On non-normal cyclic subgroups of prime order or order 4 of finite groups

Definition 1.1

[8, Definition 1.1] Let G be a group. We say that G has the N N M property – shortly: G is an N N M -group – if each non-normal subgroup of G is contained in a non-normal maximal subgroup of G .

Definition 1.2

Let G be a group. We say that G has the N C M property – shortly: G is an N C M -group – if each non-normal cyclic subgroup of prime power order in G is contained in a non-normal maximal subgroup of G .

Definition 1.3

Let G be a group. We say that G has the N L M property – shortly: G is an N L M -group – if each non-normal cyclic subgroup of prime order or order 4 in G is contained in a non-normal maximal subgroup of G .

Theorem A

A group G is a solvable T -group if and only if every subgroup of G is an N C M -group.

Theorem B

Let G be a group such that all its non-nilpotent subgroups are N L M -groups. Then G is supersolvable.

Theorem C

Let G be a non-nilpotent minimal non- N L M -group (non- N L M -group all of whose proper subgroups are N L M -groups). Then G must be one of the following types:

1. G is supersolvable.

2. G = P ⋊ Q , where P is isomorphic to the quaternion group Q 8 , Q = ⟨ z ⟩ is cyclic of order 3 r > 1 , z induces an automorphism in P such that P / Φ ( P ) is a faithful and irreducible Q -module, and z centralizes Φ ( P ) .

3. G = P ⋊ Q , where Q = ⟨ z ⟩ is cyclic of order q r > 1 , q ∤ p − 1 , and P is an irreducible Q -module over the field of p elements with kernel ⟨ z q ⟩ in Q .

4. G = P ⋊ Q , where P = ⟨ a 0 , a 1 , … , a q − 1 ⟩ is an elementary abelian p -group of order p q , Q = ⟨ z ⟩ is cyclic of order q r , q f is the highest power of q dividing p − 1 and r > f ≥ 1 . Define a j z = a j + 1 for 0 ≤ j < q − 1 and a q − 1 z = a 0 i , where i is a primitive q f th root of unity modulo p .

Lemma 2.1

Let G = P ⋊ Q , where P and Q are Sylow subgroups of G , P is normal in G and Q is abelian. Then all maximal subgroups of G are of type either P Q 1 g or P 1 Q g , where Q 1 is a maximal subgroup of Q and P 1 is such that P / P 1 is non-trivial and G -simple, where g varies in G and P Q 1 g is normal in G .

Lemma 2.2

Let G be a minimal non-supersolvable group. Then

1. G has a unique normal Sylow p -subgroup P for some prime p .

2. P / Φ ( P ) is a minimal normal subgroup of G / Φ ( P ) .

3. If p ≠ 2 , then the exponent of P is p .

4. If P is non-abelian, then Φ ( P ) = Z ( P ) = P ′ .

5. If P is non-abelian and p = 2 , then the exponent of P is 4.

6. If P is abelian, then the exponent of P is p .

Lemma 2.3

[9, Satz 4] If a group G has four maximal supersolvable subgroups of pairwise coprime indices, then G is supersolvable.

Lemma 2.4

[10, Theorem 1] A group G is a minimal non- T -group (non- T -group all of whose proper subgroups are T -groups) if and only if it is one of the following seven types.

1. The generalized quaternion group of order 16.

2. G = ⟨ x , y ∣ x p m = 1 = y p n , y − 1 x y = x 1 + p m − 1 ⟩ , where p is any prime, m > 1 and n > 0 .

3. G = ⟨ x , y , z ∣ z = [ x , y ] , 1 = x p m = y p n = z p = [ x , z ] = [ y , z ] ⟩ , where p is any prime, m > 0 and n > 0 .

4. G = P ⋊ Q , where P is the quaternion group of order 8, Q = ⟨ y ⟩ is cyclic of order 3 r > 1 , y induces an automorphism permuting cyclically the three maximal subgroups of P .

