\documentclass[12pt]{article}

\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{amsmath,amssymb,amsthm,mathtools,amscd}
\usepackage[margin=1in]{geometry}
\usepackage[hidelinks]{hyperref}
\usepackage{enumitem}

\newtheoremstyle{plainsl}%
  {\topsep}% space above
  {\topsep}% space below
  {\slshape}% body font
  {}% indent amount
  {\bfseries}% theorem head font
  {.}% punctuation after theorem head
  {5pt plus 1pt minus 1pt}% space after theorem head
  {}% theorem head specification
\theoremstyle{plainsl}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}

\newcommand{\Symm}{\mathbf{Symm}}
\newcommand{\NSym}{\mathbf{NSym}}
\newcommand{\ZZ}{\mathbb Z}
\newcommand{\QQ}{\mathbb Q}
\newcommand{\NN}{\mathbb N}
\newcommand{\CC}{\mathbb C}
\newcommand{\Cl}{\operatorname{Cl}}
\newcommand{\ch}{\operatorname{ch}}
\newcommand{\Ind}{\operatorname{Ind}}
\newcommand{\Res}{\operatorname{Res}}
\newcommand{\Des}{\operatorname{Des}}
\newcommand{\Fix}{\operatorname{Fix}}
\newcommand{\lcm}{\operatorname{lcm}}
\newcommand{\id}{\operatorname{id}}
\newcommand{\one}{\mathbf 1}
\newcommand{\boxt}{\mathbin{\boxtimes}}
\newcommand{\depth}{\operatorname{depth}}
\newcommand{\Inj}{\operatorname{Inj}}
\newcommand{\sgn}{\operatorname{sgn}}
\newcommand{\Hom}{\operatorname{Hom}}
\newcommand{\End}{\operatorname{End}}
\newcommand{\tr}{\operatorname{tr}}
\newcommand{\rad}{\operatorname{rad}}
\newcommand{\BoxProd}{\mathbin{\boxdot}}
\newcommand{\AntiProd}{\mathbin{\diamondsuit}}
\newenvironment{statement}{\begin{quote}}{\end{quote}}

\title{The Anti-Arithmetic Product of Symmetric Functions}
\author{GPT-6, edited by Darij Grinberg}
\date{\today}

\begin{document}

\maketitle

\begin{abstract}
The \emph{arithmetic product} $\boxdot$ is a bilinear operation
on the ring of symmetric functions over $\mathbb Q$ defined by
\[
 p_\lambda \ \boxdot\ p_\mu
 = \prod_{i,j}p_{\operatorname{lcm}(\lambda_i,\mu_j)}^{\gcd(\lambda_i,\mu_j)}
\]
on the power-sum basis $(p_\lambda)$.
It is known that this operation preserves the subring
$\mathbf{Symm}_{\mathbb Z}$ of symmetric functions with
integer coefficients, and is Schur-positive on pairs of Schur
functions. Indeed, it corresponds to induction of
representations from $S_n \times S_m$ to $S_{nm}$.

In this work, we consider the \emph{anti-arithmetic product}
$\diamondsuit$, which is defined by the same formula but with
$\gcd$ and $\operatorname{lcm}$ interchanged.
We show that it, too, preserves $\mathbf{Symm}_{\mathbb Z}$,
even though it lacks the Schur positivity property.
This was conjectured on MathOverflow in 2014.

Our method is a more intricate variant of the
representation-theoretical interpretation of $\boxdot$.
The anti-arithmetic product does not correspond to an
operation on actual representations; but its adjoint can still
be described as a map on characters, which (as we show) is a
$\lambda$-ring morphism from the character ring of
$S_{mn}$ to that of $S_m \times S_n$.
Now, a 1975 theorem of Boorman shows that the
representation ring of $S_n$ is generated by the natural
permutation representation $M_n = \mathbb Q^n$ as a $\lambda$-ring.
Thus, the proof of integrality boils down to only computing
the image of $M_{mn}$ under this adjoint, which can be done using
M\"obius inversion in the representation ring.

% We prove both integrality statements using representation theory.
% For the arithmetic product, the representation-theoretic interpretation
% is well-known:  The Frobenius characteristic identifies
% the $n$-th graded components of $\Symm_{\ZZ}$ and of $\Symm_{\QQ}$
% with (the additive groups of) the representation ring and
% the class function algebra of $S_n$, respectively;
% thus, $\BoxProd$ becomes a bilinear product
% on the class functions, and in this guise it turns out to be adjoint
% to the pullback of a natural group homomorphism
% $S_m \times S_n \to S_{mn}$. This yields both integrality and
% Schur positivity.

% For the anti-arithmetic product, there is no Schur positivity, and
% thus $\AntiProd$ cannot be adjoint to the pullback of a group
% homomorphism $S_m \times S_n \to S_{mn}$.  However, using the notions
% of $\lambda$-rings and Adams operations, we can define a stand-in for
% this missing homomorphism, producing virtual representations out of
% actual ones.  The key insight is that the adjoint of $\AntiProd$ is a
% $\lambda$-ring homomorphism, but -- as Boorman proved in 1975 -- the
% representation ring of $S_n$ is generated by the natural
% permutation representation $\QQ^n$ as a $\lambda$-ring.  Thus, in
% order to show that this adjoint preserves the integral structure, it
% suffices to compute its value on $\QQ^n$.  This value (a virtual
% representation) is constructed using Adams operations and M\"obius
% inversion.

This work is aimed at readers familiar with representation rings
and the representation theory of symmetric groups.  No prior
knowledge of $\lambda$-rings is presumed.
Three self-contained proofs of Boorman's theorem are given in the
appendices, two of them lifting the theorem to the noncommutative
symmetric functions (or Solomon's descent algebra), recovering
a result of Schocker.

\medskip
\textbf{Manifest.} This work was written by GPT-6 based on a proof
found by GPT-5.5. It has then been extensively edited and fully
proofread by myself.
\par
This work is in the public domain. --DG  
\end{abstract}

\tableofcontents

\section{Definitions and statements}
\label{sec:definitions}

\subsection{Integral and rational symmetric functions}

Let $\Symm_\ZZ$ denote the ring of symmetric functions over $\ZZ$, and
let $\Symm_\QQ$ be the analogous ring over $\QQ$. Thus,
\[
 \Symm_\QQ=\QQ\otimes_\ZZ\Symm_\ZZ.
\]
We write $\Symm_{\ZZ,n}$ and $\Symm_{\QQ,n}$ for the homogeneous
components of degree $n$ of $\Symm_\ZZ$ and $\Symm_\QQ$.
Our conventions for symmetric functions are the standard ones; see,
for example, Sagan~\cite[Chapter~4]{Sagan}, Stanley~\cite[Chapter~7]{Stanley}, or
Grinberg--Reiner~\cite[Section~2]{GrinbergReiner}.

For $r\geq1$, let $p_r$ be the $r$-th power-sum symmetric function, and
for a partition $\lambda=(\lambda_1,\ldots,\lambda_\ell)$ let
\[
 p_\lambda=p_{\lambda_1}\cdots p_{\lambda_\ell}.
\]
Each $p_\lambda$ belongs to $\Symm_\ZZ$, but the family
$(p_\lambda)_\lambda$ is only a $\QQ$-basis of $\Symm_\QQ$, not a
$\ZZ$-basis of $\Symm_\ZZ$.  For instance,
\[
 h_2=\frac{p_1^2+p_2}{2}.
\]
Equivalently,
\[
 \Symm_\QQ=\QQ[p_1,p_2,p_3,\ldots],
\]
whereas the integral ring is
\[
 \Symm_\ZZ=\ZZ[h_1,h_2,h_3,\ldots]
            =\ZZ[e_1,e_2,e_3,\ldots].
\]

Consequently, it is very easy to define multilinear operations on
$\Symm_\QQ$ by prescribing their values on power sums, but it is a
separate question whether such operations preserve $\Symm_\ZZ$.  Many
familiar operations do: the Kronecker product and plethysm are standard
examples\footnote{Admittedly, plethysm is not multilinear, but the
same construction principle applies with some complications.}.
This phenomenon, and several operations conveniently defined
on the power-sum basis, are discussed for example in
\cite[Exercise~2.9.4]{GrinbergReiner}; see also the symmetric-function
chapters of Stanley~\cite[Chapter~7]{Stanley} and
Macdonald~\cite[Chapter~I]{Macdonald}.  The two operations considered
below provide another illustration of precisely this integrality issue.

\subsection{The arithmetic and anti-arithmetic products}

Since the $p_\lambda$ form a $\QQ$-basis of $\Symm_\QQ$,
the following definitions make sense over $\QQ$.

\begin{definition}
The \emph{arithmetic product} $\BoxProd$ is the
$\QQ$-bilinear operation on $\Symm_\QQ$ defined by
\begin{equation}
 p_\lambda\BoxProd p_\mu
 =\prod_{i=1}^r\prod_{j=1}^s
 p_{\lcm(\lambda_i,\mu_j)}^{\gcd(\lambda_i,\mu_j)}
 \label{eq:def-arithmetic}
\end{equation}
for all pairs of partitions
$\lambda=(\lambda_1,\ldots,\lambda_r)$ and
$\mu=(\mu_1,\ldots,\mu_s)$ (written without trailing zeroes).

The \emph{anti-arithmetic product} $\AntiProd$ is the $\QQ$-bilinear
operation on $\Symm_\QQ$ defined by
\begin{equation}
 p_\lambda\AntiProd p_\mu
 =\prod_{i=1}^r\prod_{j=1}^s
 p_{\gcd(\lambda_i,\mu_j)}^{\lcm(\lambda_i,\mu_j)}
 \label{eq:def-anti}
\end{equation}
for all pairs of partitions
$\lambda=(\lambda_1,\ldots,\lambda_r)$ and
$\mu=(\mu_1,\ldots,\mu_s)$ (written without trailing zeroes).
\end{definition}

If $|\lambda|=m$ and $|\mu|=n$, then both right-hand sides have degree
$mn$, because each pair $(\lambda_i,\mu_j)$ contributes total degree
$\lambda_i\mu_j$.  Thus, for $m,n\geq0$, we have
\[
 \BoxProd,\AntiProd:\Symm_{\QQ,m}\times\Symm_{\QQ,n}
 \longrightarrow \Symm_{\QQ,mn}.
\]

The arithmetic product is the symmetric-function shadow of the
arithmetic product of combinatorial species of Maia and M\'endez
\cite{MaiaMendez}; its integrality is discussed in
\cite[Exercise~4.4.9]{GrinbergReiner} and in the MathOverflow question
\cite{ArithmeticMO}.  For a further development of the species-theoretic
side, Li~\cite{LiPrimeGraphs} introduced an exponential composition based
on the arithmetic product and used it to study prime graphs under Cartesian
product.  The anti-arithmetic product was introduced in the
MathOverflow question \cite{AntiArithmeticMO}, where its integrality was
asked for.  Unlike the arithmetic product, it is not Schur-positive on
pairs of Schur functions; for example,
$s_{(2,1,1)}\AntiProd s_{(2,1,1)}$
has Schur coefficients of both signs
\cite{AntiArithmeticMO}.\footnote{More explicitly, its coefficients of
$s_{(12,4)}$ and $s_{(14,1,1)}$ are $-3$ and $1$, respectively.}

Our main result is the following.

\begin{theorem}[Integrality]
\label{thm:main-integrality}
For all $m,n\geq0$, the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item
\label{thm:main-integrality-arithmetic}
We have
\[
 \Symm_{\ZZ,m}\BoxProd\Symm_{\ZZ,n}
 \subseteq \Symm_{\ZZ,mn}.
\]
\item
\label{thm:main-integrality-positive}
If $f\in\Symm_{\ZZ,m}$ and $g\in\Symm_{\ZZ,n}$ are Schur-positive,
then $f\BoxProd g$ is Schur-positive.
\item
\label{thm:main-integrality-anti}
We have
\[
 \Symm_{\ZZ,m}\AntiProd\Symm_{\ZZ,n}
 \subseteq \Symm_{\ZZ,mn}.
\]
\end{enumerate}
\end{theorem}

Parts (a) and (b) are known.  We include them because the
proof of (a) is parallel to our proof of part (c), except
at exactly one point.  Both proofs proceed using the classical Frobenius
correspondence between symmetric functions and representations/class functions
of symmetric groups.  For the arithmetic product $\BoxProd$, a genuine group
homomorphism\footnote{We let $S_n$ denote the $n$-th symmetric group;
this group consists of the permutations of the set $[n] := \{1,2,\ldots,n\}$.}
\[
 S_m\times S_n\longrightarrow S_{mn}
\]
provides the required pullback on representations.  For the
anti-arithmetic product no such homomorphism is available; instead, we
construct a certain \emph{virtual} representation by Adams operations
and M\"obius inversion.

Before we present these proofs, we shall develop the machinery needed
for them.  Very little of it is new, but not all of it is found in
textbooks.

\begin{remark}[Degree zero]
\label{rmk:deg0}
Theorem~\ref{thm:main-integrality} is easy when one of
$m$ and $n$ is $0$. Indeed, by the empty
product convention, each partition $\lambda$ satisfies
\[
1\BoxProd p_\lambda =
p_\lambda\BoxProd 1 = 1\AntiProd p_\lambda
= p_\lambda\AntiProd 1 = 1
= p_\lambda\left(1,0,0,0,\ldots\right).
\]
Thus, by linearity, for any $f \in \Symm_\ZZ$, we have
\[
1\BoxProd f =
f\BoxProd 1 = 1\AntiProd f
= f\AntiProd 1
= f\left(1,0,0,0,\ldots\right) \in \ZZ \subseteq \Symm_\ZZ.
\]
In particular, if $\lambda$ is a partition, then
$1 \BoxProd s_\lambda = s_\lambda \BoxProd 1$
is $1$ if $\lambda$ is a one-row partition and is $0$ otherwise.

Thus, in proving Theorem~\ref{thm:main-integrality},
we can focus on the case when $m,n\geq 1$.
We thus WLOG assume that $m,n\geq 1$ from now on.
\end{remark}

\begin{remark}
Both operations $\BoxProd$ and $\AntiProd$ are easily seen
to be commutative and associative.
Moreover, $p_1$ is a neutral element for $\BoxProd$, whereas
$\AntiProd$ has no neutral element.
\end{remark}

\section{Exterior powers, Adams operations, and class functions}
\label{sec:lambda}

We will use the basic language of $\lambda$-rings and their Adams
operations.
The words ``$\lambda$-ring'' and ``Adams operation'' can suggest a good
deal more machinery than we need.  In this section we develop the small
part of the theory used later, starting from exterior powers and the
standard relation between elementary symmetric functions and power
sums.  For systematic treatments, see Knutson~\cite[Chapter~I, especially \S\S1 and~4]{Knutson},
Yau~\cite[Chapters~1 and~3]{Yau}, or Hazewinkel~\cite[Section~16]{Hazewinkel}.  None of the
general structure theorems from these references will be used.  For the
symmetric-group representation rings specifically, Thibon~\cite{ThibonAdams}
and Scharf--Thibon~\cite{ScharfThibon} develop Adams operations and inner
plethysm directly in symmetric-function language.  A modern structural
perspective, emphasizing Adams operations as natural transformations of the
representation-ring functor, is given by Meir--Szymik~\cite{MeirSzymik}.

\subsection{Representation rings and class functions}

For every finite group $G$ below, we work with finite-dimensional
representations over $\QQ$.  The constructions in this section do not
require $\QQ$ to be a splitting field.  For symmetric groups and their
finite direct products, however, $\QQ$ is a splitting field; this will
be used in Section~\ref{sec:Sn}.

Let $\Cl_\QQ(G)$ be the $\QQ$-algebra of $\QQ$-valued class functions on
$G$, with pointwise addition and multiplication.  In particular, the
character $\chi_V$ of any finite-dimensional representation $V$ of $G$
over $\QQ$ belongs to $\Cl_\QQ(G)$.
Let $R(G)$ be the
Grothendieck ring of finite-dimensional $\QQ G$-modules
(see, e.g., \cite[Chapter 9]{Serre}).  Thus $R(G)$ is
a $\ZZ$-algebra: addition comes from direct sum and multiplication from
tensor product.  We write $[V]$ for the class of a representation $V$
(although we will occasionally drop the brackets and just write $V$).
Maschke's theorem and ordinary character theory give an injective ring
homomorphism
\begin{equation}
 \chi:R(G)\hookrightarrow\Cl_\QQ(G),
 \qquad [V]\longmapsto\chi_V,
 \label{eq:character-embedding}
\end{equation}
which we shall call the \emph{character embedding}.
We shall usually identify $R(G)$ with its image under this map.  Standard
background on these facts may be found, for example, in
Etingof--Golberg--Hensel--Liu--Schwendner--Vaintrob--Yudovina
\cite[\S 4.2]{EtingofEtAl}, Serre~\cite[Chapter 2]{Serre}, or the
lecture notes of Lassueur~\cite[\S 9]{Lassueur}.

\subsection{A weak notion of \texorpdfstring{$\lambda$}{lambda}-ring}

Our notion of a $\lambda$-ring is a weak one, imposing axioms for
$\lambda^0(x)$ and $\lambda^1(x)$ and $\lambda^n(x+y)$.
Many authors know such $\lambda$-rings under the name of
\emph{pre-$\lambda$-rings}, and only deem them worthy of the name
``$\lambda$-ring'' if they satisfy additional axioms for
$\lambda^n(xy)$ and $\lambda^n(\lambda^m(y))$.
To us, these extra axioms are unnecessary.

\begin{definition}
A \emph{$\lambda$-ring} in this note (often called a \emph{pre-$\lambda$
ring}) is a commutative unital ring $A$ equipped with (usually non-linear) maps
\[
 \lambda^r:A\longrightarrow A\qquad \text{for all } r\geq0
\]
such that all $x,y\in A$ satisfy the axioms
\begin{align}
 \lambda^0(x)&=1,\label{eq:lambda-zero}\\
 \lambda^1(x)&=x,\label{eq:lambda-one}\\
 \lambda^n(x+y) &= \sum_{i=0}^n \lambda^i(x) \lambda^{n-i}(y).
 \label{eq:lambda-additive-n}
\end{align}
We can encode these maps $\lambda^r$ into a single generating function
\begin{align}
 \lambda_t(x)=\sum_{r\geq0}\lambda^r(x)t^r\in A[[t]]
\label{eq:lambdat}
\end{align}
defined for all $x \in A$
(that is, into a map $\lambda_t : A \to A[[t]]$);
then the axioms \eqref{eq:lambda-zero} and
\eqref{eq:lambda-one} say that
\[
\lambda_t(x) = 1 + xt + \left(\text{higher powers of }t\right),
\]
whereas the axiom \eqref{eq:lambda-additive-n} can be equivalently
rewritten as
\begin{align}
\lambda_t(x+y)&=\lambda_t(x)\lambda_t(y).
 \label{eq:lambda-additive}
\end{align}
In other words, $\lambda_t$ must be a group homomorphism from
the additive group of $A$ to the multiplicative group of
formal power series with constant term $1$ over $A$, and it
must have the property that $\dfrac{d}{dt}\lambda_t(x)\mid_{t=0}\, = x$
for each $x \in A$.

A \emph{$\lambda$-ring morphism} is a unital ring homomorphism
$f : A \to B$ between two $\lambda$-rings that commutes
with every $\lambda^r$ (that is, satisfies $f \circ \lambda^r
= \lambda^r \circ f$ for each $r \geq 0$).
A \emph{$\lambda$-subring} of a $\lambda$-ring $A$ is a unital subring
closed under every $\lambda^r$.
The $\lambda$-subring of $A$ \emph{generated} by a subset $X\subseteq A$ is the
smallest $\lambda$-subring of $A$ containing $X$.
\end{definition}

Note that the operations $\lambda^r$ on a $\lambda$-ring $A$,
taken in combination, carry the same information as their
generating series $\lambda_t$.
In particular, a $\lambda$-ring can be defined by providing
$\lambda_t$ instead of the $\lambda^r$'s.

\begin{example}[The binomial $\lambda$-ring]
The ring $\ZZ$ is a $\lambda$-ring under
\[
 \lambda^r(a)=\binom ar,
 \qquad
 \lambda_t(a)=(1+t)^a.
\]
The binomial theorem shows that these two
equalities fit together with \eqref{eq:lambdat};
the equality \eqref{eq:lambda-additive-n} is the Chu--Vandermonde convolution.
\end{example}

\begin{example}[Representation rings]
Let $G$ be a group, and consider its representation ring $R(G)$.
For an actual $G$-representation $V$, put
\begin{equation}
 \lambda_t([V])
 =\sum_{r\geq0}[\textstyle\bigwedge^rV]t^r \in R(G)[[t]].
 \label{eq:lambda-rep-actual}
\end{equation}
The canonical decomposition
\[
 \bigwedge^r(V\oplus W)
 \cong\bigoplus_{i+j=r}\bigwedge^iV\otimes\bigwedge^jW
\]
gives
\[
 \lambda_t([V\oplus W])=\lambda_t([V])\lambda_t([W]).
\]
Thus, \eqref{eq:lambda-rep-actual} defines a monoid
homomorphism $x \mapsto \lambda_t(x)$
from the additive monoid of actual representations of $G$ to
the multiplicative group $1+tR(G)[[t]]$
of power series with constant term $1$.
Since $R(G)$ is the Grothendieck completion of the former monoid, this
homomorphism extends
uniquely to a group homomorphism
\[
 \lambda_t:R(G)\longrightarrow 1+tR(G)[[t]]
\]
by the rule
\[
 \lambda_t(x-y)=\frac{\lambda_t(x)}{\lambda_t(y)}.
\]
Thus $R(G)$ is a $\lambda$-ring
(the proof of \eqref{eq:lambda-one} is easy).
\end{example}

\subsection{\texorpdfstring{$\psi$}{psi}-rings}

Adams operations are even easier to axiomatize.

\begin{definition}
A \emph{$\psi$-ring} is a commutative unital ring $A$ equipped with
unital ring endomorphisms
\[
 \psi^r:A\longrightarrow A\qquad \text{ for all } r\geq1
\]
such that
\[
 \psi^1=\id
 \qquad\text{and}\qquad
 \psi^{rs}=\psi^r\circ\psi^s
 \quad\text{for all }r,s\geq1.
\]
The endomorphisms $\psi^r$ are known as the \emph{Adams operations}
of $A$.
A \emph{morphism of $\psi$-rings} is a unital ring homomorphism commuting with
all $\psi^r$.
\end{definition}

\begin{proposition}
\label{prop:class-functions-psi}
For every finite group $G$, the class-function algebra $\Cl_\QQ(G)$ is a
$\psi$-ring under the $\psi$-operations $\psi^r$ defined by
\begin{equation}
 (\psi^r f)(g)=f(g^r)
 \qquad \text{ for all } r \geq 1 \text{ and } f \in \Cl_\QQ(G)
 \text{ and } g \in G.
 \label{eq:class-psi}
\end{equation}
\end{proposition}

\begin{proof}
The function $\psi^r f$ defined in \eqref{eq:class-psi} is a class
function, because if $g$ and $h$ are two conjugate elements of $G$,
then their $r$-th powers $g^r$ and $h^r$ are also conjugate.
The map $\psi^r$ preserves sums, products, and the constant
function $1$, because all operations on class functions are pointwise.
Moreover, for all $r,s\geq 1$ and all $f \in \Cl_\QQ(G)$ and all
$g \in G$, we have
\[
 \psi^r(\psi^s f)(g)=(\psi^s f)(g^r)
 = f((g^r)^s)=f(g^{rs})=(\psi^{rs}f)(g).
\]
Thus, $\psi^{rs}=\psi^r\circ\psi^s$.
\end{proof}

\subsection{From \texorpdfstring{$\psi$}{psi} to
\texorpdfstring{$\lambda$}{lambda} over \texorpdfstring{$\QQ$}{Q}}

Here is the only general construction we need from the theory of
$\lambda$-rings.  The qualification ``over
$\QQ$'' matters: over an arbitrary ring the formula below can introduce
denominators, and their cancellation is a genuine integrality question.

\begin{proposition}
\label{prop:psi-to-lambda}
Let $A$ be a $\psi$-ring over $\QQ$
(that is, a $\QQ$-algebra equipped with a $\psi$-ring structure).
For any $x \in A$, define $\lambda_t(x) \in A[[t]]$
by
\begin{equation}
 \lambda_t(x)
 =\exp\left(
   \sum_{k\geq1}(-1)^{k-1}\psi^k(x)\frac{t^k}{k}
 \right).
 \label{eq:psi-to-lambda}
\end{equation}
Then the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item
\label{prop:psi-to-lambda-structure}
The operations $\lambda^r$ defined by \eqref{eq:lambdat} in terms of this
$\lambda_t$ make $A$ into a $\lambda$-ring.
\item
\label{prop:psi-to-lambda-morphism}
Every morphism of $\psi$-rings over $\QQ$ is a morphism of the resulting
$\lambda$-rings.
\end{enumerate}
\end{proposition}

The familiar symmetric-function identity
\begin{equation}
 \sum_{r\geq0}e_rt^r
 =\exp\left(\sum_{k\geq1}(-1)^{k-1}p_k\frac{t^k}{k}\right)
 \label{eq:e-vs-p-generating}
\end{equation}
is one way to remember the formula \eqref{eq:psi-to-lambda};
comparing the two identities shows that the
$\QQ$-algebra homomorphism $\Symm_{\QQ} \to A$ that sends all
power-sums $p_k$ to $\psi^k(x)$ (for a given $x\in A$) will
send all $e_r$ to $\lambda^r(x)$.

\begin{proof}[Proof of Proposition~\ref{prop:psi-to-lambda}.]
(a) Since $\psi^1=\id$, the coefficient
of $t$ in \eqref{eq:psi-to-lambda} is $x$, while the constant coefficient
is $1$.  Since every $\psi^k$ is additive,
\begin{align*}
 \lambda_t(x+y)
 &=\exp\left(\sum_{k\geq1}(-1)^{k-1}
     (\psi^k(x)+\psi^k(y))\frac{t^k}{k}\right)\\
 &=\exp\left(\sum_{k\geq1}(-1)^{k-1}
     \psi^k(x)\frac{t^k}{k}\right)
     \cdot \exp\left(\sum_{k\geq1}(-1)^{k-1}
     \psi^k(y)\frac{t^k}{k}\right)\\
 &=\lambda_t(x)\lambda_t(y).
\end{align*}
This proves the $\lambda$-ring axioms \eqref{eq:lambda-zero},
\eqref{eq:lambda-one} and \eqref{eq:lambda-additive}, thus showing
that $A$ is indeed a $\lambda$-ring.

