Centrally clean elements and central Drazin inverses in a ring

Li Wende Chen Jianlong

(School of Mathematics, Southeast University, Nanjing 211189, China)

Abstract：Element a in ring R is called centrally clean if it is the sum of central idempotent e and unit u. Moreover, a=e+u is called a centrally clean decomposition of a and R is called a centrally clean ring if every element of R is centrally clean. First, some characterizations of centrally clean elements are given. Furthermore, some properties of centrally clean rings, as well as the necessary and sufficient conditions for R to be a centrally clean ring are investigated. Centrally clean rings are closely related to the central Drazin inverses. Then, in terms of centrally clean decomposition, the necessary and sufficient conditions for the existence of central Drazin inverses are presented. Moreover, the central cleanness of special rings, such as corner rings, the ring of formal power series over ring R, and a direct product ∏Rα of ring Rα, is analyzed. Furthermore, the central group invertibility of combinations of two central idempotents in the algebra over a field is investigated. Finally, as an application, an example that lists all invertible, central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements of the group ring Z2S3 is given.

Key words：centrally clean element; centrally clean ring; central Drazin inverse; central group inverse

In the research on ring theory, the cleanness of a ring is a basic but important topic. Clean rings originated from the study of exchange rings, which play an important role in the cancellation of modules. The interesting characterizations and properties of clean rings have motivated many scholars to conduct further investigations. The concept of clean rings was first introduced by Nicholson[1]in 1977. Subsequently, Nicholson et al.[2]proved that the linear transformation of a countable vector space over a division ring is clean. In 1999, Nicholson[3]introduced a strongly clean ring and presented some equivalent characterizations of strongly clean elements and rings. In 2001, Han et al.[4]investigated the cleanness of group rings, the ring of formal power series over a ring, and a direct product of rings. In 2011, Hiremath et al.[5]presented some characterizations of strongly clean rings. More details concerning the cleanness of the rings can be found in Refs.[6-12].

Throughout the paper,Rdenotes an associative ring with unity 1. The center ofRis denoted byC(R)={x∈R:ax=xafor alla∈R}. The elemente∈Ris considered idempotent ife2=e. In contrast, the elemente∈Ris considered central idempotent ife2=eande∈C(R). The symbolsE(R), CE(R),U(R), andJ(R) denote the sets of all idempotents, central idempotents, invertible elements, and Jacobson radicals ofR, respectively. Recall that the elementa∈Ris considered clean ifu∈U(R) ande∈E(R) exist such thata=u+e. The elementa∈Ris considered strongly clean ifu∈U(R) ande∈E(R) exist such thata=u+eandue=eu. In this case,a=e+uis considered a strongly clean decomposition.

Drazin[13]introduced the concept of pseudo-inverse (usually called Drazin inverse) in rings and semigroups. The elementa∈Ris considered a Drazin invertible ifx∈Rand the nonnegative integerkexist such thatxax=x,ax=xa,xak+1=ak. Suchxis unique if it exists and is considered the Drazin inverse ofa. The smallest nonnegative integerksatisfying the previously presented equations is called the Drazin index ofa. Ifk=1, thenxis considered the group inverse ofa.

Further research showed that there is a close connection between clean rings and Drazin inverses. For example, Zhu et al.[14]proved thata∈Ris a Drazin invertible if and only ifu∈U(R),e∈E(R) and the positive integernexist such thatan=u+eis a strongly clean decomposition andanR∩eR=0. Moreover, many scholars investigated the Drazin invertibility of combinations of idempotents. For instance, Liu et al.[15]analyzed this topic in complex matrices, i.e., Drazin invertibility ofaP+bQ+cPQ+dQP+ePQP+fQPQ+gQPQPfor idempotent complex matricesPandQunder the conditions (PQ)2=(QP)2. More details concerning Drazin inverses can be found in Refs.[16-26].

In 2019, to analyze the commutative properties of Drazin inverses (see Example 2.8 in Ref.[27]), Wu et al.[28]introduced the concept of central Drazin inverses.

Definition1[28]Elementa∈Ris considered a central Drazin invertible ifx∈Rand the nonnegative integerkexist such thatxax=x,xa∈C(R),xak+1=ak. Suchxis unique if it exists and is considered the central Drazin inverse ofa, denoted byac. The smallest nonnegative integerksatisfying the previously presented equations is still the Drazin index ofa, denoted by ind(a). Ifk=1, thenxis called the central group inverse ofa, denoted bya?.