5. G = P ⋊ Q , where P = ⟨ a , b ⟩ is an elementary abelian p -group of order p 2 , and Q = ⟨ y ⟩ is cyclic of order q r . Define a y = a i , b y = b i j , p ≡ 1 ( mod q f ) , and r ≥ f > 0 , where i is the least positive primitive q f th root of unity modulo p , j = 1 + k q f − 1 , with 0 < k < q .

6. G = P ⋊ Q , where Q = ⟨ y ⟩ is cyclic of order q r > 1 , with q ∤ p − 1 , and P is an irreducible Q -module over the field of p elements with kernel ⟨ y q ⟩ in Q .

7. G = P ⋊ Q , where P = ⟨ a 0 , a 1 , … , a q − 1 ⟩ is an elementary abelian p -group of order p q , Q = ⟨ y ⟩ is cyclic of order q r , q f is the highest power of q dividing p − 1 and r > f ≥ 1 . Define a j y = a j + 1 for 0 ≤ j < q − 1 and a q − 1 y = a 0 i , where i is a primitive q f th root of unity modulo p .

Proof of Theorem A

It is obvious that a nilpotent group is an N C M -group if and only if it is a Dedekind group, i.e., its all subgroups are normal.

By [8, Theorem 1], we only need to prove the sufficient condition for the case that G is non-nilpotent.

Assume that G is a counterexample of minimal order and every subgroup of G is an N C M -group. Surely, the assumption is inherited by subgroups. Consequently, every proper subgroup of G is a solvable T -group by the minimality of G .

Let G be of type (4) in Lemma 2.4. By Lemma 2.1, G has two kinds of maximal subgroups Φ ( P ) ⟨ y ⟩ g and P ⟨ y 3 ⟩ , where g ∈ G , Φ ( P ) ⟨ y ⟩ g ⋬ G , and P ⟨ y 3 ⟩ ⊴ G . Since y induces an automorphism permuting cyclically the three maximal subgroups of the quaternion group, three maximal subgroups of Q 8 are not normal in G . However, they are not contained in Φ ( P ) ⟨ y ⟩ g , a contradiction.

Let G be of type (5) in Lemma 2.4. Then non-normal subgroup ⟨ a b ⟩ of G is only contained in P ⟨ y q ⟩ which is normal in G , a contradiction.

Let G be of type (6) in Lemma 2.4. In fact, G is minimal non-nilpotent and every subgroup of P is not normal in G . However, G has two kinds of maximal subgroups ⟨ y ⟩ g and P ⟨ y q ⟩ ⊴ G , where g ∈ G . This induces a contradiction.

Let G be of type (7) in Lemma 2.4. Similar arguments as above, every subgroup of P is not normal in G and only contained in P ⟨ y q ⟩ ⊴ G , a contradiction.□

Corollary 3.1

Let G be a group. Then the following conditions are equivalent:

1. G is a solvable T -group.

2. Every subgroup of G is an N N M -group.

3. Every subgroup of G is an N C M -group.

Example 3.2

Let G = ⟨ x , y ∣ x p 2 = 1 = y p 3 , y − 1 x y = x 1 + p ⟩ , where p is any prime. Then G is minimal non-abelian and not a T -group.

Lemma 4.1

Let G = P ⋊ Q be a supersolvable N L M -group, where P and Q are Sylow subgroups of G , P is elementary abelian and Q is cyclic non-normal in G . If all subgroups of G are N L M -groups, then G is a T -group.

Proof

Clearly, we only need to consider the case ∣ P ∣ ≥ p 2 for a prime p . Since G is supersolvable and P is elementary abelian, P can be regarded as a completely reducible Q -module over the field of p elements, say P = ⟨ x 1 ⟩ × ⟨ x 2 ⟩ × ⋯ × ⟨ x m ⟩ , Q = ⟨ y ⟩ with x i y = x i α i for some 1 ≤ α i < p , 1 ≤ i ≤ m . According to the structure of T -groups [11], if G is not a T -group, then there exist i , j such that α i ≠ α j . Take the subgroup H = ⟨ x i , x j , y ⟩ . The non-normal maximal subgroups of H are ⟨ x i ⟩ ⟨ y ⟩ h and ⟨ x j ⟩ ⟨ y ⟩ h , where h varies in H . The subgroup U = ⟨ x i x j ⟩ has order p , but is neither normal in H nor contained in a non-normal maximal subgroup of H . Hence, H is not an N L M -group. This contradiction induces that G is a T -group.□

