# Theoretical Researches aboutu-Maximal Subgroups and Its Applications in Charactering IntuG

Authors: Li Zhang; Zheng-Qun Cai
Journal: Journal of Chemistry
 (2019)

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## Abstract

LetG be a finite group and u be the class of all finite supersoluble groups. A supersoluble subgroup U of G is called u-maximal in G if for any supersoluble subgroup V of G containing U, V=U. Moreover, IntuG is the intersection of all u-maximal subgroups of G. This paper obtains some new criteria on IntuG, by assuming that some subgroups of G are either Φ-I-supplemented or Φ-I-embedded in G. Here, a subgroup H of G is called Φ-I-supplemented in G if there exists a subnormal subgroup T of G such that G=HT and H∩THG/HG≤ΦH/HGIntuG and Φ-I-embedded in G if there exists a S-quasinormal subgroup T of G such that HT is S-quasinormal in G and H∩THG/HG≤ΦH/HGIntuG.


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## Body

## 1. Introduction

As we know, the model of chemical substances, such as crystal, is a graph, whose change process can be represented by symmetric groups or others. Therefore, group theory plays an important role in chemistry and physics ([1, 2]). However, this paper focuses on a question in group theory, which will promote its development and, consequently, contribute chemistry and physics in many ways.Throughout this paper, all groups are finite.G always denotes a group, p denotes a prime, and πG is the set of all prime divisors of G. All unexplained notation and terminology are standard, as in [3, 4, 5].Recall that a classF of groups is called a formation if F is closed under taking homomorphic images and subdirect products. A formation F is said to be (1) saturated, if G∈F whenever G/ΦG∈F and (2) hereditary, if H∈F whenever H≤G∈F. Following ([3], Chap. III, Definition 3.1), a subgroup U of G is called F-maximal in G provided that (1) U∈F and (2) if U≤V≤G and V∈F, then U=V. Moreover, IntFG [6] denotes the intersection of all F-maximal subgroups of G. As we know, the F-hypercentre ZFG of G is the largest normal subgroup of G such that each G-chief factor H/K below ZFG satisfies H/K⋊G/CGH/K∈F. Clearly, ZFG≤IntFG for any group G (see ([6], Theorem C)).LetF be a hereditary saturated formation and N be a normal subgroup of G contained in ZFG. Then, the following holds (1) AN∈F for any subgroup A of G with A∈F and (2) T∈F for any subgroup T of G with T/N∈F. It is well known that the extensive applications of F-hypercentral subgroups are based on the above properties, and there are two main topics about F-hypercentre: 1 the influence of F-hypercentral subgroups on the structure of finite groups; 2 the criteria of F-hypercentral subgroups.However, in [6], Theorem C shows that the above two properties still hold when N≤IntFG, instead of the stronger condition N≤ZFG. Therefore, it would be rather natural and of great significance to study IntFG. In fact, some recent results in this topic can be found in, for example, [6, 7, 8–12]. Particularly, in [6, 10], the authors have shown that ZuG<IntuG in general, given the condition under which ZuG=IntuG holds for every group G.In connection with the topic ofZuG, a question naturally arises as follows:Question 1.
Can we give a condition under which a normal subgroup of G is contained inIntuG?
In [9], Chen et al. gave some conditions under which a normal subgroup of G contained in IntuG. In this paper, we still pay attention to Question 1. Furthermore, we explore new criteria by the help of the following notion.Definition 1.
LetH be a subgroup of a group G. Then, H is called(1)
Φ-I-supplemented in G if there exists a subnormal subgroup T of G such that G=HT and H∩THG/HG≤ΦH/HGIntuG/HG(2)
Φ-I-embedded in G if there exists a S-quasinormal subgroup T of G such that HT is S-quasinormal in G and H∩THG/HG≤ΦH/HGIntuG/HG
Our main results are the following:Theorem 1.
LetE be a normal subgroup of G. For every prime p∈πE and every noncyclic Sylow p-subgroup P of E, assume that all maximal subgroups of P are either Φ-I-supplemented or Φ-I-embedded in G. Then E≤IntuG.Theorem 2.
LetE be a normal subgroup of G. For every prime p∈πE and every noncyclic Sylow P-subgroup P of E, assume that all cyclic subgroups of P with order P and 4 (when P is a nonabelian 2-group) are either Φ-I-supplemented or Φ-I-embedded in G. Then, E≤IntuG.


