On complemented minimal subgroups in finite groups

J. Group Theory 6 (2003), 159–167

Journal of Group Theory ( de Gruyter 2003

On complemented minimal subgroups in finite groups Guo Xiuyun*, K. P. Shum** and A. Ballester-Bolinches (Communicated by W. J. Shi)

Abstract. We consider the structure of finite groups in which every minimal subgroup of some maximal subgroup is complemented. First we analyse the structure of a finite group in which every maximal subgroup is supersolvable with elementary abelian Sylow subgroups. Next we prove that a finite group G is solvable if it has a maximal subgroup M such that every minimal subgroup of M is complemented in G. As an application, some conditions for a finite group to be supersolvable are given.

1 Introduction We recall that a subgroup H of a finite group G is complemented in G if there exists a subgroup K of G such that G ¼ HK and H V K ¼ 1. Such a subgroup K of G is called a complement to H in G. The existence of complements for certain subgroups of a finite group provides useful structural information; see for instance [1], [5], [6]. In a recent paper, Ballester-Bolinches and Guo [2] studied finite groups for which every minimal subgroup is complemented. They proved that such groups are just the finite supersolvable groups with elementary abelian Sylow subgroups. They also studied finite groups in which all maximal subgroups or 2-maximal subgroups of Sylow subgroups are complemented. In this paper, we further analyse finite groups in which every minimal subgroup of some maximal subgroup is complemented. First we determine the structure of a finite group whose maximal subgroups are all supersolvable with elementary abelian Sylow subgroups. Next we show that a finite group G is solvable if it has a maximal subgroup M such that every minimal subgroup of M is complemented in G. As a consequence, some conditions for a finite group G to be supersolvable are obtained. For notation and conventions we refer to the book of Robinson [1]. Unless otherwise stated, G will always be a finite group. * The research of the first author is partially supported by a Natural Science Foundation grant of Shanxi Province, China ** The research of the second author is partially supported by UGC(HK) grant #2260126 (1999/2000).

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Guo Xiuyun, K. P. Shum and A. Ballester-Bolinches

2

Basic lemmas

We list the following lemmas which will be frequently used in the sequel. Lemma 2.1 ([2, Lemma 1]). Let N d G. Then the following statements hold. (a) If H c K c G and H is complemented in G, then H is complemented in K. (b) If N c H and H is complemented in G, then H=N is complemented in G=N. (c) Let p be a set of primes. If N is a p 0 -subgroup and A is a p-subgroup of G, then A is complemented in G if and only if AN=N is complemented in G=N. Lemma 2.2 ([1, Corollary 2, p. 120]). Let M be a maximal subgroup of a group G. If coreðMÞ ¼ 1 and if A, B are distinct minimal normal subgroups of G, then (a) G ¼ AM ¼ BM and A V M ¼ B V M ¼ 1, (b) A is the centralizer of B in G (and B the centralizer of A), and (c) A, B and AB V M are isomorphic non-abelian groups. Lemma 2.3. Every minimal subgroup of G is complemented in G if and only if G is a supersolvable group with elementary abelian Sylow subgroups. Proof. This follows from [2, Corollary 2] and [4, Theorem 2]. Lemma 2.4. Let N be a minimal normal subgroup of G. If every minimal subgroup of N is complemented in G, then N is cyclic of prime order. Proof. By Lemma 2.3, N is supersolvable and so is an elementary abelian p-group for some prime p. Let hzi be a minimal subgroup of N and let H be a complement to hzi in G. Since N ¼ hziðN V HÞ and N V H d H, we have N V H d G. Thus by the minimality of N we have N V H ¼ 1, and therefore N is cyclic of order p. Lemma 2.5. Let M c G. If N is a minimal normal subgroup of G such that N c M and every minimal subgroup of M has a complement in G, then every minimal subgroup of M=N has a complement in G=N. Proof. By Lemma 2.4, N is cyclic of prime order p, say. Let hxi be a minimal subgroup of M=N. If jhxij ¼ q and q 0 p, then clearly M has an element x such that jxj ¼ q and hxi ¼ hxiN=N. Now by hypothesis and Lemma 2.1 (c), we know that hxi ¼ hxiN=N has a complement in G=N. If jhxij ¼ p, then, since every Sylow subgroup of M is elementary abelian, M has a minimal subgroup hxi such that hxi ¼ hxiN=N. Hence, by our hypothesis again, hxi has a complement K in G. If N c K, then K=N is a complement of hxi in G=N. If N G K, then G ¼ NK and so by the minimality of N we have N V K ¼ 1. Thus hxiN ¼ ðhxiNÞ V ðNKÞ ¼ NðhxiN V KÞ: Since hxiN c M and N V K ¼ 1, we know that ðhxiNÞ V K is a minimal subgroup

