Theoretical Researches about -Maximal Subgroups and Its Applications in Charactering

Abstract

Let G be a finite group and be the class of all finite supersoluble groups. A supersoluble subgroup U of G is called -maximal in G if for any supersoluble subgroup V of G containing U, . Moreover, is the intersection of all -maximal subgroups of G. This paper obtains some new criteria on , by assuming that some subgroups of G are either -I-supplemented or -I-embedded in G. Here, a subgroup H of G is called -I-supplemented in G if there exists a subnormal subgroup T of G such that and and -I-embedded in G if there exists a S-quasinormal subgroup T of G such that is S-quasinormal in G and .

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, denotes a prime, and is the set of all prime divisors of . All unexplained notation and terminology are standard, as in [3, 4, 5].

Recall that a class of groups is called a formation if is closed under taking homomorphic images and subdirect products. A formation is said to be (1) saturated, if whenever and (2) hereditary, if whenever . Following ([3], Chap. III, Definition 3.1), a subgroup U of G is called -maximal in G provided that (1) and (2) if and , then . Moreover, [6] denotes the intersection of all -maximal subgroups of G. As we know, the -hypercentre of G is the largest normal subgroup of G such that each G-chief factor below satisfies . Clearly, for any group G (see ([6], Theorem C)).

Let be a hereditary saturated formation and N be a normal subgroup of G contained in . Then, the following holds (1) for any subgroup A of G with and (2) for any subgroup T of G with . It is well known that the extensive applications of -hypercentral subgroups are based on the above properties, and there are two main topics about -hypercentre: the influence of -hypercentral subgroups on the structure of finite groups; the criteria of -hypercentral subgroups.

However, in [6], Theorem C shows that the above two properties still hold when , instead of the stronger condition . Therefore, it would be rather natural and of great significance to study . 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 in general, given the condition under which holds for every group G.

In connection with the topic of , a question naturally arises as follows:

Question 1. Can we give a condition under which a normal subgroup of G is contained in ?
In [9], Chen et al. gave some conditions under which a normal subgroup of G contained in . In this paper, we still pay attention to Question 1. Furthermore, we explore new criteria by the help of the following notion.

Definition 1. Let H be a subgroup of a group G. Then, H is called(1)-I-supplemented in G if there exists a subnormal subgroup T of G such that and (2)-I-embedded in G if there exists a S-quasinormal subgroup T of G such that is S-quasinormal in G and Our main results are the following:

Theorem 1. Let E be a normal subgroup of G. For every prime and every noncyclic Sylow -subgroup of E, assume that all maximal subgroups of are either -I-supplemented or -I-embedded in G. Then .

Theorem 2. Let E be a normal subgroup of G. For every prime and every noncyclic Sylow -subgroup of E, assume that all cyclic subgroups of with order and 4 (when is a nonabelian 2-group) are either -I-supplemented or -I-embedded in G. Then, .

2. Preliminaries

Lemma 1 (see [13], Chapter 1 or [4], Chapter 1, Lemmas 5.34 and 5.35]). Assume that H is a subgroup of G, , and .(1)If H is S-quasinormal in G, then is S-quasinormal in E(2)If H is S-quasinormal in G, then is S-quasinormal in (3)Assume that H is a -group, then H is S-quasinormal in G if and only if (4)The set of S-quasinormal subgroups of G is a sublattice of the subnormal subgroup lattice of G(5)If H is S-quasinormal in G, then is nilpotent(6)If H is a π-group and H is subnormal in G, then

Lemma 2 ([6], Theorem C). Let be a nonempty hereditary saturated formation. Assume that H, E, and N are subgroups of G with .(1)(2)(3)If , then (4)If , then (5)

Lemma 3. Assume that H is a -I-supplemented (resp., -I-embedded) subgroup of G.(1)If N is a normal subgroup of G satisfying either or , then is -I-supplemented (resp., -I-embedded) in (2)If K is a subgroup of G containing H, then H is -I-supplemented (resp., -I-embedded) in K

Proof. As the proof for -I-embedded subgroups is similar, we just assume that H is -I-supplemented in G. Then, G has a subnormal subgroup T such that and .(1)Clearly, is a subnormal subgroup of G such that . Consider . If , then by the modular law. Assume that . Then , which implies that . Thus, in both cases. Note that , and there exists the isomorphism: So from ([3], Chap. A, Theorem 9.2(e)) and Lemma 2(1), it follows that . By the definition, is -I-supplemented in .(2)Assume that . Clearly, is subnormal in G and . Note that and the isomorphismTherefore, by ([3], Chap. A, Theorem 9.2(e)) and Lemma 2(2), . So H is -I-supplemented in K.

