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On the normal indices of proper subgroups of finite groups
Journal of Pure and Applied Algebra 214 (2010) 1285–1290
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On the normal indices of proper subgroups of finite groupsI Xianhua Li ∗ , Xinjian Zhang School of Mathematical Science, Soochow University, Suzhou, Jiangsu 215006, People’s Republic of China
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Article history: Received 26 October 2008 Received in revised form 7 October 2009 Available online 5 November 2009 Communicated by M. Sapir
abstract We extended the normal index from maximal subgroups to proper subgroups. We give a quantitative version of all results obtained by using c-normal subgroups and obtain some new characterizations of solvable, supersolvable and nilpotent groups by the normal indices of proper subgroups. © 2009 Elsevier B.V. All rights reserved.
MSC: 20D10 20D15
1. Introduction In this paper, all groups considered are finite. In 1954, Huppert obtained the famous theorem: a group G is supersolvable if and only if the index of every maximal subgroup of G is prime. After that, Deskins in [1] introduced the normal index of a maximal subgroup M of a group G. The normal index of a maximal subgroup M of a group G, denoted by η(G : M ), is the order of a chief factor H /K of G, where H is a minimal supplement of M in G. Several authors gave the characterizations of solvability, supersolvability, etc. of a group G by the normal indices of some maximal subgroups (see [1,4–7]). This provides some well quantitative versions and ways of characterizations of a group G. However, the definition of the normal index is confined to the maximal subgroups only and maximal subgroups are special subgroups, we cannot go further. Recall that a subgroup H of a group G is called c-normal in G if there exists a normal subgroup N of G such that NH = G and H ∩ N ≤ HG . In fact, Guo and Li in [12] pointed out that the concept of c-normality provides a better tool for us and it makes things easier and simpler than normal indices do. In our present work, we would like to say that it may not be true. In [2], we extended the concept of the normal index from maximal subgroups to proper subgroups of a group. In this paper, we prove that if the normal index of a subgroup H of G is equal to the index |G : H |, then H is c-normal in G. Hence we give a quantitative characterization of c-normality of subgroups by using the normal index. It means that the c-normality of a subgroup H is equivalent to that H has the special normal index. On the other hand, the normal index is a number, which provides a new tool and way to study the properties of groups by the quantitative information of themselves. Using the normal index of a proper subgroup, we give the characterizations of a group to be solvable, supersolvable and nilpotent. 2. Definitions and preliminary results Definition 2.1 ([2, Definition 2.2]). Let H be a proper subgroup of a group G and K a normal subgroup of G. Then |K /K ∩ HG | is called the normal index of H in G, denoted by η∗ (G : H ), if K satisfies (1) G = KH; (2) if G = TH for any normal subgroup T of G, then we have |T /T ∩ HG | ≥ |K /K ∩ HG |. I This work was supported by the National Natural Science Foundation of China (Grant Nos. 10571128, 10871032), the Natural Science Foundation of Jiangsu Province (No. BK2008156). ∗ Corresponding author. E-mail address: [email protected] (X. Li).
0022-4049/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.jpaa.2009.10.012
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Let G be a group and H be a proper subgroup of G. Let K1 and K2 be normal subgroups of G such that K1 satisfies the conditions of Definition 2.1 and K2 satisfies HG ≤ K2 , G = K2 H and |K2 | ≤ |T | for any normal subgroup T of G such that HG ≤ T and G = TH. Then G = K1 H = K2 H, |K2 /HG | = |K2 /K2 ∩ HG | ≥ |K1 /K1 ∩ HG | by Definition 2.1. On the other hand, by G = (K1 HG )H and the hypothesis of K2 , we have |K1 /K1 ∩ HG | = |K1 HG /HG | ≥ |K2 /HG |. Hence |K1 /K1 ∩ HG | = |K2 /HG |. Thus we have shown that Definition 2.1 is equivalent to the following. Definition 2.2. Let H be a proper subgroup of a group G and K a normal subgroup of G containing HG . Then |K /HG | is called the normal index of H in G, if K satisfies (1) G = KH; (2) |K | ≤ |T | for any normal subgroup T of G such that HG ≤ T and G = TH. Lemma 2.3 ([2]). Let G be a group. (1) If M is a maximal subgroup of G, then η∗ (G : M ) = η(G : M ). (2) If N E G and H is a subgroup of G such that N ≤ H, then η∗ (G/N : H /N ) = η∗ (G : H ). (3) G is solvable if and only if there exists a solvable 2-maximal subgroup L of G such that η∗ (G : L) = |G : L|. Remark 2.1. (1), (2) and (3) in Lemma 2.3 are Proposition 2.1, Lemma 2.2 and Theorem 3.1 in [2], respectively. (1) in Lemma 2.3 shows that the normal index here is a generalization of the concept defined in [1]. Lemma 2.4 ([9]). Let G be a group and H, K be subgroups of G. (1) G is solvable if and only if there exists a solvable c-normal maximal subgroup M of G. (2) If H is c-normal in G and H ≤ K ≤ G, then H is c-normal in K . Remark 2.2. (1) and (2) are Theorem 3.4 and Lemma 2.1 (3) in [9], respectively. Lemma 2.5 ([8, Lemma 3.12]). Let p be the smallest prime dividing the order of a group G and P a Sylow p-subgroup of G. If |P | ≤ p2 and G is A4 -free, then G is p-nilpotent. 3. Normal index theorems and properties of a group In the theory of groups, some index theorems have fundamental importance, for example, Lagrange’s theorem, etc.. In this section, we are interested in what will happen if we replace the index by the normal index in some index theorems. That is, if the normal index theorem is true, what properties does G have? Let H and K be subgroups of G. Firstly, since |G : H | | η∗ (G : H ), if (η∗ (G : H ), η∗ (G : K )) = 1, then (|G : H |, |G : K |) = 1 and so G = HK . Secondly, to be analogous to Lagrange’s theorem, we consider the case |G| = |H |η∗ (G : H ). This is equivalent to |G : H | = η∗ (G : H ). Theorem 3.1. Let H be a proper subgroup of a group G. Then η∗ (G : H ) = |G : H | if and only if H is c-normal in G. Proof. Assume that K is a normal subgroup of G such that η∗ (G : H ) = |K /HG |. By the definition of the normal index, G = KH. Hence |G : H | = |K : K ∩ H |. It follows from the hypothesis η∗ (G : H ) = |G : H | that K ∩ H = HG . Thus H is c-normal in G. Now we prove the sufficiency. Suppose that H is c-normal in G, then there exists a normal subgroup K of G such that G = KH and K ∩ H ≤ HG . It follows that K ∩ H = K ∩ HG . Hence |G : H | = |K : K ∩ HG | = |KHG /HG | and |G : H | > η∗ (G : H ). Since |G : H | divides η∗ (G : H ), η∗ (G : H ) = |G : H | holds. Theorem 3.1 gives a quantitative characterization of the c-normality of subgroups. Replacing the condition that H is c-normal in G by η∗ (G : H ) = |G : H |, we can obtain a quantitative version of all results obtained by using c-normal subgroups. For instance, Theorem 3.2 in [12], Theorem 4.1 in [9] and Theorem 4.2 in [13] can be written in the following forms, respectively. (1) Let G1 be a class of Frobenius groups with elementary abelian kernel N and cyclic complement M, such that |M | is square-free and the order of M is not prime. If G is G1 -free and η∗ (G : L) = |G : L| for each 2-maximal subgroup L of G, then G is supersolvable. (2) Let G be a group. Suppose η∗ (G : P1 ) = |G : P1 | for every Sylow subgroup P of G and every maximal subgroup P1 of P. Then G is supersolvable. (3) Let G be a group and p be the smallest prime divisor of |G|. Assume that G is A4 -free and η∗ (G : P2 ) = |G : P2 | for every 2-maximal subgroup P2 of the Sylow p-subgroups of G. Then G/Op (G) is p-nilpotent. Thirdly, to be parallel to the index theorem |G| = |G : K ||K : H | for H ≤ K ≤ G, we consider the question: if η∗ (G : H ) = η∗ (G : K )η∗ (K : H ) for any H ≤ K ≤ G, what properties does G have? Definition 3.2. Let G be a group. Then G is called an NIT -group if η∗ (G : H ) = η∗ (G : K )η∗ (K : H ) for any H ≤ K ≤ G. Lemma 3.3. Let G be an NIT -group. Then (1) H is an NIT -group for any subgroup H of G; (2) G/N is an NIT -group for any normal subgroup N of G.
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Proof. From Lemma 2.3 (2) and the definition of the NIT -group, the conclusions are obvious. Theorem 3.4. Let G be a group. Then the following statements are equivalent. (1) (2) (3) (4)
G is an NIT -group; G is a supersolvable NIT -group; η∗ (G : H ) = |G : H | for any subgroup H of G; H is c-normal in G for any subgroup H of G.
