Finite groups whose maximal subgroups of even order are MSN-groups

Lemma 2.1

[1, Theorem 3] Let G be an MSN-group. Then G has the Sylow tower property for some ordering of the primes in π ( G ) , and hence, G is solvable.

Lemma 2.2

A subgroup of an MSN-group must be an MSN-group.

Proof

By the definition of MSN-groups and the property of subnormal subgroups, the lemma is true.□

Lemma 2.3

[5, Theorem 2.7] Let G be a group. Then the following conditions are equivalent:

1. Maximal subgroups of Sylow subgroups of G are subnormal in G ;

2. G = H ⋊ K , where (1) H is a nilpotent normal Hall subgroup of G , and K is a group whose Sylow subgroups are cyclic and the maximal subgroups of its Sylow subgroups are normal in K ; (2) a generator x of any Sylow p-subgroup of K induces an automorphism of order 1 or p on H .

Remark 2.4

[2, Remark] In Lemma 2.3, the normality of maximal subgroups of p -Sylow subgroups of K can be extended to G .

Lemma 2.5

[2, Lemma 2.8] Let G be a solvable minimal non-MSN-group. Then ∣ G ∣ = p a q b , where p and q are distinct primes and at least one of a and b is more than 1.

Lemma 2.6

[6, Theorem 10.1.4] If a group G has a fixed-point-free automorphism of order 2, then G is abelian.

Lemma 2.7

[7, Corollary 1] Every minimal simple group is isomorphic to one of the following groups:

1. PSL ( 3 , 3 ) ;

2. The Suzuki group Sz ( 2 q ) , where q is an odd prime;

3. PSL ( 2 , p ) , where p is a prime with p > 3 and p 2 ≢ 1 ( mod 5 ) ;

4. PSL ( 2 , 2 q ) , where q is a prime;

5. PSL ( 2 , 3 q ) , where q is an odd prime.

Theorem 3.1

Let G be a solvable non-MSN-group of even order. If all maximal subgroups of G of even order are MSN-groups, then ∣ π ( G ) ∣ ≤ 3 .

Proof

Let π ( G ) = { p 1 , p 2 , … , p s } with p 1 = 2 and { P 1 , P 2 , … , P s } be a Sylow basis of G . If G is a minimal non-MSN-group, then ∣ π ( G ) ∣ = 2 by Lemma 2.5. So the conclusion holds.

Now we assume that G is not a minimal non-MSN-group. By hypothesis, G possesses a maximal subgroup M of odd order which is not an MSN-group. Without loss of generality, let M = P 2 ⋯ P s .

Since M is not an MSN-group, then there exist a positive integer j and a maximal subgroup P ∗ of P j such that P ∗ is not subnormal in M . Without loss of generality, we can let j = s . If s ≥ 4 , then P 1 P i P s ( i = 2 , … , s − 1 ) is a proper subgroup of G of even order, and hence, it is an MSN-group by hypothesis and Lemma 2.2. If P s is non-cyclic, then P s is normal in P 1 P i P s by Lemma 2.3, and so P s is normal in G = P 1 P 2 ⋯ P s . Therefore, P ∗ is subnormal in G , a contradiction. Hence, P s is cyclic and P ∗ is normal in P 1 P i P s by Remark 2.4. So P ∗ is normal in G = P 1 P 2 ⋯ P s , a contradiction. Thus, ∣ π ( G ) ∣ ≤ 3 .□

Theorem 3.2

Let G be a solvable non-MSN-group with π ( G ) = { 2 , p } . Then all maximal subgroups of G of even order are MSN-groups if and only if G is a minimal non-MSN-group.

