Generalizations of Morphic Group Rings

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International Journal of Mathematics and Mathematical Sciences Volume 2007, Article ID 50591,8pages

doi:10.1155/2007/50591

Research Article

Generalizations of Morphic Group Rings

Libo Zan, Jianlong Chen, and Qinghe Huang

Received 9 October 2006; Revised 20 December 2006; Accepted 5 February 2007

Recommended by Howard E. Bell

An elementain a ringRis called left morphic if there existsb∈Rsuch that 1R(a)=Rb and 1R(b)=Ra.Ris called left morphic if every element ofRis left morphic. An elementa in a ringRis called leftπ-morphic (resp., leftG-morphic) if there exists a positive integer nsuch thatan(resp.,anwithan=0) is left morphic.Ris called leftπ-morphic (resp., left G-morphic) if every element ofRis leftπ-morphic (resp., leftG-morphic). In this paper, theG-morphic problem andπ-morphic problem of group rings are studied.

1. Introduction

An elementain a ringRis said to be left morphic ifR/Ra∼=lR(a), which is equivalent to that there existsb∈Rsuch that lR(a)=Rband lR(b)=Ra, where lR(a) denotes the left annihilator ofainR.Ris called left morphic if every element ofRis left morphic. Right morphic elements and rings are defined analogously. Nicholson and S ´anchez Campos introduced and investigated left morphic rings in [1] (see also [2–4] for more detailed discussion).

Left morphic rings are generalized to leftπ-morphic rings and leftG-morphic rings by Huang and Chen [5]. An elementa∈Ris called leftπ-morphic (resp., leftG-morphic) if there exists a positive integernsuch thatan(resp.,anwithan=0) is left morphic.R is called leftπ-morphic (resp., leftG-morphic) if every element ofRis leftπ-morphic (resp., leftG-morphic).Ris calledπ-morphic (resp.,G-morphic) if it is left and rightπ -morphic (resp., left and rightG-morphic). Moreover, they find examples which show that leftπ-morphic rings are proper generalizations of left morphic rings, and leftG-morphic elements need not be left morphic.

Example 1.1 [5, Example 2.13]. LetR=F[x,σ]/(x2₎_{= {a}₊_xb_|_a_,_b_∈_F}_{, where}_F _{is a}

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S=R⊕R, thenλ=(1,xb)∈S(whereb∈F, butb∈F) is leftG-morphic, but not left morphic.

The question of when a group ring is morphic was studied by Chen et al. [6]. In this paper, we investigate when a group ring isπ-morphic (resp.,G-morphic). InSection 2, several general results aboutπ-morphic andG-morphic group rings are obtained. In Section 3, necessary and suﬃcient conditions forRGto be leftG-morphic are also given, whereR=Zn,G is a finite Abelian group. In particular, we prove that ifG is a finite Abelian group or a finitep-group,r1, thenZprGisπ-morphic.

All rings in this paper are associative rings with identity. LetRbe a ring and letGbe a group. We denote byRGthe group ring ofGoverR. The following concepts in group rings play very important roles in our discussion and will be used frequently later. For any elementu=Σaigi∈RG, whereai∈R,gi∈G, the augmentation ofu, denoted by(u), is defined by(u)=Σai. The augmentation ideal ofRG, denoted byΔ(G), is defined by

Δ(G)= {u∈RG|(u)=0}. IfGis a cyclic group generated byg, thenΔ(G)=RG(1−g). For any finite subgroupH ofG,H is defined to beH =∀h∈Hh. WhenH is a normal subgroup,H is a central element inRG. For any group elementg∈Gof finite order, definegbyg=1 +g+···+go(g)−1_{, where}_o₍_g_{) is the order of}_g_{. It is not hard to verify}

that ifo(g)<∞, then lRG(1−g)=RGg, and if|G|<∞, then lRG(G)=Δ(G). So ifGis a finite cyclic group, thenGis always left morphic inRG. For more background knowledge about group rings, we refer readers to [7,8].

