Algebra and Logic, Vol. 46, No. 6, 2007

FINITE GROUPS WITH SUBNORMAL SCHMIDT SUBGROUPS V. A. Vedernikov∗

UDC 512.542

Keywords: finite group, Frobenius group, Schmidt subgroup, subnormal subgroup. We give a complete description of the structure of finite non-nilpotent groups all Schmidt subgroups of which are subnormal.

INTRODUCTION O. Yu. Schmidt in [1] studied into the structure of finite non-nilpotent groups all of whose proper subgroups are nilpotent. Later on, such were called Schmidt groups. In [1], the following facts were stated: a Schmidt group G is biprimary, that is, π(G) = {p, q}; a Sylow p-subgroup Gp is normal in G; Gq =
x is a cyclic group;
xq  = Oq (G) ≤ Z(G) and |Gp /Φ(Gp )| = pm , where m is the exponent of a number p modulo q. In [2-4], additional properties of Schmidt groups were pointed out: namely, |Φ(Gp )|
pm/2 , Φ(Gp ) = [Gp , Gp ] ≤ Z(G), and exp(Gp ) is equal to p, for p > 2, and does not exceed 4 for p = 2. Schmidt groups have found numerous applications to problems in group theory (see, e.g., [5-12]). To our knowledge, possibilities for using Schmidt groups in studies of finite groups were first mentioned in [5, Chap. 4]. In [9, 11] is a review of results on Schmidt groups and their applications in group theory. Thus, [8, 12] treat of properties of a finite non-nilpotent group all Schmidt subgroups of which are subnormal. In [8], it was proved that such a group is metanilpotent, and in [12], it was stated that the derived subgroup of the group is nilpotent. Note also that [13] completely describes the structure of finite non-nilpotent groups all of whose non-nilpotent subgroups are normal, and [14, Thm. 4] provides a full-fledged characterization of the structure of such groups with subnormal biprimary non-nilpotent subgroups. The objective of the present paper is to furnish a complete description of the structure of finite nonnilpotent groups all Schmidt subgroups of which are subnormal. Throughout, we deal with finite groups only; the notation and definitions are standard and can be found in [5, 10, 15-17]. We point out just some of these. A group with a normal Sylow p-subgroup is said to be p-closed. A group whose order is divisible by a prime p is called a pd-group (cf. [5]). A group possessing a normal complement to its Sylow p-subgroup is referred to as a p-nilpotent group. We write [A]B to denote the semidirect product of a normal subgroup A and a subgroup B. Following [11, 12], we call a Schmidt group with a normal Sylow p-subgroup and a non-normal cyclic Sylow q-subgroup an S
p,q -group. For the S
p,q -group S, we write S = [Sp ]Sq , where Sp is a normal Sylow p-subgroup and Sq is a cyclic non-normal Sylow q-subgroup. An F
p,d -group is a Frobenius group whose kernel is an elementary Abelian group of order pm , with cyclic complement of order d, where m is the exponent of p modulo q, for any q ∈ π(d). Moscow City Pedagogical University, Moscow, Russia; [email protected]. Translated from Algebra i Logika, Vol. 46, No. 6, pp. 669-687, November-December, 2007. Original article submitted February 28, 2007.

c 2007 Springer Science+Business Media, Inc. 0002-5232/07/4606-0363


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Writing N  G signifies the fact that N is a normal subgroup of G. By F (G) we denote a Fitting subgroup, by Φ(G) a Frattini subgroup, and by Gp a Sylow p-subgroup of G. The last member of an upper central series of G is called the hypercenter of G and is denoted by H(G). 1. PRELIMINARY RESULTS LEMMA 1. Let G be a group, N  G, A ≤ G, B ≤ G, H(G) be the hypercenter of G, and N ≤ H(G). Then: (1) A ∩ H(G) ≤ H(A); (2) H(G/N ) = H(G)/N and F (G/N ) = F (G)/N ; (3) if N = 1 then N ∩ Z(G) = 1; (4) A is subnormal in G iff AN/N is subnormal in G/N ; (5) if G = A × B and U is a subdirect product of A and B then H(G) = H(A) × H(B) and H(U ) = U ∩ H(G). Proof. Let 1 = Z0 < Z1 < Z2 < . . . < Zk = H(G) and Zi+1 /Zi = Z(G/Zi ), i = 0, 1, . . . , k − 1. Then 1 = A0 ≤ A1 ≤ A2 ≤ . . . ≤ Ak , where Ai = A ∩ Zi ; moreover, Ai+1 /Ai = A ∩ Zi+1 /A ∩ Zi ∼ = (A ∩ Zi+1 )Zi /Zi ≤ Zi+1 /Zi , i = 0, 1, . . . , k − 1, is an ascending central chain in