Proof. (1) ⇒ (2): We use induction on |G|. Suppose that G is simple. If G is solvable, then |G| = p and the result is true. Assume that G is a non-abelian simple group. Let M be a maximal subgroup of G and L a maximal subgroup of M. Then by the definition of the NIT -group, η∗ (G : L) = η∗ (G : M )η∗ (M : L). Since G is simple, η∗ (G : L) = η∗ (G : M ) = |G|, so η∗ (M : L) = 1, that is, M = L, a contradiction. This shows that M = 1 and so G is a cyclic group of prime order, again a contradiction. Hence we may assume that G is not simple. Let N be a minimal normal subgroup of G, then |N | < |G| and |G/N | < |G|. By induction, G/N and N are both supersolvable from Lemma 3.3. Thus N is an elementary abelian group. Let |N | = pa . Suppose that a > 1. Let T be a maximal subgroup of N. Then |T | = pa−1 , T > 1 and T E N. By the minimality of N, TG = 1. Let K E G such that G = KT and η∗ (G : T ) = |K /TG | = |K |. Then by the hypothesis that G is an NIT -group, η∗ (G : T ) = η∗ (G : N )η∗ (N : T ). It follows from Lemma 2.3 that |K | = |G : N ||N : T | = |G : T |. Hence K ∩ T = 1. Note that G = NK and N ∩ K = 1 or N. If N ∩ K = 1, then |N | = |T |, a contradiction. Thus N ∩ K = N. But this case forces that T ≤ N ≤ K and so T = 1, again a contradiction. Therefore a = 1, that is, |N | = p. Thus G is supersolvable as desired. (2) ⇒ (3): We also use induction on |G|. For any maximal subgroup M of G, M is an NIT -group by Lemma 3.3. The induction implies that η∗ (M : H ) = |M : H | for any subgroup H of M. Since G is supersolvable, η∗ (G : M ) = |G : M |. Now from the hypothesis and the definition of the NIT -group, it follows that η∗ (G : H ) = η∗ (G : M )η∗ (M : H ) and so η∗ (G : H ) = |G : H |. (3) ⇔ (4): It follows immediately from Theorem 3.1. (3) ⇒ (1): Let H and K be subgroups of G and H ≤ K . We first prove that η∗ (K : H ) = |K : H |. Let T E G such that G = TH and η∗ (G : H ) = |T /HG |, then K = K ∩ TH = H (T ∩ K ) and T ∩ K E K . Since η∗ (G : H ) = |G : H | and HG ≤ T ∩ H, we have η∗ (G : H ) = |T /HG | ≥ |T : T ∩ H | = |G : H |, which implies that HG = T ∩ H. Thus ((T ∩ K )/HG )∩(H /HG ) ⊆ (T /HG )∩(H /HG ) = HG /HG , and so η∗ (K /HG : H /HG ) = |T ∩ K /HG |. Because K = H (T ∩ K ), we have |K : H | = |T ∩ K : T ∩ H | = |T ∩ K /HG |. By Lemma 2.3(2), η∗ (K /HG : H /HG ) = η∗ (K : H ). Hence η∗ (K : H ) = |T ∩ K /HG | = |K : H |. And by the hypothesis, η∗ (G : K ) = |G : K |. Hence η∗ (G : H ) = |G : H | = |G : K ||K : H | = η∗ (G : K )η∗ (K : H ). By Definition 3.2, G is an NIT -group. Theorem 3.5. Let G be a group. If for every maximal subgroup M of G and every maximal subgroup H of M, η∗ (G : H ) = η∗ (G : M )η∗ (M : H ), then G is solvable and every 2-maximal subgroup of G is c-normal in G. Proof. We first prove that G is solvable. Assume that it is false and G is a minimal counterexample. Let N be a minimal normal subgroup of G. By Lemma 2.3, G/N satisfies the hypothesis and by induction G/N is solvable. A routine argument shows that N is the unique minimal normal subgroup of G and Φ (G) = 1. Now we assume that N is non-solvable. Let p = max π (N ) , then p > 2. Let P ∈ Sylp (N ). Then NG (P ) ≤ NG (Z (J (P ))) < G. Hence there exists a maximal subgroup M of G such that NG (Z (J (P ))) ≤ M. By the Frattini argument, we have G = NG (Z (J (P )))N = MN. Now we claim that N ∩ M ≤ Φ (M ). Otherwise, let M1 be a maximal subgroup of M such that M = M1 (N ∩ M ), then G = MN = M1 N, so η∗ (G : M ) = η∗ (G : M1 ) = |N |. But this is impossible for the hypothesis η∗ (G : M1 ) = η∗ (G : M )η∗ (M : M1 ). So N ∩ M ≤ Φ (M ). By NN (Z (J (P ))) = N ∩ NG (Z (J (P ))) ⊆ N ∩ M ≤ Φ (M ), we get NN (Z (J (P ))) is nilpotent. By the Glauberman– Thompson theorem, we obtain N is p-nilpotent, which contradicts to the minimality of N. This contradiction shows that N is solvable, and so G is solvable. Let M be a maximal subgroup of G and H be a maximal subgroup of M. Let η∗ (G : M ) = |T /MG | and η∗ (M : H ) = |S /HM |, where MG ≤ T E G, HM ≤ S E M. Then T /MG is a normal complement of M /MG in G/MG and S /HM is a normal complement of H /HM in M /HM , respectively. Hence we have η∗ (G : M ) = |G/MG : M /MG | = |G : M | and η∗ (M : H ) = |M /HM : H /HM | = |M : H |. From the hypothesis, it follows that η∗ (G : H ) = η∗ (G : M )η∗ (M : H ) = |G : M ||M : H | = |G : H |. By Theorem 3.1, H is c-normal in G. Corollary 3.6. Let G be a group and F be the class of Frobenius groups with elementary abelian kernel N and cyclic complement K , such that |K | is square-free and the order of K is not prime. If G is F -free and η∗ (G : H ) = η∗ (G : M )η∗ (M : H ) for every maximal subgroup M of G and every maximal subgroup H of M, then G is supersolvable. Proof. From Theorem 3.5, we have G is solvable and each 2-maximal subgroup of G is c-normal in G. Then by [12, Theorem 3.2], G is supersolvable. 4. The normal indices of some special subgroups First, we consider the influence of the normal indices of some 2-maximal subgroups of a group G on the structure of G. Lemma 4.1. Let M be a maximal subgroup of a group G. If η(G : M ) is square-free, then η(G : M ) is a prime and η(G : M ) = |G : M |.
