text:
\section*{Introduction}

The Quantum Yang--Baxter Equation plays a crucial role in the theory of two-dimensional integrable systems of field theory
and statistical mechanics. This appeared first in the works of Yang \cite{Yang1967} and Baxter \cite{Baxter1971}.
It also lead to the theory of quantum groups and related areas \cite{Kassel1995}. Recall that a solution of the Quantum
Yang--Baxter Equation is a linear map $R\colon V\otimes V\to V\otimes  V$, where $V$ is a vector space, such that
$R_{12}R_{13}R_{23}= R_{23}R_{13}R_{12}$, where $R_{ij}$ denotes the map $V\otimes V\otimes V \to V\otimes  V\otimes V$
acting as $R$ on the $(i,j)$-th tensor factors and as the identity on the remaining factor.
Zamolodchikov \cite{Zamol1980,Zamol1981} described a generalization of this equation, the Tetrahedron Equation,
for three-dimensional integrable systems. Maillet in \cite{Ma94} shows that solutions of the Pentagon Equation lead
to solutions of the Tetrahedron Equation.  Recall that a linear map $S\colon V\otimes V\to V\otimes V$ is a solution of
the Pentagon Equation if, on $V\otimes V \otimes V$, one has $S_{23}S_{13}S_{12}=S_{12}S_{23}$. (In the literature
there is no agreement because some authors call $S_{12}S_{13}S_{23}=S_{23}S_{12}$ the Pentagon Equation. However, solutions
of both equations are in bijection via $S\mapsto\tau S\tau$, where $\tau\colon V\otimes V\to V\otimes V$ is the flip map
defined as $\tau(u\otimes v)=v\otimes u$.) Solutions of the Pentagon Equation appear now in various contexts. For example,
Kashaev in \cite{Ka96} has shown that the Pentagon Equation plays the same role for the Heisenberg double as the Quantum
Yang--Baxter Equation does for the Drinfeld double. Militaru \cite{Mi04} has shown that any finite-dimensional Hopf algebra
is characterized by an invertible solution of the Pentagon Equation. For other contexts we refer the reader to \cite{DiMu15}.
Solutions of the Pentagon Equation also appear with different terminologies. For example, for a Hilbert space $H$, a unitary
operator on $H\otimes H$ is said to be multiplicative if it is a solution of the Pentagon Equation \cite{BaSk93}; and in \cite{St98},
for a fixed braided monoidal category $\mathcal{V}$, an arrow $V\colon A\otimes A \to A \otimes A$ in $\mathcal{V}$
is said to be a fusion operator if it satisfies the Pentagon Equation.

Note that if $S$ is a basis of the vector space $V$ then a map $s\colon S\times S\to S\times S$ such that
$s_{23}s_{13}s_{12}=s_{12}s_{23}$ (respectively, $s_{12}s_{13}s_{23}=s_{23}s_{13}s_{12}$), where $s_{ij}$ denotes the map
$S\times S\times S\to S\times S\times S$ acting as $s$ on the $(i,j)$-th components and as the identity on the remaining component,
induces a solution of the Pentagon Equation (respectively, the Quantum Yang--Baxter Equation). In this case, one says that $(S,s)$
is a set-theoretic solution of the Pentagon Equation (respectively, the Quantum Yang--Baxter Equation). Drinfeld,
in \cite{Drinfeld1992}, posed the question of finding these set-theoretic solutions.

In the case of the Yang--Baxter Equation, a subclass of this type of solutions, the non-degenerate involutive solutions, has received
a lot of attention in past two decades; we give only a few references \cite{ESS,CJO2010,CJO2014,GIVdB,Lu2000,Rump2005}. This class of
solutions is not only studied for the applications of the Yang--Baxter Equation in physics, but also for its connection with other topics
in mathematics of recent interest, such as semigroups of $I$-type and Bieberbach groups, bijective $1$-cocycles, radical rings, triply
factorized groups, constructions of semisimple minimal triangular Hopf algebras, regular subgroups of the holomorph and Hopf--Galois extensions,
and groups of central type. Also general non-degenerate bijective solutions have received a lot of attention, see for example \cite{GuVe2017}.
It has been shown \cite{BCJ2016,Bach2018} that a description of all finite bijective non-degenerate solutions of the Yang--Baxter Equation
follows from a description of the algebraic structures called braces and skew braces, as introduced by Rump, and Guarnieri and Vendramin,
respectively. However the classification of these structures, and hence the construction of all finite bijective (involutive) non-degenerate
set-theoretic solutions  of the Yang--Baxter Equation is presently beyond reach.

Also the set-theoretic solutions of the Pentagon Equation received a large interest; we refer to, for example,
\cite{KaRe07,BaSk93,BaSk03,Jiang2005,Ka2011,KaSe98}. Recently, in \cite{CMM19x}, Catino, Mazzotta and Miccoli
dealt with a restrictive case of set-theoretic solutions of the Pentagon Equation. Namely, they described
all, not necessarily bijective, solutions $(S,s)$ of the Pentagon Equation of the form $s(x,y)=(x\cdot y,x*y)$,
where either $(S,\cdotp)$ or $(S,*)$ is a group. Furthermore, Catino, Mazotta and Stefanelli in \cite{CMS2019}
provided a method to obtain solutions of the Pentagon Equation on the matched product of two semigroups, that is a
semigroup including the classical Zappa--Sz\'ep product.

In this paper we continue the investigations on the description of bijective set-theoretic solutions of the Pentagon Equation. Surprisingly,
we are able to describe all such involutive solutions. This  is in strong contrast with the knowledge for involutive solutions
of the Yang--Baxter Equation.

\begin{mthm}
	Assume that $(S,s)$ is an involutive solution of the Pentagon Equation. Then there exist elementary abelian $2$-groups
	$(A,+)$ and $(G,\cdotp)$, a non-empty set $X$ and a map $\sigma\colon A\to \Sym(X)$ (not necessarily a group morphism)
	such that the set $S$ may be identified with $X\times A\times G$ and then
	\[s((x,a,g),(y,b,h))=((x,a,g\cdot h),(\sigma_{a+b}\sigma_b^{-1}(y),a+b,h))\] for all $x,y\in X$, $a,b\in A$ and $g,h\in G$.

	Moreover, all involutive solutions, up to isomorphism, of the Pentagon Equation defined on a set $S$ are in a bijective correspondence
	with decompositions of $S$ as a product $X\times A\times G$, where $X$ is a non-empty set and $A,G$ are elementary abelian $2$-groups.
	In particular, if $S$ is finite of cardinality $|S|=2^n(2m+1)$ for some $n,m\geq 0$ then there exist, up to isomorphism, exactly $\binom{n+2}{2}$
	involutive solutions of the Pentagon Equation defined on $S$.
\end{mthm}

\section{Definitions and examples}

Let $S$ be a non-empty set. A map $s\colon S\times S\to S \times S$ (or a pair $(S,s)$) is said to be a
\emph{set-theoretic solution of the Pentagon Equation} provided \[s_{23}s_{13}s_{12}=s_{12}s_{23},\]
where $s_{12}=s\times{\id}$, $s_{23}={\id}\times s$ and $s_{13}=(\tau\times{\id})({\id}\times s)(\tau\times{\id})$
are mappings from $S^3$ to itself and $\tau\colon S\times S\to S\times S$ is the flip map, i.e., $\tau (x,y) =(y,x)$.
Throughout the paper we shall simply say that $(S,s)$ is a solution of the PE. We say that the solution $(S,s)$
is \emph{finite} if $S$ is a finite set, \emph{bijective} provided $s$ is a bijective map, \emph{bijective of finite order}
if there exists a positive integer $n$ such that $s^n=\id$, and \emph{involutive} if $s^2=\id$.

Writing $s(x,y)=(x\cdot y,\theta_x(y))$ for $x,y\in S$ it is easy to see that $(S,s)$ is a solution of the PE if and only if the following equalities
\begin{align}
	(x\cdot y)\cdot z & =x\cdot (y\cdot z),\label{eq:p1}\\
	\theta_x(y)\cdot\theta_{x\cdot y}(z) & =\theta_x(y\cdot z),\label{eq:p2}\\
	\theta_{\theta_x(y)}\theta_{x\cdot y} & =\theta_y\label{eq:p3}
\end{align}
hold for all $x,y,z\in S$. In particular, $(S,\cdotp)$ must be a semigroup (dealing with a solution $(S,s)$ of the PE we shall frequently
denote the multiplication in $S$ as a concatenation, that is $x\cdot y=xy$ for $x,y\in S$). Moreover, $(S,s)$ is a solution of the PE if
and only if the map $t=\tau s \tau\colon S\times S\to S\times S$ satisfies \[t_{12}t_{13}t_{23}=t_{23}t_{12},\]
where $t_{12}=t\times{\id}$, $t_{23}={\id}\times t$ and $t_{13}=(\tau\times{\id})({\id}\times t)(\tau\times{\id})$
are mappings from $S^3$ to itself. Such a map $t$ is called a \emph{solution of the Reversed Pentagon Equation (RPE)}.
Note that $(S,s)$ is a bijective solution of the PE if and only if $(S,s^{-1})$ is a bijective solution of the RPE. In particular,
an involutive solution of the PE also is a solution of the RPE.

\begin{exa}[{cf. \cite{CMM19x}}]
	Assume that $S$ is a semigroup and $f$ an idempotent endomorphism of $S$. Then the map $s\colon S\times S\to S\times S$,
	defined as $s(x,y)=(xy,f(y))$, is a solution of the PE. In particular, if $e\in S$ is an idempotent
	then $s(x,y)=(xy,e)$ is a solution of the PE.
\end{exa}

\begin{exa}[{cf. \cite{Mi98}}]
	If $S$ is a set and $f,g\colon S\to S$ are commuting idempotent maps then the map $s\colon S\times S\to S\times S$,
	given by $s(x,y )=(f(x),g(y))$, is a solution of both the PE and the RPE.
\end{exa}

\begin{exa}
	Let $G$ be a group of finite exponent. Let $E=\{1,\dotsc,n\}$ and let $\sigma\in\Sym(n)$ be a permutation satisfying
	\begin{equation}\label{eq:sigma}
	\sigma^{\sigma(i)+1}=\sigma^i
	\end{equation}
	for each $i\in E$. Put $S=E\times G$ and let $s\colon S\times S\to S\times S$ be defined by \[s((i,a),(j,b))=((i,ab),(\sigma^i(j),b)).\]
	Then $(S,s)$ is a bijective solution of the PE of order equal to the least common multiple of the order of $\sigma$ and the exponent of $G$.
	First we shall check that $(S,s)$ is a solution of the PE. Observe that \eqref{eq:p1} is trivially satisfied because $S$
	is a semigroup for the multiplication defined by $(i,a)\cdot(j,b)=(i,ab)$. Moreover,
	\[\theta_{(i,a)}(j,b)\theta_{(i,a)(j,b)}(k,c)= (\sigma^i(j),b)\theta_{(i,ab)}(k,c)=(\sigma^i(j),b)(\sigma^i(k),c)=(\sigma^i(j),bc)\]
	and \[\theta_{(i,a)}((j,b)(k,c))=\theta_{(i,a)}(j,bc)=(\sigma^i(j),bc),\]
	i.e., condition \eqref{eq:p2} holds. To prove that condition \eqref{eq:p3} holds, first note that $\sigma^{\sigma^i(j)+i}=\sigma^j$ for all
	$i,j\in E$. Indeed, by assumption, the formula holds for $i=1$. So assume $i>1$. The result then follows by induction, because
	\[\sigma^{\sigma^i(j)+i}=\sigma^{\sigma(\sigma^{i-1}(j))+1}\sigma^{i-1}=\sigma^{\sigma^{i-1}(j)}\sigma^{i-1}=\sigma^{\sigma^{i-1}(j)+i-1}.\]
	Hence, we also have that
	\begin{align*}
		\theta_{\theta_{(i,a)}(j,b)}\theta_{(i,a)(j,b)}(k,c) &= \theta_{(\sigma^i(j),b)}\theta_{(i,ab)}(k,c)
		=\theta_{(\sigma^i(j),b)}(\sigma^i(k),c)\\ & =(\sigma^{\sigma^i(j)+i}(k),c)=(\sigma^j(k),c)=\theta_{(j,b)}(k,c).
	\end{align*}
	Thus, indeed $(S,s)$ is a solution of PE. For a positive integer $m$ we get that
	\[s^m((i,a),(j,b))=((i,ab^m),(\sigma^{mi}(j),b)).\]
	Hence $s^m=\id$ if and only if $m$ is a multiple of both the exponent of $G$ and the order of $\sigma$. Consequently, $s$ has finite
	order equal to the least common multiple of the order of $\sigma$ and the exponent of $G$. Finally, $(S,s)$ is also a solution of the
	RPE if and only if $G$ is the trivial group and $\sigma$ is either the identity or an involution.

