arXiv:1608.03353v1 [] 11 Aug 2016

Finite groups whose n-maximal subgroups are σ-subnormal

∗

Wenbin Guo
Department of Mathematics, University of Science and Technology of China,
Hefei 230026, P. R. China
E-mail: wbguo@ustc.edu.cn

Alexander N. Skiba
Department of Mathematics, Francisk Skorina Gomel State University,
Gomel 246019, Belarus
E-mail: alexander.skiba49@gmail.com

Abstract
Let σ = {σi |i ∈ I} be some partition of the set of all primes P. A set H of subgroups of G is
said to be a complete Hall σ-set of G if every member 6= 1 of H is a Hall σi -subgroup of G, for
some i ∈ I, and H contains exact one Hall σi -subgroup of G for every σi ∈ σ(G). A subgroup H
of G is said to be: σ-permutable or σ-quasinormal in G if G possesses a complete Hall σ-set set H
such that HAx = Ax H for all A ∈ H and x ∈ G: σ-subnormal in G if there is a subgroup chain
A = A0 ≤ A1 ≤ · · · ≤ At = G such that either Ai−1 E Ai or Ai /(Ai−1 )Ai is a finite σi -group for
some σi ∈ σ for all i = 1, . . . t.
If Mn < Mn−1 < . . . < M1 < M0 = G, where Mi is a maximal subgroup of Mi−1 , i =
1, 2, . . . , n, then M is said to be an n-maximal subgroup of G. If each n-maximal subgroup
of G is σ-subnormal (σ-quasinormal, respectively) in G but, in the case n > 1, some (n − 1)maximal subgroup is not σ-subnormal (not σ-quasinormal, respectively)) in G, we write mσ (G) =
n (mσq (G) = n, respectively).
In this paper, we show that the parameters mσ (G) and mσq (G) make possible to bound the
σ-nilpotent length lσ (G) (see below the definitions of the terms employed), the rank r(G) and
the number |π(G)| of all distinct primes dividing the order |G| of a finite soluble group G. We also
give conditions under which a finite group is σ-soluble or σ-nilpotent, and describe the structure
of a finite soluble group G in the case when mσ (G) = |π(G)|. Some known results are generalized.

1

Introduction

Throughout this paper, all groups are finite and G always denotes a finite group. Moreover, P is the
set of all primes, π ⊆ P and π ′ = P \ π. If n is an integer, the symbol π(n) denotes the set of all
∗

Research is supported by a NNSF grant of China (Grant # 11371335) and Wu Wen-Tsun Key Laboratory of
Mathematics of Chinese Academy of Sciences.
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Keywords: finite group, n-maximal subgroup, σ-subnormal subgroup, σ-quasinormal subgroup, σ-soluble group,
σ-nilpotent group.
0
Mathematics Subject Classification (2010): 20D10, 20D20, 20D30, 20D35.

1

primes dividing |n|; as usual, π(G) = π(|G|), the set of all primes dividing the order of G. Let A
and B be subgroups of G. We say that A forms an irreducible pair with B if AB = BA and A is a
maximal subgroup of AB.
In what follows, σ = {σi |i ∈ I ⊆ N} is some partition of P, that is, P = ∪i∈I σi and σi ∩ σj = ∅
for all i 6= j. By analogy with the notations π(n) and π(G), we put σ(n) = {σi |σi ∩ π(n) 6= ∅} and
σ(G) = σ(|G|).
In the mathematical practice, we often deal with the following two special partitions of P: σ =
{{2}, {3}, . . .} and σ = {π, π ′ }.
A group G is called: σ-primary [1] if |σ(G)| ≤ 1, that is, G is a σi -group for some i; σ-nilpotent
or σ-decomposable (Shemetkov [2]) if G = A1 × · · · × Ar for some σ-primary groups A1 , . . . , Ar ;
σ-soluble [1] if every chief factor of G is σ-primary. By the σ-nilpotent length (denoted by lσ (G)) of
a σ-soluble group G we mean the length of the shortest normal chain of G with σ-nilpotent factors.
The group G is nilpotent if and only if G is σ-nilpotent where σ = {{2}, {3}, . . .}; G is πdecomposable, that is, G = Oπ (G) × Oπ′ (G) if and only if G is σ-nilpotent where σ = {π, π ′ }.
The σ-nilpotent groups have many applications in the formation theory [2, 3, 4, 5] (see also
the recent papers [1, 6] and the survey [7]), and such groups are exactly the groups in which every
subgroup is σ-subnormal in the sense of the following definition (see Proposition 3.4 below).
Definition 1.1. A subgroup A of G is called σ-subnormal in G [1] if there is a subgroup chain
A = A0 ≤ A1 ≤ · · · ≤ At = G
such that either Ai−1 E Ai or Ai /(Ai−1 )Ai is σ-primary for all i = 1, . . . , t.
It should be noted that the idea of this concept originates from the paper of Kegel [8] where it
was discussed one another generalization of nilpotency.
There are some motivations for introducing and study of σ-subnormal subgroups. First of all,
the set of all σ-subnormal subgroups of G forms a sublattice of the lattice of all subgroups of G
(see Proposition 2.5 below). This fact is a generalizations of the classical Wielandt’s result which
states that the set of all subnormal subgroups of G forms a sublattice of the lattice of all subgroups
of G. Thus, the σ-subnormal subgroups are very convenient for applications. The first among such
applications were obtained in [1, 9].
Recall that a set H of subgroups of G is said to be a complete Hall σ-set of G if every member
6= 1 of H is a Hall σi -subgroup of G, for some i ∈ I, and H contains exact one Hall σi -subgroup of G
for every σi ∈ σ(G). A subgroup H of G is said to be σ-permutable or σ-quasinormal in G [1] if G
possesses a complete Hall σ-set set H such that HAx = Ax H for all A ∈ H and x ∈ G. In particular,
H is said to be S-permutable or S-quasinormal in G [10, 11] if H permutes with all Sylow subgroups
of G.
If H is an S-quasinormal subgroup of G, then H is subnormal in G (Kegel) and H/HG is nilpotent
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(Deskins) [10, Theorem 1.2.14]). But in general, if H is σ-quasinormal in G, then H is not necessary
subnormal and H/HG may be non-nilpotent [1]. Nevertheless, in this case when H is σ-quasinormal,
H is σ-subnormal in G and H/HG is σ-nilpotent [1, Theorem B].
Examples and some other applications of σ-subnormal subgroups and σ-quasinormal subgroups
were discussed in [1, 7, 9]. In this paper, we consider some applications of such subgroups in the
theory of n-maximal subgroups.
Recall that if Mn < Mn−1 < . . . < M1 < M0 = G (*), where Mi is a maximal subgroup of Mi−1
for all i = 1, . . . , n, then the chain (*) is said to be a maximal chain of G of length n and Mn (n > 0),
is an n-maximal subgroup of G.
If each n-maximal subgroup of G is σ-subnormal (σ-quasinormal, respectively) in G but, in the
case n > 1, some (n − 1)-maximal subgroup is not σ-subnormal (not σ-quasinormal, respectively))
in G, we write mσ (G) = n (mσq (G) = n, respectively).
Note that mσ (G) = 1 = mσq (G) if and only if G is σ-nilpotent by Proposition 3.4 below. We
also show (see Corollaries 1.7 and 1.8 below) that mσ (G) = 2 if and only if G is a Schmidt group
G with abelian Sylow subgroups such that |σ(G)| = |π(G)|, and mσq (G) = 2 if and only if G is a
supersoluble group with mσ (G) = 2. Finally, note that every group with mσ (G) = 3 is σ-soluble
(see Theorem 1.4 below), and there are non-soluble groups, for example the alternating group A5 of
degree 5, with mσq (G) = 4.
If G is soluble, the parameters mσ (G) and mσq (G) make possible to bound the σ-nilpotent length
lσ (G), the rank r(G) and the number |π(G)| of all distinct primes dividing |G|.
Recall that the rank r(G) of a soluble group G is the maximal integer k such that G has a chief
factor of order pk for some prime p (see [12, p. 685]).
Theorem 1.2. Suppose that G is σ-soluble and let H be a complete Hall σ-set of G. Then the
following statements hold.
(i) If G is soluble but it is not σ-nilpotent and r(H) ≤ r ∈ N for all H ∈ H, then r(G) ≤
mσq (G) + r − 2.
(ii) lσ (G) ≤ mσ (G).
(iii) If G is soluble but it is not σ-nilpotent, then |π(G)| ≤ mσ (G).
Now, let’s consider some applications of Theorems 1.2.
The relationship between n-maximal subgroups (where n > 1) of a group G and the structure of
G was studied by many authors (see, in particular, the recent papers [13, 14, 15, 16, 17, 18, 19, 20]
and Chapter 4 in the book [11]). One of the first results in this direction were obtained by Huppert
[21]. In fact, Huppert proved [21] that: if every 2-maximal subgroup of G is normal in G, then G is
supersoluble; if every 3-maximal subgroup of G is normal in G, then G is soluble of rank r(G) at most
two. The first of these two results was generalized by Agrawal [22] : If every 2-maximal subgroup L
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of G is S-quasinormal in G, then G is supersoluble. In the universe of all soluble groups the both
Huppert’s observations and some similar results in [23] are special cases of the following general
result (Mann [24]): If G is soluble and every n-maximal subgroup L of G (n > 1) is quasinormal
(that is, L permutes with all subgroups of G), then r(G) ≤ n − 1.
In the case σ = {{2}, {3}, . . .} we get from Theorem 1.2(i) the following generalization of the last
of these results.
Corollary 1.3. Suppose that G is soluble and each n-maximal subgroup of G (n > 1) is Squasinormal in G. Then r(G) ≤ n − 1.
The following theorem allows us to obtain the above mentioned result of Agrawal.
Theorem 1.4. (i) If in every maximal chain M3 < M2 < M1 < M0 = G of G of length 3, one of
M3 , M2 and M1 is σ-subnormal in G, then G is σ-soluble.
(ii) If 1 < mσ (G) ≤ 3, then G is soluble.
Corolary 1.5 (Spencer [25]). If in every maximal chain M3 < M2 < M1 < M0 = G of G of
length 3, one of M3 , M2 and M1 is subnormal in G, then G is soluble.
Corollary 1.6 (Mann [24]). If every 3-maximal subgroup of G is subnormal in G, then G is
soluble.
Recall that G is called a Schmidt group if G is not nilpotent but every proper subgroup of G is
nilpotent.
Corollary 1.7. The equality mσ (G) = 2 is true if and only if G is a Schmidt group G with
abelian Sylow subgroups such that |σ(G)| = |π(G)|.
Corollary 1.8. The equality mσq (G) = 2 is true if and only if G is a supersoluble group with
mσ (G) = 2.
From Corollaries 1.8 and Theorem 1.4, we get
Corollary 1.9 (see Agrawal [22] or Theorem 6.5 in [26, Ch.1])). If every 2-maximal subgroup of
G is S-quasinormal in G, then G is supersoluble. Moreover, if |π(G)| > 2, then G is nilpotent.
From Theorem 1.2(iii) we know that for every soluble but non-σ-nilpotent group G we have
|π(G)| ≤ mσ (G). In the case when |π(G)| = mσ (G) the structure of such a group G can be described
completely as follows.
Theorem 1.10. Suppose that G is soluble. Then mσ (G) = |π(G)| if and only if G is a group of
one of the following two types:
(i) G is a p-group for some prime p.
(ii) G = D ⋊ M , where D = GNσ is an abelian Hall subgroup of G, and the following hold:
(a) Every non-σ-subnormal Sylow subgroup P1 of G is cyclic and the maximal subgroup of P1
is σ-subnormal in G. Moreover, if P1 , . . . , Pn is a Sylow basis of G, then P2 , . . . , Pn are elementary
4

