Lihua ZHANGYanming XIAQinhai ZHANG

1 Introduction

A finite nonabelianp-group is said to be minimal nonabelian if its every proper subgroup is abelian.As numerous results in[1–3]show,the structure of ap-group depends essentially on its minimal nonabelian subgroups.Minimal nonabelianp-groups were classified by L.R´edei[4].More general groups,D1-groups,than minimal nonabelianp-groups were introduced and characterized in[5].A finitep-groupGis calledD1-group if the order of the derived subgroup of every maximal subgroup ofGis at mostp.On the other hand,nonmetacyclicp-groups all of whose maximal subgroups are metacyclic were classified in[6].A natural question is:What can be said about finitep-groups all of whose maximal subgroups either are metacyclic or have a derived subgroup of order≤p?This is also a question proposed by Berkovich and Janko in their joint book.

Problem 725(cf.[2])Study thep-groups all of whose maximal subgroups either are metacyclic or have a derived subgroup of order≤p.

For convenience,the groups studied in Problem 725 is calledP-groups.Since metacyclicp-groups have been studied and classified in[7–8],respectively,hence assume thatP-groups are nonmetacyclic in this paper.We classifyP-groups up to isomorphism.

2 Preliminaries

LetGbe a finitep-group.We usec(G),exp(G)andd(G)to denote the nilpotency class,the exponent and the minimal number of generators ofGrespectively.For any positive integerandLet

denote the lower central series ofG,wherec=c(G),andGndenote thenth term of the lower central series of a groupG.We useandH∗Kto denote the cyclic group of orderpm,the direct product ofncyclic groups of orderpm,and a central product ofHandKrespectively.H?Gdenotes thatHis a maximal subgroup ofG.Other notations and symbols refer to[9].

Lemma 2.1(cf.[10,Lemma 2.2])Assume that G is a finite nonabelian p-group.Then the following conditions are equivalent:

(1)G is minimal nonabelian;

(2)d(G)=2and

(3)d(G)=2andΦ(G)=Z(G).

Lemma 2.2(cf.[2,Lemma 65.1])Assume that G is a minimal nonabelian p-group.Then G is one of the following groups:

Lemma 2.3(cf.[6])Assume that G is a minimal nonmetacyclic p-group.Then G is isomorphic to one of the following groups:

(1)an elementary abelian p-group of order p3;

(2)p>2,a nonabelian p-group of order p3andexp(G)=p;

(3)a group of order34with

(4)

(5)

Lemma 2.4(cf.[6])A finite p-group G is metacyclic if and only ifis metacyclic.

Lemma 2.5(cf.[11,Theorem 5.12])Let G be a p-group.

(1)If is cyclic,then G?is cyclic;

(2)If Z(G?)is cyclic,then G?is cyclic;

(3)If Z(Φ(G))is cyclic,thenΦ(G)is cyclic.

Lemma 2.6(cf.[1,Lemma 1.1])If a nonabelian p-group G has an abelian maximal subgroup,then

Lemma 2.7(cf.[9,Theorem 2.4.4])Let p be an odd prime.Then a finite p-group G is metacyclic if and only if ω(G)≤2.

Lemma 2.8(cf.[9,Theorem 2.4.3])A two-generator2-group G is metacyclic if and only if d(M)≤2for all maximal subgroups M of G.

Lemma 2.9If G is a finite p-group with d(G)=2,then

ProofSinceis cyclic.It follow thatis cyclic.Since

Lemma 2.10Let G be a p-group,N

ProofLetandThenIt follows thatThusIt follows by Lemma 2.9 thatSinceG/Nis metacyclic andN≤M,G/Mis metacyclic.It follows from Lemma 2.4 thatGis metacyclic.

Lemma 2.11Let G be a metacyclic p-group.Then the following conclusions are true:

(1)d(G)≤2and G?is cyclic.

(2)G?≤?1(G).

(3)If H≤G,then H is metacyclic.

(4)If N?G,then G/N is metacyclic.

Proof(1)is obvious.

(2)IfGis abelian,thenIfGis not abelian,then there existssuch thatis cyclic.ThusSinceG/G?is not cyclic,Thus

(3)SinceGis metacyclic,there existssuch thatis cyclic.SinceH∩?a?charSinceis cyclic.SoHis metacyclic.

(4)SinceGis metacyclic,there existssuch thatis cyclic.Sinceis cyclic.Sinceis cyclic.Sinceis metacyclic.

