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\title[Representation theorems for rings, \today]{Representation
  theorems for rings}   
\author[D. Pierce, \today]{David Pierce}
\date{\today\ (preliminary version)}

\address{Mathematics Dept\\
Middle East Technical University\\
Ankara 06531\\
Turkey}
\email{dpierce@metu.edu.tr}
\urladdr{http://metu.edu.tr/~dpierce/}
%\subjclass[2000]{Primary 03C60; Secondary 12Hxx, 12L12.}


\newtheorem{theorem}{Theorem}
\newtheorem*{lemma}{Lemma}

\newcommand{\defn}[1]{\textbf{#1}}   % for words being defined


\newcommand{\Dee}{\mathop{D}}

\newcommand{\lto}{\Rightarrow}

\newcommand{\Der}[1]{\operatorname{Der}(#1)} % space of derivations of...
\newcommand{\qsep}{\;}                 % follows a quantified variable
%\newcommand{\Forall}[1]{\forall{#1}\qsep }
\newcommand{\Exists}[1]{\exists{#1}\qsep }

\usepackage{upgreek}
\newcommand{\varN}{\upomega}        % my usual preference for this

\newcommand{\stnd}[1]{\mathbb{#1}}
\newcommand{\Z}{\stnd{Z}}         % integers
\newcommand{\R}{\stnd{R}}         % reals
\newcommand{\C}{\stnd{C}}         % complex numbers

\newcommand{\End}[1]{\operatorname{End}(#1)}
\newcommand{\Mult}[1]{\operatorname{Mult}(#1)}
\newcommand{\Hom}[1]{\operatorname{Hom}(#1)}
\newcommand{\Sym}[1]{\operatorname{Sym}(#1)}

\newcommand{\Er}{(E,\cdot)}
\newcommand{\aerE}{(\End E,\circ)}

\newcommand{\gpgen}[1]{\left\langle#1\right\rangle}% subgroup generated by #1

\newcommand{\inv}{^{-1}}                % mult. inverse

\newcommand{\RC}{\mathscr R}


%\newcommand{\abg}{\mathcal L_{\mathrm{ab}}}
\newcommand{\vect}[1]{\bm{#1}}
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\newcommand{\brkt}{\mathsf{b}}    % Lie bracket

%\newcommand{\invo}[1]{\overset{\bullet}{#1}}
\newcommand{\invo}[1]{\breve{#1}}

\newcommand{\fmsp}{\,} % formula space
\newcommand{\myop}[1]{\mathsf{#1}} % my operation
\newcommand{\mult}{\myop m}
%\newcommand{\genprdt}[3]{#1\fmsp#2\fmsp#3}
\newcommand{\genprdt}[3]{#1(#2,#3)}
\newcommand{\prdt}[2]{\genprdt{\mult}{#1}{#2}}
%\newcommand{\genlmlt}[2]{#1(#2)}
\newcommand{\genlmlt}[2]{\lambda^{#1}(#2)}
\newcommand{\lmlt}[1]{\genlmlt{\mult}{#1}}
\newcommand{\cmpt}{\myop c}

\newcommand{\rtms}[2]{\lambda^{#1}\times\lambda^{#2}}
\newcommand{\rccc}{\cmpt\circ(\rtms{\cmpt}{\cmpt})}
\newcommand{\rcci}{\cmpt\circ(\rtms{\cmpt}{\invo{\cmpt}})}
\newcommand{\rcic}{\cmpt\circ(\rtms{\invo{\cmpt}}{\cmpt})}
\newcommand{\rcii}{\cmpt\circ(\rtms{\invo{\cmpt}}{\invo{\cmpt}})}
\newcommand{\ricc}{\invo{\cmpt}\circ(\rtms{\cmpt}{\cmpt})}
\newcommand{\rici}{\invo{\cmpt}\circ(\rtms{\cmpt}{\invo{\cmpt}})}
\newcommand{\riic}{\invo{\cmpt}\circ(\rtms{\invo{\cmpt}}{\cmpt})}
\newcommand{\riii}{\invo{\cmpt}\circ(\rtms{\invo{\cmpt}}{\invo{\cmpt}})}