In Ref.[28], a centrally clean element and a centrally clean ring are also introduced.

Definition2[28]Leta∈R. Ifu∈U(R) ande∈CE(R) exist such thata=u+e, thenais considered centrally clean. In this case,a=u+eis considered a centrally clean decomposition ofa. Thus, centrally clean is clean. If every element ofRis centrally clean, thenRis considered a centrally clean ring.

Moreover,a∈Ris a central Drazin invertible if and only ifu∈U(R),e∈CE(R) and the positive integernexist such thata=u+eis a centrally clean decomposition, andanR∩eR=0, or equivalently,u∈U(R),e∈CE(R) and the positive integernexist such thata=u+eis a centrally clean decomposition andaeis nilpotent. Subsequently, Zhao et al.[29]investigated the one-sided central Drazin inverses.

Motivated by the previous studies, we investigated centrally clean elements and central Drazin inverses inR. We first give an example and characterizations of centrally clean elements. Then, we analyze the properties of centrally clean rings and provide some equivalent characterizations. Moreover, we present the necessary and sufficient conditions for the existence of central Drazin inverses in terms of centrally clean decompositions. In addition, we investigate the central group invertibility of combinations of two central idempotents. Finally, we calculate all invertible, central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements of the group ringZ2S3.

1 Characterization of Centrally Clean Elements

First, we provide an example of centrally clean elements.

Example1

1) Units are centrally clean.

2) The elements inJ(R) are centrally clean.

3) Nilpotent elements are centrally clean.

4) Central idempotents are centrally clean.

Proof1) and 3) are obvious.

2) Letx∈J(R). Notably,J(R)={x∈R: 1-axis left invertible for anya∈R} andJ(R)={x∈R: 1-xais right invertible for anya∈R}.

Then, we takea=1, and it follows that 1-x∈U(R). Hence,xis centrally clean.

4) Lete∈CE(R). Given that (2e-1)2=1 and (1-e)2=1-e, it follows that 2e-1∈U(R) and 1-e∈CE(R). Then,e=(2e-1)+(1-e). Hence,eis centrally clean.

Leta∈R. Then, we useRaandaRto denote the left and right ideals generated bya, respectively. We usel(a) andr(a) to denote the left and right annihilators ofa, respectively. That is,

Ra={ra:r∈R},aR={ar:r∈R}

l(a)={x∈R:xa=0},r(a)={x∈R:ax=0}

Nicholson[3]proved that ife∈E(R) anda∈eReis strongly clean, thena∈Ris also strongly clean. Moreover, he provided some characterizations of strongly clean elements. Then, we investigate the relevant characterizations of centrally clean elements.

Lemma1[3]Leta∈Rande∈E(R) withea=ae. Then, the following conditions are equivalent:

1)ae∈U(eRe).

2)e∈Raandl(a)?l(e).

3)e∈aRandr(a)?r(e).

Theorem1Leta∈R. Then, the following conditions are equivalent:

1)ais centrally clean.

2)e∈CE(R) exists such thatl(a)?R(1-e)?R(1-a) andl(1-a)?Re?Ra.

3)e∈CE(R) exists such thatr(a)?(1-e)R?(1-a)Randl(1-a)?eR?aR.

4)e∈CE(R) exists such thatea∈U(eR) and (1-e)(1-a)∈U((1-e)R).

5)e∈CE(R) exists such thateais centrally clean ineRand (1-e)(1-a) is centrally clean in (1-e)R.

6)e∈CE(R) exists such thateais centrally clean ineRand (1-e)ais centrally clean in (1-e)R.

7) The decomposition 1=e1+e2+…+enexists, wherenis a positive integer,eis a centrally orthogonal idempotent, andeiais centrally clean ineiRfor each positive integeri.

Proof1)?2). Given thatais centrally clean, we can suppose thata=(1-e)+u, wheree∈CE(R) andu∈U(R). Ifra=0, thenr(1-e)+ru=0, and it follows thatr=ruu-1=[-r(1-e)]u-1∈R(1-e), i.e.,l(a)?R(1-e). Moreover, fromae=[(1-e)+u]e=ue, we derivee=u-1ae=u-1ea∈Ra, i.e.,Re?Ra.

