On finite 𝜎-tower groups

Definition 1.1

(1) A group 𝐺 is said to be a Sylow tower group if either G = 1 or 𝐺 has a normal series 1 = G 0 < G 1 < ⋯ < G t - 1 < G t = G such that G i / G i - 1 is a 𝑝-group, p ∈ π ⁢ ( G ) , and G / G i and G i - 1 are p ′ -groups for all i = 1 , … , t .

(2) We say that 𝐺 is a 𝜎-tower group if either G = 1 or 𝐺 has a normal series 1 = G 0 < G 1 < ⋯ < G t - 1 < G t = G such that G i / G i - 1 is a σ i -group, σ i ∈ σ ⁢ ( G ) , and G / G i and G i - 1 are σ i ′ -groups for all i = 1 , … , t .

Remark 1.2

(1) Let 1 = G 0 < G 1 < ⋯ < G t - 1 < G t = G be a series of 𝐺, where all the G i are normal Hall subgroups of 𝐺, and let σ i = π ⁢ ( G i / G i - 1 ) for all i = 1 , … , t . Then σ i ∩ σ j = ∅ for all i ≠ j , so σ = { σ 1 , … , σ t , ( σ 1 ∪ ⋯ ∪ σ t ) ′ } is a partition of ℙ and 𝐺 is a 𝜎-tower group.

Example 1.3

(i) In the classical case, when σ = σ 1 = { { 2 } , { 3 } , … } , 𝐺 is σ 1 -soluble (respectively σ 1 -nilpotent) if and only if 𝐺 is soluble (respectively nilpotent); 𝐺 is a σ 1 -tower group if and only if 𝐺 is a Sylow tower group. A subgroup 𝐴 of 𝐺 is σ 1 -subnormal if and only if it is subnormal in 𝐺.

(ii) In the other classical case, when σ = σ π = { π , π ′ } , 𝐺 is σ π -soluble (respectively σ π -nilpotent) if and only if 𝐺 is 𝜋-separable (respectively 𝜋-decomposable, that is, G = O π ⁢ ( G ) × O π ′ ⁢ ( G ) ); 𝐺 is a σ π -tower group if and only if 𝐺 is either 𝜋-closed or π ′ -closed. A subgroup 𝐴 of 𝐺 is σ π -subnormal in 𝐺 if and only if there is a subgroup chain A = A 0 ≤ A 1 ≤ ⋯ ≤ A t = G such that either A i - 1 ⁢ ⊴ ⁢ A i , or A i / ( A i - 1 ) A i is a 𝜋-group, or a π ′ -group for all i = 1 , … , t .

(iii) In fact, in the theory of 𝜋-soluble groups ( π = { p 1 , … , p n } ), we deal with the partition σ = σ 1 ⁢ π = { { p 1 } , … , { p n } , π ′ } of ℙ. Hence 𝐺 is σ 1 ⁢ π -soluble (respectively σ 1 ⁢ π -nilpotent) if and only if 𝐺 is 𝜋-soluble (respectively 𝜋-special [5], that is, G = O p 1 ⁢ ( G ) × ⋯ × O p n ⁢ ( G ) × O π ′ ⁢ ( G ) ); 𝐺 is a σ 1 ⁢ π -tower group if and only if 𝐺 has a Hall π ′ -subgroup 𝐻 and Sylow p i -subgroups P i , i = 1 , … , t , where p 1 , … , p t ∈ π , such that, for some 𝑘, all subgroups

are normal in 𝐺.

(iv) Let A = C 7 ≀ C 11 = L ⋊ C 11 , where C 7 and C 11 are cyclic groups of order 7 and 11, respectively, and 𝐿 is the base group of the regular wreath product 𝐴. Let G = A 5 ≀ A = K ⋊ A = K ⋊ ( L ⋊ C 11 ) , where 𝐾 is the base group of the regular wreath product 𝐺 and A 5 is the alternating group of degree 5. Then 𝐺 is not a Sylow tower group but it is a 𝜎-tower group, where

Question 1.4

Question 1.4 (see [19, Question 4.8] or [21, Question 7.22])

Let 𝐺 be a 𝜎-soluble group and | σ ⁢ ( G ) | = n . Assume that every ( n + 1 ) -maximal subgroup of 𝐺 is 𝜎-subnormal. Is it true then that 𝐺 is a 𝜎-tower group?

Theorem 1.5

Let 𝐺 be a 𝜎-soluble group. Then the following statements hold.

1. If either h σ ⁢ ( G ) < | σ ⁢ ( G ) | or h σ ⁢ ( G ) < | π ⁢ ( G ) | and 𝐺 is soluble, then 𝐺 is 𝜎-nilpotent.

2. If either h σ ⁢ ( G ) ≤ | σ ⁢ ( G ) | + 1 or h σ ⁢ ( G ) ≤ | π ⁢ ( G ) | + 1 and 𝐺 is soluble, then 𝐺 is a 𝜎-tower group.

Corollary 1.6

Let | σ ⁢ ( G ) | = n . If every ( n + 1 ) -maximal subgroup of 𝐺 is 𝜎-subnormal in 𝐺, then 𝐺 is a 𝜎-tower group.

Corollary 1.7

Corollary 1.7 (see [18, Theorem 0.1])

Let 𝐺 be a soluble group and suppose that, for all σ i ∈ σ ⁢ ( G ) , every Hall σ i -subgroup of 𝐺 is supersoluble. Then 𝐺 is a 𝜎-tower group if h σ ⁢ ( G ) ≤ | σ ⁢ ( G ) | + 1 .

