The Frobenius characteristic preserves outer products

Both sides of the asserted identity are $\mathbb{Z}$-bilinear in $(f,g)$: the outer product is bilinear by [F1], the characteristic map is linear by [F2], and multiplication in ${\mathrm{\Lambda }}_{\mathbb{C}}$ is bilinear. Since every element of $R(S_m)$ is an integral combination of honest characters, and likewise for $R(S_n)$, it suffices to prove $\mathrm{ch}(\chi \circ \psi )=\mathrm{ch}(\chi )\mathrm{ch}(\psi )$ for honest characters $\chi$ of $S_m$ and $\psi$ of $S_n$.

For honest $\chi ,\psi$, the character $\chi ⊠\psi$ of $H$ is honest, so Frobenius' formula [F3] applied to $G=S_{m+n}$ and the subgroup $H$ of [F5] gives $(\chi \circ \psi )(w)=\frac{1}{m!n!}{\sum }_{x\in S_{m+n}:x^{-1}wx\in H}(\chi ⊠\psi )(x^{-1}wx)$.

The condition $x^{-1}wx\in H$ says that $x^{-1}wx$ preserves $\{1,\dots ,m\}$, equivalently that $w$ preserves $A:=x(\{1,\dots ,m\})$. Hence the $x$ occurring in the sum are exactly those with $x(\{1,\dots ,m\})=A$ for some $m$-element $w$-invariant subset $A\subseteq \{1,\dots ,m+n\}$, and for each such $A$ there are exactly $m!n!$ permutations $x$ with $x(\{1,\dots ,m\})=A$. For all $x$ with the same $A$, the permutation $x^{-1}wx\in H$ has its $S_m$-component conjugate through $x$ to $w{\mathrm{\mid }}_A$ and its $S_n$-component conjugate to $w{\mathrm{\mid }}_{A^c}$, so $(\chi ⊠\psi )(x^{-1}wx)=\chi ({\mu }_A)\psi ({\nu }_A)$, where ${\mu }_A$ is the cycle type of $w{\mathrm{\mid }}_A$ and ${\nu }_A$ the cycle type of $w{\mathrm{\mid }}_{A^c}$; this is independent of $x$. With the factor $m!n!\mathrm{/}m!n!$ cancelling, $(\chi \circ \psi )(w)={\sum }_A\chi ({\mu }_A)\psi ({\nu }_A)$, the sum over the $m$-element $w$-invariant subsets $A$.

An $m$-element set $A$ is $w$-invariant exactly when it is a union of cycles of $w$, and then $\mathrm{\mid }{\mu }_A\mathrm{\mid }=m$ and the multisets of parts of ${\mu }_A$ and ${\nu }_A$ merge to the multiset of parts of $\rho$. Conversely, every split $\rho =\mu \uplus \nu$ with $\mathrm{\mid }\mu \mathrm{\mid }=m$ arises this way. For a fixed split, the number of $m$-element $w$-invariant $A$ with ${\mu }_A=\mu$ is ${\prod }_{i\ge 1}\left(\genfrac{}{}{0px}{}{m_i(\rho )}{m_i(\mu )}\right)$, because for each cycle length $i$ one independently chooses which $m_i(\mu )$ of the $m_i(\rho )$ cycles of $w$ of length $i$ are included in $A$. Therefore $(\chi \circ \psi )(\rho )={\sum }_{\mu \uplus \nu =\rho ,\mathrm{\mid }\mu \mathrm{\mid }=m}\left({\prod }_{i\ge 1}\left(\genfrac{}{}{0px}{}{m_i(\rho )}{m_i(\mu )}\right)\right)\chi (\mu )\psi (\nu )$, an expression depending only on the cycle type $\rho$ of $w$.

On the other side $\mathrm{ch}(\chi )\mathrm{ch}(\psi )={\sum }_{\mu ⊢m}{\sum }_{\nu ⊢n}\chi (\mu )\psi (\nu )p_{\mu }{p}_{\nu }\mathrm{/}(z_{\mu }{z}_{\nu })$, using $p_{\mu }{p}_{\nu }=p_{\mu \uplus \nu }$ from [F4], the coefficient of $p_{\rho }\mathrm{/}{z}_{\rho }$ equals ${\sum }_{\mu \uplus \nu =\rho }\chi (\mu )\psi (\nu )z_{\rho }\mathrm{/}(z_{\mu }{z}_{\nu })$; for a split $\mu \uplus \nu =\rho$ one has $\frac{z_{\rho }}{z_{\mu }{z}_{\nu }}={\prod }_{i\ge 1}\frac{m_i(\rho )!}{m_i(\mu )!m_i(\nu )!}={\prod }_{i\ge 1}\left(\genfrac{}{}{0px}{}{m_i(\rho )}{m_i(\mu )}\right)$, since $m_i(\nu )=m_i(\rho )-m_i(\mu )$. This is exactly the coefficient in step 3.1; since $\{p_{\rho }\mathrm{/}{z}_{\rho }:\rho ⊢m+n\}$ is a $\mathbb{C}$-basis of ${\mathrm{\Lambda }}_{\mathbb{C}}^{m+n}$ by scalar extension [F4] and [F2] gives the same power-sum coefficients for the characteristic, $\mathrm{ch}(\chi \circ \psi )=\mathrm{ch}(\chi )\mathrm{ch}(\psi )$.

By step 1.1 the identity holds for all virtual characters $f\in R(S_m)$ and $g\in R(S_n)$. The cycle-split formula in step 3.1 also extends to these $f,g$ by bilinearity: $(f\circ g)(\rho )={\sum }_{\mu \uplus \nu =\rho ,\mathrm{\mid }\mu \mathrm{\mid }=m}\left({\prod }_i\left(\genfrac{}{}{0px}{}{m_i(\rho )}{m_i(\mu )}\right)\right)f(\mu )g(\nu )$. If $f,g$ are rational-valued, every term is rational, so $f\circ g$ is rational-valued. By [F2], $\mathrm{ch}(f)\in {\mathrm{\Lambda }}_{\mathbb{Q}}^m$ and $\mathrm{ch}(g)\in {\mathrm{\Lambda }}_{\mathbb{Q}}^n$, while their product and $\mathrm{ch}(f\circ g)$ lie in ${\mathrm{\Lambda }}_{\mathbb{Q}}^{m+n}$. Thus the identity holds over $\mathbb{Q}$ as claimed.