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On inequalities of subgroups and the structure of finite groups
Journal of Inequalities and Applications volume 2013, Article number: 427 (2013)
Abstract
Let G be a group and H be a subgroup of G. We say that H is weakly Φsupplemented in G if G has a subgroup T such that HT = G and H\cap T\le \mathrm{\Phi}(H), where \mathrm{\Phi}(H) denotes the Frattini subgroup of H. 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.
1 Introduction
All groups in this paper are finite.
Let G be a group and H be a subgroup of G. H is said to be complemented in G if G has a subgroup K such that G=HK and H\cap K=1. 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., [1–4]). For example, a classical result of Hall is about the solubility of a group G satisfying that every Sylow subgroup of G is complemented in G [3]. A subgroup H of a group G is said to be supplemented in G if G has a subgroup T such that G=HT. It is clear that every subgroup of a group G is supplemented in G and every complemented subgroup is also a supplemented subgroup. However, supplemented subgroups may not be complemented. Based on this investigation, we introduce the following new inequalities of subgroups related closely to supplementarity of subgroups.
Definition 1.1 Let G be a group and H be a subgroup of G. H is said to be weakly Φsupplemented in G if G has a subgroup T such that G=HT and H\cap T\le \mathrm{\Phi}(H), where \mathrm{\Phi}(H) is the Frattini subgroup of H.
By the definition, a complemented subgroup is still a weakly Φsupplemented subgroup. However, the converse does not hold.
Example 1.2 Let G={Q}_{8}, the quaternion group of order 8, and let H be a subgroup of G of order 4. Then H is weakly Φsupplemented in G but not complemented in G.
This example shows that the class of all weakly Φsupplemented 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 Φsupplemented 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 3, we present a new criterion for the nilpotency of finite groups. In Section 4, a characterization of supersolubility of groups is given under the assumption that cyclic subgroups of order prime or 4 are weakly Φsupplemented.
The notation and terminology in this paper are standard and the reader is referred to [5] if necessary.
2 Preliminaries
In this section, we give some lemmas which will be useful in the sequel.
Lemma 2.1 Let G be a group. Suppose that H is a weakly Φsupplemented subgroup of G.

(1)
If H\le M\le G, then H is weakly Φsupplemented in M.

(2)
Suppose that N\u22b4G and N\le H. Then H/N is weakly Φsupplemented in G/N.

(3)
If E is a normal subgroup of G and (H,E)=1, then HE/E is weakly Φsupplemented in G/E.
Proof (1) If H is weakly Φsupplemented in G, there exists a subgroup T in G such that HT = G and H\cap T\le \mathrm{\Phi}(H). Since H\le M, H(T\cap M)=HT\cap M=M and H\cap (T\cap M)\le (H\cap T)\cap M\le \mathrm{\Phi}(H)\cap M=\mathrm{\Phi}(H), hence H is weakly Φsupplemented in M.

(2)
Since H is weakly Φsupplemented in G, there exists a subgroup T such that HT = G and H\cap T\le \mathrm{\Phi}(H). Then it is easy to see that (H/N)(TN/N)=G/N and (H/N)\cap (TN/N)=(H\cap T)N/N\le \mathrm{\Phi}(H)N/N\le \mathrm{\Phi}(H/N). Therefore, H/N is weakly Φsupplemented in G/N.

(3)
Assume that H is weakly Φsupplemented in G, and let T be a subgroup of G such that
HT=G\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}H\cap T\le \mathrm{\Phi}(H).
Then (HE/E)(TE/E)=(HT)E/E=G/E. Since (H,E)=1,
Hence, we have that
It follows that
This shows that HE/E is weakly Φsupplemented in G/E. □
Recall that a group G is called quasinilpotent if given any chief factor H/K of G, every automorphism of H/K induced by an element of G is inner; the generalized Fitting subgroup {F}^{\ast}(G) of G is the product of all normal quasinilpotent subgroups of G. The following wellknown facts about the generalized Fitting subgroup of a group G will be used in our proofs (see [[6], Chapter X] and [[7], Lemma 4]).
Lemma 2.2 Let G be a group.

(1)
If G is quasinilpotent and N is a normal subgroup of G, then N and G/N are quasinilpotent.

(2)
If N is a normal subgroup of G, then {F}^{\ast}(N)=N\cap {F}^{\ast}(G).