2. General results

In this section, several general results aboutπ-morphic andG-morphic group rings are given.

Theorem 2.1. LetRbe a ring and letGbe a locally finite group. IfRGis leftπ-morphic (resp., leftG-morphic), thenRis leftπ-morphic (resp., leftG-morphic).

Proof. For anya∈R, sinceais leftπ-morphic (resp., leftG-morphic) inRG, there exist a positive integern(resp.,an=0) andu∈RGsuch that lRG(an)=RGuand lRG(u)=RGan_. Letu=ni=1aigiandH= g1,...,gn . SinceGis a locally finite group,His a finite group. Sinceanu=uan=0, we havean(u)=(anu)=0 and(u)an=(uan)=0, where(u) is the augmentation ofu. ThusRb⊆lR(an) andRan⊆lR(b), whereb=(u). Next we show that in fact,Rb=lR(an) andRan=lR(b). Soais leftπ-morphic (resp., leftG-morphic) inR, and thusRis leftπ-morphic (resp., leftG-morphic).

Letx∈lR(an). Thenx∈lRG(an)=RGu, sox=vu,v∈RG. Taking the augmentation on both sides, we obtainx=(x)=(vu)=(v)(u)=(v)b∈Rb. Therefore, lR(an)⊆ Rb, and thus lR(an)=Rb. Next, let y∈lR(b). Then yb=0. LetH=h∈Hh. Sinceu∈ RH, we haveHu =(u)H=b_H. ThusyHu =yb_H₌0, soyH∈lRG(u)=RGan_{. Hence} y_H ₌_Σa_g_gan_{. Comparing the coe}_ﬃ_{cients of the identity on both sides, we obtain that} y=aean∈Ran, and so lR(b)⊆Ran. This implies that lR(b)=Ran. Therefore,ais left π-morphic (resp., leftG-morphic) and so isR.

Corollary 2.2. IfG=H×Kis a locally finite group andRGis leftπ-morphic (resp., left

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Proof. Note thatRG=R(H×K)∼=(RH)K. ByTheorem 2.1,RHis leftπ-morphic (resp., leftG-morphic). SimilarlyRKis leftπ-morphic (resp., leftG-morphic).

Theorem 2.3. LetGbe a locally finite group. IfRHis leftπ-morphic (resp., leftG-morphic) for every finite subgroupHofG, thenRGis leftπ-morphic (resp., leftG-morphic).

Proof. Letu=ni=1aigi. Now we show thatuis leftπ-morphic (resp., leftG-morphic) in

RG. DenoteH= g1,...,gn . SinceGis locally finite,His a finite group. By the

assump-tion,RH is leftπ-morphic (resp., leftG-morphic). Sinceu∈RH, there exist a positive integern(resp.,un=0) andc∈RHsuch that lRH(un)=RHcand lRH(c)=RHun. Since un_c₌_cun₌_{0, we have}_RGc_⊆_l_RG(_un_{) and}_RGun_⊆_l_RG(_c_{). We next show that the other} inclusions also hold.

Letv∈lRG(un) and let{1,g1,g2,...}be a left coset representative ofH inG. That is,

G=H∪g

1H∪g2H∪ ···. Nowv can be written asv=

g

ibi, wherebi∈RH. Since 0=vun₌_g

i(biun) andbiun∈RH, we obtain thatbiun=0 for alli. Sobi∈lRH(un)= RHc, and thusbi=cic for someci∈RH. It follows thatv=gibi=(gici)c∈RGc, so lRG(un)⊆RGc, and thus lRG(un)=RGc. Similarly, we can prove that lRG(c)=RGun_. This shows thatuis leftπ-morphic (resp., leftG-morphic) inRG, and thereforeRGis left

π-morphic (resp., leftG-morphic).