abelian and and P1 forms an irreducible pairs with all such subgroups; if {H1 , . . . , Ht } is a complete
Hall σ-set of G, P1 ≤ H1 and P1 is not of prime order, then H2 , . . . , Ht are normal in G.
(b) Some Sylow subgroup of M is not σ-subnormal in G. Hence M acts irreducibly on every
Sylow subgroup of D.
(c) If G possesses at least two non-σ-subnormal non-isomorphic Sylow subgroups, then all nonσ-subnormal Sylow subgroups are of prime order.
(d) If P is a non-σ-subnormal Sylow subgroup of G, P is a σi -group and V is the maximal
subgroup of P , then |G : NG (V )| is a σi -number.
In this theorem GNσ denotes the σ-nilpotent residual of G, that is, the intersection of all normal
subgroups N of G with σ-nilpotent quotient G/N .
In the case when σ = {{2}, {3}, . . .}, from Theorems 1.2 and 1.10 we get the following known
result.
Corollary 1.11 (Mann [24]). Suppose that G is a soluble group and each n-maximal subgroup
of G is subnormal. If n ≤ |π(G)|, then G is either of the following type:
(a) G is nilpotent.
(b) G = HN , where,
(i) N is a normal abelian Hall subgroup, and all Sylow subgroups of N are elementary abelian.
(ii) H is a cyclic Hall subgroup, and |H| is either a prime power or square-free number.
(iii) (|N |, |H|) = 1.
(iv) If Hp is a Sylow subgroup of H and Nq is a Sylow subgroup of N , then Hp induces in Nq an
irreducible automorphism group of order p or 1. In the latter case, |Nq | =q.
Conversely, a group of type (a) or (b) has each n-maximal subgroup subnormal.
Corollary 1.12 (Mann [24]). Suppose that G is a soluble group and each n-maximal subgroup
of G is subnormal. If n < |π(G)|, then G is nilpotent.
We prove Theorems 1.2, 1.4 and 1.10 in Sections 4, 5 and 6, respectively. But before them, as
preparatory steps, we prove in Section 2 that the set of all σ-subnormal subgroups of G forms a
sublattice of the lattice of all subgroups of G, and we collect in Section 3 some needed properties of
σ-soluble and σ-nilpotent groups.
All unexplained notation and terminology are standard. The reader is referred to [27], [5] and
[11] if necessary.

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2

The lattice Lσ (G) of all σ-subnormal subgroups

It is not difficult to show that the intersection of any two σ-subnormal subgroups of G is also σsubnormal in G (see Lemma 2.2(1)(2) below). It is well-known that any partially ordered set with
1 in which there is the greatest lower bound for each its non-empty subset is a lattice. Hence the
set Lσ (G) of all σ-subnormal subgroups of G is a lattice. In this section, we show that Lσ (G) is a
sublattice of the lattice of all subgroups of G.
We use Sσ and Nσ to denote the class of all σ-soluble groups and the class of all σ-nilpotent
groups, respectively.
Lemma 2.1 (See Lemma 2.5 in [1]). The class Nσ is closed under taking direct products,
homomorphic images and subgroups. Moreover, if H is a normal subgroup of G and H/H ∩ Φ(G) is
σ-nilpotent, then H is σ-nilpotent.
In what follows, Π is always supposed to be a non-empty subset of the set σ and Π′ = σ \ Π. We
say that: a natural number n is a Π-number if σ(n) ⊆ Π; G is a Π-group if |G| is a Π-number; G is
σ-perfect if GNσ = G.
We call the product of all normal σ-nilpotent subgroups of G the σ-Fitting subgroup of G and
denote it by Fσ (G).
Lemma 2.2. Let A, K and N be subgroups of G. Suppose that A is σ-subnormal in G and N
is normal in G.
(1) A ∩ K is σ-subnormal in K.
(2) If K is a σ-subnormal subgroup of A, then K is σ-subnormal in G.
(3) If N ≤ K and K/N is σ-subnormal in G/N , then K is σ-subnormal in G.
(4) If K ≤ A and A is σ-nilpotent, then K is σ-subnormal in G.
(5) If N is a σi -group, for some i, then N ≤ NG (Oσi (A)).
(6) If K ≤ E ≤ G, where K is σ-subnormal in E, then KN/N is σ-subnormal in N E/N .
(7) If A is σ-perfect, then A is subnormal in G.
(8) Suppose that N is the product of some minimal normal subgroups of G and N is not σ-primary.
Suppose also that G = AN , N is non-abelian and all composition factors of N are isomorphic. Then
N ≤ NG (A).
(9) If A is a σi -group, then A ≤ Oσi (G). Hence if A is σ-nilpotent, then A ≤ Fσ (G).
Proof. Statements (1)–(5) follow from Lemma 2.6 in [1].
(6) By hypothesis, there is a chain K = K0 ≤ K1 ≤ · · · ≤ Kn = E such that either Ki−1 E Ki
or Ki /(Ki−1 )Ki is σ-primary for all i = 1, . . . , n. Consider the chain
KN/N = K0 N/N ≤ K1 N/N ≤ · · · ≤ Kn N/N = EN/N.
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Assume that Ki−1 N/N is not normal in Ki N/N . Then L = Ki−1 is not normal in T = Ki and so
T /LT is σ-primary. Then
(T /LT )/(LT (T ∩ N )/LT ) = (T /LT )/((T ∩ N LT )/LT ) ≃
T /(T ∩ N LT ) ≃ T N/LT N ≃ (T N/N )/(LT N/N )
is σ-primary. But LT N/N ≤ (LN/N )T N/N . Hence (T N/N )/(LN/N )T N/N is σ-primary. This shows
that KN/N is σ-subnormal in N E/N .
By hypothesis, there is a chain A = A0 ≤ A1 ≤ · · · ≤ Ar = G such that either Ai−1 E Ai
or Ai /(Ai−1 )Ai is σ-primary for all i = 1, . . . , r. Let M = Ar−1 . We can assume without loss of
generality that M 6= G.
(7) Assume that this assertion is false and let G be a counterexample of minimal order. First we
show that A ≤ MG . This is clear if M is normal in G. Now assume that G/MG is a σi -group. Then
from the isomorphism AMG /MG ≃ A/A ∩ MG and A = ANσ we get that A = Oσi (A) ≤ A ∩ MG , so
A ≤ MG . The choice of G implies that A is subnormal in M , so A is subnormal in MG by Assertion
(1). Therefore A is subnormal in G.
(8) Assume that this assertion is false. First note that N  M since G = AN and M < G by
hypothesis.
If M is not normal in G, then G/MG is σ-primary and so N ≃ N MG /MG is σ-primary, which
contradicts the hypothesis. Hence M is normal in G and so N = (N ∩ M ) × N0 , where N ∩ M and
N0 are normal in G. Then N ∩ M ≤ Soc(M ) by [27, A, 4.13(c)] and M = M ∩ AN = A(N ∩ M ).
The choice of G implies that N ∩ M ≤ NM (A) ≤ NG (A). On the other hand, N0 ∩ M = 1 and so
N0 ≤ CG (M ) ≤ CG (A). Thus N ≤ NG (A), a contradiction. Hence we have (8).
(9) Assume that this assertion is false and let G be a counterexample of minimal order. Then
1 < A < G. Let D = Oσi (G), R be a minimal normal subgroup of G and O/R = Oσi (G/R). Then
the choice of G and Assertion (6) imply that AR/R ≤ O/R, so A ≤ O. Therefore R  D, so D = 1
and A ∩ R < R. Suppose that L = A ∩ R 6= 1. Assertion (1) implies that L is σ-subnormal in R. If
R < G, the choice of G implies that L ≤ Oσi (R) ≤ D since Oσi (R) is a characteristic subgroup of
R. But then D 6= 1, a contradiction. Hence R = G is a simple group, which is also impossible since
1 < A < G. Therefore R ∩ A = 1. If O < G, the choice of G implies that A ≤ Oσi (O) ≤ D = 1.
This contradiction shows that G/R = O/R is a σi -group. Hence R is the unique minimal normal
subgroup of G. Moreover, Lemma 2.1 implies that R  Φ(G), so CG (R) ≤ R by [27, A, 15.2].
Now we show that A ⊆ R. First assume that R is σ-primary. Then R is a σj -group for some
σj ∈ σ \ {σi } and so Oσj (A) = A. Therefore R ≤ NG (A) by Assertion (5). Consequently, A ≤
CG (R) ≤ R. Now assume that R is not σ-primary. Then R is not abelian. Hence R is the product
of some minimal normal subgroups of RA by [27, A, 4.13(c)]. Hence R ≤ NRA (A) by Assertion (8).
Then AR = A × R and so also A ≤ CG (R) ≤ R. This contradiction completes the proof the fact
that A ≤ Oσi (G). Now assume that A is σ-nilpotent. Then A = A1 × · · · × At , where A1 , . . . , At are
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σ-primary groups. Since A is σ-subnormal in G, every factor Ai is σ-subnormal in G. Hence Ai is
contained in Oσj (G) for some j = j(Ai ), so Ai ≤ Fσ (G). Thus A ≤ Fσ (G).
The lemma is proved.
Let
F0σ (G) ≤ F1σ (G) ≤ · · · ≤ Fiσ (G) ≤ · · ·
be the upper σ-nilpotent series of G, that is, F0σ (G) = 1 and
Fiσ (G)/F(i−1)σ (G) = Fσ (G/F(i−1)σ (G))
for i > 0. If n is the smallest integer such that Fnσ (G) = G, then n coincides with the σ-nilpotent
length of G.
We use Nnσ to denote the class of all σ-soluble groups G such that lσ (G) ≤ n (n > 0).
Lemma 2.3. The following holds:
(i) If a non-empty class F of groups is closed under taking direct products, homomorphic images
and subgroups, then the class Nσ F is also closed under taking direct products, homomorphic images
and subgroups. Moreover, if G/Φ(G) ∈ Nσ F, then G ∈ Nσ F.
(ii) The class Nnσ is closed under taking direct products, homomorphic images and subgroups.
Moreover, if G/Φ(G) ∈ Nnσ , then G ∈ Nnσ .
Proof. (i) This assertion can be proved by the direct calculations. For instance, if G/Φ(G) ∈
Nσ F and N/Φ(G) is a normal subgroup of G/Φ(G) such that (G/Φ(G))/(N/Φ(G)) ≃ G/N ∈ F and
N/Φ(G) is σ-nilpotent, then N is σ-nilpotent by Lemma 2.1. Hence G ∈ Nσ F.
(ii) In the case when n = 1, the assertion follows from Lemma 2.1. Now assume that n > 1 and
the assertion is true for n − 1. It is not difficult to show that
Nnσ = Nσ Nσn−1 .
Therefore this assertion is a corollary of (i).
The lemma is proved.
Proposition 2.4. Let A be a σ-subnormal subgroup of G. If A is σ-soluble and lσ (A) ≤ n, then
A ≤ Fnσ (G).
Proof. First note that in view of Lemma 2.2(9), Fσ (A) ≤ Fσ (G). Hence lσ (AFσ (G)/Fσ (G)) =
lσ (A/A ∩ Fσ (G)) ≤ n − 1 and so by induction we have AFσ (G)/Fσ (G) ≤ Fn−1σ (G/Fσ (G)) =
Fnσ (G)/Fσ (G). Therefore A ≤ Fnσ (G). The lemma is proved.
Proposition 2.5. The set of all σ-subnormal subgroups A of G with lσ (A) ≤ n forms a sublattice
of the lattice of all subgroups of G.
Proof. In view of Lemma 2.3, Proposition 2.4 and Statements (1) and (2) of Lemma 2.2, we
need only to show that if A and B are σ-subnormal subgroups of G, then hA, Bi is σ-subnormal in
G.
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Assume that this is false and let G be a counterexample of minimal order. Then A 6= 1 6= B and
hA, Bi =
6 G. Let R be a minimal normal subgroup of G.
(1) hA, BiR = G. Hence hA, BiG = 1
Suppose that L = hA, BiR 6= G. Lemma 2.2(1) implies that A and B are σ-subnormal in L. The
choice of G implies that hA, Bi is σ-subnormal in L. On the other hand,
L/R = hA, BiR/R = hAR/R, BR/Ri,
where AR/R and BR/R are σ-subnormal in G/R by Lemma 2.2(6), so the choice of G implies that
L/R is σ-subnormal in G/R and so L is σ-subnormal in G by Lemma 2.2(3). But then hA, Bi is
σ-subnormal in G by Lemma 2.2(2). This contradiction shows that we have (1).
(2) If S is a non-identity characteristic subgroup of C, where C ∈ {A, B}, then R  NG (C).
Indeed, if R ≤ NG (C), then R ≤ NG (S) and so
S G = S hA,BiR = S hA,Bi ≤ C hA,Bi ≤ hA, BiG = 1,
a contradiction.
(3) R is a σi -group for some i ∈ I.
Suppose that this is false. Then R is non-abelian, which implies that R is the product of some
minimal normal subgroups of RA by [27, A, 4.13(c)]. Hence R ≤ NRA (A) ≤ NG (C) by Lemma
2.2(8), contrary to Claim (2). Hence we have (3).
Final contradiction. First we show that A and B are σi -groups. Indeed, since R is a σi -group
by Claim (3), R ≤ NG (O σi (A)) by Lemma 2.2(5). But Oσi (A) is a characteristic subgroup of A,
so Oσi (A) = 1 by Claim (2). Hence A is a σi -group. Similarly one can get that B is a σi -group.
Therefore hA, Bi ≤ Oσi (G) by Lemma 2.2(9). Hence hA, Bi is σ-subnormal in G by Lemma 2.2(4).
But this contradicts the choice of G. The proposition is proved.
Corollary 2.6 (H. Wielandt [27, A, 14.4]). The set of all subnormal subgroups of G forms a
subllatice of the lattice of all subgroups of G.

