Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 14, 675-683 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 221116, P.R. China 2. Department of Mathematics University of Science and Technology of China Hefei 230026, P.R. China 3. School of Math. and Informational Science Shandong Institute of Business and Technology Yantai 264005, 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 #10771180) and a postgraduate innovation grant of Xuzhou Normal University.
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.
Influence of minimal subgroups 677 2 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.
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,
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
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 3.2.1 (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 3.2.2 (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 3.2.3 (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
Influence of minimal subgroups 681 G is an F-group. Corollary 3.2.4 (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 3.2.5 (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 3.2.6 (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 3.2.7 (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 3.2.8 (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
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), 335-342. [2] J. Buckley, Finite groups whose minimal subgroups are normal, Math. Z, 1970, 116: 15-17. [3] B. Huppert, Endliche Gruppen I, Berlin: Springer-Verlag, 1968. [4] B. Huppert, Endliche Gruppen I, Berlin-Heidelberg-New York, Springer- Verlag, 1967. [5] W. Guo, On F-supplemented subgroups of finite groups, manuscripta math., 127(2008), 139-150.
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, 2000. [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, 101-106. [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, 179-189. [9] D. J. S. Robinson, A Course In The Theory Of Groups, Spring-Vrelag, New York, Heidelberg, Berlin, 1982. [10] L. A. Shemetkov, A.N. Skiba, Formations of Algebraic Systems, Moscow, Nauka, 1989. [11] Y. Wang, C-normality of groups and its properties. J. Algebra, 180, 1996, 954-965. [12] Y. Wang, Finite groups with some subgroups of Sylow subgroups c- supplemented, J. Algebra, 224(2000), 467-478. Received: November, 2009