Corollary 1.8

Corollary 1.8 (see [10, Theorem 7.1])

If a soluble group 𝐺 is not 𝜎-nilpotent, then | π ⁢ ( G ) | ≤ h σ ⁢ ( G ) .

Corollary 1.9

Corollary 1.9 (see [17, Theorem 6])

Let | π ⁢ ( G ) | = n + 1 . If every 𝑛-maximal subgroup of 𝐺 is subnormal in 𝐺, then 𝐺 is nilpotent.

Corollary 1.10

Corollary 1.10 (see [17, Theorem 9])

Let | π ⁢ ( G ) | = n . If every ( n + 1 ) -maximal subgroup of 𝐺 is subnormal in 𝐺, then 𝐺 is a Sylow tower group.

Corollary 1.11

Corollary 1.11 (see [22, Theorem 2])

If h ⁢ ( G ) < | π ⁢ ( G ) | , then 𝐺 is nilpotent.

Corollary 1.12

Corollary 1.12 (see [22, Theorem 3])

If h ⁢ ( G ) ≤ | π ⁢ ( G ) | + 1 , then 𝐺 is a Sylow tower group.

Corollary 1.13

If, for every 3-maximal subgroup 𝐴 of 𝐺, there is a subgroup chain A = A 0 ≤ A 1 ≤ ⋯ ≤ A t = G , where either A i - 1 ⁢ ⊴ ⁢ A i , or A i / ( A i - 1 ) A i is a 𝜋-group, or a π ′ -group for all i = 1 , … , t , then 𝐺 is either 𝜋-closed or π ′ -closed.

Lemma 2.1

Lemma 2.1 (see [19, Lemma 2.6])

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 N ≤ K and K / N is 𝜎-subnormal in G / N , then 𝐾 is 𝜎-subnormal in 𝐺.

3. If K ≤ E ≤ G , where 𝐾 is 𝜎-subnormal in 𝐸, then K ⁢ N / N is 𝜎-subnormal in N ⁢ E / N .

4. If 𝐴 is a Hall Π-subgroup of 𝐺, then 𝐴 is normal in 𝐺.

5. If 𝐵 is a 𝜎-subnormal subgroup of 𝐺, then ⟨ A , B ⟩ and A ∩ B are 𝜎-subnormal in 𝐺.

6. If 𝐺 is 𝜎-soluble and 𝐴 is a σ i -group for some 𝑖, then A ≤ O σ i ⁢ ( G ) .

Lemma 2.2

Let n = h σ ⁢ ( G ) .

1. If 𝑀 is a non-𝜎-subnormal maximal subgroup of 𝐺, then h σ ⁢ ( M ) ≤ n - 1 .

2. If 𝑅 is a normal subgroup of 𝐺, then h σ ⁢ ( G / R ) ≤ n .

Proof

(1) Since 𝑀 is not 𝜎-subnormal in 𝐺 and n = h σ ⁢ ( G ) , some member ≠ M of any maximal chain M n - 1 < M n - 2 < ⋯ < M 1 < M 0 = M of 𝑀 of length n - 1 is 𝜎-subnormal in 𝐺 and so 𝜎-subnormal in 𝑀 by Lemma 2.1 (1). Hence h σ ⁢ ( M ) ≤ n - 1 .

(2) If M m / R < M m - 1 / R < ⋯ < M 1 / R < M 0 / R = G / R is a maximal chain in G / R none of whose members are 𝜎-subnormal in G / R , then all members M m , … , M 1 of the maximal chain M m < M m - 1 < ⋯ < M 1 < M 0 = G of 𝐺 are not 𝜎-subnormal in 𝐺 by Lemma 2.1 (2), which implies that h σ ⁢ ( G / R ) ≤ n . The lemma is proved. ∎

Lemma 2.3

Lemma 2.3 (see [10, Proposition 3.2])

If 𝐺 is 𝜎-soluble and 𝑀 is a maximal subgroup of 𝐺, then | G : M | is 𝜎-primary.

Lemma 2.4

If 𝐻 is a normal subgroup of 𝐺 and H / ( H ∩ Φ ⁢ ( G ) ) is 𝜋-closed, then 𝐻 is 𝜋-closed.

Proof

Let Φ = H ∩ Φ ⁢ ( G ) , and let V / Φ be a normal Hall 𝜋-subgroup of H / Φ . Let 𝐷 be a Hall π ′ -subgroup of Φ. Then 𝐷 is a normal Hall π ′ -subgroup of 𝑉, so 𝑉 has a Hall 𝜋-subgroup, 𝐸 say, by the Schur–Zassenhaus theorem. It is clear that 𝑉 is π ′ -soluble, so any two Hall 𝜋-subgroups of 𝑉 are conjugated in 𝑉. Therefore, by the Frattini argument, we have

Thus 𝐸 is normal in 𝐺. The lemma is proved. ∎

Lemma 2.5

The set 𝔉, of all 𝜋-closed groups, is a hereditary saturated Fitting formation.

Proof

First note that if 𝐺 is 𝜋-closed, that is, 𝐺 has a normal Hall 𝜋-subgroup 𝐻, then H ∩ E is a normal Hall 𝜋-subgroup in E ≤ G , so the class 𝔉 is hereditary. On the other hand, H ⁢ N / N is a normal Hall 𝜋-subgroup of G / N for every normal subgroup 𝑁 of 𝐺, so every homomorphic image of any group G ∈ F belongs to the class 𝔉.

