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Let T_n be the full transformation semigroup of all mappings from the set {1,...,n} to itself under composition. Let E = E(T_n) denote the set of idempotents of T_n and let e be an arbitrary idempotent satisfying |im(e)|=r < n-1. We prove…

Group Theory · Mathematics 2014-02-26 Robert Gray , Nik Ruskuc

We develop some new topological tools to study maximal subgroups of free idempotent generated semigroups. As an application, we show that the rank 1 component of the free idempotent generated semigroup of the biordered set of a full matrix…

Group Theory · Mathematics 2013-03-26 Mark Brittenham , Stuart W. Margolis , John Meakin

Denote by $\mathcal T_n$ and $\mathcal S_n$ the full transformation semigroup and the symmetric group on the set $\{1,\ldots,n\}$, and $\mathcal E_n=\{1\}\cup(\mathcal T_n\setminus \mathcal S_n)$. Let $\mathcal T(X,\mathcal P)$ denote the…

Group Theory · Mathematics 2017-12-14 Igor Dolinka , James East

The study of the free idempotent generated semigroup $\mathrm{IG}(E)$ over a biordered set $E$ began with the seminal work of Nambooripad in the 1970s and has seen a recent revival with a number of new approaches, both geometric and…

Group Theory · Mathematics 2017-12-14 Igor Dolinka , Victoria Gould , Dandan Yang

We show that the rank r component of the free idempotent generated semigroup of the biordered set of the full linear monoid of n x n matrices over a division ring Q has maximal subgroup isomorphic to the general linear group GL_r(Q), where…

Group Theory · Mathematics 2014-03-10 Igor Dolinka , Robert Gray

Gray and Ruskuc have shown that any group G occurs as the maximal subgroup of some free idempotent generated semigroup IG(E) on a biordered set of idempotents E, thus resolving a long standing open question. Given the group G, they make a…

Rings and Algebras · Mathematics 2012-09-07 Victoria Gould , Dandan Yang

With each semigroup one can associate a partial algebra, called the biordered set, which captures important algebraic and geometric features of the structure of idempotents of that semigroup. For a biordered set $\mathcal{E}$, one can…

Group Theory · Mathematics 2022-10-07 Igor Dolinka

This paper investigates the maximal subgroups of a free projection-generated regular $*$-semigroup $PG(P)$ over a projection algebra $P$, and their relationship to the maximal subgroups of the free idempotent-generated semigroup $IG(E)$…

Group Theory · Mathematics 2025-07-10 James East , Robert D. Gray , P. A. Azeef Muhammed , Nik Ruskuc

We calculate the rank and idempotent rank of the semigroup $E(X,P)$ generated by the idempotents of the semigroup $T(X,P)$, which consists of all transformations of the finite set $X$ preserving a non-uniform partition $P$. We also classify…

Group Theory · Mathematics 2017-12-14 Igor Dolinka , James East , James D. Mitchell

We use topological methods to study the maximal subgroups of the free idempotent generated semigroup on a biordered set. We use these to give an example of a free idempotent generated semigroup with maximal subgroup isomorphic to the free…

Group Theory · Mathematics 2008-08-14 Mark Brittenham , Stuart W. Margolis , John Meakin

The set of idempotents of any semigroup carries the structure of a biordered set, which contains a great deal of information concerning the idempotent generated subsemigroup of the semigroup in question. This leads to the construction of a…

Group Theory · Mathematics 2019-01-11 Yang Dandan , Igor Dolinka , Victoria Gould

We prove the following results: (1) Every group is a maximal subgroup of some free idempotent generated semigroup. (2) Every finitely presented group is a maximal subgroup of some free idempotent generated semigroup arising from a finite…

Group Theory · Mathematics 2011-08-02 Robert Gray , Nik Ruskuc

We prove that a semi-direct product of two finite rank free groups $F_k$ and $F_n$ such that $F_k$ acts on $F_n$ by polynomially growing automorphisms acts properly isometrically on a finite dimensional CAT(0) cube complex provided some…

Group Theory · Mathematics 2023-03-09 François Gautero

Given an arbitrary group $G$ we construct a semigroup of idempotents (band) $B_G$ with the property that the free idempotent generated semigroup over $B_G$ has a maximal subgroup isomorphic to $G$. If $G$ is finitely presented then $B_G$ is…

Group Theory · Mathematics 2014-03-10 Igor Dolinka , Nik Ruškuc

We provide short and direct proofs for some classical theorems proved by Howie, Levi and McFadden concerning idempotent generated semigroups of transformations on a finite set.

Group Theory · Mathematics 2011-02-01 João Araújo

Let $\mathcal{S}$ be a finite cyclic semigroup written additively. An element $e$ of $\mathcal{S}$ is said to be idempotent if $e+e=e$. A sequence $T$ over $\mathcal{S}$ is called {\sl idempotent-sum free} provided that no idempotent of…

Combinatorics · Mathematics 2020-06-02 Guoqing Wang

We prove that all maximal subgroups of the free idempotent generated semigroup over a band B are free for all B belonging to a band variety V if and only if V consists either of left seminormal bands, or of right seminormal bands.

Group Theory · Mathematics 2014-03-10 Igor Dolinka

Free idempotent generated semigroups IG$(E)$, where $E$ is a biordered set, have provided a focus of recent research, the majority of the efforts concentrating on the behaviour of the maximal subgroups. Inspired by an example of Brittenham,…

Rings and Algebras · Mathematics 2015-01-06 Yang Dandan , Victoria Gould

Totally positive (TP) and totally nonnegative (TN) matrices connect to analysis, mechanics, and to dual canonical bases in reductive groups, by well-known works of Schoenberg, Gantmacher-Krein, Lusztig, and others. TP matrices form a…

Rings and Algebras · Mathematics 2026-01-19 Projesh Nath Choudhury , Shaun Fallat , Chi-Kwong Li

With every reduced $E$-Fountain semigroup $S$ which satisfies the generalized right ample condition we associate a category with zero morphisms $\mathcal{C}(S)$. Under some assumptions we prove an isomorphism of $\Bbbk$-algebras $\Bbbk…

Representation Theory · Mathematics 2025-02-18 Itamar Stein
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