The Influence of Minimal Subgroups on the Structure of Finite Groups 1

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1 Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 14, The Influence of Minimal Subgroups on the Structure of Finite Groups 1 Honggao Zhang 1, Jianhong Huang 1,2 and Yufeng Liu 3 1. Department of Mathematics, Xuzhou Normal University Xuzhou , P.R. China 2. Department of Mathematics University of Science and Technology of China Hefei , P.R. China 3. School of Math. and Informational Science Shandong Institute of Business and Technology Yantai , P.R. China zhanghonggao2007@126.com, jhh320@126.com Abstract Let G be a finite group and F be a class of groups. A subgroup H of G is said to be F-supplemented in G if there exists a subgroup T of G such that G = TH and (H T )H G /H G is contained in the F-hypercenter Z F (G/H G)ofG/H G. In this paper, we investigate further the influence of F-supplemented subgroups on the structure of finite groups. Mathematics Subject Classification: 20D10, 20D15, 20D20 Keywords: Finite groups, F-supplemented subgroups, minimal subgroups, saturated formation 1 Introduction Throughout this paper, all groups considered are finite and G denotes a finite group. The notations and terminologies are standard, as in [6] and [9]. The relationship between the subgroups of G and the structure of G has been extensively studied in the literature. Ito [3, III, 5.3] has proved that if G 1 Research is supported by an NNSF grant of China (Grant # ) and a postgraduate innovation grant of Xuzhou Normal University.

2 676 Honggao Zhang, Jianhong Huang and Yufeng Liu is a group of odd order and all minimal subgroups of G lie in the center of G, then G is nilpotent. Buckley [2] proved that if G is a group of odd order and all minimal subgroups of G are normal in G, then G is supersoluble. Recently, by considering some special supplemented subgroups, people obtained a series of new interesting results. For example, Wang introduced the concepts of c- normal subgroup [11] and c-supplemented subgroup [12]: a subgroup H of a group G is said to be c-supplemented (c-normal) in G if there exists a subgroup (normal subgroup) K of G such that G=HK and H K H G, where H G is the maximal normal subgroup of G contained in H. They used them to study conditions for solubility and supersolubility of finite groups. Later, Guo [5] introduced the following concept. Definition 1.1([5]). Let F be a class of groups. A subgroup H of G is said to be F-supplemented in G if there exists a subgroup T of G such that G = TH and (H T )H G /H G is contained in the F-hypercenter Z F (G/H G )ofg/h G. In this case, T is called an F-supplement of H in G. Recall that, for a class F of groups, a chief factor H/K of G is called F- center if [H/K](G/C G (H/K)) F (see [6, 10]). The symbol Z F (G) denotes the F-hypercenter of G, that is, the product of all such normal subgroups H of G whose G-chief factors are F-center. A subgroup H of G is said to be F-hypercentral in G if H Z (G). F A class of groups F is said to be S-closed (S n -closed) if it contains all subgroups (all normal subgroups, respectively) of every its group. A class F of groups is called a formation if it is closed under homomorphic image and every group G has a smallest normal subgroup (called F-residual and denoted by G F ) with quotient in F. A formation F is said to be saturated if it contains every group G with G/Φ(G) F. We use N, U and S to denote the formations of all nilpotent groups, all supersoluble groups and all soluble groups, respectively. It is well known that N, U and S are all S-closed saturated formations. Obviously, all the subgroups, whether they are normal, complemented, c- normal, c-supplemented, are all F-supplemented subgroups, for any nonempty saturated formation F. However, the converse is not true (see Example 1.2, 1.3 in [5]). By using some F-supplemented subgroups, Guo [5] has given some conditions under which a finite group belongs to some formations. The purpose of this paper is to go further into influence of F-supplemented subgroups on the structure of finite groups. Some new results are obtained and a series of previously known results are generalized.

