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

Authors: Na Tang; Xianhua Li
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

LetM be a maximal subgroup of finite group G. For each chief factor H/K of G such that K≤M and G=MH, we called the order of H/K the normal index of M and M∩H/K a section of M in G. Using the concepts of normal index and c-section, we obtain some new characterizations of p-solvable, 2-supersolvable, and p-nilpotent.


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## Body

## 1. Introduction

In this paper, all groups considered are finite. Letπ(G) denote the set of prime divisors of |G|, and for p∈π(G) let Sylp(G) denote the set of Sylow p-subgroups of G. Write M⋖G to indicate that M is a maximal subgroup of G. For convenience, we cite the following relative definitions. For a fixed prime p∈π(G),(1)
Fc(G)={M∣M⋖G and |G:M| is composite},(2)
Fp(G)={M∣M⋖G and |G:M|p=1},(3)
Fpc(G)=Fp(G)∩Fc(G)},(4)
Fp(G)={M∣M⋖G and NG(P)≤M}, where P∈Sylp(G),(5)
Fpc(G)=Fp(G)∩Fc(G).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 M of a group G, the order of a chief factor H/K of G, where H is minimal in the set of normal supplements of M in G, is known as the normal index of M of G, denoted by η(G:M). If H/K is such a chief factor, then G=MH, K≤M, and |G:M|=|H/K:(H/K)∩(M/K)|, so |H/K|=|(H∩M)/K∥G:M|. The intersection (M∩H)/K is called a c-section of M. Li and Wang in [3] proved that every maximal subgroup M of G has a unique c-section up to isomorphism. Let Sec(M) denote a group which is isomorphic to a c-section of M. Then η(G:M)=|Sec(M)|·|G:M|. Deskins [2] showed that G is solvable if and only if η(G:M)=|G:M| for every maximal subgroup M of G. The investigations on the normal index have been developed by many scholars; see [3–7]. But the earlier results concern the cases where p is either the largest prime dividing |G| or an odd prime. In 2010, Zhang and Li analyzed the case when p=2 and obtained some interesting results. In particular we note the following theorems.Theorem 1 (see [8, Theorem 3.1]).
A groupG is solvable if and only if η(G:M)2=1 for every M∈F2(G).Theorem 2 (see [8, Theorem 3.4]).
A groupG is solvable if and only if Sec(M) is either a 2′-group or an abelian 2-group for every M∈F2(G).We observe that Theorems1 and 2 still hold by replacing 2 with another prime p. For example, let G=S4 and let M∈F3(G). Since the order of M is 6 or 12, η(G:M)3=1. So G satisfies the hypotheses of Theorem 1. But G is 3-solvable. It is natural to ask that the theorems above hold or not for any prime p. In part 3, we give positive answer and relative results.


## 2. Preliminary Results

Lemma 3 (see [8, Lemma 2.2]).
LetG be a group, N a normal subgroup of G, and p∈π(G). Let M be a maximal subgroup of G and N≤M.(1)
We haveη(G/N:M/N)=η(G:M) and Sec(M/N)≅Sec(M).(2)
IfM/N∈Fp(G/N), then M∈Fp(G).(3)
IfM/N∈Fp(G/N), then M∈Fp(G).(4)
Ifp=max⁡π(G), then Fp(G)=Fpc(G).Lemma 4 (see [6, Theorem 7]).
G is p-supersolvable if and only if, for each maximal subgroup M of G, η(G:M)p=|G:M|p=1 or p.