Proof of Theorem B

Assume that G is a counterexample of minimal order. Obviously, the assumption is inherited by subgroups. By the minimality of G , every proper subgroup of G is supersolvable. By Lemma 2.3, π ( G ) has cardinal 2 or 3.

Now we only need to consider the following two cases.

Case 1. Let G = P Q , where P ∈ Syl p ( G ) , Q ∈ Syl q ( G ) , and P ⊴ G , Q ⋬ G .

By Lemma 2.1, all maximal subgroups of G are either of the type P Q 1 or of the type Φ ( P ) Q g , where g ∈ G , and Q 1 is a maximal subgroup of Q . By the assumption that all non-nilpotent subgroups of G are N L M -groups and Lemma 2.2(3)(5)(6), the exponent of P is p or 4 if p = 2 , and P has a cyclic subgroup N of order p or 4, which is not normal in G and N ≰ Φ ( P ) Q g . However, N is only contained in P Q 1 which is normal in G , a contradiction to the fact that G is an N L M -group as it is non-nilpotent. So G is supersolvable.

Case 2. Let G = P Q R , where P ∈ Syl p ( G ) , Q ∈ Syl q ( G ) , R ∈ Syl r ( G ) , P ⊴ G , Q ⋬ G , and R ⋬ G .

By the minimality of G , we have easily that p is the maximal prime divisor of π ( G ) and let p > q > r . Lemma 2.2(3) implies that the exponent of P is p .

1. Assume that P is abelian.

   Since the exponent of P is p , we have that P is an elementary abelian group and it is a minimal normal subgroup of G . If P ⟨ x ⟩ is nilpotent for any x ∈ Q , then Q is also normal in G , which contradicts the fact that minimal non-supersolvable has a unique normal Sylow subgroup. So there exists some element y such that P ⟨ y ⟩ is a non-nilpotent N L M -group, and P ⟨ y ⟩ is a T -group by Lemma 4.1. Thus, every subgroup of P is normal in P Q . Similarly, every subgroup of P is normal in P R , and is normal in G as well, a contradiction.

2. Assume that P is non-abelian.

   By Lemma 2.2(4), Φ ( P ) = Z ( P ) = P ′ . Note that the exponent of P is p , we have easily that G / Φ ( P ) is minimal non-supersolvable and it satisfies the condition of the theorem, and P / Φ ( P ) is a minimal normal subgroup of G / Φ ( P ) . By induction, similar arguments to (1), G / Φ ( P ) is supersolvable, and so G / Φ ( G ) is supersolvable. Therefore, G is supersolvable, which is the final contradiction.□

The following two results are obvious by Theorem B.

Corollary 4.2

Let G be a group such that all its non-nilpotent subgroups are N C M -groups. Then G is supersolvable.

Corollary 4.3

[8, Theorem 7] Let G be a group such that all its non-nilpotent subgroups are N N M -groups. Then G is supersolvable.

Theorem D

Let G be a p -group. Then G is a minimal non- N L M -group if and only if it is one of the following types:

1. G = ⟨ a , b ∣ a p n = 1 , b p = 1 , [ a , b ] = a p n − 1 ⟩ , p an odd prime.

2. G = ⟨ a , b , c ∣ a p n = b p = c p = 1 , [ a , b ] = c , [ c , a ] = [ c , b ] = 1 ⟩ , p an odd prime.

3. G = ⟨ a , b , c ∣ a 2 n = b 2 m = c 2 = 1 , [ a , b ] = c , [ c , a ] = [ c , b ] = 1 ⟩ , where n ≥ m and 1 ≤ m ≤ 2 (if n = 2 , m = 1 , then G is isomorphic to the dihedral group D 8 ).