Lemma 2.12Let G be a p-group with.Then G?is abelian.

ProofIfG?not abelian,then|Z(G?)|=p.By Lemma 2.5(2),G?is cyclic,a contradiction.

Lemma 2.13Assume that G is a finite p-group,and M1,M2are two distinct maximalsubgroups of G.Then

ProofSincecharfor.Thenandandare abelian.Ifis abelian,thenIfis not abelian,thenSinceis the second maximal subgroup ofby Lemma 2.6.Thus

Lemma 2.14Let G be a P-group.Then

(1)if,then H is a P-group;

(2)if ,then G/N is a P-group.

Proof(1)LetThen there existssuch thatthenIfthenMis metacyclic sinceGis aP-group.By Lemma 2.11(3),Kis metacyclic.SoHis aP-group.

(2)LetThenM?G.IfthenthenMis metacyclic sinceGis aP-group.By Lemma 2.11(4),M/Nis metacyclic.

Lemma 2.15If a finite p-group G has at least one metacyclic maximal subgroup,thenΦ(G)is metacyclic.

ProofIt is straightforward by Lemma 2.11.

Lemma 2.16If G is a finite p-group withexp(G?)=p,then c(G)=2if and only ifΦ(G)≤Z(G).

Proof⇐.It is obvious.

⇒.SinceSincefor allx,y∈G.It follows thatThat is,

Lemma 2.17Assume that G is a finite p-group.If G?∼=C3p,then d(G)=2if and only if G/Φ(G?)G3∼=Mp(n,m,1).

Proof⇐.Sinceand

⇒.SinceOn the other hand,by Lemma 2.9,we get

It follows from Lemma 2.1 thatis minimal nonabelian.SinceG?is nonmetacyclic,Gis nonmetacyclic by Lemma 2.11.It follows from Lemma 2.4 thatis nonmetacyclic.Soby Lemma 2.2.

Lemma 2.18If G is a P-group,then all the proper quotient groups of G/Ω1(G?)are abelian or metacyclic.

ProofLetThenH

Lemma 2.19Let G be a P-group.If G has at least two nonmetacyclic maximal subgroups,then|G?|≤p3.

ProofLetHandKbe two distinct nonmetacyclic maximal subgroups ofG.Then|H?|≤pand|K?|≤p.Moreover,|H?K?|≤p2.It follows from Lemma 2.13 that|G?|≤p3.

3 P-Groups with Exactly One Nonmetacyclic Maximal Subgroup

Theorem 3.1Let G be a p-group with exactly one nonmetacyclic maximal subgroup,where p is an odd prime.Then G is a P-group if and only if G is isomorphic to one of the following pairwise non-isomorphic groups:

ProofObviously,andd(G)≤3.IfGis abelian,thend(G)=3.In this case,,wheren≥m≥k.We claimm=k=1.If not,thenm?=1.Thus there existM1?GandM2?Gsuch thatandare nonmetacyclic.This contradicts the hypothesis.Thus we get the group(1).

Assume thatGis not abelian.LetN≤G?with|N|=pandand

Ifis metacyclic,then,sinceGis nonmetacyclic,G?=Nby Lemma 2.10.Sinceis minimal nonabelian by Lemma 2.1.Thusby Lemma 2.2.We claimm=1.If not,then let.It is easy to see that

HenceM1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.Thus we get the group(4).

Assume thatis nonmetacyclic.Thenis minimal nonmetacyclic orhas exactly one nonmetacyclic maximal subgroup.

If|G|=p4,then,by checking the classification of groups of orderp4,we get thatGis one of the groups(1),(3),(4)whenn=2 and is group(2)whenk=2.

If|G|=p5,then the theorem is true by checking the classification of groups of orderp5given in[12].

Let|G|≥p6.Then.By the classification of minimal nonmetacyclicp-groups by Blackburn[6]and Lemma 2.3,is not minimal nonmetacyclic.Thushas exactly one nonmetacyclic maximal subgroup.By the induction hypothesis,is isomorphic to one of the groups in the theorem.

Case 1

In this case,d(G)=3 and|G?|=p.Such groups are classified by[13,Theorem 3.1],andGis isomorphic to one of the groupsand

Ifwherethen letandIt is easy to see that

It follows thatM1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.

Ifthen we claimn=1.If not,then letandWe have

It follows thatM1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.Thusn=m=1.This is the group(2).