(b) If a ring homomorphism
$\varphi:A\to B$ commutes with all $\psi^k$, then applying $\varphi$ to
\eqref{eq:psi-to-lambda} shows that it commutes with every $\lambda^r$.
\end{proof}

\begin{remark}
\label{rem:psi-to-lambda-weaker}
The proof of Proposition~\ref{prop:psi-to-lambda} uses much less than a
$\psi$-ring structure.  To construct the weak $\lambda$-ring structure
\eqref{eq:psi-to-lambda}, it is enough that $A$ be a commutative
$\QQ$-algebra and that we be given additive maps
\[
 \psi^r:A\longrightarrow A\qquad \text{ for all } r \geq 1
\]
with $\psi^1=\id$.  Neither multiplicativity of the $\psi^k$ nor the
relations $\psi^{rs}=\psi^r\circ\psi^s$ enter the proof.  (Additivity
already implies $\QQ$-linearity here.)  These stronger properties are
part of the definition of a $\psi$-ring because they hold for the Adams
operations that interest us and are what lead to the usual special
$\lambda$-ring structure; they are not needed for the weak
$\lambda$-ring axioms used in Proposition~\ref{prop:psi-to-lambda}.
\end{remark}

Differentiating the logarithm of \eqref{eq:psi-to-lambda} gives
\begin{equation}
\left(\dfrac{d}{dt} \lambda_t(x)\right) / \lambda_t(x)
= \sum_{k\geq 1} (-1)^{k-1} \psi^k(x) t^{k-1},
\end{equation}
that is,
\begin{equation}
\dfrac{d}{dt} \lambda_t(x)
= \lambda_t(x) \cdot \sum_{k\geq 1} (-1)^{k-1} \psi^k(x) t^{k-1}.
\end{equation}
Comparing $t^{r-1}$-coefficients, we obtain the Newton recurrence
\begin{equation}
 r\lambda^r(x)
 =\sum_{k=1}^r(-1)^{k-1}
   \lambda^{r-k}(x)\psi^k(x)
 \label{eq:newton-lambda-psi}
\end{equation}
for each $r\geq 1$ and $x\in A$.
This equation can be solved for $\psi^r(x)$, yielding
\begin{equation}
 \psi^r(x)-\lambda^1(x)\psi^{r-1}(x)
 +\lambda^2(x)\psi^{r-2}(x)-\cdots
 +(-1)^r r\lambda^r(x)=0.
 \label{eq:newton-integral}
\end{equation}
Thus, $\psi^r(x)$ can be expressed as a polynomial in
the inputs
$\psi^1(x),\psi^2(x),\ldots,\psi^{r-1}(x)$ and
$\lambda^1(x),\lambda^2(x),\ldots,\lambda^r(x)$.
This shows (by induction) that each $\psi^r(x)$ can be expressed
as an \emph{integral} polynomial in the elements
\[
 \lambda^1(x),\lambda^2(x),\ldots,\lambda^r(x)
\]
(a version of the usual Newton identities for power-sum
symmetric functions).
For example,
\[
 \psi^1(x)=\lambda^1(x),
 \qquad
 \psi^2(x)=\lambda^1(x)^2-2\lambda^2(x).
\]
The absence of denominators in this direction will be useful below,
as it shows that the $\lambda$-ring structure on $A$ in
Proposition~\ref{prop:psi-to-lambda} uniquely determines the
$\psi$-ring structure it originates from.

\subsection{Characters respect exterior powers}

Consider again a finite group $G$.
The class function algebra $\Cl_\QQ(G)$ of $G$ is a $\psi$-ring
over $\QQ$, and thus (by Proposition~\ref{prop:psi-to-lambda})
becomes a $\lambda$-ring.
We shall now show that $R(G)$ is a $\lambda$-subring of this
$\lambda$-ring $\Cl_\QQ(G)$;
that is, the $\lambda$-ring structure on $R(G)$ is a restriction
of that induced by the $\psi$-ring $\Cl_\QQ(G)$.

\begin{proposition}
\label{prop:character-lambda}
\begin{enumerate}
\item[(a)]
The character embedding
\[
 R(G)\hookrightarrow\Cl_\QQ(G)
\]
is a morphism of $\lambda$-rings.
\item[(b)] The operations $\psi^r$ of
\eqref{eq:class-psi} preserve $R(G)$.
\item[(c)] For every virtual
representation $x\in R(G)$ and every $g \in G$, we have
\begin{equation}
 \chi_{\psi^r(x)}(g)=\chi_x(g^r),
 \label{eq:adams-character-formula}
\end{equation}
where $\chi_y$ denotes the image of any $y \in R(G)$
under the character embedding.
\end{enumerate}
\end{proposition}

\begin{proof}
(a) Let $V$ be an actual representation of $G$,
and let $A$ be the matrix by which a given element
$g\in G$ acts on $V$.  On one hand, by the classical
formula for the characteristic polynomial of a matrix
in terms of its principal minors\footnote{To be fully
precise, the characteristic polynomial of $A$ is
$\det(tI-A)$ rather than $\det(I+tA)$. But these two
polynomials have the same coefficients up to sign and
order. Alternatively, Proposition~11 in
\cite[\S III.8.5]{Bourbaki-Alg1}, applied with
$u=A$, $\xi=1$, and $\eta=t$, gives
\eqref{eq:det-exterior} directly.}, we have
\begin{equation}
 \det(I+tA)
 =\sum_{j\geq0}\tr(\textstyle\bigwedge^j A)t^j.
 \label{eq:det-exterior}
\end{equation}
On the other hand, if the eigenvalues of $A$ in an algebraic closure are
$\alpha_1,\ldots,\alpha_d$, then
\begin{align}
 \det(I+tA)
 &=\prod_{i=1}^d(1+\alpha_i t)
 = \sum_{r\geq 0} e_r(\alpha_1,\alpha_2,\ldots,\alpha_d)t^r
 \notag\\
 &=\exp\left(
   \sum_{k\geq1}(-1)^{k-1}
   \left(\sum_i\alpha_i^k\right)\frac{t^k}{k}
 \right)
 \qquad \left(\text{by \eqref{eq:e-vs-p-generating}}\right)
 \notag\\
 &=\exp\left(
   \sum_{k\geq1}(-1)^{k-1}(\psi^k(\chi_V))(g)\frac{t^k}{k}
 \right),
 \label{eq:det-powertraces}
\end{align}
since each $k \geq 1$ satisfies
$\sum_i\alpha_i^k = \tr(A^k) = \chi_V(g^k) = (\psi^k(\chi_V))(g)$.
Comparing this with \eqref{eq:det-exterior}, we find
\[
\sum_{j\geq0}\tr({\textstyle\bigwedge^j} A)t^j
= \exp\left(
   \sum_{k\geq1}(-1)^{k-1}(\psi^k(\chi_V))(g)\frac{t^k}{k}
 \right).
\]
On the other hand, applying \eqref{eq:psi-to-lambda} to
$x = \chi_V$, and evaluating at $g$, we obtain
\[
\sum_{j\geq0}(\lambda^j(\chi_V))(g)t^j
 =\exp\left(
   \sum_{k\geq1}(-1)^{k-1}(\psi^k(\chi_V))(g)\frac{t^k}{k}
 \right)
\]
(since the algebra structure on $\Cl_\QQ(G)$ is pointwise).
Comparing these two equalities, we find
\[
\sum_{j\geq0}(\lambda^j(\chi_V))(g)t^j
= \sum_{j\geq0}\tr({\textstyle\bigwedge^j} A)t^j.
\]
Thus, for each $j \geq 0$, we obtain $\lambda^j(\chi_V)(g)
= \tr({\textstyle\bigwedge^j} A) = \chi_{\bigwedge^j V}(g)
= \chi_{\lambda^j([V])}(g)$.
Since $g$ was arbitrary, this proves that the class-function
$\lambda^j(\chi_V)$ equals $\chi_{\lambda^j([V])}$.

This shows that the character embedding $R(G) \to \Cl_\QQ(G)$
commutes with $\lambda^j$ (and thus with $\lambda_t$) at
least on actual representations.
The fact that $\lambda_t$ is a group homomorphism extends this
to all virtual representations (since $R(G)$ is generated as
an abelian group by the actual representations). Thus
the character embedding is a $\lambda$-ring morphism.

(b) Let $x \in R(G)$.
Then, by \eqref{eq:newton-integral}, we can write $\psi^r(x)$
as an integral polynomial in the exterior-power operations on
$x$.  Hence $\psi^r(x)\in R(G)$.

(c) The character embedding is a $\lambda$-ring morphism by
part (a), and thus commutes with the $\psi^r$ operations.
Thus, for any $x \in R(G)$ and $r \geq 1$, we have
$\chi_{\psi^r(x)} = \psi^r(\chi_x)$. Evaluating this at a
$g \in G$ yields \eqref{eq:adams-character-formula}.
\end{proof}

Thus we may regard $R(G)$ simultaneously as a $\lambda$-subring and a
$\psi$-subring of $\Cl_\QQ(G)$.

\subsection{A word about special \texorpdfstring{$\lambda$}{lambda}-rings}

In the usual terminology, one often reserves ``$\lambda$-ring'' for a
structure satisfying additional universal identities governing
$\lambda^r(xy)$ and $\lambda^r(\lambda^s(x))$; such rings are often
called \emph{special $\lambda$-rings}.  Representation rings are special,
and a $\psi$-ring over $\QQ$ as above yields a special $\lambda$-ring because
its Adams operations are commuting ring endomorphisms.  These stronger
axioms play no role in our proof, so we do not state them.  See
Knutson~\cite[Chapter~I]{Knutson}, Yau~\cite[Chapters~1--3]{Yau}, and
Hazewinkel~\cite[Section~16]{Hazewinkel} for the full theory.

% \subsection{Tensor products of \texorpdfstring{$\psi$}{psi}-rings}
% \label{subsec:tensor-psi}

% If $A$ and $B$ are $\psi$-rings over $\QQ$, then $A\otimes_\QQ B$ becomes a
% $\psi$-ring over $\QQ$ by
% \begin{equation}
 % \psi^r(a\otimes b)=\psi^r(a)\otimes\psi^r(b).
 % \label{eq:tensor-psi}
% \end{equation}
% This is well-defined because the $\psi^r$ are $\QQ$-linear ring
% endomorphisms, and the identities $\psi^{rs}=\psi^r\psi^s$ are inherited
% factorwise.  Proposition~\ref{prop:psi-to-lambda} then supplies the
% corresponding $\lambda$-operations.

% For finite groups $G$ and $H$, the map
% \begin{equation}
 % \Cl_\QQ(G)\otimes_\QQ\Cl_\QQ(H)
 % \longrightarrow\Cl_\QQ(G\times H),
 % \qquad
 % f\otimes h\longmapsto\bigl((g,k)\mapsto f(g)h(k)\bigr)
 % \label{eq:class-tensor}
% \end{equation}
% is an isomorphism of $\QQ$-algebras, since conjugacy classes in
% $G\times H$ are pairs of conjugacy classes
% (and since $\QQ^X \otimes_\QQ \QQ^Y \cong \QQ^{X\times Y}$
% for any two finite sets $X$ and $Y$).  It is visibly an
% isomorphism of $\psi$-rings, because
% \[
 % (g,k)^r=(g^r,k^r).
% \]
% Hence it is also an isomorphism of the induced $\lambda$-rings.

% There is also a (subtler) notion of tensor products of
% $\lambda$-rings, but we will not need it. The only
% tensor products of $\lambda$-rings that we will use are
% $\Cl_\QQ(S_m)\otimes_\QQ\Cl_\QQ(S_n)$ and
% $R(S_m)\otimes_\ZZ R(S_n)$; the former is obtained by
% tensoring two $\psi$-rings (as above), while the latter will
% be identified with the representation
% ring $R(S_m\times S_n)$ by external tensor product.  The required
% isomorphism is proved in Subsection~\ref{subsec:product-groups}, once the
% irreducible representations of the symmetric groups have been recalled.

\section{The representation ring of the symmetric group}
\label{sec:Sn}

\subsection{Frobenius characteristic}

For any partition $\lambda=(1^{m_1}2^{m_2}\cdots)\vdash n$, set
\[
 z_\lambda=\prod_{i\geq1}i^{m_i}m_i!.
\]
For a class function $f\in\Cl_\QQ(S_n)$, write $f(\lambda)$ for its value
on the conjugacy class of cycle type $\lambda$.  The
\emph{Frobenius characteristic}
(also known as the \emph{characteristic map} in \cite[\S 4.7]{Sagan},
or as the \emph{characteristic isomorphism} in \cite[(5.5)]{Wildon-sf})
is the $\QQ$-linear map
\begin{equation}
 \ch_n:\Cl_\QQ(S_n)\longrightarrow\Symm_{\QQ,n},
 \qquad
 \ch_n(f)=\sum_{\lambda\vdash n}
 f(\lambda)\frac{p_\lambda}{z_\lambda}.
 \label{eq:frob-char}
\end{equation}
It is an isomorphism of $\QQ$-vector spaces, since the $p_\lambda$
form a basis of $\Symm_{\QQ,n}$.  The basic theorem of
Frobenius (see, e.g., \cite[just before Proposition 4.7.2]{Sagan})
says that
\begin{equation}
 \ch_n(\chi^\lambda)=s_\lambda,
 \label{eq:frob-specht}
\end{equation}
where $\chi^\lambda$ is the irreducible character of the Specht module
$S^\lambda$.  Since the Specht modules are defined over $\QQ$ and form a
complete set of absolutely irreducible $\QQ S_n$-modules, restriction of
\eqref{eq:frob-char} gives an isomorphism of abelian groups
\begin{equation}
 R(S_n)\xrightarrow{\ \sim\ }\Symm_{\ZZ,n}.
 \label{eq:R-Symm-integral}
\end{equation}
Thus we have a commutative square
\[
\begin{array}{ccc}
 R(S_n)&\lhook\joinrel\longrightarrow&\Cl_\QQ(S_n)\\
 \big\downarrow\scriptstyle\ch_n&&\big\downarrow\scriptstyle\ch_n\\
 \Symm_{\ZZ,n}&\lhook\joinrel\longrightarrow&\Symm_{\QQ,n}.
\end{array}
\]
This is the bridge between the integrality problem for symmetric
functions in $\Symm_{\QQ,n}$ and the integral lattice of virtual
characters in $\Cl_\QQ(S_n)$.
Note that the product on $R(S_n)$ and $\Cl_\QQ(S_n)$ corresponds
to the so-called \emph{Kronecker product} (also known as the
\emph{internal product}) on the symmetric functions; but we will
not gain anything from this fact.

\subsection{The natural representation generates as a
\texorpdfstring{$\lambda$}{lambda}-ring}

Let
\[
 M_n=\QQ^n
\]
be the natural permutation representation of $S_n$, with basis
$e_1,\ldots,e_n$.  Its action is given by $\sigma(e_i)
= e_{\sigma(i)}$ for any $\sigma \in S_n$ and $i \in [n]$.  Let
$V_n$ be the reflection representation of $S_n$, that is, the
subrepresentation of $M_n$ consisting of vectors
whose coordinates sum to zero.  For $n\geq2$, this is the
Specht module $S^{(n-1,1)}$; for $n=1$, it is the zero representation.  Then
\begin{equation}
 M_n\cong\one\oplus V_n.
 \label{eq:natural-standard}
\end{equation}
We will need a 1975 result of
Evelyn Boorman~\cite{Boorman}: the
representation $M_n$ generates $R(S_n)$ as a $\lambda$-ring.  Marin later proved
the equivalent statement that the exterior powers of $V_n$ generate $R(S_n)$
as a ring~\cite{Marin}; his proof uses a formula of Dvir and points out
earlier equivalent symmetric-function results of Butler~\cite{Butler} and
Boorman.  We record the theorem in the form needed later.
Three self-contained proofs are given in the appendices.

\begin{theorem}[Boorman; Marin]
\label{thm:Marin}
For every $n\geq1$, the $\lambda$-subring of $R(S_n)$ generated by
$[M_n]$ is all of $R(S_n)$.
\end{theorem}

Boorman's proof is a more general categorical argument, based on a backward
induction on stabilizers of tuples and applicable also to Burnside rings.
Marin's proof filters by the number $n-\lambda_1$ of boxes below the first
row and uses Dvir's formula to identify the top-depth multiplication.  We
will give three further proofs in the appendices below.
The first proof (Appendix~\ref{app:triangular-proof}) uses triangularity
with respect to the same natural depth filtration as Marin, but replaces
Dvir's formula by an elementary Schur-functor filtration and Kostka
triangularity.  For generalizations of hook-type generating sets to
wreath products, see Harman~\cite{Harman}.
Appendix~\ref{app:NSym-generation} gives a second and a third proof,
obtained from stronger integral generation theorems for Solomon's descent
algebra (or, equivalently, for noncommutative symmetric functions).  For
a readable introduction to Solomon's descent algebra, including its
realization through the face semigroup algebra of the braid arrangement,
see Saliola~\cite[\S2, especially \S2.1]{Saliola}.

\subsection{The scalar product and integral self-duality}
\label{subsec:inner-product}

For any finite group $G$, we define the bilinear scalar
product $\langle \cdot, \cdot \rangle_G$ on $\Cl_{\QQ}(G)$ by
setting
\begin{equation}
 \langle f,g\rangle_G
 =\frac1{|G|}\sum_{x\in G}f(x)g(x^{-1})
 \qquad \text{ for all } f,g \in \Cl_\QQ(G).
 \label{eq:class-inner-product}
\end{equation}
For symmetric groups, every conjugacy class is invariant under inversion,
so this is simply the usual character scalar product without any need
for complex conjugation.
For $G=S_n$, the irreducible characters $\chi^\lambda$, with
$\lambda\vdash n$, are an orthonormal basis of $\Cl_\QQ(S_n)$.
Indeed, for any finite group $G$, it is well-known that any
two finite-dimensional $G$-representations $U$ and $V$ satisfy
\begin{equation}
 \langle \chi_U, \chi_V \rangle_G
 = \dim \Hom_G(U,V),
 \label{eq:char-scal}
\end{equation}
which (by Schur's lemma) is $0$ when $U$ and $V$ are
non-isomorphic irreducibles and is positive when $U$ and $V$
are isomorphic irreducibles.  For $G=S_n$, all Specht modules
$S^\lambda$ satisfy $\Hom_{S_n}(S^\lambda, S^\lambda) \cong \QQ$
(see, e.g., \cite[last paragraph of \S 5.13]{EtingofEtAl}) and therefore
$\langle \chi^\lambda, \chi^\lambda \rangle_{S_n} = 1$.

The Hall scalar product on $\Symm_{\QQ,n}$ is characterized by
\begin{equation}
 \langle p_\lambda,p_\mu\rangle
 =\delta_{\lambda\mu}z_\lambda.
 \label{eq:Hall-p}
\end{equation}
From \eqref{eq:frob-char} one checks immediately that Frobenius
characteristic is an isometry:
\begin{equation}
 \langle\ch_n(f),\ch_n(g)\rangle
 =\langle f,g\rangle_{S_n}
 \qquad \text{ for all } f, g \in \Cl_\QQ(S_n).
 \label{eq:frob-isometry}
\end{equation}
Equivalently, the Schur functions form an orthonormal basis:
\[
 \langle s_\lambda,s_\mu\rangle=\delta_{\lambda\mu}.
\]
In particular, the lattice $\Symm_{\ZZ,n}$ is self-dual:
\begin{lemma}
\label{lem:self-dual}
If $F\in\Symm_{\QQ,n}$ satisfies
\[
 \langle F,H\rangle\in\ZZ
 \qquad\text{for every }H\in\Symm_{\ZZ,n},
\]
then $F\in\Symm_{\ZZ,n}$.
\end{lemma}

\begin{proof}
Write $F=\sum_{\lambda\vdash n}c_\lambda s_\lambda$.
Then, if $\langle F,H\rangle\in\ZZ$ for every
$H\in\Symm_{\ZZ,n}$,
then in particular $\langle F,s_\lambda\rangle\in\ZZ$
for every $\lambda \vdash n$; but the orthonormality
of the Schur functions yields
$\langle F,s_\lambda\rangle = c_\lambda$, so that we
obtain $c_\lambda\in\ZZ$ for every $\lambda$,
and therefore $F\in\Symm_{\ZZ,n}$.
\end{proof}

We note one simple fact connecting the scalar product with
the Frobenius characteristic $\ch_n$:
If $h \in \Cl_\QQ(S_n)$ is a class function, then
\begin{equation}
 \langle p_\nu,\ch_n(h) \rangle = h(\nu)
 \label{eq:p-evaluates-class}
\end{equation}
for every partition $\nu$ of $n$.
This follows from \eqref{eq:frob-char} and \eqref{eq:Hall-p}.

\subsection{Products of symmetric groups}
\label{subsec:product-groups}

For any finite groups $G$ and $H$, the map
\begin{equation}
 \Cl_\QQ(G)\otimes_\QQ\Cl_\QQ(H)
 \longrightarrow\Cl_\QQ(G\times H),
 \qquad
 f\otimes h\longmapsto\bigl((g,k)\mapsto f(g)h(k)\bigr)
 \label{eq:class-tensor}
\end{equation}
is an isomorphism of $\QQ$-algebras, since conjugacy classes in
$G\times H$ are pairs of conjugacy classes
(and since $\QQ^X \otimes_\QQ \QQ^Y \cong \QQ^{X\times Y}$
for any two finite sets $X$ and $Y$).