Now we show that if N 1 , … , N t are normal subgroups of 𝐺 such that G / N i ∈ F for all 𝑖, then G / ( N 1 ∩ ⋯ ∩ N t ) ∈ F . We use induction on t + | G | . Therefore, since

we have only to consider the case t = 2 . Moreover, we can assume without loss of generality that N 1 ∩ N 2 = 1 . It is clear that G ≃ G / ( N 1 ∩ N 2 ) is 𝜋-separable, so 𝐺 has a Hall 𝜋-subgroup 𝐻. Then N i ⁢ H / N i is normal in G / N i since G / N i is 𝜋-closed, so N i ⁢ H is normal in 𝐺. Hence

is normal in 𝐺. Therefore, G / ( N 1 ∩ ⋯ ∩ N t ) ∈ F , so G / G F ∈ F . Hence, in view of the remarks in the previous paragraph, 𝔉 is a hereditary formation. Finally, this formation is saturated by Lemma 2.4, and if N , M ∈ F are normal 𝜋-closed subgroups of 𝐺 with normal Hall 𝜋-subgroups N π and M π , then N π ⁢ M π is a normal Hall 𝜋-subgroup of 𝐺. Therefore, 𝔉 is a Fitting class. The lemma is proved. ∎

Lemma 2.6

The class N σ , of all 𝜎-nilpotent groups, is a hereditary saturated Fitting formation.

Lemma 2.7

The following statements hold.

1. The class of all 𝜎-tower groups is closed under taking homomorphic images.

2. The class 𝔉, of all 𝜎-tower groups of type 𝜓, is a hereditary saturated Fitting formation.

Proof

(1) This part can be proved by direct checking.

(2) Let 𝔉 be the class of all 𝜎-tower groups of type 𝜓. If | σ | = 1 , that is, σ = { σ 1 } , where σ 1 = P , then 𝔉 is the class of all groups, and so, in this case, statement (2) is trivial. Let | σ | > 1 .

Let σ 0 be the set of all σ i such that σ i is not the smallest element (with respect to 𝜓) in 𝜎. The set σ 0 is not empty since | σ | > 1 . For every σ i ∈ σ 0 , we define the set π i = f ⁢ ( σ i ) := ⋃ σ k ⁢ ψ ⁢ σ i σ k .

Now, in view of Lemma 2.5, it is enough to show that G ≠ 1 is a 𝜎-tower group of type 𝜓 if and only if 𝐺 is π i -closed for all π i in Π * = { π i = f ⁢ ( σ i ) ∣ σ i ∈ σ 0 } . We can assume without loss of generality that G ≠ 1 and σ ⁢ ( G ) = { σ 1 , σ 2 , … , σ n } , where σ 1 ⁢ ψ ⁢ σ 2 ⁢ ψ ⁢ ⋯ ⁢ ψ ⁢ σ n .

First assume that 𝐺 is a 𝜎-tower group of type 𝜓, that is, 𝐺 possesses a normal series

in which, for every j = 1 , … , t , the factor G j / G j - 1 is a σ j -group and G j - 1 and G / G j are σ j ′ -groups. If π i ∩ π ⁢ ( G ) = ∅ , then the identity subgroup 1 is the Hall π i -subgroup of 𝐺.

Now assume π i ∩ π ⁢ ( G ) ≠ ∅ . Then, for σ i , we have either σ t ⁢ ψ ⁢ σ i or σ i ∈ σ ⁢ ( G ) . In the former case, we have π ⁢ ( G ) ⊆ π i , and so, in this case, 𝐺 is the Hall π i -subgroup of 𝐺.

Finally, assume that σ i ∈ σ ⁢ ( G ) , that is, σ i ∈ σ ⁢ ( G i / G i - 1 ) . Then i ≠ 1 since π i ∩ π ⁢ ( G ) ≠ ∅ and G i - 1 is a Hall π i -subgroup of 𝐺, so 𝐺 is π i -closed.

Now assume that 𝐺 is π i -closed for all π i ∈ Π * . We show by induction on | G | that in this case 𝐺 is a 𝜎-tower group of type 𝜓. If t = 1 , it is evident. Now assume that t > 1 . Then 𝐺 is π 2 -closed, where π 2 = ⋃ σ k ⁢ ψ ⁢ σ 2 σ k , by hypothesis. Hence 𝐺 has a normal Hall σ 1 -subgroup G 1 since π 2 ∩ π ⁢ ( G ) ⊆ σ 1 . Finally, the hypothesis holds for G / G 1 by Lemma 2.5, so G / G 1 is a 𝜎-tower group of type 𝜓 by induction. But then 𝐺 is a 𝜎-tower group of type 𝜓. Therefore, assertion (2) holds. The lemma is proved. ∎

Lemma 2.8

If every maximal subgroup of 𝐺 is 𝜎-subnormal in 𝐺, then 𝐺 is 𝜎-nilpotent.

Proof

If every maximal subgroup 𝑀 of 𝐺 is 𝜎-subnormal in 𝐺, then G / M G is 𝜎-primary, and so it is 𝜎-nilpotent. Therefore, G / Φ ⁢ ( G ) is 𝜎-nilpotent, and so 𝐺 is 𝜎-nilpotent by Lemma 2.6. The lemma is proved. ∎

Proof of Theorem 1.5

Let 𝑅 be a minimal normal subgroup of 𝐺. Then 𝑅 is a σ i -group for some 𝑖 since 𝐺 is 𝜎-soluble by hypothesis. Note also that 𝐺 has a Hall Π-subgroup for all Π ⊆ σ by [20, Theorem A].