If,then we claimm=k=1.If not,then,whenletand.We have

when,letandWe have

In either case,it follows thatM1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.Thus we get the group(3).

Case 2

SinceLetG=?a,b,c?.Notice thatN≤Z(G).Then

In the same way,[a,b,b]=1.It follows thatG3=1.Thusc(G)=2.Since[a,b]p=[a,bp]=1.Henceand.Let.Then

wherex,y,i,jandtare positive integers,andp|jandp|thave at most one to be true.It is easy to prove that

We discuss the two cases:(modp)andx?≡0(modp).

(i)

LetandM2=?a,c,d?.SinceM1andM2contain a subgroup,M1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.

(ii)x?≡0(modp).(iia)y≡0(modp).By the same argument as that of(i),Ghas two distinct nonmetacyclic maximal subgroups,a contradiction.

(iib)y?≡0(modp).

In this case,Let.Then

This is reduced to the case of(iia).

Case 3

By a similar argument as that of Case 2,we get thatGhas two distinct nonmetacyclic maximal subgroups.Hence this case does not occur.These details are omitted.

Case 4

Let.SinceThus|G4|=1.Hence

Notice that.Thus

It follows thatFinitep-groupsGwithandare classified by[14],and suchGare the groups(1),(2),(4),(5),(6),(7)and(8)withm=1 in[14,Theorem 11].

IfGis the group(1)or(2)in[14,Theorem 11],then it is the group(5)or(6)in this theorem.

IfGis the group(4)in[14,Theorem 11],thenGis minimal nonmetacyclic.This contradicts the hypothesis.

IfGis the group(5)in[14,Theorem 11],then letandIt is easy to verify thatM1andM2contain a subgroupThusM1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.

IfGis one of the groups(6),(7)and(8)in[14,Theorem 11],then it is easy to verify thatGhas two distinct nonmetacyclic maximal subgroups.This is also a contradiction.

Case 5

LetG=?a,b?.Since.It follows from Lemma 2.12 thatG?is abelian.ThusG4=1 andG3=?[c.a],[c,b]??=1.Notice thatThus

and

It follows that,ando(a)=pn,o(c)=p.LetThenand[c,b]=dt,where(modp).

If(modp),then

wheresandtare positive integers.Sincewe let

It is easy to verify thatM1andM2contain a subgroupHenceM1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.

If(modp),then,by replacingawithand lettingthis is reduced to the case ofj≡0(modp).

Case 6wherev=1 or a fixed square non-residue modulop.

LetG=?a,b?.SinceIt follows from Lemma 2.12 thatG?is abelian.ThusG4=1 and.Notice thatbp,cp∈N≤Z(G).Hence

and

It follows thatando(a)=pn,o(c)=p.LetThenandThus

wheres,j,tare positive integers and(modp).Sincewe let

It is easy to verify thatM1andM2contain a subgroupHenceM1andM2are two distinct nonmetacyclic maximal subgroups ofG.This contradicts the hypothesis.

We prove the groups(1)–(6)in the theorem are pairwise non-isomorphic according tod(G)=2 or 3.

Ifd(G)=2,thenGis one of the groups(1),(2)and(3).|G?|=1 for the group(1),andfor the groups(2)and(3).

Let the group(2)be isomorphic to the group(3).Then for the group(3),letand,whereandk3are positive integers with(i1,p)=1,(j2,p)=1 and(k3,p)=1.It follows thata1,b1andc1satisfy the relation of the group(2).ButHence(modpn−1),a contradiction.

Ifd(G)=3,thenGis one of the groups(4),(5)and(6).|G?|=pfor the group(4),andfor the groups(5)and(6).

Let the group(5)be isomorphic to the group(6).Then for the group(5),letandwherei1,j1,k1,i2,j2andk2are positive integers with(i1,p)=1 and(j2,p)=1.Let[a1,b1]=c1.Thena1,b1andc1satisfy the relation of the group(6).Since(modpn).On the other hand,,a

contradiction.

Finally,we prove that the groups(1)–(6)in the theorem satisfy the hypothesis by taking the group(3)for example.In this case,

Obviously,andandare all maximal subgroups ofG,where 0≤i,j,t

Due to the classification of finite 2-groups with exactly one nonmetacyclic maximal subgroup by Z.Janko[15],it is enough to check that those groups in[15]areP-groups.The following theorem lists the results and the proof is omitted.

