Siberian Mathematical Journal, Vol. 55, No. 3, pp. 415–427, 2014 c 2014 Guo W. and Revin D.O. Original Russian Text Copyright
ON THE CLASS OF GROUPS WITH PRONORMAL HALL π-SUBGROUPS W. Guo and D. O. Revin
UDC 512.542.6
Abstract: Given a set π of prime numbers, we define the class Pπ of all finite groups in which Hall πsubgroups exist and are pronormal by analogy with the Hall classes Eπ , Cπ , and Dπ . We study whether Pπ is closed under the main class-theoretic closure operations. In particular, we establish that Pπ is a saturated formation. DOI: 10.1134/S0037446614030033 Keywords: finite group, Hall subgroup, pronormal subgroup, class of finite groups, properties Eπ , Cπ , and Dπ , property Pπ , class-theoretic closure operations, formation, saturated formation, Fitting class
1. Introduction We use the term a group, meaning a finite group. Denote by π a fixed set of prime numbers. Denote by π the set of all prime numbers outside π, by π(n) the set of all prime divisors of the positive integer n, and, given a group G, by π(G) the set π(|G|). Recall that a group G satisfying π(G) ⊆ π is called a π-group. A subgroup H of a group G is called a Hall π-subgroup whenever π(H) ⊆ π and π(|G : H|) ⊆ π . In accordance with [1], say that a group G enjoys property Eπ whenever G includes at least one Hall π-subgroup. If in a group G with property Eπ every pair of Hall π-subgroups are conjugate then we say that G enjoys property Cπ . If every π-subgroup of a group G with property Cπ lies in some Hall π-subgroup then we say that G enjoys property Dπ . Refer to a group with property Eπ (Cπ , Dπ ) also as an Eπ -group (respectively Cπ -group, Dπ -group). In addition, we denote by the symbols Eπ , Cπ , and Dπ the classes of all Eπ -, Cπ -, and Dπ -groups. The reasons to study the classes Eπ , Cπ , and Dπ are both Sylow’s theorem and the results of Hall and Chunikhin showing that, firstly, given a set π, every solvable group enjoys property Dπ , and, secondly, for every nonsolvable group G there is a set π with G ∈ Eπ . In every solvable group G the conjugacy of the Hall π-subgroups implies their pronormality. Recall that in Hall’s definition a subgroup H of a group G is called pronormal if for every g ∈ G the subgroups H and H g are conjugate in H, H g . The classical examples of pronormal subgroups are: • normal subgroups; • maximal subgroups; • Sylow subgroups of finite groups; • Carter subgroups of finite solvable groups; • Hall subgroups of finite solvable groups. By the results of Vdovin [2] the Carter subgroups (that is, self-normalizable nilpotent subgroups) remain pronormal in all (not necessarily solvable) finite groups. In contrast to the Carter subgroups, the The first author was supported by the NNSF of China (Grant 11371335) and the Chinese Universities Scientific Fund (Project WK0010000029). The second author was supported by the Russian Foundation for Basic Research (Grant 13–01–00505) and the Target Program of the Siberian Division of the Russian Academy of Sciences for 2012–2014 (Integration Project No. 14). Hefei and Novosibirsk. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 55, No. 3, pp. 509–524, May–June, 2014. Original article submitted August 17, 2013. c 2014 Springer Science+Business Media, Inc. 0037-4466/14/5503–0415
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Hall subgroups of a nonsolvable group need not be pronormal; some examples appear in [3, 4] (also see Lemmas 19 and 20 below). By analogy with Hall’s notation, we say that a finite group G enjoys property Pπ , or is a Pπ -group, or belongs to the class Pπ , if G ∈ Eπ and every Hall π-subgroup of G is pronormal. It is shown in [5] that the Hall subgroups of finite simple groups are pronormal, and it is established in [4] that the Hall π-subgroups are pronormal in every group in Cπ . Thus, Cπ ⊆ Pπ ⊆ Eπ and every simple Eπ -group belongs to the class Pπ . The question [3, Problem 7.20; 6, Problem 6] is open: For which π is the inclusion Cπ ⊆ Eπ strict? It / π. At the same time, many examples of sets π is known [7, Theorem A] that Eπ = Cπ provided that 2 ∈ with Eπ = Cπ are available. Another natural problem is as follows: For each of the inclusions Cπ ⊆ Pπ and Pπ ⊆ Eπ determine for which π it is strict? The following statement reduces this question to the strictness of the inclusions Cπ ⊆ Eπ . Proposition 1. For every set π the following are equivalent: (1) Cπ = Eπ ; (2) Cπ = Pπ ; (3) Pπ = Eπ . In the theory of classes of groups and the theory of properties Eπ , Cπ , and Dπ in particular there are important questions of closedness under the class-theoretic operations s, q, sn , r0 , n0 , d0 ,e, ez , eΦ , and p (see [8, 9]) defined by their actions on a class X of finite groups as follows: sX = {G | G is isomorphic to a subgroup of some group H ∈ X }; qX = {G | G is an epimorphic image of some group H ∈ X }; of some group H ∈ X sn X = {G | G is isomorphic to a subnormal subgroup }; N = 1 (i = 1, . . . , m) ; r0 X = G | ∃Ni G such that G/Ni ∈ X and m i i=1 n0 X = {G | ∃Ni G such that Ni ∈ X and G = N1 , . . . , Nm (i = 1, . . . , m)}; d0 X = {G | G H1 × · · · × Hm for some Hi ∈ X (i = 1, . . . , m)}; eX = {G | G possesses a series 1 = G0 G1 · · · Gm = G such that Gi /Gi−1 ∈ X (i = 1, . . . , m)}; ez X = {G | ∃N G such that N ≤ Z∞ (G) and G/N ∈ X }; eΦ X = {G | ∃N G such that N ≤ Φ(G) and G/N ∈ X }; pX = {G | ∀H< · G G/HG ∈ X }. Here the expression H G means that the subgroup H is subnormal in G, while the expression H< · G means that H is maximal in G. Given H ≤ G, denote by HG the normal subgroup HG = g∈G H g . Table 1. Is the equality cX = X always satisfied X ∈ {Eπ , Cπ , Pπ }?