\newcommand{\rcc}{\lambda^{\cmpt}\circ\cmpt}
\newcommand{\rci}{\lambda^{\cmpt}\circ\invo{\cmpt}}
\newcommand{\ric}{\lambda^{\invo{\cmpt}}\circ\cmpt}
\newcommand{\rii}{\lambda^{\invo{\cmpt}}\circ\invo{\cmpt}}

\newcommand{\EE}{\End{E}}  % endomorphism-ring of E

 \begin{document}

   \begin{abstract}
Among rings, the associative rings, along with the rings that respect the
Jacobi identity, 
as well as rings with trivial multiplication,
are unique in possessing a certain kind of representation theorem.
Such rings can be obtained from an arbitrary abelian group $E$ by taking
the abelian group of endomorphisms of $E$ and introducing, as a
multiplication, the appropriate combination of composition and reverse
composition.  Other rings, for example Jordan rings, can be obtained
this way as well; but only for an associative ring, or a ring
satisfying the Jacobi identity, will the multiplication always be
respected by the canonical homomorphism from the underlying abelian
group to the endomorphism group of that group.
   \end{abstract}

   \maketitle

\section{Introduction}\label{sect:intro}

Associative rings, along with rings that respect the Jacobi identity,
as well as rings with trivial multiplication,
are unique in possessing a certain kind of representation theorem.
This observation is a consequence of Theorem~\ref{thm:c} below.

The observation arises from considering a field $K$ together with the
set $\Der K$ of its
derivations.  Then $\Der K$ has the structure of both a vector-space
over $K$ and a Lie ring.  Let $V$ be a subspace and sub-ring of $\Der
K$, and let $k$ be the constant field of $V$.  Then $V$ is what is
termed by Rinehart \cite{MR0154906} a \emph{$(k,K)$-Lie
  algebra.}  Other terms include \emph{pseudo-alg\`ebre de
    Lie} \cite{MR0055323} and \emph{Lie $d$-ring} \cite{MR0125867}, as
one may
learn from Stasheff \cite [p.~228]{MR1443334}, in whose terminology
$(V,K)$ is a \emph{Lie-Rinehart pair over $k$.}  This term applies
more generally to the situation where $k$ is
just a (commutative associative) ring and $K$ is a commutative
algebra over $k$.  In any case, reference to $k$ may distract one from
seeing the symmetry or dualism present in the pair $(V,K)$.  But it is
just this dualism that I want now to bring out.

In the pair $(V,K)$, first of all, $K$ and $V$ are abelian groups, each
acting faithfully 
as a group of endomorphisms of the other.  That is, we have, say,
\begin{align*}
  (x,\vect v)\mapsto x*\vect v&\colon K\times V\to V,\\
(\vect v,x)\mapsto \vect v\Dee x&\colon V\times K\to K,
\end{align*}
both maps being bi-additive:
\begin{subequations}
\begin{align}\label{*}
  (x+y)*\vect v&=x*\vect v+y*\vect v,& x*(\vect u+\vect v)&=x*\vect
  u+x*\vect v,\\\label{D}  
  (\vect u+\vect v)\Dee z&=\vect u\Dee z+\vect v\Dee z,& \vect v\Dee
  (x+y)&=\vect v\Dee x+\vect v\Dee y. 
\end{align}
\end{subequations}
Faithfulness of the actions is given by the formulas
\begin{subequations}\label{faith}
\begin{gather}\label{E*}
  \Exists{\vect v}(x*\vect v=0\lto x=0),\\\label{ED}
  \Exists x(\vect v\Dee x=0\lto\vect v=0).
\end{gather}
\end{subequations}
Moreover, $K$ has the multiplication $(x,y)\mapsto xy$; and $V$ has the
multiplication $(\vect u,\vect v)\mapsto[\vect u,\vect v]$, which make
the abelian groups a commutative
associative ring and a Lie ring respectively.  This means
just that the identities
\begin{subequations}\label{ring}
\begin{align}
\label{xyz}
  xy-yx&=0,& (xy)z&=x(yz)\\
\intertext{hold in $K$; and in $V$,}\label{Jacobi}
  [\vect v,\vect v]&=0,& [[\vect u,\vect v],\vect w]&=[\vect u,[\vect
      v,\vect w]]-[\vect v,[\vect u,\vect w]], 
\end{align}
\end{subequations}
where the latter equation in~\eqref{Jacobi} is the Jacobi identity.
Here 
\begin{equation}\label{eqn:Lb}
[\vect u,\vect v]=\vect u\circ\vect v-\vect
v\circ \vect u,
\end{equation}
although Kaplansky 
\cite[p.~2]{MR43:2145} also uses the notation
$\vect u\vect v$, `to emphasize all possible analogies with rings other than Lie
rings'---which is just what I want to do.