	A concrete example is the following: $E=\{1,2,3,4\}$ and $\sigma\in\{ \id, (1\,2)(3\,4), (1\,4)(2\,3), (1\,4\,3\,2)\}$.
	One can easily verify that actually on $E$ these $\sigma$ are (the unique) permutations in $\Sym(4)$ that satisfy
	condition \eqref{eq:sigma}.
\end{exa}

The following natural definition is needed to determine when solutions are considered to be isomorphic
and it also yields a category of the solutions of the PE.

\begin{dfn}\label{def:hom}
	Let $(S,s)$ and $(S',s')$ be solutions of the PE. A morphism from $(S,s)$ to $(S',s')$ is a map $f\colon S\to S'$ such that
	$(f\times f)s=s'(f\times f)$ or, in other words, if the following diagram
	\[\begin{tikzcd}[row sep=30pt,column sep=30pt]
	S\times S\ar[r,"f{\times}f"]\ar[d,"s",swap] & S'\times S'\ar[d,"s'"]\\
	S\times S\ar[r,"f{\times}f"] & S'\times S'
	\end{tikzcd}\]
	is commutative. If a bijective morphism $f$ exists then we say that the solutions $(S,s)$ and $(S',s')$ are isomorphic.
\end{dfn}

With notations as in the Definition~\ref{def:hom}, write $s(x,y)=(x\cdot y,\theta_x(y))$ and $s'(a,b)=(a*b,\theta'_a(b))$.
One easily verifies that a map $f\colon S\to S'$ is a morphism of solutions if and only if
\begin{equation}\label{eq:hom}
f(x\cdot y)=f(x)*f(y)\qquad\text{and}\qquad f\theta_x=\theta'_{f(x)}f
\end{equation}
for all $x,y\in S$. In particular, $f$ is a morphism of semigroups.

Note also that if $(S_1,s_1)$ and $(S_2,s_2)$ are solutions of the PE then $(S,s)$, where $S=S_1\times S_2$ and the map
$s\colon S\times S\to S\times S$ is defined by the following formula \[s((x_1,x_2),(y_1,y_2))=(s_1(x_1,y_1),s_2(x_2,y_2)),\]
is again a solution of the PE. Clearly, $(S,s)$ is just a product of $(S_1,s_1)$ and $(S_2,s_2)$
in the category of solutions of the PE. Hence we may write $(S,s)=(S_1,s_1)\times(S_2,s_2)$.

\section{Bijective set-theoretic solutions of the Pentagon Equation}

In this section we show some basic properties of bijective solutions of finite order.
These will be crucial to deal with arbitrary involutive solutions.

It is clear that semigroups will play a role in the description of solutions of the PE.
For background and details on this topic we refer the reader to \cite{CP}.

Recall that a semigroup $S$ is called \emph{(left) simple} provided $S$ is the only (left) ideal of $S$.
Clearly each left simple semigroup is a simple semigroup.

\begin{lem}\label{lem:S=SS}
	Assume that $(S,s)$ is a bijective solution of the PE. Then $S=SS$ and $T=\{\theta_x:x\in S\}$
	is a subsemigroup of the transformation monoid $\Map(S,S)$.
	\begin{proof}
		Let $x,y \in S$. Since $s$ is bijective there exist $u,v \in S$ such that $s(u,v)=(x,y)$, i.e., $x=uv$ and
		$y=\theta_u(v)$. Therefore, $S=SS$. Moreover, from \eqref{eq:p3} it follows that $\theta_y\theta_x=
		\theta_{\theta_u(v)}\theta_{uv}=\theta_v\in T$. Hence, $T$ is a semigroup.
	\end{proof}
\end{lem}

\begin{pro}\label{pro:sgr}
	Assume that $(S,s)$ is a solution of the PE. If $s$ is bijective of finite order then $S$ is a simple semigroup.
	In particular, $|S|=1$ if and only if $S$ has a zero element.
	\begin{proof}
		Since $s$ is of finite order there exists $n \in \mathbb{N}$ such that $s^n=\id$. Then, for each $x,y\in S$,
		there exists $z\in S$ such that $xyz=x$. Hence, for any $x,y\in S$, we have $x\in SyS$. Therefore,
		every principal ideal, and thus every ideal of $S$, equals $S$, i.e., $S$ is a simple semigroup.
	\end{proof}
\end{pro}

Let $S$ be a semigroup and $\theta_x\colon S\to S$ a map for every $x\in S$. Consider the map $s\colon S\times S \to S \times S$
defined by $s(x,y)=(xy,\theta_x(y))$. Then $s$ is a solution of the RPE if and only if the following equalities
\begin{align}
	xy & =x\theta_y(z)yz,\label{eq:rp1}\\
	\theta_{x\theta_y(z)}(yz) & =\theta_x(y)z,\label{eq:rp2}\\
	\theta_{\theta_x(y)} & =\theta_x\theta_y\label{eq:rp3}
\end{align}
hold for all $x,y,z\in S$.

Recall that a semigroup $E$ is called a \emph{left zero semigroup} if every element of $E$ is a left zero of $E$, i.e., $xy=x$ for all $x,y\in E$.
Moreover, a semigroup $S$ is said to be a \emph{left group} if $S$ is a direct product $E\times G$ of a left zero semigroup $E$ and a group $G$.
Obviously, a left group is left simple, and it is easy to see that each left group is right cancellative, whereas each idempotent of a left group
is a right identity (see \cite{CP} for more details).

\begin{pro}\label{pro:GxE}
	Let $(S,s)$ be a solution of both the PE and the RPE (for example this is the case if $(S,s)$ is an involutive solution of the PE).
	If $s$ is bijective of finite order then $S$ is a left group.
	\begin{proof}
		Since $s$ is of finite order, there exists a positive integer $n$ such that $s^n=\id$.
		Let $x \in S$. Clearly, there exists $e\in S$ such that $xe=x$. By Lemma~\ref{lem:S=SS}, there exist $u,v\in S$
		such that $x=uv$. Hence, since $(S,s)$ also is a solution of the RPE, from \eqref{eq:rp1} we get that
		$x=uv=u\theta_v(z)vz$ for every $z\in S$. Hence $S=Sz$ for each $z\in S$. So, $S$ is left simple.
		Therefore, $e\in S=Sx$ and there exists $z\in S$ such that $e=zx$. So, $e^2=(zx)e=z(xe)=zx=e$, that is, $e$ is an idempotent.
		Thus $S$ is left simple and contains an idempotent. Consequently, by \cite[Theorem I.1.27]{CP}, we conclude that $S$ is a left group.
	\end{proof}
\end{pro}

\begin{lem}\label{lem:left-0-th}
	Let $(S,s)$ be a bijective solution of the PE. If $S$ is a left zero semigroup then all maps $\theta_x$ for $x\in S$ are bijective.
	Moreover, for each $x\in S$, the map $\theta_x$ is either the identity or a fixed-point free permutation.
	\begin{proof}
		Since $S$ is a left zero semigroup, we have $s(x,y)=(x,\theta_x(y))$ for all $x,y \in S$. If $\theta_x(y)=\theta_x(z)$ for some $x,y,z\in S$
		then $s(x,y)=(x,\theta_x(y))=(x,\theta_x(z))=s(x,z)$. Since $s$ is bijective, we get $y=z$ and thus $\theta_x$ is injective.
		Moreover, let $u\in S$ be such that $s(x,u)=(x,y)$. Then $\theta_x(u)=y$ and thus $\theta_x$ also is surjective.

		Furthermore, if $\theta_x$ has a fixed point, say $z\in S$, then, by \eqref{eq:p3},
		$\theta_z\theta_x=\theta_{\theta_x(z)}\theta_{xz}=\theta_z$. Because $\theta_z$ is bijective this leads to $\theta_x=\id$, as desired.
	\end{proof}
\end{lem}

\section{Involutive solutions: a reduction}

In this section we will prove that the description of all involutive set-theoretic solutions $(S,s)$ of the PE on a semigroup $S$
can be reduced to the description of solutions on a left zero semigroup. More precisely, we will show that $S$ may be identified with
$E\times G$, where $E$ is a left zero semigroup and $G$ is a group, and then that $s$ is composed of solutions $s_E$ and $s_G$
on $E$ and $G$, respectively. The solution $s_G$ is unique as it is known that on a group there only is one bijective (not necessarily
involutive) solution of the PE. This has been shown by Kashaev and Sergeev in \cite{KaSe98}. Catino, Mazzotta and Micolli extended this
result by showing that all solutions of PE on a group are determined by normal subgroups \cite{CMM19x}. For completeness' sake we give
an easy proof that on a group there is unique bijective solution of the PE.

\begin{pro}[{see \cite{KaSe98,CMM19x}}]\label{GroupUnique}
	Assume that $(G,s)$ is a bijective solution of the PE, where $G$ is a group. Then $s(x,y)=(xy,y)$ for all $x,y\in G$.
	In particular, there is a unique solution of the PE on a group $G$, which will be denoted $s_G$. Moreover, the order of
	$s$ equals the exponent of $G$; and thus $s$ is involutive precisely when $G$ is an elementary abelian $2$-group.
	\begin{proof}
		As before write $s(x,y)=(xy,\theta_x (y))$ for $x,y\in G$. Let $1$ be the identity element of the group $G$.
		First we show that $\theta_1$ is injective. Indeed, for $x\in G$ note that, by \eqref{eq:p2},
		\[\theta_x (x^{-1}) =\theta_x (x^{-1}1)=\theta_x(x^{-1}) \theta_{xx^{-1}} (1) =\theta_x(x^{-1}) \theta_1(1).\]
		Thus, since $G$ is a group, it follows that $\theta_1(1)=1$. Hence, again by \eqref{eq:p2},
		\[1=\theta_1 (1)=\theta_{1}(xx^{-1})=\theta_{1}(x) \theta_x(x^{-1}).\]
		Consequently, $\theta_{1}(x)^{-1}=\theta_x(x^{-1})$.

		Now, assume that $\theta_{1}(x) =\theta_{1}(y)$ for some $x,y\in G$. Then, by the previous,
		\[\theta_x(x^{-1})=\theta_{1}(x)^{-1} =\theta_{1}(y)^{-1}=\theta_y(y^{-1}).\]
		Therefore, \[s(x,x^{-1}) =(1,\theta_x(x^{-1}))=(1, \theta_y(y^{-1})) =s(y,y^{-1}).\]
		Since, by assumption, $s$ is bijective, it follows that $x=y$ and thus, indeed, $\theta_{1}$ in injective.

		From \eqref{eq:p2} we also obtain that $\theta_x(1) \theta_x(1) =\theta_x(1)$ for all $x\in G$ and thus
		$\theta_x (1)=1$ for all $x\in G$. Hence, \eqref{eq:p3} yields that $\theta_1 \theta_x=\theta_{\theta_x(1)}\theta_x =\theta_1$.
		Since, by the above, $\theta_1$ is injective, we obtain that $\theta_x =\id$ for all $x\in G$. Consequently, $s(x,y)=(xy,y)$ for
		all $x,y\in G$.
	\end{proof}
\end{pro}

Recall that a solution $(S,s)$ of the PE is called involutive provided $s^2=\id$ or, equivalently, the following equalities
\begin{align}
	xy\theta_x(y) & =x,\label{eq:i1}\\
	\theta_{xy}(\theta_x(y)) & =y\label{eq:i2}
\end{align}
hold for all $x,y\in S$.

\begin{thm}\label{thm:involuiveGxI}
	Let $(S,s)$ be an involutive solution of the PE. Then there exist a left zero semigroup $E$, an involutive solution $(E,s_E)$ of the PE
	on $E$, and an elementary abelian $2$-group $G$ such that $S$ may be identified with $E\times G$ and then \[(S,s)=(E,s_E) \times (G,s_G),\]
	where $s_G$ is the unique bijective solution of the PE on the group $G$. Moreover, $T=\{\theta_x:x \in S\}$ is an elementary abelian $2$-group.
	The converse is also true, that is, if $(E,s_E)$ is an involutive solution of the PE on a left zero semigroup $E$ and $(G,s_G)$ is
	the unique bijective solution of the PE on an elementary abelian $2$-group $G$ then the product $(E,s_E)\times(G,s_G)$ is an involutive
	solution of the PE.
	\begin{proof}
		Note that $(S,s)$ also is a solution of the RPE and thus, by Proposition~\ref{pro:GxE}, $S$ may be identified with $E\times G$,
		where $E$ is a left zero semigroup and $G$ is a group. Moreover, $E(S)=E\times \{1\}$, where by $E(S)$ we denote the set of
		idempotents of the semigroup $S$. Recall also that each idempotent of $S$ is a right identity in $S$, that is if $e\in E(S)$
		then $xe=x$ for each $x\in S$. Let $x \in S$ and $e\in E(S)$. Then, by \eqref{eq:p2}, we obtain
		\[\theta_x(e)=\theta_x(e^2)=\theta_x(e)\theta_{xe}(e)=\theta_x(e)\theta_x(e),\]
		i.e., $\theta_x(e)\in E(S)$. If $e=(i,1)$ for some $i \in E$ then we denote by $\vartheta_x(i)$ the unique element of $E$
		such that $\theta_x(e)=(\vartheta_x(i),1)$. If $y=(i,g)\in S$ then \eqref{eq:p2} yields
		\[\theta_x(y)=\theta_x(ey)=\theta_x(e)\theta_{xe}(y)=\theta_x(e)\theta_x(y)=(\vartheta_x(i), 1) \theta_x(y)\]
		and thus $\theta_x(y)=(\vartheta_x(i),h)$ for some $h\in G$. Moreover, write $x=(j,a)$ for some $j\in E$ and $a\in G$.
		Then it follows, by \eqref{eq:i1}, that $(j, agh)=xy\theta_x(y)=x =(j,a)$. Hence we have $h=g^{-1}$ and consequently
		$\theta_x(i,g)=\theta_x(y)=(\vartheta_x(i), g^{-1})$. Finally, by \eqref{eq:p3},
		\[(\vartheta_{(\vartheta_x(i),1)}\vartheta_x(j),g)
		=\theta_{\theta_x(e)}\theta_x(y)=\theta_{\theta_x(e)}\theta_{xe}(y)=\theta_e(y)=(\vartheta_e(i),g^{-1}).\]
		So, $g=g^{-1}$ for all $g\in G$, and thus $G$ is an elementary abelian $2$-group. We obtained
		\[\theta_x(i,g)=(\vartheta_x(i),g).\] Moreover, note that, by \eqref{eq:i2},
		$(\vartheta^2_x(i),1)=\theta^2_x(i,1)=\theta_{x(i,1)}\theta_x(i,1) =(i,1)$, i.e., $\vartheta_x^2(i)=i$. Hence, we obtain
		$\theta_x^2(i,g)=(\vartheta^2_x(i),g)=(i,g)$, i.e., $\theta^2_x=\id$. It follows that $T=\{\theta_x:x \in S\}$ is an elementary abelian $
		2$-group, since $T$ is a semigroup by Lemma~\ref{lem:S=SS} and, by \eqref{eq:rp3}, $\id =\theta^2_x=\theta_{\theta_x(x)}\in T$.

