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Much study has been done on semigroups which are unions of groups. There are several ways in which a union of groups can be made into a semigroup in which each of the component groups arises as subgroups of the constructed semigroup. An…

Group Theory · Mathematics 2024-02-16 A. R. Rajan , S. Sheena , C. S. Preenu

Given a semigroup $S$, a diagonal subsemigroup $\rho$ is defined to be a reflexive and compatible relation on $S$, i.e. a subsemigroup of the direct square $S\times S$ containing the diagonal $\{ (s,s)\colon s\in S\}$. When $S$ is finite,…

Rings and Algebras · Mathematics 2026-02-20 Callum Barber , Nik Ruškuc

An involution on a semigroup S (or any algebra with an underlying associative binary operation) is a function f:S->S that satisfies f(xy)=f(y)f(x) and f(f(x))=x for all x,y in S. The set I(S) of all such involutions on S generates a…

Group Theory · Mathematics 2015-09-16 James East , Thomas E. Nordahl

Assume $G\prec H$ are groups and ${\cal A}\subseteq{\cal P}(G),\ {\cal B}\subseteq{\cal P}(H)$ are algebras of sets closed under left group translation. Under some additional assumptions we find algebraic connections between the Ellis…

Logic · Mathematics 2023-08-24 Adam Malinowski , Ludomir Newelski

A subsemigroup S of a semigroup Q is a local left order in Q if, for every maximal subgroup H of Q, the intersection of S with H is a local left order in the sense of group theory. That is, every q in H can be written as a#b for some a,b in…

Rings and Algebras · Mathematics 2007-05-23 Victoria Gould

In this article, we show that a group $G$ is the union of two proper subsemigroups if and only if $G$ has a nontrivial left-orderable quotient. Furthermore, if $G$ is the union of two proper semigroups, then there exists a minimum normal…

Group Theory · Mathematics 2020-02-13 Casey Donoven

A finite group $G$ is called uniformly semi-rational if there exists an integer $r$ such that the generators of every cyclic sugroup $\langle x \rangle$ of $G$ lie in at most two conjugacy classes, namely $x^G$ or $(x^r)^G$. In this paper,…

Group Theory · Mathematics 2024-10-16 Marco Vergani

Let $\mathbf G$ be a connected reductive algebraic group over an algebraically closed field, and let $s\in\mathbf G$ be a semisimple element. We show that the centraliser of $s$ is the semi-direct product of its identity component by its…

Group Theory · Mathematics 2026-04-20 François Digne , Jean Michel

We study C*-algebras associated to right LCM semigroups, that is, semigroups which are left cancellative and for which any two principal right ideals are either disjoint or intersect in another principal right ideal. If $P$ is such a…

Operator Algebras · Mathematics 2014-09-05 Charles Starling

We prove that a nonempty subset $B$ of a regular hypersemigroup $H$ is a bi-ideal of $H$ if and only if it is represented in the form $B=A*C$ where $A$ is a right ideal and $C$ a left ideal of $H$. We also show that an hypersemigroup $H$ is…

General Mathematics · Mathematics 2015-12-01 Niovi Kehayopulu

In this paper, we apply the theory of algebraic cohomology to study the amenability of Thompson's group $\mathcal{F}$. We introduce the notion of unique factorization semigroup which contains Thompson's semigroup $\mathcal{S}$ and the free…

Operator Algebras · Mathematics 2022-02-08 Linzhe Huang

Let $\E$ be a finite dimensional Hilbert space. This note finds all factorizations of the right shift semigroup $\S^\E=(S_t^\E)_{t\ge 0}$ on $L^2(\R_+,\E)$ into the product of $n$ commuting contractive semigroups, i.e., characterizes all…

Functional Analysis · Mathematics 2026-02-03 Tirthankar Bhattacharyya , Shubham Rastogi , Kalyan B. Sinha , Vijaya Kumar U

We give a new definition of the semigroup C*-algebra of a left cancellative semigroup, which resolves problems of the construction by X. Li. Namely, the new construction is functorial, and the independence of ideals in the semigroup does…

Operator Algebras · Mathematics 2019-05-07 Marat Aukhadiev

Let $S$ be an algebraic semigroup (not necessarily linear) defined over a field $F$. We show that there exists a positive integer $n$ such that $x^n$ belongs to a subgroup of $S(F)$ for any $x \in S(F)$. In particular, the semigroup $S(F)$…

Algebraic Geometry · Mathematics 2013-07-19 Michel Brion , Lex E. Renner

We present a procedure to enumerate the whole set of numerical semigroups with a given Frobenius number F, S(F). The methodology is based on the construction of a partition of S(F) by a congruence relation. We identify exactly one…

Commutative Algebra · Mathematics 2011-05-26 V. Blanco , J. C. Rosales

We study group congruences on the semigroup $\boldsymbol{B}_{\omega}^{\mathscr{F}}$ and its homomorphic retracts in the case when an ${\omega}$-closed family $\mathscr{F}$ which consists of inductive non-empty subsets of $\omega$. It is…

Group Theory · Mathematics 2023-06-05 Oleg Gutik , Mykola Mykhalenych

When a semigroup has a unary operation, it is possible to define two binary operations, namely, left and right division. In addition it is well known that groups can be defined in terms of those two divisions. The aim of this paper is to…

Group Theory · Mathematics 2012-10-01 Joao Araujo , Michael Kinyon

Let $Q$ be an inverse semigroup. A subsemigroup $S$ of $Q$ is a left I-order in $Q$ and $Q$ is a semigroup of left I-quotients of $S$ if every element in $Q$ can be written as $a^{-1}b$, where $a, b \in S$ and $a^{-1}$ is the inverse of $a$…

Rings and Algebras · Mathematics 2022-05-04 Victoria Gould , Georgia Schneider

Here we characterize regular and completely regular ordered semigroups by their minimal bi-ideals. A minimal bi-ideal is expressed as a product of a minimal right ideal and a minimal left ideal. Furthermore, we show that every bi-ideal in a…

Rings and Algebras · Mathematics 2017-01-26 Kalyan Hansda

We call a semigroup $S$ f-noetherian if every right congruence of finite index on $S$ is finitely generated. We prove that every finitely generated semigroup is f-noetherian, and investigate whether the properties of being f-noetherian and…

Group Theory · Mathematics 2020-02-13 Craig Miller