Rewritea=(1-e)+uas 1-a=e+(-u). Then, by a similar argument, we can obtainl(1-a)?ReandR(1-e)?R(1-a).

2)?4). Frome∈Re?Raandl(a)?R(1-e)=l(e), we can obtainea∈U(eR) based on Lemma 1. Similarly, we can derive (1-e)(1-a)∈U((1-e)R).

1)?3)?4) are similar to 1)?2)?4).

4)?5). This is obvious from 1) of Example 1.

5)?6). Given that (1-e)(1-a) is centrally clean in (1-e)R, it follows that (1-e)a=(1-e)-(1-e)(1-a) is also centrally clean in (1-e)R.

6)?7) Writee1=eande2=1-e. Then,e1e2=0 and 1=e1+e2. Moreover,eiais centrally clean ineiRfor eachi.

7)?1). For each positive integeri,eiais centrally clean ineiR.Then, we suppose thateia=fi+ui, wherefi∈CE(eiR) andui∈U(eiR), and it follows thatvi∈eiRexists such thatviui=uivi=ei. Given thateiis a centrally orthogonal idempotent, we derivef=∑fi∈CE(R),u=∑ui∈U(R), andu-1=∑vi. Hence,a=∑eia=∑(fi+ui)=∑fi+∑ui=f+u. Therefore,ais centrally clean.

Proposition1Leta∈R. Then, the following conditions are equivalent:

1)ais centrally clean.

2)v∈U(R) andf∈CE(R) exist such thatf=fvaand 1-f=-(1-f)v(1-a).

3)u∈U(R) andf∈CE(R) exist such thatf=fuaand 1-f=(1-f)u(1-a).

4)f∈CE(R) andx,y∈Rexist such thatf=fxa, 1-f=(1-f)y(1-a), andfx-(1-f)y∈U(R).

Proof1)?2). Leta=u+e, whereu∈U(R) ande∈CE(R). Writef=1-eandv=u-1. Then,

f=(1-e)u-1a=fva

and

1-f=-eu-1(1-a)=-(1-f)v(1-a)

2)?1). Writee=1-f. Then,v(a-e)=[fv+(1-f)v](a-1+f)=fva-fv+fv-(1-f)v(1-a)=f+1-f=1. Hence,a-e=v-1, i.e.,ais centrally clean.

2)?3). Writeu=(2f-1)v. Then,f=fva=fuaand 1-f=-(1-f)v(1-a)=(1-f)u(1-a).

3)?2) is similar to 2)?3).

2)?4). Writex=vandy=-v. Then,f=fxaand 1-f=(1-f)y(1-a).

4)?2). Writev=fx-(1-f)y∈U(R). Then,fx=fvand -(1-f)y=(1-f)v, and it follows thatf=fxa=fvaand 1-f=(1-f)y(1-a)=-(1-f)v(1-a).

2 Characterizations of Centrally Clean Rings

Recall that ifR/J(R) is a division ring, thenRis considered a local ring.

Proposition2Every local ring is a centrally clean ring.

ProofLeta∈R. Ifa∈J(R), thenais centrally clean based on 2) of Example 1. Ifa?J(R), thena+J(R)∈U(R/J(R)). Hence,x+J(R)∈R/J(R) exists such that

(a+J(R))(x+J(R))=1+J(R)

and it follows thatax+J(R)=1+J(R), i.e.,ax-1∈J(R). Therefore,ax=1-(1-ax)∈U(R). Then,ais right invertible inR. Similarly, we can deduce thatais left invertible inR. It follows thata∈U(R), and hence, it is centrally clean.

In 2004, Nicholson et al.[12]proved that ifR≠0, thenR[x] is not clean. Then, it is obvious thatR[x] is not centrally clean whenR≠0.

Proposition3The following conditions are equivalent:

1) 2∈U(R), andRis centrally clean.

2) For anya∈R,u∈U(R) andx∈C(R) exist, withx2=1, such thata=u+x.

Then, we provide some characterizations of centrally clean rings.

Theorem2The following conditions are equivalent:

1)Ris centrally clean.

