A Note on the Normal Index and the c-Section of Maximal Subgroups of a Finite Group

Abstract

Let be a maximal subgroup of finite group . For each chief factor of such that and , we called the order of the normal index of and a section of in . Using the concepts of normal index and c-section, we obtain some new characterizations of p-solvable, 2-supersolvable, and p-nilpotent.

1. Introduction

In this paper, all groups considered are finite. Let denote the set of prime divisors of , and for let denote the set of Sylow -subgroups of . Write to indicate that is a maximal subgroup of . For convenience, we cite the following relative definitions. For a fixed prime ,(1) and is composite},(2) and ,(3),(4) and , where ,(5).

The remaining notation and terminology in this paper are standard, as in Huppert [1].

In 1959, Deskins [2] introduced the concept of normal index. For a maximal subgroup of a group , the order of a chief factor of , where is minimal in the set of normal supplements of in , is known as the normal index of of , denoted by . If is such a chief factor, then , , and , so . The intersection is called a -section of . Li and Wang in [3] proved that every maximal subgroup of has a unique -section up to isomorphism. Let denote a group which is isomorphic to a -section of . Then . Deskins [2] showed that is solvable if and only if for every maximal subgroup of . The investigations on the normal index have been developed by many scholars; see [3–7]. But the earlier results concern the cases where is either the largest prime dividing or an odd prime. In 2010, Zhang and Li analyzed the case when and obtained some interesting results. In particular we note the following theorems.

Theorem 1 (see [8, Theorem 3.1]). A group is solvable if and only if for every .

Theorem 2 (see [8, Theorem 3.4]). A group is solvable if and only if is either a -group or an abelian 2-group for every .

We observe that Theorems 1 and 2 still hold by replacing 2 with another prime . For example, let and let . Since the order of is 6 or 12, . So satisfies the hypotheses of Theorem 1. But is 3-solvable. It is natural to ask that the theorems above hold or not for any prime . In part 3, we give positive answer and relative results.

2. Preliminary Results

Lemma 3 (see [8, Lemma 2.2]). Let be a group, a normal subgroup of , and . Let be a maximal subgroup of and .(1)We have and .(2)If , then .(3)If , then .(4)If , then .

Lemma 4 (see [6, Theorem 7]). is -supersolvable if and only if, for each maximal subgroup of , or .

3. Main Results

Theorem 5. is -solvable if and only if for every .

Proof. : Suppose that is -solvable and let be a minimal normal subgroup. If a maximal subgroup containing , then, by induction, it follows that . If , then we must have , since is a -group.
: Conversely, let hold for each maximal subgroup . Only we need to consider that is not simple. Otherwise, . Certainly, is -solvable.
Now let be a minimal normal subgroup of . Observe the quotient group . For every maximal subgroup , it is easy to see . By Lemma 3 and hypothesis, . Hence is -solvable by induction. Since the class of all -solvable groups is a saturated formation, we may suppose that is the unique minimal normal subgroup of . If , then is a -group. Moreover, is -solvable, and so is . Now consider . Let be a Sylow -subgroup of and . Then is a Sylow -subgroup of . Clearly, . So is contained in some maximal subgroup of . Hence . By Frattini argument, . It follows that , a contradiction, and we are done.

Corollary 6. is solvable if and only if, for every , , where is an arbitrary divisor of .

It was announced by Zhang and Li in [8, Theorem 5] that a group is solvable if and only if is a 2′-group or an abelian 2-group for . We extend this theorem by proving the following.

Theorem 7. is -solvable if and only if, for any , is an abelian -group or a -group, where is a prime divisor of .

Proof. : Suppose that is -solvable and let be a minimal normal subgroup. If a maximal subgroup containing , then, by induction, it follows that is an abelian -group or a -group in view of Lemma 3. If , then . If , then , and so is a -group. Now consider . By the -solvability of , it implies that is an elementary abelian -group. It follows that is an abelian -group.
: Conversely, suppose is an abelian -group or a -group. Let be a minimal normal subgroup of . By Lemma 3, satisfies the hypotheses of the theorem. Then by induction, is -solvability. If , then is -solvable. Now assume that , then is -solvable. Let be a Sylow -solvable of and . Then, is a Sylow -subgroup of . Obviously, . So is contained in some maximal subgroup of , and consequently, . By Frattini argument, . Then the minimal normality of shows . On the other hand, . Combining the hypothesis, is an abelian -group, and so is . It follows that . By Burnside Theorem, is -nilpotent, which contradicts the minimal normality of . Therefore, the conclusion holds.

