R E S E A R C H
Open Access
On inequalities of subgroups and the
structure of finite groups
Jinbao Li
1and Fengyan Xie
2**Correspondence:
2Mathematics and Information
Engineering Department, Humanistic Management College of Anyang Normal University, Anyang, 455000, P.R. China
Full list of author information is available at the end of the article
Abstract
LetGbe a group andHbe a subgroup ofG. We say thatHis weakly
-supplemented inGifGhas a subgroupTsuch thatHT=GandH∩T≤
(H), where
(H) denotes the Frattini subgroup ofH. In this paper, properties of this new kind of inequalities of subgroups are investigated and new characterizations of nilpotency and
supersolubility of finite groups in terms of the new inequalities are obtained. MSC: 20D10; 20D15; 20D20
Keywords: finite groups; weakly
-supplemented subgroups; nilpotent groups; supersoluble groups
1 Introduction
All groups in this paper are finite.
LetGbe a group andHbe a subgroup ofG.His said to be complemented inGifGhas a subgroupKsuch thatG=HKandH∩K= . A lot of information about the structure of finite groups can be obtained under the assumption that some families of subgroups are complemented (cf.,e.g., [–]). For example, a classical result of Hall is about the sol-ubility of a groupGsatisfying that every Sylow subgroup ofGis complemented inG[].
A subgroupHof a groupGis said to be supplemented inGifGhas a subgroupT such
thatG=HT. It is clear that every subgroup of a groupGis supplemented inGand every complemented subgroup is also a supplemented subgroup. However, supplemented sub-groups may not be complemented. Based on this investigation, we introduce the following new inequalities of subgroups related closely to supplementarity of subgroups.
Definition . Let Gbe a group and H be a subgroup of G. H is said to be weakly
-supplemented inGifGhas a subgroupTsuch thatG=HTandH∩T≤(H), where
(H) is the Frattini subgroup ofH.
By the definition, a complemented subgroup is still a weakly-supplemented subgroup. However, the converse does not hold.
Example . LetG=Q, the quaternion group of order , and letHbe a subgroup ofG
of order . ThenHis weakly-supplemented inGbut not complemented inG.
This example shows that the class of all weakly-supplemented subgroups is wider than the class of all complemented subgroups. Thus, a question arises naturally:
Can we characterize the structure of finite groups in terms of the weakly-supplemented subgroups?
In this paper, we try to study this question and characterize the structure of groups by this new kind of inequalities of subgroups. In Section , we present a new criterion for the nilpotency of finite groups. In Section , a characterization of supersolubility of groups is given under the assumption that cyclic subgroups of order prime or are weakly
-supplemented.
The notation and terminology in this paper are standard and the reader is referred to [] if necessary.
2 Preliminaries
In this section, we give some lemmas which will be useful in the sequel.
Lemma . Let G be a group.Suppose that H is a weakly-supplemented subgroup of G.
() IfH≤M≤G,thenHis weakly-supplemented inM.
() Suppose thatNGandN≤H.ThenH/Nis weakly-supplemented inG/N.
() IfEis a normal subgroup ofGand(|H|,|E|) = ,thenHE/Eis weakly
-supplemented inG/E.
Proof () IfHis weakly-supplemented inG, there exists a subgroupT inGsuch that
HT =GandH∩T≤(H). SinceH≤M,H(T∩M) =HT∩M=MandH∩(T∩M)≤ (H∩T)∩M≤(H)∩M=(H), henceHis weakly-supplemented inM.
() SinceHis weakly-supplemented inG, there exists a subgroupTsuch thatHT=G
andH∩T≤(H). Then it is easy to see that (H/N)(TN/N) =G/Nand (H/N)∩(TN/N) = (H∩T)N/N≤(H)N/N≤(H/N). Therefore,H/Nis weakly-supplemented inG/N. () Assume thatHis weakly-supplemented inG, and letT be a subgroup ofGsuch that
HT=G and H∩T≤(H).