Moreover, Z∞ (G) and Φ(G) stand respectively for the hypercenter and Frattini subgroup of G. For the classes Eπ , Cπ , and Dπ the questions of closedness under the c Eπ Cπ Pπ operations defined above were studied by many famous mathematicians s no no no for more than fifty years and play a key role in the theory of these classes q yes yes* yes* (see the surveys [3, 10]). In some cases the closedness is easy to estabsn yes no no lish; for instance, it is easy to prove that q, sn Eπ = Eπ . The equality r0 yes* yes* yes* eCπ = Cπ is harder and was verified for the first time by Chunikhin n0 no yes* no (see [11]). But in some cases the questions of closedness turn out excepd0 yes yes yes e no yes no tionally complicated. A difficult question, for instance, is the closedness ez yes yes yes of Dπ under the operations sn and e noted by Wielandt at the XIII eΦ yes yes yes International Mathematical Congress in Edinburgh [12]. The books by Suzuki [13], Shemetkov [14], Doerk and Hawks [9], and the first author [8] mention some questions of closedness. Some questions were included into The Kourovka Notebook [15]. Let us also mention the articles [1, 4, 5, 16–42] dealing with the problems of closedness of the classes Eπ , Cπ , and Dπ under the 416
operations s, q, sn , r0 , n0 , d0 , e, ez , eΦ , and p. Presently these questions are completely settled; although in some cases the proof of the closedness of some class under these operations depends essentially on the classification of finite simple groups (see [3]). In this article we study the closedness of the class Pπ under the operations s, q, sn , r0 , n0 , d0 , e, ez , eΦ , and p. We use the results of [4, 5, 16–18], which, in turn, rest on the classification of finite simple groups. We summarize the results of this activity together with the similar results on the classes Eπ and Cπ (with the exception of the operation p) in Table 1.1) More precisely, we have Theorem 1. The following hold: (A) If c ∈ {q, r0 , d0 , ez , eΦ } then cPπ = Pπ for every set π of prime numbers. (B) If c ∈ {s, sn , n0 , e} then cPπ = Pπ for some set π of prime numbers. (C) If c ∈ {sn , e} and cPπ = Pπ for some set π of prime numbers then cEπ = Eπ and cCπ = Cπ . Corollary 1. For c ∈ {s, q, sn , r0 , n0 , d0 , e, ez , eΦ } the following are equivalent: (1) cPπ = Pπ for every set π of prime numbers; (2) cEπ = Eπ and cCπ = Cπ for every set π of prime numbers. Recall that each q, r0 -closed class X of finite groups is called a formation. Furthermore, if X is eΦ -closed then X is called a saturated formation. Each q, p-closed class is called a Schunk class. It is known (see [8, p. 69] or [9, Chapter 3, Proposition 4.1] for instance) that each saturated formation is a Schunk class. Corollary 2. For each set π of prime numbers the class Pπ is a saturated formation. Corollary 3. For every set π of prime numbers, pPπ = Pπ . In particular, Pπ is a Schunk class. We can give examples of sets π such that for each operation c ∈ {s, sn , n0 , e} at least one of the classes Eπ or Cπ is not closed under c (for instance, the set {2, 3} can do). By Theorem 1, for every π of this type and each operation c on the list above the class Pπ is not c-closed. In particular, Pπ is not a Fitting class.2) Thus, in this case the class Pπ is an example of a saturated formation “in pure form” (that is, Pπ is a saturated formation and nothing more). This circumstance may prove useful in constructing various counterexamples in the theory of classes. Together with that, a trivial corollary of the inclusions Cπ ⊆ Pπ ⊆ Eπ and the closedness of at least one of the classes Eπ and Cπ under the operations q, sn , r0 , n0 , d0 , e, ez , eΦ , and p is Proposition 2. For every set π of prime numbers with Eπ = Cπ the class Pπ is q, sn , r0 , n0 , d0 , e, ez , eΦ , p-closed. Recall also that the product of two given classes X and Y of finite groups is the class X Y = {G | ∃A G such that A ∈ X , G/A ∈ Y }. A theorem of Chunikhin (see [19, 20] or [13, Chapter 5, (3.12)]) asserts that Cπ Eπ = Eπ and Cπ Cπ = Cπ . As a byproduct, in the proof of Theorem 1 we obtain some analog of this theorem for the class Pπ . Proposition 3. For every set π of prime numbers, Cπ Pπ = Pπ . Other byproducts of this article are Propositions 5 and 4 below which yield criteria for the sn closedness of the class Cπ and the e-closedness of the class Eπ in terms of the so-called almost simple groups. 2. Preliminary Results We use the standard notation of [8, 9, 13, 14]. Given a finite group G, denote by Hallπ (G) the set of all Hall π-subgroups. We express the fact that a subgroup H is pronormal in a group G as H prn G. The symbol Zn stands for the cyclic group of order n. 1) 2)
The asterisk marks each claim whose proof uses the classification of finite simple groups. Recall that every sn , n0 -closed class of groups is called a Fitting class. 417
Lemma 1. If H prn G and H ≤ M ≤ G then H prn M . Proof. Follows from the definition of pronormal subgroup. Lemma 2. If G ∈ Pπ , H ∈ Hallπ (G), and H ≤ M ≤ G then M ∈ Pπ . Proof. Follows from Lemma 1. Lemma 3. Take an arbitrary group G with a normal subgroup A. If G ∈ Eπ and H ∈ Hallπ (G) then A, G/A ∈ Eπ ; furthermore, H ∩ A ∈ Hallπ (A), In particular, sn , qEπ = Eπ . Proof. See [1, Lemma 1].
