Some fixedpoint results on (generalized) BruckReilly ∗extensions of monoids
 Eylem Guzel Karpuz^{1}Email author,
 Ahmet Sinan Çevik^{2},
 Jörg Koppitz^{3} and
 Ismail Naci Cangul^{4}
https://doi.org/10.1186/16871812201378
© Guzel Karpuz et al.; licensee Springer 2013
Received: 28 January 2013
Accepted: 21 March 2013
Published: 29 March 2013
Abstract
In this paper, we determine necessary and sufficient conditions for BruckReilly and generalized BruckReilly ∗extensions of arbitrary monoids to be regular, coregular and strongly πinverse. These semigroup classes have applications in various field of mathematics, such as matrix theory, discrete mathematics and padic analysis (especially in operator theory). In addition, while regularity and coregularity have so many applications in the meaning of boundaries (again in operator theory), inverse monoids and BruckReilly extensions contain a mixture fixedpoint results of algebra, topology and geometry within the purposes of this journal.
MSC:20E22, 20M15, 20M18.
Keywords
1 Introduction and preliminaries
In combinatorial group and semigroup theory, for a finitely generated semigroup (monoid), a fundamental question is to find its presentation with respect to some (irreducible) system of generators and relators, and then classify it with respect to semigroup classes. In this sense, in [1], the authors obtained a presentation for the BruckReilly extension, which was studied previously by Bruck [2], Munn [3] and Reilly [4]. In different manners, this extension has been considered as a fundamental construction in the theory of semigroups. In detail, many classes of regular semigroups are characterized by BruckReilly extensions; for instance, any bisimple regular wsemigroup is isomorphic to a Reilly extension of a group [4] and any simple regular wsemigroup is isomorphic to a BruckReilly extension of a finite chain of groups [5, 6]. After that, in another important paper [7], the author obtained a new monoid, namely the generalized BruckReilly ∗extension, and presented the structure of the ∗bisimple type A wsemigroup. Later on, in [8], the authors studied the structure theorem of the ∗bisimple type A ${w}^{2}$semigroups as the generalized BruckReilly ∗extension. Moreover, in a joint work [9], it has been recently defined a presentation for the generalized BruckReilly ∗extension and then obtained a GröbnerShirshov basis of this new construction. As we depicted in the abstract of this paper, BruckReilly, its general version generalized BruckReilly ∗extension of monoids and semigroup classes are not only important in combinatorial algebra but also in linear algebra, discrete mathematics and topology. So these semigroup classes, regular, coregular, inverse and strongly πinverse, are the most studied classes in algebra.
In this paper, as a next step of these above results, we investigate regularity, coregularity and strongly πinverse properties over BruckReilly and generalized BruckReilly ∗extensions of monoids. We recall that regularity and strongly πinverse properties have been already studied for some other special extensions (semidirect and wreath products) of monoids [10, 11]. We further recall that these two important properties have been also investigated for the semidirect product version of Schützenberger products of any two monoids [12, 13]. However, there are not yet such investigations concerning coregularity. As we depicted in the abstract, semigroup classes have important applications in various fields of mathematics, such as matrix theory, discrete mathematics and padic analysis (especially in operator theory). In addition, while regularity and coregularity have so many applications in the meaning of boundaries (again in operator theory), inverse monoids and BruckReilly extensions contain a mixture of algebra, topology and geometry within the purposes of this journal.
Now let us present the following fundamental material that will be needed in this paper. We refer the reader to [14–16] for more detailed knowledge.
An element a of a semigroup S is called regular if there exists $x\in S$ such that $axa=a$. The semigroup S is called regular if all its elements are regular. Groups are of course regular semigroups, but the class of regular semigroups is vastly more extensive than the class of groups (see [16]). Further, to have an inverse element can also be important in a semigroup. Therefore, we call S is an inverse semigroup if every element has exactly one inverse. The wellknown examples of inverse semigroups are groups and semilattices. An element $a\in S$ is called coregular and b its coinverse if $a=aba=bab$. A semigroup S is said to be coregular if each element of S is coregular [17]. In addition, let $E(S)$ and RegS be the set of idempotent and regular elements, respectively. We then say that S is called πregular if, for every $s\in S$, there is an $m\in \mathbb{N}$ such that ${s}^{m}\in RegS$. Moreover, if S is πregular and the set $E(S)$ is a commutative subsemigroup of S, then S is called strongly πinverse semigroup [16]. We recall that RegS is an inverse subsemigroup of a strongly πinverse semigroup S.
2 BruckReilly extensions of monoids
where $t=max(n,{m}^{\prime})$ and ${\theta}^{0}$ is the identity map on A, forms a monoid with identity $(0,{1}_{A},0)$. Then this monoid is called the BruckReilly extension of A determined by θ [2–4] and denoted by $\mathit{BR}(A,\theta )$.