the group A. Consequently, Ak = A ∩ H(G) ≤ H(A). (2) Let N ⊆ Zs and s be the least natural number with this property. Then 1 = N/N ≤ Z1 N/N ≤ Z2 N/N ≤ . . . ≤ Zs−1 N/N ≤ Zs /N ≤ . . . ≤ Zk /N = H(G)/N is an ascending central chain in G/N . Therefore, H(G)/N ≤ H(G/N ). Since (G/N )/(H(G)/N ) ∼ = G/H(G) and Z(G/H(G)) = 1, we have Z((G/N )/(H(G)/N )) = 1; hence H(G/N ) = H(G)/N . The definition of H(G) implies that N ≤ H(G) ⊆ F (G), and if C ≤ G and C/N is nilpotent, then C is also nilpotent. Since F (G)/N ⊆ F (G/N ) = R/N and R is nilpotent, we have R ⊆ F (G), which yields F (G/N ) = F (G)/N . (3) Let N be a minimal normal subgroup of G. By induction on the order of G, we show that N ⊆ Z(G). Let M be a minimal normal subgroup of G contained in Z(G). It is not hard to prove that N M/M is a minimal normal subgroup of G/M , and that N M/M ⊆ H(G)/M = H(G/M ) in view of item (2). By induction CG/M (N M/M ) = G/M , and hence CG (N M/M ) = G. The groups N M/M and N are G-isomorphic, and so CG (N ) = CG (N M/M ) = G. (4) Let A be subnormal in G. Then AN is subnormal in G, and so is AN/N in G/N . Let AN/N be subnormal in G/N . By induction on |G|, we show that A is subnormal in G. For N = 1, the statement holds true. Let N = 1 and Z = N ∩ Z(G). Then Z = 1. by item (3). In view of G/N ∼ = (G/Z)/(N/Z), the restriction AN/N ∼ = (AN/Z)/(N/Z) = (AZ/Z)(N/Z)/(N/Z) of this isomorphism implies that (AZ/Z)(N/Z)/(N/Z) is subnormal in (G/Z)/(N/Z); that is, by induction AZ/Z is subnormal in G/Z. Consequently, AZ is subnormal in G, and A  AZ. Hence A is subnormal in G. (5) Let ab ∈ U , where a ∈ A and b ∈ B. Keeping in mind that Z(G) = Z(A)×Z(B) and ab ∈ Z(U ) iff a ∈ Z(A) and b ∈ Z(B), we have Z(U ) = U ∩ Z(G). By [18, Lemma 2], U Z(G)/Z(G) is a subdirect product of groups AZ(G)/Z(G) and BZ(G)/Z(G). By induction H(U Z(G)/Z(G)) = H(G/Z(G))∩U Z(G)/Z(G), and hence H(U Z(G))/Z(G) = (H(G)∩U Z(G))/Z(G) in view of item (2). This implies that H(U )Z(G)/Z(G) = Z(G)(H(G) ∩ U )/Z(G). From H(U )Z(G)/Z(G) ∼ = H(U )/H(U ) ∩ Z(G) = H(U )/Z(U ) and Z(G)(H(G) ∩ U )/Z(G) ∼ = H(G) ∩ U/H(G) ∩ U ∩ Z(G) = H(G) ∩ U/U ∩ Z(G) = H(G) ∩ U/Z(U ), it follows that H(U )/Z(U ) ∼ = H(G) ∩ U/Z(U ). Since |H(U )| = |H(G) ∩ U |, and H(G) ∩ U ⊆ H(U ) by (1), we have H(U ) = H(G) ∩ U . The lemma is proved. LEMMA 2. Let S be an S
p,q -subgroup of G, N  G, and N ≤ H(G). Then: 364

(1) H(S) = Z(S) = Φ(S) and S/H(S) is an F
p,q -group; (2) SN/N is an S
p,q -subgroup of G/N ; (3) if K/N is an S
p,q -subgroup of G/N and T is a Schmidt subgroup of K, then T is a subnormal S
p,q -subgroup of K, K = T N , and T ∩ N ≤ Z(T ). Proof. (1) Follows directly from the structure of a Schmidt group. (2) Since SN/N ∼ = S/S ∩ N and S ∩ N ≤ H(S) (by Lemma 1(1)), in view of the structure of a Schmidt group, we conclude that S/S ∩ N is an S
p,q -group; hence SN/N is an S
p,q -subgroup of G/N . (3) By (2), T N/N is a Schmidt group and is a subgroup of the S
p,q -group K/N ; so K = T N , T is an S
p,q -group, and by Lemma 1(1), T ∩ N ≤ H(T ) = Z(T ) and N ≤ H(K). Since T N/N is subnormal in K/N , the subgroup T is subnormal in K by Lemma 1(4). The lemma is proved. LEMMA 3. Let A and B be distinct normal, respectively, S
p,q - and S
r,q -subgroups of G, and G = AB. Then: (1) if Ap ⊆ B, then G contains non-subnormal S
p,q - and S
r,q -subgroups; (2) if Ap ⊆ B, then p = r, Ap = Bp , G/H(G) is an F
p,q -group, and every Schmidt subgroup S of G is a subnormal S
p,q -subgroup of G; moreover, Sp = Ap , G = SH(G), S ∩ H(G) = Z(S), and Φ(Sp ) is a Sylow p-subgroup of H(G). Proof. Let D = A ∩ B. Then D  G and G/D = A/D × B/D. The structure of a Schmidt group implies that Aq ⊆ D and Bq ⊆ D. (1) Let Ap ⊆ D. From the structure of an S
p,q -group A, it follows that D ≤ Z(A). Assume Br ⊆ D. Then Br ⊆ A and r = p. Since B  G and Bp is characteristic in B, we have Bp  G. Consequently, Bp  A. Since Ap ⊆ D, Bp is a proper subgroup of Ap . In view of the structure of an S
p,q -group A, we conclude that Bp ≤ Φ(Ap ). By [3, Thm. 3], |Φ(Ap )|
pa/2 , where a is the exponent of p modulo q, and an S