3

Some properties of σ-soluble and σ-nilpotent groups

The direct calculations show that the following lemma is true
Lemma 3.1. The class Sσ is closed under taking direct products, homomorphic images and
subgroups. Moreover, the extension of a σ-soluble group by a σ-soluble group is a σ-soluble group.
A subgroup H of G is said to be: a Hall Π-subgroup of G if |H| is Π-number and |G : H| is
a σ-Hall subgroup of G if H is a Hall Π-subgroup of G for some Π ⊆ σ. If G has
a complete Hall σ-set H = {1, H1 , . . . , Ht } such that Hi Hj = Hj Hi for all i, j, then we say that
{H1 , . . . , Ht } is a σ-basis of G.
Π′ -number;

9

Let A, B and R be subgroups of G. Then A is said to R-permute with B [28] if for some x ∈ R
we have AB x = B x A.
The following proposition gives the basic properties of σ-soluble groups.
Proposition 3.2. Let G is σ-soluble. Then:
(i) |G : M | is σ-primary for every maximal subgroup M of G.
(ii) For every σi ∈ σ(G), G has a maximal subgroup M such that |G : M | is a σi -number.
(iii) G has a σ-basis {H1 , . . . , Ht } such that for each i 6= j every Sylow subgroup of Hi G-permutes
with every Sylow subgroup of Hj .
(iv) For any Π, G has a Hall Π-subgroup and every σ-Hall subgroup of G G-permutes with every
Sylow subgroup of G.
(v) For any Π, G has a Hall Π-subgroup E, every Π-subgroup of G is contained in some conjugate
of E and E G-permutes with every Sylow subgroup of G.
Proof. (i) If H/MG is a chief factor of G, then |(G/MG ) : (M/MG )| = |G : M | divides |H/MG |,
so it is σ-primary.
(ii) Let R be a minimal normal subgroup of G. Then R is a σk -group, for some k. If R is not a Hall
σk -subgroup of G, then G/R is a σ-soluble group such that σ(G/R) = σ(G). Hence by induction,
for every σi ∈ σ(G/R), G/R has a maximal subgroup M/R such that |(G/R) : (M/R)| = |G : M | is
a σi -number. Now suppose that R is a Hall σk -subgroup of G and let U be a complement to R in
G. Then G has a maximal subgroup M such that |G : M | divides |R|, so it is a σk -number. On the
other hand, for every σi 6= σk , σi ∈ σ(G/R) and so as above we get that G has a maximal subgroup
M such that |G : M | is a σi -number.
(iii), (iv), (v) See Theorems A and B in [9]. The proposition is proved.
Let H/K be a chief factor of G. Then we say that H/K is σ-central in G if the semidirect
product (H/K) ⋊ (G/CG (H/K)) is σ-primary. Otherwise, it is called σ-eccentric in G.
The following lemma is well-known (see for example Lemma 3.29 in [29]).
Lemma 3.3. Let R be an abelian minimal normal subgroup of G such that G = RM for a
maximal subgroup M of G. Then G/MG ≃ R ⋊ (G/CG (R)).
It is well-known that a nilpotent group can be characterized as the group in which each subgroup,
or each Sylow subgroup, or each maximal subgroup is subnormal. The following result demonstrates
that there is a quite similar relation between σ-nilpotency and σ-subnormality.
Proposition 3.4. Any two of the following conditions are equivalent:
(i) G is σ-nilpotent.
(ii) Every chief factor of G is σ-central in G.
(iii) G has a complete Hall σ-set H such that every member of H is σ-subnormal in G.
10