The abelian group $\EE$ of endomorphisms of an abelian
group $E$ becomes 
\begin{compactenum}
  \item
an associative ring, when equipped with composition; 
\item
a Lie ring, when equipped with the Lie bracket operation, $(\vect
u,\vect v)\mapsto[\vect u,\vect v]$.
\end{compactenum}
The embeddings $x\mapsto(\vect v\mapsto x*\vect v)$ of $K$ in $\End V$ and
$\vect v\mapsto(x\mapsto\vect v\Dee x)$ of $V$ in $\End K$
are in fact homomorphisms of associative rings
and Lie rings respectively:
\begin{subequations}\label{xy}
 \begin{align}\label{(xy)*z}
   (xy)*\vect v&=x*(y*\vect v),\\\label{(xy)Dz}
(\vect u\vect v)\Dee x&=\vect u\Dee(\vect v\Dee x)-\vect v\Dee(\vect u\Dee x).
\end{align}
\end{subequations}
The two actions interact according to:
\begin{subequations}\label{*D}
\begin{align}\label{x*yDz}
  (x*\vect v)\Dee y&=x(\vect v\Dee y),\\\label{xDy*z}
(\vect u\Dee x)*\vect v&=[\vect u,x*\vect v]-x*[\vect u,\vect v].
\end{align}
\end{subequations}
Indeed,~\eqref{x*yDz} is because $V$ is a space \emph{of derivations.}
Moreover,~\eqref{xDy*z} is a result of~\eqref{eqn:Lb} and the Leibniz rule of
differentiation.  The latter is \eqref{xD(yz)} below and is dual
to a rearrangement of~\eqref{xDy*z}:
\begin{subequations}\label{Dyz}
\begin{align}\label{xD(yz)}
  \vect v\Dee(xy)&=x(\vect v\Dee y)+(\vect v\Dee x)y,\\\label{x*(yz)}
  x*[\vect u,\vect v]&=[\vect u,x*\vect v]-(\vect u\Dee x)*\vect v.
\end{align}
\end{subequations}
The foregoing pairs of formulas can be taken as the definition of a
\defn{Lie--Rinehart pair.}  Model-theoretically speaking, the class of
Lie--Rinehart pairs is elementary, with $\forall\exists$ axioms.

EDITED TO HERE

\begin{theorem}\label{thm:=}
  In the foregoing definition, the identities~\eqref{ring} (except
  $xy-yx=0$) and~\eqref{Dyz} are redundant.  In fact, the
  identities~\eqref{xDy*z},~\eqref{xD(yz)}, and~\eqref{x*(yz)} 
  are equivalent.
\end{theorem}