2) Every elementx∈Rcan be written asx=u-e, whereu∈U(R) ande∈CE(R).

3) Every elementx∈Rcan be written asx=u+e, whereu∈U(R)∪0 ande∈CE(R).

4) Every elementx∈Rcan be written asx=u-e, whereu∈U(R)∪0 ande∈CE(R).

Proof1)?2). Letx∈R. Then, -x=e+v, and it follows thatx=-v-e,u=-v∈U(R) ande∈CE(R).

2)?3) and 3)?4) are similar to 1)?2).

4)?1). Letx∈R. Then, we derive -x=u-ebased on the assumption, whereu∈U(R)∪0 ande∈CE(R). Hence,x=(-u)+e. The case whenu=0 follows from 4) of Example 1.

Recall that ife∈E(R) exists such thate∈aRand 1-e∈(1-a)Rfor anya∈R, thenRis an exchange ring. Moreover, if every idempotent ofRis central, thenRis called abelian.

Theorem3The following conditions are equivalent:

1)Ris centrally clean.

2)Ris an exchange and abelian.

3)Ris clean and abelian.

4) For anya∈R,e∈CE(R) exists such thate∈aRand 1-e∈(1-a)R.

Proof1)?3). It suffices to prove that every idempotent ofRis central. Lete∈E(R). Then, we derivee=f+u, wheref∈CE(R) andu∈U(R), and it follows that iff+u=(f+u)2=f+2fu+u2, then 1=u+2f. Hence, we obtaine=f+u=f+1-2f=1-f∈CE(R).

3)?4). Given that clean rings are exchange rings, it follows that, for anya∈R,e∈E(R) exists such thate∈aRand 1-e∈(1-a)R. Given thatRis abelian, we derivee∈CE(R).

4)?2). This is enough to show thatRis abelian. Letf∈E(R). Then, according to the assumption,e∈CE(R) exists such thate∈fRand 1-e∈(1-f)R. Hence, we obtainfe=eand (1-f)(1-e)=1-e. Then,f=e∈CE(R). Therefore,Ris abelian.

2)?1). Letx∈R. Given thatRis an exchange ring,e∈E(R) exists such thate∈xRand 1-e∈(1-x)R. Lete=xa′, wherea′∈R. Then,e=e2=xa′xa′. Writea=a′xa′, and it follows thate=xaandae=a′xa′xa′=a′xa′=a. Then,axa=a. Given thatRis abelian, we deriveax=axax=xa(ax)=xaxa=xa. By a similar argument, we can obtain (1-e)=(1-x)b,b(1-e)=b, and (1-x)b=b(1-x). Furthermore, we can obtain [x-(1-e)](a-b)=xa-xb-(1-e)a+(1-e)b=e+(1-x)b=1 and (a-b)[x-(1-e)]=1. That is, [x-(1-e)]-1=a-b. Then,x=x-(1-e)+(1-e). Hence,Ris centrally clean.

Proposition4Letp∈CE(R). Then,a∈pRis centrally clean inRif and only ifais centrally clean inpR.

ProofThe necessity is clearly stated in Theorem 1. Conversely, assumea∈pRis centrally clean inR. Then,e∈CE(R) andu∈U(R) exist such thata=e+u, and it follows thatpa=pe+pu,pe∈CE(pR), andpu∈U(pR). Frompa=a, we derivea=pe+pu. Hence,ais centrally clean inpR.

Corollary1Letp∈CE(R). IfRis a centrally clean ring, then so ispR.

Han et al.[4]proved that whene∈E(R), ifeReand (1-e)R(1-e) are clean rings, then so isR. Here, we consider the case ofe∈CE(R).

Corollary2Lete∈CE(R). IfeRand (1-e)Rare centrally clean, then so isR.

ProofThis is clearly stated in Theorem 1.

Han et al.[4]also investigated the cleanness of group rings, the ring of formal power series over a ring, and a direct product of rings. Then, we analyze the relevant results ofR[[x]] and ∏Rα.

Proposition5The ringR[[x]] is centrally clean if and only ifRis centrally clean.