## 2. Preliminaries

Lemma 1 (see [13], Chapter 1 or [4], Chapter 1, Lemmas 5.34 and 5.35]).
Assume thatH is a subgroup of G, E≤G, and N⊴G.(1)
IfH is S-quasinormal in G, then H∩E is S-quasinormal in E(2)
IfH is S-quasinormal in G, then HN/N is S-quasinormal in G/N(3)
Assume thatH is a p-group, then H is S-quasinormal in G if and only if OpG≤NGH(4)
The set ofS-quasinormal subgroups of G is a sublattice of the subnormal subgroup lattice of G(5)
IfH is S-quasinormal in G, then HG/HG is nilpotent(6)
IfH is a π-group and H is subnormal in G, then H≤OπGLemma 2 ([6], Theorem C).
LetF be a nonempty hereditary saturated formation. Assume that H, E, and N are subgroups of G with N⊴G.(1)
IntFHN/N≤IntFHN/N(2)
IntFH∩E≤IntFH∩E(3)
IfH/H∩IntFG∈F, then H∈F(4)
IfN≤IntFG, then IntFG/N=IntFG/N(5)
ZFG≤IntFGLemma 3.
Assume thatH is a Φ-I-supplemented (resp., Φ-I-embedded) subgroup of G.(1)
IfN is a normal subgroup of G satisfying either N≤H or H,N=1, then HN/N is Φ-I-supplemented (resp., Φ-I-embedded) in G/N(2)
IfK is a subgroup of G containing H, then H is Φ-I-supplemented (resp., Φ-I-embedded) in KProof.
As the proof forΦ-I-embedded subgroups is similar, we just assume that H is Φ-I-supplemented in G. Then, G has a subnormal subgroup T such that G=HT and H∩THG/HG≤ΦH/HGIntuG/HG.(1)
Clearly,TN/N is a subnormal subgroup of G such that G/N=HN/N⋅TN/N. Consider HN∩TN. If N≤H, then H∩TN=H∩TN by the modular law. Assume that H,N=1. Then HN∩T:H∩T,HN∩T:N∩T=N∩HT,H∩NT=1, which implies that HN∩T=H∩TN∩T. Thus, HN∩TN=H∩TN in both cases. Note that HGN/N≤HN/NG/N=HNG/N, and there exists the isomorphism:(1)HN∩TNHNGHNG=H∩THNGHNG≅H∩THG/HG⋅HNG/HGHNG/HG.So from ([3], Chap. A, Theorem 9.2(e)) and Lemma 2(1), it follows that HN∩TNHNG/HNG≤ΦHN/HNGIntuG/HNG. By the definition, HN/N is Φ-I-supplemented in G/N.(2)
Assume thatT0=T∩K. Clearly, T0 is subnormal in G and HT0=HT∩K=K. Note that HG≤HK and the isomorphism(2)H∩THKHK≅H∩THG/HG⋅HK/HGHK/HG.
Therefore, by ([3], Chap. A, Theorem 9.2(e)) and Lemma 2(2), H∩THK/HK≤ΦH/HKIntuK/HK. So H is Φ-I-supplemented in K.Lemma 4 ([9], Lemmas 2.3 and 2.8).
(1)
Letp be a prime divisor of G with G,p−1=1. Then, IntuG≤IntNpG, where Np denotes the class of all p-nilpotent groups.(2)
Assume thatF is a nonempty hereditary saturated formation. Let N1, N2 be normal subgroups of G, I1/N1=IntFG/N1 and I2/N2=IntFG/N2. Then, I1∩I2/N1∩N2=IntFG/N1∩N2.Lemma 5.
(1)
IfT is a subnormal subgroup of G such that G:T is a power of p, then OpG≤T([13], Lemma 1.1.11).(2)
Assume thatN is a minimal normal subgroup of G. Then, N≤OpG or N=p.Proof.
IfN≰OpG, then N∩OpG=1. Note that NOpG/OpG is a minimal normal subgroup of the p-group G/OpG. So the G-isomorphism N≅NOpG/OpG shows that N=p.Lemma 6 ([3], Chap. A, Lemma 8.4).
LetN be a nilpotent normal subgroup of G and M a maximal subgroup of G satisfying N≰M. Then, N∩M is a normal subgroup of G.Lemma 7 ([5], Chap. VI, Theorem 4.7).
LetP be a Sylow p-subgroup of G and N a normal subgroup of G. If P∩N≤ΦP, then N is p-nilpotent.