\begin{proof}
  The redundancy of~\eqref{ring} follows from~\eqref{xy}
  and~\eqref{faith}.  Indeed, as noted above, by~\eqref{xy}, we have
  a \emph{homomorphism} from $K$ to an associative ring and from $V$
  to a Lie ring; by~\eqref{faith}, those homomorphisms are embeddings.
As for
  the second claim, we have already noted the equivalence
  of~\eqref{xDy*z} and~\eqref{x*(yz)}.  Moreover,
\begin{align*}
  (x\Dee(yz))*w
&=x((yz)*w)-(yz)*(xw)&&\text{[by~\eqref{xDy*z}]}\\
&=x(y*(z*w))-y*(z*(xw))&&\text{[by~\eqref{(xy)*z}]}\\
&
\begin{aligned}
=(x\Dee y)*(z*w)&+y*(x(z*w))\\
& -y*(x(z*w)-(x\Dee z)*w)
\end{aligned}
&&\text{[by~\eqref{xDy*z} or~\eqref{x*(yz)}]}\\
&=(x\Dee y)*(z*w)+y*((x\Dee z)*w)&&\text{[by~\eqref{*}]}\\
&=((x\Dee y)z)*w+(y(x\Dee z))*w,&&\text{[by~\eqref{(xy)*z}]}
\end{align*}
whence~\eqref{xD(yz)}, by~\eqref{*} and~\eqref{E*}.  So~\eqref{Dyz}
follows from~\eqref{xDy*z} (and everything else).
Conversely,
\begin{align*}
  (x(y*z))\Dee w
&=x\Dee((y*z)\Dee w)-(y*z)\Dee(x\Dee w)&&\text{[by~\eqref{(xy)Dz}]}\\
&=x\Dee(y(z\Dee w))-y(z\Dee (x\Dee w))&&\text{[by~\eqref{x*yDz}]}\\
&=y(x\Dee(z\Dee w))+(x\Dee y)(z\Dee w)-y(z\Dee (x\Dee
  w))&&\text{[by~\eqref{xD(yz)}]}\\
&=y((xz)\Dee w)+ (x\Dee y)(z\Dee w)&&\text{[by~\eqref{(xy)Dz}]}\\
&=(y*(xz))\Dee w+ ((x\Dee y)*z)\Dee w&&\text{[by~\eqref{x*yDz}]},
\end{align*}
whence~\eqref{xDy*z} and~\eqref{x*(yz)}, by~\eqref{D} and~\eqref{ED}.
\end{proof}

\section{Representation theorems}

The original representation theorem is attributed to
Cayley:
every group embeds in a symmetry group.\footnote{See \cite{MR562924}
  for discussion of the attribution of Cayley's Theorem.}
Indeed, if
$(G,{}\cdot{})$ is a group, let $\lambda$ be
the function that takes each element $g$ of $G$ to the permutation
$x\mapsto g\cdot  
x$ of $G$.  Then $\lambda$ is an
embedding of $(G,{}\cdot{})$ in $(\Sym G,{}\circ{})$.

To Cayley's Theorem,
Stone \cite{MR1501865} gives what he calls a precise analogue: every
Boolean ring embeds in 
a Boolean ring of \emph{sets,} namely the sets of prime ideals of the ring.
The embedding takes $x$ to the set of prime ideals that do \emph{not}
contain $x$. 

For abelian groups, the situation is not so neat.  There is a
\emph{prototypical} abelian group, $\Z$; but the variety of abelian
groups as such is multifarious.
However, if we fix an abelian group $A$, then, for every abelian group
$G$, we have the homomorphism $x\mapsto(y\mapsto y(x))$ from $G$ to
$\Hom{\Hom{G,A},A}$.  Suppose in particular that $A$ is divisible and
has elements of all orders: for example, $A$ might be the circle
$\R/\Z$, which features in Pontryagin duality for topological groups.
Then the homomorphism from $G$ to $\Hom{\Hom{G,A},A}$ is an embedding.
Indeed, if $x\in G$, then $\gpgen x$ 
embeds in $A$, and this embedding extends, by divisibility of $A$, to
an element $y$ of $\Hom{G,A}$, so that $y(x)\neq0$ if $x\neq0$.  Thus
every abelian group embeds in an abelian group of homomorphisms into a
particular group.

