Outer induction makes the graded symmetric-group representation group a commutative graded ring

For each $t$, define ${\beta }_t(i)=i+1$ from $\{0,\dots ,t-1\}$ to $\{1,\dots ,t\}$, with the empty bijection if $t=0$. The group isomorphism $c_t(\sigma )={\beta }_t\sigma {\beta }_t^{-1}$ preserves products and carries the zero-based ordered blocks to the one-based blocks in [F3]. Pullback along $c_t$ identifies representations and character groups. Explicitly, for a one-based subgroup $H^1$ and module $W$, set $H^0=c_{m+n}^{-1}(H^1)$ and $h{\cdot }_{0}w=c_{m+n}(h){\cdot }_{1}w$. The map $F^1\mapsto F^0=F^1\circ c_{m+n}$ has inverse composition with $c_{m+n}^{-1}$, and $F^0(gh)=c_{m+n}(h{)}^{-1}{\cdot }_1{F}^0(g)=h^{-1}{\cdot }_0{F}^0(g)$. Since $c_{m+n}(x^{-1}g)=c_{m+n}(x{)}^{-1}{c}_{m+n}(g)$, it also intertwines the transported left actions in [F5]. Thus the one-based calculation represents the same outer product in $R_S$.

First take honest characters $\chi ,\psi ,\eta$ of $S_m,S_n,S_r$. In the one-based realization let $B_1=\{1,\dots ,m\}$, $B_2=\{m+1,\dots ,m+n\}$, and $B_3=\{m+n+1,\dots ,m+n+r\}$, allowing empty blocks, and let $J$ be the permutations preserving each $B_i$. The identity, products, and inverses preserve each block, so $J\le S_{m+n+r}$ by [F12]. Restriction to the three blocks identifies $J$ with the componentwise product $S_m\times S_n\times S_r$ by [F10, F11]. Both parenthesized two-block embeddings have image $J$, and on a triple $(a,b,c)$ both parenthesized external product characters have value $\chi (a)\psi (b)\eta (c)$ by [F3]. Hence the two parenthesized external product characters on $J$ agree.

In the one-based realization define $s_{m,n}\in S_{m+n}$ by $s_{m,n}(i)=n+i$ for $1\le i\le m$ and $s_{m,n}(m+j)=j$ for $1\le j\le n$. Its two ranges are disjoint and cover $\{1,\dots ,m+n\}$, so it is a permutation, also when one block is empty. If ${\iota }_{m,n}$ denotes the block embedding in [F3], its action on the two blocks gives $s_{m,n}{\iota }_{m,n}(a,b)s_{m,n}^{-1}={\iota }_{n,m}(b,a)$. Thus $s_{m,n}$ conjugates the first block subgroup to the second.

When one block is empty, the block subgroup $S_0\times S_n$ or $S_n\times S_0$ is all of $S_n$, and its external product character identifies with the other factor by [F3]. For any $S_n$-module $V$, evaluation at the identity identifies ${\mathrm{Ind}}_{S_n}^{S_n}V$ with $V$: its inverse sends $v$ to the covariant function $g\mapsto g^{-1}v$, and both maps respect left translation by [F5]. Thus the unit identities hold for honest characters, and bilinearity with finite expansions [F3, F4] gives $e\circ f=f=f\circ e$ for every virtual character.

Apply [F8] at each outer tensor step and then [F7] to induction in stages. Both $(\chi \circ \psi )\circ \eta$ and $\chi \circ (\psi \circ \eta )$ become induction from the same subgroup $J$ to $S_{m+n+r}$ of the equal external product characters in step 1.2. By [F6] their induced characters are equal. Bilinearity and the finite integral character expansions in [F3, F4] extend this equality to all virtual $\chi ,\psi ,\eta$, proving associativity.

For honest characters $\chi ,\psi$, the representation conjugated by $s_{m,n}$ has value $\chi (a)\psi (b)=\psi (b)\chi (a)$ at ${\iota }_{n,m}(b,a)$, so it is exactly $\psi ⊠\chi$ by [F3, F9]. The conjugation-invariance result [F9] therefore gives $\chi \circ \psi =\psi \circ \chi$. Bilinearity and the finite integral expansions in [F3, F4] extend this equality to virtual characters.

Steps 1.1, 2.1, 2.2, and 1.4 establish compatibility with the library's group convention, associativity, commutativity, and the unit. The product on the direct sum is graded by [F3, F4], and all virtual characters are finite integral combinations, so the stated ring axioms follow. Every relabeling and block conjugator was given explicitly, and every extension used finite sums; no form of the axiom of choice is used.