On complemented minimal subgroups in finite groups

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of M and ðhxiNÞ V K c K. Denote ðhxiNÞ V K by h yi. Then h yi has a complement K1 in K, by our hypothesis and Lemma 2.1 (a). Thus G ¼ NK ¼ NK1 h yi and NK1 V h yi ¼ 1 (for otherwise we have h yi c NK1 and so G ¼ NK1 , a contradiction). This leads to ðhxiN=NÞðNK1 =NÞ ¼ ðh yiN=NÞðNK1 =NÞ ¼ G=N and ðhyiN=NÞ V ðNK1 =NÞ ¼ 1. Hence every minimal subgroup of M=N has a complement in G=N.

3 Minimal non-BNS-groups Knowledge of the structure of minimal non-nilpotent groups, analysed by Schmidt and Re´dei (see [10, Theorem 9.1.9 and Exercises 9.1.11]) is very useful in the study of finite groups. In this section, we determine the structure of a group in which every maximal subgroup is supersolvable with elementary abelian Sylow subgroups. We write BNS for the family of groups in which each minimal subgroup is complemented. Theorem 3.1. Let G be a group. If every maximal subgroup of G is in BNS and jpðGÞj d 4, then G is in BNS. Proof. Let jGj ¼ p1a1 p2a2 p3a3 p4a4 m where pi is prime, ðpi ; mÞ ¼ 1 and ai d 1 for each i. Since every maximal subgroup of G is in BNS, by [2, Corollary 1], every maximal subgroup of G is supersolvable and therefore G is solvable. Thus G has a maximal b subgroup Mi such that jG : Mi j ¼ pi i and bi c ai for each i. By [3, Satz 5], G is supersolvable. Clearly every Sylow subgroup of G is elementary abelian. Hence G is a BNS-group by [4, Theorem 2]. Theorem 3.2. If every maximal subgroup of the group G belongs to BNS but G does not, then one of the following statements is true. 2 (I) G ¼ ha j a p ¼ 1i, where p is a prime. (II) G ¼ ha; bi with a p ¼ b p ¼ c p ¼ 1, ½a; b ¼ c, ½a; c ¼ ½b; c ¼ 1, where p is an odd prime. (III) G ¼ P b Q is a minimal non-abelian group of order pq a where p, q are distinct primes with p a ðq 1Þ and a > 1 and P A Syl p G, Q A Sylq G. (IV) G is a minimal non-supersolvable group of order pqr p , where p, q and r are distinct primes with pq j ðr 1Þ and p j ðq 1Þ, and G ¼ hc1 ; c2 ; . . . ; cp ; a; bi satisfies the following relations: c1r ¼ c2r ¼ ¼ cpr ¼ 1, ci cj ¼ cj ci for i; j ¼ 1; 2; . . . ; p, a p ¼ b q ¼ 1, b a ¼ b u and the order of u ðmod qÞ is p, u piþ1 cia ¼ ciþ1 for i ¼ 1; 2; . . . ; p 1, cpa ¼ c1 , cib ¼ civ for i ¼ 1; . . . ; p; the order of v ðmod rÞ is q. Conversely, if G of type (I), (II), (III) or (IV), then G not belong to BNS but every maximal subgroup of G belongs to BNS.