## 3. Main Results

Theorem 5.
G is p-solvable if and only if η(G:M)p=1 for every M∈Fp(G).Proof.
⇒: Suppose that G is p-solvable and let N be a minimal normal subgroup. If a maximal subgroup M∈Fp(G) containing N, then, by induction, it follows that η(G/N:M/N)p=η(G:M)p=1. If N⊈M, then we must have |N|p=1, since N is a p′-group.
⇐: Conversely, let η(G:M)p=1 hold for each maximal subgroup M∈F2(G). Only we need to consider that G is not simple. Otherwise, |G|p=η(G:M)p=1. Certainly, G is p-solvable.
Now letN be a minimal normal subgroup of G. Observe the quotient group G/N. For every maximal subgroup M/N∈Fp(G/N), it is easy to see M∈Fp(G). By Lemma 3 and hypothesis, η(G/N:M/N)p=η(G:M)p=1. Hence G/N is p-solvable by induction. Since the class of all p-solvable groups is a saturated formation, we may suppose that N is the unique minimal normal subgroup of G. If |N|p=1, then N is a p′-group. Moreover, G/N is p-solvable, and so is G. Now consider |N|p≠1. Let P be a Sylow p-subgroup of G and K=P∩N. Then K is a Sylow p-subgroup of N. Clearly, NG(P)≤NG(K)<G. So NG(K) is contained in some maximal subgroup T of G. Hence T∈Fp(G). By Frattini argument, G=NNG(K)=NT. It follows that |N|p=η(G:T)p=1, a contradiction, and we are done.Corollary 6.
G is solvable if and only if, for every M∈Fp(G), η(G:M)=1, where p is an arbitrary divisor of |G|.It was announced by Zhang and Li in [8, Theorem 5] that a group G is solvable if and only if Sec(M) is a 2′-group or an abelian 2-group for M∈F2(G). We extend this theorem by proving the following.Theorem 7.
G is p-solvable if and only if, for any M∈Fp(G), Sec(M) is an abelian p-group or a p′-group, where p is a prime divisor of |G|.Proof.
⇒: Suppose that G is p-solvable and let N be a minimal normal subgroup. If a maximal subgroup M∈Fp(G) containing N, then, by induction, it follows that Sec(M/N)≅Sec(M) is an abelian p-group or a p′-group in view of Lemma 3. If N⊈M, then G=MN. If |N|p=1, then η(G:M)p=|N|p=1, and so Sec(M) is a p′-group. Now consider |N|p=1. By the p-solvability of G, it implies that N is an elementary abelian p-group. It follows that Sec(M)≅M∩N is an abelian p-group.
⇐: Conversely, suppose Sec(M) is an abelian p-group or a p′-group. Let N be a minimal normal subgroup of G. By Lemma 3, G/N satisfies the hypotheses of the theorem. Then by induction, G/N is p-solvability. If |N|p=1, then G is p-solvable. Now assume that |N|p≠1, then G is p-solvable. Let P be a Sylow p-solvable of G and K=P∩N. Then, K is a Sylow p-subgroup of N. Obviously, NG(P)≤NG(K)<G. So NG(K) is contained in some maximal subgroup T of G, and consequently, T∈Fp(G). By Frattini argument, G=NNG(K)=NT. Then the minimal normality of N shows Sec(M)≅M∩N. On the other hand, K≤NN(K)≤M∩N. Combining the hypothesis, Sec(M) is an abelian p-group, and so is M∩N. It follows that NN(K)=CN(K). By Burnside Theorem, N is p-nilpotent, which contradicts the minimal normality of N. Therefore, the conclusion holds.In view of Theorem7 it is natural to ask if a group G is p-solvable when |Sec(M)|p=pα or 1, for M∈Fp(G), where p is a prime divisor of |G|. The answer of the question is negative. For example, set G=PSL(2,7) and p=3; every maximal subgroup M satisfies that |Sec(M)|3=3, but G is not 3-solvable. For p-solvable, the condition that Sec(M) is an abelian p-group is crucial.It is proved in [6, Theorem 7] that a group G is p-supersolvable if and only if, for each maximal subgroup M of G, η(G:M)p=|G:M|p=1 or p. It is natural to ask if a group G is p-supersolvable when η(G:M)p=1 or p for any maximal subgroup M of G. The answer of the question is negative. For example, set G=PSL(2,7) and p=3; every maximal subgroup M satisfies that η(G:M)3=3, but G is not 3-supersolvable. But assuming that p=2, the result holds or not. For the question, we give the positive answer. Next, we prove the result.Theorem 8.
G is 2-supersolvable if and only if, for any maximal subgroup M of G, η(G:M)2=1 or 2.Proof.
⇒: Suppose that G is 2-supersolvable. Certainly, G is solvable. By Lemma 4, the necessity holds.