(i) Assume that this assertion is false, and let 𝐺 be a counterexample of minimal order, that is, every proper section H / K of 𝐺 such that either

or (in the case when 𝐺 is soluble)

is 𝜎-nilpotent.

(1) If either h σ ⁢ ( G ) < | σ ⁢ ( G ) | and 𝑅 is a Hall σ i -subgroup of 𝐺 or 𝐺 is a soluble group with h σ ⁢ ( G ) < | π ⁢ ( G ) | and 𝑅 is a Sylow subgroup of 𝐺, then every maximal subgroup of 𝐺 not containing 𝑅 is not 𝜎-subnormal in 𝐺.

Assume that 𝐺 has a 𝜎-subnormal maximal subgroup 𝑀 such that G = R ⁢ M . Then G / M G is 𝜎-primary, and so, in fact, G / M G is a σ i -group.

First suppose that h σ ⁢ ( G ) < | σ ⁢ ( G ) | and 𝑅 is a Hall σ i -subgroup of 𝐺. Then G = M G ⁢ R and R ∩ M G = 1 , so G = R × M G . Hence, since 𝐺 is not 𝜎-nilpotent by the choice of 𝐺, M G is not 𝜎-nilpotent by Lemma 2.6. Therefore, m := h σ ⁢ ( M G ) > 1 , so M G possesses a maximal chain

of length m - 1 , where M i is not 𝜎-subnormal in M G for all i = 1 , … , m - 1 and in every maximal chain L m < L m - 1 < ⋯ < L 1 < L 0 = M G of M G of length 𝑚 some member L i ≠ M G is 𝜎-subnormal in M G .

Now consider the series M m - 1 < ⋯ < M 1 < M 1 ⁢ R < G . We show that if either M 1 ⁢ R < T < G or M 1 < T ≤ R ⁢ M 1 , then the subgroup 𝑇 is not 𝜎-subnormal in 𝐺.

First assume that M 1 ⁢ R < T < G = R × M G . Then M 1 ≤ T ∩ M G ≤ M G , so either T ∩ M G = M G or M 1 = T ∩ M G . However, in the former case, we have M G ≤ T and then G = R ⁢ M G ≤ T , a contradiction. Hence M 1 = T ∩ M G , and so 𝑇 is not 𝜎-subnormal in 𝐺 by Lemma 2.1 (1) since M 1 is not 𝜎-subnormal in M G . Now assume that M 1 < T ≤ R ⁢ M 1 . Then T = M 1 ⁢ ( T ∩ R ) , so

and hence 𝑇 is not 𝜎-subnormal in 𝐺.

Therefore, 𝐺 has a maximal chain of length n > m , every non-identity member ≠ G of which is not 𝜎-subnormal in 𝐺. It follows that

so h σ ⁢ ( M G ) < | σ ⁢ ( M G ) | since | σ ⁢ ( G ) | - 1 ≤ | σ ⁢ ( M G ) | ≤ | σ ⁢ ( G ) | by Lemma 2.3. Therefore, M G is 𝜎-nilpotent by the choice of 𝐺, a contradiction.

Now assume that 𝐺 is a soluble group with h σ ⁢ ( G ) < | π ⁢ ( G ) | and 𝑅 is a Sylow 𝑝-subgroup of 𝐺 for some prime p ∈ σ i . First suppose that 𝑀 is 𝜎-nilpotent, and let 𝐻 be a Hall σ i -subgroup of 𝑀 and 𝐸 a Hall { σ i } ′ -subgroup of 𝑀. Then 𝐻 and 𝐸 are normal in 𝑀, and so R ⁢ H is a normal Hall σ i -subgroup of 𝐺 and 𝐸 is a 𝜎-subnormal Hall { σ i } ′ -subgroup of 𝐺. But then 𝐸 is normal in 𝐺 by Lemma 2.1 (4). On the other hand, 𝐸 is 𝜎-nilpotent by Lemma 2.6, so G = ( R ⁢ H ) × E is 𝜎-nilpotent, a contradiction. Hence 𝑀 is not 𝜎-nilpotent. Now, arguing as in the previous paragraph, we can obtain a contradiction. Therefore, the claim holds.

(2) The hypothesis holds for every maximal non-𝜎-subnormal subgroup 𝑀 of 𝐺. Hence 𝑀 is 𝜎-nilpotent.

First assume that h σ ⁢ ( G ) < | σ ⁢ ( G ) | . Then h σ ⁢ ( G ) - 1 < | σ ⁢ ( G ) | - 1 . On the other hand, we have h σ ⁢ ( M ) ≤ h σ ⁢ ( G ) - 1 by Lemma 2.2 (2), so h σ ⁢ ( M ) < | σ ⁢ ( G ) | - 1 . Finally, | σ ⁢ ( G ) | - 1 ≤ | σ ⁢ ( M ) | ≤ | σ ⁢ ( G ) | by Lemma 2.3, so h σ ⁢ ( M ) < | σ ⁢ ( M ) | . If 𝐺 is soluble and h σ ⁢ ( G ) < | π ⁢ ( G ) | , then, similarly, we can get that h σ ⁢ ( M ) < | π ⁢ ( M ) | . Therefore, the hypothesis holds for 𝑀; hence 𝑀 is 𝜎-nilpotent by the choice of 𝐺.