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Proof. Suppose that every maximal subgroup of G belongs to BNS but G does not. Then jpðGÞj c 3 by Theorem 3.1. Assume first that G is a p-group. Then clearly G 2 has the form ha j a p ¼ 1i if G is abelian. If G is not abelian then G is a minimal nonabelian group and every maximal subgroup of G is elementary abelian. By the main result of Re´dei [9], the group G is of type (II). Now assume that G is not a p-group. Then every Sylow subgroup of G is a proper subgroup and so is elementary abelian. By [2, Corollary 1] every maximal subgroup of G is supersolvable. If G is supersolvable, then G is in BNS by [4, Theorem 2], a contradiction. Hence G is a minimal non-supersolvable group. A result of Doerk [3] shows that G is a minimal non-nilpotent group or a Sylow tower group, and in both cases we have (1) G has exactly one normal Sylow r-subgroup R such that G ¼ M b R, R is a minimal normal subgroup of G and M is supersolvable. Since every minimal subgroup of G=R G M is complemented, Lemma 2.1 (c) yields that (2) every minimal subgroup of M is complemented in G. If G has another minimal normal subgroup N, then N is a r 0 -group and therefore N c M. It follows that N has prime order and that G has a subgroup H such that G ¼ H b N, by (2). This yields that G G G=ðN V RÞ is supersolvable, as G=N and G=R are supersolvable, a contradiction. Hence R is the unique minimal normal subgroup of G. Since also R ¼ Or ðGÞ, the centralizer of R in G is the direct product of R and an r 0 -group by the Schur–Zassenhaus theorem and so we have the following: (3) F ðGÞ ¼ CG ðRÞ ¼ R. (Here, F ðGÞ denotes the Fitting subgroup of G.) Recall that a group G is called strictly r-closed if it has a normal Sylow r-subgroup Gr with G=Gr abelian of exponent dividing r 1. If our group G is a minimal nonnilpotent group, then clearly G is a minimal non-abelian group and M has prime order p 0 r. If p j ðr 1Þ, then G is strictly r-closed and therefore is supersolvable (see [11, Theorem I.1.9]), a contradiction. Hence p a ðr 1Þ and G is of type (III). If G is a Sylow tower group and jpðGÞj ¼ 2, then M is a p-group for some prime p 0 r. We claim that jMj ¼ p. In fact, for any 1 0 x A M, it is clear that M c CG ðxÞ. If NR ðhxiÞ 0 1, then as NR ðhxiÞ V R is normal in G the minimality of R implies that NG ðhxiÞ ¼ G, in contradiction to F ðGÞ ¼ R. Thus NG ðhxiÞ ¼ M, so that every element x 0 1 of M acts as a fixed-point-free automorphism of R. By [7, Theorem 7.24] we conclude that M is cyclic and, as the Sylow p-subgroups are elementary abelian, we have jMj ¼ p. In this case too, G is a minimal non-nilpotent group and therefore G is of type (III). Now we assume that G is a Sylow tower group and that jpðGÞj ¼ 3. Let jGj ¼ p a q b r l with p < q < r. Since G is a Sylow tower group, there exist P A Syl p G and Q A Sylq G such that QR d G and PQ ¼ QP. We recall that R d G. Since QR is supersolvable, QR=R ¼ NQR ðRÞ=CQR ðRÞ is strictly r-closed by [11, Theorem I.1.12]. It follows that the exponent of Q divides r 1, that is, q j ðr 1Þ. Similarly p j ðr 1Þ and thus pq j ðr 1Þ. If PQ is abelian then G 0 ¼ R. Clearly G is strictly r-closed and therefore is supersolvable (see [11, Theorem I.1.9), a contradiction. Hence