(3)
F(G)\u2a7d{F}^{\ast}(G)={F}^{\ast}({F}^{\ast}(G)). If {F}^{\ast}(G) is soluble, then {F}^{\ast}(G)=F(G).

(4)
Let p be a prime and P be a normal psubgroup of G. Then {F}^{\ast}(G/\mathrm{\Phi}(P))={F}^{\ast}(G)/\mathrm{\Phi}(P). If P is contained in Z(G), then {F}^{\ast}(G/P)={F}^{\ast}(G)/P.
3 New characterizations of nilpotency
Theorem 3.1 Let G be a group with a normal subgroup N such that G/N is pnilpotent. Suppose that every minimal subgroup of N of order p is contained in Z(G), and every cyclic subgroup of N with order 4 (if p=2) is weakly Φsupplemented in G. Then G is pnilpotent.
Proof Suppose that the assertion is not true, and let G be a counterexample of minimal order. Then:

(1)
G is a minimal nonnilpotent group and G=P\u22caQ, where Q is a Sylow qsubgroup of G, P/\mathrm{\Phi}(P) is a chief factor of G and exp(P)=p or exp(P)=4.
Let L be a proper subgroup of G. Because L/L\cap N\simeq LN/N\le G/N and G/N is pnilpotent, L/L\cap N is pnilpotent. By the hypothesis and Lemma 2.1, every cyclic subgroup of L\cap N with order 4 (if p=2) is weakly Φsupplemented in L. Since every minimal subgroup of N of order p is contained in Z(G) and Z(G)\cap L\le Z(L), every minimal subgroup of L\cap N of order p is contained in Z(L). Therefore L satisfies the hypothesis. Hence, by the choice of G, L is pnilpotent. It follows that G is a minimal nonpnilpotent group. Then, by [[8], Chapter IV, Theorem 5.4] and [[5], Theorem 3.4.11], G has a normal Sylow psubgroup P satisfying that G=P\u22caQ, where Q is a Sylow qsubgroup of G, P/\mathrm{\Phi}(P) is a chief factor of G and exp(P)=p or exp(P)=4.

(2)
There exists an element with order 4 of P.
It is easy to see that P is contained in N. Assume that (2) is false. Then exp(P)=p by (1). By the hypothesis, P is contained in Z(G). Therefore G is nilpotent. This contradiction shows that (2) holds.