Then (HE/E)(TE/E) = (HT)E/E=G/E. Since (|H|,|E|) = ,
|HE∩T:HE∩T∩H|,|HE∩T:HE∩T∩E|= .
Hence, we have that
HE∩T= (HE∩T∩H)(HE∩T∩E) = (H∩T)(E∩T). It follows that
(HE/E)∩(TE/E) = (HE∩T)E/E= (H∩T)E/E≤(H)E/E≤(HE/E).
This shows thatHE/Eis weakly-supplemented inG/E.
well-known facts about the generalized Fitting subgroup of a groupGwill be used in our proofs (see [, Chapter X] and [, Lemma ]).
Lemma . Let G be a group.
() IfGis quasinilpotent andNis a normal subgroup ofG,thenNandG/Nare quasinilpotent.
() IfNis a normal subgroup ofG,thenF∗(N) =N∩F∗(G).
() F(G)≤F∗(G) =F∗(F∗(G)).IfF∗(G)is soluble,thenF∗(G) =F(G). () Letpbe a prime andPbe a normalp-subgroup ofG.Then
F∗(G/(P)) =F∗(G)/(P).IfPis contained inZ(G),thenF∗(G/P) =F∗(G)/P.
3 New characterizations of nilpotency
Theorem . Let G be a group with a normal subgroup N such that G/N is p-nilpotent.
Suppose that every minimal subgroup of N of order p is contained in Z(G),and every cyclic subgroup of N with order (if p= )is weakly-supplemented in G.Then G is p-nilpotent.
Proof Suppose that the assertion is not true, and letGbe a counterexample of minimal order. Then:
()Gis a minimal non-nilpotent group andG=PQ, whereQis a Sylowq-subgroup ofG,P/(P) is a chief factor ofGandexp(P) =porexp(P) = .
Let Lbe a proper subgroup of G. BecauseL/L∩NLN/N≤G/N and G/N is p -nilpotent,L/L∩Nisp-nilpotent. By the hypothesis and Lemma ., every cyclic subgroup ofL∩Nwith order (ifp=) is weakly-supplemented inL. Since every minimal sub-group ofNof orderpis contained inZ(G) andZ(G)∩L≤Z(L), every minimal subgroup ofL∩Nof orderpis contained inZ(L). ThereforeLsatisfies the hypothesis. Hence, by the choice ofG,Lisp-nilpotent. It follows thatGis a minimal non-p-nilpotent group.
Then, by [, Chapter IV, Theorem .] and [, Theorem ..], Ghas a normal Sylow
p-subgroupPsatisfying thatG=PQ, whereQis a Sylowq-subgroup ofG,P/(P) is a chief factor ofGandexp(P) =porexp(P) = .
() There exists an element with order ofP.
It is easy to see thatPis contained inN. Assume that () is false. Thenexp(P) =pby (). By the hypothesis,Pis contained inZ(G). ThereforeGis nilpotent. This contradiction shows that () holds.
() The final contradiction.
Letx∈Pand|x|= . Thenxis weakly-supplemented inG. Thus there exists a sub-groupTofGsuch thatxT=Gandx ∩T≤(x) =x. SinceP/(P)∩T(P)/(P)
is normal inG/(P), we haveP∩T(P) =Por(P). IfP∩T(P) =P, thenP≤T and thereforex ∩T=x, a contradiction. HenceP∩T(P) =(P) and soP=P∩ xT=
x(P∩T) =x. It follows from [, Chapter IV, Theorem .] thatGis nilpotent. This
contradiction completes the proof.
Theorem . Let G be a group with a normal subgroup N such that G/N is nilpotent.
Suppose that every minimal subgroup of F∗(N)is contained in Z(G)and that every cyclic subgroup of F∗(N)with orderis weakly-supplemented in G.Then G is nilpotent.