Thus, in particular, there is a ring isomorphism
\begin{align}
 \label{eq:prop:R-Cl-square:bot}
 \Cl_\QQ(S_m)\otimes_\QQ\Cl_\QQ(S_n)
 \overset{\cong}{\longrightarrow} \Cl_\QQ(S_m\times S_n)
\end{align}
which sends every $f \otimes g$ to the class function on
$S_m\times S_n$ given by $(\sigma, \tau) \mapsto f(\sigma) g(\tau)$;
we shall denote the latter class function by $f \boxtimes g$.
It is easy to see that this isomorphism preserves scalar products,
in the sense that
all $f_1, f_2 \in \Cl_\QQ(S_m)$ and $g_1, g_2 \in \Cl_\QQ(S_n)$
satisfy
\begin{equation}
 \langle f_1\otimes g_1,f_2\otimes g_2\rangle_{S_m\times S_n}
 =\langle f_1,f_2\rangle_{S_m} \langle g_1,g_2\rangle_{S_n}.
 \label{eq:product-inner-product}
\end{equation}
This follows from the definition of the scalar product using a simple
double-sum computation.

On the other hand, if $U$ is a $\QQ[S_m]$-module and $V$ is a
$\QQ[S_n]$-module, then there is a $\QQ[S_m\times S_n]$-module
$U \boxtimes V$ called the \emph{external tensor product}
of $U$ and $V$; it is simply the tensor product $U \otimes V$
on which $S_m$ acts on the first factor while $S_n$ acts on
the second. Thus we easily obtain a ring homomorphism
\begin{align}
 \label{eq:prop:R-Cl-square:top}
 R(S_m) \otimes_\ZZ R(S_n)
 \to R(S_m \times S_n)
\end{align}
that sends each $[U] \otimes [V]$ to $[U \boxtimes V]$.

There is furthermore a canonical ring homomorphism
\begin{align}
 \label{eq:prop:R-Cl-square:left}
 R(S_m) \otimes_\ZZ R(S_n)
 \to \Cl_\QQ(S_m)\otimes_\ZZ\Cl_\QQ(S_n)
 = \Cl_\QQ(S_m)\otimes_\QQ\Cl_\QQ(S_n)
\end{align}
obtained by tensoring the character embeddings of $S_m$
and $S_n$ (note that tensoring two $\QQ$-vector spaces
over $\ZZ$ is the same as tensoring them over $\QQ$).
We claim the following:

\begin{proposition} \label{prop:R-Cl-square}
The ring homomorphism \eqref{eq:prop:R-Cl-square:top}
is an isomorphism, and the ring homomorphism
\eqref{eq:prop:R-Cl-square:left} is injective. These two
homomorphisms as well as the isomorphism
\eqref{eq:prop:R-Cl-square:bot} and the character
embedding $R(S_m \times S_n) \to \Cl_\QQ(S_m \times S_n)$
fit together into a commutative diagram
\begin{align}
\label{eq:prop:R-Cl-square:comm}
\begin{CD}
 R(S_m)\otimes_\ZZ R(S_n) @>{\sim}>> R(S_m\times S_n)\\
 @VVV @VVV\\
 \Cl_\QQ(S_m)\otimes_\QQ\Cl_\QQ(S_n)
 @>{\sim}>> \Cl_\QQ(S_m\times S_n).
\end{CD}
\end{align}
\end{proposition}

\begin{proof}
Any $\QQ[S_m]$-module $U$ and any $\QQ[S_n]$-module $V$
satisfy
\[
\chi_{U\boxtimes V} = \chi_U \boxtimes \chi_V.
\]
Thus, the diagram \eqref{eq:prop:R-Cl-square:comm} is
commutative.

Moreover, this shows that any
partitions $\lambda, \nu$ of $m$ and $\mu, \rho$ of $n$ satisfy
\begin{align*}
 \left\langle \chi_{S^\lambda\boxtimes S^\mu},
 \chi_{S^\nu\boxtimes S^\rho} \right\rangle_{S_m\times S_n}
 &=\left\langle
 \chi^\lambda\boxtimes\chi^\mu,
 \chi^\nu\boxtimes\chi^\rho
 \right\rangle_{S_m\times S_n}
 =
 \left\langle\chi^\lambda,\chi^\nu\right\rangle_{S_m}
 \left\langle\chi^\mu,\chi^\rho\right\rangle_{S_n}
 \qquad \left(\text{by \eqref{eq:product-inner-product}}\right) \\
 &=\delta_{\lambda\nu}\delta_{\mu\rho}
 \qquad \left(\text{since the $\chi^\lambda$ are orthonormal}\right).
\end{align*}
In particular, each $S^\lambda\boxtimes S^\mu$ is irreducible
(since a group representation whose character has norm%
\footnote{By ``norm'' we mean its scalar product with itself.} $1$ is
always irreducible\footnote{In fact, if it were not, then it would
break into a direct sum of two nontrivial subrepresentations
(by Maschke), and thus its character would have norm $\geq 2$
by \eqref{eq:char-scal}.}),
and these irreducible $\QQ[S_m\times S_n]$-modules $S^\lambda\boxtimes S^\mu$
are non-isomorphic for distinct pairs $(\lambda, \mu)$
(since their scalar products with each other are $0$).
Thus, we have found $p(m) p(n)$ mutually orthonormal vectors
$\chi_{S^\lambda\boxtimes S^\mu}$ in the $\QQ$-vector space
$\Cl_\QQ(S_m \times S_n)$ (where $p(k)$ denotes the number of all
partitions of $k$).
Since $p(m) p(n)$ is the dimension of this $\QQ$-vector space
(because $S_m \times S_n$ has $p(m) p(n)$
conjugacy classes), these $p(m) p(n)$ orthonormal vectors must
form an orthonormal basis of $\Cl_\QQ(S_m \times S_n)$; hence,
there cannot be any further irreducible $\QQ[S_m\times S_n]$-modules
(because \eqref{eq:char-scal} shows that
any such module would have a character orthogonal to all
the $\chi_{S^\lambda\boxtimes S^\mu}$, thus causing
$\Cl_\QQ(S_m \times S_n)$ to have dimension larger than $p(m) p(n)$).
In other words, our external tensor products
$S^\lambda\boxtimes S^\mu$ are a complete enumeration
of all the irreducible representations of $S_m \times S_n$
(with no duplicates, since they are pairwise non-isomorphic).
This shows that \eqref{eq:prop:R-Cl-square:top} is an isomorphism.

So the top row of the diagram \eqref{eq:prop:R-Cl-square:comm} is
an isomorphism, while the right column is injective (being the
character embedding).
Hence, the left column must be injective as well.
In other words, \eqref{eq:prop:R-Cl-square:left} is injective.
This completes the proof.
\end{proof}

Note that we could have saved ourselves some trouble in the above
proof if we recalled the general result that if $G$ and $H$ are
two finite groups and $K$ is a field of characteristic $0$
over which both groups are
split (in particular, if $G = S_m$ and $H = S_n$, then $K$ can be
any field of characteristic $0$), then the irreducible representations
of $G \times H$ over $K$ are precisely the external tensor
products $U \boxtimes V$ where $U$ is an irreducible representation
of $G$ and $V$ is an irreducible representation of $H$.
(See, e.g., \cite[Theorem~3.10.2]{EtingofEtAl}.)
But we chose the above proof for its self-containedness.

Proposition~\ref{prop:R-Cl-square} allows us to move freely
between representations/class functions of $S_m\times S_n$
and tensor products of representations/class functions on the two factors.

\section{Arithmetic and anti-arithmetic products}
\label{sec:application}

We now prove Theorem~\ref{thm:main-integrality}.  The argument is most
transparent after passing to the adjoints of the two products.

\subsection{Two pullback maps on conjugacy classes}

First, we define two maps from $S_m \times S_n$ to $S_{mn}$:
the \emph{arithmetic rule} $\mathcal B$ and the
\emph{anti-arithmetic rule} $\mathcal A$.

To define them, we let $\sigma\in S_m$ and $\tau\in S_n$.

For the arithmetic rule, let $\mathcal B(\sigma,\tau)$ be the permutation
of $[m]\times[n]$ defined by
\begin{equation}
 \mathcal B(\sigma,\tau)(i,j)=(\sigma(i),\tau(j)).
 \label{eq:product-action}
\end{equation}
This permutation $\mathcal B(\sigma,\tau)$ is also known as
$\sigma \times \tau$, and its
cycle type can be easily described: Each pair consisting of an
$a$-cycle $\left(x_1,x_2,\ldots,x_a\right)$ of $\sigma$
and a $b$-cycle $\left(y_1,y_2,\ldots,y_b\right)$ of $\tau$
induces
\begin{align}
 \gcd(a,b)\text{ cycles of length }\lcm(a,b)
 \label{eq:arithmetic-cycles}
\end{align}
in the cycle decomposition of $\mathcal B(\sigma,\tau)$
(their union is the whole Cartesian product of the two chosen
cycles)\footnote{The reason for this is pretty simple:
The orbit of any pair
$(x_i, y_j) \in \left\{x_1,x_2,\ldots,x_a\right\}
\times\left\{y_1,y_2,\ldots,y_b\right\}$ under the
permutation $\mathcal B(\sigma,\tau) = \sigma \times \tau$
has size $\lcm(a,b)$ (because $\sigma^k(x_i) = x_i$ holds
only when $k$ is a multiple of $a$, whereas $\tau^k(y_j) = y_j$
holds only when $k$ is a multiple of $b$).
Thus, the $ab$-element set $\left\{x_1,x_2,\ldots,x_a\right\}
\times\left\{y_1,y_2,\ldots,y_b\right\}$ is partitioned into
orbits of size $\lcm(a,b)$ each. Of course, the number of these
orbits must thus be $\dfrac{ab}{\lcm(a,b)} = \gcd(a,b)$.}.
Thus the map
\begin{equation}
 \mathcal B:S_m\times S_n\longrightarrow S_{mn}
 \label{eq:B-homomorphism}
\end{equation}
is a genuine group homomorphism (after choosing an identification
$[m]\times[n]\cong[mn]$), and its cycle rule is exactly
\eqref{eq:def-arithmetic}.

For the anti-arithmetic rule, we define
$\mathcal A(\sigma,\tau) \in S_{mn}$ only up to conjugacy,
by specifying the cycle type of $\mathcal A(\sigma,\tau)$
rather than the permutation itself:
Namely, for every pair consisting of an
$a$-cycle of $\sigma$ and a $b$-cycle of $\tau$, the
permutation $\mathcal A(\sigma,\tau)$ shall get
\begin{equation}
 \lcm(a,b)\text{ cycles of length }\gcd(a,b).
 \label{eq:anti-cycles}
\end{equation}
The total number of letters contributed by this pair is again $ab$, so
these data define a partition of $mn$, hence a conjugacy class of
$S_{mn}$.  The specific value of $\mathcal A(\sigma,\tau)$ in this
conjugacy class can be chosen arbitrarily; thus, $\mathcal A$ will not
(usually) be a group homomorphism.

The two class maps define pullbacks
\begin{align}
 \Delta^{\BoxProd}_{m,n}:\Cl_\QQ(S_{mn})
 &\longrightarrow\Cl_\QQ(S_m\times S_n),
 \label{eq:Delta-box}\\
 \Delta^{\AntiProd}_{m,n}:\Cl_\QQ(S_{mn})
 &\longrightarrow\Cl_\QQ(S_m\times S_n)
 \label{eq:Delta-diamond}
\end{align}
by
\begin{align*}
 (\Delta^{\BoxProd}_{m,n}f)(\sigma,\tau)
 &=f(\mathcal B(\sigma,\tau)),\\
 (\Delta^{\AntiProd}_{m,n}f)(\sigma,\tau)
 &=f(\mathcal A(\sigma,\tau)).
\end{align*}
Both $\Delta^{\BoxProd}_{m,n}$ and $\Delta^{\AntiProd}_{m,n}$
are unital algebra homomorphisms because multiplication of class
functions is pointwise.

\subsection{These pullbacks are the adjoints of the two products}

Recall that the Frobenius characteristic is an isomorphism
$\ch_k:\Cl_\QQ(S_k)\longrightarrow\Symm_{\QQ,k}$ for each $k \geq 0$.
Thus, we can transport the maps \eqref{eq:Delta-box} and
\eqref{eq:Delta-diamond} through the Frobenius characteristic.
That is, we define two $\QQ$-linear maps
\begin{align}
 \Delta^{\BoxProd}_{m,n}:\Symm_{\QQ,mn}
 &\longrightarrow\Symm_{\QQ,m}\otimes_\QQ\Symm_{\QQ,n},
 \label{eq:Delta-box-Symm}\\
 \Delta^{\AntiProd}_{m,n}:\Symm_{\QQ,mn}
 &\longrightarrow\Symm_{\QQ,m}\otimes_\QQ\Symm_{\QQ,n}
 \label{eq:Delta-diamond-Symm}
\end{align}
(we use the same symbols as for the original two maps
\eqref{eq:Delta-box} and \eqref{eq:Delta-diamond}) so that the
following diagram commutes for each
$\circ\in\{\BoxProd,\AntiProd\}$:
\begin{align}
\begin{CD}
 \Cl_\QQ(S_{mn}) @>{\Delta^\circ_{m,n}}>>
 \Cl_\QQ(S_m\times S_n)\\
 @V{\ch_{mn}}V{\cong}V @V{}V{\cong}V\\
 \Symm_{\QQ,mn} @>{\Delta^\circ_{m,n}}>>
 \Symm_{\QQ,m}\otimes_\QQ\Symm_{\QQ,n},
\end{CD}
\label{eq:ch-square}
\end{align}
where the right vertical arrow is the composition
of the inverse of \eqref{eq:prop:R-Cl-square:bot}
with $\ch_m\otimes\ch_n : \Cl_\QQ(S_m) \otimes_\QQ
\Cl_\QQ(S_n) \to \Symm_{\QQ,m}\otimes_\QQ\Symm_{\QQ,n}$.  All tensor products of rational
symmetric-function spaces below are over $\QQ$.  Their scalar product is
specified by
\begin{align}
 \langle F_1\otimes G_1,F_2\otimes G_2\rangle
 =\langle F_1,F_2\rangle\langle G_1,G_2\rangle.
 \label{eq:scal-on-tens}
\end{align}

\begin{proposition}
\label{prop:adjointness}
For $F\in\Symm_{\QQ,m}$, $G\in\Symm_{\QQ,n}$, and
$H\in\Symm_{\QQ,mn}$, the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item
\label{prop:adjointness-box}
We have
\begin{equation}
 \langle F\BoxProd G,H\rangle
 =\langle F\otimes G,\Delta^{\BoxProd}_{m,n}H\rangle.
 \label{eq:adjoint-box}
\end{equation}
\item
\label{prop:adjointness-anti}
We have
\begin{equation}
 \langle F\AntiProd G,H\rangle
 =\langle F\otimes G,\Delta^{\AntiProd}_{m,n}H\rangle.
 \label{eq:adjoint-diamond}
\end{equation}
\end{enumerate}
\end{proposition}

\begin{proof}
(a) By bilinearity, it suffices to take $F=p_\lambda$ and $G=p_\mu$
for two partitions $\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_r)$
and $\mu = (\mu_1, \mu_2, \ldots, \mu_s)$.
Thus,
\[
F \BoxProd G = p_\lambda \BoxProd p_\mu
= \prod_{i=1}^r\prod_{j=1}^s
 p_{\lcm(\lambda_i,\mu_j)}^{\gcd(\lambda_i,\mu_j)}
= p_\omega,
\]
where $\omega$ is the partition obtained by writing down
$\gcd(\lambda_i, \mu_j)$ copies of $\lcm(\lambda_i, \mu_j)$
for each pair $(i, j)$.
By \eqref{eq:arithmetic-cycles}, this partition
$\omega$ is precisely the cycle type of $\mathcal B(\sigma, \tau)$
when $\sigma$ is a permutation with cycle type $\lambda$ and
$\tau$ is a permutation with cycle type $\mu$.
Consequently,
\begin{align}
(\Delta^{\BoxProd}_{m,n}f)(\lambda, \mu)
= f(\omega)
\qquad \text{ for any } f \in \Cl_\QQ(S_{mn})
\label{pf:prop:adjointness-anti:fomega}
\end{align}
(by the definition of $\Delta^{\BoxProd}_{m,n}$).

Now, let $h \in \Cl_\QQ(S_{mn})$ be the class function with
$\ch_{mn}(h)=H$. Then, from
\eqref{eq:p-evaluates-class} we have
\begin{equation}
 \langle p_\nu,H\rangle=h(\nu)
\end{equation}
for every partition $\nu$ of $mn$.
Applying this to $\nu=\omega$, we obtain
\begin{equation}
 \langle F \BoxProd G, H\rangle = h(\omega)
\label{eq:pf:prop:adjointness:4}
\end{equation}
(since $F \BoxProd G = p_\omega$).
Thus we have expressed the left-hand side of \eqref{eq:adjoint-box}
as $h(\omega)$.

To compute the right-hand side, write
$k=\Delta^{\BoxProd}_{m,n}h \in \Cl_\QQ(S_m \times S_n)
= \Cl_\QQ(S_m) \otimes_\QQ \Cl_\QQ(S_n)$, where the last
equality sign is really the isomorphism
\eqref{eq:prop:R-Cl-square:bot} used as an identification.
Thus, its Frobenius characteristic (i.e., the image of $k$
under $\ch_m \otimes \ch_n$) is
\[
 \Delta^{\BoxProd}_{m,n}H
 =\sum_{\alpha\vdash m}\sum_{\beta\vdash n}
 k(\alpha,\beta)\frac{p_\alpha\otimes p_\beta}
                         {z_\alpha z_\beta}
\]
(by the definitions of $\ch_m$ and $\ch_n$).
Taking the scalar product of this equality
with $F \otimes G = p_\lambda \otimes p_\mu$,
we thus obtain
\begin{align*}
\langle F\otimes G,\Delta^{\BoxProd}_{m,n}H\rangle
&= \left\langle p_\lambda\otimes p_\mu,
\sum_{\alpha\vdash m}\sum_{\beta\vdash n}
 k(\alpha,\beta)\frac{p_\alpha\otimes p_\beta}
                         {z_\alpha z_\beta}\right\rangle \\
&= \sum_{\alpha\vdash m}\sum_{\beta\vdash n}
 \dfrac{k(\alpha,\beta)}{z_\alpha z_\beta}
 \langle p_\lambda, p_\alpha\rangle
 \cdot \langle p_\mu, p_\beta\rangle
 \qquad \left(\text{by \eqref{eq:scal-on-tens}}\right)\\
&= \sum_{\alpha\vdash m}\sum_{\beta\vdash n}
 \dfrac{k(\alpha,\beta)}{z_\alpha z_\beta}
 \delta_{\lambda\alpha} z_\lambda
 \cdot \delta_{\mu\beta} z_\mu
 \qquad \left(\text{by \eqref{eq:Hall-p}}\right) \\
&= \dfrac{k(\lambda, \mu)}{z_\lambda z_\mu} z_\lambda z_\mu
= k(\lambda, \mu) = (\Delta^{\BoxProd}_{m,n}h)(\lambda, \mu)
= h(\omega) \qquad \left(\text{by \eqref{pf:prop:adjointness-anti:fomega}}\right).
\end{align*}
Comparing this with \eqref{eq:pf:prop:adjointness:4},
we obtain $\langle F\BoxProd G,H\rangle
 =\langle F\otimes G,\Delta^{\BoxProd}_{m,n}H\rangle$,
and part (a) is proved.

(b) The anti-arithmetic case is identical, with the
anti-arithmetic cycle type in place of the arithmetic one.
\end{proof}

Thus integrality of the products follows from integrality of their
adjoints.

\begin{proposition}
\label{prop:adjoint-integrality-criterion}
Let $\circ$ denote either $\BoxProd$ or $\AntiProd$.  If
\begin{equation}
 \Delta^\circ_{m,n}\bigl(\Symm_{\ZZ,mn}\bigr)
 \subseteq
 \Symm_{\ZZ,m}\otimes_\ZZ\Symm_{\ZZ,n},
 \label{eq:Delta-integral}
\end{equation}
then
\[
 \Symm_{\ZZ,m}\circ\Symm_{\ZZ,n}
 \subseteq\Symm_{\ZZ,mn}.
\]
\end{proposition}

\begin{proof}
Take $F\in\Symm_{\ZZ,m}$ and $G\in\Symm_{\ZZ,n}$.  For every
$H\in\Symm_{\ZZ,mn}$,
Proposition~\ref{prop:adjointness} gives
\begin{align*}
 \langle F\circ G,H\rangle
 &=\langle F\otimes G,\Delta^\circ_{m,n}H\rangle \\
 &\in \langle \Symm_{\ZZ,m}\otimes_\ZZ\Symm_{\ZZ,n},\ %
 \Symm_{\ZZ,m}\otimes_\ZZ\Symm_{\ZZ,n}\rangle
 \qquad \left(\text{by \eqref{eq:Delta-integral}}\right)\\
 &\subseteq\ZZ
\end{align*}
because the Schur bases on the two tensor factors are orthonormal.
Lemma~\ref{lem:self-dual} now implies $F\circ G\in\Symm_{\ZZ,mn}$.
Thus, $\Symm_{\ZZ,m}\circ\Symm_{\ZZ,n}
 \subseteq\Symm_{\ZZ,mn}$.
\end{proof}

In view of the commutative diagram \eqref{eq:ch-square},
condition \eqref{eq:Delta-integral} can be rewritten as
\begin{equation}
 \Delta^\circ_{m,n}\bigl(R(S_{mn})\bigr)
 \subseteq R(S_m\times S_n).
 \label{eq:representation-integrality-target}
\end{equation}
Indeed, the left vertical isomorphism in \eqref{eq:ch-square}
sends $R(S_{mn})$ precisely onto $\Symm_{\ZZ,mn}$, by
\eqref{eq:R-Symm-integral}.  Since the diagram \eqref{eq:ch-square}
is commutative, this entails that the right vertical isomorphism
in \eqref{eq:ch-square} sends $\Delta^\circ_{m,n}(R(S_{mn}))$
precisely onto $\Delta^\circ_{m,n}(\Symm_{\ZZ,mn})$.
On the other hand, this right vertical isomorphism also sends
$R(S_m\times S_n)$ precisely onto
$\Symm_{\ZZ,m}\otimes_\ZZ\Symm_{\ZZ,n}$: indeed, by
Proposition~\ref{prop:R-Cl-square}, the external tensor products
$S^\lambda\boxtimes S^\mu$ form a $\ZZ$-basis of $R(S_m\times S_n)$,
and their images are $s_\lambda\otimes s_\mu$, the $\ZZ$-basis of this
tensor-product lattice.
Hence, we conclude that the
right vertical isomorphism in \eqref{eq:ch-square}
transforms both sides of
\eqref{eq:representation-integrality-target} into the respective
sides of \eqref{eq:Delta-integral}.
Thus, in order to prove \eqref{eq:Delta-integral}, we need only
verify \eqref{eq:representation-integrality-target}.
We thus aim to do this for both of our pullbacks.

\subsection{Power compatibility}

The first decisive observation is that both class maps
$\mathcal B$ and $\mathcal A$ commute
(up to conjugacy) with raising permutations to powers.