## 3. Proofs of Main Theorems

The following two propositions are main steps in the proof of Theorems 1 and 2, which also have independent meanings (see Corollaries 1 and 2).Proposition 1.
LetP be a Sylow p-subgroup of G, where p is a prime divisor of G with G,p−1=1. Assume that all maximal subgroups of P are either Φ-I-supplemented or Φ-I-embedded in G. Then, G is p-nilpotent.Proof.
Suppose that the assertion is false and letG be a counterexample for which G is minimal. We proceed via the following steps:(1)
G has the unique minimal normal subgroup.LetN be a minimal normal subgroup of G. Assume that M/N is an arbitrary maximal subgroup of PN/N, which is a Sylow p-subgroup of G/N. Then M=M∩PN=M∩PN. Denote P1=M∩P. Since P:P1=PN:M=p, P1 is a maximal subgroup of P. By the hypothesis, G has a subnormal (resp., S-quasinormal) subgroup T such that P1 is Φ-I-supplemented (resp., Φ-I-embedded) in G. Note that P1∩N=P∩N is a Sylow p-subgroup of N, so P1N∩T:P1∩T=N∩P1T:P1∩N is a p′-number. However, P1N∩T:N∩T=P1∩TN:P1∩N is a p-number. So we have P1N∩T=P1∩TN∩T and then M/N∩TN/N=P1N∩TN/N=P1∩TN/N. Similarly as Lemma 3, it is easy to show that TN/N is a subnormal (resp., S-quasinormal) subgroup of G/N such that M/N is Φ-I-supplemented (resp., Φ-I-embedded) in G/N. Therefore, G/N satisfies the hypothesis. So the choice of G implies that G/N is p-nilpotent. Consequently, N is the unique minimal normal subgroup of G.(2)
Op′G=1 and IntuG=1.IfOp′G≠1, then the uniqueness of N implies that N≤Op′G. In the case, G/Op′G is p-nilpotent and so is G, a contradiction. Keep Lemmas 2(3) and 5(1) in mind. It is easy to obtain that IntuG=1.(3)
OpG=1.IfOpG>1, then N≤OpG and N>p. Clearly, N≰ΦG. Then, there exists a maximal subgroup M of G such that N≰M. Together with the uniqueness of N, G=N⋊M. Note that OpG∩M⊴G by Lemma 6, so OpG∩M=1 and, consequently, OpG=NOpG∩M=N. Here, P=N⋊P∩M and then P:P∩M=N>p. Thus, P has a maximal subgroup P1 such that P∩M<P1<P. Clearly, P1G=1 and by the hypothesis, P1 is either Φ-I-supplemented or Φ-I-embedded in G.First assume thatP1 is Φ-I-supplemented in G. Combining with (2), there exists a subnormal subgroup T of G such that G=P1T and P1∩T≤ΦP1. According to Lemma 5, we have N≤OpG≤T, which deduces that P1∩N≤P1∩T≤ΦP1. In this case, P1=P1∩NP∩M=P1∩NP∩M=P∩M, a contradiction.Now suppose thatP1 is Φ-I-embedded in G, that is, there exists a S-quasinormal subgroup T of G such that P1T is S-quasinormal in G and P1∩T≤ΦP1. If T=1, then P1 is S-quasinormal in G. From Lemma 1(3) and the choice of P1, we deduce that P1⊴G, a contradiction. So T>1. We further assume that TG=1. By Lemma 1(5), TG is nilpotent. Combining with (2), TG is a p-group. Hence T≤TG=N, which implies that P1T≤P1N=P, that is, P1T is a S-quasinormal p-subgroup of G. From Lemma 1(4)(6), it follows that P1T≤OpG=N, which shows that P1≤N and consequently N=P. In this case, by Lemma 1(3), T is normal in G. Hence, the minimality of N implies that T=N. Consequently, P1=P1∩T≤ΦP1, which shows that P1=1, a contradiction. Therefore, TG>1 and the uniqueness of N implies N≤TG≤T. Consequently, P1∩N≤P1∩T≤ΦP1. Similarly as the above, it is impossible. So we should assume that OpG=1.(4)