Recall that a groupGis called a semidirect product ofHbyK, denoted byG=HK, ifH,Kare subgroups ofGsuch that (1)HG; (2)HK=G; (3)H∩K=1.

Theorem 2.4. LetG=HK,|H|<∞. IfRGis leftπ-morphic (resp., leftG-morphic), thenRKis also leftπ-morphic (resp., leftG-morphic).

Proof. We show that for anya∈RK,ais leftπ-morphic (resp., leftG-morphic) inRK. Sinceais leftπ-morphic (resp., leftG-morphic) inRG, there exist a positive integern (resp.,an=0) andu∈RGsuch that lRG(an)=RGuand lRG(u)=RGan_{. Let}_u₌_u_i_k_i, whereui∈RH,ki∈K(sinceG=HK, the expression ofuis unique) andan=ajkj whereaj∈R. Denoteb=(ui)ki, sob∈RK. We will show that lRK(an)=RKband

lRK(b)=RKan. Soais leftπ-morphic (resp., leftG-morphic) inRK, and thusRKis left π-morphic (resp., leftG-morphic).

Letω:G→G/H be the natural group homomorphism. We extendω to a ring ho-momorphism (still denote it by ω). That is,ω:RG→R(G/H) defined byω(aigi)=

aiω(gi). Clearly, ker(ω)∩RK= {0}andω(v)=(v) for allv∈RH. Since 0=an_u_{, we} have 0=ω(an)ω(u)=ω(an)ω(uiki)=ω(an)(ui)ω(ki)=ω(an(ui)ki)=ω(anb). Sinceanb∈RK, we conclude thatanb=0. Similarly,ban=0. This shows thatRKb⊆

lRK(an) andRKan⊆lRK(b). We next show that the other inclusions also hold.

Letx∈lRK(an). Thenx∈lRG(an)=RGu. Sox=vu. Letv=vjkjandc=(vj)kj, where vj ∈RH, kj∈K. Then ω(x)=ω(v)ω(u)=(vj)ω(kj)(ui)ω(ki)=ω(cb). Thus x−cb∈kerω∩RK = {0}. Therefore x=cb∈RKb. This shows that lRK(an)⊆ RKb, and thus lRK(an)=RKb.

Let y∈ lRK(b). Then yb=0. Since HG,H=h∈Hh is central in RG. Now we haveyHu =yHuiki=y H(ui)ki=yHb =ybH=0. SoyH∈lRG(u)=RGan. Thus

H y=yH=wan_{, where}_w₌_h

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hjy=H y =wan=

hjujan. (2.1)

SinceH∩K= {1}, the expression ofwanis unique. Comparing the coeﬃcients of the identityh0=ein (2.1), we obtainy=u0an∈RKan. Thus lRK(b)⊆RKan, and therefore

lRK(b)=RKan.

From now on, we always assume thatGis a finite group.

Proposition 2.5. Assume thatp is a prime number andr >1. IfZprGis leftG-morphic,

thenpdoes not divide|G|.

Proof. Assume that p| |G|. Then there existsg∈Gsuch that o(g)=p. Letu=pr−1_G_,

whereG=g∈Gg. Sinceuis leftG-morphic inZprG, there exists a positive integernsuch thatunis left morphic inZprG. Sinceu2=0,uis left morphic inZ_prG. By Chen et al. [6, Theorem 2.7], this is impossible. Sop|G|.

Theorem 2.6. Assume that p is a prime number andGis a finite p-group.ZprGis left G-morphic if and only ifGis a cyclic group andr=1.

Proof. “⇒” It follows fromProposition 2.5thatr=1. SinceR=Zp is a field andGis a finitep-group,RGis a local ring by Nicholson theorem [9]. BecauseRGis left Artinian, the Jacobson radicalJ(RG) is nilpotent. SinceRGis leftG-morphic,RGis left special by Huang and Chen [5, Theorem 2.8]. So it is left morphic. According to Chen et al. [6, Theorem 2.9],Gis a cyclic group.