(3) G / N is 𝜎-nilpotent for every minimal normal subgroup 𝑁 of 𝐺, so 𝑅 is the unique minimal normal subgroup of 𝐺 and Φ ⁢ ( G ) = 1 . Hence a Hall σ i -subgroup 𝐻 of 𝐺 is normal in 𝐺 and O σ j ⁢ ( G ) = 1 for all j ≠ i .

If N ≤ Φ ⁢ ( G ) , then σ ⁢ ( G / N ) = σ ⁢ ( G ) and π ⁢ ( G / N ) = π ⁢ ( G ) . On the other hand, h σ ⁢ ( G / N ) ≤ h σ ⁢ ( G ) by Lemma 2.2 (2), so the hypothesis holds for G / N , and hence G / N is 𝜎-nilpotent by the choice of 𝐺. Therefore, Φ ⁢ ( G ) = 1 by Lemma 2.6 since 𝐺 is not 𝜎-nilpotent. If 𝑁 is not a Hall σ i -subgroup of 𝐺, then σ ⁢ ( G / N ) = σ ⁢ ( G ) , and if 𝐺 is soluble and 𝑁 is not a Sylow subgroup of 𝐺, then π ⁢ ( G / N ) = π ⁢ ( G ) , so in these cases, the hypothesis also holds for G / N . Therefore, G / N is 𝜎-nilpotent by the choice of 𝐺.

Finally, assume that either 𝑁 is a Hall σ j -subgroup of 𝐺 for some 𝑗 or 𝐺 is a soluble group with h σ ⁢ ( G ) < | π ⁢ ( G ) | and 𝑁 is a Sylow subgroup of 𝐺. Let 𝑀 be a maximal subgroup of 𝐺 such that G = N ⁢ M . Then 𝑀 is not 𝜎-subnormal in 𝐺 by claim (1). Hence 𝑀 is 𝜎-nilpotent by claim (2), so

is 𝜎-nilpotent by Lemma 2.6. If N ≠ R , then G ≃ G / ( R ∩ N ) is 𝜎-nilpotent by Lemma 2.6. Therefore, 𝑅 is the unique minimal normal subgroup of 𝐺, and G / R is 𝜎-nilpotent. Hence a Hall σ i -subgroup H / N of G / N is normal in G / N , so 𝐻 is a normal Hall σ i -subgroup of 𝐺.

(4) If 𝑀 is a maximal subgroup of 𝐺 such that G = R ⁢ M , then 𝑀 is 𝜎-nilpotent.

If 𝑀 is 𝜎-subnormal in 𝐺, then from claim (3), we get that G ≃ G / M G = G / 1 is 𝜎-primary, and so 𝐺 is 𝜎-nilpotent, contrary to the choice of 𝐺. Hence 𝑀 is not 𝜎-subnormal in 𝐺, and so 𝑀 is 𝜎-nilpotent by claim (2) and the choice of 𝐺.

(5) 𝑅 is not abelian.

Assume that 𝑅 is a 𝑝-subgroup of 𝐺 for some prime 𝑝. From claim (3), it follows that G = R ⋊ M for some maximal subgroup 𝑀 of 𝐺. Then 𝑀 is 𝜎-nilpotent by claim (4), and hence 𝑀 is not 𝜎-subnormal in 𝐺 by Lemma 2.1 (6) and claim (3). From Lemma 2.2 (1), it follows that h σ ⁢ ( M ) ≤ h σ ⁢ ( G ) - 1 < | σ ⁢ ( G ) | - 1 in the case when h σ ⁢ ( G ) < | σ ⁢ ( G ) | , and we have h σ ⁢ ( M ) ≤ h σ ⁢ ( G ) - 1 < | π ⁢ ( G ) | - 1 in the case when 𝐺 is soluble and h σ ⁢ ( G ) < | π ⁢ ( G ) | . Therefore, some non-identity 𝜎-primary subgroup 𝑆 of 𝑀 is 𝜎-subnormal in 𝐺. But this is impossible by Lemma 2.1 (6) and claims (3), (4), and (5) since 𝑀 is 𝜎-nilpotent. Therefore, claim (5) holds.

The final contradiction for (i). From claim (3), it follows that the subgroup 𝐻 has a complement 𝐸 in 𝐺 by the Schur–Zassenhaus theorem. Let 𝑞 be any prime dividing | H | . We show that 𝑞 does not divide | G : N G ⁢ ( E ) | .

First note that 𝐸 is a Hall { σ i } ′ -subgroup of 𝐺, so from [20, Theorem A], it follows that there is a Sylow 𝑞-subgroup 𝑄 of 𝐺 such that V := Q ⁢ E = E ⁢ Q is a subgroup of 𝐺. If Q ∩ R = 1 , then R ∩ V = 1 , and so R ⁢ V / R ≃ V = Q × E is 𝜎-nilpotent by claim (3). Hence 𝑞 does not divide | G : N G ⁢ ( E ) | . Now assume that R q = Q ∩ R ≠ 1 . Then Q ≤ N G ⁢ ( R q ) and G = R ⁢ N G ⁢ ( R q ) by the Frattini argument. Moreover, N G ⁢ ( R q ) ≠ G by claim (5), so for some maximal subgroup 𝑀 of 𝐺, we have Q ≤ N G ⁢ ( R q ) ≤ M . Hence G = R ⁢ M , and then 𝑀 is 𝜎-nilpotent by claim (4). Therefore, for a Hall { σ i } ′ -subgroup 𝑊 of 𝑀, we have [ Q , W ] = 1 and 𝑊 is a Hall { σ i } ′ -subgroup of 𝐺. Hence 𝐸 and 𝑊 are conjugate in 𝐺 by [20, Theorem A], so 𝑞 does not divide | G : N G ⁢ ( E ) | . Therefore, 𝑞 does not divide | G : N G ⁢ ( E ) | for all q ∈ π ⁢ ( H ) , and so 𝐸 is normal in 𝐺, contrary to claim (3). Hence assertion (i) holds.