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(4) PQ is non-abelian. Now we claim that P is cyclic. Each cyclic subgroup hxi of P has a complement H in G by (2). Then H d G since jG : Hj ¼ p and p is the smallest prime dividing jGj, and therefore H 0 d G. On the other hand, the supersolvability of H implies that H 0 is nilpotent. Thus H 0 c F ðGÞ ¼ R and the Hall r 0 -subgroups of H are abelian. Since jG : Hj ¼ p and QR is a normal p 0 -subgroup of G, we have QR c H and therefore ½Q; P V H c Q V H 0 c Q V R ¼ 1: Hence if P were not cyclic then PQ would be abelian, in contradiction to (4). Thus P is cyclic of order p. Next we prove that Q is cyclic. Each cyclic subgroup h yi of Q has a complement H and we may choose H such that P c H. If Q is not cyclic, then Q V H 0 1. Since Q V H acts on R, we have R ¼ CR ðQ V HÞ ½R; Q V H. On the other hand, ðPQÞ V H is supersolvable since PQ is supersolvable. Thus Q V H d ðPQÞ V H. It follows that CR ðQ V HÞ is normalized by P, and therefore CR ðQ V HÞ d G since R and Q are abelian. If CR ðQ V HÞ 0 1, then the minimality of R implies that CR ðQ V HÞ ¼ R, in contradiction to CG ðRÞ ¼ R. Hence CR ðQ V HÞ ¼ 1 and ½R; Q V H ¼ R. Since R c H, we have H 0 d R. The supersolvabilty of H implies that H 0 is nilpotent. Since CG ðRÞ ¼ R we have H 0 ¼ R, so that the group H V ðPQÞ G ðH V ðPQÞÞ=R V ðH V ðPQÞÞ G H=R is abelian. Hence if Q were not cyclic then PQ would be abelian, a contradiction. Thus Q is cyclic of order q. Let P ¼ hai and Q ¼ hbi. Since Q d PQ and PQ is non-abelian, we have b a ¼ b u , the order of u ðmod qÞ is p and p j ðq 1Þ. Also since QR is supersolvable and R d QR, there exists a subgroup hc1 i of R of order r such that hc1 i d QR. Observp1 ing that QR d G, we know that hc1a i; . . . ; hc1a i are all normal in QR. It follows i1 that hc1 ihc2 i . . . hcp i is normal in G, where ci ¼ c1a for i ¼ 2; . . . ; p. The minimality of R implies that R ¼ hc1 i . . . hcp i. Thus l c p. By [6, Theorem VI.8.1] we also have p c l and therefore l ¼ p, and R ¼ hc1 i hc2 i hcp i: By the choice of ci above, cia ¼ ciþ1 for i ¼ 1; 2; . . . ; p 1 and cpa ¼ c1 . Since hci i is normal in QR, we may assume that cib ¼ cili for i ¼ 1; 2; . . . ; p and of course the order of li ðmod rÞ is q. Now PQ acts on R and considering R as a vector space over GFðrÞ we obtain a matrix representation T of PQ such that 1 0 0 1 0 1 0 l1 0 0 B0 0 0C C B B 0 l 0 C 2 B. . B C .. C .. C B. . .. .. ; TðbÞ ¼ TðaÞ ¼ B .. C .. .C . B B. . C: C B .A . @. . @0 0 1A 0 0 lp 1 0 0 Since b a ¼ b u , we have Tðb a Þ ¼ Tðb u Þ. It follows that

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0

lp B0 B Tðb a Þ ¼ B B .. @. 0

0 l1 .. . 0

.. .

1

0

l1u C B0 C B C¼B . C B. A @. lp1 0 0 0 .. .

.. .

0 l2u .. . 0

1 0 0C C u .. C C ¼ Tðb Þ: . A lpu

Hence l1u ¼ lp , l2u ¼ l1 , l3u ¼ l2 ; . . . ; lpu ¼ lp1 . It follows that lp ¼ l1u ;

2

lp1 ¼ l1u ; . . . ; l2 ¼ l1u

p1

;