(iv) Every subgroup of G is σ-subnormal in G.
(v) Every maximal subgroup of G is σ-subnormal in G.
Proof. Since (i) ⇒ (iii) and (iv) ⇒ (v) are clear, it is enough to prove the implications (i) ⇒
(ii), (ii) ⇒ (v), (iii) ⇒ (i), (i) ⇒ (iv) and (v) ⇒ (i).
(i) ⇒ (ii) For every chief factor H/K of G, where H ≤ Hi , we have that (H/K) ⋊ (G/CG (H/K))
is a π(Hi )-group. Hence (H/K) ⋊ (G/CG (H/K)) is σ-primary. Now applaying the Jordan-Hölder
theorem [27, A, 3.2], we get that every chief factor of G is σ-central.
(ii) ⇒ (v) Let M be a maximal subgroup of G. Assume that MG 6= 1. It is clear that the
hypothesis holds for G/MG , so M/MG is σ-subnormal in G/MG by induction. Hence M is σsubnormal in G by Lemma 2.2(3). Now assume that MG = 1. By [27, A, 15.2], either G has
a unique minimal normal subgroup R or G has exactly two minimal normal subgroups R and N
and the following hold: R and N are isomorphic non-abelian groups, R ∩ M = 1 = N ∩ M and
CG (R) = N . Let C = CG (R). Suppose that R is abelian. Then C = R by [27, A, 15.2], so in this
case we have G ≃ G/MG ≃ R ⋊ (G/CG (R)) is σ-primary by Lemma 3.3(ii). Hence, for some σi ∈ σ,
G is a σi -group. But then M is σ-subnormal in G. Similarly we get that M is σ-subnormal in G in
the case when C = 1. Finally, if C = N , then G/N ≃ M ≃ G/R is σ-primary. It follows that G is a
σi -group. Thus M is σ-subnormal in G.
(iii) ⇒ (i) This follows from Proposition 2.5 since every member of H is σ-nilpotent.
(i) ⇒ (iv) This follows from the implications (i) ⇒ (ii), (ii) ⇒ (v) and the evident fact that every
subgroup of a σ-nilpotent group is σ-nilpotent (see Lemma 2.1).
(v) ⇒ (i) Assume that this is false and let G be a counterexample of minimal order.
First note that G is σ-soluble. Indeed, for any maximal subgroup M of G, G/MG is σ-primary
and so G/MG is σ-soluble. But then G/Φ(G) is a subdirect product of some σ-soluble groups, which
implies that G/Φ(G) is σ-soluble by Lemma 3.1. Hence G is σ-soluble. By Proposition 3.2, G has a
a complete Hall σ-set H = {1, H1 , . . . , Ht }.
Let H = Hi and R be a minimal normal subgroup of G. We show that H is normal in G.
Assume that is false. By Lemma 2.2(6), the hypothesis holds for G/R, so HR/R is normal in G/R
by the choice of G. Hence we can assume that R  H, so R ∩ H = 1 since G is σ-soluble. If
G has a minimal normal subgroup N 6= R, then as above we get that HN is normal in G and so
RH ∩ N H = H(R ∩ N H) = H(R ∩ N ) = H is normal in G. Therefore R is the unique minimal
normal subgroup of G. Moreover, in view of Lemma 2.1, we have R  Φ(G) since HR/R ≃ H is
σ-nilpotent and HR is normal in G. Let M be a maximal subgroup of G such that G = RM . Then
MG = 1. But M is σ-subnormal in G by hypothesis and G ≃ G/MG is σ-primary, which implies
that G is a σi -group, for some σi ∈ σ. Therefore H = G. This contradiction shows that (v) ⇒ (i).
The proposition is proved.
We say that G is Π-closed if OΠ (G) is a Hall Π-subgroup of G.
11

Lemma 3.5. Let H be a normal subgroup of G. If H/H ∩ Φ(G) is Π-closed, then H is Π-closed.
Proof. See the proof of Lemma 2.5 in [1].
The integers n and m are called σ-coprime if σ(n) ∩ σ(m) = ∅.
Lemma 3.6. If a σ-soluble group G has three Π-closed subgroups A, B and C whose indices
|G : A|, |G : B|, |G : C| are pairwise σ-coprime, then G is Π-closed.
Proof. Suppose that this lemma is false and let G be counterexample of minimal order. Let N
be a minimal normal subgroup of G. Then the hypothesis holds for G/N , so G/N is Π-closed by the
choice of G. Therefore N is not a Π-group. Moreover, N is the unique minimal normal subgroup of
G and, by Lemma 3.5, N  Φ(G). Hence CG (N ) ≤ N . Since G is σ-soluble by hypothesis, N is a
σi -group for some i. Then σi ∈ Π′ .
Since |G : A|, |G : B|, |G : C| are pairwise σ-coprime, there are at least two subgroups, say A
and B, such that N ≤ A ∩ B. Then OΠ (A) ≤ CG (N ) ≤ N , so OΠ (A) = 1. But by hypothesis, A is
Π-closed, hence A is a Π′ -group. Similarly we get that B is a Π′ -group and so G = AB is a Π′ -group.
But then G is Π-closed. This contradiction completes the proof of the lemma.
Lemma 3.7 (See [12, III, 5.2] ). If G is a Schmidt group, then G = P ⋊ Q, where P = GN is
a Sylow p-subgroup of G and Q = hxi is a cyclic Sylow q-subgroup of G. Moreover, hxq i ≤ Φ(G),
P/Φ(P ) is a chief factor of G, P is of exponent p or exponent 4 (if P is a non-abelian 2-group) and
Φ(P ) = 1 if P is abelian.
Proposition 3.8. Let G be a σ-soluble group. Suppose that G is not Π-closed but all proper
subgroups of G are Π-closed. Then G is a Π′ -closed Schmidt group.
Proof. Suppose that this proposition is false and let G be a counterexample of minimal order.
Let R be a minimal normal subgroup of G and {H1 , . . . , Ht } a complete Hall σ-set of G. Without
loss of generality we can assume that Hi is a σi -group for all i = 1, . . . , t.
(1) |σ(G)| = 2. Hence G = H1 H2 .
It is clear that |σ(G)| > 1. Suppose that |σ(G)| > 2. Then, since G is σ-soluble, there are
maximal subgroups M1 , M2 and M3 whose indices |G : M1 |, |G : M2 | and |G : M3 | are pairwise
σ-coprime. But the subgroups M1 , M2 and M3 are Π-closed by hypothesis. Hence G is Π-closed by
Lemma 3.6, a contradiction. Thus |σ(G)| = 2.
Without loss of generality we can assume that σ2 ∈ Π. Then Π ∩ σ(G) = {σ2 }.
(2) If either R ≤ Φ(G) or R ≤ H2 , then G/R is a Π′ -closed Schmidt group.
Lemma 3.5 and the choice of G imply that G/R is not Π-closed. On the other hand, every maximal
subgroup M/R of G/R is Π-closed since M is Π-closed by hypothesis. Hence the hypothesis holds
for G/R. The choice of G implies that G/R is a Π′ -closed Schmidt group.
(3) Φ(G) = 1, R is the unique minimal normal subgroup of G and R ≤ H1 .
Suppose that R ≤ Φ(G). Then Ḡ = G/R is a Π′ -closed Schmidt group by Claim (2), so Ḡ =
12

H̄1 ⋊ H̄2 = P̄ ⋊ Q̄, where H̄1 = P̄ is a p-group and H̄2 = Q̄ is a q-group for some primes p and q by
Lemma 3.7. Therefore, in fact, G is not p-nilpotent but every maximal subgroup of G is p-nilpotent.
Hence G is Π′ -closed Schmidt group by [12, IV, 5.4], a contradiction. Therefore Φ(G) = 1.
Now assume that G has a minimal normal subgroup L 6= R. Since Φ(G) = 1, there are maximal
subgroups M and T of G such that LM = G and RT = G. By hypothesis, M and T are Π-closed.
Hence G/L ≃ LM/L ≃ M/M ∩ L is Π-closed. Similarly, G/R is Π-closed and so G ≃ G/L ∩ R is
Π-closed, a contradiction. Hence R is the unique minimal normal subgroup of G and R ≤ H1 .
Final contradiction. In view of Claim (3), CG (R) ≤ R and R ≤ H1 . Hence |H2 | is a prime and
RH2 = G since every proper subgroup of G is Π-closed. Therefore R = H1 , so R is not abelian since
G is not a Π′ -closed Schmidt group. It is clear that for any prime p dividing |R| there is a Sylow
p-subgroup P of G such that P H2 = H2 P by Lemma 3.2(iv). But H2 P < G, so H2 P = H2 ⋊ P .
Therefore R ≤ NG (H2 ) and thereby G = R × H2 = H1 × H2 is σ-nilpotent. This final contradiction
completes the proof.
We say that G is σ-fiber if |σ(G)| = |π(G)|.
Corollary 3.9. Suppose that G is not σ-nilpotent but every proper subgroup of G is σ-nilpotent.
If G is σ-soluble, then G is a σ-fiber Schmidt group.
Proof. It is clear that G is σ-nilpotent if and only if G is Π-closed for all Π ⊆ σ. Hence, for
some Π, G is not Π-closed. On the other hand, every proper subgroup of G is Π-closed. Hence G is
a Schmidt group by Proposition 3.8 and clearly |σ(G)| = π(G)|.