3 Influence of minimal subgroups Preliminaries Lemma 2.1 ([5, Lemma 2.1]). Let F be a non-empty saturated formation, A G and Z = Z F (G). Then (1) If A is normal in G, then AZ/A Z F (G/A). (2) If F is S-closed, then Z A Z (A). F (3) If F is S n -closed and A is normal in G, then Z A Z (A). F (4) If G F, then Z = G. Lemma 2.2 ([5, Lemma 2.2]). Let H K G. Then (1) H is F-supplemented in G if and only if G has a subgroup T such that G = HT, H G T and (H/H G ) (T/H G ) Z F (G/H G). (2) Suppose that H is normal in G. Then K/H is F-supplemented in G/H if and only if K is F-supplemented in G. (3) Suppose that H is normal in G. Then, for every F-supplemented subgroup E in G satisfying ( H, E ) =1, HE/H is F-supplemented in G/H. (4) If H is F-supplemented in G and F is S-closed, then H is F-supplemented in K. (5) If H is F-supplemented in G, K is normal in G and F is S n -closed, then H is F-supplemented in K. (6) If G F, then every subgroup of G is F-supplemented in G. Lemma 2.3 ([4, VI, 14.3]). If G has an abelian Sylow p-subgroup P, then Z(G) G P =1. Lemma 2.4 ([1, Theorem 1]). Let F be a saturated formation and G be a minimal non-f-group such that (G F ) is a proper subgroup of G F, then G F is a soluble group. Lemma 2.5 ([6, Corollary 3.2.9]). If F is a saturated formation, then [G F,Z F (G)] = 1, for any group G. Lemma 2.6 ([5, Theorem 3.2]). Let F be a S-closed saturated formation containing U. Suppose that G has a normal subgroup E such that G/E F. If all cyclic subgroups of E of prime order and order 4 are U-supplemented in G, then G F. Lemma 2.7 ([5, Theorem 4.1]). A group G is soluble if and only if every minimal subgroup of G is S-supplemented in G.

4 678 Honggao Zhang, Jianhong Huang and Yufeng Liu 3 Main Results and applications Theorem 3.1. Let F be a S-closed saturated formation which satisfies that every minimal non-f-group is soluble. Then G is an F-group if and only if every cyclic subgroup of order 4 of G is F-supplemented in G and every minimal subgroup of G is contained in the F-hypercenter of G. Proof. The necessity is obvious, we only need to prove that sufficiency. Assume that the assertion is false and let G be a counterexample of minimal order. Let L be a proper subgroup of G. Since every cyclic subgroup of L of order 4isF-supplemented in G, we know that every cyclic subgroup of L of order 4isF-supplemented in L by Lemma 2.2(4). On the other hand, since every minimal subgroup of L is a minimal subgroup of G, every minimal subgroup of L is contained in Z (G) F L Z (L) F by Lemma 2.1. By the choice of G, L is an F-group and so G is a minimal non-f-group. By [6, Theorem 3.4.2] and the hypothesis, we know that G is soluble and G has the following properties: i) G F is a p-group, for some prime p ; ii) G F /Φ(G F ) is a chief factor of G; iii) If G F is abelian, then G F is an elementary abelian group; iv) If p>2, then the exponent of G F is p; Ifp = 2, then the exponent of G F is 2 or 4. Suppose that the exponent of G F is a prime. It follows that G F Z F (G) and so G F, a contradiction. Now assume that G F is not abelian and p = 2. We claim that there is no element of order 4 in G F \Φ(G F ). Assume that there exists an element x G F \Φ(G F ) with x = 4. Then by hypothesis, x is F-supplemented in G. Hence by Lemma 2.2(1), there exists a subgroup T of G such that G = x T and x / x G T/ x G Z F (G/ x G). We first assume that x / x G T/ x G = 1. Then x T = x G. If x G = 4, then x = x G G. Hence x Φ(G F )/Φ(G F ) G/Φ(G F ). Since G F /Φ(G F ) is a chief factor of G, x Φ(G F )=G F and so G F = x, a contradiction. If x G = 2, then G : T = x : T x = 2. Hence 1 T G. Because G is a minimal non-f-group, we have that T F. Suppose that G F = G. Then: (1) If G 2, then there exists a subgroup H of G such that H =2. By hypothesis, H Z F (G) and so 2 π(f). Since F is a saturated formation, N π(f) F and hence G F, a contradiction. (2) If G = 2, then G Z (G). F Hence G F, a contradiction. Therefore G F G and so G F F. It follows that G/T = G F /(G F T ) F. Hence G F T. It follows that G = x T = T, which contradicts G : T =2. Now assume that x G = 1. Let P 1 = G F T. Suppose that P 1 G. If P 1 Φ(G F ), then G F = G F G = G F x T = x (G F T ) = x, a contradiction. So P 1 Φ(G F ). since G F /Φ(G F ) is a chief factor of G,