LetMbe a proper normal subgroup ofG. We argue thatMsatisfies the hypothesis. Since
M/M∩NMN/N≤G/N,M/M∩Nis nilpotent. By Lemma .,F∗(M∩N)≤F∗(N). Hence every minimal subgroup ofF∗(M∩N) is containedZ(M), and every cyclic sub-group ofF∗(M∩N) of order is weakly-supplemented inMby Lemma .. Therefore
Msatisfies the hypothesis and so it is nilpotent by the minimality ofG. Furthermore, we have thatF(G) is the unique maximal normal subgroup ofGandG/F(G) is a chief factor ofG. In view of Theorem ., we also haveN=G=GN, whereGNdenotes the smallest
normal subgroup ofGsuch thatG/GNis nilpotent. SinceZ
∞(G)∩GN≤Z(GN) [, Corol-lary ..], we haveZ∞(G) =Z(G). LetF∗(G) =F, letpbe the smallest prime dividing the order ofF, and letPbe the Sylowp-subgroup ofF. ThenFis a proper normal subgroup of
Gby Theorem . andPis normal inG. LetQbe an arbitrary Sylowq-subgroup ofGwith
q=p, a prime. By Lemma . and Theorem .,PQisp-nilpotent and soQis contained in
CG(P). This implies thatOp(G)≤CG(P). HenceG=CG(P) sinceG=GN. ThenP≤Z(G).
By Lemma .,F∗(G/P) =F∗(G)/P. Obviously, does not divide the order ofF∗(G/P). By Lemma .,G/Pfulfils the condition and so it is nilpotent by the choice ofG, which shows
thatGis nilpotent, a finial contradiction completing the proof.
4 New characterizations of supersolubility
Lemma . Let p be the smallest prime dividing the order of a group G,and let P be a
Sylow p-subgroup of G.Then G is p-nilpotent if and only if every cyclic subgroup of P of order prime or (if P is a non-abelian-group)not having a supersoluble supplement in G is weakly-supplemented in G.
Proof The necessity part is obvious. We prove the sufficiency. Suppose it is false. Then
Gis non-p-nilpotent and soGcontains a minimal non-p-nilpotent subgroupA. ThenA
is a minimal non-nilpotent group and possesses the following properties: ()A= [Ap]Aq,
whereAp is the Sylowp-subgroup ofA,Aq is the Sylowq-subgroup ofAandAp is the smallest normal subgroup ofAsuch thatA/Apis nilpotent; ()Ap/(Ap) is a chief factor ofA; ()exp(Ap) =por . Without loss of generality, we suppose thatAp≤P. It is easy to
see from Lemma . that every cyclic subgroup ofAof orderpor (ifp= ) not having a supersoluble supplement inAis weakly-supplemented inA. Letxbe an element of
Ap such that x∈/ (Ap) andH=x. ThenH is of orderpor . Furthermore, one can suppose thatHdoes not have any supersoluble supplement inAbecause if every cyclic subgroup ofAof orderpor has a supersoluble supplement inA, thenAis nilpotent. Then, by the hypothesis,His weakly-supplemented inGand soAhas a subgroupTsuch thatA=HT andH∩T≤(H). SinceAp/(Ap)∩T(Ap)/(Ap) is normal inA/(Ap),
Ap∩T(Ap) =Apor(Ap) by (). If the former occurs, thenG=T andH∩T =H, a
contradiction. Suppose the latter holds, thenAp=H, which implies thatAis nilpotent, a
final contradiction completing the proof.