(3)
The final contradiction.
Let x\in P and x=4. Then \u3008x\u3009 is weakly Φsupplemented in G. Thus there exists a subgroup T of G such that \u3008x\u3009T=G and \u3008x\u3009\cap T\le \mathrm{\Phi}(\u3008x\u3009)=\u3008{x}^{2}\u3009. Since P/\mathrm{\Phi}(P)\cap T\mathrm{\Phi}(P)/\mathrm{\Phi}(P) is normal in G/\mathrm{\Phi}(P), we have P\cap T\mathrm{\Phi}(P)=P or \mathrm{\Phi}(P). If P\cap T\mathrm{\Phi}(P)=P, then P\le T and therefore \u3008x\u3009\cap T=\u3008x\u3009, a contradiction. Hence P\cap T\mathrm{\Phi}(P)=\mathrm{\Phi}(P) and so P=P\cap \u3008x\u3009T=\u3008x\u3009(P\cap T)=\u3008x\u3009. It follows from [[8], Chapter IV, Theorem 2.8] that G is nilpotent. This contradiction completes the proof. □
Theorem 3.2 Let G be a group with a normal subgroup N such that G/N is nilpotent. Suppose that every minimal subgroup of {F}^{\ast}(N) is contained in Z(G) and that every cyclic subgroup of {F}^{\ast}(N) with order 4 is weakly Φsupplemented in G. Then G is nilpotent.
Proof Suppose that the statement is not true, and let G be a counterexample of minimal order.
Let M be a proper normal subgroup of G. We argue that M satisfies the hypothesis. Since M/M\cap N\simeq MN/N\le G/N, M/M\cap N is nilpotent. By Lemma 2.2, {F}^{\ast}(M\cap N)\le {F}^{\ast}(N). Hence every minimal subgroup of {F}^{\ast}(M\cap N) is contained Z(M), and every cyclic subgroup of {F}^{\ast}(M\cap N) of order 4 is weakly Φsupplemented in M by Lemma 2.1. Therefore M satisfies the hypothesis and so it is nilpotent by the minimality of G. Furthermore, we have that F(G) is the unique maximal normal subgroup of G and G/F(G) is a chief factor of G. In view of Theorem 3.1, we also have N=G={G}^{\mathfrak{N}}, where {G}^{\mathfrak{N}} denotes the smallest normal subgroup of G such that G/{G}^{\mathfrak{N}} is nilpotent. Since {Z}_{\mathrm{\infty}}(G)\cap {G}^{\mathfrak{N}}\le Z({G}^{\mathfrak{N}}) [[5], Corollary 3.2.9], we have {Z}_{\mathrm{\infty}}(G)=Z(G). Let {F}^{\ast}(G)=F, let p be the smallest prime dividing the order of F, and let P be the Sylow psubgroup of F. Then F is a proper normal subgroup of G by Theorem 3.1 and P is normal in G. Let Q be an arbitrary Sylow qsubgroup of G with q\ne p, a prime. By Lemma 2.1 and Theorem 3.1, PQ is pnilpotent and so Q is contained in {C}_{G}(P). This implies that {O}^{p}(G)\le {C}_{G}(P). Hence G={C}_{G}(P) since G={G}^{\mathfrak{N}}. Then P\le Z(G). By Lemma 2.2, {F}^{\ast}(G/P)={F}^{\ast}(G)/P. Obviously, 2 does not divide the order of {F}^{\ast}(G/P). By Lemma 2.1, G/P fulfils the condition and so it is nilpotent by the choice of G, which shows that G is nilpotent, a finial contradiction completing the proof. □
4 New characterizations of supersolubility
Lemma 4.1 Let p be the smallest prime dividing the order of a group G, and let P be a Sylow psubgroup of G. Then G is pnilpotent if and only if every cyclic subgroup of P of order prime or 4 (if P is a nonabelian 2group) not having a supersoluble supplement in G is weakly Φsupplemented in G.
Proof The necessity part is obvious. We prove the sufficiency. Suppose it is false. Then G is nonpnilpotent and so G contains a minimal nonpnilpotent subgroup A. Then A is a minimal nonnilpotent group and possesses the following properties: (1) A=[{A}_{p}]{A}_{q}, where {A}_{p} is the Sylow psubgroup of A, {A}_{q} is the Sylow qsubgroup of A and {A}_{p} is the smallest normal subgroup of A such that A/{A}_{p} is nilpotent; (2) {A}_{p}/\mathrm{\Phi}({A}_{p}) is a chief factor of A; (3) exp({A}_{p})=p or 4. Without loss of generality, we suppose that {A}_{p}\le P. It is easy to see from Lemma 2.1 that every cyclic subgroup of A of order p or 4 (if p=2) not having a supersoluble supplement in A is weakly Φsupplemented in A. Let x be an element of {A}_{p} such that x\notin \mathrm{\Phi}({A}_{p}) and H=\u3008x\u3009. Then H is of order p or 4. Furthermore, one can suppose that H does not have any supersoluble supplement in A because if every cyclic subgroup of A of order p or 4 has a supersoluble supplement in A, then A is nilpotent. Then, by the hypothesis, H is weakly Φsupplemented in G and so A has a subgroup T such that A=HT and H\cap T\le \mathrm{\Phi}(H). Since {A}_{p}/\mathrm{\Phi}({A}_{p})\cap T\mathrm{\Phi}({A}_{p})/\mathrm{\Phi}({A}_{p}) is normal in A/\mathrm{\Phi}({A}_{p}), {A}_{p}\cap T\mathrm{\Phi}({A}_{p})={A}_{p} or \mathrm{\Phi}({A}_{p}) by (2). If the former occurs, then G=T and H\cap T=H, a contradiction. Suppose the latter holds, then {A}_{p}=H, which implies that A is nilpotent, a final contradiction completing the proof. □