Proof

Clearly, the maximal subgroup of G of odd order (if exists) is a Sylow subgroup, and so it is an MSN-group. By Lemma 2.2, the rest is clear.□

Theorem 3.3

Let G be a solvable non-MSN-group of even order with ∣ π ( G ) ∣ = 3 , where P ∈ Syl p ( G ) , Q ∈ Syl q ( G ) , and R ∈ Syl 2 ( G ) with p > q > r = 2 . Suppose that all maximal subgroups of G of even order are MSN-groups. Then one of the following statements holds:

1. G = M × R , where M is a minimal non-MSN-group and R is a cyclic group of order 2;

2. G = ⟨ a , b ∣ a p = b 2 q n = 1 , b − 1 a b = a i ⟩ , where i 2 q ≢ 1 ( mod p ) , i 2 q 2 ≡ 1 ( mod p ) , n ≥ 2 , 0 < i < p ;

3. G = M ⋊ R = ( P ⋊ Q ) ⋊ R with 1 < C M ( R ) < M , where M is a non-MSN-group with P non-cyclic and Q = ⟨ b ⟩ of order q n , n ≥ 2 , b induces an automorphism of order q 2 on P and R is a cyclic group of order 2;

4. G = M ⋊ R = ( Q ⋊ P ) ⋊ R with 1 < C M ( R ) < M , where M is a non-MSN-group with P and Q non-cyclic, P Φ ( Q ) is nilpotent, and R is a cyclic group of order 2;

5. G = M ⋊ R = ( Q ⋊ P ) ⋊ R with 1 < C M ( R ) < M , where M is a non-MSN-group with Q non-cyclic and P = ⟨ a ⟩ of order p m , m ≥ 2 , a induces an automorphism of order p 2 on Q , and R is a cyclic group of order 2;

6. G = M ⋊ R = ( P ⋊ Q ) ⋊ R with 1 < C M ( R ) < M , where M is a non-MSN-group with P and Q non-cyclic, Φ ( P ) Q is nilpotent, and R is a cyclic group of order 2;

7. G = M ⋊ R = P Q ⋊ R with 1 < C M ( R ) < M , where M is a non-MSN-group with P cyclic and Q non-cyclic, R is a cyclic group of order 2;

8. G = M ⋊ R = ( P ⋊ Q ) ⋊ R with 1 < C M ( R ) < M , where M is a non-MSN-group with P of order p and Q non-cyclic, R is a cyclic group of order 2;

9. G = R ⋊ M = R ⋊ ( P ⋊ Q ) , where M = P ⋊ Q = ⟨ a , b ∣ a p = b q n = 1 , b − 1 a b = a i ⟩ , i q ≢ 1 ( mod p ) , i q 2 ≡ 1 ( mod p ) , n ≥ 2 , 0 < i < p , P R is nilpotent and Q R is a minimal non-abelian group, and R is elementary abelian with ∣ R ∣ ≥ 4 ;

10. G = R ⋊ M = R ⋊ ( P ⋊ Q ) , where M = P ⋊ Q = ⟨ a , b ∣ a p = b q n = 1 , b − 1 a b = a i ⟩ , i q ≢ 1 ( mod p ) , i q 2 ≡ 1 ( mod p ) , n ≥ 2 , 0 < i < p , Q R is nilpotent and PR is a minimal non-abelian group, and R is elementary abelian with ∣ R ∣ ≥ 4 ;

11. G = R ⋊ M = R ⋊ ( P ⋊ Q ) , where M = P ⋊ Q = ⟨ a , b ∣ a p = b q n = 1 , b − 1 a b = a i ⟩ , i q ≢ 1 ( mod p ) , i q 2 ≡ 1 ( mod p ) , n ≥ 2 , 0 < i < p , b induces an automorphism of order q on R and P R is non-nilpotent, and R is elementary abelian with ∣ R ∣ ≥ 4 ;

12. G = R ⋊ M = R ⋊ P Q , where M is a non-MSN-group with P cyclic and Q non-cyclic, QR is nilpotent and PR is a minimal non-abelian group, and R is elementary abelian with ∣ R ∣ ≥ 4 ;

13. G = R ⋊ M = R ⋊ ( P ⋊ Q ) , where M is a non-MSN-group with P non-cyclic and Q = ⟨ b ⟩ of order q n , n ≥ 2 , b induces an automorphism of order q 2 on P , P R is nilpotent and Q R is a minimal non-abelian group, and R is elementary abelian with ∣ R ∣ ≥ 4 .