HA/A ∈ Hallπ (G/A).
Lemma 4. If G ∈ Eπ and A G then Hallπ (G/A) = {HA/A | H ∈ Hallπ (G)}. Proof. See [18, Corollary 9]. Lemma 5 (Chunikhin’s theorem). For every set π of prime numbers, we have Cπ Cπ = Cπ (equivalently, eCπ = Cπ ) and Cπ Eπ = Eπ . Proof. See [13, Chapter 5, (3.12)]. Lemma 6. If G ∈ Cπ , A G, and H ∈ Hallπ (G) then HA ∈ Cπ . Proof. See [17, Theorem 1]. Lemma 7. If G ∈ Cπ and A G then G/A ∈ Cπ . Proof. See [17, Lemma 9]. Lemma 8. Consider a finite group G = HA, where H is a Hall π-subgroup and A is a normal subgroup of G which is the direct product A = S1 × · · · × Sn of some finite simple groups. Then G ∈ Cπ if and only if AutG (Si ) ∈ Cπ for all i = 1, . . . , n. Proof. See [17, Lemma 17]. A finite group is called separable whenever it possesses a (sub)normal series whose all factors are either π- or π -groups. Lemma 9 (Chunikhin). Every π-separable group enjoys property Dπ . Proof. See [13, Chapter 5, Theorem 3.7]. Lemma 10. The Hall π-subgroups of a π-separable group are pronormal. Proof. Follows from Lemma 9 and the closedness of the class of π-separable groups under taking subgroups. Lemma 11. Suppose that X ∈ {Eπ , Cπ }. For every group G with normal subgroups M and N , if G/M ∈ X and G/N ∈ X then G/(M ∩ N ) ∈ X . Proof. See [18, Corollary 7; 16, Theorem 1]. Lemma 12. If G ∈ Eπ is a simple group then G ∈ Pπ . Proof. See [5, Theorem 1]. Lemma 13. For every set π of prime numbers, we have Cπ ⊆ Pπ . Proof. See [4, Theorem 1]. 418
Lemma 14. Given a subgroup H of a group G, take g ∈ G and y ∈ H, H g . If the subgroups H y and H g are conjugate in H y , H g then the subgroups H and H g are conjugate in H, H g . Proof. See [4, Lemma 10]. Lemma 15. Let : G → G1 be a group homomorphism and H ≤ G. If H prn G then H prn G. Proof. This is obvious. Lemma 16. Let G be a group with normal subgroups G1 , . . . , Gn such that [Gi , Gj ] = 1 for i = j and G = G1 · · · Gn . For every i = 1, . . . , n choose in Gi a pronormal subgroup Hi and put H = H1 , . . . , Hn . Then H prn G. Proof. See [4, Lemma 12]. Lemma 17. Consider a group G with H ∈ Hallπ (G) for some set π of prime numbers and A G satisfying G = HA. If (H ∩ A) prn A then H prn G. Proof. See [4, Lemma 13]. Lemma 18. Consider a group G with H ∈ Hallπ (G) for some set π of prime numbers and A G satisfying G = HA. If A ∈ Pπ then G ∈ Pπ . Proof. The assumption implies that G ∈ Eπ , |G : A| is a π-number, and G = KA for every K ∈ Hallπ (G). Since A ∈ Pπ , Lemma 3 yields (K ∩ A) prn A. Applying Lemma 17, we conclude that K prn G and G ∈ Pπ . Lemma 19. For every group G ∈ Eπ \ Cπ and every prime number p ∈ π the regular wreath product G Zp belongs to Eπ \ Pπ . The proof of this lemma appeared in [4, Proof of Theorem 3]. For the sake of completeness, we present this proof here since the construction plays an important role below. The group G includes two nonconjugate Hall π-subgroups U and V . Consider the direct product Y =G · · · × G ×G× p times
of p isomorphic copies of G. The mapping τ : Y → Y defined as (g1 , g2 , . . . , gp−1 , gp ) → (g2 , g3 , . . . , gp , g1 ), g1 , g2 , . . . , gp ∈ G, is an order p automorphism of Y . Consider the natural split extension X of Y by the group τ , which is isomorphic to the regular wreath product G Zp . Since Y is a normal subgroup of X and the index |X : Y | = p is not divisible by the numbers in π, it follows that Hallπ (X) = Hallπ (Y ). Define in Y the natural subgroups K=U H =V ×U × U × U, · · · × U ×V. ×U × × · · · p−1 times
p−1 times