Theorem 3.2Let G be a2-group with exactly one nonmetacyclic maximal subgroup.Then G is a P-group if and only if G is isomorphic to one of the following pairwise non-isomorphic groups:

4 P-Groups with at Least Two Nonmetacyclic Maximal Subgroups

For convenience of statement,we introduce the following.

Assumption 4.1A finite p-group G has at least two nonmetacyclic maximal subgroups and a metacyclic maximal subgroup.

Theorem 4.1If G is not a D1-group,then G is a P-group satisfying Assumption4.1if and only if G is isomorphic to one of the following non-isomorphic groups:

ProofLetMbe a metacyclic maximal subgroup ofG.Thusd(G)≤3.By Lemma 2.19,Moreover,we claimIf not,thenIt follows from[5,Theorem 3.1]thatGis aD1-group.This contradicts the hypothesis.Soorp3.

Case 1

By Lemma 2.12,G?is abelian.LetM1andM2be two distinct nonmetacyclic maximal subgroups ofG.It follows fromand Lemma 2.13 thatand1.Thusand

IfthenHence,a contradiction.

Ifthen,sinceGis aP-group and not aD1-group,there existssuch thatandMis metacyclic.It follows from Lemma 2.11 thatM?is cyclic.Sinceor,a contradiction.

Assume.SinceGis aP-group and not aD1-group,there existsM?Gsuch thatandMis metacyclic.

Ifd(G)=2,then let.We get thatis minimal nonabelian.ThusM?=It follows from Lemma 2.11 thatM?is cyclic.This is a contradiction.

Assumed(G)=3.Sinceis abelian,SinceThenandThusGis isomorphic to one of the groups in[13,Theorem 3.1].

Ifis isomorphic to the group(1)in[13,Theorem 3.1],that is,

thenandNotice thatThuso[a,b]=4.Since(modThis is a contradiction.

Ifis isomorphic to the group(2)in[13,Theorem 3.1],that is,

then,by the same argument as that of the above paragraph,a contradiction occurs.

Ifis isomorphic to the group(3)in[13,Theorem 3.1],that is,

thenand,thenIfk=1,thenIt follows from Lemma 2.18 thatGis not aP-group.This contradicts the hypothesis.

Ifis isomorphic to the group(4)in[13,Theorem 3.1],that is,

then,by the same argument as that of the group(3),we get thatGis not aP-group.

Ifis isomorphic to the group(5)in[13,Theorem 3.1],that is,

then.LetG=?a,b,c?.SinceThus

Notice thatThusand[a,c],[b,c]and[a,b,a]can not belong toat the same time.

Ifthen there existssuch that.ThusMis nonmetacyclic.By calculation,o([ap,b])=pandSinceIt follows thatGis not aP-group.Iforthen,by similar argument as that of the casewe get thatGis not aP-group.

To sum up,there does not exist aP-group which is not aD1-group with

Case 2

If,then,sinceGis aP-group and not aD1-group,there existssuch thatandMis metacyclic.It follows from Lemma 2.11 thatM?is cyclic.SinceThis is a contradiction.Hence

SinceGis not aD1-group,d(G)=3 by[5,Theorem 3.1].SinceG?is abelian,

ThusLetThenThusis isomorphic to one of the groups in[13,Theorem 3.1].

Ifis isomorphic to one of the groups(1)–(4)in[13,Theorem 3.1],then,by a similar argument as that of Case 1,we get thatGis not aP-group.

Assume thatis isomorphic to one of the group(5)in[13,Theorem 3.1],that is,

thenThusandIfk>1,thenIt follows from Lemma 2.18 thatGis not aP-group.Assumek=1.

SinceSince.Thus

Sinceand[b,c]are contained in Ω1(G?).HenceIt follows that

Ifn≥2,thenc(G)=2.Ifn=1,thenc(G)=3.We discuss the two cases:c(G)=2 andc(G)=3.

Case 2.1c(G)=2.

It is easy to see thato(a)=pn+2.Notice thatAssumeandwithout loss of generality.We discuss the possible value ofiandj.

(i)i=j=0.Then

(ii)andj=0.LetThen

(iii)andi=0.Let.Then

(iv)

Ifo(a)≥o(b),then,lettinga1=abt,wheret=−ij−1,we get[a1,c]=1.This is reduced to the case(iii).

Ifo(a)

We discuss the possible value ofo(b)ando(c).