⇐: Conversely, assume the result is not true and let G be a counterexample of minimal order. Now, we assert G is not simple. If not, then η(G:M)2=|G|2=1 or 2. For |G|2=1, it is clear that G is 2-supersolvable, a contradiction. Assume that |G|2=2. Then G is a cyclic group of order 2, and so G is 2-supersolvable, a contradiction. This contradiction shows G is not simple. Let N be the minimal normal subgroup of G. By Lemma 3, G/N satisfies the hypotheses of the theorem. The minimal choice of G implies that G/N is 2-supersolvable. If N is contained in each maximal subgroup M of G, then N⊆Φ(G), and consequently, G/Φ(G) is 2-supersolvable, and so is G, a contradiction. Hence there is a maximal subgroup M of G, such that G=MN. Suppose that |N|2=1. It follows that G is 2-supersolvable, a contradiction. So |N|2≠1. By hypothesis, η(G:M)2=|N|2=2. Moreover, N is solvable. Therefore, |N|=2, and so G is 2-supersolvable, which contradicts the assumption. Now the proof of theorem is completed.Theorem 9.
SupposeG is a group and p is the smallest prime divisor of |G|. Then G is p-nilpotent if and only if the following conditions are satisfied:(1)
η(G:M)p=1 or p for every maximal subgroup M of G;(2)
ifη(G:M)p=p for some maximal subgroup M, then M⊴G.Proof.
⇒: Assume that G is p-nilpotent. Then G is p-supersolvable and (1) holds by Lemma 4. Now let M be a maximal subgroup of G with η(G:M)p=p and G=PT, where P is a Sylow p-subgroup and T is a normal Hall p′-subgroup of G. Suppose T⊈M and let 1⊴⋯⊴T2⊴T1⊴⋯⊴T⊴G be a chief group series, where T1≰M and T2≤M. Then η(G:M)p=|T1/T2|p=1, a contradiction. Hence T⊆M. Since η(G:M)p=p, |G:M|p=1 or p. If |G:M|p=1, then some Sylow p-subgroup of G, say P1, is contained in M, and it follows that G=P1T⊆M, a contradiction. Therefore, |G:M|p=p. Since M=T(M∩P), |P|/|M∩P|=|G:M|p=p, which leads to P∩M⊴P. Hence M⊴PM=G.
⇐: Now suppose (1) and (2) hold. If, for each maximal subgroup M of G, η(G:M)p=1, then by Theorem 8, G is p-solvable. Combining condition (2), we have that G is not simple. Let N be a minimal normal subgroup of G. By Lemma 3, G/N satisfies the hypotheses. By induction, G/N is p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, we may regard N as the unique minimal normal subgroup of G and Φ(G)=1. So there exists a maximal subgroup M of G such that G=NM and η(G:M)p=|N|p=1 or p.
Suppose|N|p=1. Then N is a p′-group. Since G/N is p-nilpotent, G is p-nilpotent.
Assume|N|p=p. Since p is the smallest prime divisor of |G|, N is p-nilpotent, and so |N|=p. It now follows that M∩N=1 and M≅G/N is p-nilpotent. Note M=MpMp′, where Mp is a Sylow p-subgroup and Mp′ is a normal Hall p′-subgroup of M. Then by (2), Mp′ char M⊴G, and so Mp′⊴G. Consequently, G=MN=(MpN)Mp′ and Mp′ is a normal Hall p′-subgroup of G.The proof of the theorem has been done.Obviously, in Theorem9, removing the condition “p is the smallest prime divisor of |G|”, and the result does not hold.Theorem 10.
G has a p-nilpotent maximal subgroup M with prime power normal index; then G is p-solvable.Proof.
Assume that the theorem is false and letG be a minimal counterexample. Let M be a p-solvable maximal subgroup of G with η(G:M)=qα, where q is a prime. Now we assert that G is not simple. Otherwise, η(G:M)=|G|=qα, a contradiction. Let N be a minimal normal subgroup of G. Next, we consider the following two cases.
Case
1
(
N
⊆
M
). Then by Lemma 3, η(G/N:M/N)=η(G:M)=qα. Since M is p-solvable, M/N and N are p-solvable. By the minimal choice of G, it implies that G/N is p-solvable, so is G, a contradiction.
Case
2
(
N
⊈
M
). Then G=MN and G/N≅M/(M∩N) is p-solvable. On the other hand, η(G:M)=|N| is a q-group. Thus G is p-solvable, a final contradiction. This contradiction completes the proof of the theorem.Theorem 11.
IfG has a p-supersolvable maximal subgroup M such that η(G:M) is a prime and MG=1, then G is p-supersolvable.Proof.
Assume the result is not true and letG be a counterexample of minimal order. By Theorem 10, G is p-solvable. Let N be an arbitrary minimal normal subgroup of G. Then N is an abelian p-group or a p′-group. Moreover, since MG=1, G=MN. Suppose every minimal normal subgroup N of G is a p′-group. But G/N=MN/N≅M/M∩N is p-supersolvable; it follows that G is p-supersolvable, a contradiction. This contradiction shows that there exists some minimal normal subgroup K of Gwhich is an abelian p-group. Then G=MK and M∩K=1. From this it follows that η(G:M)=|K| is a prime. Since G/K≅M is p-supersolvable, G is p-supersolvable, a contradiction. Hence the result holds.



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