Lemma 4 ([9], Lemmas 2.3 and 2.8). (1)Let be a prime divisor of with . Then, , where denotes the class of all -nilpotent groups.(2)Assume that is a nonempty hereditary saturated formation. Let , be normal subgroups of G, and . Then, .

Lemma 5. (1)If T is a subnormal subgroup of G such that is a power of , then ([13], Lemma 1.1.11).(2)Assume that N is a minimal normal subgroup of G. Then, or .

Proof. If , then . Note that is a minimal normal subgroup of the -group . So the G-isomorphism shows that .

Lemma 6 ([3], Chap. A, Lemma 8.4). Let N be a nilpotent normal subgroup of G and M a maximal subgroup of G satisfying . Then, is a normal subgroup of G.

Lemma 7 ([5], Chap. VI, Theorem 4.7). Let be a Sylow -subgroup of G and N a normal subgroup of G. If , then N is -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. Let be a Sylow -subgroup of G, where is a prime divisor of with . Assume that all maximal subgroups of are either -I-supplemented or -I-embedded in G. Then, G is -nilpotent.

Proof. Suppose that the assertion is false and let G be a counterexample for which is minimal. We proceed via the following steps:(1)G has the unique minimal normal subgroup. Let N be a minimal normal subgroup of G. Assume that is an arbitrary maximal subgroup of , which is a Sylow -subgroup of . Then . Denote . Since , is a maximal subgroup of . By the hypothesis, G has a subnormal (resp., S-quasinormal) subgroup T such that is -I-supplemented (resp., -I-embedded) in G. Note that is a Sylow -subgroup of N, so is a -number. However, is a -number. So we have and then . Similarly as Lemma 3, it is easy to show that is a subnormal (resp., S-quasinormal) subgroup of such that is -I-supplemented (resp., -I-embedded) in . Therefore, satisfies the hypothesis. So the choice of G implies that is -nilpotent. Consequently, N is the unique minimal normal subgroup of G.(2) and . If , then the uniqueness of N implies that . In the case, is -nilpotent and so is G, a contradiction. Keep Lemmas 2(3) and 5(1) in mind. It is easy to obtain that .(3). If , then and . Clearly, . Then, there exists a maximal subgroup M of G such that . Together with the uniqueness of N, . Note that by Lemma 6, so and, consequently, . Here, and then . Thus, has a maximal subgroup such that . Clearly, and by the hypothesis, is either -I-supplemented or -I-embedded in G. First assume that is -I-supplemented in G. Combining with (2), there exists a subnormal subgroup T of G such that and . According to Lemma 5, we have , which deduces that . In this case, , a contradiction. Now suppose that is -I-embedded in G, that is, there exists a S-quasinormal subgroup T of G such that is S-quasinormal in G and . If , then is S-quasinormal in G. From Lemma 1(3) and the choice of , we deduce that , a contradiction. So . We further assume that . By Lemma 1(5), is nilpotent. Combining with (2), is a -group. Hence , which implies that , that is, is a S-quasinormal -subgroup of G. From Lemma 1(4)(6), it follows that , which shows that and consequently . In this case, by Lemma 1(3), T is normal in G. Hence, the minimality of N implies that . Consequently, , which shows that , a contradiction. Therefore, and the uniqueness of N implies . Consequently, . Similarly as the above, it is impossible. So we should assume that .(4)Final contradiction. Assume that , that is, G is a simple group. For any maximal subgroup of , if T is a subnormal (resp., S-quasinormal) subgroup of G such that is -I-supplemented (resp., -I-embedded) in G, then . As a result, , that is, , a contradiction. Therefore, . If , then N satisfies the hypothesis by Lemma 3(2). So the choice of G and the relationship deduce that N is -nilpotent, which contradicts (2) and (3). In general, we conclude . Let be a maximal subgroup of containing . Then and is either -I-supplemented or -I-embedded in G. First assume that is -I-supplemented in G. So there exists a subnormal subgroup T of G such that and . According to Lemma 6, we have that . Hence, . Note that and . So . However, it deduces that N is -nilpotent by Lemma 7, a contradiction. If is -I-embedded in G, then G has a S-quasinormal subgroup T such that is S-quasinormal in G and . Note that implies that is S-quasinormal in G, which contradicts (3) and Lemma 1(4)(6). Moreover, if , then by Lemma 1(5) and the uniqueness of N, and N is nilpotent, which contradicts (1) and (3). Therefore, and consequently . In this case, we finally conclude that . By Lemma 7, we also have that N is -nilpotent, a contradiction. This contradiction completes the proof.