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Proof. Let H /K be a chief factor of G, where H is a minimal supplement of M in G. By the definition of the normal index, |H /K | = η(G : M ). Hence the hypothesis implies that |H /K | is square-free and so H /K is solvable. It follows from the minimality of H /K that |H /K | is a prime. This forces that H /K ∩ M /K = K /K . Noting that G = MH, |H /K | = |G/K : M /K | = |G : M |, the conclusions are true. Lemma 4.2. Let G be a group, M a subgroup of G and H a maximal subgroup of M. Then (1) η∗ (G : H ) = aη∗ (M : H ), where a > 1 is a natural number; (2) Let η∗ (G : H ) = |K /HG | and G = KH, where HG E K E G. If M is maximal in G and K /HG is solvable, then η∗ (G : H ) = bη(G : M ), where b > 1 is a natural number. Additionally, if H is solvable, then G is solvable. (3) If H is solvable and η∗ (G : H ) is either square-free or the square of a prime, then G is solvable. (4) Let M be a maximal subgroup of a group G. If M has a maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime, then η(G : M ) is a prime. Proof. Let K be a normal subgroup of G containing HG such that G = KH and η∗ (G : H ) = |K /HG |. Then M = M ∩ KH = (M ∩ K )H and M ∩ K E M. If η∗ (G : H ) is either square-free or the square of a prime, then K /HG is solvable. If H is solvable, then K is solvable. Since G/K ∼ = H /H ∩ K is solvable, we get G is solvable. So (3) holds. Next we prove (1) and (2). Since H is maximal in M and M = (M ∩ K )H, it is obvious that HG < M ∩ K and M ∩ K * H. Let 1 < · · · < HG = M0 < M1 < · · · < Ms = M ∩ K ≤ · · · ≤ M be a chief series of M. Then there exists i ∈ {1, . . . , s} such that Mi * H but Mi−1 ⊆ H. It follows that Mi is a minimal supplement of H in M. Thus η∗ (M : H ) = η(M : H ) = |Mi /Mi−1 | and Mi−1 ≤ HM . Hence η∗ (M : H ) divides |M ∩ K /HM |. Note that HG < K ∩ M. If |K /HG | = |K ∩ M /HG |, then |K | = |K ∩ M |, K = K ∩ M, thus K ≤ M and G = KH = M, a contradiction. So |K /HG | > |K ∩ M /HG | ≥ η∗ (M : H ). (1) holds. Suppose that M is a maximal subgroup of G and K /HG is solvable. Let N /MG be a minimal normal subgroup of G/MG contained in KMG /MG , it is easy to prove that η∗ (G : M ) = |N /MG |. Thus η∗ (G : M ) divides |KMG /MG |. Since KMG /MG ∼ = K /K ∩ MG and HG ≤ K ∩ MG , we obtain η∗ (G : H ) = bη(G : M ). Now we claim that K /HG is not a minimal normal subgroup of G/HG . Otherwise, K /HG is an elementary abelian group. Since G = HK = KM, we have K /HG ∩ H /HG = HG /HG and K /HG ∩M /HG = HG /HG . Thus |H | = |M |, a contradiction. Let N /HG be a minimal normal subgroup of G/HG contained in K /HG . Then N < K and so |N /HG | < |K /HG | and |K /N | > 1. If N ≤ M, then N ≤ MG and so |KMG /MG | = |K /K ∩ MG | < |K /HG |. By G = (KMG )M and the definition of η∗ (G : M ), η∗ (G : M ) ≤ |KMG /MG | < |K /HG | = η∗ (G : H ). If N 6≤ M, then G = NM, we have η∗ (G : M ) ≤ |NMG /MG | = |N /N ∩ MG | ≤ |N /HG | < |K /HG | = η∗ (G : H ). Thus we get b > 1, and (2) holds. (4) is immediate from (2) and Lemma 4.1. Lemma 4.3. Let K be a normal subgroup of a group G. Assume that G/K is supersolvable. (1) If |K | is square-free, then G is supersolvable. (2) Let |K | be the square of a prime. If K is not a minimal normal subgroup of G, then G is supersolvable. Proof. (1) Refine 1 G K G G into a chief series G0 = 1 G G1 · · · < · · · < Gs < K = Gs+1 < · · · < Gn = G. Since |K | is square-free and G/K is supersolvable, Gi+1 /Gi are cyclic groups with prime orders for i ∈ {0, 1, . . . , s}. So G is supersolvable. (2) Let N be a minimal normal subgroup of G contained in K . Then |N | and |K /N | are primes. It is easy to see that G is supersolvable. Lemma 4.4. Let H be a supersolvable subgroup of a group G. If one of the following holds, then G/HG is supersolvable. (1) η∗ (G : H ) is square-free. (2) H is a 2-maximal subgroup of G, η∗ (G : H ) is either square-free or the square of a prime. Proof. Let K be a normal subgroup of G such that G = HK , HG ≤ K and η∗ (G : H ) = |K /HG |. (1) Since H is supersolvable, (G/HG )/(K /HG ) ∼ = G/K ∼ = H /H ∩K is supersolvable. By the hypothesis, |K /HG | is square-free. By Lemma 4.3(1), we have G/HG is supersolvable. (2) By (1), we may assume that |K /HG | is the square of a prime. Since H is a 2-maximal subgroup of G and K /HG is abelian, it is easy to see that K /HG is not a minimal normal subgroup of G/HG . Note that (G/HG )/(K /HG ) ∼ = G/K ∼ = H /H ∩ K is supersolvable. By Lemma 4.3(2), we have G/HG is supersolvable. Lemma 4.5. Let G be a group and M be a supersolvable maximal subgroup of G. If F (G) * M and |G : M | is a prime, then G is supersolvable. Proof. Suppose that the statement is false and let G be a minimal counterexample. Since F (G) * M and M is a maximal subgroup of G, G = F (G)M. Since G/F (G) ∼ = M /F (G)∩M, F (G) and G/F (G) are solvable. Hence G is solvable. If Φ (G) 6= 1, since F (G)/Φ (G) = F (G/Φ (G)) and F (G) * M, we have F (G/Φ (G)) * M /Φ (G). This shows that G/Φ (G) satisfies the hypothesis. Because G is a minimal counterexample, we have G/Φ (G) is supersolvable, thus G is supersolvable, a contradiction. Hence Φ (G) = 1. Since G is solvable, F (G) is the direct product of minimal normal subgroups of G. Hence there exists a minimal normal subgroup N of G such that N is not contained in M. Then G = NM, N ∩ M = 1 and so |N | = |G : M |. It follows that G/N ∼ = M and |N | is prime. Since M is supersolvable, we get G is supersolvable, a contradiction. Hence the statement is true.
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Theorem 4.6. Let G be a group. Suppose that G has a maximal subgroup M such that η∗ (G : L) = |G : L| for each maximal subgroup L of M. Then (1) G is solvable; (2) M /MG is a cyclic group of square-free order; (3) Assume that MG < M. Then Z (G/MG ) = 1, (G/MG )0 = F (G/MG ) is a Sylow subgroup of G/MG and a minimal normal complement subgroup of M /MG in G/MG . (4) If MG = 1, then M is a cyclic group of square-free order, F (G) is a Sylow subgroup of G and a minimal normal complement subgroup of M in G. Furthermore, for each maximal subgroup K of G, if K is not normal in G, then there exist g ∈ G such that K = M g and K ∩ M = 1 if g 6∈ M. Proof. Let L be a maximal subgroup of M. By Lemma 3.1 we obtain L is c-normal in G. From Lemma 2.4 (2), it follows that L is c-normal in M. By [9, Corollary 3.3], M is solvable and so L is solvable. That is, G has a solvable 2-maximal subgroup L such that η∗ (G : L) = |G : L|. By Lemma 2.3(3), G is solvable and then (1) holds. Now we prove (2). Without loss of generality, we may assume that MG = 1. Then Φ (G) = 1 and G0 6= 1. Since G is solvable, by [11, Theorem 15.6]), G has the unique minimal normal subgroup, N say, N = CG (N ) = F (G) and G = N o M. If Z (G) 6= 1, then N ≤ Z (G) and so MG = M, a contradiction, hence Z (G) = 1. From the first paragraph, M is solvable. So M has a composite series M = M0 B M1 B M2 B · · · B Mn = 1 such that Mi−1 /Mi are cyclic groups with prime orders for i ∈ {1, 2, . . . , n}. It follows that Mi is maximal in Mi−1 for i = 1, 2, . . . , n. Now by the hypothesis M1 is c-normal in G, there exists a normal subgroup K1 of G such that G = M1 K1 and M1 ∩ K1 ≤ (M1 )G = 1. It is obvious that G = K1 M and K1 ∩ M E M. Since M = M ∩ M1 K1 = M1 (M ∩ K ), we have |M : M1 | = |K1 ∩ M : M1 ∩ K1 |. Hence |K1 ∩ M | is a prime. Let K1 ∩ M = T1 . By T1 ∩ M1 = 1, M = T1 × M1 . Noting that M2 C M1 , T1 × M2 C M and |M : (T1 × M2 )| = |M1 : M2 | is a prime, so T1 × M2 is maximal in M. Thus there exists a normal subgroup K2 of G such that G = K2 (T1 × M2 ) and K2 ∩ (T1 × M2 ) ≤ (T1 × M2 )G ≤ MG = 1. Put K2 ∩ M = T2 , then T2 E M and |T2 | is prime, so we have M = T2 × T1 × M2 . Similarly, we consider M3 , . . . , Mn and finally conclude that M = Tn × Tn−1 × · · · × T2 × T1 , where Ti are cyclic groups of prime orders for i = 1, 2, . . . , n. So M is abelian and every Sylow subgroup of M is an elementary abelian group. Let g ∈ G \ M, then g = mn, where m ∈ M, n ∈ N and n 6= 1. If M ∩ M g 6= 1, then there exists an element a ∈ M such that an = a1 ∈ M. Hence, [a, n] = a−1 a1 ∈ N ∩ M = 1. It follows that an = a and n ∈ NG (hai). Again by M ≤ NG (hai) and the maximality of M, we have NG (hai) = G, that is, hai E G, contrary to MG = 1. Thus G is a Frobenius group with complement M. By the Frobenius theorem [10, Theorem 10.3] and the maximality of M we know that N is the Frobenius kernel. By [3, Theorem 10.5.6], we get M is cyclic and |M | is square-free. (2) is proved. Assume that |N | = pα . If p divides