Final contradiction.Assume thatN=G, that is, G is a simple group. For any maximal subgroup P1 of P, if T is a subnormal (resp., S-quasinormal) subgroup of G such that P1 is Φ-I-supplemented (resp., Φ-I-embedded) in G, then T=G. As a result, P1=P1∩T≤ΦP1, that is, P1=1, a contradiction. Therefore, N<G. If P≤N, then N satisfies the hypothesis by Lemma 3(2). So the choice of G and the relationship N<G deduce that N is p-nilpotent, which contradicts (2) and (3). In general, we conclude 1<P∩N<P. Let P1 be a maximal subgroup of P containing P∩N. Then P1G=1 and P1 is either Φ-I-supplemented or Φ-I-embedded in G.First assume thatP1 is Φ-I-supplemented in G. So there exists a subnormal subgroup T of G such that G=P1T and P1∩T≤ΦP1. According to Lemma 6, we have that N≤OpG≤T. Hence, P1∩N≤P1∩T≤ΦP1. Note that P1∩N=P∩N and ΦP1≤ΦP. So P∩N≤ΦP. However, it deduces that N is p-nilpotent by Lemma 7, a contradiction.IfP1 is Φ-I-embedded in G, then G has a S-quasinormal subgroup T such that P1T is S-quasinormal in G and P1∩T≤ΦP1. Note that T=1 implies that P1 is S-quasinormal in G, which contradicts (3) and Lemma 1(4)(6). Moreover, if TG=1, then by Lemma 1(5) and the uniqueness of N, N≤TG and N is nilpotent, which contradicts (1) and (3). Therefore, TG>1 and consequently N≤TG≤T. In this case, we finally conclude that P∩N=P1∩N≤P1∩T≤ΦP1≤ΦP. By Lemma 7, we also have that N is p-nilpotent, a contradiction. This contradiction completes the proof.Proposition 2.
LetP be a Sylow p-subgroup of G, where p is a prime divisor of G with G,p−1=1. Assume that all cyclic subgroups of P of order p and order 4 when P is a nonabelian 2-group are either Φ-I-supplemented or Φ-I-embedded in G. Then, G is p-nilpotent.Proof.
Suppose that the assertion is false and letG be a counterexample of minimal order.
LetM be a proper subgroup of G and Mp a Sylow p-subgroup of M. Then, Mp≤Pg for some g∈G. Now consider Mg, which has a Sylow p-subgroup Mpg≤P. By Lemma 3(2), Mg satisfies the hypothesis for G. So Mg is p-nilpotent by the minimality of G. Consequently, M is p-nilpotent, and G is a minimal non-p-nilpotent group. By [14], Theorem 3.4.11, the following hold (i) G=P⋊Q, where P=GN and Q is a Sylow q-subgroup of G with q≠p; (ii) P/ΦP is a noncyclic G-chief factor; (iii) the exponent of P is p or 4 (when P is a nonabelian 2-subgroup). Take x∈P/ΦP, and denote H=x. Then, H has order p or 4, HG≤ΦP, and ΦH/HG=1. Moreover, H is either Φ-I-supplemented or Φ-I-embedded in G by the hypothesis.
First assume thatH is Φ-I-embedded in G. Let T be a S-quasinormal subgroup of G such that HT is S-quasinormal in G and H∩THG/HG≤IntuG/HG. Note that P/ΦP is a G-chief factor, so we separate the proof into three cases: (1) P∩TG=P; (2)P∩TG≤ΦP; (3) P∩TG≤ΦP and P∩TG=P. If the first case holds, then P≤T and consequently H/HG=H∩THG/HG≤IntuG/HG. From Lemmas 2(1) and 4(1), we further deduce that 1<HΦP/ΦP≤IntuG/ΦP≤IntNpG/ΦP, where Np denotes the class of all p-nilpotent groups. So PΦP/ΦP≤IntNpG/ΦP by (ii). Together with Lemma 2(3), we finally have that G/ΦP is p-nilpotent, and so is G, a contradiction. In case (2), we have P∩T≤ΦP. Then HΦP/ΦP=HP∩TΦP/ΦP=P/ΦP∩HTΦP/ΦP is S-quasinormal in G by Lemma 1(2)(4). Note that P/ΦP is abelian. So HΦP/ΦP⊴G/ΦP according to Lemma 1(3). Consequently, P/ΦP=HΦP/ΦP by (ii), a contradiction. Lastly, suppose that case (3) holds, that is, P∩TG≤ΦP and P≤TG. If TG=G, then G/TG is p-nilpotent by Lemma 1(5). By (i), we have that G/P∩TG is p-nilpotent, and furthermore, G/ΦP is p-nilpotent, which deduces that G is p-nilpotent, a contradiction. So we have P≤TG<G and TG is p-nilpotent as G is a minimal non-p-nilpotent group. Note that Op′G=1, so T≤TG=P. By Lemma 1(2), we obtain that TΦP/ΦP is a S-quasinormal subgroup of G/ΦP contained in P/ΦP. Consequently, we deduce that TΦP/ΦP⊴G/ΦP from Lemma 1(3). Therefore, TΦP=ΦP or TΦP=P, that is, T≤ΦP or T=P. However, from the proof of cases (1) and (2), we know that it is impossible.