p,q -group B is structured so as to yield |Bp |  pa . Contradiction. Thus Br ⊆ D. Hence D ≤ Z(B). In view of D ≤ Z(A) and D ≤ Z(B), D ≤ Z(G). By Lemma 1(1), (5), we have H(G/D) = H(G)/D, H(A/D) = H(A)/D, H(B/D) = H(B)/D, and H(G)/D = H(A)/D × H(B)/D. From (G/D)/(H(G)/D) ∼ = (A/D)/(H(A)/D) × (B/D)/(H(B)/D), it follows that G/H(G) ∼ = A/H(A)×B/H(B), with A∩H(G) = H(A) and B ∩H(G) = H(B). By virtue of Lemma 1(4), G contains a non-subnormal Schmidt subgroup S iff G/H(G) contains a non-subnormal Schmidt subgroup SH(G)/H(G). Therefore, we may assume that H(G) = 1, G = A × B, A is an S
p,q -subgroup of order pa q, and B is an S
r,q -subgroup of order rb q, where b is the exponent of r modulo q. Let f and g be epimorphisms of the groups A and B, respectively, onto a group Zq . By [18, item 1 of Lemma 1], G contains T as a subgroup (a subdirect product of A and B) such that T = {ab | f (a) = g(b), a ∈ A, b ∈ B}, with T ∩ A = Ap , T ∩ B = Br , T /Ap ∼ = B, T /Br ∼ = A, and G = AT = BT (see also [15, Sec. 9.11, Thm. 1]). Since Br is a minimal normal subgroup of B, Br is one of T . In view of the modular identity, Br is complemented in T if it is complemented in G. It follows that T = [Br ]U , with U ∼ = A and U < T . Assume U is subnormal in G. Then U is subnormal in T ; hence T = Br × U . This entails CG (Br ) ≥
T, A = G, which is a contradiction with Z(G) = 1. Consequently, U is a non-subnormal S
p,q -subgroup of G. In a similar way, we can show that G contains a non-subnormal S
r,q -subgroup. (2) Let Ap ⊆ D. Then the fact that Ap  B and the structure of a Schmidt group B imply that r = p and Bp = Ap . Consequently, Φ(Ap ) ≤ Z(A) ∩ Z(B) ≤ Z(G). Since D = Ap × Dq and Dq ≤ Z(A) ∩ Z(B), we have Z = Φ(Ap ) × Dq ≤ Z(G). Let Gq be a Sylow q-subgroup of G. Then there are Sylow q-subgroups Aq and Bq of the groups A and B, respectively, such that Gq = Aq Bq ; moreover, Aq ∩ Bq = Dq and Gq /Dq = Aq /Dq × Bq /Dq is an Abelian q-group. Consider G = G/Z. Then G is a non-nilpotent group, Gp = Ap Z/Z

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is a minimal normal subgroup of order pa in G, and Gq = Gq Z/Z ∼ = Gq /Gq ∩ Z = Gq /Dq is an Abelian q-group. Since Gq acts irreducibly on Gp , it follows by [10, Lemma 4.1] that the group Gq /CGq (Gp ) is cyclic, with CGq (Gp ) = Z(G). Note that CGq (Gp ) = CGq Z/Z (Ap Z/Z) = CGq Z (Ap Z/Z)/Z = ZCGq (Ap /Φ(Ap ))/Z. Let C = CGq (Ap /Φ(Ap )). Since Φ(Ap ) ≤ Z(G), C ≤ CGq (Ap ) by [10, Lemma 3.10]. From CGq (Ap ) ≤ CGq (Ap /Φ(Ap )), it follows that C = CGq (Ap ). Since |Aq : (Aq ∩ C)| = |Bq : (Bq ∩ C)| = q and Gq /C is a cyclic group, we have |Gq /C| = q. Consequently, Gq /CGq (Gp ) = (Gq Z/Z)/(ZC/Z) ∼ = Gq Z/ZC = ∼ ∼ ∼ Gq Φ(Ap )/CΦ(Ap ) = Gq /C; hence G/Z(G) = G/H(G) is a Frobenius group with kernel P = Ap /Φ(Ap ) of order pa and complement Q of order q. Let S be a Schmidt subgroup of G. By Lemma 2(3), G = SH(G), and so S is a subnormal S
p,q subgroup of G. From Ap = Gp G we obtain Sp ⊆ Ap , and from G = SH(G) we derive Gp = Sp Φ(Ap ) = Sp . Since Φ(Ap ) ≤ Ap ∩ H(G) ≤ A ∩ H(G) ≤ Z(A), Φ(Ap ) is a Sylow p-subgroup of H(G). The lemma is proved. LEMMA 4. Let S be an S
p,q -subgroup of G and let all S
p,q -subgroups of G be subnormal in G. Then: (1) Sp  G, Sp is a Sylow p-subgroup of S G , Φ(Sp ) ⊆ Z(S G ), and Φ(Sp ) is a Sylow p-subgroup of H(S G ); (2) S G /H(S G ) is an F
p,q -group; (3) if T is an S
p,q -subgroup of G and C = T G S G , then Cp = Tp = Sp , C/H(C) is an F
p,q -group, C = T H(C) = SH(C), T ∩ H(C) = Z(T ), and T G ∩ H(C) = H(T G ). Proof. If S  G then S G = S, and so statements (1) and (2) hold true. Let S not be normal in G. Since S is subnormal in G, it follows by [10, Thm. 7.1] that there exists an element x ∈ G such that S x = S, S ≤ NG (S x ), and S x ≤ NG (S). Let V = SS x . By [10, Thm. 7.5], the subgroup V is subnormal in G. In G, all S
p,q -subgroups are subnormal, S  V , and S x  V . By Lemma 3(2), Vp = Sp  V , V /H(V ) is an F
