LOCALLY NILPOTENT GROUPS
COSTANTINO DELIZIA AND CHIARA NICOTERA Received 5 July 2005
We define the power-commutative kernel of a group. In particular, we describe the power- commutative kernel of locally nilpotent groups, and of finite groups having a nontrivial center.
A groupGis calledpower commutative, or aPC-group, if [xm,yn]=1 implies [x,y]=1 for allx,y∈Gsuch thatxm=1,yn=1. So power-commutative groups are those groups in which commutativity of nontrivial powers of two elements implies commutativity of the two elements. Clearly,G is aPC-group if and only ifCG(x)=CG(xn) for allx∈G and all integersnsuch thatxn=1. Obvious examples ofPC-groups are groups in which commutativity is a transitive relation on the set of nontrivial elements (CT-groups) and groups of prime exponent.
Recall that a groupGis called anR-groupifxn=ynimpliesx=yfor allx,y∈Gand for all positive integersn. In other words,R-groups are groups in which the extraction of roots is unique. A result due to Mal’cev and Cernikov (see, e.g., [3]) states that every nilpotent torsion-free group is anR-group. There is a natural connection betweenPC- groups andR-groups. For, as pointed out in [3], a torsion-free group is aPC-group if and only if it is anR-group.
In [5], Wu gave the classification of locally finitePC-groups. In particular, she proved that a finite group is aPC-group if and only if the centralizer of each nontrivial element is abelian or of prime exponent. This result implies that a finite group having a nontrivial center is aPC-group if and only if it is abelian or it has prime exponent. Moreover, the class ofPC-groups is contained in the class of groups in which the centralizer of each nontrivial element is nilpotent. This class of groups was investigated by many authors (see, e.g., [1,4]).
In analogy to what is done in [2] to define the commutative-transitive kernel of a group, we introduce an ascending series
{1} =P0(G)≤P1(G)≤ ··· ≤Pt(G)≤ ··· (1) of characteristic subgroups ofGcontained in the derived subgroupG. We defineP1(G) as
Copyright©2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:17 (2005) 2719–2722 DOI:10.1155/IJMMS.2005.2719
2720 On the power-commutative kernel of locally nilpotent groups
the subgroup ofGgenerated by those commutators [x,y] such that there exist positive integersn,m withxn=1, ym=1, and [xn,ym]=1. If t >1 thenPt(G) is defined by Pt(G)/Pt−1(G)=P1(G/Pt−1(G)). Finally, thePC-kernelofGis the subgroupP(G) ofG defined by
P(G)=
t∈N
Pt(G). (2)
Obviously, for any groupG, thePC-kernelP(G) is characteristic inG,G/P(G) is aPC- group, andGis aPC-group if and only ifP(G)= {1}.
Letᐄbe a class of groups. Then one can ask whether there exists a nonnegative integer nsuch thatPn(G)=P(G) for allG∈ᐄ. Of courseP(G)=Pn(G) if and only ifG/Pn(G) is aPC-group.
In this paper, we give affirmative answers to the previous question whenᐄis the class of locally nilpotent groups, or the class of finite groups having a nontrivial center. In both cases, we prove thatP(G)=P1(G) for allG∈ᐄ.
Our first results are concerned with the power-commutative kernel of finite nilpotent groups.
Proposition1. Letpbe a prime andGa finitep-group. ThenG/P1(G)is aPC-group.
Proof. Notice thatP1(G)≤Mfor every maximal subgroupMofGsinceP1(G)≤G≤ Φ(G), whereΦ(G) is the Frattini subgroup ofG. This implies thatM/P1(G) is a maximal subgroup ofG/P1(G) if and only ifMis a maximal subgroup ofG.
LetGbe a counterexample of least order. For any maximal subgroupMofGwe ob- tainM/P1(G)(M/P1(M))/(P1(G)/P1(M)). HenceM/P1(G) is aPC-group since it is a quotient of a finitePC-group (see [5]). It follows that a maximal subgroup ofG/P1(G) is abelian or it has exponentp.