Proof

Since ∣ π ( G ) ∣ = 3 , G is not a minimal non-MSN-group by Lemma 2.5. Then there exists a maximal subgroup M of G of order odd such that M is not an MSN-group by hypothesis. The solvability of G implies that we can let π ( G ) = { p , q , r } with p > q > r = 2 and { P , Q , R } be a Sylow basis of G . Furthermore, we can let M be a Hall 2 ′ -subgroup of G and G = M R , where M = P Q .

By hypothesis and Lemma 2.2, both P R and Q R are MSN-groups. We first prove that R is either of order 2 or elementary abelian from two cases as follows.

Case 1 R is cyclic.

Since P R and Q R are MSN-groups, then Φ ( R ) is normal in P R and Q R by Remark 2.4, and so Φ ( R ) is normal in G . If Φ ( R ) ≠ 1 , then M < M Φ ( R ) < M R = G , which contradicts that M is a maximal subgroup of G , so Φ ( R ) = 1 and R is of order 2.

Case 2 R is non-cyclic.

Since P R and Q R are MSN-groups, then R is normal in P R and Q R by Lemma 2.3, and so R is normal in G . We have that Φ ( R ) is normal in G as Φ ( R ) char R ⊴ G . If Φ ( R ) ≠ 1 , then M < M Φ ( R ) < M R = G , which contradicts that M is a maximal subgroup of G , so Φ ( R ) = 1 and R is elementary abelian.

We next complete the rest of the proof as follows.

(1) R is of order 2.

Clearly, M is the normal 2-complement.

Suppose C M ( R ) = 1 . Then an automorphism of R acting on M is both of order 2 and fixed-point-free. It follows from Lemma 2.6 that M is abelian, and so M is an MSN-group, a contradiction. Thus, C M ( R ) > 1 .

If C M ( R ) = M , then M R = M × R , and M 1 R is an MSN-group by hypothesis for any maximal subgroup M 1 of M . By Lemma 2.2, M 1 is an MSN-group, and so M is a minimal non-MSN-group. Therefore, G is of type (I).

Now we consider that 1 < C M ( R ) < M . Let 1 ⊴ ⋯ ⊴ K ⊴ P Q ⊴ G be a chief series of G . Since G is solvable, then one of P and Q is contained in K . If Q ≤ K , then K R is an MSN-group by hypothesis, and Q is either cyclic or normal in K by Lemma 2.3. If Q is normal in K , then Q is normal in G as Q char K ⊴ G . Similarly, if P ≤ K , then P is either cyclic or normal in G . Obviously, Q cannot be both cyclic and normal in G , so we discuss from the five cases as follows.

(1-a) Q is cyclic but not normal in G .

Clearly, P ⊴ G . Then P Φ ( Q ) R is an MSN-group by hypothesis, and so P Φ ( Q ) is an MSN-group by Lemma 2.2. Similarly, since Φ ( P ) Q R is contained in some maximal subgroup of G of even order, then Φ ( P ) Q is an MSN-group by Lemma 2.2.

