Outer Pieri rules for a trivial or sign factor

For $r\ge 1$, consider the one-row shape $(r)$. Every tabloid has the same single row by [F6], and each column has one node, so [F7] makes its column stabilizer trivial. Its column antisymmetrizer is therefore the identity by [F5], and its polytabloid spans the one-dimensional tabloid module, on which $S_r$ acts trivially. Hence $S^{(r)}$ is the trivial representation from [F8].

For $r\ge 1$, consider the one-column shape $(1^r)$. Every row has one node by [F6], so the tabloids $\sigma \{t\}$ are distinct as $\sigma$ ranges over $S_r$. By [F7] its column stabilizer is all of $S_r$, and the column antisymmetrizer gives a nonzero vector $e_t={\sum }_{\sigma \in S_r}\mathrm{sgn}(\sigma )\sigma \{t\}$ by [F5]. For $\gamma \in S_r$, changing the summation variable and using [F10] gives $\gamma e_t=\mathrm{sgn}(\gamma )e_t$. Every tableau of this shape is $\delta t$ for some $\delta \in S_r$; its column stabilizer is $\delta C_t{\delta }^{-1}$ and [F10] makes the corresponding antisymmetrizer $\delta {\kappa }_t{\delta }^{-1}$, so its polytabloid is $\delta e_t=\mathrm{sgn}(\delta )e_t$. Hence the Specht module is one-dimensional and has the sign action [F9].

Take $r\ge 1$ and $\lambda ⊢m+r$ with $\mu \subseteq \lambda$. By [F1], the multiplicity in the first outer product is $c_{\mu ,(r)}^{\lambda }$. A tableau of content $(r)$ has only the entry $1$, so there is exactly one possible filling; it is semistandard precisely when no two added nodes share a column by [F3, F4], and its word $1^r$ is a lattice word. Therefore $c_{\mu ,(r)}^{\lambda }=1$ exactly for horizontal $r$-strips and is zero otherwise.

For content $(1^r)$, every letter $1,\dots ,r$ occurs once. A lattice word must begin with $1$; inductively, after $1,\dots ,k$ its next letter must be $k+1$, since any larger unused letter would violate the prefix inequality for that letter and its predecessor. Thus the only possible LR word is $12\cdots r$. If a row had two nodes, their distinct entries would appear right-to-left in decreasing order, contradicting this increasing word; hence any such tableau has a vertical strip. Conversely, for a vertical strip, fill the nodes in reading order with $1,\dots ,r$: each row has at most one node, entries increase down each column, and the word is lattice. This filling is unique, so $c_{\mu ,(1^r)}^{\lambda }=1$ exactly for vertical $r$-strips and is zero otherwise.

By [F1], steps 1.3 and 1.4 give the multiplicities in the two induced modules, and steps 1.1 and 1.2 identify their factors $S^{(r)}$ and $S^{(1^r)}$ with the trivial and sign representations. Therefore the induced modules are the stated direct sums, with each listed summand appearing once.

If $r=0$, [F11] gives $\lambda =\mu$ as the only possible shape, and the empty LR tableau has coefficient one by [F2]. The empty Specht module is trivial by [F5, F8], so the outer LR rule [F1] gives the asserted induction module as $S^{\mu }$. All tableau sets and direct sums above are finite, and no representatives or bases are selected; no form of the axiom of choice is used.