“⇐” IfG= g , clearlyZpGis a special ring. Therefore it is leftG-morphic.

Theorem 2.7. Assume thatpis a prime number andGis a finitep-group,r1, thenZprG

isπ-morphic.

Proof. SinceR=Zpr _{is local and}G_{is a finite} p_-group,RG_{is a local ring by Nicholson’S} theorem [9]. BecauseRis Artinian andGis a finite group,RGis Artinian by Connell [10, Theorem 1], and so the Jacobson radicalJ(RG) is nilpotent. According to Huang and Chen [5, Lemma 2.10], every element ofRGis either nilpotent or invertible. SoRGis

π-morphic.

Remark 2.8. ByTheorem 2.6, whenr >1 andGis a finite p-group,ZprGis not leftG -morphic, but by the above theorem, it isπ-morphic.

3. Abelian group rings

In this section, we discuss when an Abelian group ringRGis leftπ-morphic (resp., left G-morphic).

Lemma 3.1 [6, Lemma 3.1]. (R1⊕R2⊕ ··· ⊕Rs)G∼= ⊕si=1RiG.

Lemma 3.2. IfR=R1⊕R2⊕ ··· ⊕Rsis leftπ-morphic (resp., leftG-morphic), then each

Riis leftπ-morphic (resp., leftG-morphic).

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a positive integern (resp., rn=0) such that lR(u)=Rrn _{and lR}₍_rn₎₌_Ru_{, so we have}

lRi(ui)=Rirni and lRi(rin)=Riui. Thenriis leftπ-morphic (resp., leftG-morphic) inRi, and thusRiis leftπ-morphic (resp., leftG-morphic).

Lemma 3.3. LetDbe a division ring ands2. The the following statements are equivalent: (1)D(Cm1× ··· ×Cms) is leftG-morphic;

(2)D(Cmi×Cmj) is leftG-morphic for any 1i=js; (3) at most one ofm1,m2,...,msis not invertible inD.

Proof. We will prove (3)⇒(1)⇒(2)⇒(3).

“(3)⇒(1)” We may assume thatm1,...,ms−1are invertible inD. So|Cm1× ··· ×Cms−1| =m1× ··· ×ms−1 is invertible inD. By Maschke’s theorem, D(Cm1× ··· ×Cms−1) is semisimple. It follows from [6, Lemma 3.5] thatD(Cm1× ··· ×Cms−1×Cms) is strongly morphic, so it isG-morphic and (2.1) holds.

“(1)⇒(2)” Note thatD(Cm1× ··· ×Cms)∼=D(Cmi×Cmj)(

k=i,jCmk) for any 1i=

js. It follows fromTheorem 2.1thatD(Cmi×Cmj) is leftG-morphic.

“(2)⇒(3)” We prove it by contradiction. We may assume thatm1,m2are not invertible

inD. Let char(D)=p >0. By assumption,pdivides bothm1andm2. So we havemi=

pri_t_{i, where (}_t_i,_p₎₌_1,_r_i_1,_i₌_{1, 2.}

Note thatCm1×Cm2∼=(Cpr1×C_pr2)×(C_t1×Ct2), soD(C_m1×C_m2)∼=D(C_pr1×C_pr2)× (Ct1×Ct2). SinceD(Cm1×Cm2) is leftG-morphic,D(Cpr1×Cpr2) is leftG-morphic by Theorem 2.1. Because Cpr1×Cpr2 is a finite p-group,D(Cpr1×Cpr2) is a local Artinian ring, so the Jacobson radical of this group ring is nilpotent. This ring is a left special ring, and then it is left morphic by Huang and chen [5, Theorem 2.8]. ThusCpr1×Cpr2 must

be cyclic, a contradiction.

Proposition 3.4. LetGbe a finite Abelian group andr >1. ThenZprGisG-morphic if and

only if (p,|G|)=1.