In the above references, the authors used $\mathit{BR}(A,\theta )$ to prove that every semigroup embeds in a simple monoid, and to characterize special classes of inverse semigroups. In [[3], Theorem 3.1], Munn showed that $\mathit{BR}(A,\theta )$ is an inverse semigroup if and only if A is inverse. So, the following result is a direct consequence of this theorem.
Corollary 1 Let A be an arbitrary monoid. Then $\mathit{BR}(A,\theta )$ is regular if and only if A is regular.
with $t=max(m,{m}^{\prime})$ and $b=(a{\theta}^{tm})({a}^{\prime}{\theta}^{t{m}^{\prime}})\in A$, the set $\{(m,a,m)a\in A,m\in {\mathbb{N}}^{0}\}$ becomes a subsemigroup of $\mathit{BR}(A,\theta )$. Thus, we further have the following lemma.
Lemma 1 Let $(m,a,n)\in \mathit{BR}(A,\theta )$. If $(m,a,n)$ is coregular then $m=n$.
where $S=max({n}^{\prime},m)$ and ${S}^{\prime}=max(nm+S,{m}^{\prime})$. This gives $m={m}^{\prime}{n}^{\prime}n+m+{S}^{\prime}$, $n={n}^{\prime}{m}^{\prime}+{S}^{\prime}$, and consequently, ${n}^{\prime}={m}^{\prime}$. Together with $m+{m}^{\prime}=n+{n}^{\prime}$, we obtain $m=n$ as required. □
Now we can present the following result.
Theorem 1 Let A be a monoid. Then ${A}^{\prime}=\{(m,a,m)a\in A,m\in {\mathbb{N}}^{0}\}\le \mathit{BR}(A,\theta )$ is coregular if and only if A is coregular.
By (1) and (2), we clearly have ${m}^{\prime}=0$, and hence, $a{a}^{\prime}a=a$ and ${a}^{\prime}a{a}^{\prime}=a$. So A is coregular.
Therefore, ${A}^{\prime}=\{(m,a,m)a\in A,m\in {\mathbb{N}}^{0}\}\le \mathit{BR}(A,\theta )$ is coregular. □
In [[3], Theorem 3.1], it is proved that:

$(m,a,n)$ is an idempotent element in $\mathit{BR}(A,\theta )$ if and only if $m=n$ and a is an idempotent element in A.
This result will be used in the proof of the following theorem.
Theorem 2 $\mathit{BR}(A,\theta )$ is strongly πinverse if and only if A is regular and the idempotents in A commute.
By the assumption given in the beginning of this section, since aθ is in the ℋclass of the element ${1}_{A}$, we obtain aθ is a group element, and so there is an inverse element ${({(a\theta )}^{r1})}^{1}$. Thus, by (3), we get $a=a{(a\theta )}^{r1}{a}^{\prime}a$; in other words, $a\in RegA$. Consequently, A is regular. Now, let us also show that the elements in $E(A)$ are commutative. But this is quite clear by the fact that the idempotents in $\mathit{BR}(A,\theta )$ commute if and only if the idempotents in A commute (see [[3], Theorem 3.1(5)]).
Conversely, let us suppose that A is regular and the idempotents in A commute. Then $\mathit{BR}(A,\theta )$ is regular, where πregular by Corollary 1. Moreover, again by [[3], Theorem 3.1(5)], $E(\mathit{BR}(A,\theta ))$ is a commutative subsemigroup, which is required to $\mathit{BR}(A,\theta )$ satisfy strongly πinverse property, hence the result. □
3 The generalized BruckReilly ∗extension of monoids
Suppose that A is an arbitrary monoid having ${H}_{1}^{\ast}$ and ${H}_{1}$ as the ${\mathcal{H}}^{\ast}$ and ℋ classes containing the identity element ${1}_{A}$ of A. Moreover, let us assume that β and γ are morphisms from A into ${H}_{1}^{\ast}$ and, for an element u in ${H}_{1}$, let ${\lambda}_{u}$ be the inner automorphism of ${H}_{1}^{\ast}$ defined by $x\mapsto ux{u}^{1}$ such that $\gamma {\lambda}_{u}=\beta \gamma $.
where $d=max(p,{n}^{\prime})$ and ${\beta}^{0}$, ${\gamma}^{0}$ are interpreted as the identity map of A, and also ${u}^{0}$ is interpreted as the identity ${1}_{A}$ of A. In [8], Yu Shung and LiMin Wang showed that S is a monoid with the identity $(0,0,{1}_{A},0,0)$. In fact, this new monoid $S={\mathbb{N}}^{0}\times {\mathbb{N}}^{0}\times A\times {\mathbb{N}}^{0}\times {\mathbb{N}}^{0}$ is denoted by ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$ and called generalized BruckReilly ∗extension of A determined by the morphisms β, γ and the element u.