Proposition 2. Let be a Sylow -subgroup of G, where is a prime divisor of with . Assume that all cyclic subgroups of of order and order 4 when is a nonabelian 2-group are either -I-supplemented or -I-embedded in G. Then, G is -nilpotent.

Proof. Suppose that the assertion is false and let G be a counterexample of minimal order.
Let M be a proper subgroup of G and a Sylow -subgroup of M. Then, for some . Now consider , which has a Sylow -subgroup . By Lemma 3(2), satisfies the hypothesis for G. So is -nilpotent by the minimality of G. Consequently, M is -nilpotent, and G is a minimal non--nilpotent group. By [14], Theorem 3.4.11, the following hold (i) , where and Q is a Sylow q-subgroup of G with ; (ii) is a noncyclic G-chief factor; (iii) the exponent of is or 4 (when is a nonabelian 2-subgroup). Take , and denote . Then, H has order or 4, , and . Moreover, H is either -I-supplemented or -I-embedded in G by the hypothesis.
First assume that H is -I-embedded in G. Let T be a S-quasinormal subgroup of G such that is S-quasinormal in G and . Note that is a G-chief factor, so we separate the proof into three cases: (1) ; (2); (3) and . If the first case holds, then and consequently . From Lemmas 2(1) and 4(1), we further deduce that , where denotes the class of all -nilpotent groups. So by (ii). Together with Lemma 2(3), we finally have that is -nilpotent, and so is G, a contradiction. In case (2), we have . Then is S-quasinormal in G by Lemma 1(2)(4). Note that is abelian. So according to Lemma 1(3). Consequently, by (ii), a contradiction. Lastly, suppose that case (3) holds, that is, and . If , then is p-nilpotent by Lemma 1(5). By (i), we have that is -nilpotent, and furthermore, is -nilpotent, which deduces that G is -nilpotent, a contradiction. So we have and is -nilpotent as G is a minimal non--nilpotent group. Note that , so . By Lemma 1(2), we obtain that is a S-quasinormal subgroup of contained in . Consequently, we deduce that from Lemma 1(3). Therefore, or , that is, or . However, from the proof of cases (1) and (2), we know that it is impossible.
Then assume that H is -I-supplemented in G, and T is a subnormal subgroup of G such that and . Clearly, by Lemma 5(1). So we have . 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 let be a counterexample for which is minimal. We proceed via the following steps.(1)E is a -group. Assume that , is the smallest prime divisor of and is a Sylow -subgroup of E. If is cyclic, then E is -nilpotent (see [15], Theorem 10.1.9). Now assume that 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 -nilpotent. Let be the normal -Hall subgroup of E. Then is a normal subgroup of G and satisfies the hypothesis. Hence by the choice of . Suppose that is cyclic. Then, for the G-isomorphism . By Lemma 2(5), . Now assume that is noncyclic. By Lemma 3(1), we can easily obtain that satisfies the hypothesis. Analogously, the choice of implies that . Therefore, in any case, . Furthermore, by Lemma 2(4), a contradiction. Thus, , that is, E is a -group.(2). Since , there exists a -maximal subgroup X of G such that . By Lemma 3(2), satisfies the hypothesis for . If , then the choice of implies that . Note that is supersoluble for the isomorphism . So by Lemma 2(3). Furthermore, the choice of X implies , that is, . This contradiction shows that and, consequently, is supersoluble as .(3)G has the unique Sylow -subgroup. Let q be the largest prime dividing and a Sylow q-subgroup of G. Assume that . Note that is supersoluble. So and . Consider , which satisfies the hypothesis by Lemma 3(2). Note that is the smallest prime divisor of and E is the Sylow -subgroup of . So by Proposition 1, is -nilpotent. Therefore, and, consequently, . Now consider , which satisfies the hypothesis by Lemma 3(1). So the choice of implies that . Moreover, the isomorphism deduces that is supersoluble. Together with Lemma 2(3), we finally obtain is supersoluble. Furthermore, G is supersoluble by the isomorphism . This contradiction shows . So (3) holds.(4)Final contradiction. Let N be a minimal normal subgroup of G contained in E. Consider , which satisfies the hypothesis by Lemma 3(1). So the choice of implies that . Note that is supersoluble by the isomorphism . Combining with Lemma 2(3), is supersoluble. Therefore, and N is the unique minimal normal subgroup of G contained in E. Note that is a normal subgroup of G contained in E. So the uniqueness of N implies that . 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, by the uniqueness of N. Note that is a nontrivial normal subgroup of G. So , that is, . Take be an arbitrary maximal subgroup of E. Clearly, , and by the hypothesis, is either -I-supplemented or -I-embedded in G. Assume that is -I-supplemented in G. Let T be a subnormal subgroup of G such that and . Then, and, consequently, by the minimality of E. In this case, , which implies by the minimality of E again, a contradiction. Now suppose that is -I-embedded in G and T is a S-quasinormal subgroup of G such that is S-quasinormal in G and . It is easy to show that the above holds if T is replaced by . So, without loss of generality, assume that . Since T is S-quasinormal in G, we have by Lemma 1(3) and the relationship . Therefore, or . If , then is S-quasinormal in G and, similarly as above, , which contradicts the minimality of E. But if , then , which also deduces a contradiction as above. So the proof is completed.

