Modular simple modules of the symmetric group

Fix a partition $\lambda ⊢n$. By [F2] applied to $F=k$: $D^{\lambda }=0$ if and only if $\lambda$ is not $p$-regular, and if $\lambda$ is $p$-regular then $D^{\lambda }\mathrm{\ne }0$ is self-dual and absolutely irreducible, $R^{\lambda }=\mathrm{rad}(S_k^{\lambda })$ is the unique maximal submodule of $S_k^{\lambda }$, and $D^{\lambda }$ is the simple head of $S_k^{\lambda }$. This is assertion 1.

By [F4] and [F5], $\sigma \in S_n$ is $p$-regular if and only if no cycle length of $\sigma$ is divisible by $p$; in the cycle-type notation of [F6] this says $c_k(\sigma )=0$ whenever $p\mid k$.

We prove the generating-function identity ${\prod }_{j\ge 1}(1+x^j+\cdots +x^{(p-1)j})={\prod }_{p\nmid j}(1-x^j{)}^{-1}$ in the formal power series ring $\mathbb{Z}[\text{\hspace{0.17em}⁣}[x]\text{\hspace{0.17em}⁣}]$, coefficient by coefficient. Fix $N\ge 1$ and use $1+x^j+\cdots +x^{(p-1)j}=(1-x^{pj})(1-x^j{)}^{-1}$ in $\mathbb{Z}[x]$: ${\prod }_{j=1}^N(1+x^j+\cdots +x^{(p-1)j})=({\prod }_{\begin{array}{c}m\le pN\\ p\mid m\end{array}}(1-x^m))({\prod }_{m\le N}(1-x^m){)}^{-1}={\prod }_{\begin{array}{c}N<m\le pN\\ p\mid m\end{array}}(1-x^m){\prod }_{\begin{array}{c}m\le N\\ p\nmid m\end{array}}(1-x^m{)}^{-1},$ because the factors $(1-x^m)$ with $p\mid m$ and $m\le N$ occur in numerator and denominator and cancel. Every factor of the first product after the cancellation has exponent $m>N$, so the product is $1$ modulo $x^{N+1}$. Therefore the two sides of the displayed identity have equal coefficients of $x^n$ for every $n$: taking $N\ge n$ and reducing the finite truncations modulo $x^{n+1}$ shows that any coefficient of $x^n$ is a finite sum of $\pm 1$'s on both sides.

By [F6] the map sending a conjugacy class to the cycle type of any representative is a bijection onto the tuples $(c_1,\dots ,c_n)$ with ${\sum }_{k}k{c}_k=n$, and by step 1.2 a class is $p$-regular exactly when its tuple satisfies $c_k=0$ for every $k$ divisible by $p$. Such tuples are exactly the partitions of $n$ all of whose parts are not divisible by $p$. Hence $\mathrm{\#}\{p\text{-regular classes of }{S}_n\}=\mathrm{\#}\{\text{partitions of }n\text{ into parts not divisible by }p\}\mathrm{.}$

The coefficient of $x^n$ in ${\prod }_{j\ge 1}(1+x^j+\cdots +x^{(p-1)j})$ is the number of tuples $(e_j{)}_{j\ge 1}$ with $0\le e_j\le p-1$ and ${\sum }_{j}j{e}_j=n$; only $j\le n$ can contribute, so this is a finite count, and such a tuple records exactly the partition of $n$ in which the part $j$ occurs $e_j<p$ times. Hence this coefficient is the number of $p$-regular partitions of $n$. The coefficient of $x^n$ in ${\prod }_{p\nmid j}(1-x^j{)}^{-1}$ is likewise the number of partitions of $n$ all of whose parts are not divisible by $p$. By step 1.3 the two coefficients are equal, so $\mathrm{\#}\{p\text{-regular }\lambda ⊢n\}=\mathrm{\#}\{\text{partitions of }n\text{ into parts not divisible by }p\}\mathrm{.}$

Every $p$-regular $\lambda ⊢n$ gives a nonzero simple module $D^{\lambda }$ by step 1.1, and distinct $p$-regular partitions give non-isomorphic modules: if $D^{\lambda }\cong D^{\mu }$ with $\lambda ,\mu$ $p$-regular, then $\lambda =\mu$ by [F3] applied to $F=k$. Hence $\lambda \mapsto [D^{\lambda }]$ is an injection from the set of $p$-regular partitions of $n$ into the set of isomorphism classes of simple $k[S_n]$-modules, and therefore the number of isomorphism classes of simple $k[S_n]$-modules is at least the number of $p$-regular partitions of $n$.

Combining steps 2.1 and 2.2 gives $\mathrm{\#}\{p\text{-regular }\lambda ⊢n\}=\mathrm{\#}\{p\text{-regular classes of }{S}_n\},$ and by [F7] applied to the finite group $S_n$ over its splitting field $k$ this common number equals the number of isomorphism classes of simple $k[S_n]$-modules.

By step 3.1 the number of isomorphism classes of simple $k[S_n]$-modules equals the number of $p$-regular partitions of $n$, while step 2.3 exhibits an injection between the same two finite sets. An injection between finite sets of equal cardinality is a bijection, so every simple $k[S_n]$-module is isomorphic to $D^{\lambda }$ for exactly one $p$-regular $\lambda ⊢n$. Combined with the self-duality and absolute irreducibility of step 1.1, this is assertions 2 and 3.

Assertion 1 is step 1.1, assertion 2 is step 2.3, and assertion 3 is step 4.1; no part of the argument assumes more about $k$ than that it is a splitting field of characteristic $p$ for $S_n$ and its subgroups. For $n=0$ there is exactly one partition, $\mathrm{\varnothing }$, of $0$, and it is $p$-regular by [F1]; $S_0$ has one element, of order $1$, so its unique class is $p$-regular and the counts $1=1=1$ hold; $D^{\mathrm{\varnothing }}\cong k$ is the one simple $k[S_0]$-module. For $p=2$ the same count applies: the $2$-regular partitions of $n$ are those with distinct parts, the $2$-regular classes of $S_n$ are those with all cycle lengths odd, and both are counted by the same coefficient. All counting is coefficient-wise finite, and no step divides by $p$, by a group order, or averages over a group.