For the arithmetic map $\mathcal B$,
this is immediate from the fact that it is a
group homomorphism:
%\eqref{eq:B-homomorphism}:
\begin{align}
 \mathcal B(\sigma^r,\tau^r)
 = \mathcal B((\sigma,\tau)^r)
 = \mathcal B(\sigma,\tau)^r.
\label{eq:Br}
\end{align}
For the anti-arithmetic map $\mathcal A$,
it is not automatic, but it is still true at
the level of conjugacy classes.
The verification relies on the following
elementary number-theoretic identity:

\begin{lemma}[A gcd identity]
\label{lem:gcd-power-identity}
For all positive integers $a,b,r$, one has
\begin{equation}
 \gcd\left(\frac{a}{\gcd(a,r)},
           \frac{b}{\gcd(b,r)}\right)
 =\frac{\gcd(a,b)}{\gcd(\gcd(a,b),r)}.
 \label{eq:gcd-power-identity}
\end{equation}
\end{lemma}

\begin{proof}
A straightforward proof can be done using $p$-valuations:
Fix a prime $p$, and put
\[
 \alpha=v_p(a),\qquad \beta=v_p(b),\qquad \rho=v_p(r).
\]
It is classical that any two positive integers $s$
and $t$ satisfy $v_p(\gcd(s,t)) = \min(v_p(s), v_p(t))$.
Thus, every positive integer $m$ satisfies
\[
v_p\left(m/\gcd(m,r)\right) = \left(v_p(m)-\rho\right)_+,
\]
where $x_+=\max(x,0)$.   Hence,
the $p$-adic valuation of the left-hand side of
\eqref{eq:gcd-power-identity} is\footnote{We are
using the fact that $\min(x_+,y_+)=(\min(x,y))_+$
for any reals $x$ and $y$.}
\[
 \min\bigl((\alpha-\rho)_+,(\beta-\rho)_+\bigr)
 =(\min(\alpha-\rho,\beta-\rho))_+
 =(\min(\alpha,\beta)-\rho)_+.
\]
The valuation of the right-hand side is
\[
 (v_p(\gcd(a,b))-\rho)_+
 =(\min(\alpha,\beta)-\rho)_+.
\]
Thus the two sides have the same $p$-adic valuation for every prime $p$,
so they are identical.
\end{proof}

\begin{lemma}
\label{lem:anti-power-compatible}
For all $r\geq1$, we have
\begin{equation}
 \mathcal A(\sigma^r,\tau^r)
 \sim \mathcal A(\sigma,\tau)^r,
 \label{eq:anti-power-compatible}
\end{equation}
where $\sim$ denotes conjugacy in $S_{mn}$.
\end{lemma}

\begin{proof}
Consider one $a$-cycle of $\sigma$ and one $b$-cycle of
$\tau$.  We shall show that they contribute the same amount
of cycles to $\mathcal A(\sigma^r, \tau^r)$
as they do to $\mathcal A(\sigma, \tau)^r$, and that these
cycles all have the same length.

Set
\[
 g=\gcd(a,b),
 \qquad
 c=\gcd(g,r).
\]
The definition of $\mathcal A(\sigma,\tau)$ shows that our
two chosen cycles of $\sigma$ and $\tau$ produce
$\lcm(a,b)$ cycles of length $g$ in $\mathcal A(\sigma,\tau)$.
Taking the $r$-th power splits each such $g$-cycle into
$c$ cycles of length $g/c$ (since the $r$-th power of a
$g$-cycle is a permutation of cycle type $(g/c,g/c,\ldots,g/c)$).
Thus the permutation $\mathcal A(\sigma,\tau)^r$ gains
\begin{equation}
 \lcm(a,b)c
 \quad\text{cycles of length }g/c
 \label{eq:power-right-count}
\end{equation}
from our two cycles.

Now take the $r$-th powers of $\sigma$ and $\tau$ first.
The $a$-cycle of $\sigma$ splits into $\gcd(a,r)$ many cycles of $\sigma^r$,
each having length $a/\gcd(a,r)$.
The $b$-cycle of $\tau$ splits into $\gcd(b,r)$ many cycles of $\tau^r$,
each having length $b/\gcd(b,r)$.
Each pair consisting of one of the $a/\gcd(a,r)$-cycles of $\sigma^r$
and one of the $b/\gcd(b,r)$-cycles of $\tau^r$ then gives rise to
a number of cycles of $\mathcal A(\sigma^r,\tau^r)$, each having
length
\begin{align*}
 \gcd\left(\frac{a}{\gcd(a,r)},
           \frac{b}{\gcd(b,r)}\right)
 &= \frac{\gcd(a,b)}{\gcd(\gcd(a,b),r)}
    \qquad \left(\text{by Lemma~\ref{lem:gcd-power-identity}}\right) \\
 &= \frac{g}{\gcd(g,r)} =\frac{g}{c}.
\end{align*}
Since the block coming from our initial two cycles of $\sigma$ and $\tau$
has $ab$ letters in total, the number of such cycles is
\[
 \frac{ab}{g/c}
 =\frac{ab}{g}c
 =\frac{ab}{\gcd(a,b)}c
 =\lcm(a,b)c.
\]
So $\mathcal A(\sigma^r,\tau^r)$ receives
\[
 \lcm(a,b)c
 \quad\text{cycles of length }g/c
\]
from our two cycles.
This agrees with \eqref{eq:power-right-count}.
So the permutations $ \mathcal A(\sigma^r,\tau^r)$
and $\mathcal A(\sigma,\tau)^r$ have the same
cycles of each size, and consequently are conjugate.
\end{proof}

\begin{corollary}
\label{cor:Delta-psi}
On rational class functions, the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item
\label{cor:Delta-psi-box}
The map $\Delta^{\BoxProd}_{m,n}:\Cl_\QQ(S_{mn})
 \longrightarrow\Cl_\QQ(S_m\times S_n)$ is a morphism of $\psi$-rings, and
hence of $\lambda$-rings.
\item
\label{cor:Delta-psi-anti}
The map $\Delta^{\AntiProd}_{m,n}:\Cl_\QQ(S_{mn})
 \longrightarrow\Cl_\QQ(S_m\times S_n)$ is a morphism of $\psi$-rings, and
hence of $\lambda$-rings.
\end{enumerate}
\end{corollary}

\begin{proof}
For either part, let $\mathcal C$ denote the relevant class map
($\mathcal B$ for part (a), and $\mathcal A$ for part (b)).
Then, any $\sigma \in S_m$ and $\tau \in S_n$ satisfy
\begin{align}
\mathcal C(\sigma,\tau)^r
\sim \mathcal C(\sigma^r,\tau^r)
\label{pf:cor:Delta-psi:CsC}
\end{align}
(where $\sim$ means conjugacy in $S_{mn}$).
Indeed, for part~(a), this follows from \eqref{eq:Br};
for part~(b), it follows from
Lemma~\ref{lem:anti-power-compatible}.

Let $\Delta$ denote the map $\Delta^{\BoxProd}_{m,n}$ in part (a), or
the map $\Delta^{\AntiProd}_{m,n}$ in part (b).
We must show that $\Delta$ is a $\psi$-ring morphism and hence
a $\lambda$-ring morphism.

For any
class function $f \in \Cl_\QQ(S_{mn})$ and any $r \geq 1$ and any $\sigma \in S_m$
and $\tau \in S_n$, we have
\begin{align*}
 (\Delta\psi^r f)(\sigma,\tau)
 &=(\psi^r f)(\mathcal C(\sigma,\tau))
   \qquad\left(\text{by the definition of }\Delta\right)\\
 &=f(\mathcal C(\sigma,\tau)^r)
   \qquad\left(\text{by the definition of }\psi^r\right)\\
 &=f(\mathcal C(\sigma^r,\tau^r))
   \qquad \left(\text{by \eqref{pf:cor:Delta-psi:CsC}}\right) \\
 &=(\Delta f)(\sigma^r,\tau^r)
   \qquad\left(\text{by the definition of }\Delta\right)\\
 &=(\psi^r\Delta f)(\sigma,\tau)
   \qquad\left(\text{by the definition of }\psi^r\right).
\end{align*}
Thus, $\Delta\psi^r f = \psi^r\Delta f$ for any $f$ and $r$.
Hence, $\Delta$ is a $\psi$-ring morphism (since $\Delta$ is a
ring morphism).
Consequently, Proposition~\ref{prop:psi-to-lambda}(b) makes $\Delta$
a $\lambda$-ring morphism.
\end{proof}

At this point, Theorem~\ref{thm:Marin} reduces the entire integrality
problem to one calculation: we need only check that the pullback
$\Delta([M_{mn}])$ (where $\Delta$ is $\Delta^{\BoxProd}_{m,n}$
or $\Delta^{\AntiProd}_{m,n}$) of the
natural representation $M_{mn}$ is an integral virtual representation
of $S_m\times S_n$.
Once this is proved, it will follow by
Corollary~\ref{cor:Delta-psi} that $\Delta$ (being a $\lambda$-ring
morphism, with $R(S_m\times S_n)$ a $\lambda$-subring of its target)
sends the entire $\lambda$-subring of $R(S_{mn})$
generated by $[M_{mn}]$ to $R(S_m \times S_n)$;
but Theorem~\ref{thm:Marin} shows that this $\lambda$-subring is
the whole $R(S_{mn})$, and thus it will follow that
$\Delta(R(S_{mn})) \subseteq R(S_m \times S_n)$.
That is, \eqref{eq:representation-integrality-target} will follow.
This, in turn, will yield \eqref{eq:Delta-integral} (since
\eqref{eq:representation-integrality-target} is just a restatement
of \eqref{eq:Delta-integral}), and therefore (by
Proposition~\ref{prop:adjoint-integrality-criterion}) we will obtain
$\Symm_{\ZZ,m}\circ\Symm_{\ZZ,n}
 \subseteq\Symm_{\ZZ,mn}$,
which will prove parts (a) and (c) of Theorem~\ref{thm:main-integrality}.

\subsection{The arithmetic product: an actual pullback}
\label{subsec:proof-ab}

For the arithmetic map there is nothing to construct: Since
the map $\mathcal B$ is a group homomorphism, its pullback
$\Delta^{\BoxProd}_{m,n}$ is just restriction of
characters/representations along this homomorphism.
Restricting the
natural permutation representation $M_{mn}$ of $S_{mn}$ along
the group homomorphism
\eqref{eq:B-homomorphism} (and then identifying $[mn]$ with
$[m]\times[n]$) gives the permutation representation
of $S_m \times S_n$ on $[m]\times[n]$
(or, rather, an isomorphic copy thereof).
On the basis vector $e_i\otimes e_j$ of
$M_m\boxtimes M_n$, the element $(\sigma,\tau) \in S_m \times S_n$ acts by
\[
 e_i\otimes e_j\longmapsto e_{\sigma(i)}\otimes e_{\tau(j)},
\]
which is exactly the product action on $[m]\times[n]$.  Therefore
\begin{equation}
 \Delta^{\BoxProd}_{m,n}(M_{mn})
 =M_m\boxtimes M_n
 \in R(S_m\times S_n).
 \label{eq:arithmetic-natural-image}
\end{equation}
Corollary~\ref{cor:Delta-psi} says that
$\Delta^{\BoxProd}_{m,n}$ is a $\lambda$-ring morphism on rational class
functions.  Since $R(S_m\times S_n)$ is a $\lambda$-subring of
$\Cl_\QQ(S_m\times S_n)$, equation~\eqref{eq:arithmetic-natural-image},
together with the fact that $M_{mn}$ generates $R(S_{mn})$ as a
$\lambda$-ring (by Theorem~\ref{thm:Marin}), gives
\[
 \Delta^{\BoxProd}_{m,n}(R(S_{mn}))
 \subseteq R(S_m\times S_n).
\]
That is, \eqref{eq:representation-integrality-target} holds for
${\circ} = {\BoxProd}$.
Hence, \eqref{eq:Delta-integral} holds for ${\circ} = {\BoxProd}$ (since
\eqref{eq:representation-integrality-target} is just a restatement
of \eqref{eq:Delta-integral}).
Proposition~\ref{prop:adjoint-integrality-criterion} thus proves the
arithmetic integrality assertion
\[
 \Symm_{\ZZ,m}\BoxProd\Symm_{\ZZ,n}
 \subseteq\Symm_{\ZZ,mn}.
\]
In other words, Theorem~\ref{thm:main-integrality} (a) is proved.

There is also a stronger conclusion.  Since
$\mathcal B:S_m\times S_n\hookrightarrow S_{mn}$ is a genuine
group embedding, $\Delta^{\BoxProd}_{m,n}$ on representation rings is
ordinary restriction.  Its adjoint is induction.  That is, the map
$\BoxProd$ on the class function algebras is induction (since
\eqref{eq:adjoint-box} shows that this map is the adjoint of
$\Delta^{\BoxProd}_{m,n}$). Thus, for
$\lambda\vdash m$ and $\mu\vdash n$, we have
\begin{equation}
 s_\lambda\BoxProd s_\mu
 =\ch_{mn}\left(
 \Ind_{S_m\times S_n}^{S_{mn}}
 (S^\lambda\boxtimes S^\mu)
 \right),
 \label{eq:arithmetic-induction}
\end{equation}
where $S_m\times S_n$ is embedded by the product action.  Hence,
$s_\lambda \BoxProd s_\mu$ is Schur positive.
This yields Theorem~\ref{thm:main-integrality} (b).

The same product-action restriction
$S_{mn}\downarrow S_m\times S_n$ has recently been studied by
Ryba~\cite{Ryba} from the viewpoint of stable symmetric-group characters;
his Kronecker-comultiplication formulas and stability results give a close
representation-theoretic companion to the arithmetic side of the present
note.

\subsection{Detecting exact cycle lengths by Adams operations}
\label{subsec:exact-cycles}

For the anti-arithmetic map, the pullback of $M_{mn}$ is not supplied by
an actual restriction functor.  We now construct it as a virtual
representation.

For any $1\leq a\leq m$, define
\begin{equation}
 C_{m,a}
 =\sum_{d\mid a}\mu(a/d)\,\psi^d(M_m)
 \in R(S_m),
 \label{eq:Cma-definition}
\end{equation}
where $\mu$ is the number-theoretic M\"obius function.  This is an
integral virtual representation because the Adams operations preserve
$R(S_m)$ by Proposition~\ref{prop:character-lambda}(b).

If $\sigma\in S_m$ is any permutation, we shall write
$m_a(\sigma)$ for its number of $a$-cycles.

\begin{lemma}
\label{lem:Cma-character}
Let $\sigma\in S_m$ and $1 \leq a \leq m$ be arbitrary.
Then,
\begin{equation}
 \chi_{C_{m,a}}(\sigma)=a\,m_a(\sigma).
 \label{eq:Cma-character}
\end{equation}
\end{lemma}

\begin{proof}
The character $\chi_{M_m}$ of the natural permutation representation $M_m$ of $S_m$
is the class function that sends each permutation to its number
of fixed points.  Hence,
each $d\geq 1$ satisfies $\chi_{M_m}(\sigma^d)
 =|\Fix(\sigma^d)|$.
Since \eqref{eq:adams-character-formula} yields
$\chi_{\psi^d(M_m)}(\sigma)
 =\chi_{M_m}(\sigma^d)$, we can rewrite this as
\begin{equation}
 \chi_{\psi^d(M_m)}(\sigma)
 %=\chi_{M_m}(\sigma^d)
 =|\Fix(\sigma^d)|
 =\sum_{b\mid d}b\,m_b(\sigma).
 \label{eq:fixed-points-power}
\end{equation}
% Here, the middle equality holds because $M_m$ is a permutation representation,
% whose character counts fixed basis vectors.  For the last equality,
Here, the last equality sign is because the fixed points of $\sigma^d$
are precisely the points that lie on $b$-cycles of $\sigma$ that
satisfy $b \mid d$,
% (indeed, a $b$-cycle of $\sigma$ is fixed pointwise by $\sigma^d$ exactly when
% $b\mid d$), and
and because each such $b$-cycle contributes exactly $b$
fixed points for $\sigma^d$.
% in that case it contributes all of its $b$ letters, and otherwise
% it contributes none.

%Substituting \eqref{eq:fixed-points-power} into
Taking characters in
\eqref{eq:Cma-definition} gives
\begin{align}
 \chi_{C_{m,a}}(\sigma)
 &= \sum_{d\mid a}\mu(a/d)\,\chi_{\psi^d(M_m)}(\sigma) \nonumber\\
 &=\sum_{d\mid a}\mu(a/d)
   \sum_{b\mid d}b\,m_b(\sigma)
   \qquad\left(\text{by \eqref{eq:fixed-points-power}}\right) \nonumber\\
 &=\sum_{b\mid a}b\,m_b(\sigma)
   \sum_{\substack{d:\ b\mid d\mid a}}\mu(a/d).
   \label{pf:lem:Cma-character:5}
\end{align}
The inner sum is $1$ for $b=a$ and $0$ otherwise, by M\"obius
inversion.  Thus, \eqref{pf:lem:Cma-character:5} simplifies to
$\chi_{C_{m,a}}(\sigma)=a\,m_a(\sigma)$, which is precisely
\eqref{eq:Cma-character}.
\end{proof}

Thus $C_{m,a}$ is a virtual representation whose character counts the
letters lying in cycles of \emph{exactly} length $a$.

\begin{example}
For the first few values of $a$, formula~\eqref{eq:Cma-definition} gives
(whenever the displayed indices are at most $m$)
\[
 C_{m,1}=M_m,
 \qquad
 C_{m,2}=\psi^2(M_m)-M_m,
\]
and
\[
 C_{m,6}=\psi^6(M_m)-\psi^3(M_m)-\psi^2(M_m)+M_m.
\]
Accordingly, their characters are respectively
$m_1(\sigma)$, $2m_2(\sigma)$, and $6m_6(\sigma)$.
\end{example}

\begin{remark}[Prior work on $C_{m,a}$]
\label{rem:Cma-GLLV}
The Frobenius characteristic $\ch_m$ sends the virtual character
$C_{m,a}$ to the symmetric function $h_{m-a}p_a$.
This is not hard to prove using the Murnaghan--Nakayama rule.
It also follows easily from
Giannelli--Law--Long--Vallejo~\cite[Definition~3.7 and
Theorem~3.9]{GiannelliLawLongVallejo}.  For a partition $\lambda$ and a
positive integer $e$, they define a virtual character $V^\lambda[e]$ by
a signed sum over all ways of adding an $e$-hook to $\lambda$.
The sign is exactly the usual Murnaghan--Nakayama sign
$(-1)^{\ell}$, where $\ell$ is the leg length of the added hook; indeed,
their proof starts from the power-sum Murnaghan--Nakayama identity
\[
 s_\lambda p_e
 =\sum_\alpha (-1)^{\ell(\alpha/\lambda)}s_\alpha.
\]
Their Theorem~3.9 says that, if a permutation $\sigma$
has exactly $k$ cycles of length
$e$, then
\[
 V^\lambda[e](\sigma)=ke\,\chi^\lambda(\tau),
\]
where $\tau$ is obtained from $\sigma$
by deleting one such $e$-cycle (and the value is
$0$ when $k=0$).  Taking $\lambda=(m-a)$ and $e=a$ (with $\lambda$ the
empty partition when $a=m$) gives
\[
 V^{(m-a)}[a](\sigma)=a\,m_a(\sigma).
\]
Hence Lemma~\ref{lem:Cma-character} shows that
\[
 C_{m,a}=V^{(m-a)}[a]
 \qquad\text{in }R(S_m).
\]
Thus the virtual character $C_{m,a}$ has appeared before; formula
\eqref{eq:Cma-definition} gives a different Adams--M\"obius expression
for this special case.
\end{remark}

\begin{remark}[Cycle-counting functions and character polynomials]
The functions
\[
 X_a(\sigma)=m_a(\sigma)
\]
are the classical cycle-counting functions that
(taken over all $a \in \{1,2,\ldots,n\}$ together) can serve
as an alternative encoding of the cycle type of $\sigma$.
For each partition $\mu$,
Garsia and Goupil \cite[equation~I.2]{GarsiaGoupil} express the
character value $\chi^{(n-|\mu|,\mu)}(\sigma)$ (where
$n\geq \mu_1+|\mu|$ is arbitrary) as a polynomial
$q_\mu(X_1(\sigma),X_2(\sigma),\ldots,X_n(\sigma))$ in these
functions. (They write $a_i$ for $X_i(\sigma)$.)
They give an explicit umbral formula for $q_\mu$
in \cite[Proposition~I.1]{GarsiaGoupil}.

% The notation in Garsia--Goupil is different: on page~1,
% they write a permutation's cycle type as $\alpha=1^{a_1}2^{a_2}\cdots n^{a_n}$,
% so that $a_i=X_i(\sigma)$.  On page~2, their equation~(I.2) reads
% \[
 % \chi^{(n-|\mu|,\mu)}(\sigma)
 % =q_\mu(a_1,a_2,\ldots,a_n)
 % =q_\mu(X_1(\sigma),X_2(\sigma),\ldots,X_n(\sigma))
% \]
% whenever $n-|\mu|\geq\mu_1$
% \cite[p.~2, equation~(I.2)]{GarsiaGoupil}.
% Thus their formal variable $x_i$ is evaluated at the number of
% $i$-cycles; they do not introduce the notation $X_i$ for the counting
% function itself.  Their Proposition~I.1, also on page~2, gives an
% explicit umbral formula for $q_\mu$.
% The historical paragraph on that page, between equations~(I.3)
% and~(I.4), attributes the implicit use of character polynomials to
% Murnaghan and their later identification to Specht.  All page numbers
% here are the printed page numbers of the article.

A particularly close symmetric-function precedent for the calculation
above is the evaluation at permutation eigenvalues used by
Orellana--Zabrocki \cite{OrellanaZabrocki}.
Their Section~2.1 defines $\Xi_\mu$ as the multiset of eigenvalues of a
permutation matrix of cycle type $\mu$.  Their Section~8,
equation~(66), gives exactly
\[
 p_d[\Xi_\mu]=\sum_{b\mid d}b\,m_b(\mu)
\]
\cite[Section~8, equation~(66)]{OrellanaZabrocki}.
This is equation~\eqref{eq:fixed-points-power} in the present note.
Their Section~5, equation~(22), defines character polynomials in the
cycle-counting variables, and their Proposition~12 identifies them with
symmetric functions under the mutually inverse substitutions
\[
 p_k\longmapsto\sum_{d\mid k}dX_d,
 \qquad
 X_k\longmapsto\frac1k\sum_{d\mid k}\mu(k/d)p_d
\]
\cite[Section~5, equation~(22) and Proposition~12]{OrellanaZabrocki}.
The inverse substitution is also written in Section~8, in the discussion
between equations~(67) and~(69).
\footnote{All section, proposition, and equation numbers cited here for
Orellana--Zabrocki refer to arXiv:1605.06672v5.}

Thus both the cycle-counting identity and its M\"obius inversion occur
explicitly in this literature.  Lemma~\ref{lem:Cma-character} applies
that inversion to the Adams operations of $M_m$, while
Proposition~\ref{prop:character-lambda} ensures that the resulting
function $aX_a$ is an \emph{integral virtual character}.
\end{remark}

\subsection{The anti-arithmetic substitute for the natural representation}
\label{subsec:proof-c}

Define
\begin{equation}
 W_{m,n}
 =\sum_{\substack{1\leq a\leq m,\ 1\leq b\leq n;\\
                  \gcd(a,b)=1}}
 C_{m,a}\boxtimes C_{n,b}
 \in R(S_m\times S_n)
 \label{eq:W-definition}
\end{equation}
(this holds because $C_{m,a} \in R(S_m)$ and $C_{n,b} \in R(S_n)$).

\begin{proposition}
\label{prop:anti-natural-image}
Using the character embedding to identify each representation
ring $R(G)$
with a subring of $\Cl_\QQ(G)$, the following statements hold.
\begin{enumerate}[label=(\alph*)]
\item
\label{prop:anti-natural-image-equality}
%As class functions on $S_m\times S_n$,
In $\Cl_\QQ(S_m\times S_n)$, we have
\begin{equation}
 \Delta^{\AntiProd}_{m,n}(M_{mn})=W_{m,n}.
 \label{eq:anti-natural-image}
\end{equation}
\item
\label{prop:anti-natural-image-integral}
We have
$\Delta^{\AntiProd}_{m,n}(M_{mn})\in R(S_m\times S_n)$.
\end{enumerate}
\end{proposition}

\begin{proof}
For $(\sigma,\tau)\in S_m\times S_n$, Lemma~\ref{lem:Cma-character}
gives
\begin{equation}
 \chi_{W_{m,n}}(\sigma,\tau)
 =\sum_{\substack{1\leq a\leq m,\ 1\leq b\leq n;\\
                  \gcd(a,b)=1}}
   a\,m_a(\sigma)\,b\,m_b(\tau).
 \label{eq:W-character}
\end{equation}
Indeed, characters multiply under external tensor products:
\[
 \chi_{C_{m,a}\boxtimes C_{n,b}}(\sigma,\tau)
 =\chi_{C_{m,a}}(\sigma)\chi_{C_{n,b}}(\tau),
\]
and Lemma~\ref{lem:Cma-character} evaluates these two factors as
$a\,m_a(\sigma)$ and $b\,m_b(\tau)$, respectively.