First,we claimo(b)≥p2ando(c)=p.If not,theno(b)=p.Thusa contradiction.Ifo(c)=p2,thenLet.Thenwhich satisfies the relations.Thuso(c)=p2can be reduced to the case ofo(c)=p.Hence we only need to discuss the possible value ofo(b).

Ifo(b)=pm,thenm≥2.

Ifa,bandcare of the relation(4.1),then

wherem≥2 andn≥2.This is the group(1).

Ifa,bandcare of the relation(4.2),then

wheren≥2.Ifm=2,then,we get[a,c1]=1.This is reduced to the group(1).Ifm≥3,then we get the group(2).

Ifa,bandcare of the relation(4.3),then

wherem≥2 andn≥2.This is the group(3).

If,thenm≥1.Assume,where(p,s)=1 without loss of generality.

Ifa,bandcare of the relation(4.1),then,letifn+1≥m.We getThis is reduced to the case wheno(b)=pm.Ifn+1

This is the group(4).

Ifa,bandcare of the relation(4.2),then,letifn+1>m.We get

This is reduced to the case whenIfn+1≤m,then

This is the group(5).

Ifa,bandcare of the relation(4.3),then,letifn+1≥m.We get

This is reduced to the case wheno(b)=pm.Ifn+1

This is the group(6).

Case 2.2c(G)=3.

In this case,o(a)=p3.

First,we claimo(b)≥p2ifp>2.If not,theno(b)=p.Thus

a contradiction.Notice thatWithout loss of generality,we assume

Ifp=2,then[a,b]=a2ora6.Without loss of generality,we assume

We discuss the possible value ofiandj.

(i)i=j=0.Then

and

(ii)andj=0.

Ifp>2,then letcj−1replacec.We get

Ifp=2 and[a,b]=a6,then letb1=bcifo(b)≥o(c).We get[a,b1]=a2.This is reduced to the relation(4.6).Assumeo(b)

(iii)andi=0.

Ifp>2,then letcj−1replacec.We get

Ifp=2 and[a,b]=a6,then leta1=acifo(c)=p.We get.This is reduced to the relation(4.6).Ifo(c)=p2,then

(iv)and

Ifthen leta1=abt,wheret=−ij−1.We get[a1,c]=1.This is reduced to(iii).

Ifthen letb1=bat,wheret=−ji−1.We get[b1,c]=1.This is reduced to(ii).

We discuss the possible value ofo(b)ando(c).

Case 2.2.1o(b)=pmando(c)=p.

In this case,m≥2 ifp>2 andm≥1 ifp=2.

Ifa,bandcare of the relation(4.4),then

This is the group(7).

Ifa,bandcare of the relation(4.5),then

This is the group(8).

Ifa,bandcare of the relation(4.6),then,letc1=cb−pifp>2 andm=2.We geto(c1)=o(c)and[a,c1]=1.This is reduced to the group(7).Ifp>2 andm≥3,then

This is the group(9).

Ifa,bandcare of the relation(4.7),then

This is the group(10).

Case 2.2.2ando(c)=p.

In this case,m≥1.Without loss of generality,we assume,where(p,s)=1.

Ifa,bandcare of the relation(4.4),then letifp>2 andm≤2.We getThis is reduced to the group(7).Ifp=2 andm=1,then letWe getThis is reduced to the group(7).Ifp>2 andm≥3 orp=2 andm≥2,then

This is the group(11).

Ifa,bandcare of the relation(4.5),then letifm≥2.We geto(a1)=o(a).This is reduced to the group(11).Thusm=1 and

This is the group(12).

Ifa,bandcare of the relation(4.6),then letifp>2 andm=1.We getThis is reduced to the group(10).Ifp>2 andm=2,then letb1=aanda1=b.By symmetry,the relation keeps invariant.Let.ThenThis is reduced to the group(11).Thus,ifp>2 andm≥3 orp=2 andm≥2,then

This is the group(13).

Ifa,bandcare of the relation(4.8),then letifp>2 andm≤2.We getThis is reduced to the group(10).Ifp=2 andm=1,then letWe getThis is reduced to the group(10).Ifp=2 andm≥1 orp>2 andm≥2,then letWe get.This is reduced to the group(11).

In the case ofo(b)=pmando(c)=p2oro(b)=pm+1ando(c)=p2,by a similar argument as that of Cases 2.2.1 and 2.2.2,we get the groups(7)–(13).There is no new group to occur.

Those groups listed in Theorem 4.1 are pairwise non-isomorphic.To prove this is a tedious but not trivial work,so the details are omitted.

