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diff --git a/library/algebra/semigroup.tex b/library/algebra/semigroup.tex new file mode 100644 index 0000000..e090a56 --- /dev/null +++ b/library/algebra/semigroup.tex @@ -0,0 +1,82 @@ +\import{algebra/magma.tex} + +\section{Semigroups} + +\begin{struct}\label{semigroup} + A semigroup $A$ is a magma such that + \begin{enumerate} + %\item for all $a, b, c\in \carrier[A]$ we have $\mul[A](a,\mul[A](b,c)) = \mul[A](\mul[A](a,b),c)$. + \item\label{semigroup_assoc} for all $a, b, c$ we have $\mul[A](a,\mul[A](b,c)) = \mul[A](\mul[A](a,b),c)$. + \end{enumerate} +\end{struct} + + + +\section{Regular semigroups} + + +\begin{struct}\label{regularsemigroup} + A regular semigroup $A$ is a semigroup such that + \begin{enumerate} + %\item for all $a\in \carrier[A]$ there exists $b\in\carrier[A]$ such that $\mul[A](a, \mul[A](b, a)) = a$. + \item\label{regularsemigroup_regular} for all $a$ there exists $b\in\carrier[A]$ such that $\mul[A](a, \mul[A](b, a)) = a$. + \end{enumerate} +\end{struct} + + + +\section{Inverse semigroups} + +\begin{struct}\label{inversesemigroup} + An inverse semigroup $A$ is a regular semigroup such that + \begin{enumerate} + \item\label{inversesemigroup_comm} for all $a,b\in\idempotents{A}$ we have $\mul[A](a, b) = \mul[A](b, a)$. + \end{enumerate} +\end{struct} + +\begin{proposition}\label{inversesemigroup_is_semigroup} + Suppose $A$ is an inverse semigroup. + Then $A$ is a semigroup. +\end{proposition} + + +\begin{proposition}\label{inversesemigroup_is_regularsemigroup} + Suppose $A$ is an inverse semigroup. + Then $A$ is a regular semigroup. +\end{proposition} + +\begin{proposition}\label{idempotentelems_eq_iff_orbits_eq} + Let $A$ be an inverse semigroup. + Let $e,f\in\idempotents{A}$. + % TODO use Orbits explicitly? + Suppose for all $x\in\carrier[A]$ there exists $y\in\carrier[A]$ + such that $x\cdot e = y\cdot f$. + Suppose for all $x\in\carrier[A]$ there exists $y\in\carrier[A]$ + such that $x\cdot f = y\cdot e$. + Then $e = f$. +\end{proposition} +\begin{proof} + Take $x, y\in\carrier[A]$ such that $e = x\cdot f$ and $f = y\cdot e$ by \cref{idempotents}. + % + \begin{align*} + e + &= x \cdot f + \explanation{by assumption}\\ + &= x\cdot (f\cdot f) + \explanation{by \cref{idempotents}}\\ + &= (x\cdot f)\cdot f + \explanation{by \cref{semigroup_assoc,inversesemigroup_is_semigroup}}\\ + &= e\cdot f + \explanation{by assumption}\\ + &= f\cdot e + \explanation{by \hyperref[inversesemigroup_comm]{commutativity of idempotent elements}}\\ + &= (y\cdot e)\cdot e + \explanation{by assumption}\\ + &= y\cdot (e\cdot e) + \explanation{by \cref{semigroup_assoc,inversesemigroup_is_semigroup}}\\ + &= y \cdot e + \explanation{by \cref{idempotents}}\\ + &= f + \explanation{by assumption} + \end{align*} +\end{proof} |