(ii) Assume that this assertion is false, and let 𝐺 be a counterexample of minimal order, that is, every proper section H / K of 𝐺 such that either

or (in the case when 𝐺 is soluble)

is a 𝜎-tower group.

(1) If either h σ ⁢ ( G ) ≤ | σ ⁢ ( G ) | + 1 and 𝑅 is a Hall σ i -subgroup of 𝐺 or 𝐺 is a soluble group with h σ ⁢ ( G ) ≤ | π ⁢ ( G ) | + 1 and 𝑅 is a Sylow subgroup of 𝐺, then every maximal subgroup of 𝐺 not containing 𝑅 is not 𝜎-subnormal in 𝐺 (see claim (1) in the proof of part (i)).

(2) Every non-𝜎-subnormal maximal subgroup 𝑀 of 𝐺 is a 𝜎-tower group.

First assume that 𝐺 is soluble and that h σ ⁢ ( G ) ≤ | π ⁢ ( G ) | + 1 . Then we have h σ ⁢ ( G ) - 1 ≤ | π ⁢ ( G ) | . On the other hand, we have h σ ⁢ ( M ) ≤ h σ ⁢ ( G ) - 1 by Lemma 2.2 (1), so h σ ⁢ ( M ) ≤ | π ⁢ ( G ) | . Finally, | π ⁢ ( G ) | - 1 ≤ | π ⁢ ( M ) | ≤ | π ⁢ ( G ) | since 𝐺 is soluble, so h σ ⁢ ( M ) ≤ | π ⁢ ( M ) | + 1 . Now, if h σ ⁢ ( G ) ≤ | σ ⁢ ( G ) | + 1 , then we can get similarly that h σ ⁢ ( M ) ≤ | σ ⁢ ( M ) | + 1 . Therefore, the hypothesis holds for 𝑀, so 𝑀 is a 𝜎-tower group by the choice of 𝐺.

(3) If h σ ⁢ ( G ) ≤ | σ ⁢ ( G ) | + 1 and 𝐻 is a Hall σ j -subgroup of 𝐺 for some σ j ∈ σ ⁢ ( G ) , then 𝐻 is not 𝜎-subnormal in 𝐺.

Assume that 𝐻 is 𝜎-subnormal in 𝐺. Then 𝐻 is normal in 𝐺 by Lemma 2.1 (4). Since σ j ∈ σ ⁢ ( G ) , H ≠ 1 , and so, for some minimal normal subgroup of 𝐺, for 𝑅 say, we have R ≤ H .

First suppose that R < H . Then σ ⁢ ( G / R ) = σ ⁢ ( G ) and h σ ⁢ ( G / R ) ≤ h σ ⁢ ( G ) by Lemma 2.2 (2). Therefore, the hypothesis holds for G / R , so G / R is a 𝜎-tower group. Hence G / H is a 𝜎-tower group by Lemma 2.7 (1). But then 𝐺 is 𝜎-tower group, a contradiction.

Therefore, R = H , so for some maximal subgroup 𝑀 of 𝐺, we have G = R ⁢ M . If 𝑀 is a 𝜎-tower group, then G / H = G / R ≃ M / ( M ∩ R ) is a 𝜎-tower group by Lemma 2.7 (1), and so 𝐺 is a 𝜎-tower group since 𝐻 is a Hall σ j -subgroup of 𝐺, a contradiction. Therefore, 𝑀 is not a 𝜎-tower group, and so 𝑀 is 𝜎-subnormal in 𝐺 by claim (2). But this is impossible by claim (1). This contradiction completes the proof of the claim.

(4) G / N is a 𝜎-tower group for every minimal normal subgroup 𝑁 of 𝐺, so Φ ⁢ ( G ) = 1 .

In view of claim (3), σ ⁢ ( G / N ) = σ ⁢ ( G ) . Therefore, if h σ ⁢ ( G ) ≤ | σ ⁢ ( G ) | + 1 , then the hypothesis holds for G / N by Lemma 2.2 (2), and so G / N is a 𝜎-tower group by the choice of 𝐺.

Now assume that 𝐺 is soluble and h σ ⁢ ( G ) ≤ | π ⁢ ( G ) | + 1 . If 𝑁 is not a Sylow subgroup of 𝐺, then π ⁢ ( G / N ) = π ⁢ ( G ) , and hence the hypothesis holds for G / N , so G / N is a 𝜎-tower group. Finally, if 𝑁 is a Sylow subgroup of 𝐺 and 𝑀 is a maximal subgroup of 𝐺 such that G = N ⁢ M , then 𝑀 is not 𝜎-subnormal in 𝐺 by claim (1), and so 𝑀 is a 𝜎-tower group by claim (2), which implies that G / N ≃ M / ( N ∩ M ) is a 𝜎-tower group.