		Next, we claim that the map $\vartheta_x$ does not depend on $a$ (recall that $x=(j,a)$), that is, $\vartheta_{(j,a)}=\vartheta_{(j,1)}$.
		To prove this, first note that $\theta_{\theta_x(y)}=\theta_x\theta_y$ by \eqref{eq:rp3} and also $\theta_{\theta_x(y)}\theta_{xy}=\theta_y$
		by \eqref{eq:p3}. Therefore, we get $\theta_{xy}=(\theta_x\theta_y)^{-1}\theta_y=\theta_x$, since $T$ is an elementary abelian $2$-group.
		In particular, \[ (\vartheta_x(i),1)=\theta_x(i,1)=\theta_{(j,1)x}(i,1) =\theta_{(j,1)}(i,1)=(\vartheta_{(j,1)}(i),1),\]
		i.e., $\vartheta_{(j,a)}(i)=\vartheta_{(j,1)}(i)$. This proves the claim.

		For the sake of simplicity we may identify the elements of $E(S)$ with the elements of $E$ and write $\vartheta_{(j,1)}$ simply as
		$\vartheta_j$. Therefore, \[\theta_{(i,a)}(j,g)=\theta_{(i,1)}(j,g)=(\vartheta_i(j),g).\]
		Now consider $s_E$, the restriction of $s$ to $E\times E$. We claim that $(E,s_E)$ is an involutive solution of the PE. First,
		note that $s_E(i,j)=(i,\vartheta_i(j))$ for $i,j\in E$. Hence, \eqref{eq:p1} follows since $E$ is a subsemigroup of $S$,
		\eqref{eq:p2} trivially follows since $E$ is a left zero semigroup (i.e., $ij=i$ for all $i,j \in E$). Let $i,j,k \in E$. Then
		\[(\vartheta_{\vartheta_i(j)}\vartheta_i(k),1) =( \vartheta_{\vartheta_i(j)}\vartheta_{ij}(k),1)=
		\theta_{\theta_{(i,1)}(j,1)}\theta_{(i,1)(j,1)}(k,1)=\theta_{(j,1)}(k,1)=( \vartheta_j(k),1),\]
		which means that $\vartheta_{\vartheta_i(j)}\vartheta_i=\vartheta_j$, that is \eqref{eq:p3} holds. Moreover, since $\vartheta_i^2=\id$,
		it holds that $s_E^2(i,j)=s_E(i,\vartheta_i(j))=(i,\vartheta_i^2(j))=(i,j)$. Therefore, $(E,s_E)$ is an involutive solution, as claimed.

		Finally, from Proposition~\ref{GroupUnique} we know that on $G$ there is a unique bijective solution defined by $s_G(g,h)=(gh,h)$
		for $g,h\in G$. Since $G$ is an elementary abelian $2$-group, $s_G$ is involutive. Then,
		\[s((i,g),(j,h))=((i,g)(j,h),\theta_{(i,g)}(j,h))=((i,gh),(\vartheta_i(j),h))=(s_E\times s_G)((i,j),(g,h)).\]
		Since the converse readily can be verified, the result follows.
	\end{proof}
\end{thm}

It follows from the previous result that if $(S,s)$ is an involutive solution then $S= E\times G$ and $G$ is an elementary
abelian $2$-group. Hence, $E\times\{1\}=E(S)=\{x^2:x\in S\}$ and $G$ is isomorphic to $x^2Sx^2$ for any $x \in S$.

\begin{cor}\label{cor:basicFact}
	Let $(S,s)$ be an involutive solution of the PE. Then:
	\begin{enumerate}
		\item $\theta_{\theta_x(y)}=\theta_x\theta_y$, $\theta^2_x=\id$ and $\theta_{xy}=\theta_x$ for all $x,y\in S$.
		\item $x\theta_y(z)=xz$ for all $x,y,z\in S$.
		\item $\theta_x$ is an automorphism of $S$ for each $x\in S$.
	\end{enumerate}
	\begin{proof}
		From Theorem~\ref{thm:involuiveGxI} we know that $S=E\times G$ with $G$ an elementary abelian $2$-group and
		$E= E(S)=\{x^2:x\in S\}$. Moreover, we have $x^3=x$ for each $x\in S$. Furthermore, from the proof of
		Theorem~\ref{thm:involuiveGxI}, we know that $\theta_x=\theta_{x^2}$ and $\theta_x^{2}=\id$.

		Since $(S,s)$ is an involutive solution of the PE and thus also of the RPE, it follows from from \eqref{eq:rp3} that
		$\theta_{\theta_x(y)}=\theta_x\theta_y$ for all $x,y\in S$. From \eqref{eq:p3} we thus also get that
		\[\theta_{xy}=\theta_{\theta_x(y)}^{-1}\theta_y=\theta_{\theta_x(y)}\theta_y=\theta_x\theta_y\theta_y=\theta_x\] for all $x,y \in S$.
		This proves part (1).

		To prove part (2), note that Theorem~\ref{thm:involuiveGxI} yields that $y^2$ and $z^2$ are idempotents that are right identities in $S$.
		It then follows from $\theta_y=\theta_{y^2}$ and \eqref{eq:rp1} that
		\[x\theta_y(z)=x\theta_{y^2}(z)y^2z^2=(x\theta_{y^2}(z)y^2z)z=xy^2z=xz.\]

		Part (3) follows from \eqref{eq:p2} and part (1). Indeed, $\theta_x(yz)=\theta_x(y)\theta_{xy}(z)=\theta_x(y)\theta_x(z)$.
	\end{proof}
\end{cor}

Since the so-called commutative and cocommutative solutions of the PE play a role, especially in the context
of locally compact groups and Hilbert spaces (see \cite{BaSk93}), we recall definitions of these notions and give
a simple proof that involutive solutions are of this type. A solution $(S,s)$ of the PE is said to be \emph{commutative}
if $s_{12}s_{13}=s_{13}s_{12}$, i.e., if the following equalities
\begin{align}
	xyz&=xzy,\label{eq:c1}\\
	\theta_x&=\theta_{xy}\label{eq:c2}
\end{align}
hold for all $x,y,z \in S$. Similarly, $(S,s)$ is said to be \emph{cocommutative} provided
$s_{13}s_{23}=s_{23}s_{13}$, i.e., if the following equalities
\begin{align}
	x\theta_y(z)&=xz,\label{eq:cc1}\\
	\theta_x\theta_y&=\theta_y\theta_x\label{eq:cc2}
\end{align}
hold for all $x,y,z \in S$.

\begin{cor}
	Assume that $(S,s)$ is an involutive solution of the PE. Then $(S,s)$ is both commutative and cocommutative.
	\begin{proof}
		By Theorem~\ref{thm:involuiveGxI} we may assume that $S=E\times G$ is the direct product of a left zero semigroup $E$
		and an elementary abelian $2$-group $G$. Let $(e_i,g_i)\in S$ for $1\leq i\leq 3$. Then
		\[(e_1,g_1)(e_2,g_2)(e_3,g_3)=(e_1,g_1g_2g_3)=(e_1,g_1g_3g_2)=(e_1,g_1)(e_3,g_3)(e_2,g_2),\]
		i.e., \eqref{eq:c1} holds in $S$. Moreover, by Corollary~\ref{cor:basicFact}(1), \eqref{eq:c2} also holds.
		Hence $(S,s)$ is commutative. Furthermore, by Corollary~\ref{cor:basicFact}(2), we have that \eqref{eq:cc1} holds.
		Finally, \eqref{eq:cc2} holds since, by Theorem~\ref{thm:involuiveGxI}, $T=\{\theta_x:x\in S\}$ is an abelian group.
		Therefore, $(S,s)$ is cocommutative.
	\end{proof}
\end{cor}

\section{Retraction of an involutive solution and irretractable solutions}\label{subsec:ret}

In this section we introduce the notion of an irretractable involutive solution of the PE and we prove that
for a given cardinality there is unique such solution and we give an actual construction of this solution.

Assume that $(S,s)$ is an involutive solution of the PE. We define the equivalence relation $\sim$ on $S$,
called the {\em retract}, as follows. For $x,y\in S$, \[x\sim y\iff\theta_x=\theta_y.\]

Put $\overline{S}=S/{\sim}$ and let $\overline{x}$ denote the $\sim$-class of $x\in S$, called the retract class of $x$.
From Corollary~\ref{cor:basicFact}(1) we know that $\theta_{xy}= \theta_x$ for all $x,y\in S$ or, in other words,
\begin{equation}\label{eq:star}
xy\sim x.
\end{equation}
In particular, if $x_1\sim x_2$ and $y_1\sim y_2$ for some elements $x_1,x_2,y_1,y_2\in S$ then $x_1y_1\sim x_1\sim x_2 \sim x_2y_2$.
Therefore, $\sim$ is a congruence on $S$ and $\overline{S}$ is a semigroup. Moreover, again by Corollary~\ref{cor:basicFact}(1),
we get \[\theta_{\theta_{x_1}(y_1)}=\theta_{x_1}\theta_{y_1}=\theta_{x_2}\theta_{y_2}=\theta_{\theta_{x_2}(y_2)}.\]
Hence, $\theta_{x_1}(y_1)\sim\theta_{x_2}(y_2)$ and thus the map $\overline{\theta}_{\overline{x}}\colon\overline{S}\to\overline{S}$,
given by $\overline{\theta}_{\overline{x}}(\overline{y})=\overline{\theta_x(y)}$, is well defined.
This allows us to define a map $\overline{s}\colon\overline{S}\times \overline{S}\to \overline{S}\times \overline{S}$ by the formula
\[\overline{s}(\overline{x},\overline{y})=(\overline{x},\overline{\theta}_{\overline{x}}(\overline{y})).\]
Note that \eqref{eq:star} assures that $\overline{S}$ is a left zero semigroup.
Hence the \emph{retract} \[\Ret(S,s)=(\overline{S},\overline{s})\] of $(S,s)$ satisfies both \eqref{eq:p1} and \eqref{eq:p2}. Further,
\[\overline{\theta}_{\overline{\theta}_{\overline{x}}(\overline{y})}\overline{\theta}_{\overline{x}}(\overline{z})
=\overline{\theta}_{\overline{\theta_x(y)}}(\overline{\theta_x(z)})=\overline{\theta_{\theta_x(y)}\theta_x(z)}=\overline{\theta_y(z)}
=\overline{\theta}_{\overline{y}}(\overline{z})\]
for all $x,y,z\in S$, i.e., the equality \eqref{eq:p3} also holds for $\Ret(S,s)$. So, $\Ret (S,s)$ is a solution of the PE.
Finally, by Corollary~\ref{cor:basicFact}(1), $\theta_x^2=\id$ for each $x\in S$; hence
also $\overline{\theta}_{\overline{x}}^2=\id$. It follows that $\Ret(S,s)$ satisfies both equalities \eqref{eq:i1}
and \eqref{eq:i2}. Thus we have proved the first part of Proposition~\ref{pro:RII}.
For the second part we need the following definition.

\begin{dfn}
	An involutive solution $(S,s)$ of the PE is said to be \emph{irretractable} if $(S,s)=\Ret(S,s)$.
\end{dfn}

It is clear that a finite involutive solution $(S,s)$ of the PE is irretractable if and only if $|S|=|T|$, where
$T=\{\theta_x:x \in S\}$. In this case, by Theorem~\ref{thm:involuiveGxI}, $|S|$ is a power of $2$.

\begin{pro}\label{pro:RII}
	Let $(S,s)$ be an an involutive solution of the PE. Then $\Ret(S,s)$ is an irretractable involutive solution of the PE.
	\begin{proof}
		We only need to show the second part. This is shown as follows. If $x,y\in S$ then by Corollary~\ref{cor:basicFact}(1) we get
		\begin{align*}
			\overline{\theta}_{\overline{x}}=\overline{\theta}_{\overline{y}}
			& \iff \overline{\theta_x(z)}=\overline{\theta_y(z)}\text{ for all }z\in S\\
			& \iff \theta_{\theta_x(z)}=\theta_{\theta_y(z)}\text{ for all }z\in S\\
			& \iff \theta_x\theta_z=\theta_y\theta_z\text{ for all }z\in S\\
			& \iff \theta_x=\theta_y\\
			& \iff \overline{x} =\overline{y}.
		\end{align*}
		Hence the result is proved.
	\end{proof}
\end{pro}

\begin{lem}\label{lem:irr-l-0}
	Assume that $(S,s)$ is an involutive solution of the PE. Then $\theta_x(y)\sim\theta_y(x)$ for all $x,y\in S$. If in addition
	the solution $(S,s)$ is irretractable then $\theta_x(y)=\theta_y(x)$ for all $x,y\in S$ and $S$ is a left zero semigroup.
	\begin{proof}
		Let $x,y \in S$. By Theorem~\ref{thm:involuiveGxI} we know that $T=\{\theta_x:x\in S\}$ is an abelian group. Then,
		by Corollary~\ref{cor:basicFact}(1), we get $\theta_{\theta_x(y)}=\theta_x\theta_y=\theta_y\theta_x=\theta_{\theta_y(x)}$.
		Hence $\theta_x(y) \sim \theta_y(x)$. This proves the first part of the statement.