Then assume thatH is Φ-I-supplemented in G, and T is a subnormal subgroup of G such that G=HT and H∩THG/HG≤IntuG/HG. Clearly, P≤OpG≤T by Lemma 5(1). So we have H/HG=H∩THG/HG≤IntuG/HG. Similarly as the first case above, we know that it is impossible. Thus, the assertion holds.
Now we true to prove Theorems 1 and 2.Proof of Theorem 1.
Suppose that the result is false and letG,E be a counterexample for which G+E is minimal. We proceed via the following steps.(1)
E is a p-group.Assume thatπE>1, p is the smallest prime divisor of E and Ep is a Sylow p-subgroup of E. If Ep is cyclic, then E is p-nilpotent (see [15], Theorem 10.1.9). Now assume that Ep is noncyclic. From the hypothesis and Lemma 3(2), it follows that E satisfies the hypothesis of Proposition 1. So we have that E is still p-nilpotent. Let Ep′ be the normal p′-Hall subgroup of E. Then Ep′ is a normal subgroup of G and G,Ep′ satisfies the hypothesis. Hence Ep′≤IntuG by the choice of G,E.Suppose thatEp is cyclic. Then, E/Ep′≤ZuG/Ep′ for the G-isomorphism E/Ep′≅Ep. By Lemma 2(5), E/Ep′≤IntuG/Ep′. Now assume that Ep is noncyclic. By Lemma 3(1), we can easily obtain that G/Ep′,E/Ep′ satisfies the hypothesis. Analogously, the choice of G,E implies that E/Ep′≤IntuG/Ep′. Therefore, in any case, E/Ep′≤IntuG/Ep′. Furthermore, E≤IntuG by Lemma 2(4), a contradiction. Thus, πE=1, that is, E is a p-group.(2)
G/E∈u.SinceE≰IntuG, there exists a u-maximal subgroup X of G such that E≰X. By Lemma 3(2), EX,E satisfies the hypothesis for G,E. If EX<G, then the choice of G,E implies that E≤IntuEX. Note that EX/E is supersoluble for the isomorphism EX/E≅X/E∩X. So EX∈u by Lemma 2(3). Furthermore, the choice of X implies EX=X, that is, E≤X. This contradiction shows that G=EX and, consequently, G/E is supersoluble as G/E≅X/E∩X.(3)
G has the unique Sylow p-subgroup.Letq be the largest prime dividing G and Gq a Sylow q-subgroup of G. Assume that q>p. Note that G/E is supersoluble. So GqE/E⊴G/E and GqE⊴G. Consider GqE,E, which satisfies the hypothesis by Lemma 3(2). Note that p is the smallest prime divisor of GqE and E is the Sylow p-subgroup of GqE. So by Proposition 1, GqE is p-nilpotent. Therefore, Gq⊴GqE and, consequently, Gq⊴G. Now consider G/Gq,GqE/Gq, which satisfies the hypothesis by Lemma 3(1). So the choice of G,E implies that GqE/E≤IntuG/E. Moreover, the isomorphism G/Gq/GqE/Gq≅G/E/GqE/E deduces that G/Gq/GqE/Gq is supersoluble. Together with Lemma 2(3), we finally obtain G/Gq is supersoluble. Furthermore, G is supersoluble by the isomorphism G≅G/E∩Gq. This contradiction shows q=p. So (3) holds.(4)
Final contradiction.LetN be a minimal normal subgroup of G contained in E. Consider G/N,E/N, which satisfies the hypothesis by Lemma 3(1). So the choice of G,E implies that E/N≤IntuG/N. Note that G/N/E/N is supersoluble by the isomorphism G/N/E/N≅G/E. Combining with Lemma 2(3), G/N is supersoluble. Therefore, N≰ΦG and N is the unique minimal normal subgroup of G contained in E.Note thatE∩ΦG is a normal subgroup of G contained in E. So the uniqueness of N implies that E∩ΦG=1. Consequently, E is the direct product