p-regular and p-restricted labels under transpose and sign

Suppose $\mu$ is $p$-restricted, so that ${\mu }^{\mathrm{\prime }}$ is $p$-regular by [F1]. Applying [F2] with $\lambda :={\mu }^{\mathrm{\prime }}$ gives a $k[S_n]$-isomorphism $S_k^{{\mu }^{\mathrm{\prime }}}\otimes \mathrm{sgn}\cong (S_k^{({\mu }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}}{)}^{\ast }=(S_k^{\mu }{)}^{\ast }=S(\mu )$, because transposition is an involution.

Let $M$ be a $k[S_n]$-module and let $L$ be a one-dimensional $k[S_n]$-module with basis $w$ and character $\epsilon (\sigma )\in k^{\times }$, so that $\sigma \cdot (x\otimes w)=\epsilon (\sigma )(\sigma x)\otimes w$. Identifying $M\otimes L$ with $M$ by the linear isomorphism $x\otimes w\mapsto x$, the action becomes $\sigma \cdot x=\epsilon (\sigma )\sigma x$. Since every $\epsilon (\sigma )$ is a nonzero scalar, a subspace $W\le M$ is $S_n$-stable for the twisted action if and only if it is $S_n$-stable for the original action; hence the two actions have the same submodule lattice, the same maximal submodules and the same radical, the radical being their intersection and the head the quotient by it [F5]. Consequently $\mathrm{hd}(M\otimes L)\cong \mathrm{hd}(M)\otimes L$, and if $M$ is simple (respectively absolutely irreducible) then so is $M\otimes L$. For $L=\mathrm{sgn}$ one has ${\epsilon }^2=1$ by [F4], so twisting twice returns the original action and $(M\otimes \mathrm{sgn})\otimes \mathrm{sgn}\cong M$.

By [F3] the $p$-regular partition ${\mu }^{\mathrm{\prime }}$ satisfies $\mathrm{hd}(S_k^{{\mu }^{\mathrm{\prime }}})=D^{{\mu }^{\mathrm{\prime }}}$ with $R^{{\mu }^{\mathrm{\prime }}}=\mathrm{rad}(S_k^{{\mu }^{\mathrm{\prime }}})$ the unique maximal submodule. Using step 1.1 and step 1.2 with $L=\mathrm{sgn}$, $D(\mu )=\mathrm{hd}(S(\mu ))\cong \mathrm{hd}(S_k^{{\mu }^{\mathrm{\prime }}}\otimes \mathrm{sgn})\cong \mathrm{hd}(S_k^{{\mu }^{\mathrm{\prime }}})\otimes \mathrm{sgn}=D^{{\mu }^{\mathrm{\prime }}}\otimes \mathrm{sgn},$ which is nonzero, simple, self-dual and absolutely irreducible by [F3] and step 1.2. This proves assertion 1.

Let $\lambda$ be $p$-regular. By [F1] the conjugate ${\lambda }^{\mathrm{\prime }}$ is $p$-restricted, so step 2.1 applies to $\mu :={\lambda }^{\mathrm{\prime }}$ and gives $D({\lambda }^{\mathrm{\prime }})\cong D^{({\lambda }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}}\otimes \mathrm{sgn}=D^{\lambda }\otimes \mathrm{sgn}$. By step 1.2, tensoring this isomorphism with $\mathrm{sgn}$ and using $(M\otimes \mathrm{sgn})\otimes \mathrm{sgn}\cong M$ yields $D({\lambda }^{\mathrm{\prime }})\otimes \mathrm{sgn}\cong D^{\lambda }$. This is assertion 2.

For assertion 3, first note that $\mu \mapsto D(\mu )$ is injective on $p$-restricted partitions: if $D({\mu }_1)\cong D({\mu }_2)$, then step 2.1 gives $D^{{\mu }_1^{\mathrm{\prime }}}\otimes \mathrm{sgn}\cong D^{{\mu }_2^{\mathrm{\prime }}}\otimes \mathrm{sgn}$, and tensoring with $\mathrm{sgn}$ and using step 1.2 gives $D^{{\mu }_1^{\mathrm{\prime }}}\cong D^{{\mu }_2^{\mathrm{\prime }}}$; by [F3] the $p$-regular partitions ${\mu }_1^{\mathrm{\prime }},{\mu }_2^{\mathrm{\prime }}$ are equal, hence ${\mu }_1={\mu }_2$ by [F1]. It is surjective as well: if $X$ is a simple $k[S_n]$-module, then $X\otimes \mathrm{sgn}$ is simple by step 1.2, so by [F3] it is isomorphic to $D^{\lambda }$ for some $p$-regular $\lambda$; taking $\mu :={\lambda }^{\mathrm{\prime }}$, which is $p$-restricted by [F1], step 3.1 gives $X\cong (X\otimes \mathrm{sgn})\otimes \mathrm{sgn}\cong D^{\lambda }\otimes \mathrm{sgn}\cong D(\mu )$, where the first isomorphism is the canonical one of step 1.2. Hence $\mu \mapsto D(\mu )$ is a bijection.

Assertions 1, 2 and 3 are steps 2.1, 3.1 and 4.1. For $n=0$ the unique partition $\mathrm{\varnothing }$ is $p$-restricted and $p$-regular by [F1], $S(\mathrm{\varnothing })\cong k^{\ast }\cong k$ is the trivial module, and $D(\mathrm{\varnothing })\cong D^{\mathrm{\varnothing }}\otimes \mathrm{sgn}\cong k$ is the unique simple module, so all statements hold. In characteristic $2$ the sign representation is trivial by [F4], so assertions 1-3 read $D(\mu )\cong D^{{\mu }^{\mathrm{\prime }}}$ and $D^{\lambda }\cong D({\lambda }^{\mathrm{\prime }})$, with the transposition still present; if additionally $\mu ={\mu }^{\mathrm{\prime }}$ then the two labellings agree on $\mu$, and otherwise they differ. The argument nowhere divides by $p$ or by a group order and never averages over $S_n$.