		For the second part, assume that $(S,s)$ also is irretractable. By Theorem~\ref{thm:involuiveGxI} we may assume that
		$S= E\times G$, where $E$ is a left zero semigroup and $G$ is an elementary abelian $2$-group. Let $e\in E$ and $g\in G$.
		Then Corollary~\ref{cor:basicFact}(1) yields $\theta_{(e,g)}=\theta_{(e,1)(e,g)}=\theta_{(e,1)}$.
		Hence, the irretractability of $(S,s)$ gives $g=1$. Thus $|G|=1$ and $S= E\times\{1\}\cong E$ is a left zero semigroup
	\end{proof}
\end{lem}

In next two propositions we completely describe the structure of all irretractable involutive solutions of the PE.

\begin{pro}\label{pro:irretract-2-abelian}
	Let $(A,+)$ be an elementary abelian $2$-group. Define the map $t\colon A\times A\to A\times A$ by $t(x,y)=(x,x+y)$.
	Then $(A,t)$ is an irretractable involutive solution of the PE. Conversely, if $(S,s)$ an irretractable involutive
	solution of the PE then there exists a natural structure $(S,+)$ of an elementary abelian $2$-group on $S$ such that
	$s(x,y)=(x,x+y)$ for all $x,y\in S$.
	\begin{proof}
		First, note that we have $t=\tau s_A \tau$, where $s_A$ is the unique solution of the PE on the group $A$
		(see Proposition~\ref{GroupUnique}) and $\tau\colon A\times A\to A\times A$ is the flip map. Hence, $t$
		is an involutive solution of the RPE and the PE. It remains to prove that $(A,t)$ is irretractable. If $x,y\in A$ then
		\begin{align*}
			\theta_x=\theta_y & \iff \theta_x(z)=\theta_y(z)\text{ for all }z\in A\\
			& \iff x+z=y+z\text{ for all }z\in A\\
			&\iff x=y.
		\end{align*}
		Therefore, $\Ret(A,t)=(A,t)$ and the solution $(A,t)$ is irretractable.

		Conversely, assume that $(S,s)$ is an irretractable involutive solution of the PE.
		By Lemma~\ref{lem:irr-l-0} we know that $s$ is of the form $s(x,y)=(x,\theta_x(y))$.
		We claim that the operation $x+y=\theta_x(y)$ for $x,y\in S$ makes $(S,+)$ an elementary
		abelian $2$-group. Observe first that, by Corollary~\ref{cor:basicFact}(1), we get
		\[x+(y+z)=x+\theta_y(z)=\theta_x\theta_y(z)=\theta_{\theta_x(y)}(z)=\theta_{x+y}(z)=(x+y)+z.\]
		Hence $(S,+)$ is a semigroup. Further, by Lemma~\ref{lem:irr-l-0}, \[x+y=\theta_x(y)=\theta_y(x)=y+x,\]
		i.e., the semigroup $(S,+)$ is commutative. Since the solution $(S,s)$ is irretractable, there exists a unique element $0\in S$
		such that $\theta_0=\id$ (it is enough to put $0=\theta_a(a)$ for any $a\in S$ because
		$\theta_{\theta_a(a)}=\theta_a^2=\id$). Then \[0+x=\theta_0(x)=x\qquad\text{and}\qquad x+0=0+x=x.\]
		Hence $0$ is the identity of $(S,+)$.
		Finally, $x+x=\theta_x(x)=0$. Therefore, $(S,+)$ is a elementary abelian $2$-group, as claimed, and $s(x,y)=(x,x+y)$.
	\end{proof}
\end{pro}

The solution $(A,t)$ on an elementary abelian $2$-group $(A,+)$ will be denoted as $(A,t_A)$.
Recall also that $t_A=\tau s_A\tau$, where $\tau\colon A\times A\to A\times A$ is the flip map
and $s_A$ is the unique bijective solution of the PE defined on the group $(A,+)$.

Next we show that irretractable involutive solutions of the PE are uniquely, up to isomorphism, determined by their cardinalities.

\begin{pro}\label{dim-arg}
	Assume that $(S,s)$ and $(S',s')$ are irretractable involutive solutions of the PE.
	Then the solutions $(S,s)$ and $(S',s')$ are isomorphic if and only if $|S|=|S'|$.
	\begin{proof}
		Clearly if $(S,s)$ and $(S',s')$ are isomorphic then $|S|=|S'|$. Conversely, assume that $|S|=|S'|$.
		Since $(S,s)$ and $(S',s')$ are irretractable involutive solutions, Proposition~\ref{pro:irretract-2-abelian} implies
		that $S$ and $S'$ admit natural structures of elementary abelian $2$-groups, let us denote them by $(S,+)$
		and $(S',+)$ respectively, such that $s(x,y)=(x,x+y)$ for all $x,y\in S$ and $s'(a,b)=(a,a+b)$ for all $a,b\in S'$.
		We claim that the groups $(S,+)$ and $(S',+)$ are isomorphic. This is obvious in case $|S|=|S'|$ is finite, as then,
		$|S|=|S'|=2^n$ and $(S,+)\cong C_2^n\cong(S',+)$ for some non-negative integer $n$. So, assume that $S$ and $S'$ are infinite.
		Considering $S$ and $S'$ as vector spaces over the field $\mathbb{F}_2$ with two elements, we get
		\[\dim_{\mathbb{F}_2}S=|S|=|S'|=\dim_{\mathbb{F}_2}S'.\]
		Hence, it follows that groups $(S,+)$ and $(S',+)$ are isomorphic, and the claim is proved.

		Therefore, there exists a group isomorphism $f\colon(S,+)\to(S',+)$. Moreover,
		\begin{align*}
			(f\times f)s(x,y) & =(f\times f)(x,x+y)=(f(x),f(x+y))\\
			& =(f(x),f(x)+f(y))=s'(f(x),f(y))=s'(f\times f)(x,y),
		\end{align*}
		which means that $f$ is also a morphism of solutions. This finishes the proof.
	\end{proof}
\end{pro}

\section{A description of all involutive solutions of the PE}

In this section we give a description of all involutive solutions of the PE. We shall start with the following construction.

\begin{pro}\label{pro:retrsol-prime}
	Let $(A,t_A)$ be the irretractable involutive solution of the PE defined on an elementary abelian $2$-group $(A,+)$.
	Assume that $X$ is a non-empty set and $\sigma\colon A\to\Sym(X)$. Put $S=X\times A$ and define the map
	$s\colon S\times S\to S\times S$ by the formula \[s((x,a),(y,b)) =((x,a),(\sigma_{a+b}\sigma_b^{-1}(y),a+b)).\]
	Then $(S,s)$ is an involutive solution of the PE. Moreover, $\Ret(S,s)=(A,t_A)$.
	\begin{proof}
		Write $\theta_{(x,a)}(y,b)=(\sigma_{a+b}\sigma_b^{-1}(y),a+b)$. Define $(x,a)\cdot(y,b)=(x,a)$ for $(x,a)\in S$ and $(y,b)\in S$.
		Then $(S,\cdotp)$ is a left zero semigroup. Hence both equalities \eqref{eq:p1} and \eqref{eq:p2} hold for $(S,s)$. Moreover, we have
		\begin{align*}
			\theta_{\theta_{(x,a)}(y,b)}\theta_{(x,a)\cdot(y,b)}(z,c)
			& =\theta_{(\sigma_{a+b}\sigma_b^{-1}(y),a+b)}\theta_{(x,a)}(z,c)\\
			& =\theta_{(\sigma_{a+b}\sigma_b^{-1}(y),a+b)}(\sigma_{a+c}\sigma_c^{-1}(z),a+c)\\
			& =(\sigma_{(a+b)+(a+c)}\sigma_{a+c}^{-1}\sigma_{a+c}\sigma_c^{-1}(z),(a+b)+(a+c))\\
			& =(\sigma_{b+c}\sigma_c^{-1}(z),b+c)\\
			& =\theta_{(y,b)}(z,c).
		\end{align*}
		Thus, also the equality \eqref{eq:p3} holds for $(S,s)$. Furthermore,
		\begin{align*}
			s^2((x,a),(y,b)) &=s((x,a),(\sigma_{a+b}\sigma_b^{-1}(y),a+b))\\
			& =((x,a),(\sigma_{a+(a+b)}\sigma_{a+b}^{-1}\sigma_{a+b}\sigma_b^{-1}(y),a+(a+b)))\\
			& =((x,a),(y,b)).
		\end{align*}
		Therefore, $(S,s)$ is an involutive solution of the PE.

		Finally, if $(x,a)\in S$ and $(y,b)\in S$ then $(x,a)\sim(y,b)$ or, in other words, $\theta_{(x,a)}=\theta_{(y,b)}$
		if and only if $a=b$. Hence the quotient $S/{\sim}$ may be identified with $A$. It remains to observe that under this
		identification we get $\Ret(S,s)=(A,t_A)$. Thus the result is proved.
	\end{proof}
\end{pro}

The solution $(S,s)$ constructed in Proposition~\ref{pro:retrsol-prime} will be called the
\emph{extension of $(A,t_A)$ by $X$ and $\sigma$} and denoted by \[\Ext_X^\sigma(A,t_A).\]

\begin{lem}\label{lem:classes}
	Let $(S,s)$ be an involutive solution of the PE. Then all retract classes on $S$ have the same cardinality.
	\begin{proof}
		Fix $x,y\in S$. Define $X=\{z\in S:\theta_z=\theta_x\}$ and $Y=\{z\in S:\theta_z=\theta_y\}$, the retract classes of
		$x$ and $y$, respectively. Consider the map $f\colon X\to Y$ given by the formula $f(z)=\theta_y\theta_x(z)$.
		Note that $f$ is well defined, since for $z\in X$ we have, by Corollary~\ref{cor:basicFact}(1) and Theorem~\ref{thm:involuiveGxI}, that
		\[\theta_{f(z)}=\theta_{\theta_y\theta_x(z)}=\theta_y\theta_{\theta_x(z)}= \theta_y\theta_x\theta_z=\theta_y\theta_x^2=\theta_y,\]
		and thus $f(z)\in Y$. Since $\theta_y\theta_x$ is a bijective map, clearly $f$ is injective.
		Moreover, if $w\in Y$ then put $z=\theta_x\theta_y(w)$. Then, again by Corollary~\ref{cor:basicFact}(1) and Theorem~\ref{thm:involuiveGxI},
		\[\theta_z= \theta_{\theta_x\theta_y(w)} =\theta_x\theta_{\theta_y(w)}=\theta_x\theta_y\theta_w=\theta_x\theta_y^2=\theta_x.\]
		So $z \in X$. Finally, $f(z)=\theta_y\theta_x\theta_x\theta_y(w)=\theta_x^2\theta_y^2(w)=w$. Thus $f$ is also surjective.
	\end{proof}
\end{lem}

The following result shows that solutions constructed in Proposition~\ref{pro:retrsol-prime}
are actually all solutions of the PE defined on left zero semigroups.

\begin{pro}\label{pro:semidir-prime}
	Assume that $(S,s)$ is an involutive solution of the PE defined on a left zero semigroup $S$.
	Then there exists an abelian elementary $2$-group $(A,+)$ and a non-empty set $X$ such that
	$S$ may be identified with $X\times A$ and then $(S,s)=\Ext_X^\sigma(A,t_A)$ for some $\sigma\colon A\to\Sym(X)$.
	\begin{proof}
		Write $s(x,y)=(x,\theta_x(y))$. By Proposition~\ref{pro:irretract-2-abelian} we know that if
		$\Ret(S,s)=(A,\overline{s})$ then $A$ is an elementary abelian $2$-group such that
		\[\overline{s}(a,b)=(a,\overline{\theta}_a(b))=(a,a+b)\] for all $a,b\in A$. Observe also that $(A,\overline{s})=(A,t_A)$.

