\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{\Cl}{\operatorname{Cl}} \newcommand{\ch}{\operatorname{ch}} \newcommand{\Ind}{\operatorname{Ind}} \newcommand{\Res}{\operatorname{Res}} \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{Arithmetic and Anti-Arithmetic Products of Symmetric Functions:\\ A Representation-Theoretic Proof of Integrality} \author{GPT-6, partly edited by Darij Grinberg, not yet proofread} \date{unproofread draft, \today} \begin{document} \maketitle \begin{abstract} The power sums form a basis of the algebra of symmetric functions over $\QQ$, but not over $\ZZ$. Thus an operation defined by a simple rule on power sums need not preserve integral symmetric functions. Yet, many of them do, each time requiring proof. We study two such operations. The \emph{arithmetic product} $\BoxProd$ is given by \[ p_\lambda\BoxProd p_\mu =\prod_{i,j}p_{\lcm(\lambda_i,\mu_j)}^{\gcd(\lambda_i,\mu_j)}, \] and the \emph{anti-arithmetic product} by the same formula with $\gcd$ and $\lcm$ interchanged. The arithmetic product is known to preserve integral symmetric functions, and is in fact Schur-positive on pairs of Schur functions. The analogous integrality of the anti-arithmetic product was posed as an open question on MathOverflow in 2014. 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 representation ring and the class function algebra of the symmetric group $S_n$; 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 note is aimed at readers familiar with representation rings and the representation theory of symmetric groups. No prior knowledge of $\lambda$-rings is presumed. Boorman's theorem is proved in two different ways, one of which lifts it to the descent algebra of $S_n$. \medskip \textbf{Manifest.} This note was written by GPT-6 based on a proof found by GPT-5.5. I have then edited it. \\ This note 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{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}. 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} The statement about $\BoxProd$ is classical. We include it because its proof is almost the same as the proof of the $\AntiProd$ statement, 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. \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{Knutson}, Yau~\cite{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 simplicity, whenever a general finite group $G$ occurs below, we assume that the representations under discussion are realizable over $\QQ$. This includes all symmetric groups and their finite direct products, which are the only groups needed in the proof. Alternatively, one may replace $\QQ$ throughout this section by any characteristic-zero splitting field of $G$. 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 $V$. 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 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 $\lambda^1(x),\lambda^2(x),\ldots,\lambda^{r-1}(x)$. 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 \cite{Knutson,Yau,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 Frobenius characteristic 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. The basic theorem of Frobenius 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 \[ V_n=S^{(n-1,1)} \] be the reflection 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 and give a self-contained proof using triangularity (leading term analysis). \begin{theorem}[Boorman; Marin] \label{thm:Marin} For every $n\geq1$, the smallest $\lambda$-subring of $R(S_n)$ containing $M_n$ is all of $R(S_n)$. \end{theorem} The proof below is neither Boorman's nor Marin's original proof; it is a triangular argument using the number of boxes below the first row. It is included so that Theorem~\ref{thm:main-integrality} does not depend on any nonstandard representation-theoretic black box. For generalizations of hook-type generating sets to wreath products, see Harman~\cite{Harman}. Appendix~\ref{app:NSym-generation} gives a second proof, obtained from a stronger integral generation theorem 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{A triangular proof of Theorem~\ref{thm:Marin}} 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 \[ \lambda=(n-d,\alpha), \qquad \text{where } \alpha\vdash d \text{ with } \alpha_1\leq n-d. \] 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 \[ \alpha'=(c_1,c_2,\ldots,c_r) \] 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. For $\beta\vdash d$, let $\mathbb S^\beta(M_n)$ denote the Schur functor corresponding to $\beta$, applied to $M_n$. One convenient definition in characteristic zero is \begin{equation} \mathbb S^\beta(M_n) =\Hom_{S_d}(S^\beta,M_n^{\otimes d}), \label{eq:Schur-functor} \end{equation} where the tensor power $M_n^{\otimes d}$ has two commuting actions of $S_d$ and $S_n$ (in the obvious ways: $S_d$ permutes tensor positions and $S_n$ acts diagonally on $M_n^{\otimes