Left orders in completely \(0\)-simple semigroups
An element \(a\) of a semigroup \(S\) is said to be square-cancellable if, for all \(x,y\in S^1\), \(a^2x=a^2y\) implies \(ax=ay\) and \(xa^2=ya^2\) implies \(xa=ya\). Let \(S\) be a subsemigroup of \(Q\). Then \(Q\) is a semigroup of left quotients of \(S\) and \(S\) is a left order in \(Q\) if every square-cancellable element of \(S\) lies in a subgroup of \(Q\) and every \(q\in Q\) can be written as \(q=a^\# b\) where \(a\) is square-cancellable, \(b\in S\), and \(a^\#\) is the inverse of \(a\) in a subgroup of \(S\) which contains \(a\) [see \textit{J. Fountain, M. Petrich}, J. Algebra 101, 365-402 (1986; Zbl 0589.20041)]. A semigroup \(S\) with zero is called prime if for any \(a\neq 0\neq b\) in \(S\) there is an \(x\in S\) with \(axb\neq 0\). Let \(\lambda\) and \(\rho\) be binary relations on \(S\) such that \(a\lambda b\) iff \(Sa\cap Sb\neq 0\), and \(a\rho b\) iff \(aS\cap bS\neq 0\). Denote by \(\lambda^t\) and \(\rho^t\) the transitive closure of \(\lambda\) and \(\rho\), respectively. The main result says that a semigroup \(S\) with zero is a left order in a completely 0-simple semigroup iff (i) \(S\) is prime and \(S^1\) categorical at zero, (ii) there exists an \(a\in S\) such that \(aSa\) is a left order in a group with zero, (iii) if \(x(\lambda^t\cap\rho^t)y\) in \(S\) then, for any \(z\in S\), \(zx=zy\neq 0\) implies \(x=y\) and \(xz=yz\neq 0\) implies \(x=y\). This theorem is used to obtain several alternative characterizations of left orders in completely 0-simple semigroups.
scientific article; zbMATH DE number 895855