In the most
general sense, a \defn{multiplication} on an abelian group is a
bi-additive binary operation.
Then a \defn{ring} is just an abelian group
with a multiplication.  
As noted in \S~\ref{sect:intro},
if $E$ is an abelian group, then $\aerE$ is
an associative ring.  Indeed, if symmetry groups are the prototypical
groups, then $\aerE$ might be the prototypical associative ring.
Suppose a multiplication $\cdot$ is given on $E$.  We have
$\lambda$ as a homomorphism from $E$ to $\End E$ by the same formula
as in the discussion of Cayley's Theorem above.
Then $\Er$
is associative if and only if
$\lambda$ is a homomorphism from $\Er$ into $\aerE$, since each of
these conditions is just that the second equation~\eqref{xyz} is an
identity on $E$.
Note however that, even if it is a homomorphim, $\lambda$ need not be
an \emph{embedding} of $\Er$ in $\aerE$.

Lie rings have a corresponding representation, the so-called adjoint
representation.
The Jacobi identity in~\eqref{Jacobi} means simply that 
$\lambda$ is a homomorphism from $\Er$ to $(\EE,\brkt)$,
where $\brkt$ is the bracket.  If $\lambda$ is
indeed a homomorphism here, then
we have also $x^2y=0$ on $E$, but perhaps not~$x^2=0$, the other identity
in~\eqref{Jacobi}, unless
$\lambda$ is an embedding.  
\begin{comment}


The ring
$(\EE,\brkt)$ is itself a Lie ring, again as noted in the proof of
Theorem~\ref{thm:=}, by a simple computation:
\begin{equation*}%\label{eqn:comp}
  [[x,y],z]-[x,[y,z]]+[y,[x,z]]=\sum\{\text{six
  terms}\}-\sum\{\text{same six terms}\}=0.  
\end{equation*}



\end{comment}

\section{More rings}

The possible multiplications on an abelian group $E$ compose another
abelian group, $\Mult E$.  This group has an involution,
$\mult\mapsto\invo{\mult}$, where $\invo{\mult}(x,y)=\mult(y,x)$.  If
we denote functional composition by $\cmpt$, so that $x\circ
y=\cmpt(x,y)$, then the bracket on $\EE$ is $\cmpt-\invo{\cmpt}$.

Other combinations of composition and its opposite are of interest.
In particular,
if $xy$ is now $(\cmpt+\invo{\cmpt})(x,y)$,
% $(\EE,\cmpt+\invo{\cmpt})$ is a Jordan ring,
%since if $xy$ now denotes $x\circ y+y\circ x$, 
then we obtain the defining
identities 
\begin{align*}%\label{eqn:J3}
  xy&=yx,&
(xy)x^2&=x(yx^2)
\end{align*}
of a \defn{Jordan ring.}  However, if $\Er$ is an arbitrary Jordan ring,
then  $\lambda$ is \emph{not} a homomorphism
from $\Er$ to $(\EE,\circ+\invo{\circ})$, unless
$x(yx^2)=x(yx^2)+yx^3$ identically on $E$.

Supposing $\lambda$ is a homomorphism from an arbitrary ring $\Er$ to
$(\EE,p\cmpt-q\invo{\cmpt})$ for some integers $p$ and $q$,
let us refer to $\Er$ as a
\defn{$(p,q)$-ring.}  
So a ring is a $(p,q)$-ring
just in case the equation
\begin{equation}\label{eqn:pqz}
  (xy)z=px(yz)-qy(xz)
\end{equation}
is an identity on the ring.
Then the associative rings are just the $(1,0)$-rings.
All Lie rings are $(1,1)$-rings, and a $(1,1)$-ring is a Lie ring if
$\lambda$ is injective on it.  A Jordan ring is not generally a
$(1,-1)$-ring.  In particular,
$(\EE,p\cmpt-q\invo{\cmpt})$ itself is a $(p,q)$-ring for all abelian
groups $E$ when $(p,q)$ is
$(1,0)$ or $(1,1)$ or $(0,0)$; the theorem below shows that this
condition is necessary.