4

Proof of Theorem 1.2

In this section, we need the following
Lemma 4.1 (See Lemmas 2.8, 3.1 and Theorem B(1) in [1]). Let H, K and R be subgroups of
a σ-soluble group G. Suppose that H is σ-quasinormal in G and R is normal in G. Then:
(1) If H ≤ E ≤ G, then H is σ-quasinormal in E.
(2) The subgroup HR/R is σ-quasinormal in G/R.
(3) If R ≤ H and H/R is σ-quasinormal in G/R, then H is σ-quasinormal in G.
(4) H is σ-subnormal in G.
(5) If H is a σi -group, then Oσi (G) ≤ NG (H) .
Lemma 4.2. The following statements hold:
(1) mσ (G) ≤ mσq (G).
(2) If M is a non-σ-subnormal maximal subgroup of G, then mσ (M ) ≤ mσ (G) − 1.
(3) If R is a normal subgroup of G, then mσ (G/R) ≤ mσ (G).
13

(4) If R is a normal subgroup of G and G is σ-soluble, then mσq (G/R) ≤ mσq (G).
Proof. (1) This follows from Lemma 4.1(4).
(2) Since M is not σ-subnormal in G, mσq (G) > 1. Moreover, for n = mσ (G), each (n − 1)maximal subgroup of M is σ-subnormal in M by Lemma 2.2(1). Hence mσ (M ) ≤ mσ (G) − 1.
(3) If each maximal chain of G/R has length r < mσ (G), it is clear. Otherwise, this follows from
Lemma 2.2(3)(6).
(4) This is a corollary of Lemma 4.1(3)
The lemma is proved.
The following properties of the rank of a soluble group are useful in our proof.
Lemma 4.3 (See [12, VI, Lemma 5.3]). Let G be soluble. Then:
(1) r(G/R) ≤ r(G) for all normal subgroups R of G
(2) r(E) ≤ r(G) for all subgroups E of G
(3) r(A × B) = Max{r(A), r(B)}.
Lemma 4.4 (See Huppert [21, Lemma 11]). Let G be soluble and R be a minimal normal
subgroup of G. Let H be a minimal supplement to CG (R) in G. Then H ∩ R = 1.
Lemma 4.5. The following statements hold:
(i) If each n-maximal subgroup of G is σ-subnormal and n > 1, then each (n − 1)-maximal
subgroup is σ-nilpotent.
(ii) If each n-maximal subgroup of G is σ-subnormal, then each (n + 1)-maximal subgroup is
σ-subnormal.
Proof. (i) Let H be an (n − 1)-maximal subgroup of G and K a maximal subgroup of H.
Then K is an n-maximal subgroup of G, so it is σ-subnormal in G. Then, by Lemma 2.2(1), K is
σ-subnormal in H. Therefore each maximal subgroup of H is σ-subnormal in H. It follows from
Proposition 3.4 that H is σ-nilpotent.
(ii) Let L ≤ M ≤ G, where M is an n-maximal subgroup of G and L is a maximal subgroup
of M . If n = 1, G is σ-nilpotent and so L is σ-subnormal in G by Proposition 3.4. On the other
hand, in the case when n > 1 Statement (i) implies that each (n − 1)-maximal subgroup of G in
σ-nilpotent. Then M is σ-nilpotent by Lemma 2.1, so L is σ-subnormal in G by Lemma 2.2(4).
The lemma is proved.
Proof of Theorem 1.2. Suppose that this theorem is false and let G be a counterexample of
minimal order. Let H = {1, H1 , . . . , Ht } be a complete Hall σ-set of G. Then t > 1.
(i) Suppose that this is false. Let R be a minimal normal subgroup of G and |R| = pm . Without
loss of generality we can assume that R ≤ H1 . Let n = mσq (G). Since G is not σ-nilpotent, some
maximal subgroup M of G is not σ-subnormal in G by Proposition 3.4 and so M is not σ-quasinormal
14

in G by Lemma 4.1(4). Thus n > 1.
(1) r(G/R) ≤ n + r − 2.
Assume that r(G/R) > n + r − 2. Note that {H1 R/R, . . . , Ht R/R} is a complete Hall σ-set of
G/R and r(Hi R/R) = r(Hi /Hi ∩ R) ≤ r(Hi ) ≤ r for all i = 1, . . . , t by Lemma 4.3(1). Assume
that G/R is σ-nilpotent. Then G/R = (H1 R/R) × · · · × (Ht R/R), so r(G/R) ≤ r ≤ n + r − 2 since
n > 1 by Lemma 4.3(3). This contradiction shows that G/R is not σ-nilpotent. Moreover, G/R is
σ-soluble by Lemma 3.1. By Lemma 4.1(2)(3), mσq (G/R) ≤ mσq (G) = n. The Choice of G implies
that r(G/R) ≤ mσq (G/R) + r − 2 ≤ n + r − 2, a contradiction. Hence we have (1).
(2) m > n + r − 2. Hence R is the only minimal normal subgroup of G.
First note that in view of the Jordan-Hölder theorem, Claim (1) and the choice of G we have
m > n + r − 2. If G has a minimal normal subgroup N 6= R, then r(G/N ) ≤ n + r − 2 by Claim (1),
so in view of the G-isomorphism R ≃ RN/N we get that m ≤ n + r − 2, a contradiction. Hence R
is the only minimal normal subgroup of G.
(3) If M is a proper subgroup of G, then r(M ) ≤ n + r − 2.
It is enough to consider the case when M is a maximal subgroup of G. Assume that r(M ) >
n + r − 2. Then M is not σ-nilpotent (see the proof of Claim (1)). Therefore n > 2 by Lemmas
4.1(1) and Proposition 3.4. Moreover, since G is σ-soluble, M possesses a complete Hall σ-set
{M1 , . . . , Mt } such that Mi = Hi ∩ M for all i = 1, . . . , t by Lemma 3.2(v). Hence r(Mi ) ≤ r(Hi ) ≤ r
for all i = 1, . . . , t by Lemma 4.3(2). Therefore, M satisfies the hypothesis, with n − 1 instead of
n, by Lemmas 4.1(1) and so the choice of G implies that r(M ) ≤ n − 1 + r − 2 ≤ n + r − 2, a
contradiction. Hence we have (3).
(4) R  Φ(G).
Suppose that R ≤ Φ(G). Then for a minimal supplement H to CG (R) in G we have H ∩ R = 1
by Lemma 4.4, so RH 6= G and R is a minimal normal subgroup of RH. But Claim (3) implies that
r(RH) ≤ n + r − 2 and so m ≤ n + r − 2, contrary to Claim (2). Hence we have (4).
Final contradiction for (i). Claim (4) implies that there is a maximal subgroup M of G such that
G = RM and H2 ≤ M . Then MG = 1 by Claim (2), so CG (R) = CG (R)∩RM = R(CG (R)∩M ) = R.
Let H2 = Ms be a member of a maximal chain 1 = Ml < Ml−1 < . . . < M1 < M0 = M of M . Then
R is an l-maximal subgroup of G. First suppose that l > n − 1. Assume also that H2 ≤ Mn−1 .
By hypothesis Mn−1 is σ-quasinormal in G. Hence H2x ≤ Mn−1 for all x ∈ G. It follows that
(H2 )G ≤ MG = 1, so H2 = 1, a contradiction. Therefore n ≤ s, that is, for the n-maximal subgroup
H = Mn−1 of G contained in H2 we have H 6= 1. Then H ≤ Oπ(H2 ) (G) by Lemma 2.2(9). But
R ∩ Oπ(H2 ) (G) = 1 since R ≤ H1 , so 1 < H ≤ Oπ(H2 ) (G) ≤ CG (R) = R, a contradiction.
Therefore n − 1 ≤ l, so M possesses a maximal chain 1 = Mk < Mk−1 < · · · M1 < M0 = M ,
where k < n. Then R is a k-maximal subgroup of G. Therefore every l-maximal subgroup of R is a
(k + l)-maximal subgroup of G. Let R0 be a minimal normal subgroup of H1 contained in R with
15

|R0 | = pa . Let L be an (n − k)-maximal subgroup of R with |L| = pb such that L ≤ R0 in the case
when b < a and R0 ≤ L if a ≤ b. Then L is an n-maximal subgroup of G, so L is σ-quasinormal in
G.
First suppose that L ≤ R0 . Then
LG = LH1 NG (L) = LH1 ≤ (R0 )G = R
by Lemma 4.1(5). Hence R0 = R. Then m = a ≤ r and so m ≤ r + n − 2 since n > 1, contrary to
(2). Thus R0 ≤ L, so
H N (L)
N (L)
R0G = R0 1 G
= R0 G ≤ L,
which implies that L = R, a contradiction also. Hence Assertion (i) is true.
(ii) Let n = mσ (G). Suppose that n > lσ (G). Then n > 1. Indeed, if n = 1, G is σ-nilpotent
by Proposition 3.4 and so lσ (G) = 1 = mσ (G), a contradiction. The choice of G and Lemma 4.2(3)
imply that lσ (G/Fσ (G)) ≤ mσ (G/Fσ (G)) ≤ n. Hence Fσ (G)  Φ(G) by Lemma 2.3(2). Let M
be a maximal subgroup of G such that Fσ (G)M = G. Then lσ (G/Fσ (G)) = lσ (M Fσ (G)/Fσ (G) =
lσ (M/M ∩ Fσ (G)) = lσ (G) − 1. Lemma 2.1 implies that M ∩ Fσ (G) ≤ Fσ (M ). Therefore lσ (G) ≤
lσ (M ) + 1. Note that mσ (M ) ≤ mσ (G) − 1. Indeed, since n > 1, each (n − 1)-maximal subgroup
of M is σ-quasinormal in G. Hence each (n − 1)-maximal subgroup of M is σ-quasinormal in M by
Lemma 4.1(1). Hence mσ (M ) ≤ n − 1. But the choice of G we have lσ (M ) ≤ mσ (M ), and then
lσ (G) ≤ lσ (M ) + 1 ≤ mσ (M ) + 1 ≤ mσ (G). This contradiction completes the proof of (ii).
(iii) Suppose that mσ (G) < |π(G)|. Let P1 , . . . , Pn be a Sylow basis of G and H a complete Hall
σ-set of G. Then for any i, i = 1 say, we have P1 < P1 P2 < · · · < P1 P2 · · · Pn = G, so P1 is at
least an (n − 1)-maximal subgroup of G. Therefore P1 is σ-subnormal in G by Lemma 4.5(ii) since
mσ (G) < |π(G)|. Hence every Sylow subgroup of G is σ-subnormal in G and so every member of H is
σ-subnormal in G by Proposition 2.5. But then G is σ-nilpotent by Proposition 3.4, a contradiction.
Hence |π(G)| ≤ mσ (G).
The theorem is proved.