In view of Theorem 7 it is natural to ask if a group is -solvable when or 1, for , where is a prime divisor of . The answer of the question is negative. For example, set and ; every maximal subgroup satisfies that , but is not 3-solvable. For -solvable, the condition that is an abelian -group is crucial.

It is proved in [6, Theorem 7] that a group is -supersolvable if and only if, for each maximal subgroup of , or . It is natural to ask if a group is -supersolvable when or for any maximal subgroup of . The answer of the question is negative. For example, set and ; every maximal subgroup satisfies that , but is not 3-supersolvable. But assuming that , the result holds or not. For the question, we give the positive answer. Next, we prove the result.

Theorem 8. is 2-supersolvable if and only if, for any maximal subgroup of , or 2.

Proof. : Suppose that is 2-supersolvable. Certainly, is solvable. By Lemma 4, the necessity holds.
: Conversely, assume the result is not true and let be a counterexample of minimal order. Now, we assert is not simple. If not, then or 2. For , it is clear that is 2-supersolvable, a contradiction. Assume that . Then is a cyclic group of order 2, and so is 2-supersolvable, a contradiction. This contradiction shows is not simple. Let be the minimal normal subgroup of . By Lemma 3, satisfies the hypotheses of the theorem. The minimal choice of implies that is 2-supersolvable. If is contained in each maximal subgroup of , then , and consequently, is 2-supersolvable, and so is , a contradiction. Hence there is a maximal subgroup of , such that . Suppose that . It follows that is 2-supersolvable, a contradiction. So . By hypothesis, . Moreover, is solvable. Therefore, , and so is 2-supersolvable, which contradicts the assumption. Now the proof of theorem is completed.

Theorem 9. Suppose is a group and is the smallest prime divisor of . Then is -nilpotent if and only if the following conditions are satisfied:(1) or for every maximal subgroup of ;(2)if for some maximal subgroup , then .

Proof. : Assume that is -nilpotent. Then is -supersolvable and (1) holds by Lemma 4. Now let be a maximal subgroup of with and , where is a Sylow -subgroup and is a normal Hall -subgroup of . Suppose and let be a chief group series, where and . Then , a contradiction. Hence . Since , or . If , then some Sylow -subgroup of , say , is contained in , and it follows that , a contradiction. Therefore, . Since , , which leads to . Hence .
: Now suppose (1) and (2) hold. If, for each maximal subgroup of , , then by Theorem 8, is -solvable. Combining condition (2), we have that is not simple. Let be a minimal normal subgroup of . By Lemma 3, satisfies the hypotheses. By induction, is -nilpotent. Since the class of all -nilpotent groups is a saturated formation, we may regard as the unique minimal normal subgroup of and . So there exists a maximal subgroup of such that and or .
Suppose . Then is a -group. Since is -nilpotent, is -nilpotent.
Assume . Since is the smallest prime divisor of , is -nilpotent, and so . It now follows that and is -nilpotent. Note , where is a Sylow -subgroup and is a normal Hall -subgroup of . Then by (2), char , and so . Consequently, and is a normal Hall -subgroup of .

The proof of the theorem has been done.

Obviously, in Theorem 9, removing the condition “ is the smallest prime divisor of ”, and the result does not hold.

Theorem 10. has a -nilpotent maximal subgroup with prime power normal index; then is -solvable.

Proof. Assume that the theorem is false and let be a minimal counterexample. Let be a -solvable maximal subgroup of with , where is a prime. Now we assert that is not simple. Otherwise, , a contradiction. Let be a minimal normal subgroup of . Next, we consider the following two cases.
Case 1. Then by Lemma 3, . Since is -solvable, and are -solvable. By the minimal choice of , it implies that is -solvable, so is , a contradiction.
Case 2. Then and is -solvable. On the other hand, is a -group. Thus is -solvable, a final contradiction. This contradiction completes the proof of the theorem.

Theorem 11. If has a -supersolvable maximal subgroup such that is a prime and , then is -supersolvable.

Proof. Assume the result is not true and let be a counterexample of minimal order. By Theorem 10, is -solvable. Let be an arbitrary minimal normal subgroup of . Then is an abelian -group or a -group. Moreover, since , . Suppose every minimal normal subgroup of is a -group. But is -supersolvable; it follows that is -supersolvable, a contradiction. This contradiction shows that there exists some minimal normal subgroup of which is an abelian -group. Then and . From this it follows that is a prime. Since is -supersolvable, is -supersolvable, a contradiction. Hence the result holds.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

Research of the authors is supported by NNSF Grant of China (Grants 11171243 and 11001098), Natural Science Foundation of Jiangsu (Grant BK20140451), and University Natural Science Foundation of Jiangsu (Grant 14KJB110002). The authors thank the referees and editors for their many valuable comments and suggestions.