Transformation Semigroups Which Are Disjoint Union of Symmetric Groups

Abstract

Let $X$ be a nonempty set and $T(X)$ the full transformation semigroup on $X$. For any equivalence relation $E$ on $X$, define a subsemigroup $T_{E^{\ast }}(X)$ of $T(X)$ by $$T_{E^{\ast }}(X)=\{\alpha \in T(X):\text{for all}x,y\in X,(x,y)\in E\iff (x\alpha ,y\alpha )\in E\}.$$ We have the regular part of $T_{E^{\ast }}(X)$, denoted by $\mathrm{Reg}(T)$, is the largest regular subsemigroup of $T_{E^{\ast }}(X)$. Defined the subsemigroup $Q_{E^{\ast }}(X)$ of $T_{E^{\ast }}(X)$ by $$Q_{E^{\ast }}(X)=\{\alpha \in T_{E^{\ast }}(X):|A\alpha |=1\text{and}A\cap X\alpha \ne \varnothing \text{for all}A\in X/E\}.$$ Then we can prove that this subsemigroup is the (unique) minimal ideal of $\mathrm{Reg}(T)$ which is called the kernel of $\mathrm{Reg}(T)$. In this paper, we will compute the rank of $Q_{E^{\ast }}(X)$ when $X$ is finite and prove an isomorphism theorem. Finally, we describe and count all maximal subsemigroups of $Q_{E^{\ast }}(X)$ where $X$ is a finite set.