On the other hand, the character of $M_{mn}$ at a permutation is its
number of fixed points.  Hence, $\chi_{M_{mn}}(\mathcal A(\sigma,\tau))$
is the number of fixed points of $\mathcal A(\sigma,\tau)$.  Let us
compute this number.  A pair consisting of an $a$-cycle of $\sigma$
and a $b$-cycle of $\tau$ contributes, under the anti-arithmetic rule,
$\lcm(a,b)$ cycles of length $\gcd(a,b)$ to $\mathcal A(\sigma,\tau)$.
These are fixed points exactly when $\gcd(a,b)=1$; in that case there are
\[
 \lcm(a,b)=ab
\]
of them.  Thus, the given pair of cycles
contributes exactly $ab$ fixed points
to $\mathcal A(\sigma,\tau)$ if $\gcd(a,b)=1$; otherwise it contributes
none.
Summing over all pairs of cycles gives
\[
\chi_{M_{mn}}(\mathcal A(\sigma,\tau))
= \sum_{\substack{1\leq a\leq m,\ 1\leq b\leq n;\\
                  \gcd(a,b)=1}}
   m_a(\sigma)\,m_b(\tau)\cdot ab
= \sum_{\substack{1\leq a\leq m,\ 1\leq b\leq n;\\
                  \gcd(a,b)=1}}
   a\,m_a(\sigma)\,b\,m_b(\tau).
\]
Comparing this with \eqref{eq:W-character}, we find
\[
 \chi_{W_{m,n}}(\sigma,\tau)
 =\chi_{M_{mn}}(\mathcal A(\sigma,\tau))
 =\left(\Delta^{\AntiProd}_{m,n}(\chi_{M_{mn}})\right)(\sigma,\tau).
\]
Since $\sigma$ and $\tau$ were arbitrary, this shows that
$\chi_{W_{m,n}}=\left(\Delta^{\AntiProd}_{m,n}(\chi_{M_{mn}})\right)$,
which proves part~(a).  Part~(b) follows from part~(a) and the
fact that $W_{m,n}\in R(S_m\times S_n)$ by its definition
\eqref{eq:W-definition}.
\end{proof}

We can now finish the proof of Theorem~\ref{thm:main-integrality}
in a few lines of $\lambda$-ring reasoning.

\begin{proposition}
\label{prop:anti-Delta-integral}
For all $m,n\geq1$, we have
\begin{equation}
 \Delta^{\AntiProd}_{m,n}(R(S_{mn}))
 \subseteq R(S_m\times S_n).
 \label{eq:anti-Delta-integral}
\end{equation}
\end{proposition}

\begin{proof}
By Corollary~\ref{cor:Delta-psi} (b), the map
$\Delta^{\AntiProd}_{m,n}$ is a $\lambda$-ring morphism
\[
 \Cl_\QQ(S_{mn})\longrightarrow \Cl_\QQ(S_m\times S_n).
\]
By Proposition~\ref{prop:anti-natural-image} (b), it sends the
natural representation $M_{mn}$ into the $\lambda$-subring
$R(S_m\times S_n)\subseteq\Cl_\QQ(S_m\times S_n)$.
By Theorem~\ref{thm:Marin} (applied to $mn$ instead of $n$),
this representation $M_{mn}$ generates
$R(S_{mn})$ as a $\lambda$-ring.  Therefore the whole of $R(S_{mn})$ is
sent into $R(S_m\times S_n)$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main-integrality}]
We have already proved parts (a) and (b)
in Subsection~\ref{subsec:proof-ab}.

% (a) The arithmetic integrality statement follows from
% \eqref{eq:arithmetic-natural-image}, Theorem~\ref{thm:Marin}, and
% Proposition~\ref{prop:adjoint-integrality-criterion}.

% (b) The Schur-positivity statement is exactly
% \eqref{eq:arithmetic-induction}.

% (c) The anti-arithmetic integrality statement follows from
% Proposition~\ref{prop:anti-Delta-integral} and
% Proposition~\ref{prop:adjoint-integrality-criterion}.

(c) Proposition~\ref{prop:anti-Delta-integral} shows that
\eqref{eq:representation-integrality-target} holds for
$\circ = \AntiProd$.
Thus, \eqref{eq:Delta-integral} holds for $\circ = \AntiProd$ (since
\eqref{eq:representation-integrality-target} is just a restatement
of \eqref{eq:Delta-integral}). Consequently, by
Proposition~\ref{prop:adjoint-integrality-criterion}, we find
$\Symm_{\ZZ,m}\AntiProd\Symm_{\ZZ,n}
 \subseteq\Symm_{\ZZ,mn}$,
and Theorem~\ref{thm:main-integrality} (c) is proved.
\end{proof}

\subsection{What the proof is really using}

It may be useful to isolate the short core of the above
argument for the anti-arithmetic product.  There are four ingredients.

\begin{enumerate}[label=(\roman*)]
\item The anti-arithmetic map on conjugacy classes is compatible with
powers:
\[
 \mathcal A(\sigma^r,\tau^r)\sim\mathcal A(\sigma,\tau)^r.
\]
Therefore its pullback is a $\psi$-ring morphism (i.e., a ring
morphism respecting the Adams operations) and hence a
$\lambda$-ring morphism (i.e., a ring morphism respecting the
exterior-power operations) on rational class functions.

\item The Adams operations of the natural $S_m$-representation
detect fixed points of powers:
\[
 \chi_{\psi^d(M_m)}(\sigma)=|\Fix(\sigma^d)|.
\]
M\"obius inversion therefore produces the exact-cycle virtual
representations $C_{m,a}$.

\item Coprime pairs of cycle lengths are exactly the pairs that produce
fixed points under the anti-arithmetic rule.  This gives the virtual
representation $W_{m,n}$ in \eqref{eq:W-definition}, which is the
anti-arithmetic pullback of $M_{mn}$.

\item The single representation $M_{mn}$ generates $R(S_{mn})$ as a
$\lambda$-ring.
\end{enumerate}

Some of these ingredients have their natural habitat in the Burnside
ring of $G$ more than in the representation ring $R(G)$ (or, to stay
categorical, in the category of $G$-sets rather than of
representations).  However, the Frobenius characteristic is
an isomorphism from $R(S_n)$ rather than from the Burnside ring,
and so we would not have had much of an advantage by working in the
Burnside ring.

\subsection{A general class-map criterion}

The proof also isolates a general mechanism that may be useful
elsewhere.  Let $G$ and $H$ be finite groups.  We shall call any map
\[
 a:\{\text{conjugacy classes of }G\}
 \longrightarrow
 \{\text{conjugacy classes of }H\}
\]
a \emph{class map}.  Pullback gives an algebra homomorphism
\[
 a^*:\Cl_\QQ(H)\longrightarrow\Cl_\QQ(G).
\]
For a conjugacy class $C$ of $H$ and $r\geq1$, write $C^{[r]}$
for the conjugacy class containing $h^r$, where $h$ is any element of
$C$.  If
\begin{equation}
 a([g^r])=a([g])^{[r]}\qquad
 \text{ for all } g\in G\text{ and } r\geq1,
 \label{eq:general-power-compatible}
\end{equation}
then $a^*$ commutes with all Adams operations, hence is a $\lambda$-ring
morphism on rational class functions.  To restrict this morphism to the
integral representation rings, one needs the additional arithmetic
condition
\[
 a^*(R(H))\subseteq R(G).
\]
The anti-arithmetic proof establishes this condition by checking the
image of one $\lambda$-generator.

This viewpoint explains both the similarity and the difference between
the two products.  The arithmetic class map comes from a group
homomorphism, so integrality of pullback is automatic.  The
anti-arithmetic class map only has the weaker power-compatibility
property \eqref{eq:general-power-compatible};
integrality has to be manufactured separately, and
Proposition~\ref{prop:anti-natural-image} is exactly the missing step.

\begin{example}
A basic non-homomorphic example is the power class map
\[
 [g]\longmapsto[g^q]
 \qquad \text{ for a given }q\geq1
\]
from the conjugacy classes of a finite group $G$ to themselves.  It
satisfies \eqref{eq:general-power-compatible}, and its pullback on class
functions is exactly the Adams operation $\psi^q$.  Proposition~
\ref{prop:character-lambda}(b) says that this pullback preserves $R(G)$.
\end{example}

\begin{remark}
Single $\lambda$-generation of the representation ring
(i.e., it being generated as a $\lambda$-ring by a single element)
is not entirely peculiar to symmetric groups, but (particularly
because of the integral structure) is less common than it may appear.
For instance, it can be shown that the representation ring
$R(S_2 \times S_2)$ of the Klein four-group $S_2 \times S_2$
is not generated by a single element as a $\lambda$-ring.
One may wonder what groups $G$ have the property.
To avoid
field-of-definition issues in this remark, let $R_{\CC}(G)$ denote the
complex representation ring of a finite group $G$.  If $G=C_m$ is cyclic,
then a faithful one-dimensional character generates $R_{\CC}(G)$
already as a ring.

More generally, let $V$ be a finite-dimensional complex representation of
$G$, and let $A_V$ be the $\lambda$-subring of $R_{\CC}(G)$ generated
by $V$.  Then
\[
 A_V\otimes_{\ZZ}\CC
 =R_{\CC}(G)\otimes_{\ZZ}\CC
\]
if and only if the characteristic polynomials
\[
 \det(t-V(g)) \qquad (g\in G)
\]
separate the conjugacy classes of $G$.  Indeed, their coefficients are,
up to signs, the characters of the exterior powers $\bigwedge^jV$, while
$R_{\CC}(G)\otimes_{\ZZ}\CC \cong \Cl_{\CC}(G)$ is the algebra of all
complex-valued functions on the finite set of conjugacy classes.  Thus the
algebra generated by these coefficients is the whole function algebra
exactly when they separate its points.
Integral $\lambda$-generation asks in addition that this full-rank
subring have index $1$ in $R_{\CC}(G)$.

For the necessity in the separation assertion, suppose $V(g)$ and $V(h)$
have the same characteristic polynomial.  Their eigenvalue multisets,
and thus those of $V(g^r)$ and $V(h^r)$, agree for every $r\geq1$.
By \eqref{eq:psi-to-lambda}, the class functions constant on each
such spectral equivalence class form a $\lambda$-subring containing
$V$.  Thus every element of $A_V$ takes the same value at $g$ and $h$,
so $A_V\otimes\CC$ cannot be the full class-function algebra
unless spectral equivalence separates conjugacy classes.

For comparison, Adams and Conway found a related phenomenon for compact,
simply connected Lie groups: along each arm of the Dynkin diagram, the
fundamental representations can be recovered successively from exterior
powers of the representation at the end of the arm.  Guillot
\cite{Guillot} gives an elementary proof.  Thus his construction replaces
the usual set of fundamental representations by a smaller set indexed by
the arms of the Dynkin diagram; for example, his discussion of $E_6$ uses
three $\lambda$-generators.  What is particularly convenient for $S_n$ is
that the very elementary permutation representation $M_n$ already works
integrally.
\end{remark}

\subsection{Further precedents and nearby literature}

Several parts of the proof have close relatives in the literature, although
we do not know a previous occurrence of the anti-arithmetic integrality
argument itself.

First, the use of Adams operations on $R(S_n)$ is classical.  In
symmetric-function language it is precisely inner plethysm by a power sum:
under the Frobenius characteristic, the $r$-th Adams operation corresponds
to
\[
 f\longmapsto p_r\{f\},
\]
where $\{\,\}$ denotes inner plethysm (with the power sum in the outer
slot).  Thibon~\cite{ThibonAdams} and
Scharf--Thibon~\cite{ScharfThibon} use the Hopf algebra of symmetric
functions to study precisely these Adams operators
and to recover Littlewood's formulas for inner plethysm.  The present proof
uses only the easiest part of that theory, namely the power-trace identity
$\chi_{\psi^r V}(g)=\chi_V(g^r)$ and Newton's formulas.  Meir--Szymik
\cite{MeirSzymik} give a useful modern account of Adams operations on finite
group representation rings and characterize them as natural operations on
the representation-ring functor.

Second, the cycle-length arithmetic behind the ordinary arithmetic product
belongs to a broader gcd/lcm circle of ideas.  The necklace ring of
Metropolis--Rota~\cite{MetropolisRota} has multiplication whose structure
constants involve gcd and lcm, and Dress--Siebeneicher~\cite{DressSiebeneicher}
identify closely related necklace and Burnside-ring constructions with the
big Witt vectors and $\lambda$-rings.  These works are not needed for the
proof above, but they provide a conceptual home for the same arithmetic on
cycle lengths.  On the species side, Maia--M\'endez~\cite{MaiaMendez} and
Li~\cite{LiPrimeGraphs} show how the ordinary arithmetic product interacts
with Dirichlet series, Cartesian products, and prime decompositions of
combinatorial structures.  Ryba~\cite{Ryba} studies the corresponding
restriction $S_{mn}\downarrow S_m\times S_n$ in the stable-character basis
and proves stability results for its multiplicities.

Finally, the generation result for $R(S_n)$ has a substantial history.
Murnaghan~\cite{MurnaghanGeneration} studied generation of irreducible
representations under Kronecker products already in 1955.  Butler
\cite{Butler} and especially Boorman~\cite{Boorman} obtained forms of the
one-generator result in the language of $S$-operations and $\lambda$-rings;
Marin~\cite{Marin} later gave a short proof based on Dvir's formula.  Harman
\cite{Harman} extends this circle of results to representation rings of
certain wreath products.


\appendix

\section{A triangular proof of Theorem~\ref{thm:Marin}}
\label{app:triangular-proof}

As we promised, we shall now give three proofs of
Theorem~\ref{thm:Marin}, both to keep this paper self-contained
and to explore the ``roads less traveled'' around this result.

\subsection{Depth and tail}

For a partition $\lambda\vdash n$, define its \emph{depth} by
\begin{equation}
 \depth(\lambda)=n-\lambda_1.
 \label{eq:depth}
\end{equation}
Thus, if $d=\depth(\lambda)$, we can write uniquely%
\footnote{If a partition $\beta$ is empty, then we interpret
its first part $\beta_1$ as $0$.}
\[
 \lambda=(n-d,\alpha),
 \qquad \text{where }
 \alpha\vdash d \text{ with }
 \alpha_1\leq n-d.
\]
Here $(n-d,\alpha)$ means the partition consisting of $n-d$
followed by the entries of $\alpha$.  We call $\alpha$ the
\emph{tail} of $\lambda$.
Let
\begin{align}
 \alpha'=(c_1,c_2,\ldots,c_r)
 \label{eq:alpha'=}
\end{align}
be the conjugate partition of $\alpha$, and define
\begin{equation}
 T_\lambda
 =\bigotimes_{j=1}^r\bigwedge^{c_j}M_n.
 \label{eq:T-lambda}
\end{equation}
Clearly $[T_\lambda]$ belongs to the $\lambda$-subring of $R(S_n)$
generated by $M_n$.

We shall prove that $T_\lambda$ contains $S^\lambda$ once, and that all
other constituents $S^\mu$ of $T_\lambda$
are triangularly smaller: either they have smaller
depth, or they have the same depth and a strictly smaller tail in
dominance order.

\subsection{Reminder on Schur functors}

We will use Schur functors to create new representations of $S_n$ from
old.
We first explain the two definitions of Schur functors that we shall use.
Let $V$ be a $\QQ$-vector space and $\beta\vdash d$.
The tensor power $V^{\otimes d}$ carries the right $S_d$-action
given by
\[
 (v_1\otimes\cdots\otimes v_d)\cdot g
 =v_{g(1)}\otimes\cdots\otimes v_{g(d)}
 \qquad\text{for all } g\in S_d
\]
(that is, the place-permutation action).
It also carries the corresponding left action $g \cdot w = w \cdot g^{-1}$.
Using the left action, we can define the Schur functor as a multiplicity space:
\begin{equation}
 \mathbb S^\beta_{\mathrm{Hom}}(V)
 :=\Hom_{S_d}(S^\beta,V^{\otimes d}).
 \label{eq:Schur-functor}
\end{equation}
Using the right action, we can instead define it by a balanced tensor product:
\begin{equation}
 \mathbb S^\beta_{\otimes}(V)
 :=V^{\otimes d}\otimes_{\QQ[S_d]}S^\beta.
 \label{eq:Schur-functor-tensor}
\end{equation}
The latter is the construction used by Fulton~\cite[\S8.3]{Fulton}.
Both constructions are functorial in $V$, and thus
carry any action on $V$ that acts diagonally on the tensor
power (and thus commutes with place permutations).
In particular, when $V=M_n$, they are $S_n$-representations.
The following lemma explains their equivalence, including the role of duality.

\begin{lemma}[Multiplicity spaces and balanced tensor products]
\label{lem:Hom-tensor-equivalence}
\ \ %
\begin{enumerate}
\item[(a)]
Let $G$ be a finite group, and let $W$ and $E$ be two left
$\QQ[G]$-modules, where $E$ is finite-dimensional.
Give $W$ the right $G$-action $w\cdot g=g^{-1}w$.
There is a canonical isomorphism
\begin{equation}
 W\otimes_{\QQ[G]}E^*
 \longrightarrow\Hom_G(E,W),
 \qquad
 w\otimes\varphi\longmapsto
 \left(e\longmapsto\frac1{|G|}\sum_{g\in G}\varphi(g^{-1}e)\,gw\right).
 \label{eq:Hom-tensor-average}
\end{equation}
It is natural in $W$ and $E$, and respects every action on $W$ commuting
with $G$.
\item[(b)] Consequently, a choice of $G$-equivariant isomorphism
$E\cong E^*$ gives an isomorphism
$W\otimes_{\QQ[G]}E\cong\Hom_G(E,W)$ that is natural in $W$.
\item[(c)] In particular, if $G=S_d$, then, after choosing an isomorphism
$S^\beta\cong(S^\beta)^*$, we have
\[
 \mathbb S^\beta_{\otimes}(V)
 \cong\mathbb S^\beta_{\mathrm{Hom}}(V)
\]
naturally in $V$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) This is a combination of two standard isomorphisms in group representation
theory (over fields of characteristic $0$).

For any $G$-representation $U$, we define its \emph{invariant space}
\[
U^{G}:=\left\{  u\in U\ \mid\ gu=u\text{ for all }g\in G\right\}
\]
and its \emph{coinvariant space}
\[
U_{G}:=U\diagup\operatorname*{span}\nolimits_{\QQ}\left\{
gu-u\ \mid\ u\in U\text{ and } g \in G\right\}  .
\]
Then, the \emph{averaging operator} $P_{U}:U\rightarrow U$ given by
\[
P_{U}\left(  u\right)  =\dfrac{1}{\left\vert G\right\vert }\sum_{g\in G}gu
\]
is a projection onto $U^{G}$; furthermore it kills every difference $gu-u$ and
thus factors through the coinvariant space $U_{G}$. Thus, it induces a linear
map
\[
\overline{P}_{U}:U_{G}\rightarrow U^{G},
\]
which is easily seen to be a
vector space isomorphism (its inverse simply sends each $u\in U^{G}$ to its
projection onto $U_{G}$). This isomorphism is the first ingredient we need.

The second is even more basic (and holds over any field): The ordinary tensor
product $W\otimes_{\QQ}E^{\ast}$ has the diagonal left $G$-action,
where the $G$-action on $E^{\ast}$ is given by $(g\varphi)(e)=\varphi
(g^{-1}e)$. The Hom-space $\Hom_{\QQ}\left(  E,W\right)
$ also has a canonical left $G$-action, given by $(gf)(e)=gf(g^{-1}e)$ for all
$f\in\Hom_{\QQ}\left(  E,W\right)  $ and $g\in G$ and
$e\in E$. The standard vector-space isomorphism
\[
Q:W\otimes_{\QQ}E^{\ast}\longrightarrow
\Hom_{\QQ} \left(  E,W\right)  ,
\qquad w\otimes\varphi\longmapsto(e\mapsto\varphi(e)w)
\]
is $G$-equivariant. Thus, it restricts to a vector space isomorphism
\[
Q^{G}:\left(  W\otimes_{\QQ}E^{\ast}\right)  ^{G}\longrightarrow\left(
\Hom_{\QQ}\left(  E,W\right)  \right)  ^{G}
\]
on the invariant spaces.

Now, we combine the two ingredients. Applying the above-constructed
isomorphism $\overline{P}_{U}:U_{G}\rightarrow U^{G}$ to $U=W\otimes
_{\QQ}E^{\ast}$, and composing it with the isomorphism $Q^{G}$, we
obtain an isomorphism
\[
\begin{CD}
\left(W \otimes_\QQ E^*\right)_G
@>{\overline P_{W \otimes_\QQ E^*}}>{\cong}>
\left(W \otimes_\QQ E^*\right)^G
@>{Q^G}>{\cong}>
\left(\Hom_\QQ\left(E,W\right)\right)^G \ .
\end{CD}
\]
% Old xymatrix version:
% \[
% \xymatrixcolsep{5pc}\xymatrix{
% \left(W \otimes_\QQ E^*\right)_G \ar[r]_\cong^{\overline P_{W \otimes_\QQ E^*}} &
% \left(W \otimes_\QQ E^*\right)^G \ar[r]_\cong^{Q^G} &
% \left(\Hom_\QQ\left(E,W\right)\right)^G
% }\ \ .
% \]
But the coinvariant space $\left(  W\otimes_{\QQ}E^{\ast}\right)  _{G}$
is precisely the balanced tensor product $W\otimes_{\QQ[G]}E^{\ast}$
(since the balancing relations $(g^{-1}w)\otimes\varphi=w\otimes(g\varphi)$ in
the definition of  $W\otimes_{\QQ[G]}E^{\ast}$ are precisely the
coinvariant relations $gu=u$, after applying them to pure tensors
$u=(g^{-1}w)\otimes\varphi$), whereas the invariant space $\left(
\Hom_{\QQ}\left(  E,W\right)  \right)  ^{G}$ is precisely
the space $\Hom_{G}\left(  E,W\right)  $ of $G$-equivariant maps
(since an $f\in\Hom_{\QQ}\left(  E,W\right)  $
is $G$-invariant if and only if $gf(g^{-1}e)=f\left(  e\right)  $ for all
$g\in G$ and $e\in E$, but this is equivalent to $f$ being $G$-equivariant).
Hence, the isomorphism we just obtained is an isomorphism $W\otimes
_{\QQ[G]}E^{\ast}\longrightarrow\Hom_{G}(E,W)$. Moreover,
it is given by the exact formula \eqref{eq:Hom-tensor-average} (this follows
from the definitions of $P_{U}$ and $Q$).

These constructions are natural and commute with any additional action on $W$
commuting with $G$.

(b) This is automatic.

(c) Rational Specht modules are self-dual. One can see this directly by taking
a positive definite rational bilinear form on $S^{\beta}$ and averaging it
over $S_{d}$: the resulting form remains positive definite and is
$S_{d}$-invariant, so it yields an $S_{d}$-equivariant isomorphism
from $S^{\beta}$ to $(S^{\beta})^{\ast}$.\ \ \ \ %
\footnote{See \cite[Theorem 5.19.35]{sga} for
a proof under more minimalistic assumptions.} Thus, setting $G=S_{d}$,
$E=S^{\beta}$ and $W=V^{\otimes d}$ in part (b), we obtain $V^{\otimes
d}\otimes_{\QQ[S_{d}]}S^{\beta}\cong\Hom_{S_{d}}
(S^{\beta},V^{\otimes d})$. That is, $\mathbb{S}_{\otimes}^{\beta}
(V)\cong\mathbb{S}_{\mathrm{Hom}}^{\beta}(V)$.
\end{proof}

We write $\mathbb S^\beta(V)$ for the multiplicity-space Schur functor
$\mathbb S^\beta_{\mathrm{Hom}}(V)$ defined in
\eqref{eq:Schur-functor}, and use Lemma~\ref{lem:Hom-tensor-equivalence} (c)
to pass to the tensor definition
\eqref{eq:Schur-functor-tensor} when convenient.  The isomorphism between
them is natural in $V$ once the self-duality of $S^\beta$ has been fixed;
the isomorphism with the dual in \eqref{eq:Hom-tensor-average} requires
no such choice.