of the minimal normal subgroups of G contained in E (see [14], Chap. 1, Lemma 1.8.17). Furthermore, E=N by the uniqueness of N.Note thatE∩ZGp is a nontrivial normal subgroup of G. So E∩ZGp=E, that is, E≤ZGp. Take P1 be an arbitrary maximal subgroup of E. Clearly, P1G=1, ΦP1=1 and by the hypothesis, P1 is either Φ-I-supplemented or Φ-I-embedded in G.Assume thatP1 is Φ-I-supplemented in G. Let T be a subnormal subgroup of G such that G=P1T and P1∩T≤IntuG. Then, 1<E∩T⊴G and, consequently, E∩T=E by the minimality of E. In this case, P1=P1∩T≤IntuG, which implies E≤IntuG by the minimality of E again, a contradiction.Now suppose thatP1 is Φ-I-embedded in G and T is a S-quasinormal subgroup of G such that P1T is S-quasinormal in G and P1∩T≤IntuG. It is easy to show that the above holds if T is replaced by T∩E. So, without loss of generality, assume that T≤E. Since T is S-quasinormal in G, we have T⊴G by Lemma 1(3) and the relationship E≤ZGp. Therefore, T=1 or T=E. If T=1, then P1 is S-quasinormal in G and, similarly as above, P1⊴G, which contradicts the minimality of E. But if T=E, then P1=P1∩T≤IntuG, which also deduces a contradiction as above. So the proof is completed.Proof of Theorem 2.
Suppose that the result is false and letG,E be a counterexample for which G+E is minimal. Then, G is not supersolvable. Similarly as steps (1) and (2) in the proof of Theorem 1, assume that Gu≤E and E is a p-group.
LetM be any proper subgroup of G. Consider M,E∩M, which satisfies the hypothesis for G,E by Lemma 3(2). So the minimality of G,E deduces that E∩M≤IntuM. Note that M/E∩M is supersolvable by the isomorphism M/E∩M≅ME/E≤G/E. So M is supersolvable by Lemma 2(3). Consequently, G is a minimal nonsupersolvable group and from ([14], Theorem 3.4.2), we deduce that (i) E=Gu is a p-subgroup of G; (ii) E/ΦE is a noncyclic G-chief factor; (iii) the exponent of E is p or 4 (when E is a non-abelian 2-group). Similarly, as step (3) of the proof of Theorem 1, the Sylow p-subgroup Gp of G is normal in G. Note that E/ΦE∩ZGp/ΦE is a nontrivial normal subgroup of G/ΦE, so E/ΦE≤ZGp/ΦE.
Takex∈E/ΦE, and denote H=x. Then, H has order p or 4, HG≤ΦE, and ΦH/HG=1. By the hypothesis, H is either Φ-I-supplemented or Φ-I-embedded in G.
Assume thatH is Φ-I-embedded in G. Let T be a S-quasinormal subgroup of G such that HT is S-quasinormal in G and H∩THG/HG≤IntuG/HG. Clearly, T∩E is another S-quasinormal subgroup of G such that H is Φ-I-embedded in G. So without loss of generality, assume that T≤E. Then, TΦE/ΦE is a S-quasinormal subgroup of G/ΦE contained in E/ΦE. Together with Lemma 1(3) and the relationship E/ΦE≤ZGp/ΦE, we have TΦE/ΦE⊴G/ΦE. Thus, T≤ΦE or T=E. If T≤ΦE, then HΦE/ΦE=HTΦE/ΦE is S-quasinormal in G and then HΦE/ΦE⊴G/ΦE. The choice of H shows that E/ΦE=HΦE/ΦE, which contradicts (ii). Assume that T=E. Then, H/HG=H∩THG/HG≤IntuG/HG and by Lemma 2(2), HΦE/ΦE≤IntuG/ΦE. Together with (ii), we have E/ΦE≤IntuG/HG. Recall that G/E is supersoluble. Therefore, G/ΦE is supersoluble by the isomorphism G/ΦE/E/ΦE≅G/E and Lemma 2(3). Furthermore, we have G is supersoluble, a contradiction.
Now assume thatH is Φ-I-supplemented in G and T is a subnormal subgroup of G such that G=HT and H∩THG/HG≤IntuG/HG. It is easy to show that E∩TΦE⊴G. So E∩T≤ΦE or E≤T. Similarly as the above, E≤T is impossible. However, E∩T≤ΦE implies that E=E∩HT=HE∩T=H, which contradicts (ii). Then, we complete the proof.