p,q -group of order pa q, where a is the exponent of p modulo q, Φ(Sp ) is a Sylow p-subgroup of H(V ), S ∩ H(V ) = Z(S), and S x ∩ H(V ) = Z(S x ). The fact that Φ(Sp ) ≤ Z(S) ∩ Z(S x ) entails Φ(Sp ) ≤ Z(V ). If V  G then S G = V , and so (1) and (2) hold true. Let V not be normal in G. By [10, Thm. 7.1], G contains an element y such that V y = V , V y ≤ NG (V ), and V ≤ NG (V y ). Let W = V V y and D = V ∩ V y . Then D  W . Assume Vp ⊆ V y . Consider a group W/D = V /D × V y /D. Since Sp = Vp ⊆ D, we have Sp D/D ∼ = Sp /Sp ∩ D = 1. In view of S ∩ D  S, the structure of a Schmidt group S implies Sq ⊆ D; hence SD/D is a Schmidt S
p,q -group. Similarly, S y D/D is a Schmidt S
p,q -group, with SD/D × S y D/D ≤ W/D. By Lemma 3(1), W/D contains a non-subnormal S
p,q -subgroup. In W , all S
p,q -subgroups are subnormal, and by [12, Lemma 4], all S
p,q -subgroups are subnormal in W/D. Contradiction. We have Vp ≤ V y . From Sp = Vp  V and V y ∼ = V , it follows that V y is also p-closed and Sp is a Sylow p-subgroup of y V . Consequently, Wp = Sp  W . Since W = SS x S y S xy and Sp is a Sylow p-subgroup of each of the four Schmidt groups, Φ(Sp ) is contained in the center of each of these groups; that is, Φ(Sp ) ≤ Z(W ). Proceeding further with this process, in n steps, we will see that S G is a product of 2n Schmidt S
p,q subgroups isomorphic to S, with a common Sylow p-subgroup Sp . This implies that Sp is a normal Sylow p-subgroup of S G ; hence Sp is normal in G, Φ(Sp ) ≤ Z(S G ), and S G is a {p, q}-group. Keeping in mind that Φ(Sp ) ≤ Sp ∩ H(S G ) ≤ S ∩ H(S G ) ≤ Z(S), we conclude that Φ(Sp ) is a Sylow p-subgroup of H(S G ). We claim that Z(S G ) = 1, if S is not normal in G. Let Φ(Sp ) = 1; then Φ(Sp ) ≤ Z(S G ) and the statement holds true. Let Φ(Sp ) = 1 and U = S G . Then |U | = pa q b , Sp is a minimal normal subgroup of U , and b > 1. Assume CU (Sp ) = Sp . Then every subgroup of Uq acts irreducibly on Sp , and by [10, Lemma 4.1],

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every Abelian subgroup of Uq is a cyclic group. This implies that Uq contains only one subgroup of order q; by [16, Thm. 12.5.2], Uq is a cyclic or generalized quaternion group. If q = 2 then a = 1 and |Sp | = p. Since Aut(Sp ) is a cyclic group, the group U/Sp ∼ = Uq is also cyclic. It follows that Uq is a cyclic group in any case. Hence U contains a single S
p,q -subgroup, which is a contradiction with S = S x . Consequently, CU (Sp ) = Sp . For Q = CU (Sp ) ∩ Uq , we have Q  Uq . Let D1 = Q ∩ Z(Uq ); then D1 = 1 and D1 ≤ Z(U ). Hence Z = Z(U ) = 1. By induction on |G|, we show that U/H(U ) is an F
p,q -group of order pa q. Since Z is characteristic in U , and U  G, we have Z  G. It follows that |G/Z| < |G| and U/Z = S G /Z = (SZ/Z)G/Z . Since SZ/Z ∼ = S/S ∩ Z is a subnormal S
p,q -subgroup of G/Z, and all S
p,q -subgroups of G/Z are subnormal by [12, Lemma 4], it follows by induction that (U/Z)/H(U/Z) is an F
p,q -group of order pa q. The fact that H(U/Z) = H(U )/Z implies that U/H(U ) is an F
p,q -group of order pa q. Let T be an S
p,q -subgroup of G. If T ≤ S G then C = T GS G = S G , and so item (3) follows from (2). Let T ⊆ S G . By (1), Tp is a normal Sylow p-subgroup of T G. Assume Tp ⊆ S G . Then Tp ⊆ Sp . Let F = T G ∩ S G . Consider a group C/F = T G /F × S G /F . Suppose Sp ⊆ F . Then Sp ⊆ Tp and Sp  Tp . Since |Sp |  pa and |Tp |
p3a/2 , the structure of a Schmidt group would imply that Tp = Sp ⊆ F , which is a contradiction. We have Sp ⊆ F . In view of the structure of a Schmidt group, T F/F ∼ = T /T ∩ F and SF/F ∼ = S/S ∩ F are S
p,q -groups, with T F/F × SF/F ≤ C/F . By [12, Lemma 4], all S
p,q -subgroups of C/F are subnormal. On the other hand, T F/F × SF/F by Lemma 3(1); hence C/F does also contain a non-subnormal S
p,q -subgroup. Contradiction. Consequently, Tp ⊆ S G . Item (1) implies that Tp ⊆ Sp and Tp is a normal Sylow p-subgroup of T G . Hence Tp is a normal Sylow p-subgroup of F . Since F is normal in S G , Tp = Fp is normal in S G , and so is Tp in S. The structure of a Schmidt group implies Tp = Sp . In view of (1), Φ(Sp ) ≤ Z(T G ) ∩ Z(S G ), that is, Φ(Sp ) ≤ Z(C). Now, in the same way as for U = S G , we can show that Z(C) = 1 and C/H(C) is an F
p,q -group. By Lemma 2(3), C = T H(C) = T G H(C). From Z(C/H(C)) = 1, it follows that T ∩ H(C) = H(T ) = Z(T ) and T G ∩ H(C) = H(T G ). The lemma is proved. LEMMA 5. Let A be a subnormal S
p,q -subgroup of G, B be a subnormal S