Proof. “⇐” By Chen et al. [6, Corollary 3.13], if (p,|G|)=1,ZprGis morphic, so it is G-morphic.

“⇒” ByProposition 2.5, ifr >1 andZprGisG-morphic, thenp|G|, that is, (p,|G|)=1. Theorem 3.5. LetG be a finite Abelian group.ZnG isG-morphic if and only if for each

prime numberpifp|(n,|G|), thenp2_n_{and the Sylow}_p_-subgroup_G

pofGis cyclic.

Proof. LetG=C_qt1

1 × ··· ×Cqtmm,ti1 be a finite Abelian group and letα=q1···qm. Suppose thatZnGisG-morphic. Let (n,|G|)=pr11 ···p

rs

s. Ifri>1 for somei(i.e.,p2i |n), thenn=pisin1, wheresiri>1 and (n1,pi)=1. ThusZnG∼=ZpisiG⊕Zn1G. SinceZnG isG-morphic,ZpsiiGis alsoG-morphic byLemma 3.2. ByProposition 3.4, (pi,|G|)=1. However,pi|(n,|G|). This leads to a contradiction. Thusri1 for alli. Next we show thatpi2α. Otherwise, assume thatp2i |α. There existsk=lsuch thatqk=ql=pi. Hence G∼=C_qtk

k ×Cqtll ×H. Since pi|nand p

2

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ZpiG⊕Zn1G. ByLemma 3.2,ZpiGisG-morphic. SinceZpiG∼=Zpi(C_qtk

k ×Cqtll)H, we con-clude thatZpi(C_qtk

k ×Cqtll)=Zpi(Cpitk×Cptli) isG-morphic. This contradicts the result of

Theorem 2.6. Therefore,p2i α, and thusGpiis cyclic.

Remark 3.6. According toProposition 3.4andTheorem 3.5, the following group rings are notG-morphic:

Z4C2, Z4C4, Z4(C2×C2), Z2C2×C2, Z2C2×C4. (3.1)

But byTheorem 2.7, the above group rings are allπ-morphic.

Lemma 3.7. LetRbe a ring and letGbe a group. Ifa∈Ris left morphic inR, thenais left morphic inRG.

Proof. Ifa∈Ris left morphic, there existsb∈Rsuch that lR(a)=Rband lR(b)=Ra. Sinceba=ab=0, we haveRGb⊆lRG(a) andRGa⊆lRG(b). We next show that the other inclusions also hold.

Letx∈lRG(a),x=rjgj, whererj∈R,gj∈G. Thenrjgja=0 or(rja)gj=0, so allrja=0. Thusrj∈Rbandrj=rjb,rj∈R. Therefore,x=(rjb)gj=rjgjb∈RGb. This shows that lRG(a)⊆RGb, and thus lRG(a)=RGb.

Using a similar proof, we can show that lRG(b)⊆RGa, and thus lRG(b)=RGa. Soais

left morphic inRG.

Recall that ifn=pu_n

1, (n1,p)=1, we denote thatpun.

Lemma 3.8. Letpbe a prime number,r1,prm, and 1nm.

(1) If (p,n)=1, thenpr|Cn m. (2) Ifptn,rt, thenpr−t|Cnm.

Proof. Letm=m1pr, (m1,p)=1. Then

Cn

m=m(m−1)···

m−(n−1) 1···(n−1)n =

m nCnm−−11=

m1pr

n Cmn−−11. (3.2)

(1) If (p,n)=1, then (pr,n)=1, sopr|Cn m.

(2) Ifptn,tr, thenn=n1pt, where (p,n1)=1, so

Cn m=m1p

r n Cnm−−11=

m1pr

n1pt C

n−1

m−1=

m1pr−t

n1 C

n−1

m−1. (3.3)

We havepr−t|Cn

mn1. Since (p,n1)=1, (pr−t,n1)=1, sopr−t|Cmn.