It is clear that H, K ∈ Hallπ (Y ) = Hallπ (X) and, in particular, X ∈ Eπ . Since U and V are not conjugate in G, the subgroups H and K are not conjugate in Y and, consequently, in H, K ≤ Y . Meanwhile, the definition of τ yields H τ = K. Thus, H and K are not pronormal subgroups of X, and X ∈ Eπ \ Pπ . Observe that it is easy to justify a slightly more general claim. Lemma 20. For every group G ∈ Eπ \Cπ and every nontrivial π -group F the regular wreath product G F belongs to Eπ \ Pπ . Take A, B, H ≤ G with B A. Put NH (A/B) = NH (A) ∩ NH (B) and refer to this group as the normalizer H of the quotient A/B. Every x ∈ NH (A/B) induces on A/B an automorphism by acting as Ba → Bx−1 ax. Therefore, we have a homomorphism NH (A/B) → Aut(A/B). Denote its image by AutH (A/B) and refer to it as the group of H-induced automorphisms of A/B. In the particular case B = 1 we write AutH (A) instead of AutH (A/B). 419
Lemma 21. Consider a composition series 1 = G0 < G1 < · · · < Gn = G of a finite group G which refines a principal series. The following are equivalent: (1) G ∈ Eπ ; (2) AutG (Gi /Gi−1 ) ∈ Eπ for all i = 1, . . . , n. Proof. See [18, Corollary 5]. Recall that the socle Soc(G) of a finite group G is the subgroup generated in G by all minimal normal subgroups. A finite group is called almost simple whenever its socle is a nonabelian simple group. In other words, a group is almost simple whenever it is isomorphic to a subgroup G of the automorphism group Aut(S) of a nonabelian simple group S and, moreover,3) S∼ = Inn(S) ≤ G ≤ Aut(S). Consider the direct product X = G1 × · · · × Gn of groups with coordinate projections ρi : X → Gi . A subgroup G of X is called a complete diagonal subgroup if for every i = 1, . . . , n the restriction of ρi to G is an isomorphism between G and Gi (in particular, the existence of a complete diagonal subgroup requires that all factors Gi in the direct product are isomorphic to each other). Lemma 22. Let X ∗ = G1 × · · · × Gn be the direct product of isomorphic almost simple groups G1 , . . . , Gn . Put Si = Soc(Gi ) for all i = 1, . . . , n and Y = Soc(X ∗ ) = S1 × · · · × Sn . Choose in X ∗ a complete diagonal subgroup G and put X = Y G. Then AutX (Si ) ∼ = G for all i = 1, . . . , n. Proof. First of all, the subgroup Si is normal in X for every i = 1, . . . , n; hence, AutX (Si ) ∼ = NX (Si )/CX (Si ) = X/CX (Si ). Put Ti = S1 , . . . , Si−1 , Si+1 , . . . , Sn for every i = 1, . . . , n. Since all Sj for j = 1, . . . , n are nonabelian simple groups, it is clear that CY (Si ) = Ti . Verify that CX (Si ) ≤ Y and, consequently, CX (Si ) = Ti . Take x ∈ X \ Y and verify that x ∈ / CX (Si ). Indeed, x = yg for some y ∈ Y and g ∈ G; furthermore, g∈ / Soc(G) ≤ Y . Consider the coordinate projections ρi : X ∗ → Gi and put gi = g ρi and yi = y ρi for / Soc(Gi ) = Si . For brevity. Since the restriction to G of each ρi is an isomorphism, it follows that gi ∈ every h ∈ Si we have hg = hgi , hy = hyi . −1
The mapping h → hyi is an inner automorphism of the group Si , while the automorphism h → hgi −1 to be inner. Therefore, these mappings are distinct, and there exists h ∈ Si with hyi = hgi and
fails
hx = hyg = hyi g = hyi gi = h. Thus, x ∈ / CX (Si ), which shows that CX (Si ) = CY (Si ) = Ti . The subgroup Ti is normal in X, and it is easy to see that G ∩ Ti = 1. Verify that X = GTi . Put Soc(G) = S. Then S = G ∩ Y and |X| = |GY | =
|G||S|n |G||Y | = = |G||S|n−1 . |G ∩ Y | |S|
On the other hand, GTi =
|G||Ti | = |G||S|n−1 . |G ∩ Ti |
Hence, X = GTi . Finally, AutX (Si ) ∼ = G/(G ∩ Ti ) ∼ = X/CX (Si ) = X/Ti = GTi /Ti ∼ = G. 3)
420
Below we identify the isomorphic groups S and Inn(S).
Lemma 23. Consider two normal subgroups M and N of a group G with M ∩ N = 1. For all subgroups A and B of M the following hold: (1) N ≤ CG (A); (2) if AN = BN then A = B; (3) NG (A)/N = NG/N (AN/N ). Proof. See [16, Lemma 14].