		Next, choose $w\in S$ and define $X=\{x\in S:\theta_x=\theta_w\}$, the retract class of $w$.
		We know from Lemma~\ref{lem:classes} that there exist a bijection between $X$ and each retract
		class $\overline{x}$ of $x\in S$. Hence, we may identify $S$ with $X\times A$, and under this
		identification elements $(x,a)\in S$ and $(y,b)\in S$ lie in the same retract class if and only if
		$a=b$. Furthermore, it is clear that we may write \[s((x,a),(y,b))=((x,a),(\pi_{x,a,b}(y),a+b)),\]
		i.e., $\theta_{(x,a)}(y,b)=(\pi_{x,a,b}(y),a+b)$ for some maps $\pi_{x,a,b}\in\Map(X,X)$.
		We claim that $\pi_{x,a,b}$ does not depend on $x$. To prove this choose $y\in X$.
		Since $(x,a)$ and $(y,a)$ lie in the same retract class, we get $\theta_{(x,a)}=\theta_{(y,a)}$. Hence
		\[(\pi_{x,a,b}(z),a+b)=\theta_{(x,a)}(z,b)=\theta_{(y,a)}(z,b)=(\pi_{y,a,b}(z),a+b)\]
		for all $z\in X$. Therefore, we obtain $\pi_{x,a,b}= \pi_{y,a,b}$, as claimed. Hence we may write $\pi_{a,b}=\pi_{x,a,b}$.
		Next, by Corollary~\ref{cor:basicFact}(1) we know that $\theta_{(x,a)}^2=\id$. This leads to
		\[(y,b)=\theta_{(x,a)}^2(y,b)=\theta_{(x,a)}(\pi_{a,b}(y),a+b)=(\pi_{a,a+b}\pi_{a,b}(y),a+(a+b))=(\pi_{a,a+b}\pi_{a,b}(y),b).\]
		Hence $\pi_{a,a+b}\pi_{a,b}=\id$ for all $a,b\in A$. Replacing $b$ by $a+b$ in the last equality we get
		$\pi_{a,b}\pi_{a,a+b}=\id$. Therefore, $\pi_{a,b}\in\Sym(X)$ and $\pi_{a,b}^{-1}=\pi_{a,a+b}$ for all $a,b\in A$.
		Furthermore, Corollary~\ref{cor:basicFact}(1) assures also that $\theta_{\theta_{(x,a)}(y,b)}=\theta_{(x,a)}\theta_{(y,b)}$,
		which yields
		\begin{align*}
			(\pi_{a+b,c}(z),(a+b)+c) & =\theta_{(\pi_{a,b}(y),a+b)}(z,c)=\theta_{\theta_{(x,a)}(y,b)}(z,c)\\
			& =\theta_{(x,a)}\theta_{(y,b)}(z,c)=\theta_{(x,a)}(\pi_{b,c}(z),b+c)\\
			& =(\pi_{a,b+c}\pi_{b,c}(z),a+(b+c)).
		\end{align*}
		Hence $\pi_{a,b+c}\pi_{b,c}=\pi_{a+b,c}$ for all $a,b,c\in A$. In particular, plugging $c=0$ to the last
		equality, we obtain $\pi_{a,b}=\pi_{a+b,0}\pi_{b,0}^{-1}$. Therefore, defining the map $\sigma\colon A\to\Sym(X)$
		by the formula $\sigma_a=\pi_{a,0}$, we conclude that $\theta_{(x,a)}(y,b)=(\sigma_{a+b}\sigma_b^{-1}(y),a+b)$
		and thus \[s((x,a),(y,b))=((x,a),(\sigma_{a+b}\sigma_b^{-1}(y),a+b)),\] which shows that $(S,s)=\Ext_X^\sigma(A,t_A)$.
	\end{proof}
\end{pro}

The following result shows that all solutions of the form $\Ext_X^\sigma(A,t_A)$ are, in fact, isomorphic.

\begin{pro}\label{pro:isom-of-sol-prime}
	Assume that $(A,+)$ is an elementary abelian $2$-groups and $X$ is a non-empty set. If $\sigma\colon A\to\Sym(X)$
	and $\rho\colon A\to\Sym(X)$ then the solutions $\Ext_X^\sigma(A,t_A)$ and $\Ext_X^\rho(A,t_A)$ of the PE are isomorphic.
	\begin{proof}
		Note first that if $\varphi\in\Sym(X)$ is an arbitrary permutation then defining $\pi\colon A\to\Sym(X)$ by the formula
		$\pi_a=\sigma_a\varphi$, we get $\pi_{a+b}\pi_b^{-1}=\sigma_{a+b}\sigma_b^{-1}$ for all $a,b\in A$. In particular,
		$\Ext_X^\sigma(A,t_A)=\Ext_X^\pi(A,t_A)$. Hence we may assume that $\sigma_0=\id=\rho_0$. We claim that
		the map $f\colon\Ext_X^\sigma(A,t_A)\to\Ext_X^\rho(A,t_A)$, defined as $f(x,a)=(\rho_a\sigma_a^{-1}(x),a)$,
		is an isomorphism of solutions. Clearly, $f$ is a bijective map. Moreover, put $\Ext_X^\sigma(A,t_A)=(X\times A,s_\sigma)$
		and $\Ext_X^\rho(A,t_A)=(X\times A,s_\rho)$. Then
		\begin{align*}
			(f\times f)s_\sigma((x,a),(y,b)) & =(f\times f)((x,a),(\sigma_{a+b}\sigma_b^{-1}(y),a+b))\\
			& =((\rho_a\sigma_a^{-1}(x),a),(\rho_{a+b}\sigma_{a+b}^{-1}\sigma_{a+b}\sigma_b^{-1}(y),a+b))\\
			& =((\rho_a\sigma_a^{-1}(x),a),(\rho_{a+b}\sigma_b^{-1}(y),a+b))\\
			& =((\rho_a\sigma_a^{-1}(x),a),(\rho_{a+b}\rho_b^{-1}\rho_b\sigma_b^{-1}(y),a+b))\\
			& =s_\rho((\rho_a\sigma_a^{-1}(x),a),(\rho_b\sigma_b^{-1}(y),b))\\
			& =s_\rho(f\times f)((x,a),(y,b)).
		\end{align*}
		Therefore, $f$ is a morphism of solutions and the result is proved.
	\end{proof}
\end{pro}

Combining Theorem~\ref{thm:involuiveGxI} and Proposition~\ref{pro:retrsol-prime} we obtain a complete description
of all involutive solutions of the PE. Surprisingly, it turns out that all such solutions are uniquely determined,
up to isomorphism, by two elementary abelian $2$-groups.

\begin{thm}\label{thm:main-prime}
	Assume that $(S,s)$ is an involutive solution of the PE. Then there exist elementary abelian $2$-groups
	$A$ and $G$ and a non-empty set $X$ such that $S$ may be identified with $X\times A\times G$ and then
	\[(S,s)=\Ext_X^\sigma(A,t_A)\times(G,s_G)\] for some $\sigma\colon A\to\Sym(X)$, where $(A,t_A)$ is the
	unique irretractable involutive solution of the PE on $A$ and $(G,s_G)$ is the unique bijective solution of the PE on $G$.
	Moreover, $\Ret(S,s)=(A,t_A)$ and $(S,s)$ is isomorphic to the product $(X,\id)\times(A,t_A)\times(G,s_G)$ of solutions.
	\begin{proof}
		In view of what was already proved, it remains to show that the solution $(S,s)$ is isomorphic to
		the product $(X,\id)\times(A,t_A)\times(G,s_G)$. But Proposition~\ref{pro:isom-of-sol-prime}
		yields $\Ext_X^\sigma(A,t_A)\cong\Ext_X^\rho(A,t_A)$, where $\rho\colon A\to\Sym(X)$ is defined as $\rho_a=\id$
		for all $a\in A$. Since we clearly have $\Ext_X^\rho(A,t_A)\cong(X,\id)\times(A,t_A)$, the result is proved.
	\end{proof}
\end{thm}

\begin{thm}\label{thm:main2}
	All involutive solutions, up to isomorphism, of the PE defined on a non-empty set $S$ are in a bijective correspondence
	with decompositions of $S$ as a product $X\times A\times G$, where $X$ is a non-empty set and $A,G$ are elementary abelian $2$-groups.
	\begin{proof}
		Let $(S_1,s_1)$ and $(S_2,s_2)$ be involutive solutions of the PE corresponding to decompositions
		of $S$ of the form $S_1=X_1\times A_1\times G_1$ and $S_2=X_2\times A_2\times G_2$, respectively (see Theorem~\ref{thm:main-prime}).

		If $|X_1|=|X_2|$, $|A_1|=|A_2|$ and $|G_1|=|G_2|$ then there exist a bijection $\alpha\colon X_1\to X_2$ and morphisms of groups
		$\beta\colon(A_1,+)\to(A_2,+)$ and $\gamma\colon(G_1,\cdotp)\to(G_2,\cdotp)$ (see the proof of Proposition~\ref{dim-arg}).
		But then it is easy to check that the map $f\colon(S_1,s_1)\to(S_2,s_2)$, defined as $f(x,a,g)=(\alpha(x),\beta(a),\gamma(g))$
		for $x\in X_1$, $a\in A_1$ and $g\in G_1$, is an isomorphism of solutions.

		Conversely, suppose that $f\colon(S_1,s_1)\to(S_2,s_2)$ is an isomorphism of solutions. Then \eqref{eq:hom} assures that
		$f\colon S_1\to S_2$ is a morphism of semigroups (recall that multiplication in $S_i$ for $1\leq i\leq 2$ is given as
		$(x,a,g)\cdot(y,b,h)=(x,a,g\cdot h)$ for $x,y\in X_i$, $a,b\in A_i$ and $g,h\in G_i$). Since $G_i=e_iS_ie_i$ for
		any $e_i\in E(S_i)$ (see the comment before Corollary~\ref{cor:basicFact}), we get $f(G_1)=f(eS_1e)=f(e)S_2f(e)=G_2$
		for each $e\in E(S_1)$. Hence $|G_1|=|G_2|$. Next, by Theorem~\ref{thm:main-prime}, we obtain
		\[(A_1,t_{A_1})=\Ret(S_1,s_1)\cong\Ret(S_2,s_2)=(A_2,t_{A_2})\] and thus $|A_1|=|A_2|$. Finally, to prove that $|X_1|=|X_2|$,
		write $f(x,0,1)=(\varphi(x),\psi(x),1)$ for $x\in X_1$, where $\varphi\colon X_1\to X_2$ and $\psi\colon X_1\to A_2$
		(we know that $f(x,0,1)$ is of that form because $(x,0,1)\in E(S_1)$ implies $f(x,0,1)\in E(S_2)=X_2\times A_2\times\{1\}$). Then
		\begin{align*}
			((\varphi(x),\psi(x),1),(\varphi(y),\psi(y),1)) & =(f\times f)s_1((x,0,1),(y,0,1))\\
			& =s_2(f\times f)((x,0,1),(y,0,1))\\
			& =s_2((\varphi(x),\psi(x),1),(\varphi(y),\psi(y),1)\\
			& =((\varphi(x),\psi(x),1),(\varphi(y),\psi(x)+\psi(y),1)
		\end{align*}
		for all $x,y\in X_1$. This implies that $\psi(x)=0$ for all $x\in X_1$.
		Hence $f(x,0,1)=(\varphi(x),0,1)$. Since the inverse $f^{-1}$ of $f$ has a similar property,
		we obtain that $\varphi$ is a bijection and thus $|X_1|=|X_2|$. Hence the proof is finished.
	\end{proof}
\end{thm}

As an immediate consequence of Theorem~\ref{thm:main2} we get the following result.

\begin{cor}\label{cor:binom}
	Assume that $S$ is a finite set of cardinality $|S|=2^n(2m+1)$ for some $n,m\geq 0$. Then there exist,
	up to isomorphism, exactly $\binom{n+2}{2}$ involutive solutions of the PE defined on $S$.
\end{cor}

As an illustration of the main results of this section we provide the form of all six, up to isomorphism,
involutive solutions of the PE defined on a set of cardinality $12$.

\begin{exa}
	Let $S$ be a set of cardinality $12$. The involutive solutions $(S,s)$ of the PE presented below correspond to the decomposition
	$S=X\times A\times G$, where $(A,+)$ and $(G,\cdotp)$ are elementary abelian $2$-groups. Clearly, we have
	$A,G\in\{0,C_2,C_2\times C_2\}$. Moreover, if one of the factors $A$ or $G$ is the trivial group $0$ then this factor
	will be omitted and thus we write $S=X$ if $A=G=0$ or $S=X\times A$ if $A\ne 0$ but $G=0$ or $S=X\times G$ if $A=0$ but $G\ne 0$.
	\begin{enumerate}
		\item Let $X=\{1,\dotsc,12\}$. Then $s(x,y)=(x,y)$.
		\item Let $X=\{1,\dotsc,6\}$ and $A=C_2$. Then $s((x,a),(y,b))=((x,a),(y,a+b))$.
		\item Let $X=\{1,\dotsc,6\}$ and $G=C_2$. Then $s((x,g),(y,h))=((x,g\cdot h),(y,h))$.
		\item Let $X=\{1,2,3\}$ and $A=C_2\times C_2$. Then $s((x,a),(y,b))=((x,a),(y,a+b))$.
		\item Let $X=\{1,2,3\}$ and $G=C_2\times C_2$. Then $s((x,g),(y,h))=((x,g\cdot h),(y,h))$.
		\item Let $X=\{1,2,3\}$ and $A=G=C_2$. Then $s((x,a,g),(y,b,h))=((x,a,g\cdot h),(y,a+b,h))$.
	\end{enumerate}
\end{exa}

\section{Structure monoids and algebras of solutions of the PE}

Suppose $S$ is a non-empty finite set and $s\colon S\times S\to S\times S$ is a bijective map. One simply says that $(S,s)$
is a quadratic set. Then, over a field $K$, one associates a quadratic algebra, called the structure algebra of $(S,s)$ over $K$,
\[A(K,S,s)=K\free{S \mid xy=uv\text{ if }s(x,y)=(u,v)}.\] This algebra has a natural gradation $A(K,S,s)=\bigoplus_{n\geq 0}A_n$
and is generated by $A_1=\Span_K(S)$. So $A(K,S,s)$ is a connected graded $K$-algebra and each $A_n$ is finite-dimensional. Quadratic
algebras have received a lot of attention and for more information we refer to \cite{PolPos}. Again because the defining relations
are homogeneous, \[A(K,S,s)=K[M(S,s)],\] the monoid $K$-algebra of the so-called structure monoid $M(S,s)$ of $(S,s)$ defined as
\[M(S,s)=\free{S \mid xy=uv\text{ if }s(x,y)=(u,v)}.\]

In the last two decades there has been an intense study of monoids, groups and algebras associated with quadratic sets,
see for example \cite{CedOkn,GI2012,GI2018,JesOknBook}. In case $(S,s)$ is a finite non-degenerate unitary solution of the
Quantum Yang--Baxter Equation, i.e., $(S,\tau s)$ is a finite non-degenerate involutive solution of the Yang--Baxter Equation
then it has been shown in \cite{GIVdB} that the structure algebra $A(K,S,\tau s)$ has a very rich structure that shares many
properties with polynomial algebras in finitely many commuting variables. In particular, it is a (left and right) Noetherian
algebra that satisfies a polynomial identity (abbreviated, PI) and, moreover, it is a domain that has finite Gelfand--Kirillov
dimension. Hence, $M(S,\tau s)$ has a group of quotients that is torsion-free and it is abelian-by-finite. The latter also has
been investigated in \cite{ESS}. Note that if $s=\id$ then $K[M(S,s)]=K\free{X}$, the free $K$-algebra in variables
in the set $X=\{x_s:x\in S\}$, while $K[M(S,\tau s)]=K[X]$, the polynomial algebra in commuting variables in $X$.
In case $(S,s)$ is an arbitrary finite, bijective and non-degenerate solution of the Yang--Baxter Equation, it has been
shown in \cite{JesKubVan,JesKubVan2} that the algebraic structure of $A(K,S,\tau s)$ determines when $s$ is involutive.