Lemma . Let P be a non-trivial normal p-subgroup of G,where p is a prime.Ifexp(P) =p and every minimal subgroup of G not having a supersoluble supplement in G is weakly -supplemented in G,then every chief factor of G below P is cyclic.
subgroup ofP/L. ThenH/L=xL/Lfor somex∈H\L. Then, by the hypothesis,|x|=p
and soxeither has a supersoluble supplement inGor is weakly-supplemented inG. If
xhas a supersoluble supplementTinG, thenTL/Lis a supersoluble supplement ofH/L
inG/L. Ifxis weakly-supplemented inG, thenGhas a subgroupTsuch thatG=xT
andx ∩T≤(x) = . Therefore
G/L=xL/L(TL/L) = (H/L)(TL/L) and H/L∩TL/L≤(H/L) = ,
implying thatH/Lis weakly-supplemented inG/L. HenceP/Lsatisfies the hypothesis and consequently, by induction, every chief factor ofG/LbelowP/Lis cyclic provided thatL= . Thus, every chief factor ofGbelowPis cyclic. Now suppose thatL=(P) = . ThenPis elementary abelian of exponentp. LetNbe a minimal subgroup ofP. Suppose thatN has a supersoluble supplementT inG. IfN≤T,G=T is supersoluble and the conclusion follows. IfN∩T= , thenP=P∩NT=N(P∩T). SincePis abelian,P∩Tis normal inGand so every chief factor ofGbelowP∩T is cyclic by induction. It follows that the result holds. IfNis weakly-supplemented inG, thenGhas a subgroupT such thatN∩T≤(N) = . As above, we have thatP∩Tis normal inGand every chief factor ofGbelowP∩T is cyclic. SinceP=N(P∩T), every chief factor ofGbelowPis cyclic.
Thus, the proof is complete.
Theorem . Let G be a group.Then G is supersoluble if and only if G has a normal subgroup E such that for each non-cyclic Sylow subgroup P of E,every cyclic subgroup of P of order prime or (if P is a non-abelian-group)without any supersoluble supplement in G is weakly-supplemented in G.
Proof The necessity is clear and we only need to prove the sufficiency. We proceed the proof by induction. By Lemmas . and .,Eisp-nilpotent, wherepis the smallest prime dividing the order ofE. LetHbe the Hallp-subgroup ofE. IfHis non-trivial, thenG/H
satisfies the hypothesis by Lemma . and consequentlyG/His supersoluble. By induction again, we have thatGis supersoluble. Hence one can assume thatEis ap-group andEis non-cyclic. LetGUdenote the smallest normal subgroup ofGsuch thatG/GUis
super-soluble. IfGU
<E, thenGis supersoluble by Lemma . and induction. Hence we suppose thatE=GU.
We assert thatGis supersoluble. If not, thenGis a minimal non-supersoluble group by Lemma .. Therefore,Ghas the following properties: ()E/(E) is a non-cyclic chief factor ofG; ()exp(E) =por (ifp= ). Ifexp(E) =p, then every chief factor ofGbelow
Eis cyclic by Lemma . and soGis supersoluble, a contradiction. This shows thatp= andexp(P) = . Pickx∈E\(E) such that|x|= . SetL=xandR=(E). First suppose that Lhas a supersoluble supplementT in G. ThenE/R∩TR/Ris normal inG/Rand soE/R∩TR/R= orE/R. IfE/R∩TR/R= , thenE=L, which means thatEis cyclic, a contradiction. IfE/R∩TR/R=E/R, thenG=T is supersoluble, a contradiction. Hence
Lhas no supersoluble supplement inGand so it is weakly-supplemented inGby the
hypothesis. Then there exists a subgroupTofGsuch thatG=LTandL∩T≤(L) =x.
IfE/R∩TR/R= , thenEis cyclic as above, a contradiction. IfE/R∩TR/R=E/R, thenG=T
Theorem . A group G is supersoluble if and only if G has a normal subgroup E such that G/E is supersoluble,and for each non-cyclic Sylow subgroup P of F∗(E),every cyclic subgroup of P of order prime or (if P is a non-abelian-group)without any supersoluble supplement in G is weakly-supplemented in G.
Proof The necessity is obvious and we only need to prove the sufficiency. Suppose this is not true and let Gbe a counterexample of with|G|+|E|minimal. We will derive a contradiction through the following steps.
()F=F∗(E) =F(E) <E.
LetF=F∗(E). ThenFis soluble by Lemmas . and .. ThereforeF=F∗(E) =F(E) by Lemma .. IfF=E, thenGis supersoluble by Theorem .. This contradiction implies thatF<E.