the order of M, let P ∈ Sylp (M ). Since M is cyclic, M ≤ NG (P ). Noting that N ∩ P = 1, P is not a Sylow p-subgroup of G. So M < NG (P ). The maximality of M forces that NG (P ) = G and so P E G, which contradicts to MG = 1. So N is a Sylow subgroup of G. Certainly, G/F (G) ∼ = M is cyclic, so G0 ≤ N = F (G). By G0 6= 1 and the minimality 0 of N, G = F (G) = N. So (3) holds. If MG = 1, then by (2) and (3), we have M is a cyclic group of square-free order, G0 = F (G) is a Sylow subgroup of G and a minimal normal complement of M in G. Let N = F (G) and K be a maximal subgroup of G but K is not normal in G. If N ≤ K , then K E G, a contradiction. So N K , G = NK and N ∩ K = 1. Noting that N is a Sylow subgroup of G, K and M are both complements of N in G, so there exist g ∈ G such that K = M g . From the proof of Theorem 4.6, we have M is a Frobenius complement of G. So if g 6∈ M, then K ∩ M = 1. The proof is completed. Corollary 4.7. Let G be a group. Then G is solvable if η∗ (G : L) = |G : L| for every 2-maximal subgroup L of G. Theorem 4.8. A group G is solvable if and only if there exists a solvable 2-maximal subgroup L of G such that |π (η∗ (G : L))| ≤ 2. Proof. Suppose that L is a solvable 2-maximal subgroup of G such that |π (η∗ (G : L))| ≤ 2. Let K be a normal subgroup of G such that G = KL and η∗ (G : L) = |K /LG |. The hypothesis implies that |π (K /LG )| ≤ 2 and therefore K /LG is solvable. By Lemma 4.2, G is solvable. Conversely, assume that G is solvable. By Lemma 2.3(3) there exists a 2-maximal subgroup L of G such that η∗ (G : L) = |G : L|. Let M be a maximal subgroup of G such that L is maximal in M. Then |G : L| = |G : M ||M : L|. Now applying the solvability of G, we can get |π (|G : M |)| = 1 and |π (|M : L|)| = 1. Hence |π (η∗ (G : L))| = |π (|G : L|)| ≤ 2. Theorem 4.9. Let G be a group. If G has a maximal subgroup M satisfying one of the following conditions, then G is supersolvable: (1) η∗ (G : L) is either square-free or the square of a prime for any maximal subgroup L of M; (2) M is supersolvable, MG = 1 and M has a maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime; (3) M is supersolvable , MG = 1 and η(G : M ) is square-free; (4) M is supersolvable, F (G) M and M has a maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime; (5) M is supersolvable, F (G) M and η(G : M ) is square-free;
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(6) M is cyclic and η(G : M ) is square-free; (7) M has a cyclic maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime. Proof. Assume that M satisfies (1). Let L be any maximal subgroup of M. By the hypothesis, η∗ (G : L) is either square-free or a prime square. From Lemma 4.2, it follows that η∗ (G : L) = aη∗ (M : L) = bη∗ (G : M ), where a > 1, b > 1. Hence η∗ (G : M ) and η∗ (M : L) are square-free. Now Lemma 4.1 and Lemma 2.3 force that η∗ (G : M ) and η∗ (M : L) are primes. Since |M : L| | η∗ (M : L), the index of every maximal subgroup of M in M is prime and so M is supersolvable by Huppert’s theorem. From Lemma 4.2, we may get G is solvable. Suppose that N1 , N2 are two distinct minimal normal subgroups of G. If N1 M, then G = N1 M, N1 ∩ M = 1 and η∗ (G : M ) = |N1 |. By the hypothesis, |N1 | is a prime and G/N1 ∼ = M is supersolvable. It implies that G is supersolvable. Hence we may assume that Ni ≤ M for i = 1, 2. Then Lemma 2.3 (2) shows that the hypothesis holds for G/N1 and G/N2 . By induction on |G|, G/N1 and G/N2 are supersolvable. Since the class of supersolvable groups is a saturated formation, G∼ = G/N1 ∩ N2 is supersolvable. Hence we may let N be the unique minimal normal subgroup of G and N ≤ M. By induction, G/N is supersolvable. If N ⊆ Φ (M ), then by [3, 5.2.13], N ⊆ Φ (G). It follows that G is supersolvable. Thus we suppose that N * Φ (M ). Then there exists a maximal subgroup T of M such that N * T . It follows that M = NT . By the unique minimality of N, TG = 1. Let A be a normal subgroup of G such that G = AT and η∗ (G : T ) = |A/TG | = |A|. By the hypothesis, |A| is either square-free or the square of a prime. By the unique minimality of N, N ≤ A. If N = A, then G = AT = NT = M, a contradiction, so N < A. Thus |N | is a proper divisor of |A| and so |N | is a prime. This implies that G is supersolvable and so (1) holds. If M is supersolvable and MG = 1, then L is supersolvable and LG = 1 for each maximal subgroup L of M. So if M satisfies (2), then by Lemma 4.4, G is supersolvable, which implies that (2) holds. If M satisfies one of the conditions (3), (6) and (7), then by Lemma 4.4 again, we have G is supersolvable. Now we prove (4). From Lemma 4.2, η(G : M ) is a prime. So by Lemma 4.4, G/MG is supersolvable and by the hypothesis G is solvable. It follows from Lemma 4.5 that G is supersolvable. And by (4), (5) is obvious. The proof is completed. Corollary 4.10. Let G be a group. Then G is supersolvable if G satisfies one of the following conditions: (1) η∗ (G : L) is either square-free or the square of a prime for every 2-maximal subgroup L of G. (2) η∗ (G : L) is square-free for every 2-maximal subgroup L of G. Now we consider the influence of the normal indices of some Sylow subgroups of a group G on the structure of G. Theorem 4.11. Let p be a prime dividing the order of a group G and P a Sylow p-subgroup of G. Assume that η∗ (G : P ) = pα |G : P | where α 6 2. Then the following conclusions are true: (1) If p = 2 and G is A4 -free, then G/O2 (G) is 2-nilpotent. Furthermore, if |O2 (G)| 6 22 , then G is 2-nilpotent. (2) If p 6= 2 and (|G|, p2 − 1) = 1, then G/Op (G) is p-nilpotent. Furthermore, if |Op (G)| 6 p2 , then G is p-nilpotent. Proof. Let G = G/Op (G) and P = P /Op (G), then P G = 1. Hence there exists a normal subgroup K of G such that G = K P and η∗ (G : P ) = |K |. By Lemma 2.3(2), η∗ (G : P ) = η∗ (G : P ), hence |K | = pα |G : P |. On the other hand,
|G : P | = |G : P | = |K P : P | =
|K | . |K ∩P |
It follows that |K ∩ P | = pα . That is, the order of the Sylow p-subgroups of K is pα .
(1) By Lemma 2.5, K is 2-nilpotent. Let H be the normal Hall 20 -subgroup of K . Then H is a characteristic subgroup of K and H is a normal Hall 20 -subgroup of G. Hence G is 2-nilpotent. Now by Lemma 2.5, the later statement is obvious. (2) If α = 0, then K is a normal Hall p0 -subgroup of G. So we may only consider the case α = 1 and α = 2. Let R ∈ Sylp (K ). Then R ≤ CK (R) and R is either a cyclic group of order pα or an elementary abelian group of order p2 . Hence |Aut (R)| | p(p − 1)2 (p + 1). Now by [3, 1.6.13], NK (R)/CK (R) is isomorphic to a subgroup of Aut (R). From the hypothesis (|G|, p2 − 1) = 1, it follows that NK (R) = CK (R). By Burnside’s theorem, K is p-nilpotent. Using the similar proof as (1), G/Op (G) is p-nilpotent. Theorem 4.12. Let G be a group and p ∈ π (G). Assume that there exists a Sylow p-subgroup P of G such that η∗ (G : P ) = pα |G : P |, where α 6 1. If either p = min π (G) or (|G|, p − 1) = 1 , then G/Op (G) is p-nilpotent. Specifically, G is solvable. Proof. The proof is similar to Theorem 4.11 and omitted here.
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
W.E. Deskins, On maximal subgroups, in: Proc. Sympos. Pure Math., vol.1, Amer. Math. Soc, Providence R.I., 1959, pp. 100–104. Xia Yin, Xianhua Li, The normal index of subgroups in finite groups (To appear). D.J.S. Robinson, A Course in the Theory of Groups, Springer, New York, Berlin, 1993. J.C. Beidleman, A.E. Spencer, The normal index of maximal subgroups in finite groups, Illinois J. Math. 16 (1972) 95–101. N.P. Mukherjee, Prabir Bhattacharya, The normal index of a finite group, Pacific J. Math. 132 (1) (1988) 143–148. N.P. Mukherjee, Prabir Bhattacharya, The normal index of a maximal subgroup of a finite group, Proc. Amer. Math. Soc. 106 (1) (1989) 25–32. A. Ballester-Bolinches, On the normal index of maximal subgroups in finite groups, J. Pure Appl. Algebra 64 (1990) 113–118. Guo Xiuyun, K.P. Shum, Cover-avoidance properties and the structure of finite groups, J. Pure Appl. Algebra 181 (2003) 297–308. Yanming Wang, C -normality of groups and its properties, J. Algebra 180 (1996) 954–965. H. Kurzweil, Endliche Grupper, Springer, Berlin, 1977. Klaus Doerk, Trevor Hawkes, Finite Soluble Groups, Walter de Gruyter, Berlin, New York, 1992. Li Duyu, Guo xiuyun, The influence of c-normality of subgroups on the structure of finite groups, J. Pure and Appl. Algebra 150 (2000) 53–60. Yanming Wang, Finite groups with some subgroups of Sylow subgroups c-supplemented, J. Algebra 224 (2000) 467–478.