(5) Let 𝑁 be a minimal normal subgroup of 𝐺 and 𝑀 a maximal subgroup of 𝐺 such that G = N ⁢ M . Suppose that 𝑁 is a σ j -group. Then 𝑀 has a normal series M n - 1 < ⋯ < M 1 < M 0 = M , where M k / M k + 1 is a σ i k -group of prime order and M k + 1 and M / M k are σ i k ′ -groups for all k = 0 , 1 , … , n - 2 . Moreover, M n - 1 is a Hall cyclic σ s -subgroup of order r m for some r ∈ σ s ∈ σ ⁢ ( G ) ∖ { σ j } , and the maximal subgroup of M n - 1 is 𝜎-subnormal in 𝐺. Hence 𝑀 and G / N are supersoluble.

Assume that 𝑀 is 𝜎-subnormal in 𝐺. Then G / M G is a 𝜎-primary group, and so G / M G is a 𝜎-tower group of type 𝜓 for every linear ordering 𝜓 on 𝜎. Then G ≃ G / ( N ∩ M G ) is a 𝜎-tower group by claim (4) and Lemma 2.7. This contradiction shows that 𝑀 is not 𝜎-subnormal in 𝐺. Therefore, 𝑀 is a 𝜎-tower group by claim (2).

First assume that h σ ⁢ ( G ) ≤ | σ ⁢ ( G ) | + 1 . From claim (3), it follows that

Then 𝑀 has a subgroup series 1 = M n < M n - 1 < ⋯ < M 1 < M 0 = M , where M k is a normal subgroup of 𝑀 and M k / M k + 1 is a σ i k -group and M k + 1 and M / M k are σ i k ′ -groups for all k = 0 , 1 , … , n - 1 .

If M n - 1 is a σ j -group, then N ⁢ M n - 1 is a normal Hall σ j -subgroup of 𝐺, contrary to claim (3). Hence M n - 1 is a Hall σ s -subgroup of 𝐺 for some s ≠ j . Hence M n - 1 is not 𝜎-subnormal in 𝐺 by claim (3). Therefore, every subgroup 𝑋 of 𝑀 containing M n - 1 is not 𝜎-subnormal in 𝐺 since M n - 1 is normal in 𝑋. In particular, M k is not 𝜎-subnormal in 𝐺 for all k < n - 1 . Hence, in fact,

Assume that, for some l < n - 1 , the factor group M l / M l + 1 is not of prime order. Then, for some 𝜎-subnormal subgroup 𝑊 of 𝐺 and some 𝑛-maximal subgroup 𝑉 of 𝑀, we have M n - 1 ≤ V ≤ W < M . But this is impossible by the remark at the and of the previous paragraph. Therefore, M k / M k + 1 is of prime order for all k = 0 , … , n - 2 , and so M n - 1 < ⋯ < M 1 < M 0 = M < G is a maximal chain of 𝐺 of length 𝑛. It follows that every maximal subgroup of M n - 1 is 𝜎-subnormal in 𝐺 since h σ ⁢ ( G ) = | σ ⁢ ( G ) | + 1 = n + 1 . Therefore, M n - 1 is a cyclic group of prime power order by Lemma 2.1 (5) since M n - 1 is not 𝜎-subnormal in 𝐺. Hence 𝑀, and so G / N ≃ M / ( M ∩ N ) are supersoluble. Hence (5) holds.

Similarly, we can also show that claim (5) holds in the case when 𝐺 is soluble and h σ ⁢ ( G ) ≤ | π ⁢ ( G ) | + 1 .

(6) 𝑅 is a unique minimal normal subgroup of 𝐺, so C G ⁢ ( R ) ≤ R and every maximal subgroup 𝑀 of 𝐺 with G = R ⁢ M has square-free order.

Suppose that 𝐺 has a minimal normal subgroup N ≠ R . Then, for some maximal subgroup 𝑀 of 𝐺, we have G = N ⁢ M by claim (4). Moreover, we have that G ≃ G / ( R ∩ N ) is supersoluble by claim (5), so 𝑅 and 𝑁 are groups of prime order. If | R | ≠ | N | , then R ≤ M . Let | R | = p ∈ σ i , | N | = q ∈ σ j . Suppose that i = j . Then, since the order of a Hall σ i -subgroup of 𝑀 is a prime by claim (5), R ⁢ N is a normal Hall σ i -subgroup of 𝐺. The latter contradicts claim (3). Hence i ≠ j . Then 𝑅 is contained in some Sylow 𝑝-subgroup 𝐻 of 𝑀 and 𝐻 is a Hall σ i -subgroup of 𝑀 by claim (5). On the other hand, for the same reasons, for some maximal subgroup 𝐿 of 𝐺, we have G = R ⁢ L , N ≤ L , 𝑁 is contained in some Sylow 𝑞-subgroup 𝐷 of 𝐿, and 𝐷 is a Hall σ j -subgroup of 𝐿. Therefore, any Hall σ s -subgroup of 𝐺 is its Sylow subgroup. But then 𝐺 is a 𝜎-tower group since 𝐺 is supersoluble. A contradiction.

Therefore, | R | = | N | and R ⁢ N is a normal Hall σ i -subgroup of 𝐺 since the order of a Hall σ i -subgroup of 𝑀 is a prime by claim (5). This contradiction shows that 𝑅 is a unique minimal normal subgroup of 𝐺 and C G ⁢ ( R ) ≤ R by [6, Chapter A, 15.6]. Finally, let 𝐿 be a maximal subgroup of 𝐺 with G = R ⁢ L . Then, in view of claim (5), some subgroup L n - 1 of 𝐿 is a Hall cyclic σ s -group of order r m for some r ∈ σ s ∈ σ ⁢ ( G ) ∖ { σ i } and the maximal subgroup 𝑆 of L n - 1 is 𝜎-subnormal in 𝐺. From Lemma 2.1 (6), it follows that S ≤ O σ s ⁢ ( G ) = 1 since 𝑅 is a unique minimal normal subgroup of 𝐺. Therefore, | L n - 1 | = r . For every Sylow 𝑡-subgroup 𝑇 of 𝐿, where t ≠ r , we also have | T | = t by claim (5). Therefore, claim (6) holds.