5 Influence of minimal subgroups 679 P 1 Φ(G F )/Φ(G F )=G F /Φ(G F ). It follows that P 1 = G F and so G F T.Thus G = x T = G F T = T. This contradiction shows that P 1 G. Clearly, G : N G (P 1 ) = 2 and hence N G (P 1 ) G. Since G = x T = G F N G (P 1 ), G/N G (P 1 ) = G F /G F N G (P 1 ) F. Hence G F N G (P 1 ). Since G is a minimal non-f-group, N G (P 1 ) F. This shows that G = N G (P 1 ) F, a contradiction. The contradiction as above shows that x / x G T/ x G 1. Then x G < x T x and so x G 2. If x G = 1, then x T =2 or 4. Assume that x T = 2. By the same discussion of the case x G =2,we can obtain the contradiction. Hence x T = 4. Then x T and so G = T. It follows that x Z F (G). By hypothesis, GF Z F (G) and consequently G F, a contradiction. If x G = 2, then x T = 4. So x T and G = T. This shows that x / x G Z (G/ x F G ). Since x G =2, x G Z F (G) and so Z F (G/ x G)=Z F (G)/ x G. Hence x Z F (G). This implies that G F Z F (G). Consequently G F, a contradiction. This completes the proof. Theorem 3.2. Let F be a S-closed saturated formation which satisfies that every minimal non-f-group is soluble. Then G is an F-group if and only if G has a normal subgroup N such that G/N is an F-group and every cyclic subgroup of N of order 4 is F-supplemented in G and every minimal subgroup of N is contained in the F-hypercenter of G. Proof. The necessity is obvious, we only need to prove that sufficiency. Assume that the assertion is false and choose G to be a counterexample of minimal order. Then, obviously N 1. Let L be a proper subgroup of G. Then L/L N = LN/N G/N F, which implies that L/L N F. Since N L N, by hypothesis, every cyclic subgroup of N L of order 4 is F-supplemented in G. Then by Lemma 2.2(4), every cyclic subgroup of N L of order 4 is F-supplemented in L. On the other hand, by Lemma 2.1, every minimal subgroup of N L is contained in Z (G) F L Z (L). F This shows that (L, N L) satisfies the hypothesis. Hence L F and so G is a minimal non-f-group. By [6, Theorem 3.4.2] and the hypothesis, we know that G is soluble and G has the following properties: i) G F ia a p-group, for some prime p; ii) G F /Φ(G F ) is a chief factor of G; iii) If G F is abelian, then G F is an elementary abelian group; iv) If p>2, then the exponent of G F is p; Ifp = 2, then the exponent of G F is 2 or 4. Suppose that the exponent of G F is a prime. Since G/N F, G F N. It follows from the given condition that G F Z (G). F Thus G is an F-group, a contradiction. Now assume that G F is not abelian and p = 2. We claim that there is no element of order 4 in G F \Φ(G F ). Assume that there exists an element x G F \Φ(G F ) with x = 4. Then by the hypothesis, x is F-supplemented in G. Hence by Lemma 2.2(1), there exists a subgroup T of G such that