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Journal of Pure and Applied Algebra journal homepage: www.elsevier.com/locate/jpaa
On the normal indices of proper subgroups of finite groupsI Xianhua Li ∗ , Xinjian Zhang School of Mathematical Science, Soochow University, Suzhou, Jiangsu 215006, People’s Republic of China
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Article history: Received 26 October 2008 Received in revised form 7 October 2009 Available online 5 November 2009 Communicated by M. Sapir
abstract We extended the normal index from maximal subgroups to proper subgroups. We give a quantitative version of all results obtained by using c-normal subgroups and obtain some new characterizations of solvable, supersolvable and nilpotent groups by the normal indices of proper subgroups. © 2009 Elsevier B.V. All rights reserved.
MSC: 20D10 20D15
1. Introduction In this paper, all groups considered are finite. In 1954, Huppert obtained the famous theorem: a group G is supersolvable if and only if the index of every maximal subgroup of G is prime. After that, Deskins in [1] introduced the normal index of a maximal subgroup M of a group G. The normal index of a maximal subgroup M of a group G, denoted by η(G : M ), is the order of a chief factor H /K of G, where H is a minimal supplement of M in G. Several authors gave the characterizations of solvability, supersolvability, etc. of a group G by the normal indices of some maximal subgroups (see [1,4–7]). This provides some well quantitative versions and ways of characterizations of a group G. However, the definition of the normal index is confined to the maximal subgroups only and maximal subgroups are special subgroups, we cannot go further. Recall that a subgroup H of a group G is called c-normal in G if there exists a normal subgroup N of G such that NH = G and H ∩ N ≤ HG . In fact, Guo and Li in [12] pointed out that the concept of c-normality provides a better tool for us and it makes things easier and simpler than normal indices do. In our present work, we would like to say that it may not be true. In [2], we extended the concept of the normal index from maximal subgroups to proper subgroups of a group. In this paper, we prove that if the normal index of a subgroup H of G is equal to the index |G : H |, then H is c-normal in G. Hence we give a quantitative characterization of c-normality of subgroups by using the normal index. It means that the c-normality of a subgroup H is equivalent to that H has the special normal index. On the other hand, the normal index is a number, which provides a new tool and way to study the properties of groups by the quantitative information of themselves. Using the normal index of a proper subgroup, we give the characterizations of a group to be solvable, supersolvable and nilpotent. 2. Definitions and preliminary results Definition 2.1 ([2, Definition 2.2]). Let H be a proper subgroup of a group G and K a normal subgroup of G. Then |K /K ∩ HG | is called the normal index of H in G, denoted by η∗ (G : H ), if K satisfies (1) G = KH; (2) if G = TH for any normal subgroup T of G, then we have |T /T ∩ HG | ≥ |K /K ∩ HG |. I This work was supported by the National Natural Science Foundation of China (Grant Nos. 10571128, 10871032), the Natural Science Foundation of Jiangsu Province (No. BK2008156). ∗ Corresponding author. E-mail address: [email protected] (X. Li).
0022-4049/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.jpaa.2009.10.012
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Let G be a group and H be a proper subgroup of G. Let K1 and K2 be normal subgroups of G such that K1 satisfies the conditions of Definition 2.1 and K2 satisfies HG ≤ K2 , G = K2 H and |K2 | ≤ |T | for any normal subgroup T of G such that HG ≤ T and G = TH. Then G = K1 H = K2 H, |K2 /HG | = |K2 /K2 ∩ HG | ≥ |K1 /K1 ∩ HG | by Definition 2.1. On the other hand, by G = (K1 HG )H and the hypothesis of K2 , we have |K1 /K1 ∩ HG | = |K1 HG /HG | ≥ |K2 /HG |. Hence |K1 /K1 ∩ HG | = |K2 /HG |. Thus we have shown that Definition 2.1 is equivalent to the following. Definition 2.2. Let H be a proper subgroup of a group G and K a normal subgroup of G containing HG . Then |K /HG | is called the normal index of H in G, if K satisfies (1) G = KH; (2) |K | ≤ |T | for any normal subgroup T of G such that HG ≤ T and G = TH. Lemma 2.3 ([2]). Let G be a group. (1) If M is a maximal subgroup of G, then η∗ (G : M ) = η(G : M ). (2) If N E G and H is a subgroup of G such that N ≤ H, then η∗ (G/N : H /N ) = η∗ (G : H ). (3) G is solvable if and only if there exists a solvable 2-maximal subgroup L of G such that η∗ (G : L) = |G : L|. Remark 2.1. (1), (2) and (3) in Lemma 2.3 are Proposition 2.1, Lemma 2.2 and Theorem 3.1 in [2], respectively. (1) in Lemma 2.3 shows that the normal index here is a generalization of the concept defined in [1]. Lemma 2.4 ([9]). Let G be a group and H, K be subgroups of G. (1) G is solvable if and only if there exists a solvable c-normal maximal subgroup M of G. (2) If H is c-normal in G and H ≤ K ≤ G, then H is c-normal in K . Remark 2.2. (1) and (2) are Theorem 3.4 and Lemma 2.1 (3) in [9], respectively. Lemma 2.5 ([8, Lemma 3.12]). Let p be the smallest prime dividing the order of a group G and P a Sylow p-subgroup of G. If |P | ≤ p2 and G is A4 -free, then G is p-nilpotent. 3. Normal index theorems and properties of a group In the theory of groups, some index theorems have fundamental importance, for example, Lagrange’s theorem, etc.. In this section, we are interested in what will happen if we replace the index by the normal index in some index theorems. That is, if the normal index theorem is true, what properties does G have? Let H and K be subgroups of G. Firstly, since |G : H | | η∗ (G : H ), if (η∗ (G : H ), η∗ (G : K )) = 1, then (|G : H |, |G : K |) = 1 and so G = HK . Secondly, to be analogous to Lagrange’s theorem, we consider the case |G| = |H |η∗ (G : H ). This is equivalent to |G : H | = η∗ (G : H ). Theorem 3.1. Let H be a proper subgroup of a group G. Then η∗ (G : H ) = |G : H | if and only if H is c-normal in G. Proof. Assume that K is a normal subgroup of G such that η∗ (G : H ) = |K /HG |. By the definition of the normal index, G = KH. Hence |G : H | = |K : K ∩ H |. It follows from the hypothesis η∗ (G : H ) = |G : H | that K ∩ H = HG . Thus H is c-normal in G. Now we prove the sufficiency. Suppose that H is c-normal in G, then there exists a normal subgroup K of G such that G = KH and K ∩ H ≤ HG . It follows that K ∩ H = K ∩ HG . Hence |G : H | = |K : K ∩ HG | = |KHG /HG | and |G : H | > η∗ (G : H ). Since |G : H | divides η∗ (G : H ), η∗ (G : H ) = |G : H | holds. Theorem 3.1 gives a quantitative characterization of the c-normality of subgroups. Replacing the condition that H is c-normal in G by η∗ (G : H ) = |G : H |, we can obtain a quantitative version of all results obtained by using c-normal subgroups. For instance, Theorem 3.2 in [12], Theorem 4.1 in [9] and Theorem 4.2 in [13] can be written in the following forms, respectively. (1) Let G1 be a class of Frobenius groups with elementary abelian kernel N and cyclic complement M, such that |M | is square-free and the order of M is not prime. If G is G1 -free and η∗ (G : L) = |G : L| for each 2-maximal subgroup L of G, then G is supersolvable. (2) Let G be a group. Suppose η∗ (G : P1 ) = |G : P1 | for every Sylow subgroup P of G and every maximal subgroup P1 of P. Then G is supersolvable. (3) Let G be a group and p be the smallest prime divisor of |G|. Assume that G is A4 -free and η∗ (G : P2 ) = |G : P2 | for every 2-maximal subgroup P2 of the Sylow p-subgroups of G. Then G/Op (G) is p-nilpotent. Thirdly, to be parallel to the index theorem |G| = |G : K ||K : H | for H ≤ K ≤ G, we consider the question: if η∗ (G : H ) = η∗ (G : K )η∗ (K : H ) for any H ≤ K ≤ G, what properties does G have? Definition 3.2. Let G be a group. Then G is called an NIT -group if η∗ (G : H ) = η∗ (G : K )η∗ (K : H ) for any H ≤ K ≤ G. Lemma 3.3. Let G be an NIT -group. Then (1) H is an NIT -group for any subgroup H of G; (2) G/N is an NIT -group for any normal subgroup N of G.