## 4. Some Applications

The following result follows directly from Lemma 2(3) and Theorems 1 and 2.Corollary 1.
LetE be a normal subgroup of G such that G/E is supersoluble. Then, G is supersoluble if and only if for every prime p∈πE and every noncyclic Sylow p-subgroup P of E, one of the following holds (1)
All maximal subgroups of P are either Φ-I-supplemented or Φ-I-embedded in G(2)
All cyclic subgroups of P with order p and 4 (when P is a non-abelian 2-group) are either Φ-I-supplemented or Φ-I-embedded in GRecall also that a subgroupH of G is called as follows: c-normal in G [16] if G has a normal subgroup T such that G=HT and H∩T≤HG; uc-normal in G [17] if G has a subnormal subgroup T such that G=HT and H∩THG/HG≤ZuG/HG; SΦ-supplemented in G [18] if G has a subnormal subgroup T such that G=HT and H∩T≤ΦH. Obviously, c-normal subgroups, uc-normal subgroups, and SΦ-supplemented subgroups of G are all Φ-I-supplemented in G. Hence, we have the following.Corollary 2.
G is supersoluble, if one of the following holds(a)
Every maximal subgroup of every Sylow subgroup ofG is uc-normal in G ([17], Corollary 1.3)(b)
All cyclic subgroups ofG with prime order or order 4 are uc-normal in G ([17], Corollary 1.5)(c)
Every maximal subgroup of every Sylow subgroup ofG is c-normal in G ([16], Theorem 4.1)(d)
All cyclic subgroups ofG with prime order or order 4 are c-normal in G ([16], Theorem 4.2)From Proposition 2, we obtain the following.Corollary 3 ([19], Lemma 3.1).
Letp be the smallest prime dividing G and P a Sylow p-subgroup of G. If all subgroups of P with order p or order 4 are c-normal in G, then G is p-nilpotent.Corollary 4 ([18], Theorem 3.1).
LetP be a Sylow p-subgroup of G, where p is a prime dividing G such that G,p−1=1. If every maximal subgroup of P is SΦ-supplemented in G, then G is p-nilpotent.Moreover, Theorem 3 in [20] and Theorems 3.3 and 3.4 in [21] follow directly from Theorem 1.



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