Given $\mult$ in $\Mult E$ and $a$ in $E$, let 
us denote by $\lambda^{\mult}(a)$ the endomorphism
$x\mapsto\mult(a,x)$ of $E$.  So $\lambda^{\mult}$ is a homo\-morphism
from $E$ to $\EE$.  We can now eliminate the variable $z$
from~\eqref{eqn:pqz}, getting
\begin{equation}\label{eqn:pq}
  \lambda^{\mult}(\prdt xy)=
  p\cdot\lambda^{\mult}(x)\circ\lambda^{\mult}(y)-
  q\cdot\lambda^{\mult}(y)\circ\lambda^{\mult}(x).
\end{equation}
We can go further, eliminating $(x,y)$ to get
\begin{equation}\label{eqn:pq2}
  \lambda^{\mult}\circ\mult=
  (p\cmpt-q\invo{\cmpt})\circ(\lambda^{\mult}\times\lambda^{\mult}).
\end{equation}
As a special case of this identity, we have
\begin{equation}\label{eqn:m}
  \lambda^{\mult}\circ\mult=\mult\circ(\lambda^{\mult}\times\lambda^{\mult})
\end{equation}
on $\EE$, when $\mult$ is $\cmpt$ or $\brkt$ throughout; the theorem
below is that $\cmpt$ and $\brkt$ are the
\emph{only} combinations of $\cmpt$ and $\invo{\cmpt}$ for which
\eqref{eqn:m} holds on every group $\EE$.  The
result can be obtained at length by replacing $\mult$ in \eqref{eqn:m} with
$p\cmpt-q\invo{\cmpt}$, applying each member to $(x,y)$ and then $z$,
and simplifying.  Alternatively, one can proceed more systematically
as follows. 

 \begin{lemma}\label{lem:ccc}
For every abelian group $E$, on $\EE$, the following six identities
hold: 
\begin{align*}
%\label{eqn:cc=ccc}
\rcc&=\rccc,\\%\notag
\rci&=\ricc,\\%\notag
\ric&=\riii,\\%\notag
\rii&=\rcii,\\%\notag
\rcci&=\rici,\\%\notag
\rcic&=\riic.
\end{align*}
 \end{lemma}

 \begin{proof}
The first identity expresses the associativity of $(\EE,\cmpt)$.
Continuing down the line, we have
\begin{gather*}
  \rci(x,y)(z)=\lambda^{\cmpt}(y\circ x)(z)=y\circ x\circ z,\\
\ricc(x,y)(z)
=\invo{\cmpt}(\lambda^{\cmpt}(x),\lambda^{\cmpt}(y))(z)
=(\lambda^{\cmpt}(y)\circ\lambda^{\cmpt}(x))(z)
=y\circ x\circ z,
\end{gather*}
and so on:
\begin{align*}
\rci\fmsp(x,y)(z)&=y\circ x\circ z=\ricc\fmsp(x,y)(z),\\
\ric\fmsp(x,y)(z)&=z\circ x\circ y=\riii\fmsp(x,y)(z),\\
\rii\fmsp(x,y)(z)&=z\circ y\circ x=\rcii\fmsp(x,y)(z),\\
\rcci\fmsp(x,y)(z)&=x\circ z\circ y=\rici\fmsp(x,y)(z),\\
\rcic\fmsp(x,y)(z)&=y\circ z\circ x=\riic\fmsp(x,y)(z).\qedhere
\end{align*}
\end{proof}