4. G = ⟨ a , b ∣ a 2 n = 1 , b 2 m = 1 , [ a , b ] = a 2 n − 1 ⟩ , n ≥ 2 , m ≥ 2 , and at least one of n and m is 2.

5. G = ⟨ a , b ∣ a 8 = 1 , b 2 = a 4 , [ a , b ] = a − 2 ⟩ .

Proof of Theorem C

Suppose that G is a minimal non- N L M -group, then by Theorem B, every proper subgroup of G is supersolvable. Therefore, G is either supersolvable as type (1) or minimal non-supersolvable. In the following, we assume that G is minimal non-supersolvable.

By Lemma 2.2(1)(3)(5), G has a unique normal Sylow p -subgroup P , and the exponent of P is p or 4 if P is a non-abelian 2-group. Since all proper subgroups of G are N L M -groups, P is either elementary abelian or isomorphic to the quaternion group Q 8 .

First, we examine the classification of minimal non-supersolvable groups in [15, Theorem 9]. Groups of types 5, 7, 9, and 12 in [15, Theorem 9] all contain an extraspecial subgroup P of order p 3 , where p > 2 . Note that P is not an N L M -group. Those groups are not satisfied with minimal non- N L M -groups. Let P be Q 8 as type 3 in [15, Theorem 9]. Note that G / C G ( P ) ≲ S 4 , where S 4 is the symmetric group of degree 4. It makes Q is a Sylow 3-subgroup of G . Furthermore, G is also a minimal non-nilpotent group, thus G is of type (2).

Next, we assume that G is isomorphic to one of types 2, 4 in [15, Theorem 9]. It is easy to see that all subgroups of G are P ⟨ z q ⟩ g and ⟨ z ⟩ g , where g ∈ G . In type 2, G is also a minimal non-nilpotent group, G is of type (3). In type 4, G = P ⋊ Q , where P = ⟨ a 0 , a 1 , … , a q − 1 ⟩ is elementary abelian, Q = ⟨ z ⟩ is of order q r , a j z = a j + 1 for 0 ≤ j < q − 1 and a q − 1 z = a 0 i , q f is the highest power of q dividing p − 1 and r > f ≥ 1 , and i is a primitive q f th root of unity modulo p . Since a j z q = a j i for 0 ≤ j ≤ q − 1 , z q induces a power automorphism with fixed-point-free on P . Certainly, all subgroups of P ⟨ z q ⟩ g are N L M -groups. Thus, G is of type (4).

Then, we assume that G is isomorphic to one of types 6, 8, and 10 in [15, Theorem 9]. For arbitrary element x of E , by the assumption that P ⟨ x ⟩ is an N L M -group and Lemma 4.1, P ⟨ x ⟩ is a T -group. The arbitrariness of x leads to the normality of every subgroup of P in G , a contradiction.

Finally, let G be type 11 in [15, Theorem 9]. Namely, G = P Q R , where P ∈ Syl p ( G ) , P ⊴ G , Q ∈ Syl q ( G ) , R ∈ Syl r ( G ) , R is a cyclic subgroup of order r s + t , with r a prime number and s and t integers such that s ≥ 1 and t ≥ 0 , normalizing Q , Q / Φ ( Q ) is an irreducible R -module over the field of q elements, with kernel the subgroup D of order r t of R , and P is an irreducible Q R -module over the field of p elements, where q ∣ p − 1 , r s ∣ p − 1 , and r ∣ q − 1 . In this case, Φ ( G ) p ′ , the Hall p ′ -subgroup of Φ ( G ) , coincides with Φ ( Q ) × D and centralizes P . By the same arguments as above, P R is a T -group. Therefore, every subgroup of P is normal in P R . However, the supersolvability of P Q implies that there exists a non-trivial subgroup of P which is normal in P Q . This contradicts the fact P is an irreducible Q R -module over the field of p elements. Therefore, G does not coincide with a minimal non- N L M -group.□