The outer product of $S^{(3,2)}$ and $S^{(2)}$

For each $t$, put $G_t^0=\mathrm{Sym}(\{0,\dots ,t-1\})$ and $G_t^1=\mathrm{Sym}(\{1,\dots ,t\})$, and let ${\beta }_t(i)=i+1$ and $c_t(\sigma )={\beta }_t\sigma {\beta }_t^{-1}$, with the unique empty bijection for $t=0$. The map $c_t$ is a group isomorphism because it is bijective and $c_t(\sigma \tau )=c_t(\sigma )c_t(\tau )$ by [F6]. For $a+b=t$, $c_t^{-1}$ carries the one-based block subgroup in [F2] to the zero-based block subgroup in [F5]. Relabeling every entry of a tableau by ${\beta }_t$ sends its tabloid to the corresponding tabloid and conjugates its row and column stabilizers; inversion pairs, hence permutation signs, are preserved by the shift. Thus it sends each polytabloid to the relabeled polytabloid by [F8] and intertwines the Specht actions. If $H^1\le G_t^1$ and $H^0=c_t^{-1}(H^1)$, pull an $H^1$-module $W$ back by $h{\cdot }_{0}w=c_t(h){\cdot }_{1}w$. The induced-function map $F^1\mapsto F^0$, $F^0(g)=F^1(c_t(g))$, preserves covariance since $F^0(gh)=c_t(h{)}^{-1}{\cdot }_1{F}^0(g)=h^{-1}{\cdot }_0{F}^0(g)$, and it intertwines left translation because $c_t(g_0^{-1}g)=c_t(g_0{)}^{-1}{c}_t(g)$. Hence the one-based outer-product and Specht-module calculation transports to the global zero-based convention.

Let $f^{\lambda }$ count standard tableaux of shape $\lambda$. In a nonempty standard tableau the largest entry lies in a removable corner: a node with a node to its right or below would have to carry a larger entry by [F12]. Deleting that largest entry gives a bijection with the disjoint union of standard tableaux on the shapes obtained by deleting one removable corner; conversely, appending the largest entry at any removable corner reverses the deletion. Hence $f^{\mathrm{\varnothing }}=1$ and $f^{\lambda }={\sum }_{x\in \mathrm{Rem}(\lambda )}{f}^{\lambda -x}$. The one-row and one-column shapes each have one standard tableau, and the recurrence gives $f^{(2,1)}=f^{(2)}+f^{(1,1)}=1+1=2$, $f^{(3,1)}=f^{(2,1)}+f^{(3)}=2+1=3$, $f^{(4,1)}=f^{(3,1)}+f^{(4)}=3+1=4$, $f^{(5,1)}=f^{(4,1)}+f^{(5)}=4+1=5$, $f^{(2,2)}=f^{(2,1)}=2$, $f^{(3,2)}=f^{(2,2)}+f^{(3,1)}=2+3=5$, $f^{(4,2)}=f^{(3,2)}+f^{(4,1)}=5+4=9$, $f^{(5,2)}=f^{(4,2)}+f^{(5,1)}=9+5=14$, $f^{(3,3)}=f^{(3,2)}=5$, $f^{(4,3)}=f^{(3,3)}+f^{(4,2)}=5+9=14$, $f^{(1,1,1)}=1$, $f^{(2,1,1)}=f^{(1,1,1)}+f^{(2,1)}=1+2=3$, $f^{(3,1,1)}=f^{(2,1,1)}+f^{(3,1)}=3+3=6$, $f^{(4,1,1)}=f^{(3,1,1)}+f^{(4,1)}=6+4=10$, $f^{(2,2,1)}=f^{(2,1,1)}+f^{(2,2)}=3+2=5$, $f^{(3,2,1)}=f^{(2,2,1)}+f^{(3,1,1)}+f^{(3,2)}=5+6+5=16$, $f^{(2,2,2)}=f^{(2,2,1)}=5$, $f^{(4,2,1)}=f^{(3,2,1)}+f^{(4,1,1)}+f^{(4,2)}=16+10+9=35$, $f^{(3,3,1)}=f^{(3,2,1)}+f^{(3,3)}=16+5=21$, and $f^{(3,2,2)}=f^{(3,2,1)}+f^{(2,2,2)}=16+5=21$. By [F9], the input dimensions are $5$ and $1$, and the five output dimensions are $14,14,35,21,21$.

The second factor $S^{(2)}$ is the trivial Specht module, so the outer Pieri rule [F1], with the block induction convention [F2] and the label bridge in step 1.1, expresses the induced module as the multiplicity-one sum over all partitions $\lambda ⊢7$ containing $(3,2)$ for which $\lambda \mathrm{/}(3,2)$ is a horizontal $2$-strip.

In the one-based block model of [F2], $H=S_5\times S_2$ stabilizes $B=\{6,7\}$. The map from left cosets $gH$ to two-element subsets $g(B)$ is a bijection: $H$ is exactly the setwise stabilizer of $B$, and permutations act transitively on two-element subsets. Hence $[S_7:H]=\left(\genfrac{}{}{0px}{}{7}{2}\right)=21$. By [F10] and [F11], the dimension of the induced module is $21\cdot 5\cdot 1=105$ using the input dimensions from step 1.2; the output dimensions in that step sum to $14+14+35+21+21=105$.

If $\lambda ⊢7$ contains $(3,2)$, then ${\lambda }_1\ge {\lambda }_2\ge 2$ and ${\lambda }_1\ge 3$. The case ${\lambda }_2\ge 4$ is impossible because the first two rows would contain at least eight nodes. If ${\lambda }_2=3$, the only possibilities are $(4,3)$ and $(3,3,1)$. If ${\lambda }_2=2$, the remaining nodes give $(5,2)$, $(4,2,1)$, $(3,2,2)$, or $(3,2,1,1)$. The added nodes in these six cases are respectively in columns $\{4,3\}$, $\{3,1\}$, $\{4,5\}$, $\{4,1\}$, $\{1,2\}$, and $\{1,1\}$. By [F4], exactly the last shape fails the horizontal-strip condition. Thus the five valid shapes are precisely $(5,2),(4,3),(4,2,1),(3,3,1),(3,2,2)$, with the column pairs stated.

Step 2.1 gives the induction decomposition over horizontal strips, and step 3.1 lists exactly the five such strips, each with multiplicity one. This proves the displayed module isomorphism; step 2.2 verifies its dimension independently and the result agrees with James's classical computation.