It will also be useful to note that the various  combinations of
binary operations into ternary
(or higher) operations can be related by means of permutations.
Indeed, for any abelian group $E$, for any
$n$ in $\varN$, there is a left action
%\begin{equation*}
%(\sigma, \tuple x)\longmapsto\sigma(\tuple x)
%\end{equation*}
of $\Sym n$ on $E^n$ by permutation of coordinates:
Elements $\tuple x$ of $E^n$ are functions $i\mapsto x_i$ on $n$, that
is, on $\{0,\dots,n-1\}$; elements of $\Sym n$ permute this set; then   
\begin{equation*}
  \sigma(\tuple x)=\tuple x\circ\sigma\inv.
\end{equation*}
Each function $\tuple x\mapsto\sigma(\tuple x)$ is an endomorphism of
$E^n$.
Thus $\Sym n$ can be considered as a subset of
$\End{E^n}$, though not a sub\emph{group}: $\Sym n$ is a subgroup of
 the multiplicative semigroup of $\aerE$.  For example, if
 $\mult\in\Mult E$, then
 $\invo{\mult}$ can be understood as $\mult\circ(0\;1)$.  Treating the
 equations in the lemma as identities of ternary operations on $\EE$,
 we have for example
 \begin{equation*}
\ric=\rcc\circ(0\;1\;2).
 \end{equation*}

\begin{theorem}\label{thm:c}
Suppose $\RC$ is a class of rings that, for some integers $p$ and $q$,
for all abelian groups $E$, contains $(\EE,p\cmpt+q\invo{\cmpt})$.
The following are equivalent.
\begin{compactenum}[1.]
\item
For every $\Er$ in $\RC$, the homomorphism $\lambda$ from $E$ to $\EE$
is a homomorphism from $\Er$ to $(\EE,p\cmpt+q\invo{\cmpt})$.
\item
 $(p,q)\in\{(1,0),(1,1),(0,0)\}$. 
\end{compactenum}
\end{theorem}

\begin{proof}
  Suppose $(p,q)\in\Z^2$.  It suffices to show that
  $(\EE,p\cmpt+q\invo{\cmpt})$ is a 
  $(p,q)$-ring for all $E$ if and only if
  $(p,q)\in\{(1,0),(1,1),(0,0)\}$. 
Considering Identity~\eqref{eqn:pq}, reformulated as~\eqref{eqn:pq2},
we have to show that the identity
\begin{equation*}
  (p\lambda^{\cmpt}-q\lambda^{\invo{\cmpt}})\circ(p\cmpt-q\invo{\cmpt})=
  (p\cmpt-q\invo{\cmpt})\circ
   [(p\lambda^{\cmpt}-q\lambda^{\invo{\cmpt}})\times(p\lambda^{\cmpt}-q\lambda^{\invo{\cmpt}})] 
\end{equation*}
holds on all endomorphism-groups $\EE$ if and only if
$(p,q)$ is $(1,0)$, $(1,1)$, or~$(0,0)$.
We compute:
\begin{equation*}
(p\lambda^{\cmpt}-q\lambda^{\invo{\cmpt}}) 
\circ(p\cmpt-q\invo{\cmpt})-
  (p\cmpt-q\invo{\cmpt})\circ
   [(p\lambda^{\cmpt}-q\lambda^{\invo{\cmpt}})\times
(p\lambda^{\cmpt}-q\lambda^{\invo{\cmpt}})]
\end{equation*}
\begin{multline*}
=p^2\rcc-pq\rci-pq\ric+q^2\rii-{}\\
{}-  (p\cmpt-q\invo{\cmpt})\circ
(p^2\cmpt\times\cmpt-pq\cmpt\times\invo{\cmpt}-
   pq\invo{\cmpt}\times\cmpt+q^2\invo{\cmpt}\times\invo{\cmpt})
\end{multline*}
\begin{multline*}
=p^2\rcc-pq\rci-pq\ric+q^2\rii-{}\\
{} -p^3\rccc+ p^2q\rcci+ p^2q\rcic- pq^2\rcii+{}\\
{}+p^2q\ricc- pq^2\rici- pq^2\riic+  q^3\riii)
\end{multline*}
\begin{multline*}
=p^2\rcc-pq\rci-pq\ric+q^2\rii-{}\\
{} -p^3\rcc+ p^2q\rcci+ p^2q\rcic- pq^2\rii+{}\\
{}+p^2q\rci- pq^2\rcci- pq^2\rcic+  q^3\ric)
\end{multline*}
\begin{multline*}
=(p^2-p^3)\rcc +(p^2q-pq)\rci +(q^3-pq)\ric+(q^2-pq^2)\rii +{}\\
{}+(p^2q-pq^2)\rcci+(p^2q-pq^2)\riic
\end{multline*}
by the lemma.
The five distinct coefficients here are:
\begin{equation*}
p^2(1-p),\quad
pq(p-1),\quad
q(q^2-p),\quad 
q^2(1-p),\quad 
pq(p-q).
\end{equation*}
They are all $0$ if
$(p,q)$ is $(1,0)$, $(1,1)$, or~$(0,0)$.
Conversely if they are all $0$, then either $p=0=q$, or $p=1$ and $q\in
\{0,1\}$.  This completes the proof, provided there is at least one
abelian group $E$ such that no non-trivial combination of
the six ternary operations $\rcc$, $\rci$, $\ric$,
$\rii$, $\riic$, and $\rcci$ on $\EE$ is $0$.