Lemma 4.2 Let P be a nontrivial normal psubgroup of G, where p is a prime. If exp(P)=p and every minimal subgroup of G not having a supersoluble supplement in G is weakly Φsupplemented in G, then every chief factor of G below P is cyclic.
Proof Denote \mathrm{\Phi}(P) by L. Consider the factor group P/L. We verify that P/L is a normal subgroup of G/L satisfying the hypothesis. Clearly, exp(P/L)=p. Let H/L be a minimal subgroup of P/L. Then H/L=\u3008x\u3009L/L for some x\in H\setminus L. Then, by the hypothesis, x=p and so \u3008x\u3009 either has a supersoluble supplement in G or is weakly Φsupplemented in G. If \u3008x\u3009 has a supersoluble supplement T in G, then TL/L is a supersoluble supplement of H/L in G/L. If \u3008x\u3009 is weakly Φsupplemented in G, then G has a subgroup T such that G=\u3008x\u3009T and \u3008x\u3009\cap T\le \mathrm{\Phi}(\u3008x\u3009)=1. Therefore
implying that H/L is weakly Φsupplemented in G/L. Hence P/L satisfies the hypothesis and consequently, by induction, every chief factor of G/L below P/L is cyclic provided that L\ne 1. Thus, every chief factor of G below P is cyclic. Now suppose that L=\mathrm{\Phi}(P)=1. Then P is elementary abelian of exponent p. Let N be a minimal subgroup of P. Suppose that N has a supersoluble supplement T in G. If N\le T, G=T is supersoluble and the conclusion follows. If N\cap T=1, then P=P\cap NT=N(P\cap T). Since P is abelian, P\cap T is normal in G and so every chief factor of G below P\cap T is cyclic by induction. It follows that the result holds. If N is weakly Φsupplemented in G, then G has a subgroup T such that N\cap T\le \mathrm{\Phi}(N)=1. As above, we have that P\cap T is normal in G and every chief factor of G below P\cap T is cyclic. Since P=N(P\cap T), every chief factor of G below P is cyclic. Thus, the proof is complete. □
Theorem 4.3 Let G be a group. Then G is supersoluble if and only if G has a normal subgroup E such that for each noncyclic Sylow subgroup P of E, every cyclic subgroup of P of order prime or 4 (if P is a nonabelian 2group) without any supersoluble supplement in G is weakly Φsupplemented in G.
Proof The necessity is clear and we only need to prove the sufficiency. We proceed the proof by induction. By Lemmas 2.1 and 4.1, E is pnilpotent, where p is the smallest prime dividing the order of E. Let H be the Hall {p}^{\prime}subgroup of E. If H is nontrivial, then G/H satisfies the hypothesis by Lemma 2.1 and consequently G/H is supersoluble. By induction again, we have that G is supersoluble. Hence one can assume that E is a pgroup and E is noncyclic. Let {G}^{\mathfrak{U}} denote the smallest normal subgroup of G such that G/{G}^{\mathfrak{U}} is supersoluble. If {G}^{\mathfrak{U}}<E, then G is supersoluble by Lemma 2.1 and induction. Hence we suppose that E={G}^{\mathfrak{U}}.
We assert that G is supersoluble. If not, then G is a minimal nonsupersoluble group by Lemma 2.1. Therefore, G has the following properties: (1) E/\mathrm{\Phi}(E) is a noncyclic chief factor of G; (2) exp(E)=p or 4 (if p=2). If exp(E)=p, then every chief factor of G below E is cyclic by Lemma 4.2 and so G is supersoluble, a contradiction. This shows that p=2 and exp(P)=4. Pick x\in E\setminus \mathrm{\Phi}(E) such that x=4. Set L=\u3008x\u3009 and R=\mathrm{\Phi}(E). First suppose that L has a supersoluble supplement T in G. Then E/R\cap TR/R is normal in G/R and so E/R\cap TR/R=1 or E/R. If E/R\cap TR/R=1, then E=L, which means that E is cyclic, a contradiction. If E/R\cap TR/R=E/R, then G=T is supersoluble, a contradiction. Hence L has no supersoluble supplement in G and so it is weakly Φsupplemented in G by the hypothesis. Then there exists a subgroup T of G such that G=LT and L\cap T\le \mathrm{\Phi}(L)=\u3008{x}^{2}\u3009. If E/R\cap TR/R=1, then E is cyclic as above, a contradiction. If E/R\cap TR/R=E/R, then G=T and so L\cap T=L, contradicting that L\cap T\le \u3008{x}^{2}\u3009. Hence G is supersoluble. □
Theorem 4.4 A group G is supersoluble if and only if G has a normal subgroup E such that G/E is supersoluble, and for each noncyclic Sylow subgroup P of {F}^{\ast}(E), every cyclic subgroup of P of order prime or 4 (if P is a nonabelian 2group) without any supersoluble supplement in G is weakly Φsupplemented in G.
Proof The necessity is obvious and we only need to prove the sufficiency. Suppose this is not true and let G be a counterexample of with G+E minimal. We will derive a contradiction through the following steps.