PutG=G/P1(G) andH=H/P1(G) for allP1(G)≤H≤G. If every maximal subgroup MofGhas exponentp, thenGis cyclic or of exponent p. In any caseGis aPC-group, that is a contradiction. So we may assume thatGhas a maximal subgroupMsuch thatM is abelian andMp=1. Considerg∈G\M, soG= M,g. Moreover|G:M| =p.
If there existsa∈Msuch that (ga)p=1, then (ga)p∈M\ {1}. So, for ally∈Mwe get [y, (ga)p]=1, hence [y,g]=[y,ga]=1. It follows thatGis abelian, a contradiction. Thus (ga)p=1 for alla∈M, and in particulargp=1. It follows thatagp−1+···+g+1=(ga)p= 1 for alla∈M. This impliesap=1 for alla∈CM(g), so (CM(g))p=CMp(g)=1. But
Mp∩Z(G)=1 sinceMp=1, that is a contradiction.
Proposition2. LetGbe a finite nilpotent group of ordern=pα11···pαtt(p1,...,ptdistinct primes). Ift >1thenG/P1(G)is abelian.
Proof. Let Gpi be the Sylow pi-subgroup of G for all i∈ {1,...,t}; we will prove that (Gpi)≤P1(G) for alli∈ {1,...,t}. Letx,y∈Gpi\ {1},a∈Gp1× ··· ×Gpi−1×Gpi+1×
··· ×Gpt. Put|a| =mand|x| =pir. Now|ax| =mpiras (m,pir)=1. Since (ax)pir=apir has ordermwe get [(ax)pir,y]=[apir,y]=1. Thus [ax,y]=[x,y]∈P1(G).
Corollary3. LetGbe a finite nilpotent group; thenG/P1(G)is abelian or it has exponent p. In both casesG/P1(G)is aPC-group.
Proof. The result is an immediate consequence of the previous propositions and [5, The-
orem 4].
Now we prove that the equalityP(G)=P1(G) holds for every nilpotent groupG.
Theorem4. LetGbe a nilpotent group. ThenG/P1(G)is aPC-group.
Proof. IfGis torsion-free thenGis aPC-group (see [3]), soP1(G)= {1}and the result is true. So we may suppose that the torsion subgroupTofGis nontrivial.
First of all, notice that if for elementsx,y∈G\ {1}there exists a positive integernsuch thatxn=1 and [xn,y]=1, then [x,y]∈T. This is obvious ifx∈T ory∈T, so we may assumex,y /∈T. Thenx,yT/T≤G/T. SoxT,yTis torsion-free, and [(xT)n,yT]=T implies [x,y]∈T. This means thatP1(G)⊆T.
If for anyx,y∈Gthe commutator [x,y] is periodic, then it is easy to see that there exists a positive integermsuch that [x,ym]=1. In fact, x,y is aFC-group sincex, y/Z(x,y) is finite, and therefore the set{xyt|t∈Z}is finite.
Now notice that ifx∈Tthen [x,g]∈P1(G) for allg∈G\T. In fact, [x,g]∈Timplies that there exists a positive integermsuch that [x,gm]=1. So we get [x,g]∈P1(G) because gm=1.
Finally, letx,y∈G\P1(G) such thatxn∈/ P1(G) and [xn,y]∈P1(G). Ifx,y∈Tthen x,yis a finite nilpotent group andCorollary 3implies thatx,y/P1(x,y) is a finite PC-group. Hencex,y/P1(G)∩ x,yis aPC-group and [x,y]∈P1(G). Ifx∈Tory∈ Tthen [x,y]∈P1(G), as noticed before. So we may supposex,y∈G\T. Since [xn,y]∈ P1(G)⊆T, we get [xn,y]∈Tand so there exists a positive integermsuch that [xn,ym]= 1. Therefore [x,y]∈P1(G), and the proof is complete.
Theorem5. LetGbe a locally nilpotent group. ThenP(G)=P1(G).
Proof. Letx,y∈G\P1(G) such thatxn∈/ P1(G) and [xn,y]∈P1(G). Then xn,y=
r i=1
ai,bi
, (3)
whereai,bi∈Gfor alli=1, 2,...,r, and [aαii,biβi]=1 for some positive integersαiandβi
such thataαii=1 andbiβi=1.