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Proof. From Lemma 2.3 (2) and the definition of the NIT -group, the conclusions are obvious. Theorem 3.4. Let G be a group. Then the following statements are equivalent. (1) (2) (3) (4)
G is an NIT -group; G is a supersolvable NIT -group; η∗ (G : H ) = |G : H | for any subgroup H of G; H is c-normal in G for any subgroup H of G.
Proof. (1) ⇒ (2): We use induction on |G|. Suppose that G is simple. If G is solvable, then |G| = p and the result is true. Assume that G is a non-abelian simple group. Let M be a maximal subgroup of G and L a maximal subgroup of M. Then by the definition of the NIT -group, η∗ (G : L) = η∗ (G : M )η∗ (M : L). Since G is simple, η∗ (G : L) = η∗ (G : M ) = |G|, so η∗ (M : L) = 1, that is, M = L, a contradiction. This shows that M = 1 and so G is a cyclic group of prime order, again a contradiction. Hence we may assume that G is not simple. Let N be a minimal normal subgroup of G, then |N | < |G| and |G/N | < |G|. By induction, G/N and N are both supersolvable from Lemma 3.3. Thus N is an elementary abelian group. Let |N | = pa . Suppose that a > 1. Let T be a maximal subgroup of N. Then |T | = pa−1 , T > 1 and T E N. By the minimality of N, TG = 1. Let K E G such that G = KT and η∗ (G : T ) = |K /TG | = |K |. Then by the hypothesis that G is an NIT -group, η∗ (G : T ) = η∗ (G : N )η∗ (N : T ). It follows from Lemma 2.3 that |K | = |G : N ||N : T | = |G : T |. Hence K ∩ T = 1. Note that G = NK and N ∩ K = 1 or N. If N ∩ K = 1, then |N | = |T |, a contradiction. Thus N ∩ K = N. But this case forces that T ≤ N ≤ K and so T = 1, again a contradiction. Therefore a = 1, that is, |N | = p. Thus G is supersolvable as desired. (2) ⇒ (3): We also use induction on |G|. For any maximal subgroup M of G, M is an NIT -group by Lemma 3.3. The induction implies that η∗ (M : H ) = |M : H | for any subgroup H of M. Since G is supersolvable, η∗ (G : M ) = |G : M |. Now from the hypothesis and the definition of the NIT -group, it follows that η∗ (G : H ) = η∗ (G : M )η∗ (M : H ) and so η∗ (G : H ) = |G : H |. (3) ⇔ (4): It follows immediately from Theorem 3.1. (3) ⇒ (1): Let H and K be subgroups of G and H ≤ K . We first prove that η∗ (K : H ) = |K : H |. Let T E G such that G = TH and η∗ (G : H ) = |T /HG |, then K = K ∩ TH = H (T ∩ K ) and T ∩ K E K . Since η∗ (G : H ) = |G : H | and HG ≤ T ∩ H, we have η∗ (G : H ) = |T /HG | ≥ |T : T ∩ H | = |G : H |, which implies that HG = T ∩ H. Thus ((T ∩ K )/HG )∩(H /HG ) ⊆ (T /HG )∩(H /HG ) = HG /HG , and so η∗ (K /HG : H /HG ) = |T ∩ K /HG |. Because K = H (T ∩ K ), we have |K : H | = |T ∩ K : T ∩ H | = |T ∩ K /HG |. By Lemma 2.3(2), η∗ (K /HG : H /HG ) = η∗ (K : H ). Hence η∗ (K : H ) = |T ∩ K /HG | = |K : H |. And by the hypothesis, η∗ (G : K ) = |G : K |. Hence η∗ (G : H ) = |G : H | = |G : K ||K : H | = η∗ (G : K )η∗ (K : H ). By Definition 3.2, G is an NIT -group. Theorem 3.5. Let G be a group. If for every maximal subgroup M of G and every maximal subgroup H of M, η∗ (G : H ) = η∗ (G : M )η∗ (M : H ), then G is solvable and every 2-maximal subgroup of G is c-normal in G. Proof. We first prove that G is solvable. Assume that it is false and G is a minimal counterexample. Let N be a minimal normal subgroup of G. By Lemma 2.3, G/N satisfies the hypothesis and by induction G/N is solvable. A routine argument shows that N is the unique minimal normal subgroup of G and Φ (G) = 1. Now we assume that N is non-solvable. Let p = max π (N ) , then p > 2. Let P ∈ Sylp (N ). Then NG (P ) ≤ NG (Z (J (P ))) < G. Hence there exists a maximal subgroup M of G such that NG (Z (J (P ))) ≤ M. By the Frattini argument, we have G = NG (Z (J (P )))N = MN. Now we claim that N ∩ M ≤ Φ (M ). Otherwise, let M1 be a maximal subgroup of M such that M = M1 (N ∩ M ), then G = MN = M1 N, so η∗ (G : M ) = η∗ (G : M1 ) = |N |. But this is impossible for the hypothesis η∗ (G : M1 ) = η∗ (G : M )η∗ (M : M1 ). So N ∩ M ≤ Φ (M ). By NN (Z (J (P ))) = N ∩ NG (Z (J (P ))) ⊆ N ∩ M ≤ Φ (M ), we get NN (Z (J (P ))) is nilpotent. By the Glauberman– Thompson theorem, we obtain N is p-nilpotent, which contradicts to the minimality of N. This contradiction shows that N is solvable, and so G is solvable. Let M be a maximal subgroup of G and H be a maximal subgroup of M. Let η∗ (G : M ) = |T /MG | and η∗ (M : H ) = |S /HM |, where MG ≤ T E G, HM ≤ S E M. Then T /MG is a normal complement of M /MG in G/MG and S /HM is a normal complement of H /HM in M /HM , respectively. Hence we have η∗ (G : M ) = |G/MG : M /MG | = |G : M | and η∗ (M : H ) = |M /HM : H /HM | = |M : H |. From the hypothesis, it follows that η∗ (G : H ) = η∗ (G : M )η∗ (M : H ) = |G : M ||M : H | = |G : H |. By Theorem 3.1, H is c-normal in G. Corollary 3.6. Let G be a group and F be the class of Frobenius groups with elementary abelian kernel N and cyclic complement K , such that |K | is square-free and the order of K is not prime. If G is F -free and η∗ (G : H ) = η∗ (G : M )η∗ (M : H ) for every maximal subgroup M of G and every maximal subgroup H of M, then G is supersolvable. Proof. From Theorem 3.5, we have G is solvable and each 2-maximal subgroup of G is c-normal in G. Then by [12, Theorem 3.2], G is supersolvable. 4. The normal indices of some special subgroups First, we consider the influence of the normal indices of some 2-maximal subgroups of a group G on the structure of G. Lemma 4.1. Let M be a maximal subgroup of a group G. If η(G : M ) is square-free, then η(G : M ) is a prime and η(G : M ) = |G : M |.