In this section we investigate the structure algebra $A(K,S,\tau s)=K[M(S,\tau s)]$ corresponding to an involutive finite solution $(S,s)$
of the PE. The description of the solutions obtained in the previous section allows to show that $M(S,\tau s)$ is a finite extension of a
free abelian submonoid and hence one obtains that $K[M(S,\tau s)]$ is a Noetherian PI-algebra.

Let $(S,s)$ be a solution of the PE. So, as before, we write $s(s,y)= (xy,\theta_{x}(y))$. To avoid possible confusion between
the multiplication in the semigroup $S$ and the product in $M(S,\tau s)$, we write the product in $M(S,\tau s)$ as $\circ$.
Thus \[M(S,\tau s)=\free{S\mid x\circ y=\theta_x(y)\circ xy\text{ for all }x,y\in S}.\] For $x\in S$ and $n\geq 1$
we shall write $x^{(n)}=x\circ\dotsm\circ x$ (the product of $n$ copies of $x$ in $M(S,\tau s)$) to distinguish this element
from the power $x^n$ of $x$ in $S$. Moreover, we put $x^{(0)}=1$, the identity element of $M(S,\tau s)$.

In order to state the main result of this section we recall some well-known notation. The rank $\rk S$ of a semigroup $S$
is defined as the supremum of the ranks of free abelian subsemigroups of $S$. The classical Krull dimension of an algebra $A$
is denoted by $\clK A$. Moreover, $\GK A$ stands for the Gelfand--Kirillov dimension of $A$. For more information and background
on ring theory we refer the reader to \cite{MR}.

\begin{thm}\label{algrep}
	Assume that $(S,s)$ is a finite involutive solution of the PE. Then the structure monoid $M=M(S,\tau s)$ is (free abelian)-by-finite.
	More precisely, there exist a free abelian submonoid $C\subseteq M$ of rank $n=|E(S)|\cdot|{\Ret(S,s)}|^{-1}$ and finite subsets
	$L,R\subseteq M$ such that $M=\bigcup_{f\in L}C\circ f=\bigcup_{f\in R}f\circ C$. In particular, if $K$ is a field then the structure
	algebra $A(K,S,\tau s)=K[M]$ is a module finite extension of the polynomial algebra $K[C]$ in $n$ commuting variables. Hence $K[M]$ is
	a Noetherian PI-algebra satisfying \[\clK K[M]=\GK K[M]=\rk M=n\leq|S|\] and the equality holds if and only if $s=\id$.
	\begin{proof}
		By Theorem~\ref{thm:main-prime} we may assume that $S=X\times A\times G$, where $X$ is a non-empty set,
		$(A,+)$ and $(G,\cdotp)$ are elementary abelian $2$-groups and \[s((x,a,g),(y,b,h))=((x,a,g\cdot h),(y,a+b,h)).\]
		Then $S$ is a semigroup with multiplication given by the formula $(x,a,g)\cdot(y,b,h)=(x,a,g\cdot h)$. Since
		$E(S)=X\times A\times\{1\}$ and $\Ret(S,s)=(A,t_A)$ (see Theorem~\ref{thm:main-prime}), we get $n=|E(S)|\cdot|{\Ret(S,s)}|^{-1}=|X|$.

		Write $X=\{x_1,\dotsc,x_n\}$. Since \[(x,a,g)\circ(y,b,h)=(y,a+b,h)\circ(x,a,g\cdot h)\] for all $x,y\in X$, $a,b\in A$
		and $g,h\in G$, it follows that each element $w\in M$ may be written in the following form $w=w_1\circ\dotsb\circ w_n$, where
		$w_i\in\free{(x_i,a,g):a\in A\text{ and }g\in G}\subseteq M$ for $1\leq i\leq n$. Moreover, writing $A=\{a_1,\dotsc,a_m\}$
		(clearly $m=2^r$ for some $r\geq 0$) and using the fact that \[(x,a,g)\circ(x,b,h)=(x,a+b,h)\circ(x,a,g\cdot h)=(x,b,h)\circ(x,a+b,g)\]
		for all $x\in A$, $a,b\in A$ and $g,h\in G$, we obtain that each $w_i$ may be written as $w_i=w_{i1}\circ\dotsb\circ w_{im}$,
		where $w_{ij}\in\free{(x_i,a_j,g):g\in G}$ for $1\leq j\leq m$. Finally, the identity
		\[(x,a,g)\circ(x,a,h)=(x,0,h)\circ(x,a,g\cdot h),\] holding for all $x\in X$, $a\in A$ and $g,h\in G$, yields
		$w_{ij}\in\free{(x_i,a_j,g):g\in G}\subseteq\bigcup_{g\in G}N\circ(x_i,a_j,g)$, where
		\[N=\free{(x,0,g):x\in X\text{ and }g\in G}\subseteq M.\]
		Furthermore, we have \[(x,0,g)\circ(y,b,h)=(y,b,h)\circ(x,0,g\cdot h)\] for all $x,y\in X$, $b\in A$ and $g,h\in G$.
		Replacing $g$ by $g\cdot h$ in the above formula, we get \[(y,b,h)\circ(x,0,g)=(x,0,g\cdot h)\circ(y,b,h)\] as well.
		Hence $w\circ N=N\circ w$ for each $w\in M$, that is, $N$ is a normal submonoid of $M$. Therefore,
		\begin{align*}
			w_i & =w_{i1}\circ\dotsb\circ w_{im}\in\bigcup_{\substack{g_1,\dotsc,g_m\in G\\ \varepsilon_1,\dotsc,\varepsilon_m\in\{0,1\}}}
			N\circ(x_i,a_1,g_1)^{(\varepsilon_1)}\circ\dotsb\circ(x_i,a_m,g_m)^{(\varepsilon_m)},
		\end{align*}
		which leads to $w=w_1\circ\dotsb\circ w_n\in\bigcup_{f\in F_1}N\circ f$, where $F_1\subseteq M$ is a finite subset of $M$
		consisting of elements of the form
		\[(x_1,a_1,g_{11})^{(\varepsilon_{11})}\circ\dotsb\circ(x_1,a_m,g_{1m})^{(\varepsilon_{1m})}
		\circ\dotsb\circ(x_n,a_1,g_{n1})^{(\varepsilon_{n1})}\circ\dotsb\circ(x_n,a_m,g_{nm})^{(\varepsilon_{nm})},\]
		where $g_{ij}\in G$ and $\varepsilon_{ij}\in\{0,1\}$ for $1\leq i\leq n$ and $1\leq j\leq m$
		(in particular, $|F_1|\leq(2|G|)^{nm}$).
		Thus we have proved that
		\begin{equation}\label{sum-1}
		M=\bigcup_{f\in F_1}N\circ f=\bigcup_{f\in F_1}f\circ N.
		\end{equation}
		Next, write $G=\{g_1,\dotsc,g_k\}$ (clearly $k=2^t$ for some $t\geq 0$). Since $w\circ N=N\circ w$ for each
		$w\in M$, we get that $N$ consists of elements of the form
		\[(x_1,0,g_1)^{(l_{11})}\circ\dotsb\circ(x_1,0,g_k)^{(l_{1k})}\circ\dotsb\circ(x_n,0,g_1)^{(l_{n1})}\circ\dotsb\circ(x_n,0,g_k)^{(l_{nk})},\]
		where $l_{ij}\geq 0$ for $1\leq i\leq n$ and $1\leq j\leq k$. But it is easy to see that $(x,0,g)^{(l+1)}=(x,0,1)^{(l)}\circ(x,0,g)$
		for each $l\geq 0$. Since $(x,0,g)\circ(y,0,1)=(y,0,1)\circ(x,0,g)$ for all $x,y\in X$ and $g\in G$, it follows that
		$N=\bigcup_{f\in F_2}C\circ f$, where \[C=\free{(x,0,1):x\in X}\subseteq\Cen(N),\]
		$\Cen(N)$ is the center of $N$, and $F_2\subseteq N$ is a finite subset consisting of elements of the form
		\[(x_1,0,g_1)^{(\varepsilon_{11})}\circ\dotsb\circ(x_1,0,g_k)^{(\varepsilon_{1k})}\circ\dotsb\circ
		(x_n,0,g_1)^{(\varepsilon_{n1})}\circ\dotsb\circ(x_n,0,g_k)^{(\varepsilon_{nk})},\]
		where $\varepsilon_{ij}\in\{0,1\}$ for $1\leq i\leq n$ and $1\leq j\leq k$ (in particular $|F_2|\leq 2^{nk}$). Therefore, we have
		\begin{equation}\label{sum-2}
		N=\bigcup_{f\in F_2}C\circ f=\bigcup_{f\in F_2}f\circ C.
		\end{equation}
		Combining \eqref{sum-1} and \eqref{sum-2} we conclude that $M=\bigcup_{f\in L}C\circ f=\bigcup_{f\in R}f\circ C$, where
		\[L=\{f_2\circ f_1:f_1\in F_1\text{ and }f_2\in F_2\},\qquad R=\{f_1\circ f_2:f_1\in F_1\text{ and }f_2\in F_2\}.\]
		Clearly, $L,R\subseteq M$ are finite subsets of $M$. Moreover, because $(x,0,1)\circ(y,0,1)=(y,0,1)\circ(x,0,1)$
		for all $x,y\in X$, we conclude that $C$ is a free abelian monoid of rank $|X|=n$, which finishes the proof of first
		part of the theorem.

		Since $K[M]=\sum_{f\in L}K[C]\circ f=\sum_{f\in R}f\circ K[C]$ is a module finite extension of $K[C]\cong K[t_1,\dotsc,t_n]$,
		the polynomial algebra in $n$ commuting variables, it follows that $K[M]$ is a Noetherian PI-algebra (see \cite{MR}).
		Hence, a result of Ananin \cite{Ana1989} implies that $M$ is a linear monoid and then
		\cite[Proposition~1, p.~221, Proposition~7, p.~280--281, and Theorem~14, p.~284]{Okn1991} yield
		\[\clK K[M]=\GK K[M]=\rk M=n\leq|S|.\]

		Clearly, we have equality in the above if and only if $S=X$ (i.e., $A=G=0$, the trivial group). Hence, the last
		part of the theorem follows as well.
	\end{proof}
\end{thm}

Note that if $(S,s)$ is an infinite solution of the PE then the structure algebra $A=A(K,S,\tau s)$ is not
Noetherian as the ideal $I=\sum_{x\in S}AxA$ is not finitely generated neither as a left nor right ideal.