(7) 𝑅 is a 𝑝-group for some prime p ∈ σ i . Hence R = C G ⁢ ( R ) = O p ⁢ ( G ) .

Assume that 𝑅 is not abelian and let 𝑃 be a Sylow 𝑝-subgroup of 𝑅, where 𝑝 is the smallest prime dividing | R | . Then, for some Sylow 𝑝-subgroup G p and some maximal subgroup 𝑀 of 𝐺, we have G = R ⁢ M and P ≤ G p ≤ N G ⁢ ( P ) ≤ M . From claim (5), it follows that | P | = p , so 𝑅 has a normal 𝑝-complement 𝑉 by [13, IV, 2.8]. It is clear that 𝑉 is normal in 𝐺, so the minimality of 𝑅 implies that V = 1 , and hence R = P is abelian, a contradiction. Therefore, 𝑅 is a 𝑝-group for some prime p ∈ σ i . Hence R = C G ⁢ ( R ) = O p ⁢ ( G ) by claim (6).

(8) For every maximal subgroup 𝑀 of 𝐺 with G = R ⁢ M and for a Sylow 𝑞-subgroup 𝑄 of 𝑀, where 𝑞 divides | M | and q ∈ σ i , we have q ≠ p .

Assume that q = p . From claims (3) and (5), it follows that | Q | = p and V = R ⁢ Q is a non-𝜎-subnormal Hall σ i -subgroup of 𝐺. Moreover, from claims (5), (6) and [6, Chapter A, 1.6 (b)], it follows that 𝑉 is the intersection of all maximal subgroups of 𝐺 containing 𝑉. Hence some maximal subgroup 𝐿 of 𝐺 containing 𝑉 is not 𝜎-subnormal in 𝐺 by Lemma 2.1 (5). Therefore, 𝐿 is a 𝜎-tower group by claim (2), and then 𝑉 is normal in 𝐿 since C G ⁢ ( R ) = R by claim (7). It follows that every subgroup of 𝐿 containing 𝑉 is not 𝜎-subnormal in 𝐺. But 𝑉 is an ( n - 1 ) -maximal subgroup of 𝐺, so every 2-maximal subgroup of 𝑉 is contained in some such proper subgroup 𝑊 of 𝑉 which is 𝜎-subnormal in 𝐺. If | V | > p 2 , then 𝑄 is contained in some 𝜎-subnormal subgroup 𝐺 contained in 𝑉, and so 𝑄 and 𝑉 are 𝜎-subnormal in 𝐺, a contradiction. Therefore, | V | = p 2 , so | R | = p , and hence G / R = G / C G ⁢ ( R ) is a cyclic group of order p - 1 . This contradiction completes the proof of (8).

(9) C R ⁢ ( Q ) = 1 .

Assume that C = C R ⁢ ( Q ) ≠ 1 . Then C ≠ R by claim (7), so from 1 < C < R , it follows that Q < C × Q < R ⁢ Q , and hence 𝑄 is contained in some 2-maximal subgroup 𝐿 of R ⁢ Q and 𝐿 is 𝜎-subnormal in 𝐺 (see claim (8)). But then a Hall σ i -subgroup R ⁢ Q = R ⁢ L of 𝐺 is 𝜎-subnormal in 𝐺 by Lemma 2.1 (5), contrary to claim (3). Hence we have (9).

The final contradiction for (ii). As V = R ⁢ Q is not 𝜎-subnormal in 𝐺 by claim (3), 𝑄 is not normal in 𝑀. Let F ⁢ ( M ) be the Fitting subgroup of 𝑀. Since the order of every Sylow subgroup of 𝑀 is a prime by claim (6), Q ≰ F ⁢ ( M ) and there is a prime 𝑟 dividing | F ⁢ ( M ) | such that a Sylow 𝑟-subgroup 𝐿 of F ⁢ ( M ) is normal in 𝑀 and Q ⁢ L ≠ Q × L since C M ⁢ ( F ⁢ ( M ) ) ≤ F ⁢ ( M ) by [6, Chapter A, Theorem 10.6 (a)].

Let E = ( R ⁢ L ) ⋊ Q . We show that C := C R ⁢ L ⁢ ( Q ) = 1 . Assume that 𝑟 divides | C | . Then C ⁢ Q has a Hall { q , r } -subgroup 𝑉 and 𝑉 is abelian. But L ⁢ Q is a Hall { q , r } -subgroup of 𝐺, and so V x = L ⁢ Q for some x ∈ G , so L ⁢ Q is abelian. This contradiction shows that | C | is a 𝑝-group, and then, in view of claim (9), | C | = 1 . Now let Q = ⟨ a ⟩ . Then 𝑎 induces in R ⁢ L a fixed-point-free automorphism, and hence R ⁢ L is nilpotent by the Thompson’s theorem [8, Chapter 10, Theorem 2.1]. But then L ≤ C G ⁢ ( R ) = R , a contradiction. This final contradiction completes the proof of the result. ∎