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Proof. Let H /K be a chief factor of G, where H is a minimal supplement of M in G. By the definition of the normal index, |H /K | = η(G : M ). Hence the hypothesis implies that |H /K | is square-free and so H /K is solvable. It follows from the minimality of H /K that |H /K | is a prime. This forces that H /K ∩ M /K = K /K . Noting that G = MH, |H /K | = |G/K : M /K | = |G : M |, the conclusions are true. Lemma 4.2. Let G be a group, M a subgroup of G and H a maximal subgroup of M. Then (1) η∗ (G : H ) = aη∗ (M : H ), where a > 1 is a natural number; (2) Let η∗ (G : H ) = |K /HG | and G = KH, where HG E K E G. If M is maximal in G and K /HG is solvable, then η∗ (G : H ) = bη(G : M ), where b > 1 is a natural number. Additionally, if H is solvable, then G is solvable. (3) If H is solvable and η∗ (G : H ) is either square-free or the square of a prime, then G is solvable. (4) Let M be a maximal subgroup of a group G. If M has a maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime, then η(G : M ) is a prime. Proof. Let K be a normal subgroup of G containing HG such that G = KH and η∗ (G : H ) = |K /HG |. Then M = M ∩ KH = (M ∩ K )H and M ∩ K E M. If η∗ (G : H ) is either square-free or the square of a prime, then K /HG is solvable. If H is solvable, then K is solvable. Since G/K ∼ = H /H ∩ K is solvable, we get G is solvable. So (3) holds. Next we prove (1) and (2). Since H is maximal in M and M = (M ∩ K )H, it is obvious that HG < M ∩ K and M ∩ K * H. Let 1 < · · · < HG = M0 < M1 < · · · < Ms = M ∩ K ≤ · · · ≤ M be a chief series of M. Then there exists i ∈ {1, . . . , s} such that Mi * H but Mi−1 ⊆ H. It follows that Mi is a minimal supplement of H in M. Thus η∗ (M : H ) = η(M : H ) = |Mi /Mi−1 | and Mi−1 ≤ HM . Hence η∗ (M : H ) divides |M ∩ K /HM |. Note that HG < K ∩ M. If |K /HG | = |K ∩ M /HG |, then |K | = |K ∩ M |, K = K ∩ M, thus K ≤ M and G = KH = M, a contradiction. So |K /HG | > |K ∩ M /HG | ≥ η∗ (M : H ). (1) holds. Suppose that M is a maximal subgroup of G and K /HG is solvable. Let N /MG be a minimal normal subgroup of G/MG contained in KMG /MG , it is easy to prove that η∗ (G : M ) = |N /MG |. Thus η∗ (G : M ) divides |KMG /MG |. Since KMG /MG ∼ = K /K ∩ MG and HG ≤ K ∩ MG , we obtain η∗ (G : H ) = bη(G : M ). Now we claim that K /HG is not a minimal normal subgroup of G/HG . Otherwise, K /HG is an elementary abelian group. Since G = HK = KM, we have K /HG ∩ H /HG = HG /HG and K /HG ∩M /HG = HG /HG . Thus |H | = |M |, a contradiction. Let N /HG be a minimal normal subgroup of G/HG contained in K /HG . Then N < K and so |N /HG | < |K /HG | and |K /N | > 1. If N ≤ M, then N ≤ MG and so |KMG /MG | = |K /K ∩ MG | < |K /HG |. By G = (KMG )M and the definition of η∗ (G : M ), η∗ (G : M ) ≤ |KMG /MG | < |K /HG | = η∗ (G : H ). If N 6≤ M, then G = NM, we have η∗ (G : M ) ≤ |NMG /MG | = |N /N ∩ MG | ≤ |N /HG | < |K /HG | = η∗ (G : H ). Thus we get b > 1, and (2) holds. (4) is immediate from (2) and Lemma 4.1. Lemma 4.3. Let K be a normal subgroup of a group G. Assume that G/K is supersolvable. (1) If |K | is square-free, then G is supersolvable. (2) Let |K | be the square of a prime. If K is not a minimal normal subgroup of G, then G is supersolvable. Proof. (1) Refine 1 G K G G into a chief series G0 = 1 G G1 · · · < · · · < Gs < K = Gs+1 < · · · < Gn = G. Since |K | is square-free and G/K is supersolvable, Gi+1 /Gi are cyclic groups with prime orders for i ∈ {0, 1, . . . , s}. So G is supersolvable. (2) Let N be a minimal normal subgroup of G contained in K . Then |N | and |K /N | are primes. It is easy to see that G is supersolvable. Lemma 4.4. Let H be a supersolvable subgroup of a group G. If one of the following holds, then G/HG is supersolvable. (1) η∗ (G : H ) is square-free. (2) H is a 2-maximal subgroup of G, η∗ (G : H ) is either square-free or the square of a prime. Proof. Let K be a normal subgroup of G such that G = HK , HG ≤ K and η∗ (G : H ) = |K /HG |. (1) Since H is supersolvable, (G/HG )/(K /HG ) ∼ = G/K ∼ = H /H ∩K is supersolvable. By the hypothesis, |K /HG | is square-free. By Lemma 4.3(1), we have G/HG is supersolvable. (2) By (1), we may assume that |K /HG | is the square of a prime. Since H is a 2-maximal subgroup of G and K /HG is abelian, it is easy to see that K /HG is not a minimal normal subgroup of G/HG . Note that (G/HG )/(K /HG ) ∼ = G/K ∼ = H /H ∩ K is supersolvable. By Lemma 4.3(2), we have G/HG is supersolvable. Lemma 4.5. Let G be a group and M be a supersolvable maximal subgroup of G. If F (G) * M and |G : M | is a prime, then G is supersolvable. Proof. Suppose that the statement is false and let G be a minimal counterexample. Since F (G) * M and M is a maximal subgroup of G, G = F (G)M. Since G/F (G) ∼ = M /F (G)∩M, F (G) and G/F (G) are solvable. Hence G is solvable. If Φ (G) 6= 1, since F (G)/Φ (G) = F (G/Φ (G)) and F (G) * M, we have F (G/Φ (G)) * M /Φ (G). This shows that G/Φ (G) satisfies the hypothesis. Because G is a minimal counterexample, we have G/Φ (G) is supersolvable, thus G is supersolvable, a contradiction. Hence Φ (G) = 1. Since G is solvable, F (G) is the direct product of minimal normal subgroups of G. Hence there exists a minimal normal subgroup N of G such that N is not contained in M. Then G = NM, N ∩ M = 1 and so |N | = |G : M |. It follows that G/N ∼ = M and |N | is prime. Since M is supersolvable, we get G is supersolvable, a contradiction. Hence the statement is true.
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Theorem 4.6. Let G be a group. Suppose that G has a maximal subgroup M such that η∗ (G : L) = |G : L| for each maximal subgroup L of M. Then (1) G is solvable; (2) M /MG is a cyclic group of square-free order; (3) Assume that MG < M. Then Z (G/MG ) = 1, (G/MG )0 = F (G/MG ) is a Sylow subgroup of G/MG and a minimal normal complement subgroup of M /MG in G/MG . (4) If MG = 1, then M is a cyclic group of square-free order, F (G) is a Sylow subgroup of G and a minimal normal complement subgroup of M in G. Furthermore, for each maximal subgroup K of G, if K is not normal in G, then there exist g ∈ G such that K = M g and K ∩ M = 1 if g 6∈ M. Proof. Let L be a maximal subgroup of M. By Lemma 3.1 we obtain L is c-normal in G. From Lemma 2.4 (2), it follows that L is c-normal in M. By [9, Corollary 3.3], M is solvable and so L is solvable. That is, G has a solvable 2-maximal subgroup L such that η∗ (G : L) = |G : L|. By Lemma 2.3(3), G is solvable and then (1) holds. Now we prove (2). Without loss of generality, we may assume that MG = 1. Then Φ (G) = 1 and G0 6= 1. Since G is solvable, by [11, Theorem 15.6]), G has the unique minimal normal subgroup, N say, N = CG (N ) = F (G) and G = N o M. If Z (G) 6= 1, then N ≤ Z (G) and so MG = M, a contradiction, hence Z (G) = 1. From the first paragraph, M is solvable. So M has a composite series M = M0 B M1 B M2 B · · · B Mn = 1 such that Mi−1 /Mi are cyclic groups with prime orders for i ∈ {1, 2, . . . , n}. It follows that Mi is maximal in Mi−1 for i = 1, 2, . . . , n. Now by the hypothesis M1 is c-normal in G, there exists a normal subgroup K1 of G such that G = M1 K1 and M1 ∩ K1 ≤ (M1 )G = 1. It is obvious that G = K1 M and K1 ∩ M E M. Since M = M ∩ M1 K1 = M1 (M ∩ K ), we have |M : M1 | = |K1 ∩ M : M1 ∩ K1 |. Hence |K1 ∩ M | is a prime. Let K1 ∩ M = T1 . By T1 ∩ M1 = 1, M = T1 × M1 . Noting that M2 C M1 , T1 × M2 C M and |M : (T1 × M2 )| = |M1 : M2 | is a prime, so T1 × M2 is maximal in M. Thus there exists a normal subgroup K2 of G such that G = K2 (T1 × M2 ) and K2 ∩ (T1 × M2 ) ≤ (T1 × M2 )G ≤ MG = 1. Put K2 ∩ M = T2 , then T2 E M and |T2 | is prime, so we have M = T2 × T1 × M2 . Similarly, we consider M3 , . . . , Mn and finally conclude that M = Tn × Tn−1 × · · · × T2 × T1 , where Ti are cyclic groups of prime orders for