Let Q = ⟨ z ⟩ with ∣ z ∣ = q n , n ≥ 2 . If P is cyclic, then let P = ⟨ a ⟩ with ∣ a ∣ = p m . As Φ ( P ) Q is an MSN-group, Φ ( P ) Φ ( Q ) is nilpotnet by Remark 2.4. If Φ ( P ) ≠ 1 , then Ω 1 ( P ) ≤ Φ ( P ) , and hence, P Φ ( Q ) is nilpotnet by a result in [8]. Therefore, M = P Q is an MSN-group by Lemma 2.3, a contradiction. Thus, Φ ( P ) = 1 and P is cyclic of order p . Since 1 < C M ( R ) < M and G is non-abelian with all Sylow subgroups cyclic, we have that G = P ⋊ ( Q × R ) by [9, 10.1.10]. Let Q × R = ⟨ b ⟩ and b − 1 a b = a i . It follows that ( b 2 q ) − 1 a b 2 q = a i 2 q ≠ a from ⟨ b 2 q ⟩ is not subnormal in P Q . Therefore, i 2 q ≢ 1 ( mod p ) . Since P Φ ( Q ) is an MSN-group, ⟨ b 2 q 2 ⟩ is normal in P Φ ( Q ) by Remark 2.4. Hence, ( b 2 q 2 ) − 1 a b 2 q 2 = a i 2 q 2 = a , i 2 q 2 ≡ 1 ( mod p ) , and G is of type (II). If P is non-cyclic, using similar arguments as mentioned earlier, ⟨ z q 2 ⟩ is normal in P Φ ( Q ) . Hence, G is of type (III).

(1-b) Q is non-cyclic and Q ⊴ G .

Since P Φ ( Q ) R is contained in some maximal subgroup of G of even order, P Φ ( Q ) is an MSN-group by hypothesis and Lemma 2.2. Similarly, Φ ( P ) Q is an MSN-group. If P is non-cyclic, then P Φ ( Q ) is nilpotent by Lemma 2.3, so G is of type (IV). If P is cyclic, then by Lemma 2.3 let P = ⟨ a ⟩ with ∣ a ∣ = p m and m ≥ 2 . Since Φ ( P ) Q is an MSN-group, ⟨ a p 2 ⟩ is normal in Φ ( P ) Q by Remark 2.4. So G is of type (V).

(1-c) P is non-cyclic and P ⊴ G .

If Q is cyclic, then G is of type (III) by the same arguments as in (1-a). If Q is non-cyclic, then Φ ( P ) Q R is contained in some maximal subgroup of G of even order, so Φ ( P ) Q is an MSN-group by hypothesis and Lemma 2.2. Therefore, Φ ( P ) Q is nilpotent by Lemma 2.3, and G is of type (VI).

(1-d) P is cyclic but not noraml in G .

Clearly, Q is non-cyclic, so G is of type (VII).

(1-e) P is cyclic and P ⊴ G .

If Q is cyclic, then G is of type (II) by the same arguments as in (1-a). If Q is non-cyclic, we have that P is of order p . Otherwise, Ω 1 ( P ) ≤ Φ ( P ) and Φ ( P ) Q is an MSN-group by hypothesis and Lemma 2.2. By Lemma 2.3, Φ ( P ) Q is nilpotent, and so P Q is nilpotent by a result in [8], a contradiction. Thus, G is of type (VIII).

(2) R is normal in G , and it is elementary abelian.

Assume that neither P nor Q is cyclic. Since P R and Q R are MSN-groups, P R and Q R are nilpotent by Lemma 2.3. Therefore, M R = M × R . By hypothesis and Lemma 2.2, M R 1 is an MSN-group, where R 1 < R . By Lemma 2.2 again, M is an MSN-group, a contradiction. Hence, either P or Q is cyclic.

(2-a) Both P and Q are cyclic.