d}$). \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){\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} \] where the vertical arrows are the character embeddings. Indeed, $\chi_{V\boxtimes W}(g,h)=\chi_V(g)\chi_W(h)$ for all $S_m$-representations $V$, all $S_n$-representations $W$, and all $g\in S_m$ and $h\in S_n$. Under these identifications, the scalar product factors: \begin{equation} \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:product-inner-product} \end{equation} We will therefore move freely between class functions on $S_m\times S_n$ and tensor products of 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 maps on conjugacy classes} We now 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} The cycle type of this permutation $\mathcal B(\sigma,\tau)$ 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 \[ \gcd(a,b)\text{ cycles of length }\lcm(a,b) \label{eq:arithmetic-cycles} \] in the cycle decomposition of $\mathcal B(\sigma,\tau)$ (their union is the whole Cartesian product of the two chosen cycles). 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 are unital algebra homomorphisms because multiplication of class functions is pointwise. \subsection{These pullbacks are the adjoints of the two products} Transport the maps \eqref{eq:Delta-box} and \eqref{eq:Delta-diamond} through Frobenius characteristic. We use the same symbols for the resulting maps \[ \Symm_{\QQ,mn} \longrightarrow \Symm_{\QQ,m}\otimes_\QQ\Symm_{\QQ,n}. \] \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 $G=p_\mu$. If $h$ is the class function with $\ch_{nm}(h)=H$, 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 $mn$. The left-hand side of \eqref{eq:adjoint-box} is therefore $h$ evaluated on the arithmetic cycle type associated with $(\lambda,\mu)$. By definition of the pullback, this is also \[ \langle p_\lambda\otimes p_\mu, \Delta^{\BoxProd}_{m,n}H\rangle. \] (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} and \eqref{eq:Delta-integral} give \[ \langle F\circ G,H\rangle =\langle F\otimes G,\Delta^\circ_{m,n}H\rangle\in\ZZ, \] because the Schur bases on the two tensor factors are orthonormal. Lemma~\ref{lem:self-dual} now implies $F\circ G\in\Symm_{\ZZ,mn}$. \end{proof} Under Frobenius characteristic, condition \eqref{eq:Delta-integral} is exactly \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} We now prove this for both pullbacks. \subsection{Power compatibility} The first decisive observation is that both class maps commute with raising permutations to powers. For the arithmetic map this is immediate from the group homomorphism \eqref{eq:B-homomorphism}: \[ \mathcal B(\sigma^r,\tau^r)=\mathcal B(\sigma,\tau)^r. \] For the anti-arithmetic map, it is not automatic, but it is still true at the level of conjugacy classes. We first record the elementary number-theoretic identity behind the verification. \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 easy to see that every nonzero integer $m$ satisfies $v_p\left(m/\gcd(m,r)\right) = \left(v_p(m)-\rho\right)_+$, where $x_+=\max(x,0)$. Thus, the $p$-adic valuation of the left-hand side of \eqref{eq:gcd-power-identity} is \[ \min\bigl((\alpha-\rho)_+,(\beta-\rho)_+\bigr) =(\min(\alpha,\beta)-\rho)_+. \] The valuation of the right-hand side is \[ \min(\alpha,\beta) -\min\bigl(\min(\alpha,\beta),\rho\bigr) =(\min(\alpha,\beta)-\rho)_+. \] Thus the two sides have the same $p$-adic valuation for every prime $p$. \end{proof} \begin{lemma} \label{lem:anti-power-compatible} For all $r\geq1$, \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 $g/c$. Thus the right-hand side of \eqref{eq:anti-power-compatible} 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 splits into $\gcd(a,r)$ cycles, each of length $a/\gcd(a,r)$. The same holds for the $b$-cycle. Lemma~\ref{lem:gcd-power-identity} gives \[ \gcd\left(\frac{a}{\gcd(a,r)}, \frac{b}{\gcd(b,r)}\right) =\frac{g}{c}, \] so every cycle of $\mathcal A(\sigma^r,\tau^r)$ obtained from our two cycles has length $g/c$. Since the block has $ab$ letters in total, the number of such cycles is \[ \frac{ab}{g/c} =\frac{ab}{g}c =\lcm(a,b)c, \] which agrees with \eqref{eq:power-right-count}. \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 $\Delta^{\BoxProd}_{m,n}$ is a morphism of $\psi$-rings, and hence of $\lambda$-rings. \item \label{cor:Delta-psi-anti} The map $\Delta^{\AntiProd}_{m,n}$ 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 ($\mathcal B$ or $\mathcal A$). For any class function $f$ 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) &=f(\mathcal C(\sigma,\tau)^r)\\ &=f(\mathcal C(\sigma^r,\tau^r))\\ &=(\psi^r\Delta f)(\sigma,\tau). \end{align*} For part~(a), the middle equality follows from the homomorphism property of $\mathcal B$; for part~(b), it follows from Lemma~\ref{lem:anti-power-compatible}. Thus each $\Delta$ is a $\psi$-ring