ProofLetf=a+bx+cx2+…∈R[[x]]. Given thatRis centrally clean, we can suppose thata=u+e, wheree∈CE(R) andu∈U(R). Then,f=e+(u+bx+cx2+…),e∈CE(R[[x]]), andu+bx+cx2+…∈U(R[[x]]). Therefore,R[[x]] is centrally clean.

Conversely, we know thatR[[x]]/(x) is centrally clean becauseR[[x]] is centrally clean. Hence,R?R[[x]]/(x) is centrally clean.

Lemma2LetR,Sbe two rings andφ:R→Sbe a surjective ring homomorphism. IfRis centrally clean, then so isS.

ProofIt is obvious.

Proposition6A direct productR=∏Rαis centrally clean if and only ifRαis centrally clean.

ProofGiven thatπα: ∏Rα→Rαis a surjective ring homomorphism, it follows thatRαis centrally clean based on Lemma 2.

Conversely, suppose thatRαis centrally clean. Letx=(xα)∈∏Rα. Then, for eachα, we derivexα=uα+eα, whereuα∈U(Rα) andeα∈CE(Rα). Hence, we obtainx=e+u,u=(uα)∈U(∏Rα) ande=(eα)∈CE(∏Rα), and it follows thatR=∏Rαis centrally clean.

LetLbe a two-sided ideal ofR. We suppose that the idempotents can be lifted moduloLif, given thatx∈E(R/L),e∈E(R) exists such thate-x∈L. Similarly, we can define the concept that the central idempotents can be lifted moduloLife∈CE(R) exists such thate-x∈Lforx∈CE(R/L).

Proposition7Ris centrally clean if and only ifR/J(R) is centrally clean, and the central idempotent can be lifted moduloJ(R).

ProofBased on Lemma 2, we confirm that the factor ring of a centrally clean ring is centrally clean. Then,R/J(R) is centrally clean. Given that a centrally clean ring is exchange, it follows that the idempotents can be lifted moduloJ(R). Based on Theorem 3, we determine that the idempotents ofRare central. Then, the sufficiency is proven.

3 Characterizations of Central Drazin Inverses

In this section, we mainly provide some characterizations for the existence of central Drazin inverses.

Theorem4Leta∈R. Then, the following conditions are equivalent:

1)ais central Drazin invertible.

2)u∈U(R),e∈CE(R), and the positive integermexist such thatam=euandau=ua.

3)v∈U(R) andf∈CE(R) exist such thata=f+vandaf∈Rnil.

4)p∈CE(R) exists such thatap∈U(pR) anda(1-p)∈Rnil.

Proof1)?2). Writee=aac. Then,e∈CE(R). Given thatais central Drazin invertible, and it follows that the positive integermexists such thatam=amaac=ame. Writeu=am+(1-e). Then, [am+(1-e)][(ac)me+(1-e)]=am(ac)me+am(1-e)+(1-e)(ac)me+(1-e)2=e+1-e=1. Hence, we haveu∈U(R) andu-1=(ac)me+(1-e), and it follows thatam=ame=[u-(1-e)]e=euandau=a[am+(1-e)]=am+1+a(1-e)=am+1+(1-e)a=[am+(1-e)]a=ua.

2)?3). Writef=1-e. Then,f∈CE(R). Given that (am-f)(u-1e-f)=1, we deriveam-f∈U(R). Then, (a-f)(am-1+am-2f+…+af+f)=am-f∈U(R). Hence, we obtainv=a-f∈U(R) and (af)m=amf=eu(1-e)=0, i.e.,af∈Rnil.

3)?4). Writep=1-f. Then,p∈CE(R),ap=pa=pv∈U(pR) anda(1-p)=af∈Rnil.

4)?1). Based on this assumption, it follows thatw∈U(pR) exists such thatapw=pwa=p. Frompw=w, we deriveaw=wa=p∈C(R),waw=pw=w, anda-a2w=a(1-aw)=a(1-p)∈Rnil. Hence,ais central Drazin invertible.

Zhu et al.[14]showed thatais Drazin invertible if and only ifu∈U(R),e∈E(R), and the positive integermexist such thatam=u+eis strongly clean decomposition andamR∩eR=0. Here, we investigate the relevant results of central Drazin inverses.

Theorem5Leta∈R. Then, the following conditions are equivalent:

1)ais central Drazin invertible.

2)u∈U(R),e∈CE(R), and the positive integernexist such thatan=u+eandanR∩eR=0.