() Letpbe the smallest prime dividing the order ofF. Thenpis odd.
Suppose thatp= . LetPbe a Sylow -subgroup ofF, and letQbe an arbitrary Sylow
q-subgroup ofE, whereqis an odd prime. ThenPis normal inGandPQis -nilpotent by the hypothesis, Lemmas . and .. HenceQ≤CE(P) and soO(E)≤CE(P). SetV/P=
F∗(E/P) andW=O(V)P. ThenW is a normal subgroup ofG. SinceO(E)≤CE(P), it
is immediate that every chief factor ofW belowPis central inW. It follows thatW is quasinilpotent and soW≤F∗(E) =F(E). This shows thatW is nilpotent and thereforeV
is soluble. Thus,V/P=F∗(E/P) is nilpotent. Again, sinceO(E)≤CE(P), we see thatVis
nilpotent. ThereforeV=F(E) =F∗(E) and soF∗(E/P) =F∗(E)/P=F(E)/P. By Lemma . and the choice ofG,G/Psatisfies the hypothesis and therefore is supersoluble. It follows from Theorem . thatGis supersoluble, a contradiction. Hencepis an odd prime.
Now, letPbe the Sylowp-subgroup ofF.
() LetDbe a normal subgroup ofGcontained inPsuch that every chief factor ofG
belowDis cyclic. Suppose that =D≤D≤ · · · ≤Dt=Dis a chief series ofGbelowD
andC=ti=Ci, whereCi=CG(Di/Di–). ThenE≤C.
It is easy to see thatG/Cis abelian. SinceF∗(E) =F(E)≤E∩C,F∗(E∩C) =F∗(E). If
E∩C=E, then the pair (G,E∩C) satisfies the hypothesis and|G|+|E∩C|is less that
|G|+|E|. Thus,Gis supersoluble by Lemma . and the choice ofG. Hence,E≤C, as desired.
()Pis non-cyclic.
Suppose thatPis cyclic. ThenEstabilizes a chain of subgroups ofPby (), which im-plies thatE/CE(P) is ap-group. HenceOp(E)≤CE(P). Arguing as in (), we conclude that G/Psatisfies the hypothesis and therefore is supersoluble by the choice ofG. In view of Theorem .,Gis supersoluble, contrary to the choice ofG.
Final contradiction.
By (),Pis non-cyclic. Sincepis an odd prime by (),Pcontains a characteristic sub-groupDof exponentpsuch that every non-trivialp-automorphism ofPinduces a non-trivial automorphism ofD. By Lemma ., every chief factor ofGbelowDis cyclic. Hence, by (),E/CE(D) is ap-group, which implies thatE/CE(P) is still ap-group by the property ofD. HenceOp(E)≤CE(P). Analogously to the discussion in (), one can deduce thatG/P
satisfies the hypothesis and consequentlyG/Pis supersoluble. Now, by Theorem .,Gis
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
JL carried out the new characterizations of supersolubility. FX conceived of the study and carried out the new characterizations of nilpotency. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Chongqing University of Arts and Science, Chongqing, 402160, P.R. China.2Mathematics
and Information Engineering Department, Humanistic Management College of Anyang Normal University, Anyang, 455000, P.R. China.
Acknowledgements
The authors are grateful to the referees for their helpful suggestions. This work was supported by the National Natural Science Foundation of China (Grant Nos. 11271301, 11171364, 11001226), the Scientific Research Foundation of Chongqing Municipal Education Committee (Grant No. KJ131204), the Science Fund for Creative Research Groups of Chongqing (Grant No. KJTD201321) and the Scientific Research Foundation of Chongqing University of Arts and Sciences (Grant Nos. R2012SC21, Z2012SC25).
Received: 12 April 2013 Accepted: 22 August 2013 Published: 10 September 2013
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doi:10.1186/1029-242X-2013-427