The following lemmas were established in [8].
Lemma 2 If $(m,n,v,p,q)\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$, then $(m,n,v,p,q)$ is an idempotent if and only if $m=q$, $n=p$ and v is idempotent.
if and only if ${v}^{\prime}$ is an inverse of v in A while ${m}^{\prime}=q$, ${n}^{\prime}=p$, ${p}^{\prime}=n$ and ${q}^{\prime}=m$.
Then we have an immediate consequence as in the following.
Corollary 2 Let A be a monoid. Then ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$ is regular if and only if A is regular.
In this section, we mainly characterize the properties coregularity and strongly πinverse over the generalized BruckReilly ∗extensions of monoids. More specifically, for a given monoid A, we determine the maximal submonoid of ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$, which can be held coregularity if A satisfies particular properties.
Our first observation is the following.
Lemma 4 The set $\mathcal{L}:=\{(m,n,v,n,m)v\in A,m,n\in {\mathbb{N}}^{0}\}$ is a submonoid of ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$.
Proof By considering the multiplication in (4), the proof can be seen easily. □
It turns out that all coregular elements in ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$ belong to the submonoid ℒ .
Lemma 5 Let $(m,n,v,p,q)\in {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$. If $(m,n,v,p,q)$ is coregular then $m=q$ and $n=p$.
By writing the equality (8) in (7), we get ${n}^{\prime}={p}^{\prime}$. Together with $n+{n}^{\prime}=p+{p}^{\prime}$, we obtain $n=p$. By assuming $q={m}^{\prime}$, we also get $m=q$.
Now let $(x,y,t,z,w)({x}^{\prime},{y}^{\prime},{t}^{\prime},{z}^{\prime},{w}^{\prime})=({x}^{\u2033},{y}^{\u2033},{t}^{\u2033},{z}^{\u2033},{w}^{\u2033})$. Then it is easy to verify that ${x}^{\u2033}\ge {x}^{\prime},x$ and ${y}^{\u2033}\ge {y}^{\prime},y$. If $w\ne {x}^{\prime}$, we can easily see that ${x}^{\u2033}>{x}^{\prime},x$ or ${y}^{\u2033}>{y}^{\prime},y$. This shows that $(m,n,a,p,q)({m}^{\prime},{n}^{\prime},{a}^{\prime},{p}^{\prime},{q}^{\prime})(m,n,a,p,q)\ne (m,n,a,p,q)$ if $q\ne {m}^{\prime}$ or ${q}^{\prime}\ne m$. Hence, $q\ne {m}^{\prime}$ or ${q}^{\prime}\ne m$ is not possible. □
Then we have the following result.
Theorem 3 Let A be a monoid. Then the submonoid ℒ of ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$ is coregular if and only if A is coregular.
By (9) and (10), we obtain ${n}^{\prime}=0$, and hence $v{v}^{\prime}v=v$ and ${v}^{\prime}v{v}^{\prime}=v$. So, A is coregular.
Hence, $\mathcal{L}\le {\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$ is a coregular monoid, as desired. □
In the final theorem, we consider strongly πinverse property.
Theorem 4 ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$ is strongly πinverse if and only if A is regular and the idempotents in A commute.
So, ${v}_{1}{v}_{2}={v}_{2}{v}_{1}$, as required.
Conversely, let us suppose that A is regular. Then ${\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u)$ is regular, where πregular by Corollary 2. Now we need to show that the elements in $E({\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u))$ commute. To do that, let us take $(m,n,e,n,m),({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})\in E({\mathit{GBR}}^{\ast}(A;\beta ,\gamma ;u))$, and thus $e{e}^{\prime}={e}^{\prime}e$ by Lemma 2. Now, by considering the multiplication $(m,n,e,n,m)({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})$ as defined in (4), we have the following cases.
Thus, $(m,n,e,n,m)({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})=({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})(m,n,e,n,m)$.
respectively. Since $(({u}^{n}(e\gamma ){u}^{n}){\gamma}^{{m}^{\prime}m1}){\beta}^{{n}^{\prime}},(({u}^{{n}^{\prime}}({e}^{\prime}\gamma ){u}^{{n}^{\prime}}){\gamma}^{m{m}^{\prime}1}){\beta}^{n}\in E(A)$, we clearly obtain $(m,n,e,n,m)({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})=({m}^{\prime},{n}^{\prime},{e}^{\prime},{n}^{\prime},{m}^{\prime})(m,n,e,n,m)$.
Hence, the result. □
Declarations
Acknowledgements
Dedicated to Professor Hari M Srivastava.
The second and fourth authors are partially supported by Research Project Offices (BAP) of Selcuk (with Project No. 13701071) and Uludag (with Project No. 201215 and 201219) Universities, respectively.
Authors’ Affiliations
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