Proposition 3.9. Letpbe a prime number and letGbe a finite Abelian group. If for some

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Proof. For x∈Zpr(C_pt×G)=(Z_prG)C_pt =(Z_prG)g , x=r₀+r₁g+···+r_pt₋₁gpt−1, whereri∈ZprG. Since

x1+x2+···+xsk

= k

k1=0

k1

k2=0

··· ks−2

ks−1=0

C_kk1Ck2_k1···Cks−1 ks−2x

k−k1

1 x2k1−k2···xsks−−12−ks−1xkss−1,

xpr

=r0+r1g+···+rpt−1gp

t₋₁pr

= pr

n1=0

n1

n2=0

··· npt−2

npt−1=0

Cn1prC_n1n2···Cnnptpt−−21r pr₋_n1

0

r1g

n1−n2

···rpt₋₁gpt−1npt−1. (3.4)

Claim 3.10. Letni be the first number inn1,...,npt₋₁ such thatn_iis not divisible by p. Thenpr|Cn1prC_n1n2···Cnini−1.

Proof. Ifi=1, then (n1,p)=1, and byLemma 3.8,pr|Cn1pr.

Now we seti >1. Letnk=nkpuk, 1ki−1, where (nk,p)=1. SinceCnknk−1=C nk−1−nk nk−1 , we can assume thatukuk−1. ByLemma 3.8, we havepuk−1−uk|Cnknk−1, 1ki−1, and

pui−1|C_nini

−1because (p,ni)=1. So

p(r−u1)+(u1−u2)+···+(ui−2−ui−1)+ui−1|Cn1

prC_n1n2···Cni_ni₋

1. (3.5)

Hence,pr|Cn1prC_n1n2···Cni_ni₋1.

By the above claim, if there existsnisuch thatpni, thenCn1prCn2_n1···C_nini₋₁=0 inZ_pr. So assume thatp|nj,j=1,...,pt−1, and then we have

xpr

=

p|n1, 0n1pr

p|n2, 0n2n1

···

p|npt−1, 0npt−1npt−2

×Cn1prC_n1n2···Cn_nptpt₋−1₂rp r₋_n1

0

r1g

n1−n2

···rpt₋₁gpt−1npt−1

=cigpi∈ZprGgp =Z_pr_GC_pt−1.

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Theorem 3.11. Ifpis a prime number,r1, andGis a finite Abelian group, thenZprG_is π-morphic.

Proof

Case 1. If (p,|G|)=1, then (pr,|G|)=1. By Chen et al. [6, Corollary 3.13],ZprGis mor-phic, soZprG_isπ_-morphic.

Zpr(C_pt2× ··· ×Cpts×H)Cpt1, thenxp r

∈Zpr(C_pt2× ··· ×Cpts×H)Cpt1−1byProposition 3.9. So we havexk1∈Zpr(C_pt2× ··· ×C_pts×H) for somek1. Continuing the process, we

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morphic inZprH. Thusxnis morphic inZ_prGbyLemma 3.7. Hencexisπ-morphic in

ZprG.

Acknowledgments

The authors would like to thank the referee for careful reading of the paper and for point-ing out some typos and minor errors. The research was supported by the National Natural Science Foundation of China (no. 10571026), the Natural Science Foundation of Jiangsu Province (no. BK2005207), and the Teaching and Research Award Program for Outstand-ing Young Teachers in Higher Education Institutes of MOE, China. The authors would like to thank Professor Yuanlin Li for reading of the paper and making some valuable comments and suggestions.

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[7] C. Polcino Milies and S. K. Sehgal, An Introduction to Group Rings, vol. 1 of Algebras and Appli-cations, Kluwer Academic Publishers, Dordrecht, The Netherlands, 2002.

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Libo Zan: Department of Mathematics, Southeast University, Nanjing 210096, China Email address:[email protected]

Jianlong Chen: Department of Mathematics, Southeast University, Nanjing 210096, China Email address:[email protected]