i = 1, 2, . . . , n. So M is abelian and every Sylow subgroup of M is an elementary abelian group. Let g ∈ G \ M, then g = mn, where m ∈ M, n ∈ N and n 6= 1. If M ∩ M g 6= 1, then there exists an element a ∈ M such that an = a1 ∈ M. Hence, [a, n] = a−1 a1 ∈ N ∩ M = 1. It follows that an = a and n ∈ NG (hai). Again by M ≤ NG (hai) and the maximality of M, we have NG (hai) = G, that is, hai E G, contrary to MG = 1. Thus G is a Frobenius group with complement M. By the Frobenius theorem [10, Theorem 10.3] and the maximality of M we know that N is the Frobenius kernel. By [3, Theorem 10.5.6], we get M is cyclic and |M | is square-free. (2) is proved. Assume that |N | = pα . If p divides the order of M, let P ∈ Sylp (M ). Since M is cyclic, M ≤ NG (P ). Noting that N ∩ P = 1, P is not a Sylow p-subgroup of G. So M < NG (P ). The maximality of M forces that NG (P ) = G and so P E G, which contradicts to MG = 1. So N is a Sylow subgroup of G. Certainly, G/F (G) ∼ = M is cyclic, so G0 ≤ N = F (G). By G0 6= 1 and the minimality 0 of N, G = F (G) = N. So (3) holds. If MG = 1, then by (2) and (3), we have M is a cyclic group of square-free order, G0 = F (G) is a Sylow subgroup of G and a minimal normal complement of M in G. Let N = F (G) and K be a maximal subgroup of G but K is not normal in G. If N ≤ K , then K E G, a contradiction. So N K , G = NK and N ∩ K = 1. Noting that N is a Sylow subgroup of G, K and M are both complements of N in G, so there exist g ∈ G such that K = M g . From the proof of Theorem 4.6, we have M is a Frobenius complement of G. So if g 6∈ M, then K ∩ M = 1. The proof is completed. Corollary 4.7. Let G be a group. Then G is solvable if η∗ (G : L) = |G : L| for every 2-maximal subgroup L of G. Theorem 4.8. A group G is solvable if and only if there exists a solvable 2-maximal subgroup L of G such that |π (η∗ (G : L))| ≤ 2. Proof. Suppose that L is a solvable 2-maximal subgroup of G such that |π (η∗ (G : L))| ≤ 2. Let K be a normal subgroup of G such that G = KL and η∗ (G : L) = |K /LG |. The hypothesis implies that |π (K /LG )| ≤ 2 and therefore K /LG is solvable. By Lemma 4.2, G is solvable. Conversely, assume that G is solvable. By Lemma 2.3(3) there exists a 2-maximal subgroup L of G such that η∗ (G : L) = |G : L|. Let M be a maximal subgroup of G such that L is maximal in M. Then |G : L| = |G : M ||M : L|. Now applying the solvability of G, we can get |π (|G : M |)| = 1 and |π (|M : L|)| = 1. Hence |π (η∗ (G : L))| = |π (|G : L|)| ≤ 2. Theorem 4.9. Let G be a group. If G has a maximal subgroup M satisfying one of the following conditions, then G is supersolvable: (1) η∗ (G : L) is either square-free or the square of a prime for any maximal subgroup L of M; (2) M is supersolvable, MG = 1 and M has a maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime; (3) M is supersolvable , MG = 1 and η(G : M ) is square-free; (4) M is supersolvable, F (G) M and M has a maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime; (5) M is supersolvable, F (G) M and η(G : M ) is square-free;
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(6) M is cyclic and η(G : M ) is square-free; (7) M has a cyclic maximal subgroup L such that η∗ (G : L) is either square-free or the square of a prime. Proof. Assume that M satisfies (1). Let L be any maximal subgroup of M. By the hypothesis, η∗ (G : L) is either square-free or a prime square. From Lemma 4.2, it follows that η∗ (G : L) = aη∗ (M : L) = bη∗ (G : M ), where a > 1, b > 1. Hence η∗ (G : M ) and η∗ (M : L) are square-free. Now Lemma 4.1 and Lemma 2.3 force that η∗ (G : M ) and η∗ (M : L) are primes. Since |M : L| | η∗ (M : L), the index of every maximal subgroup of M in M is prime and so M is supersolvable by Huppert’s theorem. From Lemma 4.2, we may get G is solvable. Suppose that N1 , N2 are two distinct minimal normal subgroups of G. If N1 M, then G = N1 M, N1 ∩ M = 1 and η∗ (G : M ) = |N1 |. By the hypothesis, |N1 | is a prime and G/N1 ∼ = M is supersolvable. It implies that G is supersolvable. Hence we may assume that Ni ≤ M for i = 1, 2. Then Lemma 2.3 (2) shows that the hypothesis holds for G/N1 and G/N2 . By induction on |G|, G/N1 and G/N2 are supersolvable. Since the class of supersolvable groups is a saturated formation, G∼ = G/N1 ∩ N2 is supersolvable. Hence we may let N be the unique minimal normal subgroup of G and N ≤ M. By induction, G/N is supersolvable. If N ⊆ Φ (M ), then by [3, 5.2.13], N ⊆ Φ (G). It follows that G is supersolvable. Thus we suppose that N * Φ (M ). Then there exists a maximal subgroup T of M such that N * T . It follows that M = NT . By the unique minimality of N, TG = 1. Let A be a normal subgroup of G such that G = AT and η∗ (G : T ) = |A/TG | = |A|. By the hypothesis, |A| is either square-free or the square of a prime. By the unique minimality of N, N ≤ A. If N = A, then G = AT = NT = M, a contradiction, so N < A. Thus |N | is a proper divisor of |A| and so |N | is a prime. This implies that G is supersolvable and so (1) holds. If M is supersolvable and MG = 1, then L is supersolvable and LG = 1 for each maximal subgroup L of M. So if M satisfies (2), then by Lemma 4.4, G is supersolvable, which implies that (2) holds. If M satisfies one of the conditions (3), (6) and (7), then by Lemma 4.4 again, we have G is supersolvable. Now we prove (4). From Lemma 4.2, η(G : M ) is a prime. So by Lemma 4.4, G/MG is supersolvable and by the hypothesis G is solvable. It follows from Lemma 4.5 that G is supersolvable. And by (4), (5) is obvious. The proof is completed. Corollary 4.10. Let G be a group. Then G is supersolvable if G satisfies one of the following conditions: (1) η∗ (G : L) is either square-free or the square of a prime for every 2-maximal subgroup L of G. (2) η∗ (G : L) is square-free for every 2-maximal subgroup L of G. Now we consider the influence of the normal indices of some Sylow subgroups of a group G on the structure of G. Theorem 4.11. Let p be a prime dividing the order of a group G and P a Sylow p-subgroup of G. Assume that η∗ (G : P ) = pα |G : P | where α 6 2. Then the following conclusions are true: (1) If p = 2 and G is A4 -free, then G/O2 (G) is 2-nilpotent. Furthermore, if |O2 (G)| 6 22 , then G is 2-nilpotent. (2) If p 6= 2 and (|G|, p2 − 1) = 1, then G/Op (G) is p-nilpotent. Furthermore, if |Op (G)| 6 p2 , then G is p-nilpotent. Proof. Let G = G/Op (G) and P = P /Op (G), then P G = 1. Hence there exists a normal subgroup K of G such that G = K P and η∗ (G : P ) = |K |. By Lemma 2.3(2), η∗ (G : P ) = η∗ (G : P ), hence |K | = pα |G : P |. On the other hand,
|G : P | = |G : P | = |K P : P | =
|K | . |K ∩P |
It follows that |K ∩ P | = pα . That is, the order of the Sylow p-subgroups of K is pα .
(1) By Lemma 2.5, K is 2-nilpotent. Let H be the normal Hall 20 -subgroup of K . Then H is a characteristic subgroup of K and H is a normal Hall 20 -subgroup of G. Hence G is 2-nilpotent. Now by Lemma 2.5, the later statement is obvious. (2) If α = 0, then K is a normal Hall p0 -subgroup of G. So we may only consider the case α = 1 and α = 2. Let R ∈ Sylp (K ). Then R ≤ CK (R) and R is either a cyclic group of order pα or an elementary abelian group of order p2 . Hence |Aut (R)| | p(p − 1)2 (p + 1). Now by [3, 1.6.13], NK (R)/CK (R) is isomorphic to a subgroup of Aut (R). From the hypothesis (|G|, p2 − 1) = 1, it follows that NK (R) = CK (R). By Burnside’s theorem, K is p-nilpotent. Using the similar proof as (1), G/Op (G) is p-nilpotent. Theorem 4.12. Let G be a group and p ∈ π (G). Assume that there exists a Sylow p-subgroup P of G such that η∗ (G : P ) = pα |G : P |, where α 6 1. If either p = min π (G) or (|G|, p − 1) = 1 , then G/Op (G) is p-nilpotent. Specifically, G is solvable. Proof. The proof is similar to Theorem 4.11 and omitted here.
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