morphism. Proposition~\ref{prop:psi-to-lambda}(b) makes each 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 of the natural representation $M_{mn}$ is an integral virtual representation of $S_m\times S_n$. \subsection{The arithmetic product: an actual pullback} For the arithmetic map there is nothing to construct. Restricting the natural permutation representation of $S_{mn}$ along \eqref{eq:B-homomorphism} gives the permutation representation on $[m]\times[n]$. On the basis vector $e_i\otimes e_j$ of $M_m\boxtimes M_n$, the element $(\sigma,\tau)$ 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, gives \[ \Delta^{\BoxProd}_{m,n}(R(S_{mn})) \subseteq R(S_m\times S_n). \] Proposition~\ref{prop:adjoint-integrality-criterion} proves the arithmetic integrality assertion \[ \Symm_{\ZZ,m}\BoxProd\Symm_{\ZZ,n} \subseteq\Symm_{\ZZ,mn}. \] 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. Thus for $\lambda\vdash m$ and $\mu\vdash n$, \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. This proves Schur positivity. 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 $1\leq a\leq m$, 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 $\sigma\in S_m$, write $m_a(\sigma)$ for its number of $a$-cycles. \begin{lemma} \label{lem:Cma-character} For every $\sigma\in S_m$ and $1 \leq a \leq m$, we have \begin{equation} \chi_{C_{m,a}}(\sigma)=a\,m_a(\sigma). \label{eq:Cma-character} \end{equation} \end{lemma} \begin{proof} The character of the natural permutation representation is the number of fixed points. Hence, by \eqref{eq:adams-character-formula}, each $d\geq 1$ satisfies \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, a $b$-cycle of $\sigma$ is fixed pointwise by $\sigma^d$ exactly when $b\mid d$; in that case it contributes all of its $b$ letters, and otherwise it contributes none. Substituting \eqref{eq:fixed-points-power} into \eqref{eq:Cma-definition} gives \begin{align*} \chi_{C_{m,a}}(\sigma) &=\sum_{d\mid a}\mu(a/d) \sum_{b\mid d}b\,m_b(\sigma)\\ &=\sum_{b\mid a}b\,m_b(\sigma) \sum_{\substack{d:\ b\mid d\mid a}}\mu(a/d). \end{align*} The inner sum is $1$ for $b=a$ and $0$ otherwise, by M\"obius inversion. This yields \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 \[ C_{m,1}=M_m, \qquad C_{m,2}=\psi^2(M_m)-M_m, \] 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}[A direct precedent] \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 has exactly $k$ cycles of length $e$, then \[ V^\lambda[e](\sigma)=ke\,\chi^\lambda(\tau), \] where $\tau$ is obtained 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 variables used in character polynomials; see, for example, Garsia--Goupil~\cite{GarsiaGoupil} for a combinatorial account and references to the earlier literature. A particularly close symmetric-function precedent for the calculation above is the permutation-eigenvalue viewpoint of Orellana--Zabrocki~\cite{OrellanaZabrocki}. If $\Xi_\mu$ denotes the multiset of eigenvalues of a permutation matrix of cycle type $\mu$, then \[ p_d[\Xi_\mu]=\sum_{b\mid d}b\,m_b(\mu). \] This is exactly equation~\eqref{eq:fixed-points-power}. Thus Lemma~\ref{lem:Cma-character} is M\"obius inversion of this familiar cycle-counting identity, with Proposition~\ref{prop:character-lambda} providing the additional fact needed here: the resulting function $aX_a$ is an \emph{integral virtual character}. \end{remark} \subsection{The anti-arithmetic substitute for the natural representation} 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} \begin{proposition} \label{prop:anti-natural-image} The following statements hold. \begin{enumerate}[label=(\alph*)] \item \label{prop:anti-natural-image-equality} As class functions on $S_m\times S_n$, \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_{\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. 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. Summing over all pairs of cycles gives exactly \eqref{eq:W-character}. Hence \[ \chi_{W_{m,n}}(\sigma,\tau) =\chi_{M_{mn}}(\mathcal A(\sigma,\tau)), \] 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 in one line 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}, $\Delta^{\AntiProd}_{m,n}$ is a $\lambda$-ring morphism \[ \Cl_\QQ(S_{mn})\longrightarrow \Cl_\QQ(S_m\times S_n). \] By Proposition~\ref{prop:anti-natural-image}, 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}, $M_{mn}$ 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}] (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}. \end{proof} \subsection{What the proof is really using} It may be useful to isolate the short core of the anti-arithmetic argument. 