5

Proofs of Theorem 1.4 and Corollaries 1.7 and 1.8

Proof of Theorem 1.4. Let R be a minimal normal subgroup of G.
(i) Suppose that this assertion is false and let G be a counterexample of minimal order. First
note that G/R is σ-soluble. Indeed, if R is a maximal subgroup or a 2-maximal subgroup of G, it
is clear. Otherwise, the hypothesis holds for G/R by Lemma 2.2(6), so the choice of G implies that
G/R is σ-soluble. Hence R is the unique minimal normal subgroup of G by Lemma 3.1 and R is not
σ-primary. Hence R is not abelian.
Let p be any odd prime dividing |R| and Rp a Sylow p-subgroup of R. The Frattini argument
implies that there is a maximal subgroup M of G such that NG (Rp ) ≤ M and G = RM . It is clear
16

that MG = 1, so M is not σ-subnormal in G since G/MG ≃ G is not σ-primary. Let D = M ∩ R.
Then Rp is a Sylow p-subgroup of D.
(1) D is not nilpotent. Hence D  Φ(M ) and D is not a p-group.
Assume that D is a nilpotent. Then Rp is normal in M . Hence Z(J(Rp )) is normal in M . Since
MG = 1, it follows that NG (Z(J(Rp ))) = M and so NR (Z(J(Rp ))) = D is nilpotent. This implies
that R is p-nilpotent by Glauberman-Thompson’s theorem on the normal p-complements. But then
R is a p-group, a contradiction. Hence we have (1).
(2) R < G.
Suppose that R = G is a simple non-abelian group. Assume that some proper non-identity
subgroup A of G is σ-subnormal in G. Then there is a subgroup chain A = A0 ≤ A1 ≤ · · · ≤ An = G
such that either Ai−1 E Ai or Ai /(Ai−1 )Ai is σ-primary for all i = 1, . . . , t. Without loss of generality,
we can assume that M = An−1 < G. Then MG = 1 since G = R is simple, so G ≃ G/1 is σ-primary,
a contradiction. Hence every proper σ-subnormal subgroup of G is trivial.
Let P be a Sylow p-subgroup of G, where p is the smallest prime dividing |G|, and let L be a
maximal subgroup of G containing P . Then, in view of [12, IV, 2.8], |P | > p. Let V be a maximal
subgroup of P . If |V | = p, then P is abelian, so P < L by [12, IV, 7.4]. Hence there is a 3maximal subgroup W of G such that V ≤ W . But then some proper non-identity subgroup of G is
σ-subnormal in G by hypothesis, a contradiction. Therefore |V | > p, which again implies that some
proper non-identity subgroup of G is σ-subnormal in G. This contradiction shows that we have (2).
(3) M is σ-soluble.
Let L < T < M , where L is a maximal subgroup of T and T is a maximal subgroup of M . Since
M is not σ-subnormal in G, either L or T is σ-subnormal in G and so it is σ-subnormal in M by
Lemma 2.2(1). Hence the hypothesis holds for M , so M is σ-soluble by the choice of G.
(4) M = D ⋊ T , where T is a maximal subgroup of M of prime order.
In view of Claim (1), there is a maximal subgroup T of M such that M = DT . Then G =
RM = R(DT ) = RT and so, in view of (2), T 6= 1. Assume that |T | is not a prime and let V be
a maximal subgroup of T . Since M is not σ-subnormal in G, at least one of the subgroups T or V
is σ-subnormal in G by hypothesis. Claim (3) implies that V and T are σ-soluble. Consider, for
example, the case when V is σ-subnormal in G. Since V 6= 1 and V is σ-soluble, for some i we have
Oσi (V ) 6= 1. But Oσi (V ) ≤ Oσi (G) by Lemma 2.2(9), so R is σ-primary, a contradiction. Hence |T |
is a prime, so M = D ⋊ T .
Final contradiction for (i). Since T is a maximal subgroup of M and it is cyclic, M is soluble
and so |D| is a prime power, which contradicts (1). Hence Assertion (i) is true.
(ii) Suppose that this false. Then 2 ∈ π(G). Part (i) implies that G is σ-soluble. Let H =
{H1 , . . . , Ht } be a σ-basis of G. Without loss of generality we can assume that H1 is a σ1 -group and
2 ∈ π(H1 ). Then H1 is not soluble, so |π(H1 )| > 1. Let p ∈ π(H1 ).
17

(1) t = 2 and H2 is a Sylow subgroup of G.
By Proposition 3.2(v), G has a Hall σ1′ -subgroup E and E permutes with some Sylow p-subgroup
P of G for each p ∈ π(H1 ). It is clear that EP < G. We show that P E is soluble. In fact, if P E
is σ-nilpotent, then P E = P × E, where 2 ∤ |E|, so P E is soluble. Now assume that P E is not
σ-nilpotent. Then the hypothesis holds for P E, so P E is soluble by the choice of G. Hence P E has
a Sylow basis P = {P, P1 , . . . Pn }. If t > 2 or H2 is not a Sylow subgroup of G, then every member
of P is at least 3-maximal subgroup of G. Hence every member of P is σ-subnormal in G by Lemma
4.5(ii). This shows that every Sylow subgroup of G is σ-subnormal in G. Therefore all members of
H are normal in G by Lemma 2.2(9), which implies that G is σ-nilpotent. This contradiction shows
that we have (1).
(2) Oσ1 (G) 6= 1.
Suppose that this is false. Since G is σ-soluble, R is σ-primary. Hence R ≤ H2 . Let P be a
Sylow p-subgroup of H1 and P ≤ M , where M is a maximal subgroup of H1 . Since H1 is not soluble,
P < M < H1 < G by [12, IV, 7.4]. Therefore there is a 3-maximal subgroup W of G such that
P ≤ W ≤ H1 . Then W is σ-subnormal in G, so 1 < W ≤ Oσ1 (G) by Lemma 2.2(9), a contradiction.
Hence we can assume that R ≤ Oσ1 (G).
(3) H2 is normal in G.
First suppose that R is a p-group for some prime p and let Q be a Sylow q-subgroup of H1 , where
q 6= p. By Proposition 3.2, there is x ∈ G such that H2 Qx = Qx H2 . Hence we have a subgroup
chain H2 < H2 Qx < RH2 Qx < G. It follows from Lemma 4.5(ii) that H2 is σ-subnormal in G, so it
is normal in G by Lemma 2.2(9).
Now assume that R is not abelian. Then, for any odd prime p dividing |R|, the subgroup R
is not p-nilpotent. Hence by Glauberman-Thompson’s theorem on the normal p-complement, we
have that P < NR (Z(J(P ))) < R, where P is a Sylow p-subgroup of R. Since (|R|, H2 )) = 1, H2
normalizes some Sylow p-subgroup of R, say P . Hence H2 ≤ NG (Z(J(P ))). But then we have a chain
H2 < H2 P < H2 NR (Z(J(P )))) < G, so H2 is σ-subnormal in G by Lemma 4.5(ii). Consequently
H2 is normal in G by Lemma 2.2(9).
Final contradiction for (ii) . Let P be a Sylow p-subgroup of H1 and V be a maximal subgroup of
P . Since |π(|H1 |)| > 1, V is σ-subnormal in G by Lemma 4.5(ii). Suppose that P  Oσ1 (G). Then
P  R and P is not σ-subnormal in G by Lemma 2.2(9), so P is cyclic by Proposition 2.5. Hence
R ∩ P ≤ Φ(R), and so R is p-nilpotent by the Tate theorem [12, IV, 4.7]. But then R is a p-group
or a p′ -group. Assume that R is a p-group. Then G/R is not σ-nilpotent. Otherwise, H1 /R ⊳ G/R.
It follows from Claim (3) that G = H1 × H2 , which contradicts mσ (G) > 1. Hence G/R is not
σ-nilpotent, and so 1 < mσ (G/R). But then G/R satisfies the hypothesis by Lemma 2.2(6). Hence
G/R is soluble by induction. Consequently, G is soluble, a contradiction. Now assume that R is a
p′ -group. Then by the Frittini argument, P normalizes some Sylow q-subgroup Q of R, where q 6= p
18

divides R. Hence we have a subgroup chain P < P Q < H2 P Q < G, so P is σ-subnormal in G by
Lemma 4.5(ii). This shows that every Sylow subgroup of H1 is σ-subnormal in G. It follows Lemma
2.2(9) that H1 is σ-subnormal in G. Thus H1 is normal in G by Lemma 2.2(9), so G = H1 × H2 is
σ-nilpotent. It follows from Proposition 3.4 that mσ (G) = 1. This final contradiction completes the
proof of (ii).
The theorem is proved.
Proof of Corollary 1.7. The sufficiency is clear. We only need to prove the necessity. By
Theorem 1.4, G is σ-soluble. Since mσq (G) = 2, G is not σ-nilpotent by Proposition 3.4. On the
other hand, if M is a maximal subgroup of G, then every maximal subgroup of M is σ-subnormal
in G and so it is σ-subnormal in M by Lemma 2.2(1). Therefore M is σ-nilpotent by Proposition
3.4. Hence G is a Schmidt group such that |π(G)| = |σ(G)| by Corollary 3.9. Then by Lemma 3.7,
G = P ⋊ Q, where P = GN is a Sylow p-subgroup of G and Q = hxi is a cyclic Sylow q-subgroup of
G. Moreover, P is of exponent p or exponent 4 (if P is a non-abelian 2-group) and P/Φ(P ) is a chief
factor of G. If Φ(P ) 6= 1, then there exists a maximal subgroup M of G such that Q < M < G. By
the hypothesis, Then Q is σ-subnormal in G by Lemma 4.5(ii). It follows from Lemma 2.2(9) that
Q is normal in G. This contradiction shows that every Sylow subgroup of G is abelian.
Proof of Corollary 1.8. In view of Corollary 1.7 and Lemma 3.7, G = P ⋊Q is a Schmidt group
with |π(G)| = |σ(G)|, where P is a minimal normal subgroup of G and Q is cyclic. Let |P | = pn
and |Q| = q m . Suppose that n > 1. Then G has a 2-maximal subgroup L such that |G : L| = pq.
By hypothesis L is σ-quasinormal in G, so it is, in fact, S-quasinormal in G and hence LQ = QL
is a subgroup of G with G : LQ| = p. But then LQ ∩ P is normal in LQ and |P : (LQ ∩ P )| = p,
so LQ ∩ P is normal in G and hence LQ ∩ P = 1 in view of the minimality of P . It follows that
|P | = p, a contradiction. Hence |P | = p, so G is supersoluble. The corollary is proved.