Finally,we prove that the groups(1)–(13)in Theorem 4.1 satisfy the hypothesis by taking the group(1)for example.LetGbe the group(1),that is,

By calculation,we getand

are all maximal subgroups ofG,where 0≤i,j,t

For the maximal subgroupM3,ifp=2,then,andandare all maximal subgroups ofM3,where 0≤t,j,m<2.It is easy to prove thatd(H1)=d(H2)=2.It follows from Lemma 2.8 thatM3is metacyclic.Ifp>2,thenThusIt follows from Lemma 2.7 thatM3is metacyclic.On the other hand,d(G)=3 and|G?|=p2.It follows from[5,Theorem 3.1]thatGis not aD1-group.So the group(1)satisfies the hypothesis.

Theorem 4.2If G is a D1-group,then G is a P-group satisfying Assumption4.1if and only if G is isomorphic to one of the following non-isomorphic groups:

ProofSinceGhas one metacyclic maximal subgroup,d(G)≤3.By Lemma 2.19,SinceGis aD1-group,G?is one of the following possible cases by[5,Theorem 3.1]:

Case 1|G?|=1.

In this case,d(G)=3 and.Thus,whereWe claimk=1.If not,thenandare all maximal subgroups ofG,andM1,M2andM3are nonmetacyclic,a contradiction.Thus we get the group(1).Conversely,the group(1)satisfies the theorem’s hypothesis.

Case 2

Ifd(G)=2,thenGis minimal nonabelian by Lemma 2.1.SinceGis nonmetacyclic,by Lemma 2.2.Ifm=1,thenGis the group(4)in Theorem 3.1.Obviously,Gis not the required group by the theorem’s hypothesis.Ifm>1,thenandd(Φ(G))=3.It follows from Lemma 2.11 that Φ(G)is nonmetacyclic.This contradicts Lemma 2.15.

Assumed(G)=3.ThenGis isomorphic to one of the groups in[13,Theorem 3.1].

IfGis isomorphic to the group(1)or(2)in[13,Theorem 3.1],thenGis the group(3)or(4)in Theorem 3.2.SoGis not the required group by the theorem’s hypothesis.

IfGis isomorphic to the group(3)in[13,Theorem 3.1],that is,thenifm>1.Notice thatd(Φ(G))≥3.Then it follows from Lemma 2.11 that Φ(G)is nonmetacyclic.This contradicts Lemma 2.15.Thusm=1.

Ifthen,by the same argument as that of the case whenm>1,a contradiction occurs.Ifk=1,then,it is easy to prove that all maximal subgroups ofGare nonmetacyclic.This contradicts the theorem’s hypothesis.

IfGis isomorphic to the group(4)in[13,Theorem 3.1],that is,then,by the same argument as that of the case whenm>1 above,a contradiction occurs.Thusm=1.Ifn=1,thenGis the group(2)in Theorem 3.1.SoGis not the required group.Ifn>1,then,and

are all maximal subgroups ofG,where 0≤i,j,t

IfGis isomorphic to the group(5)in[13,Theorem 3.1],that is,then,ifm=k=1,thenGis the group(3)in Theorem 3.1.SoGis not the required group.

Ifm=1 andthenand

are all maximal subgroups ofG.It is easy to verify thatM1andM2are nonmetacyclic,andM3is metacyclic.On the other hand,sinceandThus we get the group(3).

Ifandk=1,then Φ(G)=?ap,bp,d?,and

are all maximal subgroups ofG,where 0≤i,j,t

for the group(3),andfor the group(4),so the group(3)is not isomorphic to the group(4).

Case 3

Letand|N|=p.Thend(G/N)=2 andfollows from Lemma 2.1 thatG/Nis minimal nonabelian.

Ifthenby[14,Lemma 8(2)and Lemma 9].ThusGis isomorphic to one of the groups in[14,Theorem 11].IfthenGis isomorphic to one of the groups in[14,Theorems 10 and 12](notice that there is a typographical error in[14],where Theorem 12 is printed to Theorem 11).By checking the list of groups in[14,Theorems 10–12],we get thatGis metacyclic in[14,Theorem 10].This is not the required group.Those groups in[14,Theorems 11 and 12]do not satisfy the theorem’s hypothesis.The details are omitted.So this case does not occur.

Case 4

Sinceis nonmetacyclic.It follows from Lemma 2.11 that Φ(G)is nonmetacyclic.This contradicts Lemma 2.15.So this case does not occur.