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 morphism for Adams operations and hence for 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} No $G$-sets or Burnside rings are needed for the proof. They can be a useful way to discover the cycle formulas, but once the natural permutation representations have been introduced, all integrality takes place inside ordinary linear representation rings. \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(g\in G,\ 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; 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 \[ [g]\longmapsto[g^q] \qquad(q\geq1) \] 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 is not peculiar to symmetric groups, but the integral statement is stronger than it may first appear. To avoid field-of-definition issues in this remark, let $R_{\mathbb C}(G)$ denote the complex representation ring of a finite group $G$. If $G=C_m$ is cyclic, then a faithful one-dimensional character generates $R_{\mathbb C}(G)$ 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 $R_{\mathbb C}(G)$ generated by $V$. Then \[ A_V\otimes_{\mathbb Z}\mathbb C =R_{\mathbb C}(G)\otimes_{\mathbb Z}\mathbb C \] 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 $R_{\mathbb C}(G)\otimes_{\mathbb Z}\mathbb C$ 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 $R_{\mathbb C}(G)$. 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 descent-algebra proof of the Boorman--Marin theorem} \label{app:NSym-generation} This appendix records another proof of Theorem~\ref{thm:Marin}. Its main ingredient is a slightly stronger integral statement: the degree-$n$ component of the algebra of noncommutative symmetric functions is generated, for its internal product, by the complete functions with two parts. Equivalently, Solomon's descent algebra is generated by the sums of permutations that are increasing on two consecutive blocks. After abelianization this says that $R(S_n)$ is generated, as an ordinary ring, by the permutation representations on $k$-subsets. A short argument with symmetric powers then shows that all of these permutation representations already lie in the $\lambda$-subring generated by $M_n$. Schocker~\cite{Schocker} proves a related generation theorem. His main theorem generates Solomon's descent algebra by sums of descent classes associated with hook shapes. Those generators are different from the two-block complete elements used below, but the two results belong to the same circle of descent-algebra generation phenomena. \subsection{Two-block generators for noncommutative symmetric functions} Let $\NSym_\ZZ$ be the free associative $\ZZ$-algebra on generators $H_1,H_2,\ldots$, with $\deg H_i=i$, and 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 $\NSym_{\ZZ,n}$. Besides its usual product, $\NSym_{\ZZ,n}$ carries an \emph{internal product} $*$, corresponding to multiplication in Solomon's descent algebra. We shall only need the standard matrix rule of Gelfand--Krob--Lascoux--Leclerc--Retakh--Thibon \cite[Proposition~5.1]{GelfandEtAlNCSF}. There is also a useful set-theoretic interpretation of this rule. Blessenohl--Schocker \cite[Appendix~B, especially B.1 and B.5; see also \S\S12.11--12.12]{BlessenohlSchocker} realize the relevant Young characters using ordered set partitions. If two ordered set partitions have block sizes $I$ and $J$, then the matrix entry $m_{uv}$ records the cardinality of the intersection of the $u$-th block of the first partition with the $v$-th block of the second. Thus the row sums and column sums are $I$ and $J$, while reading the nonzero intersection sizes row by row gives precisely the composition occurring in Solomon's multiplication rule. In particular, the matrix formula below is the noncommutative-symmetric-function form of the same Mackey-type calculation. 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, \[ \sum_{v=1}^q m_{uv}=i_u \quad(1\leq u\leq p), \qquad \sum_{u=1}^p m_{uv}=j_v \quad(1\leq v\leq q). \] The composition $\operatorname{comp}(M)$ is obtained by reading the entries of $M$ row by row and deleting the zero entries. \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} \begin{theorem} \label{thm:NSym-two-block-generation} Fix $n\geq1$. Under the internal product, the $\ZZ$-algebra $\NSym_{\ZZ,n}$ is generated by \begin{equation} H_{(k,n-k)}=H_kH_{n-k} \qquad (1\leq k\leq n), \label{eq:NSym-two-block-generators} \end{equation} where a zero part is omitted. \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. Apply \eqref{eq:NSym-matrix-rule} to $H_{I'}*H_J$. Every matrix 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 $dm$. 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, 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. No stronger ``unitriangular basis'' statement is needed here. \end{remark} Under the usual identification of $\NSym_{\ZZ,n}$ with Solomon's descent algebra (up to the harmless opposite-algebra convention for the internal product), $H_I$ corresponds to the sum of all permutations whose descent set is contained in the set of partial sums of $I$. Hence $H_{(k,n-k)}$ corresponds to \[ a_k=\sum_{\substack{w\in S_n;\\ w(1)<\cdots