3,1)$ for $(4,2,1)$ yield $c_{(2,1),(3,1)}^{(4,2,1)}=2$.

By [F10] the global $S_7$ acts on $\{0,1,\dots ,6\}$, while [F5] gives the one-based block realization. The shift $\beta (i)=i+1$ is a bijection from $\{0,1,\dots ,6\}$ to $\{1,2,\dots ,7\}$; $\sigma \mapsto \beta \sigma {\beta }^{-1}$ preserves products and carries the initial three-letter block to $A=\{1,2,3\}$, so it preserves the subgroup index. In the one-based realization the block subgroup $H$ stabilizes $A$, and the map from its left cosets $gH$ to the images $g(A)$ is a bijection with the three-element subsets: every such subset is an image of $A$ under a permutation (extend bijections on $A$ and its four-element complement), and if $g(A)=g^{\mathrm{\prime }}(A)$ then $g^{-1}{g}^{\mathrm{\prime }}$ preserves $A$ and lies in $H$, so $gH=g^{\mathrm{\prime }}H$. Hence $[S_7:S_3\times S_4]=\left(\genfrac{}{}{0px}{}{7}{3}\right)=\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}=35$. By [F8] and [F9], the induced module has dimension $35\cdot (2\cdot 3)=210$. Step 4.1 and the dimensions in step 1.2 give the right-hand side dimension $14+15+14+2\cdot 35+20+21+21+35=210$. The LR enumeration and coset bijection are finite; any left transversal in [F8] is obtained by finitely many selections, which does not require the axiom of choice.