r,q -subgroup of G, and F = AG B G . If all S
p,q - and S
r,q -subgroups of G are subnormal, then Ap = Bp , p = r, F/H(F ) is an F
p,q -group, and every Schmidt subgroup S of F is a subnormal S
p,q -subgroup of G. Proof. Suppose that in G, all S
p,q - and S
r,q -subgroups are subnormal. By Lemma 4, Ap and Br are normal Sylow p- and r-subgroups of the groups AG and B G , respectively; moreover, AG /H(AG ) is an F
p,q -group of order pa q and B G /H(B G ) is an F
r,q -group of order rb q. Let N = H(AG )H(B G ). Then N  F and AG N/N ∼ = AG /AG ∩ N = AG /(H(AG )(AG ∩ H(B G ))). Assume Ap ⊆ B. Then Br ⊆ A; otherwise r = p, a = b, Bp  A, and the structure of a Schmidt group A would imply that Bp = Ap ⊆ B. By Lemma 4(1), Φ(Ap ) is a Sylow p-subgroup of H(AG ). Suppose Ap ⊆ N . Then Ap ⊆ Np = Φ(Ap )(H(B G ))p , whence Ap ⊆ (H(B G ))p ⊆ H(B G ) ⊂ B. Contradiction. We have Ap ⊆ N . Similarly, Br ⊆ N . Consequently, AG ∩ N/H(AG ) = H(AG )(AG ∩ H(B G ))/H(AG ) is a normal subgroup of AG /H(AG ) which does not contain Ap H(AG )/H(AG ) as a Sylow p-subgroup of AG /H(AG ). In view of the structure of an F
p,q -group AG /H(AG ), we have H(AG )(AG ∩H(B G )) = H(AG ). This implies AG ∩H(B G ) ⊆ H(AG ). Similarly, B G ∩H(AG ) ⊆ H(B G ). By Lemma 4(2), AG N/N ∼ = AG /H(AG ) and B G N/N ∼ = B G /H(B G ) are normal F
p,q - and F
r,q -subgroups of G/N , and F/N = (AG N/N )(B G N/N ). Assume Ap N/N ⊆ B G N/N . Then Ap N ⊆ B G N = B G H(AG ), that is, Ap ⊆ (B G )p (H(AG ))p =

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(B G )p Φ(Ap ). This implies that Ap ⊆ (B G )p = Bp ⊂ B, which leads us to a contradiction. We have Ap N/N ⊆ B G N/N . By Lemma 3(1), the group F/N contains non-subnormal S
p,q - and S
r,q -subgroups. In F , all S
p,q -subgroups are subnormal; by [12, Lemma 4], all S
p,q -subgroups of F/N are subnormal. Contradiction. We have Ap ⊆ B. By Lemma 4(3), F/H(F ) is an F
p,q -group, with F = AH(F ) = BH(F ). Let S be a Schmidt subgroup of F . By Lemma 2(3), S is a subnormal S
p,q -subgroup of F , and hence S is subnormal in G. The lemma is proved. 2. BASIC RESULTS THEOREM 1. In a non-nilpotent group G, all Schmidt subgroups are subnormal if and only if G/H(G) = G1 × . . . × Gn , where Gi is an F
pi ,di  -group, and moreover, (di , dj ) = 1 for any i = j, i, j ∈ {1, 2, . . . , n}. Proof. Necessity. Assume that G is a counterexample of minimal order. Let Z = Z(G) = 1. Consider G/Z. The group G/Z is non-nilpotent, every Schmidt subgroup of G/Z is subnormal in G/Z by [12, Lemma 4], and |G/Z| < |G|; by induction, therefore, the statement holds true for G/Z. By Lemma 1(2), H(G/Z) = H(G)/Z, and hence (G/Z)/H(G/Z) ∼ = G/H(G). Consequently, our statement is true also for G. Contradiction. We have Z(G) = 1. Let Φ = Φ(G). Suppose Φ = 1. Assume M is a minimal normal subgroup of G contained in Φ and M is an elementary Abelian p-group. By [10, Lemma 3.9], Op (G/CG (M )) = 1. Since Z(G) = 1, we have CG (M ) = G. Let x ∈ G\CG (M ) and x be a q-element, where q = p. Then [M ]
x is a non-nilpotent group. Let V be a Schmidt subgroup of [M ]
x. Then V is an S
p,q -group, with Vp ≤ M ; by Lemma 4(1), Vp is a Sylow p-subgroup of W = V G , with Vp
G; W is a {p, q}-group by [10, Cor. 7.7.2]. Hence Vp = M . On the Frattini lemma, G = W NG (Wq ) = M NG (Wq ). Since M ⊆ Φ, we have G = NG (Wq ). Consequently, Wq
G, which yields Vq = V ∩ Wq
V . Hence V is a nilpotent group, a contradiction. Thus Φ(G) = 1. Let F = F (G). By [8, Cor. 1.2], G/F is a nilpotent group. Since Φ(G) = 1, it follows by [10, Lemma 7.9] that G = [F ]H, F is a direct product of minimal normal subgroups of G, and H is a nilpotent group. Let K = NG (H). Assume H = K. Then K = H(F ∩ K) = H × (F ∩ K) and F ∩ K = 1. By virtue of CG (F ∩ K) ≥
F ∪ H = F H = G, Z(G) = 1, a contradiction. Consequently, NG (H) = H. Let q ∈ π(H) and Hq be a Sylow q-subgroup of H. Consider a group T = [F ]Hq . Suppose T = G. From T
G, it follows that F (T ) ⊆ F (G). Consequently, F = F (T ). By induction T /H(T ) = T1 × . . . × Ts , where Ti = [Ci ]Di is an F