\subsection{A triangular decomposition of $\mathbb S^\beta(M_n)$}

\begin{lemma}
\label{lem:Schur-functor-depth}
Let $\beta\vdash d$, and assume $\beta_1\leq n-d$.  Then, in $R(S_n)$,
we have
\begin{equation}
 [\mathbb S^\beta(M_n)]
 =[S^{(n-d,\beta)}]
 +\sum_{\substack{\mu\vdash n;\\\depth(\mu)<d}}
 a_{\beta\mu}[S^\mu]
 \label{eq:Schur-functor-triangular}
\end{equation}
for some nonnegative integers $a_{\beta\mu}$.
\end{lemma}

\begin{proof}
We will use the notation $\Inj(X,Y)$ for the set of all
injections from a set $X$ to a set $Y$.
This set has a left action by the symmetric group $S_Y$
(by composition) and a right action by the symmetric group
$S_X$ (also by composition).

\emph{Step 1: the lower part of the filtration.}
Filter $M_n^{\otimes d}$ by the number of distinct basis vectors that
occur in a pure tensor.  More precisely, let $F_r$ be the span of the
basis tensors
\[
 e_{i_1}\otimes\cdots\otimes e_{i_d}
 \qquad \text{for which $|\{i_1,\ldots,i_d\}|\leq r$.}
\]
This span $F_r$ is stable under
the commuting actions of $S_n$ and $S_d$. Thus, we obtain a
filtration
\[
0 = F_{-1} \subseteq F_0 \subseteq F_1 \subseteq \cdots \subseteq F_d
= M_n^{\otimes d}
\]
of $M_n^{\otimes d}$ by $S_n$-subrepresentations.

For a tuple $\mathbf{i}=(i_1,\ldots,i_d)\in[n]^d$, let
$\pi(\mathbf{i})$ be the set partition of $[d]$ whose blocks are the fibers
of the map $j\mapsto i_j$; equivalently, $j$ and $k$ lie in the same
block of $\pi(\mathbf{i})$ if and only if $i_j=i_k$.  We call
$\pi(\mathbf{i})$ the \emph{kernel partition} of $\mathbf{i}$.  For example, the
tuple $(3,7,3,3,7)$ has kernel partition
$\{\{1,3,4\},\{2,5\}\}$.

Now fix a set partition $\pi$ of $[d]$ with exactly $r$ blocks.  The
basis tensors $e_{i_1}\otimes\cdots\otimes e_{i_d}$ with
$\pi(\mathbf{i})=\pi$ are naturally in $S_n$-equivariant bijection with the
injections from the $r$-element set of blocks of $\pi$ into $[n]$: an
injection records the common value $i_j$ on each block.  This
$S_n$-set $\Inj(\pi,[n])$ is transitive, and the stabilizer of one such injection is
isomorphic to $S_{n-r}$ (since a permutation in the stabilizer must fix
all the $r$ elements of the image of the injection, but can permute the
remaining $n-r$ elements of $[n]$ arbitrarily).
Hence its permutation representation $\QQ[\Inj(\pi,[n])]$ is
\[
 \Ind_{S_{n-r}}^{S_n}\one.
\]
By the branching rule\footnote{In terms of symmetric functions, this
is just the Pieri rule. Under the Frobenius correspondence $\ch$,
the representation $\Ind_{S_{n-r}}^{S_n}\one$ corresponds to the
symmetric function $h_{n-r} h_1^r = s_{(n-r)} h_1^r$, which (by
repeated application of the Pieri rule) is obtained from $s_{(n-r)}$
by adding $r$ cells to the Young diagram. Obviously, adding cells
cannot make the first row any shorter, so the resulting Schur
functions $s_\mu$ all satisfy $\mu_1 \geq n-r$.},
every Specht module $S^\mu$ occurring in this representation
satisfies
\[
 \mu_1\geq n-r,
\]
and hence $\depth(\mu)\leq r$.

Thus, as an $S_n$-representation,
\[
 F_r
 =
 \bigoplus_{\substack{\pi\text{ a set partition of }[d];\\|\pi|\leq r}}
 \QQ[\Inj(\pi,[n])],
\]
where $|\pi|$ is understood in the literal sense (i.e., it is the
number of blocks of $\pi$).
%where $\Inj(\pi,[n])$ denotes the set of injections from the set of
%blocks of $\pi$ into $[n]$.
The summand indexed by $\pi$ is isomorphic,
as an $S_n$-representation, to
\[
 \Ind_{S_{n-|\pi|}}^{S_n}\one.
\]
By the branching rule, every Specht module $S^\mu$ occurring in this
summand satisfies
\[
 \depth(\mu)\leq |\pi|\leq r.
\]
Hence every $S_n$-constituent of $F_r$ has depth at most $r$
(where the depth of a Specht module $S^\nu$ is defined to be the
depth of the partition $\nu$).
It follows that every $S_n$-constituent
of $F_{d-1}$ has depth at most $d-1$.
Consequently, every $S_n$-constituent of
$\Hom_{S_d}(S^\beta, F_{d-1})$
has depth at most $d-1$ as well
(since $\Hom_{S_d}(S^\beta, F_{d-1})$ is an $S_n$-subrepresentation
of $\Hom_\QQ(S^\beta, F_{d-1}) \cong F_{d-1}^{\oplus \dim S^\beta}$).

\smallskip
\noindent
\emph{Step 2: the top quotient.}
The top quotient $F_d/F_{d-1}$ of our filtration
has a basis consisting of tensors with pairwise distinct indices.
Thus, a variant of our above reasoning shows that
\begin{equation}
 F_d/F_{d-1}\cong\QQ[\Inj([d],[n])]
 \label{eq:top-injections}
\end{equation}
equivariantly for both commuting actions of $S_n$ and $S_d$.
% , where $\Inj(X,Y)$ denotes the set of all
% injections from $X$ to $Y$.
We compute its $S^\beta$-multiplicity space using the tensor definition.
Define the subgroup $H=S_d\times S_{n-d}\subseteq S_n$, where $S_d$ permutes
$1,\ldots,d$ and $S_{n-d}$ permutes $d+1,\ldots,n$.
The injection $f_0:[d]\hookrightarrow[n]$ given by $f_0(i)=i$
identifies $\Inj([d],[n])$ with the left $S_n$-set
$S_n/S_{n-d}$ via $gS_{n-d}\mapsto g f_0$.
% Give injections the right action $f\cdot a=f\circ a$ for $a\in S_d$;
% this is the right action obtained by inverting the left place-permutation
% action in \eqref{eq:top-injections}, so that the isomorphism
% \eqref{eq:top-injections} becomes $S_d$-equivariant.
Hence there is an isomorphism of $(S_n,S_d)$-bimodules
\begin{equation}
 \QQ[S_n]\otimes_{\QQ[H]}\QQ[S_d]
 \cong\QQ[\Inj([d],[n])],
 \qquad g\otimes a\longmapsto g f_0 a,
 \label{eq:injection-bimodule}
\end{equation}
where $(b,c)\in H$ acts on $\QQ[S_d]$ by left multiplication by $b$,
and $S_d$ acts on it by right multiplication.
(The map is well-defined since $(b,c)f_0=f_0b$.  It is an isomorphism
because coset representatives for $S_n/H$, together with $a\in S_d$,
parametrize the cosets of $S_{n-d}$ in $S_n$.)
We transform the right $S_d$-action on $\QQ[\Inj([d],[n])]$
into a left $S_d$ action by the familiar rule $g \cdot f = f \cdot g^{-1}$
for all $g \in S_d$ and $f \in \QQ[\Inj([d],[n])]$.

Now, we have a chain of left $S_n$-module isomorphisms
\begin{align}
\Hom_{S_d}\bigl(S^\beta,F_d/F_{d-1}\bigr)
 &\,\cong \Hom_{S_d}\bigl(S^\beta,\QQ[\Inj([d],[n])]\bigr)
    \qquad\left(\text{by \eqref{eq:top-injections}}\right)\nonumber\\
 &\,\cong\QQ[\Inj([d],[n])]\otimes_{\QQ[S_d]}S^\beta
    \qquad\left(\text{by Lemma~\ref{lem:Hom-tensor-equivalence} (b)}\right)\nonumber\\
 &\,\cong\bigl(\QQ[S_n]\otimes_{\QQ[H]}\QQ[S_d]\bigr)
                  \otimes_{\QQ[S_d]}S^\beta
                  \qquad\left(\text{by \eqref{eq:injection-bimodule}}\right)\nonumber\\
 &\,\cong\QQ[S_n]\otimes_{\QQ[H]}
                  \bigl(\QQ[S_d]\otimes_{\QQ[S_d]}S^\beta\bigr)
                  \qquad\left(\text{by associativity of $\otimes$}\right)\nonumber\\
 &\,\cong\QQ[S_n]\otimes_{\QQ[H]}(S^\beta\boxtimes\one)\nonumber\\
 &\qquad\qquad\ \left(\begin{array}{c}\text{where $\one$ is the trivial representation of $S_{n-d}$,}\\ \text{so that $S^\beta\boxtimes\one$ is a representation of $S_d\times S_{n-d}=H$}\end{array}\right)\nonumber\\
 &\,=\Ind_{S_d\times S_{n-d}}^{S_n}(S^\beta\boxtimes\one).
 \label{eq:injection-multiplicity}
\end{align}
Here, the last $\cong$ sign relied on the $H$-equivariant isomorphism $\QQ[S_d]\otimes_{\QQ[S_d]}S^\beta\cong S^\beta\boxtimes\one$ given by $a\otimes s\mapsto as\otimes 1$; the $S_{n-d}$-factor acts trivially.
By Pieri's rule, the right-hand side of \eqref{eq:injection-multiplicity}
is the multiplicity-free sum of
$S^\nu$ over partitions $\nu\vdash n$ such that $\nu/\beta$ is a
horizontal strip of size $n-d$.

The strip has $n-d$ cells in distinct columns, and $\nu$ has
$\nu_1$ columns.  Thus every such $\nu$ satisfies $\nu_1\geq n-d$, hence
$\depth(\nu)\leq d$.  Suppose equality holds.  Then $\nu_1=n-d$.
Together with $\beta\subseteq\nu$, the horizontal-strip condition
is equivalent to the interlacing inequalities
\[
 \nu_{i+1}\leq\beta_i\qquad \text{ for all } i\geq1.
\]
Since
\[
 \sum_{i\geq1}\nu_{i+1}=n-\nu_1=d=|\beta|=\sum_{i\geq1}\beta_i,
\]
all these inequalities must be equalities.  Hence
\[
 \nu=(n-d,\beta).
\]
Thus, $S^{(n-d,\beta)}$ occurs with multiplicity $1$ in the
$S_n$-representation
\[
 \Hom_{S_d}(S^\beta,F_d/F_{d-1}),
\]
whereas all other constituents of this multiplicity space have depth $<d$.
(Here it is important to take the $S^\beta$-multiplicity space: in the
whole quotient $F_d/F_{d-1}$, the multiplicity of
$S^{(n-d,\beta)}$ is $\dim S^\beta$, which can exceed $1$.)

\smallskip
\noindent
\emph{Step 3: take the $S^\beta$-multiplicity space.}
We have
\[
M_n^{\otimes d} = F_d \cong F_{d-1} \oplus (F_d / F_{d-1})
\]
as $S_n \times S_d$-representations, by Maschke's theorem.
Applying the functor $\Hom_{S_d}(S^\beta,-)$ (which clearly
respects direct sums) to this decomposition, we see that
\[
\Hom_{S_d}(S^\beta, M_n^{\otimes d})
\cong \Hom_{S_d}(S^\beta,F_{d-1}) \oplus
\Hom_{S_d}(S^\beta,F_d / F_{d-1}).
\]
The first addend here consists entirely of
$S_n$-constituents of depth at most $d-1$ (by Step 1),
while the second addend contains $S^{(n-d,\beta)}$ with
multiplicity $1$ and, apart from that, only constituents
of depth $<d$ as well (by Step 2).
Thus, altogether,
$\Hom_{S_d}(S^\beta, M_n^{\otimes d})$
is a direct sum of $S^{(n-d,\beta)}$ with
a number of Specht modules of depth $<d$.
Since $\mathbb S^\beta(M_n) = \Hom_{S_d}(S^\beta, M_n^{\otimes d})$,
this proves \eqref{eq:Schur-functor-triangular}.
\end{proof}

Lemma~\ref{lem:Schur-functor-depth}
is a small triangular piece of the classical restriction
problem from $GL_n$ to $S_n$.  Indeed, regarding $S_n$ as the group of
permutation matrices in $GL_n$, the $S_n$-representation
$\mathbb S^\beta(M_n)$ is the restriction of the polynomial
$GL_n$-representation $\mathbb S^\beta(\QQ^n)$.  Orellana--Zabrocki
\cite[Introduction]{OrellanaZabrocki} discuss this restriction problem and
encode such restrictions by evaluating symmetric functions at the
eigenvalues of permutation matrices.  We only need the top-depth
triangular statement above.

\subsection{A triangular decomposition of $T_\lambda$}

We now use the standard Kostka-triangular decomposition of a tensor
product of exterior powers.  Recall \eqref{eq:alpha'=}.
The symmetric-function identity
\begin{equation}
 e_{\alpha'}
 =\prod_{j=1}^r e_{c_j}
 =\sum_{\beta\vdash d}K_{\beta',\alpha'}s_\beta
 \label{eq:e-Kostka}
\end{equation}
(where $K_{\lambda,\mu}$ are the Kostka numbers)%
\footnote{See, e.g., \cite[Corollary 7.15.3]{Stanley}
(setting $\nu=\varnothing$)
for a proof of \eqref{eq:e-Kostka}.}
translates, by Schur--Weyl theory, to
\begin{equation}
 T_\lambda
 \cong
 \bigoplus_{\beta\vdash d}
 K_{\beta',\alpha'}\,\mathbb S^\beta(M_n)
 \label{eq:T-Schur-functors}
\end{equation}
(for an explicit reference,
apply \cite[\S 8.3, Corollary 2(b)]{Fulton} to $\alpha'$
instead of $\mu$ and restrict to the symmetric group
$S_n \subseteq GL_n$).
Thus, in $R(S_n)$, we have
\begin{equation}
 [T_\lambda]
 =
 \sum_{\beta\vdash d}
 K_{\beta',\alpha'}\,[\mathbb S^\beta(M_n)].
 \label{eq:T-Schur-functors2}
\end{equation}
However, it is well-known (see, e.g., \cite[Proposition 7.10.5]{Stanley})
that the Kostka number $K_{\beta',\alpha'}$ is nonzero only if
$\beta'\unrhd\alpha'$ (the symbol $\unrhd$ means ``dominates
or equals''), equivalently $\beta\unlhd\alpha$.
In particular, whenever a nonzero addend occurs in
\eqref{eq:T-Schur-functors2}, we have $\beta\unlhd\alpha$ and thus
$\beta_1\leq\alpha_1\leq n-d$,
so Lemma~\ref{lem:Schur-functor-depth} applies.  Furthermore, the
addend for $\beta = \alpha$ occurs with coefficient $1$, since
(again by \cite[Proposition 7.10.5]{Stanley}) we have
\[
 K_{\alpha',\alpha'}=1.
\]
We obtain the promised triangularity:

\begin{proposition}
\label{prop:T-triangular}
Let $\lambda=(n-d,\alpha)\vdash n$.  Then
\begin{equation}
 [T_\lambda]
 =[S^\lambda]
 +\sum_{\substack{\beta\lhd\alpha}}
   K_{\beta',\alpha'}[S^{(n-d,\beta)}]
 +\sum_{\substack{\mu\vdash n;\\\depth(\mu)<d}}
   b_{\lambda\mu}[S^\mu],
 \label{eq:T-triangular}
\end{equation}
where the $b_{\lambda\mu}$ are nonnegative integers, and
$\beta\lhd\alpha$ means strict dominance.
\end{proposition}

\begin{proof}
As we observed, all the nonzero addends in \eqref{eq:T-Schur-functors2}
satisfy $\beta \unlhd \alpha$ and $\beta_1 \leq n-d$, so that
we can rewrite them using Lemma~\ref{lem:Schur-functor-depth}.
This yields
\begin{align*}
 [T_\lambda]
 &=
 \sum_{\substack{\beta\vdash d;\\ \beta \unlhd \alpha}}
 K_{\beta',\alpha'}\left([S^{(n-d,\beta)}]
 +\sum_{\substack{\mu\vdash n;\\\depth(\mu)<d}}
 a_{\beta\mu}[S^\mu]\right) \\
 &= \sum_{\substack{\beta\vdash d;\\ \beta \unlhd \alpha}}
 K_{\beta',\alpha'}\,[S^{(n-d,\beta)}]
 +\sum_{\substack{\mu\vdash n;\\\depth(\mu)<d}}
   b_{\lambda\mu}[S^\mu]
\end{align*}
for some nonnegative integers $b_{\lambda\mu}$.
In light of the fact that $K_{\alpha',\alpha'} = 1$,
we can rewrite the first sum
$\sum_{\substack{\beta\vdash d;\\ \beta \unlhd \alpha}}
 K_{\beta',\alpha'}\,[S^{(n-d,\beta)}]$ as
$[S^{(n-d,\alpha)}]
 +\sum_{\substack{\beta\vdash d;\\ \beta \lhd \alpha}}
   K_{\beta',\alpha'}[S^{(n-d,\beta)}]$; and the proposition
is proved (since $(n-d,\alpha) = \lambda$).
\end{proof}

\subsection{First proof of Theorem~\ref{thm:Marin}}

\begin{proof}[First proof of Theorem~\ref{thm:Marin}]
Let $A_n$ be the $\lambda$-subring of $R(S_n)$ generated by $M_n$.  We
prove $[S^\lambda]\in A_n$ for every $\lambda\vdash n$, by induction first
on $d=\depth(\lambda)$ and then, for fixed $d$, upward along dominance of
the tail $\alpha$ in $\lambda=(n-d,\alpha)$.

The element $[T_\lambda]$ belongs to $A_n$ (by its construction).  In
\eqref{eq:T-triangular}, every constituent in the last sum has smaller
depth, so belongs to $A_n$ by the outer induction.  Every constituent in
the middle sum has the same depth but strictly smaller tail, so belongs
to $A_n$ by the inner induction.  Subtracting these already-known
classes from $[T_\lambda]$ gives $[S^\lambda]\in A_n$.
This completes the induction.

Since the Specht classes $[S^\lambda]$ form a $\ZZ$-basis of $R(S_n)$, this
proves $A_n=R(S_n)$.
\end{proof}

\begin{remark}
The two parts of the triangular order can be seen concretely already for
$n=6$ and $d=3$.  First consider the Schur functor corresponding to
$\beta=(3)$.  Sorting the basis tensors of $\operatorname{Sym}^3(M_6)$ according to
whether they involve three, two, or one distinct basis vectors (as in the
proof of Lemma~\ref{lem:Schur-functor-depth}) gives
\begin{equation*}
 \mathbb S^{(3)}(M_6)=\operatorname{Sym}^3(M_6)
 \cong
 S^{(3,3)}
 \oplus
 \underbrace{S^{(4,1,1)}
 \oplus 2S^{(4,2)}
 \oplus 4S^{(5,1)}
 \oplus 3S^{(6)}}_{\text{constituents of depth $<3$}}.
\end{equation*}
Thus the lower-depth terms need not be lower in dominance order.  For
instance, $(3,3)$ and $(4,1,1)$ are incomparable: the first partial sum
favors $(4,1,1)$, whereas the first two partial sums favor $(3,3)$.
This is why dominance order alone is not sufficient.

The second part of the induction---dominance among tails of the same
depth---is visible if we take $\lambda=(3,2,1)$, so that $d=3$,
$\alpha=(2,1)$, and
\[
 T_\lambda=\bigwedge^2M_6\otimes M_6.
\]
Since $e_2e_1=s_{(2,1)}+s_{(1,1,1)}$, we have
\[
 T_\lambda
 \cong \mathbb S^{(2,1)}(M_6)
       \oplus \mathbb S^{(1,1,1)}(M_6),
\]
and the same calculation gives
\begin{equation*}
 T_{(3,2,1)}
 \cong
 S^{(3,2,1)}
 \oplus
 \underbrace{S^{(3,1,1,1)}}_{\substack{\text{same depth $3$,}\\
                         (1,1,1)\lhd(2,1)}}
 \oplus
 \underbrace{3S^{(4,1,1)}
 \oplus 2S^{(4,2)}
 \oplus 3S^{(5,1)}
 \oplus S^{(6)}}_{\text{constituents of depth $<3$}}.
\end{equation*}
Thus this one example displays exactly the two kinds of already-known
terms that occur in \eqref{eq:T-triangular}: smaller-depth constituents,
and same-depth constituents whose tails are strictly smaller in dominance
order.
\end{remark}

\begin{remark}[Relation with Marin's formulation]
Marin states his theorem \cite[Theorem 1.1]{Marin}
by saying that the exterior powers of
the standard representation $V_n$ generate $R(S_n)$ as a ring
(not just as a $\lambda$-ring).
This implies Theorem~\ref{thm:Marin}.  Conversely, the triangular proof
above establishes this ordinary-ring statement directly: every
$T_\lambda$ is a tensor product of exterior powers of $M_n$, and its
triangular elimination uses only addition, subtraction, and tensor
product.  Thus the exterior powers of $M_n$ generate $R(S_n)$ as a ring.
To pass between the two families of exterior powers, use
$M_n=\one\oplus V_n$, and therefore
\[
 \bigwedge^k M_n
 \cong \bigwedge^kV_n\oplus\bigwedge^{k-1}V_n.
\]
This decomposition expresses the exterior powers of $M_n$ in terms
of those of $V_n$.  Solving it recursively for $\bigwedge^kV_n$
expresses those of $V_n$ in terms of those of $M_n$.  Hence these two
families generate the same ordinary subring.  Merely knowing that a
$\lambda$-subring contains $V_n$ would not, by itself, prove ordinary
ring generation.
\end{remark}


\section{Two descent-algebra lifts of Boorman--Marin}
\label{app:NSym-generation}

Theorem~\ref{thm:Marin} says that the representation ring $R(S_n)$ is
$\lambda$-generated by the natural permutation representation $M_n$.  If we
replace $M_n$ by the reflection representation $V_n=M_n-\one$, then the
representations obtained immediately by applying the $\lambda$-operations are
\[
 \bigwedge^k V_n\cong S^{(n-k,1^k)}
 \qquad \text{for } 0\leq k\leq n-1
\]
(see, e.g., \cite[Proposition 2]{Burman-cex} for a proof).
Thus the most direct ordinary-ring form of the Boorman--Marin theorem says
that the hook representations $S^{(n-k,1^k)}$ generate $R(S_n)$ as a $\ZZ$-algebra.
Schocker~\cite{Schocker} found a remarkably literal lift of this statement to
Solomon's descent algebra: certain exact-descent-class sums attached to hook
shapes already generate the whole descent algebra.  In fact, his theorem uses
only about half of the hooks.

We begin by recalling the noncommutative-symmetric-function language in which
this lift is especially transparent.  We then translate Schocker's theorem
carefully into our conventions and give an independent proof.
First Solomon's Mackey formula proves generation by all hooks; then
multiplication by the longest permutation reduces this family to its upper half.
Afterwards we establish a second, different generation theorem: the complete
functions $H_kH_{n-k}$ also generate the descent algebra.  Their commutative
images are not hook characters, but rather the permutation characters on
$k$-subsets.  Thus the two results lift two different generating families in
$R(S_n)$.

\subsection{Complete and ribbon bases, and Solomon's Mackey formula}

Let $\NSym_\ZZ$ be the free associative $\ZZ$-algebra on generators
$H_1,H_2,\ldots$, graded in such a way that each $H_i$ is homogeneous
with $\deg H_i=i$. Also, put $H_0=1$.  If
$I=(i_1,\ldots,i_r)$ is a composition, write
\[
 H_I=H_{i_1}\cdots H_{i_r}.
\]
The elements $H_I$, for $I\models n$, form a $\ZZ$-basis of the homogeneous
component $\NSym_{\ZZ,n}$.  The algebra $\NSym_\ZZ$ is known as the
algebra of \emph{noncommutative symmetric functions} over $\ZZ$;
it was first introduced in \cite{GelfandEtAlNCSF} (see
\cite[\S 5.4]{GrinbergReiner} for a modern introduction).