3. Proof of Proposition 1
We need to establish the equivalence of the claims (1) Cπ = Eπ ; (2) Cπ = Pπ ; (3) Pπ = Eπ . It is clear that the inclusions Cπ ⊆ Pπ ⊆ Eπ (see Lemma 13) yield the implications (1) ⇒ (2) and (1) ⇒ (3). Suppose that claim (1) is false. Then there exists a group G ∈ Eπ \ Cπ . We claim that there exists a simple group S ∈ Eπ \ Cπ . Indeed, Lemma 3 implies that all composition factors of G are Eπ -groups. If all composition factors of G were Cπ -groups then by Chunikhin’s theorem (Lemma 5) so would G itself do. Thus, among the composition factors of G there is a simple group S ∈ Eπ \ Cπ . But Lemma 12 yields S ∈ Pπ ; thus, Cπ = Pπ . In addition, the set π is distinct from the set of all prime numbers and hence p ∈ π exists. Lemma 19 yields X ∈ Eπ \ Pπ for the regular wreath product X = G Zp . Therefore, Pπ = Eπ . Thus, the implications (2) ⇒ (1) and (3) ⇒ (1) hold, the equivalence of (1)–(3) is established, and Proposition 1 is proved. 4. Proof of Proposition 3 Consider a group G with A G and the natural epimorphism : G → G/A. Assuming that G ∈ Pπ and A ∈ Cπ , we need to show that G ∈ Pπ . Chunikhin’s theorem (Lemma 5) yields G ∈ Eπ . Take arbitrary H ∈ Hallπ (G) and g ∈ G. We are to show that the subgroups H and H g are conjugate in H, H g . y g Since G ∈ Pπ , there is y ∈ H, H g with H = H . By Lemma 4 we can replace H by H y and g assume that H = H ; i.e., HA = H g A. By Lemma 2 we can also replace G by NG (HA) and assume that HA G. In particular, this means that the group G is π-separable because one of the quotients, HA, of the normal series G Ha 1 is a π-group, while the other, G/HA, is a π -group. Since A ∈ Cπ and H ∩ A, H g ∩ A ∈ Hallπ (A) by Lemma 3, there is a ∈ A with H a ∩ A = H g ∩ A. Furthermore, since Cπ ⊆ Pπ by Lemma 13, there exists z ∈ H ∩ A, H a ∩ A ≤ H, H g with H g ∩ A = H a ∩ A = H z ∩ A. Applying Lemma 14 again, we may assume that H g ∩ A = H ∩ A; hence, g ∈ NG (H ∩ A). Frattini’s argument and the fact that A ∈ Cπ yield G = ANG (H ∩ A). This implies that the quotient NG (H ∩ A)/NA (H ∩ A) = NG (H ∩ A)/(A ∩ NG (H ∩ A)) ∼ = ANG (H ∩ A)/A = G/A = G is π-separable. The group NA (H ∩ A) is π-separable as well because it possesses the normal series NA (H ∩ A) H ∩ A 1, whose quotients are π - and π-groups. Consequently, NG (H ∩ A) is also a πseparable group. Now g ∈ NG (H ∩ A) and H, H g ∈ Hallπ (NG (H ∩ A)). In addition, the group H, H g is also π-separable as a subgroup of NG (H ∩ A). Hence, H, H g ∈ Dπ by Lemma 9. In particular, the subgroups H and H g are conjugate in H, H g . 421
5. Closedness Criteria for the Classes Eπ and Cπ Under Taking Normal Subgroups and Extensions By Lemmas 3 and 5, the class Eπ is sn -closed, while the class Cπ is e-closed. At the same time, there are examples of the sets π for which eEπ = Eπ and sn Cπ = Cπ . To prove Theorem 1, we need criteria for the e-closedness of Eπ and the sn -closedness of Cπ . Proposition 4. Given a set π of prime numbers, the following are equivalent: (1) eEπ = Eπ ; (2) G ∈ Eπ for each almost simple group G such that Soc(G) ∈ Eπ and G/ Soc(G) is a π-group. Proof. The implication (1) ⇒ (2) is obvious. Let us verify (2) ⇒ (1). Suppose that claim (2) holds, but claim (1) fails. Then there exists a group X ∈ / Eπ with a normal subgroup Y such that Y ∈ Eπ and X/Y ∈ Eπ . Furthermore, it is clear that 1 < Y < X. Refine the series 1 < Y < X firstly to a principal series, and then to a composition series 1 = X0 < X1 < · · · < Xn = X. Since each quotient