pi ,di  -group, and moreover, (di , dj ) = 1 for any i = j, i, j ∈ {1, 2, . . . , s}. Since H(T ) ≤ F (T ), F (T /H(T )) = F (T )/H(T ) = C1 × . . . × Cs , and Hq H(T )/H(T ) ∼ = T /F (T ) ∼ = (T /H(T ))/(F/H(T )) ∼ = D1 × . . . × Ds and Di is a = Hq , we conclude that Hq ∼ q-group for any i ∈ {1, . . . , s}; hence s = 1 and Hq ∼ = D1 . Therefore, Hq is a cyclic group for any q ∈ π(H), from which it follows that H is cyclic. Let H =
h and π(H) = {q1 , . . . , qr }. Then h = h1 h2 . . . hr , where hi is a qi -element, with i = 1, 2, . . . , r. Let F = M1 ×. . .×Mn , where Mj is a minimal normal pj -subgroup of G, with j ∈ {1, 2, . . . , n}. Suppose n = 1. Then G = [M1 ]H, M1 is a minimal normal subgroup of G, H is a cyclic group, NG (H) = H, and Z(G) = 1. Let g ∈ G\H and D = H ∩ H g . Assume D = 1. Then CG (D) ⊇
H ∪ H g  = G, which leads us to a contradiction. Consequently, H ∩ H g = 1 for any g ∈ G\H, G is a Frobenius group, and the theorem holds true for G. Contradiction. Let n > 1. Note that CF (h) = 1, in which case g ∈ CG (h) iff g ∈ CG (hi ) for any i ∈ {1, 2, . . . , r}. Since Z(G) = 1, we have F ⊆ CG (hi ). Therefore, Mk ⊆ CG (hi ) for some k ∈ {1, 2, . . . , n}. Suppose 368

Ms ⊆ CG (hi ), where s ∈ {1, 2, . . . , n} and k = s. Now [Mk ]
hi  and [Ms ]
hi  are non-nilpotent groups, and G contains S
pk ,qi  - and S
ps ,qi  -subgroups with normal Sylow subgroups Mk and Ms , respectively. By Lemma 5, Mk = Ms , which is impossible. Consequently, hi does not centralize the sole direct factor Mk of the group F . Denote by dk a product of all hi which do not centralize Mk . Note that for any direct factor Mk of the group F , there exists at least one factor hi of an element h which does not centralize Mk . Otherwise h centralizes Mk , and Mk ⊆ Z(G), which is impossible. Consider Gk = [Mk ]
dk . Let h = dk tk , where tk is a product of all factors hj of h which are not entered into the product dk . Every factor hi of tk centralizes Mk , and so tk , too, centralizes Mk . Consequently, Gk is distinguished in G by a direct factor: namely, G = Gk × Tk , where Tk = [M1 × . . . × Mk−1 × Mk+1 × . . . × Mn ]
tk . This implies G = G1 × . . . × Gn , where Gk = [Mk ]
dk , k = 1, 2, . . . , n. It is not hard to show that Gk is a Frobenius group for any k = 1, 2, . . . , n, and so our theorem holds true for G. Contradiction. Thus G = T . In this event G = [F ]H, where H is a q-group, NG (H) = H, F = M1 × . . . × Mn , Mi is a minimal normal subgroup of G, with i ∈ {1, 2, . . . , n}, and Z(G) = 1. Since NG (H) = H, H is a Sylow q-subgroup of G. We have NMi (H) = 1, and so [Mi ]H is a non-nilpotent group. In view of Lemma 5, n = 1 and G = [M ]H, where M = M1 is a an elementary Abelian p-group. Assume CH (M ) = C = 1. By virtue of C
H, C ∩ Z(H) = 1. Let c ∈ C ∩ Z(H). Then CG (c) ⊇
M ∪ H = M H = G, and hence Z(G) = 1, which is a contradiction. We have CH (M ) = 1. Let S be an S
p,q -subgroup of G. Then Sp ⊆ M , and by Lemma 4, Sp
G. Hence M = Sp , |M | = pa , and a is the exponent of p modulo q. This implies that every non-trivial subgroup of H acts irreducibly on M . By [10, Lemma 4.1], every Abelian subgroup of H is cyclic, and if we apply [16, Thm. 12.5.2], as in the proof of Lemma 4, we come to the conclusion that H is a cyclic group. Now H ∩ H g = 1 for any g ∈ G\H, and hence G is a Frobenius group. Contradiction. Sufficiency. Let G = G/H(G) = F1 × . . . × Fn , where Fi is an F
pi ,di  -group, with (di , dj ) = 1 for any i = j, i, j ∈ {1, 2, . . . , n}. Suppose S is an S
p,q -subgroup of G. We claim that S is subnormal in G. Note that by Lemma 2(2), S = SH(G)/H(G) ∼ = S/S ∩ H(G) is an S
p,q -group. Since (di , dj ) = 1 for any i = j, i, j ∈ {1, 2, . . . , n}, there exists at most one number di which is divisible by q. Assume di is not divisible by q, for any i ∈ {1, 2, . . . , n}. Then the structure of groups Fi implies that the group G, and hence its subgroup S, is q-closed, which is impossible. Consequently, there exists a unique number, for instance, dk , such that q divides dk . Let Fk = [Mk ]
dk . Since G/Mk is q-closed, SMk /Mk is also q-closed. Hence S p ≤ Mk . An S
p,q group S and an F