As an illustration  of Theorem~\ref{algrep} we give two applications in case $(S,s)$ is a finite involutive solution
of the PE based on a left zero semigroup (i.e., $G=0$; see Theorem~\ref{thm:main2}). First we show that the prime ideals
of the structure algebra $A=A(K,S,\tau s)$ over a field $K$ are very special and second we show that certain ring-theoretical
properties of $A$ encode information about the map $s$.

\begin{pro}\label{alg-1}
	Assume that $(S,s)$ is a finite involutive solution of the PE, where $S$ is a left zero semigroup. Let $M=M(S,\tau s)$.
	If $P$ is a prime ideal of the structure algebra $A(K,S,\tau s)=K[M]$ over a field $K$ then the algebra $K[M]/P$ is a commutative
	affine domain. In particular, each prime ideal of $K[M]$ is completely prime and semiprimitive and $K[M]$ is a Jacobson ring.
	\begin{proof}
		We use the same notation as in Theorem~\ref{algrep} and its proof. Because $S$ is a left zero semigroup, it easily
		is verified that each element $w$ of $M$ is normal in $M$, i.e., $w\circ M=M\circ w$. Let $P$ be a prime ideal of $K[M]$.
		Since $x\circ x=\theta_x(x)\circ x$ for each $x\in S$, we get $(x-\theta_x(x))\circ x=0$ in $K[M]$. Thus, by normality of
		elements of $M$, it follows that $(x-\theta_x(x))\circ K[M]\circ x=0$. This yields $x-\theta_x(x)\in P$ or $x\in P$.
		Define $X=S\cap P$ and $Y=S\setminus X$. By what we have already shown, it follows that $y-\theta_y(y)\in P$ for each $y\in Y$.
		Moreover, if $x,y\in Y$ then $\theta_x(y)\in Y$. Indeed, otherwise $x\circ y=\theta_x(y)\circ x\in P$ and thus, again by normality
		of elements of $M$, we get $x\in P$ or $y\in P$, a contradiction. Since $\theta_y(y)\circ x=x\circ\theta_y(y)$, we obtain
		$(\theta_x(y)-\theta_y(y))\circ x=x\circ(y-\theta_y(y))\in P$, which implies $\theta_x(y)-\theta_y(y)\in P$ because $x\notin P$.
		Thus, in consequence, $y-\theta_x(y)=(y-\theta_y(y))-(\theta_x(y)-\theta_y(y))\in P$ for all $x,y\in Y$.
		Therefore, if $P_0$ is the ideal of $K[M]$ generated by $X$ and elements of the form $y-\theta_x(y)$ for all $x,y\in Y$
		then $P_0\subseteq P$ and
		\[K[M]/P_0\cong\frac{K\free{Y\mid x\circ y=\theta_x(y)\circ x\text{ for all }x,y\in Y}}{(y-\theta_x(y):x,y\in Y)}
		\cong K[t_1,\dotsc,t_k],\]
		the polynomial algebra in $k$ commuting variables, where $k$ is equal to the number of orbits of the action of the
		group $\{\theta_x:x\in Y\}\subseteq\Sym(Y)$ on the set $Y$. In particular, the ideal $P_0$ is prime and semiprimitive
		(note also that if $P$ is a minimal prime ideal then $P=P_0$ and we have a description of $P$ in terms of generators).
		Since $K[M]/P$ is a homomorphic image of the commutative affine algebra $K[M]/P_0$, it follows that the ideal $P$
		is completely prime and semiprimitive (by Nullstellensatz, see \cite{MR}). Hence the fact that $K[M]$ is a Jacobson ring
		also follows.
	\end{proof}
\end{pro}

It is worth to add that (under the assumptions and notation from Proposition~\ref{alg-1}) the Jacobson radical $J=\mathcal{J}(K[M])$ of $K[M]$
coincides with the prime radical $\mathcal{B}(K[M])$ of $K[M]$ and that it is a nilpotent ideal, say of index $n$. But then the fact that the
algebra $K[M]/J$ is commutative leads to a conclusion that $(xy-yx)^n$ is a polynomial identity satisfied by $A(K,S,\tau s)=K[M]$.

As a consequence of Proposition~\ref{alg-1} we get the following result.

\begin{cor}\label{alg-2}
	Assume that $(S,s)$ is a finite involutive solution of the PE, where $S$ is a left zero semigroup. Let $M=M(S,\tau s)$.
	If $K$ is a field then the structure algebra $A(K,S,\tau s)=K[M]$ is a domain if and only if it is prime if and only if $s=\id$.
	\begin{proof}
		Clearly $s=\id$ implies that $K[M]$ is a domain (because in this case $K[M]\cong K[t_1,\dotsc,t_n]$, the polynomial algebra
		in $n=|S|$ commuting variables), and if $K[M]$ is a domain then $K[M]$ is prime. So, it remains to show that
		primeness of $K[M]$ implies that $s=\id$ or, equivalently, that $\theta_x=\id$ for each $x\in S$. But if $K[M]$ is prime,
		Proposition~\ref{alg-1} implies that $K[M]$ is a commutative domain. In particular, $M$ is a commutative and cancellative monoid.
		But then $x\circ y=\theta_x(y)\circ x=x\circ\theta_x(y)$ for $x,y\in S$ implies $\theta_x(y)=y$ in $M$. Since the defining
		relations of $M$ are quadratic, it follows that $S$ embeds in $M$ and thus $\theta_x(y)=y$ in $S$, which finishes the proof.
	\end{proof}
\end{cor}

We finish with an example that shows that the previous corollary cannot be extended to the semiprime case.
Indeed, we give an example that is semiprime (or equivalently semiprimitive) but not a domain.

\begin{exa}
	Let $S=C_2$ be the additively written cyclic group of order two and $s(x,y)=(x,x+y)$ for $x,y\in S$. Then
	\[M=M(S,\tau s)=\free{x,y\mid x\circ x=x\circ y=y\circ x}\] and thus, for a field $K$, we get $A(K,S,\tau s)=K[M]\cong K[u,v]/(u^2-uv)$.
	Denoting the images of variables $u,v$ in $R=K[u,v]/(u^2-uv)$ again by $u,v$, we get $R=K[v]\oplus uK[v]$. Suppose that $r\in R$
	satisfies $r^2=0$. Write $r=a+ub\in R$ for some $a,b\in K[v]$. Then $0=r^2=a^2+u(2ab+b^2v)$ implies $a^2=0$ and $2ab+b^2v=0$.
	Thus $a=0$ and then $b^2v=0$, which leads to $b=0$ (because $K[v]$, as a polynomial algebra, is a domain). Therefore,
	$r=0$ and thus the algebra $R\cong K[M]$ is reduced. Since $K[M]$ is commutative, it is semiprime but not a domain.
\end{exa}

redpajama_set_name: RedPajamaArXiv


This article will certainly tell you some good news!
Namely, a recent medical research pointed out the existence of 7 ordinary, and what's more- delicious, foods and drinks which can effectively combat cancer!
These beneficial foods do not feed cancer cells, and on the contrary, they prevent the development of tumors. Nevertheless, these foods and drinks are probably some of your favorite!
They include red wine, tomatoes, blueberries, curry, black chocolate and green tea. Moreover, experts even believe that they are even more powerful than chemo!
So, which hidden qualities give them this power?
The role of activators and inhibitor molecules is the regulation of this process. In most cases, the inhibitors are more important, but then the activators of angiogenesis become necessary in order to stimulate the increased vascular cell growth for the formation of new blood vessels.
To be more specific, what these 7 foods do is impeding the blood supply to the tumor cells, since they have anti-angiogenesis properties.
Namely, they need additional blood vessels in order to be supplied with nutrients and spread throughout the body (metastasis), through the blood. This means that their growth is conditioned by the tumor angiogenesis.
Fully- established cancer cells release their activators of angiogenesis – molecules that "cheat" the body and trigger the growth of new blood vessels that supply the tumor to nutrients and oxygen.
For this reason, the modern medicine has created drugs called inhibitors of angiogenesis, and their function is completely opposite to chemotherapy, as they do not strike cancer cells and prevent the growth of blood vessels to the tumor.
Regarding all this, scientists suggest that the best way is to consume these 7 foods and thus regulate angiogenesis. By doing so, you will prevent the growth and spread of the microscopic tumors.
Tomatoes are one of the most powerful foods for stopping angiogenesis. In fact, according to a Harvard study, people who consumed cooked tomato or tomato sauce more than four times a month, had up to 50% reduction in the risk of prostate cancer.
This is the explanation: tomatoes are a rich source of lycopene. This substance is anti-angiogenic, meaning that it dissolves in the fat and is thus absorbed through the digestive tract.
Most importantly, contrary to vitamin C, the amount of lycopene is increased when these foods are processed under high temperature or cooked with oil.
Furthermore, dishes with curries contain other beneficial substances as well.
Numerous studies claim that resveratrol kills bacteria, viruses and fungi, prolongs the life of animals, it improves the energy production of cells, glucose tolerance to diabetes, eliminates the dangerous free radicals, prevents cell damages caused by nuclear radiation, improves heart function, repairs the damaged DNA, it improves the physical and the mental status and also the concentration.
According to some testimonies red wine kill bacteria, viruses and fungi, it inhibits cancer, increases lifespan in animals, it increases it.
Only 225 ml of red wine delivers around 640 mcg of resveratrol. Resveratrol supplements (often combined with extracts from grape or other antioxidants) are usually taken in doses of 200-600 mcg per day.
Research suggests that some types of red wines such as, Pinot Noir and Bordeaux, contain larger amounts of this valuable polyphenol.


Landscaping in downtown Glen Ellyn is easier than you think. The lush plants, flowers and greenery available and native to the midwest climate are plentiful and beautiful. If you are looking for curb appeal or to accent your new patio or walk ways, Kozak Custom Landscapes can help. Our landscaping staff can help you plan, organize, shop, prep, and plant all of your flowers, bushes, trees, plants and fillers. If you prefer a more minimal clean landscape, our landscaping staff will absolutely do mulches and stone fillers to break up the plantings.
Most of what goes into good landscaping is upkeep and maintenance. Clearing weeds and fertilization will help keep a great landscape design looking wonderful for years when done regularly. The initial prep and planting of a full landscaping design can be done in a matter of days or over a few weekends if needed. Our landscaping staff is flexible and can work with you around your schedule if you want to be as involved as possible with the landscaping project. Choosing the plants, flowers, trees, or shrubbery can be done electronically, at a nursery, or even with a visit to the Arboretum just down the road from Glen Ellyn.
When it comes to landscaping, most people have a hard time visualizing a finished concept so we can provide photos of other projects to give inspiration. Every landscaping project will be custom to your home and yard but we can recommend materials and plants that work well together and are classic in all gardens. Please feel free to ask our landscapers questions about maintaining, feeding and watering schedules for best results. Since Glen Ellyn does experience all four seasons, be sure to ask about winterizing and spring fertilization for your lawns, plants, and flowers.
Please remember, landscaping can be an accent or a focal point on your property. Kozak Custom Landscapes also specializes in hardscapes, outdoor kitchen areas, grading and excavation, and drainage & irrigation systems to help improve your landscaping. You can bring new life to the entire exterior of your Glen Ellyn property with just one call to Kozak Custom Landscapes.


Wat een leuk lijstje en de kleuren zijn erg mooi gekozen.
Super gorgeous. Love your colours and design and image is just so beautiful. Nice work.
This is so beautiful Thank you for sharing with us over here today at the Classic Open Design Team Challenge and best of luck with your entry.
Love your take on our color challenge! Thanks for playing with us at Crafting When We Can!


Beerfoot Brewery features over 100 craft and mainstream beers, full bar, small brewery producing a Beer of the Month, and a beer enthusiast club appropriately named "Beer Nuts."
The wildly-successful brewery offers homemade beer by the pint, and by the the growler to go.
Have your wedding at Haak Winery and let them cater to your every need.
Texas Tail Distillery has three products that are distilled right on the island: Texas Tail Vodka, Coastline Whiskey, and multiple flavors of their Southern Marsh Moonshine.


text:
News Norwich City F.C. Things to do Lifestyle Business Support Us
Norwich Evening News > News
Temporary mortuary being used at former air base
Published: 11:08 AM January 13, 2021 Updated: 3:11 PM January 13, 2021
The main entrance to Scottow Enterprise Park at the former RAF Coltishall site. Picture: Neil Perry / Archant - Credit: Archant
A temporary mortuary which was set up - but not required - during the first wave of the Covid-19 pandemic has now been forced into use, it has been confirmed.
In early stages of the pandemic, Norfolk County Council set up the temporary facility in a former air hangar at Scottow to cope with an expected increase in the rate of deaths in the region's hospitals.
Council bosses had always hoped the facility would not be needed and during the first wave of the pandemic, it was not.
However, it has now been confirmed that the Norfolk and Norwich University Hospital has recently started making use of the facility, in a stark reminder of the severity of the pandemic.
Tom McCabe, head of paid service at Norfolk County Council is leading the local response to the coronavirus crisis. Photo: Archant - Credit: Archant Norfolk
Tom McCabe, chairman of Norfolk's Covid-19 strategic co-ordination group, said: "This short-term mortuary at the 600-acre Scottow Enterprise Park site in the village of Badersfield, was put in place to ensure that we treat people who have died with the dignity and respect they deserve in these exceptional circumstances for us all.
"It was always anticipated that during challenging periods there would be extra pressures on mortuaries, undertakers and crematoria."
An NNUH spokesman said it had been called into use due to a combination of factors - the Covid-19 pandemic, existing winter pressure and as a temporary measure while new equipment is installed on site.
NNUH chief of division clinical support services Dr Richard Goodwin, said: "Our priority is to always ensure that the deceased are treated with respect and dignity, and therefore we work with our partners to ensure every year we can use additional mortuary space, if needed."
Mr McCabe added: "This temporary mortuary provides additional capacity to help make sure the county's hospitals have enough flexibility of space in their own mortuaries and to ensure we can provide the most respectful and dignified way to look after both those who have died and their families, over this difficult periods.
2 In photos: Norwich transformed but deserted in lockdown snowfall
5 Drag Race star kicks off BBC show stint with Norwich City theme
6 Pizza and Yorkshire pudding wrap takeaway opening in Norwich
8 'Extraordinary' outbreak of Covid in Norwich prison
9 'Village would be worse without it' - Owner on plans for 17th century pub
10 Drivers face non-essential travel fines after spate of snow crashes
"Anyone whose loved one is moved to the Scottow location will be informed and we can reassure people that we have a dedicated, trained, team of staff who care for those who have died and been taken there. The temporary mortuary also received a blessing from the Bishop of Norwich back in April last year, with him making a prayer of dedication in a special address from his home."
The existing mortuary at the NNUH is also currently in use.
Staff lose jobs at retailer Outfit with plans to close permanently
Full list of Norwich Market stalls open or delivering during lockdown
£250,000 of cannabis found in two cars on A11
Military personnel deployed to help N&N cope with Covid pressures