Xi /Xi−1 is a composition factor of one of the Eπ -groups Y and X/Y , we have / Eπ for some i ∈ {1, . . . , n}. Xi /Xi−1 ∈ Eπ . In addition, Lemma 21 implies that G∗ = AutX (Xi /Xi−1 ) ∈ Put S = Xi /Xi−1 for this i. By the validity of Schreier’s conjecture, the quotient G∗ /S is a solvable / Eπ since otherwise the group. Take G satisfying S ≤ G ≤ G∗ and G/S ∈ Hallπ (G∗ /S). Observe that G ∈ ∗ ∗ / Eπ . But then we arrive at a contradiction Hall π-subgroup of G is a Hall π-subgroup of G despite G ∈ with claim (2) because G is an almost simple group, Soc(G) = S ∈ Eπ , and G/ Soc(G) is a π-group. Proposition 5. Given a set π of prime numbers, the following are equivalent: (1) sn Cπ = Cπ ; (2) Soc(G) ∈ Cπ for each almost simple group G ∈ Cπ such that G/ Soc(G) is a π-group. Proof. The implication (1) ⇒ (2) is obvious. Verify (2) ⇒ (1). Suppose that sn Cπ = Cπ . Among all groups in Cπ which include a normal subgroup outside Cπ , choose a group X of the smallest order. Take the minimal by inclusion normal subgroup Y of X satisfying Y ∈ / Cπ . Check firstly that Y is the minimal normal subgroup of X in the usual sense. Indeed, if there exists a normal subgroup N of X with 1 < N < Y then Y /N ∈ Cπ by the choice of X and the fact that X/N ∈ Cπ by Lemma 7. In addition, N ∈ Cπ by the choice of Y . Therefore, Y ∈ Cπ by Chunikhin’s theorem (Lemma 5); this is a contradiction. / Cπ , the Thus, Y = S1 × · · · × Sn for some simple subgroups S1 , . . . , Sn ; furthermore, since Y ∈ subgroups S1 , . . . , Sn are nonabelian. Lemma 6 implies that HY ∈ Cπ for every Hall π-subgroup H of X and the minimality of the order of X means that X = HY . Lemma 8 yields AutX (Si ) ∈ Cπ for all i = 1, . . . , n. But then Si ∈ Cπ by claim (2) of this lemma. Consequently, Y ∈ Cπ ; this is a contradiction. 6. Proof of Theorem 1 We divide this proof into a series of statements. (1) For every set π of prime numbers, qPπ = Pπ . Indeed, consider G ∈ Pπ with A G and put G = G/A. Verify that G ∈ Pπ . Taking H ∈ Hallπ (G), we need to show that H prn G. By Lemma 4 there is a subgroup H ∈ Hallπ (G) with H = HA/A. Since G ∈ Pπ , it follows that H prn G. Lemma 15 yields H = HA/A prn G/A = G. (2) For every set π of prime numbers, eΦ Pπ = Pπ and ez Pπ = Pπ . Consider a finite group G and A ∈ {Φ(G), Z∞ (G)}. Take N G and N ≤ A. Since A is a nilpotent group and, consequently, so is N , we have N ∈ Cπ . If G/N ∈ Pπ then G ∈ Pπ by Proposition 3. Therefore, the equalities eΦ Pπ = Pπ and ez Pπ = Pπ are justified. (3) For every set π of prime numbers, r0 Pπ = Pπ . 422
Suppose that the class Pπ not is closed under taking (finite) semidirect products. Then there exists / π. a group G with normal subgroups M and N such that G/M and G/N belong to Pπ , but G/(M ∩N )∈P Assume that G is of the smallest possible order among all these groups. Verify that the assumption of the existence of this group leads to a contradiction. Divide this proof into a series of steps. Lemma 11 implies that (i) G/(M ∩ N ) ∈ Eπ . The minimality of the order of G means that (ii) M ∩ N = 1 and G ∼ = G/(M ∩ N ). In particular, [M, N ] = 1. Therefore, (iii) G ∈ Eπ and G ∈ Pπ . This means that (iv) G includes a Hall π-subgroup H and contains an element g such that H x = H g for every x ∈ H, H g . (v) M H G and N H G. Indeed, suppose that N H G; i.e., NG (N H) < G. Observe that HN/N ≤ NG (N H)/N ≤ G/N,
HM/M ≤ NG (N H)M/M ≤ G/M,
whence by Lemmas 2 and 3 we infer that NG (N H)/N ∈ Pπ ,
NG (N H)/NM (N H) ∼ = NG (N H)M/M ∈ Pπ .