6

Proof of Theorem 1.10

Lemma 6.1. Suppose that G is σ-soluble and let H = {H1 , . . . , Ht } be a σ-basis of G. If Hi forms
an irreducible pair with Hj , then Hj is an elementary abelian Sylow subgroup of G.
Proof. Without loss of generality we can assume that G = Hi Hj . By Proposition 3.2(iv), for
each prime p dividing |Hj |, there is a Sylow p-subgroup P of Hj such that Hi P = P Hi . Hence
G = Hi P . Let R be a minimal normal subgroup of G. If R ≤ P , then the maximality of Hi implies
that R = P is elementary. On the other hand, if R ≤ Hi , then Hi /R is a maximal subgroup of G/R,
and so P ≃ P R/R is elementary by induction.
The following lemma can be proved similarly to [27, I, Proposition 4.16].
Lemma 6.2. Suppose that G 6= 1 is σ-soluble and let L be a subgroup of G. Then for each
σ-basis L = {L1 , . . . , Lr } of L, there is a σ-basis H = {H1 , . . . , Ht } of G such that Li = L ∩ Hi for
all i = 1, . . . r.
19

Lemma 6.3. Suppose that G is σ-soluble and let H = {H1 , . . . , Hn } be a σ-basis of G. If H1
forms an irreducible pair with Hix for all i > 1 and x ∈ G such that H1 Hix = Hix H1 , then every
subgroup K of G containing H1 is a σ-Hall subgroup of G.
Proof. Suppose that this is false. Without loss of generality we can assume that Hi is a σi -group.
Let K = {H1 , K2 , . . . , Kr } be a σ-basis of K. By Lemma 6.2 , there is a σ-basis {H1 , H2x2 , . . . , Hnxn }
x
x
of G such that Ki = K ∩ Hixi for all i = 2, . . . , r. Hence H1 Hj j ∩ K = H1 (Hj j ∩ K) = H1 Kj = Kj H1
x
x
and so there is a subgroup chain H1 ≤ H1 Kj ≤ H1 Hj j . The maximality of H1 in H1 Hj j implies
x
that Kj = Hj j . Thus K is a σ-Hall subgroup of G.
Lemma 6.4. Suppose that G is σ-soluble and let K be a subgroup of G. If every subgroup of G
containing K is a σ-Hall subgroup of G, then K is a k-maximal subgroup of G, where k = |σ(|G : K|)|,
and K is not a r-maximal subgroup of G for all r > k.
Proof. Let {H1 , . . . , Ht } be a σ-basis of G. The assertion follows from the fact that in any
maximal chain K = Mk < Mk−1 < . . . < M1 < M0 = G, |Mi : Mi+1 | is an order of some Hi since
both Mi and Mi+1 are σ-Hall subgroups of G. The lemma is proved.
Proof of Theorem 1.10. Let P1 , . . . , Pn be a Sylow basis of G and {H1 , . . . , Ht } a complete
Hall σ-set of G We can assume without loss of generality that each Pi is contained in some Hj and
Pi is a pi -group.
Necessity. First note that if G is σ-nilpotent, then mσ (G) = |π(G)| = 1, and so G is a p-group
for some prime p. Now we show, assuming that G is not σ-nilpotent, then G is a group of type (ii).
Assume that this is false and let G be a counterexample of minimal order. Let R be a minimal
normal subgroup of G. Without loss of generality we can assume that R ≤ H1 and Hk is a σik -group
for all k = 1, . . . , t.
(1) If G/R is not σ-nilpotent, then G/R is a group of the type (ii).
Suppose that G/R is not σ-nilpotent. We show that the hypothesis holds for G/R. Indeed, if
R < Pi , it is clear. We may, therefore, assume that R = Pi . Then R has a complement M in G such
that G = R ⋊ M . Since |π(M )| = n − 1, M satisfies the same assumptions as G, with n − 1 replacing
n, by Lemma 2.2(1). The choice of G implies that G/R ≃ M is a group of the type (ii).
(2) If Vi is a maximal subgroup of Pi , then Vi is σ-subnormal in G. Hence every non-σ-subnormal
Sylow subgroup of G is cyclic.
Since P1 , . . . , Pn is a Sylow basis of G, Vi is at m-maximal subgroup of G, where m > n. Hence
Vi is σ-subnormal in G by Lemma 4.5(ii) Therefore, if Pi is not σ-subnormal in G, then it is cyclic
by Proposition 2.5.
(3) If R is the only minimal normal subgroup of G, then each Sylow subgroup Pi of Hk has prime
order and it is not σ-subnormal in G for all k > 1.
Indeed, let V be a maximal subgroup of Pi . Then V is σ-subnormal in G by Claim (2). Hence
20

V ≤ Oσik (G) by Lemma 2.2(9). But since R ≤ H1 , we have that Oσik (G) = 1 and so V = 1.
Moreover, if Pi is σ-subnormal, then Pi ≤ Oσik (G) = 1, a contradiction. Hence we have (3).
(4) For some i, i = 1 say, Pi = P1 is not σ-subnormal in G. Hence P1 forms an irreducible pair
with Pi for all i > 1.
If Pi is σ-subnormal in G for all i = 1, . . . , n, then Hk is normal in G for all k = 1, . . . , t by
Lemma 2.2(9) and Proposition 2.5, which means that G is σ-nilpotent. Hence the first assertion of
(4) is true. Finally, note that if, for example, P1 is not a maximal subgroup of P1 P2 , then the chain
P1 < P1 P2 < · · · < P1 · · · Pn = G can be refined to a maximal chain of G of length n, at least. Hence
P1 is σ-subnormal in G by Lemma 4.5(ii). This contradiction shows that P1 forms an irreducible
pair with Pi for all i > 1.
(5) The following assertions hold.
(a) Pi is elementary abelian for all i > 1. Hence if G possesses at least two non-σ-subnormal nonisomorphic Sylow subgroups, then all non-σ-subnormal Sylow subgroups are of prime order (This
follows from Lemma 6.1 and Claims (2) and (4)).
(b) If P1 ≤ Hk and P1 is not of prime order, then H1 , . . . , Hk−1 , Hk+1 , . . . , Ht are normal in G.
Indeed, since P1 is not σ-subnormal in G, P1 is cyclic by Claim (2). Hence, if i 6= k and Pj ≤ Hi ,
then Pj does not form an irreducible pair with P1 by Claim (a) and Lemma 6.1. Therefore Claim (4)
implies that Pj is σ-subnormal in G. This shows that every Sylow subgroup of Hi is σ-subnormal in
G. Hence Hi is normal in G by Lemma 2.2(9) and Proposition 2.5 for all i 6= k.
(c) If P1 ≤ Hk and V is the maximal subgroup of P1 , then |G : NG (V )| is a σik -number.
If |P1 | is a prime, it is trivial. Assume that |P1 | is not a prime and let i 6= k. Then Hi is normal
in G by Claim (b). On the other hand, Claim (2) implies that V is σ-subnormal in G. Hence
Hi ≤ NG (V ) by Lemma 2.2(5). Hance we have (c).
(6) D is a Hall subgroup of G. Hence D has a complement M in G.
Suppose that this is false and let U be a Sylow pj -subgroup of D such that 1 < U < Pj ≤ Hk .
Lemma 2.1 implies that
(G/N )Nσ = GNσ N/N = DN/N
for any minimal normal subgroup N of G.
Let L be a minimal normal subgroup of G contained in D. Then G/L is a group of type (ii).
Indeed, assume that G/L is σ-nilpotent and so L = D. Then L < Pj and so P1 does not form
an irreducible pair with Pj . Hence L < P1 = Pj by Claim (4). From Claim (5)(b) it follows that
H1 · · · Hk−1 Hk+1 · · · Ht is normal in G, so D ≤ H1 · · · Hk−1 Hk+1 · · · Ht since G/H1 · · · Hk−1 Hk+1 · · · Ht ≃
Hk is σ-nilpotent, a contradiction. Hence G/L is not σ-nilpotent, and so G/L is a group of type
(ii) by Claim (1). Hence D/L is a Hall subgroup of G/L. If U L/L 6= 1, then U L/L is a Sylow pj subgroup of D/L and so U L/L = Pj L/L. Hence Pj ≤ U L ≤ D and so U = Pj . This contradiction
21