Case 5

Ifthen,by the same argument as that of Case 4,Gis not the required group.AssumeBy Lemma 2.16,Φ(G)≤Z(G).ThusGis the group in[13,Theorem 4.7].It is enough to check that those groups satisfy the theorem’s hypothesis.

IfGis one of the groups(A1)–(A3)and(A7)–(A8)in[13,Theorem 4.7],then Φ(G)=whenl>1.In this case,d(Φ(G))=3.Thus Φ(G)is nonmetacyclic.This contradicts Lemma 2.15.Sol=1.It is easy to get that,and

are all maximal subgroups ofG.It is easy to check thatM1andM2are nonmetacyclic,andM3is metacyclic.On the other hand,sincewe getandThus we get the groups(5)–(7)and(11)–(12),respectively.

By the same argument as that of the above paragraph,we get the following conclusions:

IfGis one of the groups(A4)–(A6)in[13,Theorem 4.7],thenl=1,and we get the groups(8)–(10),respectively.

IfGis the group(A9)in[13,Theorem 4.7],thenm=1,and we get the group(13).

IfGis the group(A10)in[13,Theorem 4.7],thenn=1,and we get the group(14).

IfGis one of the groups(A12)or(D)in[13,Theorem 4.7],thenn=1,and we get the group(15).

IfGis one of the groups(A11),(B1)–(B4)and(C)in[13,Theorem 4.7],then there is no group to be the required.

5 P-Groups All of Whose Maximal Subgroups are Nonmetacyclic

For the convenience of classifyingP-groups,here we give the classification ofD1-groups.It is not difficult to do this by[5,Theorem 3.1].

Theorem 5.1Assume that G is a finite p-group.Then G is a D1-group if and only if G is isomorphic to one of the following pairwise non-isomorphic groups:

(I)metacyclic groups

(1)an abelian group with d(G)≤2;

(2)Mp(n,m);

(3)one of the groups in[14,Theorem10];

(II)nonmetacyclic groups

(4)an abelian group with d(G)>2;

(5)(A1∗A2∗A3···∗As)Z(G),where A1,A2,···,Asare minimal nonabelian groups.In particular,if s=1and d(G)=2,then G∼=Mp(n,m,1),where n≥m≥2;

(6)one of the groups in[14,Theorems11and12];

(7)one of the groups in[16,Theorem4.9]with p>2;

(8)one of the groups in[13,Theorem4.8];

(9)one of the groups in[17,Theorem3.1].

ProofIfGis metacyclic,thend(G)≤2 andG?is cyclic.It follows from[5,Theorem 3.1]thatorIfG?=1,then we get the group(1).Ifthenby Lemmas 2.1–2.2.We get the group(2).If,then there existssuch thatand|N|=p.LetThenis minimal nonabelian.SinceGis metacyclic,is metacyclic.Henceby Lemma 2.2.Such groups withandare classified in[14],andGis one of the groups in[14,Theorem 10].By a simple verification,we get the group(3).

Assume thatGis nonmetacyclic.SinceGis aD1-group,G?is one of the following possible cases by[5,Theorem 3.1]:

IfG?=1,thenGis abelian withd(G)>2.This is the group(4).

If|G?|=p,then such groups are characterized in[18]andwhereA1,A2,···,Asare minimal nonabelian groups.This is the group(5).

Ifd(G)=2 and|G?|=p2,then there existsN

If,thenGis isomorphic to one of the groups in[14,Theorem 11].IfthenGis isomorphic to one of the groups in[14,Theorem 12].By a simple veri fi cation,we get the group(6).

Ifd(G)=2,c(G)=3 and,wherep>2,then,by Lemmas 2.9 and 2.17,Gis isomorphic to one of the groups in[16,Theorem 4.9]withp>2.This is the group(7).

Ifd(G)=3,c(G)=2 andthen,by Lemma 2.16,Gis isomorphic to one of the groups in[13,Theorem 4.8].This is the group(8).

Ifd(G)=3,c(G)=2 and,then,by Lemma 2.16,Gis isomorphic to one of the groups in[17,Theorem 3.1].This is the group(9).

Conversely,the groups in the theorem areD1-groups.The details are omitted.