pk ,dk  -group Fk are structured so as to yield pk = p and S p = Mk . Therefore, S/Mk is subnormal in G/Mk , hence S is subnormal in G, and so is S in G by Lemma 1(4). The theorem is proved. COROLLARY 1. If all Schmidt subgroups of a non-nilpotent group G are subnormal, then G/F (G) is a cyclic group. Proof. By Theorem 1 and Lemma 1(2), the group G/F (G) is a direct product of cyclic groups with mutually coprime orders d1 , d2 , . . . , dn . Hence G/F (G) is a cyclic group. The corollary is proved. COROLLARY 2. In a non-nilpotent group G, all non-nilpotent subgroups are subnormal if and only if G/H(G) is an F
p,d -group. Remark 1. Corollary 1 lends force to item (5) of a theorem in [12], and Corollary 2 generalizes the main result proved in [13]. COROLLARY 3 (see also [14, Thm. 4]). In a non-nilpotent group G, all biprimary non-nilpotent subgroups are subnormal if and only if G/H(G) = F1 × . . . × Fn , where Fi is an F
pi ,di  -group, with (di , |Fj |) = 1 for any i, j ∈ {1, 2, . . . , n}, i = j. 369

Proof. Necessity. Suppose that all biprimary non-nilpotent subgroups of a non-nilpotent group G are subnormal. By Theorem 1, G = G/H(G) = F1 × . . . × Fn , where Fi is an F
pi ,di  -group and (di , dj ) = 1 for any i, j ∈ {1, . . . , n}, i = j. Assume q|(di , |Fj |). Let Qi be a Sylow q-subgroup of Fi and Hj a biprimary non-nilpotent {pj , q}-subgroup of Fj . Consider a group H = Qi Hj . Since H is a biprimary non-nilpotent subgroup of G = G/H(G), applying Lemma 1(4), it is not hard to show that H is subnormal in G. Hence Qi = Fi ∩ H is subnormal in Fi , which is impossible. Consequently, (di , |Fj |) = 1 for any i, j ∈ {1, . . . , n}, i = j. Sufficiency. Let B be a biprimary non-nilpotent {p, q}-subgroup of G, S an S
p,q -subgroup of B, and G = G/H(G) = F1 × . . . × Fn , where Fi is an F
pi ,di  -group and (di , |Fj |) = 1 for any i, j ∈ {1, . . . , n}, i = j. Using Lemma 2(2), we then see that S = SH(G)/H(G) is an S
p,q -subgroup of B = BH(G)/H(G) and B is a biprimary non-nilpotent {p, q}-subgroup of G, where q divides a unique number dk , for some k ∈ {1, . . . , n}. As in the proof of Theorem 1 (sufficiency), it is not hard to show that S p = Mk is a pk -Sylow subgroup of Fk and pk = p. Since G/Fk is a q  -group, we have B q ≤ Q, where Q is some Sylow q-subgroup of Fk . Hence Mk < B ∩ Fk
Fk , and so B ∩ Fk is subnormal in G. In view of (pk , d1 d2 . . . dn ) = 1, the group G is pk -closed, and hence B p is subnormal in G. Since B = (B ∩ Fk )B p , the subgroup B is subnormal in G by [10, Thm. 7.5], and so is B in G by Lemma 1(4). The corollary is proved. Remark 2. Corollary 3 complements [14, Thm. 4]: it provides another characterization for nonnilpotent BS-groups, that is, groups every biprimary non-nilpotent subgroup of which is subnormal. As distinct from Theorem 4 in [14], note, Corollary 3 does not insist on orders of direct factors being mutually coprime, which allows us to furnish a more transparent description of the structure of BS-groups with trivial center. Let G = A × B, where A is a Schmidt subgroup of order 6 and B is a 2-closed Schmidt subgroup of order 12. Then G contains H = A×B3 as a non-subnormal biprimary non-nilpotent subgroup. By Theorem 1, all Schmidt subgroups of G are subnormal. Therefore, the class of all groups in which each Schmidt subgroup is subnormal properly contains the class of all non-nilpotent BS-groups. In [12], it was stated that if all p-closed Schmidt pd-subgroups of G are subnormal then G/Op (G) is p-nilpotent. Below is a theorem in which we give some extra information on properties of such groups. THEOREM 2. Let S be an S
p,q -subgroup of G and let all p-closed Schmidt pd-subgroups of G be subnormal. Then: (1) Φ(Sp ) ≤ H(G); (2) if R is an S
p,r -subgroup of G, q = r, F = S G RG , and Sp ≤ R, then Sp = Rp , Φ(Sp ) ≤ Z(F ), F/H(F ) is an F
p,qr -group, and Φ(Sp ) is a Sylow p-subgroup of H(F ); (3) if R is an S
p,r -subgroup of G, q = r, F = S G RG , and Sp ⊆ R, then D = S G ∩ RG = Φ(Sp ) ∩ Φ(Rp ) ≤ Z(F ), F/D = S G /D × RG /D, H(F ) ∩ S G = H(S G ), H(F ) ∩ RG = H(RG ), and F/H(F ) ∼ = S G /H(S G ) × RG /H(RG ) is a direct product of F
p,q - and F