shows that U L/L = 1, so U = L. Therefore L < Pj . But then, as above, we get that for any i 6= k
the subgroup Hi is normal in G. Let N be a minimal normal subgroup of G contained in Hi . Then
G/N is not σ-nilpotent since i 6= k, and so DN/N ≃ D is a Hall subgroup of G/N by Claim (1),
which implies that U = Pj since pj 6∈ π(N ) ⊆ π(Hi ). This contradiction completes the proof of (6).
(7) Some Sylow subgroup Pix of G contained in M is not σ-subnormal in G.
Suppose that every Sylow subgroup Pix of G contained in M is σ-subnormal in G. Then M is
σ-subnormal in G by Proposition 2.5. Hence there is a subgroup chain
M = M0 < M1 < · · · < Mr = G
such that either Mi−1 E Mi or Mi /(Mi−1 )Mi is σ-primary for all i = 1, . . . , r. Since G/D =
G/GNσ ≃ M is σ-nilpotent and G is not σ-nilpotent, M 6= G. Hence we can assume without loss
of generality that Mt−1 < G. If Mt−1 is normal in G, then there is a normal maximal subgroup
T of G containing Mt−1 such that G/T is nilpotent since G is soluble. Then D ≤ T , and so
G = DM = DT = T < G, a contradiction. Therefore G/(Mt−1 )G is σ-primary and thereby it is
σ-nilpotent. But then D ≤ (Mt−1 )G , so G = M D ≤ Mt−1 < G. This contradiction shows M is not
σ-subnormal in G. Hence we have (7).
(8) D is nilpotent.
Suppose that D is not nilpotent. Assume that G has a minimal normal subgroup N 6= R. Then
in view of Claim (1), D ≃ D/1 = D/(R ∩ N ) is nilpotent. Therefore R is the unique minimal normal
subgroup of G and R  Φ(G) by Lemma 2.1. It follows that R = CG (R) ≤ D.
Since D/R is nilpotent by Claim (1), there is a normal subgroup E/R of D/R such that R ≤
E < D and D = E ⋊ Pi for some i. Assume that Pi is σik -group. The Frattini argument implies
that for some x ∈ G we have M x ≤ NG (Pi ).
Suppose that for each r 6= k and for each Sylow subgroup P of G contained in M x , where P is
a σir -group, we have [P, Pi ] = 1. Then G/E ≃ Pi M x is σ-nilpotent and so D ≤ E, a contradiction.
Hence G has a Sylow subgroup Pjy satisfying the following conditions: [Pi , Pjy ] 6= 1, Pjy ≤ M x and
Pjy is a σir -group for some r 6= k.
Then Pjy Pi is σ-fiber and so Pjy is not σ-subnormal by Lemma 2.2(5) since Pjy ≤ NG (Pi ) and
[Pi , Pjy ] 6= 1. Now we show that Pjy is of prime order. Assume that this is false. Then Claim (3)
implies that R is a σir -group, so G has a minimal normal subgroup N 6= R by Claim (5)(b), a
contradiction. Hence Pjy is of prime order.
Note that since Pjy is not σ-subnormal, Pjy forms an irreducible pair with R and Pi by Claim
(4). Then CR (Pjy ) = 1 since CG (R) = R. Note also that CPi (Pjy ) = 1. Indeed, since Pjy forms an
irreducible pair with Pi and Pjy ≤ CG (Pjy ), we have either CPi (Pjy ) = Pi or CPi (Pjy ) = 1. But the
former case is impossible since [Pi , Pjy ] 6= 1, so CPi (Pjy ) = 1.
Let C = CG (Pjy ) ∩ RPi . Suppose that C 6= 1. Then C = (R ∩ C)U = U ≤ Pia for some
a ∈ R, so U × Pjy ≤ (Pi Pjy )b ≃ (Pi Pjy )b for some b ∈ R since G is soluble. It follows that C = 1.
22

Consequently, Pjy ∩ CG (RPi ) = 1. Hence RPi is nilpotent by the Thompson theorem [12, V, 8.14].
Thus Pi ≤ CG (R) = R, a contradiction. Therefore we have (8).
Claims (2)–(8) show that the necessity is true.
Sufficiency. If G is of the type (i), it is clear. Now let G be a group of the type (ii). Then G
is not σ-nilpotent by (ii)(b) and Proposition 3.4. Hence |π(G)| ≤ mσ (G) by Theorem 1.2(iii). Let
n = |π(G)|.
In order ro prove that mσ (G) ≤ |π(G)|, we only need to prove that that every n-maximal subgroup
of G is σ-subnormal in G. Assume that this is false and let E be an n-maximal subgroup of G such
that E is not σ-subnormal in G. Then some Sylow subgroup E1 of E is not σ-subnormal in G
by Proposition 2.5. We can assume that without loss of generality that E1 ≤ P1 . Then P1 is not
σ-subnormal in G by Lemma 2.2(4). If i > 1 and P1 Pix = Pix P1 , then P1 and Pix are members of
some Sylow basis of G [27, I, 4.16]. By the hypothesis, P1 forms an irreducible pair with Pix , P1 is
cyclic and the maximal subgroup of P1 is σ-subnormal in G. Hence E1 = P1 , so every subgroup of G
containing E1 is a Hall subgroup of G by Lemma 6.3. Then E is exactly a k-maximal subgroup of G,
where k = |π(|G : E|)|, by Lemma 6.4. Hence k = n = |π(G)|. But then E = 1, so E is σ-subnormal
in G. This contradiction completes the proof of the sufficiency.
The theorem is proved.

7

Final remarks and some open questions

1. We say that G is a group of σ-Spencer height hσ (G) = n if every maximal chain of G of length
n contains a proper σ-subnormal entry and there exists at least one maximal chain of G of length
n − 1 which contains no any proper σ-subnormal entry.
In particular, if σ = {{2}, {3}, · · · }, we write h(G) instead of hσ (G).
It is clear that hσ (G) ≤ mσ (G), and in general mσ (G) 6= hσ (G).
Theorem 1.2(ii)(iii) can be improved.
Theorem 7.1 [30]. Suppose that G is σ-soluble Then the following statements hold.
(i) lσ (G) ≤ hσ (G).
(ii) If a soluble group G is not σ-nilpotent, then |π(G)| ≤ hσ (G).
Corollary 7.2 (Spencer [25]). Suppose that G is a soluble group. Then:
(i) l(G) ≤ h(G), where l(G) is the nilpotent length of G.
(ii) If h(G) < |π(G)|, then G is nilpotent.
2. In view of Theorems 1.2, 1.10 and 7.1, the following questions seam natural.
Question 7.1. What is the structure of a soluble group G provided |π(G)| = mσ (G) + 1?
23

Question 7.2. What is the structure of a soluble group G provided |π(G)| = hσ (G)?
Question 7.3. What is the structure of a soluble group G provided |π(G)| = hσ (G) + 1?
Note that in the case when σ = {{2}, 3, . . .} the complete answers to these three questions are
known [24, 25].

References
[1] A. N. Skiba, On σ-subnormal and σ-permutable subgroups of finite groups, J. Algebra, 436
(2015), 1-16.
[2] L. A. Shemetkov, Formations of finite groups, Nauka, Main Editorial Board for Physical and
Mathematical Literature, Moscow, 1978.
[3] A. Ballester-Bolinches, K. Doerk, M.D. Pèrez-Ramos, On the lattice of F-subnormal subgroups,
J. Algebra, 148 (1992), 42–52.
[4] A. F. Vasil’ev, A. F. kamornikov, V. N. Semenchuk, On lattices of subgroups of finite groups,
In N.S. Chernikov, Editor, Infinite groups and related algebraic structures, pages 27–54, Kiev,
1993. Institum Matemayiki AN Ukrainy (in Russian).
[5] A. Ballester-Bolinches, L. M. Ezquerro, Classes of Finite Groups, Springer, Dordrecht, 2006.
[6] W. Guo, A. N. Skiba, Finite groups with permutable complete Wielandt sets of subgroups, J.
Group Theory, 18 (2015), 191-200.
[7] A. N. Skiba, On some results in the theory of finite partially soluble groups, Commun. Math.
Stat., 4(3) (2016), 281-312.
[8] O. H. Kegel, Untergruppenverbande endlicher Gruppen, die den subnormalteilerverband each
enthalten, Arch. Math., 30(3) (1978), 225–228.
[9] A. N. Skiba, A generalization of a Hall theorem, J. Algebra and its Application, 15(4) (2015),
21-36.
[10] A. Ballester-Bolinches, R. Esteban-Romero, M. Asaad, Products of Finite Groups, Walter de
Gruyter, Berlin-New York, 2010.
[11] W. Guo, Structure Theory for Canonical Classes of Finite Groups, Springer, Heidelberg-New
York-Dordrecht-London, 2015.
[12] B. Huppert, Endliche Gruppen I, Spronger-Verlag, Berlin-Heidelberg-New York, 1967.
[13] S. Li, Finite non-nilpotent groups all of whose second maximal subgroups are T I-groups, Mathematical Proccedings of the Royal Irish Academy, 100 (2000), 65-71.
24

[14] X. Guo, K. P. Shum, Cover-avoidance properties and the structure of finite groups, Journal of
Pure and Applied Algebra, 181 (2003), 297–308.
[15] B. Li, A. N. Skiba, New characterizations of finite supersoluble groups, Science in China Series
A: Mathematics, 50 (2008), 827–841.
[16] V. N. Kniahina, V. S. Monakhov, On the permutability of n-maximal subgroups with Schmidt
subgroups, Trudy Inst. Mat. i Mekh. UrO RAN, 18(3) (2012), 125–130.
[17] V. S. Monakhov, V. N. Kniahina, Finite groups with P-subnormal subgroups, Ricerche di
Matematica, 62(2) (2013), 307-322
[18] V. A. Kovaleva, A. N. Skiba, Finite soluble groups with all n-maximal subgroups F-subnormal,
J. Group Theory, 17 (2014), 273–290.
[19] V. A. Kovaleva, Finite groups with generalized P-subnormal second maximal subgroups, AsianEuropean Journal of Mathematics, 7 (2014), 1450047-1-1450047-8.
[20] V. A. Kovaleva, X. Yi, Finite biprimary groups with all 3-maximal subgroups U-subnormal,
Acta Matematica Hungarica, 146 (2015), 47-55.
[21] B. Huppert, Normalteiler und maximale Untergruppen endlicher Gruppen, Math. Z., 60 (1954)
409-434.
[22] R. K. Agrawal, Generalized center and hypercenter of a finite group, Proc. Amer. Math. Soc.,
54 (1976), 13-21.
[23] Z. Janko, Finite groups with invariant fourth maximal subgroups, Math. Z., 83 (1963), 82-89.
[24] A. Mann, Fnite groups whose n-maximal subgroups are subnormal, Trans Amer. Math. Soc.,
132 (1968), 395-409.
[25] A. E. Spencer, Maximal nonnormal chains in finite groups, Pacific J Math., 27 (1968), 167-173.
[26] M. Weinstein ed., Between Nilpotent and Solvable, Polygonal Publishing House, Passaic N.J.,
1982.
[27] K. Doerk, T. Hawkes, Finite Soluble Groups, Walter de Gruyter, Berlin-New York, 1992.
[28] W. Guo, K. P. Shum, A. N. Skiba, X-semipermutable subgroups of finite groups, J. Algebra,
315 (2007), 31-41.
[29] L. A. Shemetkov, A. N. Skiba, Formations of Algebraic Systems, Nauka, Main Editorial Board
for Physical and Mathematical Literature, Moscow, 1989.
[30] W. Guo, A. N. Skiba Finite groups with given maximal chains, Preprint.
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