l1 ¼ l1u

p

Writing v ¼ l1 , we see that G is of type (IV). Conversely, it is clear that the groups type (I), (II), (III) and (IV) do not belong to BNS but that their maximal subgroups do belong to BNS by Lemma 2.3. Thus the proof of our theorem is complete. Theorem 3.3. If NG ðPÞ is a BNS-group for every Sylow subgroup P of the group G, then G is a BNS-group. Proof. Let p be the smallest prime dividing jGj and P a Sylow p-subgroup of G. By our hypotheses and [2, Corollary 1], NG ðPÞ is supersolvable and therefore NG ðPÞ has a Hall p 0 -subgroup H such that NG ðPÞ ¼ P H. By Burnside’s theorem [10, Theorem 10.1.8], G has a normal p-complement M. Clearly M satisfies the hypotheses of the theorem. By induction M has a Sylow tower and therefore so does G. Let q be the largest prime dividing jGj and Q a Sylow q-subgroup of G. Then Q d G and therefore G ¼ NG ðQÞ A BNS by the hypotheses. The proof is complete. Corollary 3.4. If NG ðPÞ is a BNS-group for every non-normal Sylow subgroup P of G, then G BNS is nilpotent, where G BNS is the BNS-residual of G. Proof. Let p be the set of primes p such that the Sylow p-subgroups of G are nontrivial and normal in G. By the Schur–Zassenhaus theorem G ¼ Gp 0 b Gp , where Gp is a Hall p-subgroup of G and Gp 0 is a Hall p 0 -subgroup. Clearly Gp is nilpotent and NGp 0 ðPÞ is a BNS-group for every Sylow subgroup P of Gp 0 . By Theorem 3.3, Gp 0 A BNS and therefore GBNS is nilpotent.

4 Solvability of certain complemented finite groups In this section, we investigate the solvability of some finite groups. The following theorem has some similarity to the well-known theorem of Thompson (see [10, Theorem 10.4.2]). The present proof of Theorem 4.1 is due to the referee. Theorem 4.1. Let G be a group and let M be a maximal subgroup of G. If every minimal subgroup of M has a complement in G, then G is solvable. Proof. We assume that the theorem is false and let G be a counter-example of minimal order. By Lemma 2.5, we have coreðMÞ ¼ 1.


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Let T be a minimal normal subgroup of M. By Lemma 2.1 and Lemma 2.3, M is solvable. It follows that T is an elementary abelian p-group for some prime p. Let N be a minimal normal subgroup of G. Then, as G ¼ MN and M is solvable, N is nonsolvable. If N V M ¼ 1, then, as NG ðT Þ ¼ M, we have CN ðT Þ ¼ 1 and so Sylow’s theorem implies that N is a p 0 -group. By [7, Theorem 7.5], for every prime divisor r of jNj there is a unique T-invariant Sylow r-subgroup R of N. But on the other hand, for any g A M, we have ðR g ÞT ¼ R Tg ¼ R g , so that R g ¼ R, by the uniqueness of R. Therefore R is normalized by M. So by the maximality of M, MR ¼ G or MR ¼ R. In the former case R is normal in G and so N ¼ R is solvable, a contradiction. In the latter case we have R c N V M ¼ 1, another contradiction. Hence we may assume that M V N > 1. Since M V N d M, we may choose T c M V N. Let hxi be a minimal subgroup of T. Then hxi is a minimal subgroup of M and so it has a complement H in G. Hence N ¼ hxiðN V HÞ. Therefore jG : Hj ¼ p ¼ jN : N V Hj. Let K ¼ coreG ðHÞ. Then N V K ¼ 1, by the minimality of N and the fact that N G H. Thus NK=K is isomorphic to a subgroup of Sp , the symmetric group of degree p, and hxi A Syl p ðNÞ as hxiK=K A Syl p ðSp Þ. Consequently N is simple (since otherwise it could not have cyclic Sylow p-subgroups) and T ¼ hxi. As T is not normal in G, we have NG ðTÞ ¼ M and so NN ðTÞ ¼ M V N. If NN ðTÞ ¼ T, then T is in the centre of its normalizer and Burnside’s theorem implies that N has a normal p-complement, a contradiction as N is simple. Hence M V N > T and, as T A Syl p ðNÞ, there exists a minimal subgroup h yi of M V N of order r with r 0 p. But h yi has a complement J in G by our hypotheses and jG : Jj ¼ r ¼ jN : N V Jj. Let m be the smallest of r and p and n be the largest of r and p. Then, as N is simple, N is isomorphic to a subgroup of Sm , the symmetric group of degree m. But n > m and so n does not divide jSm j whereas n does divide jNj. This contradiction proves that G is solvable and the proof of the theorem is complete. Remark 4.2. Let M be a maximal subgroup of a group G. If every minimal subgroup of M has a complement in G, then every minimal subgroup of M has a complement in M and therefore, by Lemma 2.3, M is supersolvable and every Sylow subgroup of M is elementary abelian. We note that a group G which has a supersolvable maximal subgroup M with elementary abelian Sylow subgroups need not be soluble: the simple groups A5 and G ¼ PSLð2; 7Þ illustrate this. The following results give some conditions for a finite group G to be supersolvable. Corollary 4.3. Let M be a maximal subgroup of a group G with prime index. If every minimal subgroup of M has a complement in G, then G is supersolvable. Proof. Let jG : Mj ¼ p (where p is a prime). By Theorem 4.1, G is solvable. Let N be a minimal normal subgroup of G. Then N is an elementary abelian q-group for some prime q. Two cases arise:

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Case 1. N c M. In this case, N is cyclic of order q by Lemma 2.4. By Lemma 2.5, the group G=N satisfies the hypothesis of the corollary, and so G=N is supersolvable by induction. Consequently G is supersolvable. Case 2. N G M. In this case, we can write G ¼ MN. By the minimality of N, we have N V M ¼ 1. It follows that N is cyclic of order p. Since every minimal subgroup of M has a complement, M is supersolvable by [2, Corollary 1]. Hence G is supersolvable, as required. Theorem 4.4. Let M1 , M2 be two non-conjugate maximal subgroups of G. If for i ¼ 1; 2 every minimal subgroup of Mi has a complement in G, then G is supersolvable. Proof. By Theorem 4.1, G is solvable, and by [2, Corollary 1], M1 and M2 are supersolvable. If coreðM1 V M2 Þ 0 1, then G has a minimal normal subgroup N1 such that N1 c coreðM1 V M2 Þ. By Lemma 2.5, we know that G=N1 , M1 =N1 and M2 =N1 satisfy the hypothesis of Theorem 4.4 and thus, by induction, G=N1 is supersolvable. By Lemma 2.4, we see that N1 is cyclic of prime order. It follows that G is supersolvable. Now suppose that coreðM1 V M2 Þ ¼ 1 and let N be a minimal normal subgroup of G. Then N must be an elementary abelian p-group for some prime p. Since coreðM1 V M2 Þ ¼ 1, we may assume that N G M1 . If N c M2 , then N is cyclic of order p by Lemma 2.4. It follows that G is supersolvable since G=N G M1 is supersolvable. If N G M2 , we can assume that N is the unique minimal normal subgroup of G. Indeed, if N1 is another minimal normal subgroup of G, then N1 G M1 or N1 G M2 since coreðM1 V M2 Þ ¼ 1. Without loss of generality we may assume that N1 G M1 . It follows that G is supersolvable since G=N G M1 and G=N1 G M1 are supersolvable and N V N1 ¼ 1. Hence we assume that N is the unique minimal normal subgroup of G. As N G M1 , we have FðGÞ ¼ 1 and therefore F ðGÞ ¼ N (see [8, Lemma 2.3]). Let H ¼ H=N be a minimal normal subgroup of G=N. Then H is an elementary abelian q-group with q 0 p since F ðGÞ ¼ N. This leads to H V M1 A Sylq ðHÞ and H V M2 A Sylq ðHÞ. By Sylow’s theorem, there exists g A H such that ðH V M1 Þ g ¼ H V M2 and therefore ðNG ðH V M1 ÞÞ g ¼ NG ðH V M2 Þ. Since H V M1 is normal in the maximal subgroup M1 but not in G we have NG ðH V M1 Þ ¼ M1 . Similarly, NG ðH V M2 Þ ¼ M2 . Hence M1g ¼ M2 , which contradicts the assumption that M1 and M2 are nonconjugate. The proof of the theorem is thus complete. The following corollary is immediate from Theorem 4.4. Corollary 4.5. Let M1 , M2 be maximal subgroups of a group G. If jG : M1 j 0 jG : M2 j and for i ¼ 1; 2 every minimal subgroup of Mi ði ¼ 1; 2Þ has a complement in G, then G is supersolvable.


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Acknowledgement. The authors would like to thank the referee for his valuable suggestions and useful comments.

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