We shall also use the \emph{ribbon basis}
(see \cite[\S 3.2, \S 4.4]{GelfandEtAlNCSF}).  If $I\models n$, define
\begin{equation}
 R_I
 =\sum_{J\text{ coarsens }I}
   (-1)^{\ell(I)-\ell(J)}H_J.
 \label{eq:NSym-ribbon-definition}
\end{equation}
Here a \emph{coarsening} of $I$ is obtained by repeatedly replacing two
consecutive parts by their sum.  M\"obius inversion on the refinement poset
gives
\begin{equation}
 H_I=\sum_{J\text{ coarsens }I}R_J.
 \label{eq:NSym-H-ribbon-inversion}
\end{equation}
For example,
\[
 R_{(2,1,1)}
 =H_{(2,1,1)}-H_{(3,1)}-H_{(2,2)}+H_{(4)}.
\]

For $w\in S_n$, write
\[
 \Des(w)=\{i\in[n-1]:w(i)>w(i+1)\}.
\]
For $I=(i_1,\ldots,i_r)\models n$, put
\[
 D(I)=\{i_1,i_1+i_2,\ldots,i_1+\cdots+i_{r-1}\},
 \qquad
 x_I=\sum_{\substack{w\in S_n;\\\Des(w)\subseteq D(I)}}w
 \in \ZZ[S_n].
\]
The \emph{(integral) Solomon descent algebra} is
\begin{equation}
 \mathcal D(S_n):=\bigoplus_{I\models n}\ZZ x_I\subseteq\ZZ[S_n],
 \label{eq:descent-algebra-definition}
\end{equation}
with the usual group-algebra product.  Solomon's Mackey formula
shows that this span is closed under multiplication
(see, e.g., \cite[Theorem 3.8.1]{Bidigare-thesis},
\cite[proof of Theorem 2.1]{Saliola},
\cite[Theorem 4.12]{GrinbergParlett},
or many other places).
The \emph{descent-class correspondence} is the $\ZZ$-linear map
\begin{equation}
 \iota:\NSym_{\ZZ,n}\longrightarrow\mathcal D(S_n),
 \qquad H_I\longmapsto x_I.
 \label{eq:descent-iota}
\end{equation}
It is an isomorphism of $\ZZ$-modules.  By M\"obius inversion,
\[
 \iota(R_I)=\sum_{\substack{w\in S_n;\\\Des(w)=D(I)}}w
 =:\Delta_I,
\]
the so-called \emph{exact-descent-class sum} corresponding to $I$.
So the ribbon basis of $\NSym$
corresponds to the exact-descent-class basis of $\mathcal D(S_n)$.

Besides its usual product, $\NSym_{\ZZ,n}$ carries an \emph{internal product}
$*$, corresponding to multiplication in $\mathcal D(S_n)$.
We use the row-reading matrix convention below.  With the usual
composition of permutations, $(uv)(i)=u(v(i))$, the map $\iota$
is an anti-isomorphism:
\[
 \iota(F*G)=\iota(G)\iota(F).
\]
Thus our internal product corresponds to the opposite of the usual
group-algebra product.  This distinction does not affect generation,
but fixes the order in the hook identity below.  See, for example,
Blessenohl--Schocker~\cite[Introduction and Chapter~14]{BlessenohlSchocker}.

We shall use the standard matrix form of Solomon's Mackey formula, due in this
language to Garsia--Remmel
(\cite[Proposition 1.1]{GarsiaReutenauer}, \cite[Corollary 2.4]{Saliola},
\cite[Proposition 4.3]{BlessenohlLaue}, \cite[Theorem 2]{Willigenburg98},
\cite[Remark B.5]{BlessenohlSchocker}).
% Gelfand--Krob--Lascoux--Leclerc--Retakh--Thibon
% \cite[Proposition~5.1]{GelfandEtAlNCSF}.
% Blessenohl--Schocker
% \cite[Appendix~B, especially B.1 and B.5; see also
% \S\S12.11--12.12]{BlessenohlSchocker} give the same rule directly in terms of
% ordered set partitions.
If $I=(i_1,\ldots,i_p)$ and
$J=(j_1,\ldots,j_q)$ are compositions of $n$, then
\begin{equation}
 H_I*H_J
 =\sum_M H_{\operatorname{comp}(M)},
 \label{eq:NSym-matrix-rule}
\end{equation}
where $M=(m_{uv})$ ranges over all $p\times q$ matrices with nonnegative
integer entries whose row-sum vector is $I$ and whose column-sum vector is
$J$; explicitly, this means that
\[
 \sum_{v=1}^q m_{uv}=i_u \quad\text{ for all } 1\leq u\leq p,
 \qquad\qquad
 \sum_{u=1}^p m_{uv}=j_v \quad\text{ for all } 1\leq v\leq q.
\]
The composition $\operatorname{comp}(M)$ is obtained by reading the entries
of $M$ row by row and deleting the zero entries; we call it the
\emph{row-reading composition} of $M$.
% Set-theoretically,
% $m_{uv}$ is the cardinality of the intersection of the $u$-th block of one
% ordered set partition with the $v$-th block of the other.  This explains both
% the row and column sums and the appearance of the matrix rule.

\begin{example}
Take $I=(3,1)$ and $J=(2,2)$.  There are exactly two nonnegative
$2\times2$ matrices with row-sum vector $I$ and column-sum vector $J$:
\[
 \begin{pmatrix}1&2\\1&0\end{pmatrix},
 \qquad
 \begin{pmatrix}2&1\\0&1\end{pmatrix}.
\]
Their row-reading compositions are $(1,2,1)$ and $(2,1,1)$.
Thus \eqref{eq:NSym-matrix-rule} gives
\[
 H_{(3,1)}*H_{(2,2)}=H_{(1,2,1)}+H_{(2,1,1)}.
\]
\end{example}

Finally, recall the abelianization map
\begin{equation}
 \pi:\NSym_\ZZ\longrightarrow\Symm_\ZZ,
 \qquad H_r\longmapsto h_r.
 \label{eq:NSym-abelianization}
\end{equation}
It satisfies
\begin{equation}
 \pi(R_I)=r_I,
 \label{eq:ribbon-abelianization}
\end{equation}
where $r_I$ is the ribbon Schur function of ribbon shape $I$.  On each
homogeneous component, $\pi$ also respects the internal products: applying
$\pi$ to \eqref{eq:NSym-matrix-rule} gives the corresponding matrix rule for
the Kronecker product of complete symmetric functions
(see, e.g., \cite[\S I.7, Example 23 (e)]{Macdonald}
or \cite[Appendix B, proof of the $\xi^r \xi^q$ formula]{BlessenohlSchocker};
this is essentially an application of the Mackey formula for tensor products
of permutation characters).

In these conventions, \emph{Solomon's epimorphism} is the surjective
ring homomorphism
\begin{equation}
 \theta:\mathcal D(S_n)\longrightarrow R(S_n),
 \qquad
 x_I\longmapsto[\Ind_{S_I}^{S_n}\one],
 \label{eq:Solomon-epimorphism}
\end{equation}
where (for any composition $I = (i_1, i_2, \ldots, i_r)$ of $n$)
we let $S_I=S_{i_1}\times\cdots\times S_{i_r}$ denote the Young subgroup
permuting the consecutive blocks of sizes $i_1,\ldots,i_r$
(alternatively, $\Ind_{S_I}^{S_n}\one$ can be described as the
Young permutation module of tabloids
for a Young diagram with rows of lengths $i_1,\ldots,i_r$).
Equivalently, if $\pi_n$ denotes the degree-$n$ restriction of $\pi$, then
\begin{equation}
 \theta \circ \iota = \ch^{-1} \circ \,\pi_n.
 \label{eq:Solomon-abelianization}
\end{equation}
Indeed, $\ch([\Ind_{S_I}^{S_n}\one])=h_I=\pi(H_I)$,
so that $\ch \circ \theta \circ \iota = \pi_n$.
The compatibility of $\pi_n$ with internal products, together with the
commutativity of $R(S_n)$, shows that $\theta$ is a ring homomorphism
even though $\iota$ reverses products.  It is surjective because the
$h_\lambda$ for $\lambda\vdash n$ form a $\ZZ$-basis of $\Symm_{\ZZ,n}$.
One can equally view its values as virtual characters; in particular,
\[
 \theta(\iota(R_I))=\ch^{-1}(r_I).
\]

\subsection{Schocker's hook generators: the direct lift}

For $0\leq k\leq n-1$, define the \emph{noncommutative hook}
\begin{equation}
 Q_{n,k}:=R_{(n-k,1^k)}\in\NSym_{\ZZ,n}.
 \label{eq:hook-ribbon-generator}
\end{equation}
Thus, under the correspondence $\iota$ of $\NSym_{\ZZ,n}$ with Solomon's
descent algebra, $Q_{n,k}$ becomes
the sum of all permutations whose descent set is
\[
 \{n-k,n-k+1,\ldots,n-1\}
\]
(this is the empty set when $k=0$).
The associated ribbon is the $180^\circ$-rotated ordinary hook
shape $(n-k,1^k)$.  Hence \eqref{eq:ribbon-abelianization} gives
\begin{equation}
 \pi(Q_{n,k})=s_{(n-k,1^k)}
 \label{eq:hook-ribbon-abelianization}
\end{equation}
(since $180^\circ$-rotation of skew Young diagrams does not change
the Schur function).
Under the Frobenius characteristic, this is the character of
$\bigwedge^kV_n$.  Thus the elements $Q_{n,k}$ are exactly the descent-algebra
lifts one would naturally expect from the $\lambda$-ring formulation of the
Boorman--Marin theorem.

Schocker's theorem~\cite[p.~152, Theorem]{Schocker} is stated using the
exact-descent-class sums $\Delta_{\{1,\ldots,k\}}$ for
$(n-1)/2\leq k\leq n-1$.  Conjugation by the longest permutation $w_0$
sends a descent set $D$ to $\{n-i:i\in D\}$, and hence sends
$\Delta_{\{1,\ldots,k\}}$ to the exact-descent-class sum with descent set
$\{n-k,\ldots,n-1\}$, which is $\iota(Q_{n,k})$ in our conventions.  Thus
Schocker proves the following stronger statement.

\begin{theorem}[Schocker]
\label{thm:Schocker-hook-generation}
Fix $n\geq1$.  The integral Solomon descent algebra
$\mathcal D(S_n)$ is generated under its product by
the elements
\[
 \iota(Q_{n,k})
 \qquad
 \text{for all $k$ satisfying }\left\lceil\frac{n-1}{2}\right\rceil\leq k\leq n-1.
\]
Under Solomon's epimorphism, these generators map to the irreducible hook
characters $\chi^{(n-k,1^k)}$; see \cite[p.~152, Corollary~1]{Schocker}.
\end{theorem}

We first prove the weaker statement that all the $Q_{n,k}$ generate,
directly and integrally from Solomon's Mackey formula.  We then recover
the full theorem by multiplying hook descent classes by the longest
permutation.

\begin{proposition}
\label{prop:full-hook-generation}
Fix $n\geq1$.  Under the internal product, the $\ZZ$-algebra
$\NSym_{\ZZ,n}$ is generated by
\[
 Q_{n,0},Q_{n,1},\ldots,Q_{n,n-1}.
\]
\end{proposition}

\begin{proof}
Let $B_n$ be the $\ZZ$-subalgebra generated by the $Q_{n,k}$.  We prove
that every complete-basis element $H_I$ belongs to $B_n$.  We use a finite
double induction: first downward on the first part of $I$, and, among
compositions having the same first part, downward on the length.

The composition $(n)$ causes no difficulty, since
$H_{(n)}=R_{(n)}=Q_{n,0}$.  Now let
\[
 I=(m,a,a_3,\ldots,a_r),
 \qquad r\geq2,
\]
and assume that $H_L\in B_n$ whenever either\footnote{We write $L_1$
for the first entry of the composition $L$.} $L_1>m$, or
$L_1=m$ and $\ell(L)>r$.  Set
\[
 I'=(m+a,a_3,\ldots,a_r).
\]
Since the first part of $I'$ is greater than $m$, we have $H_{I'}\in B_n$
by induction.  We shall multiply it by the hook generator
\[
 Q_{n,a}=R_{(n-a,1^a)}.
\]

\emph{Step 1: locate the possible leading parts.}
By \eqref{eq:NSym-ribbon-definition}, $Q_{n,a}$ is an alternating sum of
$H_J$, where $J=(j_1,\ldots,j_q)$ runs over the coarsenings of
$(n-a,1^a)$.  Every such $J$ satisfies
\begin{equation}
 j_1\geq n-a,
 \qquad \text{ hence } \qquad
 \sum_{v=2}^q j_v\leq a.
 \label{eq:hook-coarsening-bounds}
\end{equation}
Consider a matrix $M=(m_{uv})$ occurring in the Mackey formula for
$H_{I'}*H_J$.  Its first row has sum $m+a$.  Its first entry therefore
satisfies
\begin{align*}
 m_{11}
 &=m+a-\sum_{v=2}^q m_{1v} \\
 &\geq m+a-\sum_{v=2}^q j_v
 \qquad \left(\text{since each $v\geq 2$ satisfies
 $m_{1v} \leq \sum_{u=1}^{r-1}m_{uv} = j_v$}\right)
 \\
 &\geq m
 \qquad
 \left(\text{since $\sum_{v=2}^q j_v\leq a$}\right).
\end{align*}
In particular, $m_{11}>0$, so $m_{11}$ is the first part of
$\operatorname{comp}(M)$.  Thus every complete-basis term occurring in
$H_{I'}*Q_{n,a}$ has first part at least $m$.

\emph{Step 2: analyze the equality case.}
Suppose that $m_{11}=m$.  Then equality must hold throughout the preceding
inequalities.  Consequently
\[
 j_1=n-a,
 \qquad
 \sum_{v=2}^q j_v=a,
\]
and every entry below the first row in columns $2,\ldots,q$ is zero.  Hence
$J$ has the form
\[
 J=(n-a,b_1,\ldots,b_s)
 \qquad \text{ with }
 \qquad b_1+\cdots+b_s=a,
\]
and there is exactly one matrix $M$ with first entry $m$, namely
\[
 \begin{pmatrix}
 m&b_1&b_2&\cdots&b_s\\
 a_3&0&0&\cdots&0\\
 a_4&0&0&\cdots&0\\
 \vdots&\vdots&\vdots&&\vdots\\
 a_r&0&0&\cdots&0
 \end{pmatrix}.
\]
Its row-reading composition is
\[
 (m,b_1,\ldots,b_s,a_3,\ldots,a_r),
\]
which has length $r+s-1\geq r$.  Equality of lengths occurs only for
$s=1$.  In that case $J=(n-a,a)$ and the row-reading composition is exactly
$I$.  The coefficient of $H_{(n-a,a)}$ in
$Q_{n,a} = R_{(n-a,1^a)}$ is
\[
 (-1)^{(a+1)-2}=(-1)^{a-1}.
\]
Therefore
\begin{equation}
 H_{I'}*Q_{n,a}
 =(-1)^{a-1}H_I
 +\sum_L c_LH_L,
 \label{eq:Schocker-Mackey-triangular}
\end{equation}
where $c_L\in\ZZ$, and every $L$ in the sum satisfies either $L_1>m$, or
$L_1=m$ and $\ell(L)>r$.

All terms in the sum in \eqref{eq:Schocker-Mackey-triangular} belong to
$B_n$ by induction, as does the left-hand side.  Since the coefficient of
$H_I$ is $\pm1$, we conclude that $H_I\in B_n$.  This completes the double
induction.
\end{proof}

\begin{remark}
The proof is an integral triangular elimination.  The relevant order on
compositions is deliberately simple: first compare the first parts, and if
they agree, compare the lengths.  Formula
\eqref{eq:Schocker-Mackey-triangular} expresses $H_I$, up to sign, in terms
of elements that are earlier in this induction.
\end{remark}

\begin{proof}[Proof of Theorem~\ref{thm:Schocker-hook-generation}]
Let $w_0 \in S_n$ be the longest permutation, given by $w_0(i)=n+1-i$.
The only permutation with every position a descent is $w_0$, so
\[
 \iota(Q_{n,n-1})=w_0.
\]
For any $w\in S_n$, right multiplication by $w_0$ reverses its
one-line notation.  In particular,
\[
 \Des(ww_0)
 =\{n-i:i\in[n-1]\setminus\Des(w)\}.
\]
If $\Des(w)=\{n-k,\ldots,n-1\}$, then this set is
$\{k+1,\ldots,n-1\}$, the descent set defining $Q_{n,n-1-k}$.
Since $w\mapsto ww_0$ is a bijection, summing over this exact descent
class gives
\[
 \iota(Q_{n,k})w_0=\iota(Q_{n,n-1-k}).
\]
That is,
$\iota(Q_{n,k})\iota(Q_{n,n-1})=\iota(Q_{n,n-1-k})$
(since $\iota(Q_{n,n-1})=w_0$).
Using the anti-isomorphism \eqref{eq:descent-iota}, this yields
\begin{equation}
 Q_{n,n-1}*Q_{n,k}=Q_{n,n-1-k}
 \qquad \text{ for all }0\leq k\leq n-1.
 \label{eq:hook-complement}
\end{equation}

Now let $B_n^+$ be the $\ZZ$-subalgebra generated by the hooks
$Q_{n,k}$ with $\lceil(n-1)/2\rceil\leq k\leq n-1$.
For any $j<\lceil(n-1)/2\rceil$, set $k=n-1-j$.
Then $k\geq\lceil(n-1)/2\rceil$, so both $Q_{n,k}$ and
$Q_{n,n-1}$ belong to $B_n^+$.  Equation~\eqref{eq:hook-complement}
therefore puts $Q_{n,j}$ in $B_n^+$ as well
(since $j=n-1-k$).
Thus the subalgebra $B_n^+$ contains each of $Q_{n,0}, Q_{n,1}, \ldots, Q_{n,n-1}$.
By Proposition~\ref{prop:full-hook-generation}, it thus equals all of
$\NSym_{\ZZ,n}$.  In other words, $\NSym_{\ZZ,n}$ under the internal
product is generated by the $Q_{n,k}$ with $\lceil(n-1)/2\rceil\leq k\leq n-1$.
By applying $\iota$, this entails that the descent algebra $\mathcal D(S_n)$
is generated by the $\iota(Q_{n,k})$ with $\lceil(n-1)/2\rceil\leq k\leq n-1$.
This proves the full upper-half generation theorem (Theorem~\ref{thm:Schocker-hook-generation})
over $\ZZ$, and hence over any commutative coefficient ring by base change.
\end{proof}

\subsection{Second proof of Theorem~\ref{thm:Marin}}

Proposition~\ref{prop:full-hook-generation} now gives a direct integral
descent-algebra proof of Boorman--Marin.

\begin{proof}[Second proof of Theorem~\ref{thm:Marin}]
Applying the surjective ring morphism $\pi$ to
Proposition~\ref{prop:full-hook-generation} and recalling
\eqref{eq:hook-ribbon-abelianization}, we see that
the hook Schur functions
\[
 s_{(n-k,1^k)} \qquad \text{ for all } 0\leq k\leq n-1
\]
generate $(\Symm_{\ZZ,n},*)$.
Subsequently applying the inverse Frobenius characteristic map $\ch^{-1}$,
we conclude that the hook Specht modules
$\bigwedge^kV_n$ generate $R(S_n)$ under tensor product.  Since
$V_n=M_n-\one$ belongs to the $\lambda$-subring generated by $M_n$, so does
$\lambda^k(V_n)=\bigwedge^kV_n$ for every $k$.  Therefore that
$\lambda$-subring contains a set of ordinary ring generators of $R(S_n)$,
and must be all of $R(S_n)$.
This proves Theorem~\ref{thm:Marin}.
\end{proof}

\subsection{A second lift: two-block complete functions}

The hook ribbons above are the natural lifts of the exterior powers appearing
in Boorman--Marin.  There is, however, another pleasantly small generating
family upstairs.  The following theorem is independent of Schocker's theorem
and will lead to a different family of generators downstairs, as well as
eventually to a different -- third -- proof of Theorem~\ref{thm:Marin}.

\begin{theorem}
\label{thm:NSym-two-block-generation}
Fix $n\geq1$.  Under the internal product, the $\ZZ$-algebra
$\NSym_{\ZZ,n}$ is generated by the $n$ elements
\begin{equation}
 H_{(k,n-k)}=H_kH_{n-k}
 \qquad \text{ with $1\leq k\leq n$},
 \label{eq:NSym-two-block-generators}
\end{equation}
where a zero part in a composition is understood to be omitted
by default.
\end{theorem}

\begin{proof}
Let $A_n$ be the $\ZZ$-subalgebra of $\NSym_{\ZZ,n}$ (with the
internal product) generated by the elements in
\eqref{eq:NSym-two-block-generators}.  We prove that $H_I\in A_n$ for every
composition $I\models n$, by downward induction on the first part of $I$.
The initial case is $I=(n)$, for which $H_I=H_{(n,0)}$ is one of the
generators.

Now let
\[
 I=(m,a_2,a_3,\ldots,a_r),
 \qquad r\geq2,
\]
and assume that $H_K\in A_n$ for every composition $K\models n$ whose first
part is strictly larger than $m$.  Set
\[
 I'=(m+a_2,a_3,\ldots,a_r)
 \qquad\text{and}\qquad
 J=(n-a_2,a_2).
\]
The first part of $I'$ is $m+a_2>m$, so $H_{I'}\in A_n$ by induction, while
$H_J$ is one of the generators and thus belongs to $A_n$ as well.
Thus, $H_{I'} * H_J \in A_n$.

Apply \eqref{eq:NSym-matrix-rule} to $H_{I'}*H_J$.  Every matrix $M$
that occurs has two columns.  Write its first row as
\[
 (m+a_2-d,d).
\]
Since the second column has total sum $a_2$, we have $0\leq d\leq a_2$.
If $d<a_2$, then the first part of $\operatorname{comp}(M)$ is
$m+a_2-d>m$.  If $d=a_2$, all later entries in the second column must vanish,
and there is exactly one possible matrix, namely
\[
 \begin{pmatrix}
 m&a_2\\
 a_3&0\\
 a_4&0\\
 \vdots&\vdots\\
 a_r&0
 \end{pmatrix}.
\]
Its row-reading composition is $I$.  Consequently
\begin{equation}
 H_{I'}*H_J
 =H_I+\sum_{\substack{K\models n;\\K_1>m}}c_KH_K
 \label{eq:NSym-unitriangular-step}
\end{equation}
for some $c_K\in\NN$, where $K_1$ denotes the first entry of $K$.
Every term in the sum belongs to $A_n$ by induction,
as does $H_{I'} * H_J$;
so \eqref{eq:NSym-unitriangular-step} yields $H_I\in A_n$.  This completes the
induction.
\end{proof}

\begin{remark}
The proof is triangular in a precise elementary sense: when
\eqref{eq:NSym-unitriangular-step} is solved for $H_I$, every other basis
element that occurs has first part strictly larger than $I_1$.  Thus the
downward induction eliminates the standard basis elements one first-part
level at a time.
\end{remark}

Under the descent-algebra identification $\iota$, the element
$H_{(k,n-k)} \in \NSym_{\ZZ, n}$ corresponds to
\[
 a_k = x_{(k,n-k)}
 = \sum_{\substack{w\in S_n;\\ \Des(w) \subseteq \{ k \} }} w
 = \sum_{\substack{w\in S_n;\\
 w(1)<\cdots<w(k);\\ w(k+1)<\cdots<w(n)}}w \in \mathcal D(S_n),
\]
with empty chains omitted;
this is a sum over the so-called \emph{Grassmannian
permutations} (or \emph{inverse riffle shuffles}) with their
only possible descent (if any) at $k$.
Thus, Theorem~\ref{thm:NSym-two-block-generation} takes the
following equivalent form:

\begin{corollary}
\label{cor:descent-two-block-generation}
For every $n\geq1$, the integral Solomon descent algebra $\mathcal D(S_n)$ is generated
by $a_1,a_2,\ldots,a_n$.
\end{corollary}

This is somewhat similar, but not identical, to
Theorem~\ref{thm:Schocker-hook-generation}.  Schocker's generators are exact
descent-class sums and map to irreducible hook characters.  The elements
$a_k$ are sums over all descent sets contained in a one-element set and map to
Young permutation characters.