Proof of Theorem 2. Suppose that the result is false and let be a counterexample for which is minimal. Then, G is not supersolvable. Similarly as steps (1) and (2) in the proof of Theorem 1, assume that and E is a -group.
Let M be any proper subgroup of G. Consider , which satisfies the hypothesis for by Lemma 3(2). So the minimality of deduces that . Note that is supersolvable by the isomorphism . 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) is a -subgroup of G; (ii) is a noncyclic G-chief factor; (iii) the exponent of E is or 4 (when E is a non-abelian 2-group). Similarly, as step (3) of the proof of Theorem 1, the Sylow -subgroup of G is normal in G. Note that is a nontrivial normal subgroup of , so .
Take , and denote . Then, H has order or 4, , and . By the hypothesis, H is either -I-supplemented or -I-embedded in G.
Assume that H is -I-embedded in G. Let T be a S-quasinormal subgroup of G such that is S-quasinormal in G and . Clearly, is another S-quasinormal subgroup of G such that H is -I-embedded in G. So without loss of generality, assume that . Then, is a S-quasinormal subgroup of contained in . Together with Lemma 1(3) and the relationship , we have . Thus, or . If , then is S-quasinormal in G and then . The choice of H shows that , which contradicts (ii). Assume that . Then, and by Lemma 2(2), . Together with (ii), we have . Recall that is supersoluble. Therefore, is supersoluble by the isomorphism and Lemma 2(3). Furthermore, we have G is supersoluble, a contradiction.
Now assume that H is -I-supplemented in G and T is a subnormal subgroup of G such that and . It is easy to show that . So or . Similarly as the above, is impossible. However, implies that , 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. Let E be a normal subgroup of G such that is supersoluble. Then, G is supersoluble if and only if for every prime and every noncyclic Sylow -subgroup of E, one of the following holds (1)All maximal subgroups of are either -I-supplemented or -I-embedded inG(2)All cyclic subgroups of with order and 4 (when is a non-abelian 2-group) are either -I-supplemented or -I-embedded inG

Recall also that a subgroup H of G is called as follows: c-normal in G [16] if G has a normal subgroup T such that and ; -normal in G [17] if G has a subnormal subgroup T such that and ; -supplemented in G [18] if G has a subnormal subgroup T such that and . Obviously, c-normal subgroups, -normal subgroups, and -supplemented subgroups of G are all -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 of G is -normal in G ([17], Corollary 1.3)(b)All cyclic subgroups of G with prime order or order 4 are -normal in G ([17], Corollary 1.5)(c)Every maximal subgroup of every Sylow subgroup of G is c-normal in G ([16], Theorem 4.1)(d)All cyclic subgroups of G 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). Let be the smallest prime dividing and a Sylow -subgroup of G. If all subgroups of with order or order 4 are c-normal in G, then G is -nilpotent.

Corollary 4 ([18], Theorem 3.1). Let be a Sylow -subgroup of G, where is a prime dividing such that . If every maximal subgroup of is -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.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare no conflicts of interest regarding the content and implications of this manuscript.

Acknowledgments

This work was supported by the Start-up Scientific Research Foundation of Anhui Jianzhu University (2017QD20) and the Key Research Projects of Natural Science in Anhui Province (KJ2019A0784).