6 680 Honggao Zhang, Jianhong Huang and Yufeng Liu G = x T and x / x G T/ x G Z F (G/ x G). We first assume that x / x G T/ x G = 1. Then x T = x G. If x G = 4, then x = x G G. Hence x Φ(G F )/Φ(G F ) G/Φ(G F ). Since G F /Φ(G F ) is a chief factor of G, x Φ(G F )=G F and so G F = x, a contradiction. If x G = 2, then G : T = x : T x = 2. Hence 1 T G. Because G is a minimal non-f-group, we have that T F. If N = G, then G F by Theorem 3.1, a contradiction. Hence 1 <N<G. Since G/N F, G F N and so G F F. It follows that G/T = G F /(G F T ) F. Hence G F T. Consequently G = x T = T, which contradicts G : T =2. Now assume that x G = 1. Let P 1 = G F T. Assume that P 1 G. If P 1 Φ(G F ), then G F = G F G = G F x T = x (G F T ) = x, a contradiction. So P 1 Φ(G F ). Since G F /Φ(G F ) is a chief factor of G, P 1 Φ(G F )/Φ(G F ) = G F /Φ(G F ). It follows that P 1 = G F and so G F T. Thus G = x T = G F T = T, which is impossible since x T = x G =1. If P 1 G, then clearly G : N G (P 1 ) = 2 and hence N G (P 1 ) G. Since G = x T = G F N G (P 1 ), G/N G (P 1 ) = G F /G F N G (P 1 ) F. Hence G F N G (P 1 ). Since G is a minimal non-f-group, N G (P 1 ) F. This shows that G = N G (P 1 ) F, a contradiction. The contradiction as above shows that x / x G T/ x G 1. Then x G < x T x and so x G 2. If x G = 1, then x T =2 or 4. Assume that x T = 2. By the same discussion of the case x G =2, we can obtain the contradiction. Hence x T = 4. Then x T and G = T. Thus x Z F (G). By hypothesis, GF Z F (G) and so G F, a contradiction. If x G = 2, then x T =4. So x T and G = T. This shows that x / x G Z (G/ x F G ). Since x G =2, x G Z (G) F and so Z F (G/ x G)=Z F (G)/ x G. Hence x Z F (G). This implies that G F Z (G). F Consequently G F, a contradiction. This completes the proof. Corollary (Miao, Guo [7]). Let F be a S-closed saturated formation which satisfies that a minimal non-f-group is soluble and its F-residual is a Sylow subgroup. If every cyclic subgroup of order 4 of G is c-normal in G and every minimal subgroup of G is contained in the F-hypercenter of G, then G is an F-group. Corollary (Miao, Guo [7]). Let F be a S-closed local formation which satisfies that a minimal non-f-group is soluble and its F-residual is a Sylow subgroup. Let N be a normal subgroup of G and G/N be an F-group. If every cyclic subgroup of order 4 of N is c-normal in G and every minimal subgroup of N is contained in the F-hypercenter of G, then G is an F-group. Corollary (Miao, Guo [8]). Let F be a S-closed local formation with the following properties: a minimal non-f-group is soluble and its F-residual is a Sylow subgroup. If any cyclic subgroup of G of order 4 is c-supplemented in G and every minimal subgroup of G is contained in the F-hypercenter of G, then

7 Influence of minimal subgroups 681 G is an F-group. Corollary (Miao, Guo [8]). Let F be a S-closed local formation with the following properties: a minimal non-f-group is soluble and its F-residual is a Sylow subgroup. Let N be a normal subgroup of G and G/N be an F-group. If any minimal subgroup of N is contained in the F-hypercenter of G and any cyclic subgroup of N of order 4 is c-supplemented in G, then G is an F-group. As we all know, the class of N of all nilpotent groups and the class of U of all supersoluble groups satisfy the condition of Theorem 3.1 and Theorem 3.2. So we have the following corollary: Corollary (Miao, Guo [8]). If any cyclic subgroup of G of order 4 is c-supplemented in G and any minimal subgroup is contained in Z (G) (in particular, if any minimal subgroup of G is contained in Z(G)), then G is nilpotent. Corollary (Miao, Guo [8]). If any cyclic subgroup of G of order 4 is c-supplemented in G and any minimal subgroup is contained in Z (G), U then G is supersoluble. Corollary (Miao, Guo [8]). Let N be a normal subgroup of G and G/N be a nilpotent group. If any cyclic subgroup of G of order 4 is c-supplemented in G and any minimal subgroup of N is contained in Z (G), then G is nilpotent. Corollary (Miao, Guo [8]). Let N be a normal subgroup of G and G/N be a supersoluble group. If any cyclic subgroup of G of order 4 is c- supplemented in G and any minimal subgroup of N is contained in Z (G), U then G is supersoluble. Theorem 3.3. Let F be a S-closed saturated formation containing U and G a group. Then G F if and only if there exists a normal subgroup N of G such that G/N F and all elements of N of odd prime order are U-supplemented in G and N has an abelian Sylow 2-subgroup and every subgroup of N of order 2 is contained in Z F (G). Proof. The necessity is clear. We only need to prove the sufficiency. Assume that the theorem is false and let G be a counterexample of minimal order. First we show that M F for every maximal subgroup M of G. IfN M, then G = MN and M/M N = MN/N F. Since F is S-closed, M Z (G) F Z F (M) by Lemma 2.1(2). Then by Lemma 2.2, we see that (M,M N) satisfies the hypothesis. Hence M F by the choice of G. Therefore G is a minimal non-f-group. Let R = G F. Then R N. Assume that R <R, where R is the derived subgroup of R. Then R is soluble by Lemma 2.4. Hence, by [6, Theorem 3.4.2] and since N has an abelian Sylow 2-subgroup, R is a p-group of exponent p. Ifp 2, then G F by Lemma 2.6, a contradiction. Suppose that p = 2, then R is a elementary abelian 2-group. Thus, by hypothesis, R Z (G) F and so G F, a contradiction. Now assume that R = R. Let T be a Sylow 2-group of R. Then T is