The choice of G yields NG (N H) ∈ Pπ . Since G/N ∈ Pπ , there is y ∈ H, H g with N H y = N H g ; consequently, gy −1 ∈ NG (N H). −1 Then, by the facts already established, the subgroup H, H gy ≤ H, H g contains an element z −1 with H z = H gy ; hence, H zy = H g , and moreover zy ∈ H, H g ; this is a contradiction with the choice of H and g. Thus, N H G. Similarly, M H G. (vi) G/M and G/N are π-separable. This assertion follows since all factors of the normal series G > M H > M and G > N H > N are either π- or π -groups. (vii) NG (M ∩ H) and NG (N ∩ H) are π-separable. Indeed, (vi) implies that NG (M ∩ H)M/M is a π-separable group since it is a subgroup of the πseparable group G/M . The group NM (M ∩ H) is π-separable as well since it possesses the normal series NM (M ∩H) > M ∩H > 1 whose quotients are π - and π-groups. (Recall that M ∩ H is a Hall π-subgroup of M by Lemma 3; hence, it is a Hall π-subgroup of NM (M ∩H).) Now NG (M ∩ H) is a π-separable group as an extension of the π-separable group NM (M ∩H) by the π-separable group NG (M ∩H)/NM (M ∩H) ∼ = NG (M ∩ H)M/M . Similarly we can prove that NG (N ∩ H) is a π-separable group. (viii) Every Hall π-subgroup K of G satisfies (M ∩ K)N/N = (KN/N ) ∩ (M N/N ). Indeed, Lemma 3 implies that M ∩ K is a Hall π-subgroup of M . Therefore, (M ∩ K)N/N is a Hall π-subgroup of M N/N . Since KN/N is a Hall π-subgroup of G/N and M N/N G/N , Lemma 3 implies that (KN/N ) ∩ (M N/N ) is a Hall π-subgroup of M N/N . Since it is obvious that (M ∩ K)N/N ≤ (KN/N ) ∩ (M N/N ), from the equalities of the orders of the right- and left-hand sides we infer that (M ∩ K)N/N = (KN/N ) ∩ (M N/N ). Claim (viii) is justified. 423
(ix) g ∈ NG (M ∩ H). Indeed, (v) and (viii) yield (H g ∩ M )N/N = H g N/N ∩ M N/N = HN/N ∩ M N/N = (H ∩ M )N/N. Consequently, (H g ∩M )N/N = (H∩M )N/N . Applying Lemma 23(2), we conclude that H g ∩M = H∩M . (x) H, H g is π-separable. Indeed, H ∩ M H, and so H ≤ NG (M ∩ H). Taking (ix) into account, we obtain H, H g ≤ NG (M ∩ H); furthermore, the subgroup NG (M ∩ H) is π-separable according to (vii). Thus, H, H g is a subgroup of a π-separable group; consequently, it is π-separable itself. (xi) A contradiction. The subgroups H and H g are Hall π-subgroups of the π-separable group H, H g , and by Lemma 9 they are conjugate in H, H g despite the choice of H and g. Claim (3) is justified. (4) For every set π of prime numbers, we have d0 Pπ = Pπ . This claim is a corollary to (3). In addition, it is also straightforward from Lemmas 3 and 23. (5) If sn Pπ = Pπ then sn Cπ = Cπ for every set π of prime numbers. Suppose that the class Cπ is not closed under taking normal subgroups. Then, by Proposition 5 and Lemma 3, there exist a nonabelian simple group S and a group G such that S ≤ G ≤ Aut(S), S ∈ Eπ \Cπ , G/S is a π-group, and G ∈ Cπ . Since Eπ \ Cπ = ∅, there exists p ∈ π . Consider the direct product Y ∗ = G × G × · · · × G, p times
the automorphism τ of Y ∗ acting as τ : (g1 , . . . , gp−1 , gp ) → (g2 , . . . , gp , g1 ) for all g1 , . . . , gp ∈ G, and the natural semidirect product X ∗ = Y ∗ τ . In X ∗ consider the subgroups Y = Soc(X ∗ ) = S · · · × S, ×S×
B0 = τ ,
p times
the τ -invariant complete diagonal subgroup A0 = {(g, g, . . . , g) | g ∈ G} ∼ =G of Y ∗ , and put A = A0 Y , B = B0 Y , and X = AB = A0 B0 Y . Furthermore, since [A0 , B0 ] = 1, it follows that A X and B X. In addition, A ∩ B0 = 1 and X = AB0 . Verify that A ∈ Cπ . Choose a Hall π-subgroup H of G. Then H ∩ S ∈ Hallπ (S). Put T = {(h1 , h2 , . . . , hp ) | h1 , h2 , . . . , hp ∈ H ∩ S} ∈ Hallπ (Y ), U = {(h, h, . . . , h) | h ∈ H} ∈ Hallπ (A0 ). The subgroup T is U -invariant and K = T U ∈ Hallπ (A). Since AY /Y is a π-group, we have A = KY = U T Y = U Y . For i ∈ {1, . . . , p} denote by Si the subgroup of Y consisting of the elements of the form (1, . . . , 1, s, 1, . . . , 1), where s ∈ S appears in slot i. Then Si ∼ = S is a nonabelian simple group for all i, and Y = S1 × · · · × Sp . Lemma 22 yields AutA (Si ) ∼ G ∈ C = π for every i = 1, . . . , p. Applying Lemma 8, we conclude that A ∈ Cπ . Furthermore, X ∈ Cπ as an extension of the Cπ -group A by the π -group B0 (see Lemma 5). 424