3)u∈U(R),e∈CE(R), and the positive integernexist such thatan=u-eandanR∩eR=0.

Proof1)?2). Given thatais central Drazin invertible, we deriveu=an-1+aac∈U(R) for any positive integern>ind(a). Writee=1-aac. Then,an=u+eis the centrally clean decomposition. Letx∈anR∩eR. Then,y,z∈Rexist such thatx=any=ez=eany=0. Hence,anR∩eR=0.

2)?1). Frome∈CE(R), it follows that the positive integermexists such that:

(ane)m=(an)me=e(an)m∈anR∩eR=0

i.e.,ane∈Rnil. Letmbe the nilpotent index ofane. Then, (an)m=um(1-e). In fact,

(an)m=(u+e)m=

um(1-e)+(an)me=

um(1-e)+(ane)m=um(1-e)

Hence, (an)mis central group invertible derived by Theorem 3.6 in Ref.[28]. Then,ais central Drazin invertible derived by Theorem 3.3 in Ref.[28].

1)?3). This is similar to the proof of 1)?2).

From Theorem 5, we derive the following corollary.

Proposition8Leta∈R. Then, the following conditions are equivalent:

1)ais central Drazin invertible.

2)e∈CE(R) and the positive integernexist such thatane=0 andan-e∈U(R).

3)e∈CE(R) and the positive integernexist such thatane=0 andan+e∈U(R).

Proof1)?2). Given thatais central Drazin invertible, we deriveu=an-1+aac∈U(R) for any positive integern>ind(a). Writee=1-aac. Then,ane=0 andan-e∈U(R).

2)?1). Letx∈anR∩eR. Then,y,z∈Rexist such thatx=any=ez=eany=aney=0. Hence,anR∩eR=0. According to Theorem 5, the proof is completed.

1)?3). This is similar to the proof of 1)?2).

For the central group inverses, we also obtain the following relevant results.

Proposition9Leta∈R. Then, the following conditions are equivalent:

1)ais central group invertible.

2)u∈U(R) ande∈CE(R) exist such thata=u+eandaR∩eR=0.

3)v∈U(R) andf∈CE(R) exist such thatf=fva, 1-f=(1-f)v(1-a), andaf=a.

Proof1)?2). This is given in Corollary 4.6 in Ref.[28].

2)?3). Writef=1-eandv=u-1(1-2e). Then,v∈U(R) andf∈CE(R), and it follows thatfva=(1-e)u-1(1-2e)(u+e)=(1-e)(1-2e)=1-e=fanda(1-f)=(u+e)e∈aR∩eR=0. Hence,af=aand

(1-f)v(1-a)=eu-1(1-2e)(1-e-u)=

eu-1(1-2e)(-u)=e=1-f

3)?2). Writeu=v-1(2f-1) ande=1-f. Given that 1-f=(1-f)v(1-a)=(1-f)v-(1-f)va=(1-f)v-va+fva=(1-f)v-va+f, it follows thata=v-1(1-f)v+v-1(2f-1)=1-f+v-1(2f-1)=e+u. Letx∈aR∩eR. Then,r,t∈Rexist such thatx=ar=et, and it follows thatfr=fvar=fv(et)=fv(1-f)t=0, i.e.,r=(1-f)r. Then,x=ar=a(1-f)r=(a-af)r=0. Therefore,aR∩eR=0.

4 Central Group Invertibility of Combinations of Two Central Idempotents

Motivated by the study conducted by Liu et al.[15], we investigate the central group invertibility of combinations of two central idempotents in this section.

In this section,Fdenotes a field andAdenotes the algebra overF.

Theorem6Letp,q∈Abe the central idempotent anda=d1p+d2q+d3pq, wheredi∈F,i=1,2,3. Then,ais central group invertible, and

where

Given thatp,q∈CE(R), we derivexa∈C(R). By computation, it follows that

Then, we obtain

and

[d1dd?-d1+d2dd?-d2+d3dd?]pq=

Letd1=1,d2=1,d3=0. Then, we obtain the following results according to Theorem 6.

Corollary3Let 2∈U(R) andp,q∈Abe the central idempotents. Then,p+qis central group invertible, and

Ifpq=p, then we obtain the following results according to Theorem 6. That is, we taked3=0 in Theorem 6.