Theorem 5.2If all maximal subgroups of a finite p-group G are nonmetacyclic,then G is a P-group if and only if G is isomorphic to one of the following non-isomorphic groups:

(1)an abelian group with d(G)>3;

(2)(A1∗A2∗A3···∗As)Z(G),where A1,A2,···,Asare minimal nonabelian groups.In particular,if s=1and d(G)=2,then where n≥m≥2.In addition,the following groups are excepted:The groups(2)–(4)and(6)in Theorem3.2,the group(4)in Lemma2.3,the groups(2)–(4)in Theorem4.1and the groups(2)–(4)in Theorem3.1;

(3)one of the groups in[14,Theorems11and12],where the following groups are excepted:The group(1)with p>2,groups(7)and(8)with p=2and m≤2,and groups(2)–(4)in[14,Theorem11];the groups(5)–(6)with p=2and m=2and group(11)in[14,Theorem12];

(4)one of the groups in[16,Theorem4.9]with p>2;

(5)one of the groups in[13,Theorem4.8],except for the groups(A1)–(A8)with l=1,(A9)with m=1,(A10)with n=1,(A12)and(D)in[13,Theorem4.8];

(6)one of the groups in[17,Theorem3.1].

ProofSinceGis aP-group,Gis a nonmetacyclicD1-group by the hypothesis.ThusGis one of the groups of type II in Theorem 5.1.SinceGhas no metacyclic maximal subgroup,we get the groups in the theorem by checking those groups in Theorems 3.1–3.2,4.1–4.2 and Lemma 2.3.The details are omitted.

AcknowledgementThe authors cordially thank the referee for valuable comments.All these help them improve the whole paper substantially.

[1]Berkovich,Y.,Groups of Prime Power Order,Vol.1,Walter de Gruyter,Berlin,New York,2008.

[2]Berkovich,Y.and Janko,Z.,Groups of Prime Power Order,Vol.2,Walter de Gruyter,Berlin,New York,2008.

[3]Berkovich,Y.and Janko,Z.,Groups of Prime Power Order,Vol.3,Walter de Gruyter,Berlin,New York,2011.

[4]R´edei,L.,Das schiefe produkt in der Gruppentheorie,Comment.Math.Helvet,20,1947,225–267.

[5]Zhang,J.Q.and Li,X.H.,Finitep-groups all of whose proper subgroups have small derived subgoups,Sci.China Ser.A,53,2010,1357–1362.

[6]Blackburn,N.,Generalizations of certain elementary theorems onp-groups,Proc.London Math.Soc.,11(3),1961,1–22.

[7]Newman,M.F.and Xu,M.Y.,Metacyclic groups of prime power oder,Chin.Adv.Math.,17,1988,106–107.

[8]Xu,M.Y.and Zhang,Q.H.,A classi fi cation of metacyclic 2-groups,Algebra Colloq.,13(1),2006,25–34.

[9]Xu,M.Y.and Qu,H.P.,Finitep-Groups(in Chinese),Beijing University Press,Beijing,2010.

[10]Xu,M.Y.,An,L.J.and Zhang,Q.H.,Finitep-groups all of whose non-abelian proper subgroups are generated by two elements,J.Algebra,319,2008,3603–3620.

[11]Xu,M.Y.,An Introduction to Finite Groups I(in Chinese),2nd edition,Scientific Press,Beijing,1999.

[12]Zhang,Q.H.,Song,Q.W.and Xu,M.Y.,The classi fi cation of some regularp-groups and its applications,Sci.China Ser.A,49(3),2006,366–386.

[13]An,L.J.,Li,L.L.,Qu,H.P.and Zhang,Q.H.,Finitep-groups with a minimal non-abelian subgroup of indexp(II),Sci.China Ser.A,57(4),2014,737–753.

[14]Li,L.L.,Qu,H.P.and Chen,G.Y.,Central extension of inner abelianp-groups(I),Acta Math.Sinica,Chinese Series,53(4),2010,675–684.

[15]Janko,Z.,Finite 2-groups with exactly one nonmetacyclic maximal subgroup,Israel J.Math.,166,2008,313–347.

[16]An,L.J.,Hu,R.F.and Zhang,Q.H.,Finitep-groups with a minimal non-abelian subgroup of indexp(IV),J.Algebra Appl.,to appear.

[17]Qu,H.P.,Xu,M.Y.and An,L.J.,Finitep-groups with a minimal non-abelian subgroup of indexp(III),Sci.China Ser.A,to appear.

[18]Berkovich,Y.,On subgroups of finitep-groups,J.Algebra,224,2000,198–240.