Finally, B ∼ / Pπ . Since B X ∈ Cπ ⊆ Pπ , we = S Zp , and since S ∈ Eπ \ Cπ , Lemma 19 yields B ∈ must have B ∈ sn Pπ = Pπ ; this is a contradiction. (6) There exist sets π of prime numbers with sn Pπ = Pπ . The set π = {2, 3} is an example, since it is known (see [3, Example 5.3]) that the class Cπ is not sn -closed for this π. (7) If sn Cπ = Cπ then sPπ = Pπ for every set π of prime numbers. In particular, the class Pπ is not s-closed in general. This claim is also a simple corollary to (5). (8) If ePπ = Pπ then eEπ = Eπ for every set π of prime numbers. Suppose that ePπ = Pπ , but eEπ = Eπ . According to Proposition 4, there exists an almost simple group G ∈ / Eπ such that S = Soc(G) ∈ Eπ and G/S is a π-group. It is clear that G/S ∈ Pπ . In addition, since the group S is simple, we obtain S ∈ Pπ by Lemma 12. But then G ∈ ePπ = Pπ ⊆ Eπ ; this is a contradiction. (9) There exist sets π with ePπ = Pπ . Indeed, examples of the sets π are available for which the class Eπ is not e-closed. For instance, π = {2, 3} works (see [3, Example 4.1]). Hence, (8) implies the claim. (10) There exist sets π with n0 Pπ = Pπ . Use the construction of [21, Theorem 1]. Suppose that the set π of prime numbers satisfies 2 ∈ π and there exists a nonabelian simple group S which has exactly two classes K1 and K2 of conjugate Hall π-subgroups and admits an automorphism ι of order 2 permuting these classes.4) Take as G the natural semidirect product Sι and put Y ∗ = G × G. The group Y ∗ admits the order 2 automorphism τ acting as τ : (g1 , g2 ) → (g2 , g1 ). Denote by X ∗ the natural semidirect product Y ∗ τ . Put Y = Soc(X ∗ ) = Soc(Y ∗ ) = S × S, σ = (ι, ι) ∈ Y ∗ , and X = Y, τ, σ. Take the natural epimorphism : X → X/Y . Observe that [τ, σ] = 1; thus, X is the elementary abelian group of order 4. The group X is the product of normal subgroups A = τ and B = τ σ. Their full preimages A and B are also normal in X, and X = AB. By [21, / Eπ . Since |A : Y | = |B : Y | = 2 ∈ π, it follows that A = HY Theorem 1], we have A, B ∈ Eπ and X ∈ and B = KY , where H ∈ Hallπ (A) and K ∈ Hallπ (B). Lemma 12 and claim (4) yield Y ∈ Pπ . Now / Eπ , and so X ∈ / Pπ . A, B ∈ Pπ by Lemma 18, and X = AB ∈ n0 Pπ , while X ∈ Thus, we have shown that for the chosen set π the classes Pπ and n0 Pπ are distinct. (11) Theorem 1 holds. Claim (A) of Theorem 1 follows from (1), (3), (4), and (2). Claim (B) follows from (7), (6), (10), and (9). Claim (C) follows from (5) and (8) together with Lemmas 3 and 5. 7. Remarks and Open Problems In view of the results of this article, the following open questions seem natural. Problem 1. Is it true that if sPπ = Pπ for some set π of prime numbers then sEπ = Eπ and sCπ = Cπ ? 4)
We can take as π the set {2, 3}. Then the group S = SL3 (3) = GL3 (2) has exactly two classes of conjugate Hall π-subgroups permuted by some outer automorphism ι of order 2 (see [3, Example 4.1]). 425
Problem 2. Is it true that if n0 Pπ = Pπ for some set π of prime numbers then n0 Eπ = Eπ ? Observe that it is simpler to obtain claim (8) from the proof of Theorem 1 basing on the criterion for the e-closedness of the class Pπ . Proposition 6. For every set π of prime numbers, ePπ = Pπ if and only if Eπ = Cπ . Proof. Suppose that ePπ = Pπ and Eπ = Cπ . As in the proof of Proposition 1, we verify that / π. Denote by there exists a simple group S with S ∈ / Cπ . Consider the group X = S Zp , where p ∈ Y =S · · × S × · p times
the base of the wreath product. Lemma 12 and claim (4) in the proof of Theorem 1 yield Y ∈ Pπ . Furthermore, X/Y ∼ / Pπ by Lemma 19; this is a contradiction. = Zp ∈ Pπ . Therefore, X ∈ ePπ . But X ∈ Hence, ePπ = Pπ ⇒ Eπ = Cπ . The converse implication follows from the inclusions Cπ ⊆ Pπ ⊆ Eπ and Lemma 5. In view of Lemma 5, claim (8) in the proof of Theorem 1 is obvious. Nevertheless, for two reasons we keep the proof of claim (8) as is. Firstly, this proof in our opinion emphasizes a certain duality in the behavior of the classes Eπ and Cπ under the operations sn and e. This duality is seen particularly clearly when we compare Lemmas 3 and 5, Propositions 4 and 5, and claims (5) and (8) in the proof of Theorem 1. Secondly, our proof rests on Proposition 4, which is of interest in its own right. Observe that the converse to (8) in the proof of Theorem 1 fails. Indeed, take π = {2, 7}. In this case Eπ = Cπ since G = 2 G2 (27) ∈ Eπ \ Cπ (see [22, Lemma 5.1]). In particular, ePπ = Pπ by Proposition 6. /π But the results of [22] imply that eEπ = Eπ . (More precisely, it is shown in [22] that in the case 3 ∈ a finite group G enjoys property Eπ if and only if so do all composition factors of G.) In this regard we wish to understand whether the converse to (5), dual to (8), holds. Problem 3. Is it true that if sn Cπ = Cπ for some set π of prime numbers then sn Pπ = Pπ ? The duality in the behavior of the classes Eπ and Cπ also makes the following problem natural. Problem 4. Is it true that for every set π of prime numbers sn Pπ = Pπ if and only if Eπ = Cπ ? References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
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[email protected] D. O. Revin Sobolev Institute of Mathematics Novosibirsk State University, Novosibirsk, Russia E-mail address:
[email protected]
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