Corollary4Letp,q∈Abe the central idempotent andpq=p. Then,d1p+d2qis central group invertible, and

Ifpq=q, then we obtain the following results according to Theorem 6.

Corollary5Letp,q∈Abe the central idempotent andpq=q. Then,d1p+d2qis central group invertible, and

5 An Example

By computation, we obtain the following results:

E(Z2S3)={0,g1,g5+g6,g1+g5+g6,g2+g3+g5,g2+

g3+g6,g2+g4+g5,g2+g4+g6,g3+g4+g5,g3+

g4+g6,g1+g2+g3+g5,g1+g2+g3+g6,g1+g2+

g4+g5,g1+g2+g4+g6,g1+g3+g4+g5,g1+g3+

g4+g6}

C(Z2S3)={0,g1,g5+g6,g1+g5+g6,g2+g3+g4,

g1+g2+g3+g4,e+g1,e}

CE(Z2S3)={0,g1,g5+g6,g1+g5+g6}

Example2All invertible, central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements ofZ2S3are listed as follows.

For convenience, we use CG(Z2S3),G(Z2S3), CD(Z2S3), and CC(Z2S3) to denote the sets of all central group invertible, group invertible, central Drazin invertible elements, and centrally clean elements ofZ2S3, respectively.

U(Z2S3)={g1,g2,g3,g4,g5,g6,e+g1,e+g2,e+g3,

e+g4,e+g5,e+g6}

CG(Z2S3)={0,U(Z2S3),g2+g3,g2+g4,g3+g4,

g5+g6,g1+g5,g1+g6}

G(Z2S3)={CG(Z2S3),g1+g2+g5,g1+g2+g6,

g1+g3+g5,g1+g3+g6,g1+g4+g5,g1+g4+g6,

g2+g3+g5,g2+g3+g6,g2+g4+g5,g2+g4+g6,

g3+g4+g5,g3+g4+g6,e+g4+g6,e+g4+g5,

e+g3+g6,e+g3+g5,e+g2+g6,e+g2+g5,e}

CD(Z2S3)={CG(Z2S3),g1+g2,g1+g3,g1+g4,

g1+g2+g3,g1+g2+g4,g1+g3+g4,g2+g5+g6,

g3+g5+g6,g4+g5+g6,e+g3+g4,e+g2+g4,

e+g2+g3,e+g1+g4,e+g1+g3,e+g1+g2,e}

CC(Z2S3)={0,U(Z2S3),g1+g2,g1+g3,g1+g4,

g1+g5,g1+g6,g5+g6,g2+g3,g2+g4,g3+g4,

g1+g2+g3,g1+g2+g4,g1+g3+g4,g2+g3+g4,

g1+g5+g6,g2+g5+g6,g3+g5+g6,g4+g5+g6,

g2+g3+g4,e+g1+g6,e+g1+g5,e+g1+g4,

e+g1+g3,e+g1+g2,e+g3+g4,e+g2+g4,

e+g2+g3,e+g6,e+g5,e}

ProofGiven thatZ2S3is finite, it follows thatZ2S3is strongly π-regular. Hence, the elements inZ2S3are Drazin invertible.

Then, we calculate the units ofZ2S3.

Letα=x1g1+x2g2+x3g3+x4g4+x5g5+x6g6andβ=y1g1+y2g2+y3g3+y4g4+y5g5+y6g6. Fromαβ=g1, we can obtain

which has a unique solution.

(1+x4+x6)(1+x4+x5)+(1+x1+x3)(1+x1+x2)

Then, from |A+B|≠0, it follows thatxi=0 for certaini∈{1,2,…,6} and the others are 1, orxi=1 for certaini∈{1,2,…,6} and the others are 0. Hence, we have

U(Z2S3)={g1,g2,g3,g4,g5,g6,e+g1,e+g2,e+g3,

e+g4,e+g5,e+g6}

Therefore,based on Theorem 3.6 in Ref.[28], Proposition 8.24 in Ref.[30], Theorem 4.5 in Ref.[28], and the definition of centrally clean elements, we can present the sets of CG(Z2S3),G(Z2S3), CD(Z2S3), and CC(Z2S3), respectively.