redpajama_set_name: RedPajamaCommonCrawl


MTW 3-4pm in Maxey 313, or by request. I will also try to have informal "office hours" in the Prentiss dining hall at 6pm on Tuesdays.
Taught Fall Semester of 2004 at Whitman College; note that this is an archive page and that some links may not work.
Note that some of the readings below link to restricted access websites such as JSTOR. Whitman folks should be able to access these from any computer on campus, or from any computer off campus once properly configured with the Whitman proxy server.
Monday (Part II): Refresher on consumer preferences and utility functions; here are some class notes (.pdf). If you could use a refresher on calculus and derivatives, try this excerpt (.pdf) from some of my class notes.
Here is Problem Set #1, on calculus, cost-minimization, and supply curves. You do not have to turn in the homework, or even do the homework, but you should expect similar problems on the exam. I'm happy to talk with you about the problem set; you can also try to make sense of the answer key.
Monday (Part I): Uncertainty and certainty equivalence. Here are some lecture notes.
This article basically started the field of health economics.
Kenneth Arrow, still alive and kicking at Stanford, shared the 1972 Nobel prize in economics (with John Hicks, a.k.a. Sir John "Hicksian demand curve" Hicks) "for their pioneering contributions to general economic equilibrium theory and welfare theory." The press releases on the Nobel prize website (linked above) are worth a read to get a sense of economic history &etc. Just about anybody who's anybody in economics has won the Nobel prize (which makes some folks wonder if they should stop giving it out or turn it into a prize for social sciences more broadly).
Do pay special attention to the material on pp. 941-943 about the Fundamental Welfare Theorems (which Arrow calls the Optimality Theorems). This is what won Arrow a share of the 1972 Nobel prize, and it also won Gerard Debreu (who worked both independently and in collaboration with Arrow) the 1983 Nobel prize in economics "for having incorporated new analytical methods into economic theory and for his rigorous reformulation of the theory of general equilibrium", and I'll do my best to stop you from graduating if you don't internalize it.
Also pay special attention to the potential political implications of Arrow's article. Consider the year of publication (1963) and think about what was happening in U.S. politics around that time.
Wednesday (Part II): Dennis S. Lees and Robert G. Rice, "Uncertainty and the Welfare Economics of Medical Care: Comment", The American Economic Review, Vol. 55, No. 1/2. (Mar. 1965), pp. 140-154. And: Kenneth J. Arrow, "Uncertainty and the Welfare Economics of Medical Care: Reply (The Implications of Transaction Costs and Adjustment Lags)", The American Economic Review, Vol. 55, No. 1/2. (Mar. 1965), pp. 154-158. For the Lees and Rice article, you should skim sections II and III, with a goal of getting the flavor of their argument. The rest of it, and Arrow�s response, make pretty straight-forward reading; during your reading you should think about the political implications of what the authors are writing.
Monday (Part I): Moral hazard and the prisoners' dilemma.
Monday (Part II): Mark V. Pauly, "The Economics of Moral Hazard: Comment", The American Economic Review, Vol. 58, No. 3, Part 1. (Jun. 1968), pp. 531-537. And: Kenneth J. Arrow, "The Economics of Moral Hazard: Further Comment", The American Economic Review, Vol. 58, No. 3, Part 1. (Jun., 1968), pp. 537-539.
Here is Problem Set #2 on expected value and insurance. You do not have to turn in the homework, or even do the homework, but you should expect similar problems on the exam. I'm happy to talk with you about the problem set; you can also try to make sense of the answer key.
Monday: George A. Akerlof, "The Market for "Lemons": Quality Uncertainty and the Market Mechanism", The Quarterly Journal of Economics, Vol. 84, No. 3. (Aug., 1970), pp. 488-500. (Do not get bogged down in the mathematical example that Akerlof does; instead, I recommend the example on p. 190 of the textbook.) Akerlof shared the 2001 Nobel Prize in economics (with A. Michael Spence and Joseph E. Stiglitz) "for their analyses of markets with asymmetric information"; the Nobel committee calls Akerlof's Lemons paper "the single most important study in the literature on economics of information". A historical side note: Akerlof's paper was rejected three times before it was finally published by the QJE. For details and other stories, read Joshua S. Gans and George B. Shepherd, "How Are the Mighty Fallen: Rejected Classic Articles by Leading Economists", The Journal of Economic Perspectives, Vol. 8, No. 1. (Winter, 1994), pp. 165-179.
Here is Problem Set #3 on demand curves. You do not have to turn in the homework, or even do the homework, but you should expect similar problems on the exam. I'm happy to talk with you about the problem set; you can also try to make sense of the answer key.
Wednesday: Michael Rothschild and Joseph Stiglitz, "Equilibrium in Competitive Insurance Markets: An Essay on the Economics of Imperfect Information", The Quarterly Journal of Economics, Vol. 90, No. 4. (Nov., 1976), pp. 629-649. Stiglitz shared the 2001 Nobel Prize in economics (with George A. Akerlof and A. Michael Spence) "for their analyses of markets with asymmetric information". Rothschild did not get a part of the Nobel prize, but he did (like me) go to Reed College. [Note: This article is difficult. Here�s what you should do with it: First, read sections I.1 through I.5; pay particular attention to the definition of equilibrium in I.4. Throughout, you should focus on the ideas, but if the math makes sense then so much the better. (It should all be pretty familiar, just with different variables than you�re used to.) Next, read the first paragraph of section II, the one dealing with the three principal conclusions. Finally, having examined the set-up and the punch line, do your best to understand I.6, which is the heart of the article. We�ll talk about this a bunch in class, so don�t drive yourself to distraction over it.
Monday (Part I): Amos Tversky and Daniel Kahneman, "The Framing of Decisions and the Psychology of Choice", Science, New Series, Vol. 211, Issue 4481 (Jan. 30, 1981), pp. 453-458. This article is available from JSTOR, but only from the computers by the Penrose reserve desk. I have hard copies for anyone who needs one.
Monday (Part II): Matthew Rabin and Richard H. Thaler, "Anomalies: Risk Aversion", Journal of Economic Perspectives, Vol. 15, No. 1 (Winter, 2001), pp. 219-232. This is a fabulous article; it is not yet available on JSTOR, but I have placed a PDF in the Course Documents section of Blackboard.
These two articles fit into a general theme that was expressed nicely by Mark J. Machina (in a difficult optional reading, Choice Under Uncertainty: Problems Solved and Unsolved, Journal of Economic Perspectives, Vol. 1, No. 1 (Summer, 1987), pp. 121-154): "Fifteen years ago [in 1972], the theory of choice under uncertainty could be considered one of the 'success stories' of economic analysis: it rested on solid axiomatic foundations, it had seen important breakthroughs in the analytics of risk, risk aversion, and their applications to economic issues, and it stood ready to provide the theoretical underpinnings for the newly emerging 'information revolution' in economics. Today choice under uncertainty is a field in flux: the standard theory is being challenged on several grounds from both within and outside economics."
Kahneman (a psychologist!) shared the 2002 Nobel Prize in economics "for having integrated insights from psychological research into economic science, especially concerning human judgment and decision-making under uncertainty". Tversky almost certainly would have shared in the prize had he not died in 1996. (Nobel prizes are not awared posthumously.) Instead, Kahneman shared the Nobel prize with Vernon L. Smith, who won "for having established laboratory experiments as a tool in empirical economic analysis, especially in the study of alternative market mechanisms."
The semi-regular "Anomalies" column in the Journal of Economic Perspectives is sort of like the X-Files for economics. This particular column is based largely on the following article, which is an optional reading: Matthew Rabin, Risk Aversion and Expected-Utility Theory: A Calibration Theorem, Econometrica, Vol. 68, No. 5 (Sep., 2000), pp. 1281-1292. Another optional reading (I like the Science article better) is Daniel Kahneman, Jack L. Knetsch, and Richard H. Thaler, Anomalies: The Endowment Effect, Loss Aversion, and Status Quo Bias, Journal of Economic Perspectives, Vol. 5, No. 1 (Winter, 1991), pp. 193-206.
A side comment on the Allais paradox: note that Maurice Allais, who won the 1988 Nobel Prize in economics "for his pioneering contributions to the theory of markets and efficient utilization of resources", was also responsible for an ongoing astronomical puzzle known as the Allais Effect. See An Invisible Hand?, The Economist, August 19, 2004.
Wednesday: We'll discuss the reading from John A. Nyman, The Theory of Demand for Health Insurance (Stanford Univ. Press, 2003). You can pick up a copy outside my door (Maxey 313) if you didn't get the handout in class. The first chapter in the reading (chapter 2) provides an excellent overview of the conventional approach to insurance, so it should be a pretty quick read; pay particular attention to the anomalies discussed at the end of the chapter. The second chapter in the reading (chapter 3) is just there for you to skim so that you can see what a cutting-edge/alternative approach to insurance theory looks like.
Question: What anomalies does Nyman identify and how serious of a threat do you think they pose to conventional theory?
Question: On another heretical (but otherwise unrelated) issue, Victor R. Fuchs, one of the giants in the field of health economics, has this to say on p. 135 of Who Shall Live? Health, Economics, and Social Choice (expanded edition, World Scientific: 1998): "It should be noted that although major-risk [i.e., catastrophic] insurance in various forms is now available from private insurance companies, the demand for it is less than overwhelming. If major-risk insurance is really what people desire in the way of medical care coverage, why don't they buy it now? And why do union leaders and representatives of other groups seek more coverage? I believe one reason is because people want an easy, convenient, systematic way of paying for medical care. It is a great mistake to view the purchase of health insurance as simply the result of the desire to avoid risk." What do you make of this argument?
Today (Wednesday, October 13) is the last day to withdraw from classes or the College without record.
Monday: Pharmaceuticals (class presentations): Keith, Laura B., Laura M.
Today (Monday, November 8) is the last day to withdraw from classes or the College with W grades; today is also the close of the P-D-F registration period.
Wednesday: Class visit from Sarah Michelson, Whitman alum who worked at Amgen!
Wednesday: Katie's presentation; class feedback; class visit from Cindy Waring to discuss Whitman's health care policy.
Monday: Social Security and Medicare overview.
Wednesday: Topic #2 (Laura M/Suraya v. Bill/Derek on Wednesday, December 1): Resolved, that the United States should adopt the Kerry/Edwards plan for catastrophic coverage as a step towards a government-centered health care system such as Canada�s single-payer system.
Monday: Debates #3 (Grant/Keith v. Julia/Shawnee on Monday, December 6): Resolved, that the United States should adopt the Bush/Cheney plan for Medical Savings Accounts (MSAs) as a step towards an individual-centered health care system.
Wednesday: Debate #4 (Spencer/Andrew v. Kathryn/Will on Wednesday, December 8): Resolved, that the United States should make prescription drugs significantly more affordable for third-world countries by changing patent laws, international treaties, and/or providing subsidies.
Topic #1 (Laura B/Jesse v. Quincy/Rusty on Monday, November 29): Resolved, that the state of Washington prohibit smoking inside buildings and vehicles open to the public and places of employment.
Topic #5 (Brett/Katie v. Jill/Jeremy on Monday, December 13 at 9am): Resolved, that the United States should take steps to significantly reduce the rate of growth of Medicare spending.
By the beginning of week 3: Pick a topic.
By the beginning of week 4: Complete background reading and other groundwork and track down secondary readings from the library/Summit/etc as necessary.
By the beginning of week 5: Complete secondary readings, write your anotated bibliography, and draft the outline of your presentation.
By the beginning of week 6: Finish the outline of your presentation, determine the readings you�ll assign, and meet with me to discuss your plan.
By the beginning of week 7: Put the finishing touches on everything and be ready to go!
Below are some ideas and leads in the area of pharmaceuticals; I'm sure there are lots that I'm leaving out. Pick a topic that looks interesting to you and then email me to let me know what it is. First come, first served.
Pharmaceutical pricing in the developed world. Here's a curious fact for you: it turns out that when brand-name prescription drugs go off-patent, meaning that other companies can make generic versions, the company making the brand-name drug often raises its price. What's going on there? [Spencer] And while we're at it: Are drugs really cheaper in Canada or Europe than in the U.S.? [Jill] And how do HMOs use drug formularies to bargain with manufacturers? (For that matter, how do HMOs decide which drugs to cover?) [Jeremy] And how does the CPI (Consumer Price Index, a measure of inflation) deal with prescription drugs? [Brett] Two good articles to start out with (both in the Course Documents section of Blackboard) are (1) Ernst R. Berndt, "Pharmaceuticals in U.S. Health Care: Determinants of Quantity and Price", Journal of Economic Perspectives, 16(4):45-66 (2002) and (2) Nancy T. Gallini, "The Economics of Patents: Lessons from Recent U.S. Patent Reform", Journal of Economic Perspectives, 16(2):131-154 (2002).
Topic #2 (Laura M/Suraya v. Bill/Derek on Wednesday, December 1): Resolved, that the United States should adopt the Kerry/Edwards plan for catastrophic coverage as a step towards a government-centered health care system such as Canada�s single-payer system.
Topic #3 (Grant/Keith v. Julia/Shawnee on Monday, December 6): Resolved, that the United States should adopt the Bush/Cheney plan for Medical Savings Accounts (MSAs) as a step towards an individual-centered health care system.
Topic #4 (Spencer/Andrew v. Kathryn/Will on Wednesday, December 8): Resolved, that the United States should make prescription drugs significantly more affordable for third-world countries by changing patent laws, international treaties, and/or providing subsidies.
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