(1)
F={F}^{\ast}(E)=F(E)<E.
Let F={F}^{\ast}(E). Then F is soluble by Lemmas 2.1 and 4.1. Therefore F={F}^{\ast}(E)=F(E) by Lemma 2.2. If F=E, then G is supersoluble by Theorem 4.3. This contradiction implies that F<E.

(2)
Let p be the smallest prime dividing the order of F. Then p is odd.
Suppose that p=2. Let P be a Sylow 2subgroup of F, and let Q be an arbitrary Sylow qsubgroup of E, where q is an odd prime. Then P is normal in G and PQ is 2nilpotent by the hypothesis, Lemmas 2.1 and 4.1. Hence Q\le {C}_{E}(P) and so {O}^{2}(E)\le {C}_{E}(P). Set V/P={F}^{\ast}(E/P) and W={O}^{2}(V)P. Then W is a normal subgroup of G. Since {O}^{2}(E)\le {C}_{E}(P), it is immediate that every chief factor of W below P is central in W. It follows that W is quasinilpotent and so W\le {F}^{\ast}(E)=F(E). This shows that W is nilpotent and therefore V is soluble. Thus, V/P={F}^{\ast}(E/P) is nilpotent. Again, since {O}^{2}(E)\le {C}_{E}(P), we see that V is nilpotent. Therefore V=F(E)={F}^{\ast}(E) and so {F}^{\ast}(E/P)={F}^{\ast}(E)/P=F(E)/P. By Lemma 2.1 and the choice of G, G/P satisfies the hypothesis and therefore is supersoluble. It follows from Theorem 4.3 that G is supersoluble, a contradiction. Hence p is an odd prime.
Now, let P be the Sylow psubgroup of F.

(3)
Let D be a normal subgroup of G contained in P such that every chief factor of G below D is cyclic. Suppose that 1={D}_{0}\le {D}_{1}\le \cdots \le {D}_{t}=D is a chief series of G below D and C={\bigcap}_{i=1}^{t}{C}_{i}, where {C}_{i}={C}_{G}({D}_{i}/{D}_{i1}). Then E\le C.
It is easy to see that G/C is abelian. Since {F}^{\ast}(E)=F(E)\le E\cap C, {F}^{\ast}(E\cap C)={F}^{\ast}(E). If E\cap C\ne E, then the pair (G,E\cap C) satisfies the hypothesis and G+E\cap C is less that G+E. Thus, G is supersoluble by Lemma 2.1 and the choice of G. Hence, E\le C, as desired.

(4)
P is noncyclic.
Suppose that P is cyclic. Then E stabilizes a chain of subgroups of P by (3), which implies that E/{C}_{E}(P) is a pgroup. Hence {O}^{p}(E)\le {C}_{E}(P). Arguing as in (2), we conclude that G/P satisfies the hypothesis and therefore is supersoluble by the choice of G. In view of Theorem 4.3, G is supersoluble, contrary to the choice of G.
Final contradiction.
By (4), P is noncyclic. Since p is an odd prime by (2), P contains a characteristic subgroup D of exponent p such that every nontrivial {p}^{\prime}automorphism of P induces a nontrivial automorphism of D. By Lemma 4.2, every chief factor of G below D is cyclic. Hence, by (3), E/{C}_{E}(D) is a pgroup, which implies that E/{C}_{E}(P) is still a pgroup by the property of D. Hence {O}^{p}(E)\le {C}_{E}(P). Analogously to the discussion in (2), one can deduce that G/P satisfies the hypothesis and consequently G/P is supersoluble. Now, by Theorem 4.3, G is supersoluble, a final violation finishing the proof. □
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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).
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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.
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Li, J., Xie, F. On inequalities of subgroups and the structure of finite groups. J Inequal Appl 2013, 427 (2013). https://doi.org/10.1186/1029242X2013427
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DOI: https://doi.org/10.1186/1029242X2013427