On the group of automorphisms of the Brandt $\lambda^0$-extension of a monoid with zero

TL;DR

This paper characterizes the automorphism group of the Brandt -extension of a monoid with zero, showing it is isomorphic to a homomorphic image of a structured product involving permutations, automorphisms, and units.

Contribution

It provides a detailed description of the automorphism group of the Brandt -extension, including an explicit isomorphism to a structured product involving known groups.

Findings

Automorphism group is isomorphic to a homomorphic image of a structured product.

Explicit binary operation defining the group structure.

Includes the group of all bijections of the index set and automorphisms of the monoid.

Abstract

The group of automorphisms of the Brandt $\lambda^0$-extension $B^0_\lambda(S)$ of an arbitrary monoid $S$ with zero is described. In particular we show that the group of automorphisms $\mathbf{Aut}(B_{\lambda}^0(S))$ of $B_{\lambda}^0(S)$ is isomorphic to a homomorphic image of the group defines on the Cartesian product $\mathscr{S}_{\lambda}\times \mathbf{Aut}(S)\times H_1^{\lambda}$ with the following binary operation: \begin{equation*} [\varphi,h,u]\cdot[\varphi^{\prime},h^{\prime},u^{\prime}]= [\varphi\varphi^{\prime},hh^{\prime},\varphi u^{\prime}\cdot uh^{\prime}], \end{equation*} where $\mathscr{S}_{\lambda}$ is the group of all bijections of the cardinal $\lambda$, $\mathbf{Aut}(S)$ is the group of all automorphisms of the semigroup $S$ and $H_1^{\lambda}$ is the direct $\lambda$-power of the group of units $H_1$ of the monoid $S$.