Let H= x,y,a1,...,ar,b1,...,br. ThenH is nilpotent, soH/P1(H) is aPC-group byTheorem 4. Since [ai,bi]∈P1(ai,bi)≤P1(H) for alli=1, 2,...,r, we get [xn,y]∈ P1(H). Thus [x,y]∈P1(H), and therefore [x,y]∈P1(G).
Now it is possible to prove thatP(G)=P1(G) for any finite groupGsuch thatZ(G)= {1}.
Proposition6. LetGbe a finite group such thatZ(G)= {1}. Then[a,b]∈P1(G)for all a,b∈G\ {1}such that(|a|,|b|)=1.
Proof. Put|a| =nand|b| =m. Then there existsz∈Z(G)\ {1}such that|z|does not dividenorm. Suppose|z|does not dividen. Then [(az)n,b]=[anzn,b]=[zn,b]=1.
Moreover (az)n=zn=1 and this yields [az,b]=[a,b]∈P1(G).
2722 On the power-commutative kernel of locally nilpotent groups
Proposition7. LetGbe a finite group such thatZ(G)= {1}. ThenG/P1(G)is nilpotent.
Proof. We may assume that the order ofG/P1(G) is not a prime power. Letpbe any prime divisor of|G/P1(G)|. Then p divides|G| andPP1(G)/P1(G) is a Sylow p-subgroup of G/P1(G) wheneverPis a Sylowp-subgroup ofG. We are going to show thatPP1(G)/P1(G) is normal inG/P1(G). Letq=pbe any prime dividing|G/P1(G)|, and letQbe a Sylow q-subgroup ofG. ThenQP1(G)/P1(G) centralizesPP1(G)/P1(G), byProposition 6. Thus the normalizer inG/P1(G) ofPP1(G)/P1(G) contains a Sylowq-subgroup ofG/P1(G) for all prime divisors of its order. Therefore this normalizer is actuallyG/P1(G), and the result
follows.
Theorem8. LetGbe a finite group such thatZ(G)= {1}. ThenG/P1(G)is abelian or it has exponentp.
Proof. SinceG/P1(G) is nilpotent byProposition 7, by [5] it suffices to show thatG/P1(G) is aPC-group. Suppose not, and letGbe a counterexample of least order. We may as- sumeG is not nilpotent, hence P1(G)≤Φ(G). Thus there exists a maximal subgroup M ofGsuch thatP1(G)⊆M. In particularG⊆M. IfZ(G)⊆M, then there existsz∈ Z(G)\M. SinceMis maximal, it follows thatzM=G. HenceMis normal inG, and G/Mis cyclic. This in turn implies thatG⊆M, a contradiction. ThusZ(G)⊆M, and so Z(M)= {1}. ThenM/P1(M) is aPC-group and thereforeG/P1(G)(M/P1(M))/((M∩ P1(G))/P1(M)) is aPC-group, the final contradiction.
References
[1] W. Feit, M. Hall Jr., and J. G. Thompson,Finite groups in which the centralizer of any non- identity element is nilpotent, Math. Z.74(1960), 1–17.
[2] B. Fine, A. M. Gaglione, G. Rosenberger, and D. Spellman,The commutative transitive kernel, Algebra Colloq.4(1997), no. 2, 141–152.
[3] A. G. Kurosh,The Theory of Groups, Chelsea, New York, 1960.
[4] M. Suzuki,Finite groups with nilpotent centralizers, Trans. Amer. Math. Soc.99(1961), 425–
470.
[5] Y.-F. Wu,On locally finite power-commutative groups, J. Group Theory3(2000), no. 1, 57–65.
Costantino Delizia: Dipartimento di Matematica e Informatica, Universit`a di Salerno, via Ponte don Melillo, 84084 Fisciano (SA), Italy
E-mail address:[email protected]
Chiara Nicotera: Dipartimento di Matematica e Informatica, Universit`a di Salerno, via Ponte don Melillo, 84084 Fisciano (SA), Italy
E-mail address:[email protected]
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