\subsection{The two-block generators downstairs}

Applying the abelianization map $\pi$ to
Theorem~\ref{thm:NSym-two-block-generation} gives another integral generation
result for the representation ring.

\begin{corollary}
\label{cor:k-subsets-generate-RSn}
For every $n\geq1$, the ring $(\Symm_{\ZZ,n},*)$ is generated by
the $n$ elements
\[
 h_kh_{n-k} \qquad \text{ with $1\leq k\leq n$}.
\]
Equivalently, $R(S_n)$ is generated under tensor product by the $n$
permutation representations
\begin{equation}
 P_{n,k}:=\QQ\!\left[\binom{[n]}{k}\right]
 \qquad
 \text{ with $1\leq k\leq n$}
 \label{eq:k-subset-module}
\end{equation}
on the $k$-subsets of $[n]$.
\end{corollary}

\begin{proof}
Under the Frobenius characteristic,
\[
 \ch_n(P_{n,k})
 =h_kh_{n-k},
\]
since $P_{n,k}\cong
\Ind_{S_k\times S_{n-k}}^{S_n}\one$.
Equivalently, its character is the two-part Young character of type
$(k,n-k)$.  In the notation of Blessenohl--Schocker
\cite[Chapter~12 and Appendix~B]{BlessenohlSchocker}, this is
$\xi^{(k,n-k)}$: their ordered set partitions of type $(k,n-k)$ are simply
pairs $(A,A^c)$, hence are naturally identified with the $k$-subsets $A$ of
$[n]$.
\end{proof}

\subsection{Third proof of Theorem~\ref{thm:Marin}}

For curiosity's sake, we shall finally outline how these
permutation modules $P_{n,k}$ can themselves be manufactured from the single
$\lambda$-generator $M_n=P_{n,1}$.  This gives a third proof of
Theorem~\ref{thm:Marin}, logically independent of Schocker's hook-generation
theorem once Theorem~\ref{thm:NSym-two-block-generation} is known,
although it may be argued that it is merely a remix of the first two
proofs.

We extend the definition \eqref{eq:k-subset-module} to $k=0$.
Of course, $P_{n,0}$ and $P_{n,n}$ are trivial $S_n$-modules.

For positive integers $c_1,\ldots,c_r$ with
$d=c_1+\cdots+c_r\leq n$, define the permutation module
\begin{equation}
 Y_{c_1,\ldots,c_r}
 =\QQ\bigl[\{(A_1,\ldots,A_r):
 A_i\subseteq[n],\ |A_i|=c_i\text{ and the }A_i\text{ are pairwise disjoint}\}\bigr].
 \label{eq:disjoint-subset-module}
\end{equation}
This is isomorphic to a Young permutation module of tabloids
(encoding each $A_i$ as the $i$-th row of a tabloid, and the
complement $[n] \setminus (A_1 \cup A_2 \cup \cdots \cup A_r)$
as the $(r+1)$-th row of this tabloid).
We call $d$ the \emph{support size} of this permutation
module.
Note that $Y_k=P_{n,k}$.

\begin{lemma}
\label{lem:disjoint-subset-modules}
Let $L_n$ be the $\lambda$-subring of $R(S_n)$ generated by $M_n$.  Suppose
that $P_{n,j}\in L_n$ for every $j<d$.  Then every
$Y_{c_1,\ldots,c_r}$ of support size $d$ with $r\geq2$ belongs to $L_n$.
\end{lemma}

\begin{proof}[Proof sketch.]
We prove the assertion simultaneously for all support sizes $d$ by strong
induction on $d$.  For $d=1$ there is nothing to prove, since $r\geq2$.
Fix $d\geq2$, assume the assertion known for all smaller support sizes,
and assume, as in the statement, that $P_{n,j}\in L_n$ for every $j<d$.

Since $r\geq2$, each $c_i<d$, so each $P_{n,c_i}$ belongs to $L_n$.  Their
tensor product $P_{n,c}^\otimes
:= P_{n,c_1} \otimes P_{n,c_2} \otimes \cdots \otimes P_{n,c_r}$
is the permutation representation on all tuples
$(A_1,\ldots,A_r)$ of subsets of $[n]$ with $|A_i|=c_i$, with no disjointness condition.
Let $\mathfrak P_{n,c}$ be the $S_n$-set consisting of these tuples.
For each nonempty $B\subseteq[r]$, define the \emph{membership region}
\[
 R_B=\left(\bigcap_{i\in B}A_i\right)
 \setminus\left(\bigcup_{i\notin B}A_i\right)
\]
(so that the $R_B$ for different $B$'s are disjoint and their
union is $A_1 \cup A_2 \cup \cdots \cup A_r$;
visually speaking, they are the regions of the Venn diagram of
$A_1,A_2,\ldots,A_r$).
Decompose the $S_n$-set $\mathfrak P_{n,c}$
into orbits according to the cardinalities
$|R_B|$ of all these membership regions.  Exactly one orbit is the
pairwise-disjoint locus
\eqref{eq:disjoint-subset-module}.  Each of the other orbits has union of size
strictly smaller than $d$, and is itself a permutation module of the form
$Y_{b_1,\ldots,b_s}$ with support size $<d$, after the nonempty membership
regions are listed as labeled blocks.

For every such smaller-support orbit module, let $e<d$ be its support
size.  All $P_{n,j}$ with $j<e$ belong to $L_n$ because $j<e<d$.  If an orbit has at least two nonempty membership regions, the induction
hypothesis puts its module in $L_n$.  If it has exactly one such region,
its module is $Y_e=P_{n,e}$, which belongs to $L_n$ by the assumption
$e<d$.  Hence all these other orbit modules belong to $L_n$.  Subtracting them from
$P_{n,c_1}\otimes\cdots\otimes P_{n,c_r}$ leaves
$Y_{c_1,\ldots,c_r}$.
\end{proof}

\begin{proposition}
\label{prop:k-subsets-in-lambda-ring}
For every $0\leq k\leq n$, one has $P_{n,k}\in L_n$.
\end{proposition}

\begin{proof}[Proof sketch]
We use strong induction on $k$.  The cases $k=0$ and $k=1$ are clear.  Assume
$k\geq2$ and that $P_{n,j}\in L_n$ for all $j<k$.

\emph{Step 1: decompose the symmetric power into permutation orbits.}
The symmetric power $\operatorname{Sym}^k(M_n)$ has a basis consisting of the
monomials of total degree $k$ in the basis vectors $e_1,\ldots,e_n$, and
$S_n$ permutes this basis.  Its orbits are indexed by partitions
$\lambda\vdash k$: the partition records the positive multiplicities of the
basis vectors occurring in a monomial.  If $m_r(\lambda)$ denotes the number
of parts of $\lambda$ equal to $r$, then the orbit of type $\lambda$ affords
the permutation module
\[
 Y_{m_1(\lambda),m_2(\lambda),\ldots},
\]
with zero entries omitted.  Its support size is
$\ell(\lambda)=\sum_r m_r(\lambda)$.

For $\lambda=(1^k)$ this orbit module is $P_{n,k}$.  For each of
the remaining partitions $\lambda \vdash k$, we have $\ell(\lambda)<k$, so
Lemma~\ref{lem:disjoint-subset-modules} puts its orbit module in $L_n$
when there are at least two nonzero multiplicities $m_r(\lambda)$.
If there is only one, the orbit module is $P_{n,\ell(\lambda)}$ and
belongs to $L_n$ directly by the induction hypothesis.  Hence
\begin{equation}
 [\operatorname{Sym}^k(M_n)]
 =[P_{n,k}]
 +\sum_{\substack{\lambda\vdash k;\\\lambda\ne(1^k)}}
   [Y_{m_1(\lambda),m_2(\lambda),\ldots}],
 \label{eq:symmetric-power-orbit-decomposition}
\end{equation}
and all addends in the sum belong to $L_n$.

\emph{Step 2: isolate the $k$-subset orbit.}
The left-hand side belongs to $L_n$.  Indeed, in the ring $R(S_n)[[t]]$,
we have (see, e.g., \cite[Lemma 3.29 and the preceding derivation]{EliaKimSupina}
for an elementary proof using eigenvalues\footnote{A more conceptual
proof can be found in \cite[Corollary 2.14 (applied to $A = \operatorname{Sym}(V)$
and $A^! = \wedge(V^*)$)]{ARS}.})
\[
 \sum_{j\geq0}[\operatorname{Sym}^j(M_n)]t^j
 =\frac{1}{\lambda_{-t}([M_n])}
 =\lambda_{-t}(-[M_n]),
\]
so each symmetric-power class $[\operatorname{Sym}^j(M_n)]$
belongs to the $\lambda$-subring $L_n$.
All terms in the
sum in \eqref{eq:symmetric-power-orbit-decomposition} belong to $L_n$, so
$[P_{n,k}]\in L_n$ as well.
\end{proof}

\begin{proof}[Third proof of Theorem~\ref{thm:Marin} (sketched)]
By Corollary~\ref{cor:k-subsets-generate-RSn}, the classes $[P_{n,k}]$
generate $R(S_n)$ as an ordinary ring.  By
Proposition~\ref{prop:k-subsets-in-lambda-ring}, every one of these classes
belongs to the $\lambda$-subring $L_n$ generated by $M_n$.  Hence
$R(S_n)\subseteq L_n$.  The reverse inclusion is tautological, so
$L_n=R(S_n)$.
\end{proof}

\begin{remark}
Let us summarize the differences between our three proofs
of Theorem~\ref{thm:Marin}:

\begin{enumerate}
\item
The first proof worked directly in $R(S_n)$, showing that the
hook Specht modules $\bigwedge^k V_n$ generate $R(S_n)$.
As an intermediate step, it used the $\mathbb S^\beta(M_n)$
(Schur functors evaluated on $M_n$) and the $T_\lambda$
(tensor products of exterior powers of $M_n$).

\item
The second proof worked in the algebra $\NSym_{\ZZ,n}$,
showing that the noncommutative hooks $Q_{n,k}$ (which are
Schocker's hook descent classes, conjugated by the longest
permutation) generate $\NSym_{\ZZ,n}$ under the Kronecker
product $*$. Upon projecting down
onto $R(S_n)$ via the chain of surjective ring morphisms
\[
%\mathcal D(S_n) \overset{\iota^{-1}}{\longrightarrow}
\NSym_{\ZZ,n} \overset{\pi}{\longrightarrow}
\Symm_{\ZZ,n} \overset{\ch^{-1}}{\longrightarrow}
R(S_n),
\]
these hooks became the hook Specht modules $\bigwedge^k V_n$.

\item
The third proof again worked in $\NSym_{\ZZ,n}$.
But instead of lifting the hook Specht modules $\bigwedge^k V_n$,
it showed that the two-block complete functions
$H_kH_{n-k}$ generate $\NSym_{\ZZ,n}$ under $*$.
Under the same ring surjections as before, these became the
$k$-subset permutation modules $P_{n,k}$.
Then, another triangularity argument was needed to
show that these $P_{n,k}$ belong to the $\lambda$-subring
generated by $M_n$, so that Theorem~\ref{thm:Marin}
again followed.
\end{enumerate}
\end{remark}

\begin{thebibliography}{99}

\bibitem{ARS}
A.~Almousa, V.~Reiner, S.~Sundaram,
\textit{Koszulity, supersolvability and Stirling Representations},
Annals of Representation Theory \textbf{2} (2025), pp. 173--247,
\url{https://doi.org/10.5802/art.23}.

\bibitem{Bidigare-thesis}
P.~Bidigare,
\emph{Hyperplane arrangement face algebras and their associated Markov chains},
PhD thesis at Univ. Michigan, 1997.

\bibitem{Boorman}
E.~H.~Boorman,
\emph{$S$-operations in representation theory},
Trans. Amer. Math. Soc. \textbf{205} (1975), 127--149,
\url{https://doi.org/10.1090/S0002-9947-1975-0364424-3}.

\bibitem{BlessenohlLaue}
D.~Blessenohl and H.~Laue,
\emph{Algebraic combinatorics related to the free {L}ie algebra},
{S\'{e}minaire Lotharingien de Combinatoire} \textbf{B29e} (1992),
\url{https://radon.mat.univie.ac.at/~slc/s/s29laue.pdf}.

\bibitem{BlessenohlSchocker}
D.~Blessenohl and M.~Schocker,
\emph{Noncommutative Character Theory of the Symmetric Group},
Imperial College Press, London, 2005,
\url{https://doi.org/10.1142/P369}.

\bibitem{Bourbaki-Alg1}
N.~Bourbaki,
\emph{Algebra I: Chapters 1--3},
English translation, Addison--Wesley, 1974; reprinted by Springer,
\url{https://link.springer.com/book/9783540642435}.

\bibitem{Butler}
P.~H.~Butler,
\emph{$S$-functions and symmetry in physics},
J. Physique Colloq. \textbf{31} (1970), C4-47--C4-50,
\url{https://doi.org/10.1051/jphyscol:1970408}.

\bibitem{DressSiebeneicher}
A.~W.~M.~Dress and C.~Siebeneicher,
\emph{The Burnside ring of the infinite cyclic group and its relations to the
necklace algebra, $\lambda$-rings, and the universal ring of Witt vectors},
Adv. Math. \textbf{78} (1989), no.~1, 1--41,
\url{https://doi.org/10.1016/0001-8708(89)90027-3}.

\bibitem{EtingofEtAl}
P.~Etingof, O.~Golberg, S.~Hensel, T.~Liu, A.~Schwendner,
D.~Vaintrob, and E.~Yudovina,
\emph{Introduction to Representation Theory},
Student Mathematical Library 59, American Mathematical Society, 2011;
also arXiv:0901.0827,
\url{https://doi.org/10.1090/stml/059}.

\bibitem{EliaKimSupina}
S.~Elia, D.~Kim, and M.~Supina,
\emph{Techniques in equivariant Ehrhart theory},
Annals of Combinatorics \textbf{28} (2024), 819--870,
\url{https://doi.org/10.1007/s00026-023-00673-z}.

\bibitem{Fulton}
W.~Fulton,
\textit{Young Tableaux},
London Mathematical Society Student Texts \textbf{35},
Cambridge University Press 1999.
See \url{https://mathoverflow.net/questions/456463/} for unofficial errata.

\bibitem{GarsiaGoupil}
A.~M.~Garsia and A.~Goupil,
\emph{Character polynomials, their $q$-analogs and the Kronecker product},
Electron. J. Combin. \textbf{16} (2009), no.~2, Research Paper R19, 40~pp.,
\url{https://doi.org/10.37236/85}.
The article PDF is available at
\url{https://www.combinatorics.org/ojs/index.php/eljc/article/download/v16i2r19/pdf/}.

\bibitem{GarsiaReutenauer}
A.~Garsia and C.~Reutenauer,
\emph{A decomposition of {S}olomon's descent algebra},
{Advances in mathematics} \textbf{77} (1989), {189--262},
\url{https://doi.org/10.1016/0001-8708(89)90020-0}.

\bibitem{GelfandEtAlNCSF}
I.~M.~Gelfand, D.~Krob, A.~Lascoux, B.~Leclerc, V.~S.~Retakh, and J.-Y.~Thibon,
\emph{Noncommutative symmetric functions},
Adv. Math. \textbf{112} (1995), no.~2, 218--348,
\url{https://doi.org/10.1006/aima.1995.1032}.

\bibitem{GiannelliLawLongVallejo}
E.~Giannelli, S.~Law, J.~Long, and C.~Vallejo,
\emph{Sylow branching coefficients and a conjecture of Malle and Navarro},
Bull. Lond. Math. Soc. \textbf{54} (2022), no.~2, 552--567,
\url{https://doi.org/10.1112/blms.12584}.
See \url{https://eugeniomaths.wordpress.com/wp-content/uploads/2022/01/5.-gllv22.pdf} for a preprint.

\bibitem{AntiArithmeticMO}
D.~Grinberg,
\emph{Anti-arithmetic product of symmetric functions: (why) is it integral?},
MathOverflow question 182083 (2014),
\url{https://mathoverflow.net/questions/182083/anti-arithmetic-product-of-symmetric-functions-why-is-it-integral}.

\bibitem{ArithmeticMO}
D.~Grinberg,
\emph{Arithmetic product of symmetric functions: why is it integral?},
MathOverflow question 138148 (2013),
\url{https://mathoverflow.net/questions/138148/arithmetic-product-of-symmetric-functions-why-is-it-integral}.

\bibitem{sga}
D.~Grinberg,
\emph{An introduction to the symmetric group algebra
[Math 701, Spring 2024 lecture notes]},
12 June 2026,
\url{https://www.cip.ifi.lmu.de/~grinberg/t/24s/sga.pdf}.

\bibitem{Burman-cex}
GPT-5.6 Sol, edited by D.~Grinberg,
\emph{A counterexample to the Burman–Kulishov
conjecture on Lie elements},
16 August 2026,
\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/burman-cex-gpt.pdf}.

\bibitem{GrinbergParlett}
D.~Grinberg, J.~Parlett,
\emph{Top to random and reverse: analysis of a new descent algebra shuffle},
arXiv:2508.06740v2,
\url{https://arxiv.org/abs/2508.06740v2}.

\bibitem{GrinbergReiner}
D.~Grinberg and V.~Reiner,
\emph{Hopf algebras in combinatorics},
arXiv:1409.8356v7 (2020),
\url{https://arxiv.org/abs/1409.8356}.

\bibitem{Guillot}
P.~Guillot,
\emph{The representation ring of a simply connected Lie group as a
$\lambda$-ring},
Comm. Algebra \textbf{35} (2007), no.~3, 875--883;
arXiv:math/0510283,
\url{https://doi.org/10.1080/00927870601115799}.

\bibitem{Harman}
N.~Harman,
\emph{Generators for the representation rings of certain wreath products},
J. Algebra \textbf{445} (2016), 125--135;
arXiv:1410.1786,
\url{https://doi.org/10.1016/j.jalgebra.2015.09.003}.

\bibitem{Hazewinkel}
M.~Hazewinkel,
\emph{Witt vectors. Part 1},
in \emph{Handbook of Algebra}, Vol.~6, Elsevier, 2009, pp.~319--472;
arXiv:0804.3888,
\url{https://doi.org/10.1016/S1570-7954(08)00207-6}.

\bibitem{Knutson}
D.~Knutson,
\emph{$\lambda$-Rings and the Representation Theory of the Symmetric Group},
Lecture Notes in Mathematics 308, Springer, 1973,
\url{https://doi.org/10.1007/BFb0069217}.

\bibitem{Lassueur}
C.~Lassueur,
\emph{Character Theory of Finite Groups},
lecture notes, RPTU Kaiserslautern--Landau, Summer Semesters 2022 and 2023,
version of June~30, 2023,
\url{https://classueur.github.io/maths/teaching/skripte/CharaktertheorieSS23.pdf}.

\bibitem{LiPrimeGraphs}
J.~Li,
\emph{Prime graphs and exponential composition of species},
J. Combin. Theory Ser. A \textbf{115} (2008), no.~8, 1374--1401;
arXiv:0705.0038,
\url{https://doi.org/10.1016/j.jcta.2008.02.008}.

\bibitem{Macdonald}
I.~G.~Macdonald,
\emph{Symmetric Functions and Hall Polynomials},
2nd ed., Oxford Mathematical Monographs, Oxford University Press, 1995,
\url{https://doi.org/10.1093/oso/9780198534891.001.0001}.

\bibitem{MaiaMendez}
M.~Maia and M.~M\'endez,
\emph{On the arithmetic product of combinatorial species},
Discrete Math. \textbf{308} (2008), no.~23, 5407--5427;
arXiv:math/0503436,
\url{https://doi.org/10.1016/j.disc.2007.09.062}.

\bibitem{Marin}
I.~Marin,
\emph{Hooks generate the representation ring of the symmetric group},
Expositiones Math. \textbf{30} (2012), no.~3, 268--276;
arXiv:1112.3127,
\url{https://doi.org/10.1016/j.exmath.2012.03.005}.

\bibitem{MeirSzymik}
E.~Meir and M.~Szymik,
\emph{Adams operations and symmetries of representation categories},
Indiana Univ. Math. J. \textbf{70} (2021), no.~2, 501--523;
arXiv:1704.03389,
\url{https://doi.org/10.1512/iumj.2021.70.8377}.

\bibitem{MetropolisRota}
N.~Metropolis and G.-C.~Rota,
\emph{Witt vectors and the algebra of necklaces},
Adv. Math. \textbf{50} (1983), no.~2, 95--125,
\url{https://doi.org/10.1016/0001-8708(83)90035-X}.

\bibitem{MurnaghanGeneration}
F.~D.~Murnaghan,
\emph{On the generation of the irreducible representations of the symmetric group},
Proc. Natl. Acad. Sci. USA \textbf{41} (1955), no.~7, 514--515,
\url{https://doi.org/10.1073/pnas.41.7.514}.

\bibitem{OrellanaZabrocki}
R.~Orellana and M.~Zabrocki,
\emph{Symmetric group characters as symmetric functions},
Adv. Math. \textbf{390} (2021), Paper No.~107943, 34~pp.;
arXiv:1605.06672v5,
\url{https://doi.org/10.1016/j.aim.2021.107943}.
The arXiv version is available at \url{https://arxiv.org/abs/1605.06672v5}.

\bibitem{Ryba}
C.~Ryba,
\emph{Kronecker comultiplication of stable characters and restriction from
$S_{mn}$ to $S_m\times S_n$},
J. Algebra \textbf{638} (2024), 720--738,
\url{https://doi.org/10.1016/j.jalgebra.2023.07.043}.

\bibitem{Sagan}
B.~E.~Sagan,
\emph{The Symmetric Group: Representations, Combinatorial Algorithms,
and Symmetric Functions},
2nd ed., Graduate Texts in Mathematics 203, Springer, 2001,
\url{https://doi.org/10.1007/978-1-4757-6804-6}.

\bibitem{Saliola}
F.~V.~Saliola,
\emph{Hyperplane arrangements and descent algebras},
lecture notes, version of January~10, 2006,
\url{https://saliola.github.io/maths/publications/LectureNotes/DesAlgLectureNotes.pdf};
corrections at
\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/saliola-errata.pdf}.

\bibitem{ScharfThibon}
T.~Scharf and J.-Y.~Thibon,
\emph{A Hopf-algebra approach to inner plethysm},
Adv. Math. \textbf{104} (1994), no.~1, 30--58,
\url{https://doi.org/10.1006/aima.1994.1019}.

\bibitem{Schocker}
M.~Schocker,
\emph{A generating set of Solomon's descent algebra},
J. Algebra \textbf{263} (2003), no.~1, 151--158,
\url{https://doi.org/10.1016/S0021-8693(03)00069-3}.

\bibitem{Serre}
J.-P.~Serre,
\emph{Linear Representations of Finite Groups},
Graduate Texts in Mathematics 42, Springer, 1977,
\url{https://doi.org/10.1007/978-1-4684-9458-7}.

\bibitem{Stanley}
R.~P.~Stanley,
\emph{Enumerative Combinatorics}, Vol.~2,
2nd ed., Cambridge Studies in Advanced Mathematics, Cambridge University
Press, 2023,
\url{https://doi.org/10.1017/9781009262538}.

\bibitem{ThibonAdams}
J.-Y.~Thibon,
\emph{Fonctions sym\'etriques et op\'erations d'Adams de l'anneau de
Grothendieck d'un groupe sym\'etrique},
C. R. Acad. Sci. Paris S\'er. I Math. \textbf{313} (1991), no.~12, 829--832.
% listed on the author's publication page,
% \url{https://igm.univ-mlv.fr/~jyt/articles.html}.

\bibitem{Wildon-sf}
M.~Wildon,
\emph{An involutive introduction to symmetric functions},
26 November 2021,
\url{https://www.ma.rhul.ac.uk/~uvah099/Maths/Sym/SymFuncs2020.pdf}.
Unofficial errata at \url{https://www.cip.ifi.lmu.de/~grinberg/algebra/symfuncs2017-2020-05-08-errata.pdf}.

\bibitem{Willigenburg98}
S.~van Willigenburg,
\emph{A proof of {Solomon}'s rule},
{J. Algebra} \textbf{206} (1998), 693--698,
\url{https://doi.org/10.1006/jabr.1998.7427}.

\bibitem{Yau}
D.~Yau,
\emph{Lambda-Rings},
World Scientific, 2010,
\url{https://doi.org/10.1142/7664}.

\end{thebibliography}



\end{document}