8 682 Honggao Zhang, Jianhong Huang and Yufeng Liu abelian and so T Z(R) = 1 by Lemma 2.3. Assume that T 1. Then there exists an element r T with r = 2. Hence r Z F (G). Since Z F (G) R is contained in Z(R) by Lemma 2.5, Z(R) T 1. This contradiction shows that R is of odd order. Therefore by Feit-Thompson theorem, R is soluble, which contradicts R = R. These contradictions show that the counterexample of minimal order does not exist. Therefore the theorem holds. Theorem 3.4. Let F be a S-closed saturated formation containing U. Suppose that G is a group with a normal subgroup H such that G/H F. Then G F if one of the following conditions holds: (a) G is 2-nilpotent and every element x of odd prime order of H is U- supplemented in G. (b) H has an abelian Sylow 2-subgroup and every subgroup of prime order of H is U-supplemented in G. Proof. (a) If G is 2-nilpotent, then H is 2-nilpotent. Let K be the 2- complement of H. Then K G. Since (G/K)/(H/K) = G/H F and H/K is a 2-group, H/K has no element of odd order. Hence G/K F by induction on G. Since K is a 2-complement of H, K has no cyclic subgroup of order 4. Thus G F by Lemma 2.6. (b) Let E = G F. Then, obviously, E H and E has abelian Sylow 2-subgroups. By hypotheses, every subgroup x of prime order of E is U- supplemented in G. Hence, by Lemma 2.2(2), x is also U-supplemented in E. It follows from Lemma 2.7 that E is soluble. Let M be a maximal subgroup of G such that E M. Then M/M E = ME/E F. It is easy to see that (M,M E) satisfies the hypothesis. Therefore M F by induction. Then, applying [6, Theorem 3.4.2], we see that E is a p-group of exponent p. Thus G F by Lemma 2.6. References [1] A. Ballester-Belinches and M.C.Pedraza-Aguilera, On minimal subgroups of finite groups, Acta Math., 73(4)(1966), [2] J. Buckley, Finite groups whose minimal subgroups are normal, Math. Z, 1970, 116: [3] B. Huppert, Endliche Gruppen I, Berlin: Springer-Verlag, [4] B. Huppert, Endliche Gruppen I, Berlin-Heidelberg-New York, Springer- Verlag, [5] W. Guo, On F-supplemented subgroups of finite groups, manuscripta math., 127(2008),

9 Influence of minimal subgroups 683 [6] W. Guo, The Theory of Classes of Groups, Science Press-Kluwer Academic Publishers, Beijing-New York-Dorlrecht-Boston-London, [7] L. Miao and W. Guo, The influence of c-normality of some subgroups on the structure of a finite group, Problem in Algebra, 3(16), 2000, [8] L. Miao and W. Guo, The influence of c-supplements of minimal subgroups on the structure of finite groups, Journal of Applied Algebra and Discrete Structures, 2(2004), No.3, [9] D. J. S. Robinson, A Course In The Theory Of Groups, Spring-Vrelag, New York, Heidelberg, Berlin, [10] L. A. Shemetkov, A.N. Skiba, Formations of Algebraic Systems, Moscow, Nauka, [11] Y. Wang, C-normality of groups and its properties. J. Algebra, 180, 1996, [12] Y. Wang, Finite groups with some subgroups of Sylow subgroups c- supplemented, J. Algebra, 224(2000), Received: November, 2009