p,r -groups, respectively. Proof. (1) By Lemma 4(1), Sp  G, Sp is a Sylow p-subgroup of S G , and Φ(Sp ) ≤ Z(S G ). Let Φ = Φ(Sp ) = 1. Since Φ is characteristic in Sp , and Sp  G, we have Φ  G. In view of [3, Thm. 3], |Sp |
p3a/2 and |Sp /Φ| = pa , where a is the exponent of p modulo q. Let M be a minimal normal subgroup of G which is contained in Φ. Consider a group C = CG (M ). Let C = G. By [10, Lemma 3.9], Op (G/C) = 1 and G\C contains a primary p -element x. Consider T = M
x. Since T is a non-nilpotent p-closed group, T contains an S
p,r -subgroup K, Kp is an elementary Abelian group of order pb , where b is the exponent of p modulo r, and Kr =
y. In view of b < a, we have r = q. Consider KSp /Φ = (Sp /Φ)(KΦ/Φ). This group is not nilpotent, since otherwise a p -element

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y would induce an identity automorphism on Sp /Φ, and on Sp by [18, Cor. A 12.7], which is impossible. Let L/Φ be an S
p,r -subgroup of KSp /Φ. Since Sp /Φ is an elementary Abelian Sylow p-subgroup of KSp /Φ, the Sylow p-group P/Φ of L/Φ has order pb < pa = |Sp /Φ|, and hence Φ < P < Sp . By [12, Lemma 4], all p-closed Schmidt pd-subgroups of G/Φ are subnormal in G/Φ, and Lemma 4(1) implies that P/Φ
G/Φ. In this event P
S, which clashes with properties of a Schmidt group S. Consequently, C = G. It follows that M ≤ Z(G). By [12, Lemma 4], all p-closed Schmidt pd-subgroups of G/M are subnormal in G/M . Appealing to properties of Schmidt groups, we see that S/M is an S
p,q -subgroup of G/M . By induction Φ(Sp /M ) = Φ(Sp )/M ≤ H(G/M ) = H(G)/M and Φ(Sp ) ≤ H(G). (2) By Lemma 4(1), S G and RG are p-closed groups with Sylow p-subgroups Sp and Rp , respectively; moreover, Φ(Sp ) ≤ Z(S G ) and Φ(Rp ) ≤ Z(RG ). By item (1), Φ(Rp ) ≤ H(G). In view of Lemmas 1 and 2, S ∩ H(G) ≤ H(S) = Z(S) and Sp ⊆ H(G). Hence Sp ⊆ Φ(Rp ). By virtue of Sp
R, the structure of a Schmidt group R implies Sp = Rp and Φ(Sp ) ≤ Z(S G ) ∩ Z(RG ) ≤ Z(F ). Since S G is a {p, q}-group and RG is a {p, r}-group, with q = r, we conclude that F/Sp = S G /Sp × RG /Sp is a nilpotent {q, r}-group. On the Schur theorem, F = [Sp ]T , Sp ∩ T = 1, T = Fq × Fr , S G = [Sp ]Fq , and RG = [Sp ]Fr . By Lemma 4(1), Φ(Sp ) is a Sylow p-subgroup of H(S G ); hence Fr centralizes H(S G ), and H(S G ) ⊆ H(F ). Since Fq centralizes H(RG ), we have H(RG ) ⊆ H(F ). Therefore, H(F ) ⊇ H(S G )H(RG ), and moreover, H(S G ) ∩ H(RG ) = Φ(Sp ) ≤ Z(F ). Let M = H(S G )H(RG ). Since F/M = S G RG M/M = S G M/M RG M/M ∼ = S G /H(S G )RG /H(RG ), applying Lemma 4(2), we see that Z(F/M ) = 1. Consequently, H(F ) = M . Now T M/M is a cyclic group of order qr, F/M = [S p M/M ]T M/M , and T M/M contains the normalizer in F/M of each of its Sylow subgroups; therefore, F/M is an F
p,qr -group by [19, Lemma 2]. In view of Φ(Sp ) ≤ Sp ∩ H(F ) ≤ S ∩ H(F ) = Z(S), Φ(Sp ) is a Sylow p-subgroup of H(F ). (3) Since D = S G ∩ RG
S G , D < Sp , properties of a Schmidt group S imply D ≤ Φ(Sp ) = Φ. By Lemma 4(1), Φ ≤ Z(S G ), Φ
G, and by item (1), Φ ≤ H(G). Similarly, D
RG and D ≤ Rp . Assume D = Rp . Since D is an elementary Abelian p-group, properties of a Schmidt group R entail |D| ≡ 1 (mod r). Consider L = [Sp ]Rr , where Rr =
y is a Sylow r-subgroup of the Schmidt group R. By [12, Lemma 4], all S
p,r -subgroups of G/Φ are subnormal in G/Φ. Assume L/Φ is nilpotent. Then y induces an identity automorphism on S p /Φ, and on Sp by [17, Cor. A 12.7]. Therefore, R is a nilpotent group. Contradiction. Thus L/Φ is a non-nilpotent p-closed group. Consequently, L/Φ contains K/Φ as an S
p,r -subgroup. Let P/Φ be a Sylow p-subgroup of K/Φ. By Lemma 4(1), P/Φ
G/Φ, and hence P
G. Since P/Φ ≤ S p /Φ is an elementary Abelian p-group, |P/Φ| = |D| = pc , where c is the exponent of p modulo r. By [3, Thm. 3], |D|
|Φ|
1/2|S p /Φ|, whence |Φ| < |P | = |Φ||D|
2|Φ| < |S p |, which clashes with properties of an S
p,q -group S. Consequently, D < Rp . In view of D
R, we have D ≤ Φ(Rp ). By Lemma 4(1), D ≤ Φ(Sp ) ∩ Φ(Rp ) ≤ Z(S G ) ∩ Z(RG ) ≤ Z(F ). Since F/D = S G /D × RG /D, applying Lemma 1(2), (5), we see that H(F ) = H(S G )H(RG ). By Lemma 4(2), F/H(F ) ∼ = S G /H(S G ) × RG /H(RG ) is a direct product of F
p,q and F
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