Such a group will be $\Z^4$.
If $k<3$, let
$f_k$ be the element $(k\;3)$ of $\Sym 4$, considered as an element of
$\End{\Z^4}$.  
If $\sigma\in\Sym 3$, considered as an element of
$\End{(\End{\Z^4})^3}$, then  
\begin{equation*}
  \sigma(f_0,f_1,f_2)=(f_i,f_j,f_k)
\end{equation*}
for some distinct $i$, $j$, and $k$ in $3$;
and
\begin{equation*}
  \rcc\circ\sigma(f_0,f_1,f_2)=
\rcc\fmsp(f_i,f_j,f_k)=
(i\;3)\circ(j\;3)\circ(k\;3)=(3\;k\;j\;i).
\end{equation*}
No non-trivial $\Z$-linear combination of these elements of
$\End{\Z^4}$ is $0$.  Indeed, these elements can be listed as
\begin{align*}
&(3\;2\;1\;0),&
&(3\;1\;2\;0),&
&(3\;2\;0\;1),&
&(3\;0\;2\;1),&
&(3\;1\;0\;2),&
&(3\;0\;1\;2).
\end{align*}
A combination of them takes $(x_0,x_1,x_2,x_3)$ to
\begin{multline*}
  a(x_1,x_2,x_3,x_0)+
  b(x_2,x_3,x_1,x_0)+
  c(x_2,x_0,x_3,x_1)+{}\\
 {}+ d(x_3,x_2,x_0,x_1)+
  e(x_1,x_3,x_0,x_2)+
  f(x_3,x_0,x_1,x_2),
\end{multline*}
which is
\begin{multline*}
  ((a+e)x_1+(b+c)x_2+(d+f)x_3,\
(c+f)x_0+(a+d)x_2+(b+e)x_3,\\
(d+e)x_0+(b+f)x_1+(a+c)x_3,\
(a+b)x_0+(c+d)x_1+(e+f)x_2).
\end{multline*}
If this is always $0$, then in particular the coefficients $a+e$,
$a+d$, and $d+e$ are $0$, so $a$, $d$, and $e$ are $0$; likewise
$b+c$,
$c+f$, and $b+f$ are $0$, so $b$, $c$, and $f$ are $0$.
But each of the six functions $\rcc$, $\rci$, $\ric$, $\rii$,
$\rcci$, and $\riic$ on $\End{\Z^4}$ is $\cmpt\circ\cmpt\circ\sigma$
for some $\sigma$ in $\Sym 3$.  Thus $\Z^4$ is as desired.
\end{proof}

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