By hypothesis and Lemma 2.2, P Φ ( Q ) and Φ ( P ) Q are MSN-groups, and so P Q is a minimal non-MSN-group. Let P = ⟨ a ⟩ and Q = ⟨ b ⟩ with ∣ a ∣ = p m and ∣ b ∣ = q n . Then by examining [2, Theorem 3.2], we conclude that M = P Q = ⟨ a , b ∣ a p = b q n = 1 , b − 1 a b = a i ⟩ , i q ≢ 1 ( mod p ) , i q 2 ≡ 1 ( mod p ) , n ≥ 2 , 0 < i < p . Clearly, P R and Q R are not all nilpotent. If P R is nilpotent, then there does not exist a non-trivial subgroup R 1 of R such that R 1 Q = Q R 1 . Otherwise, if R 1 Q = Q R 1 , where R 1 < R , we have that R 1 P Q is a proper subgroup of G of even order. Then R 1 P Q is an MSN-group by hypothesis and Lemma 2.2, and so P Q is an MSN-group by Lemma 2.2, a contradiction. Since Q R is an MSN-group, then Φ ( Q ) R is nilpotent by Remark 2.4. Thus, Q R is a minimal non-abelian group, and G is of type (IX). If Q R is nilpotent, then by a similar argument as earlier, there does not exist a non-trivial subgroup R 1 of R such that R 1 P = P R 1 . Therefore, P R is a minimal non-abelian group, and G is of type (X). If both P R and Q R are non-nilpotent, then b induces an automorphism of order q on R by Lemma 2.3, and so G is of type (XI).

(2-b) P is cyclic and Q is non-cyclic.

Since P R and Q R are MSN-groups, Φ ( P ) R and Q R are nilpotent by Remark 2.4 and Lemma 2.3, respectively. By the similar arguments as in (2-a), there does not exist a non-trivial subgroup R 1 of R such that R 1 P = P R 1 . Hence, P R is a minimal non-abelian group since Φ ( P ) R is nilpotent. Therefore, G is of type (XII).

(2-c) P is non-cyclic and Q is cyclic.

Using similar arguments as in (2-b), Φ ( Q ) R and P R are nilpotent, and so P is normal in G . Then P Φ ( Q ) R is an MSN-group by hypothesis, so P Φ ( Q ) is an MSN-group by Lemma 2.2. Let Q = ⟨ b ⟩ with ∣ b ∣ = q n , n ≥ 2 . Then ⟨ b q 2 ⟩ is normal in P Φ ( Q ) by Remark 2.4. By the similar arguments as in (2-a), there does not exist a non-trivial subgroup R 1 of R such that R 1 Q = Q R 1 . We have that Q R is a minimal non-abelian group since Φ ( Q ) R is nilpotent. Thus, G is of type (XIII).□

Corollary 3.4

Let G be a solvable non-MSN-group of even order and suppose that all maximal subgroups of G of even order are MSN-groups. If 4 ∣ ∣ G ∣ , then G is a minimal non-MSN-group or ∣ π ( G ) ∣ = 3 and G possesses a normal Sylow 2-subgroup which is elementary abelian.

Proof

It is obvious by Theorems 3.2 and 3.3.□

Theorem 3.5

Let G be a non-abelian simple group. If all maximal subgroups of G of even order are MSN-groups, then G is isomorphic to either A 5 or PSL ( 2 , 2 q ) , where A 5 is the alternating group of degree 5 and 2 q − 1 is square-free for an odd prime q.

Proof

Let M be a maximal subgroup of G . If M is a group of odd order, then M is solvable by Feit-Thompson Theorem [10] on the solvability of group of odd order. If M is a group of even order, then M is an MSN-group by hypothesis. By applying Lemma 2.1, M is solvable. So G is a minimal simple group. Using similar arguments as the proof in [2, Theorem 3.1], the required result holds.□

Theorem 3.6

Let G be a minimal simple group all of whose second maximal subgroups are MSN-groups. Then G is isomorphic to one of the following types:

1. A 5 ;

2. The Suzuki group Sz ( 2 3 ) ;

3. PSL ( 2 , p ) , where p is an odd prime with p > 5 , 5 ∤ p 2 − 1 , only one of ( p + 1 ) / 4 and ( p − 1 ) / 4 is an odd prime if p ≥ 13 , and ( p − 1 ) / 2 is square-free or r 2 for an odd prime r if ( p + 1 ) / 4 is an odd prime;

4. PSL ( 2 , 2 q ) , where q is an odd prime and 2 q − 1 is square-free or r 2 for an odd prime r ;

5. PSL ( 2 , 3 q ) , where q and ( 3 q + 1 ) / 4 are odd primes, ( 3 q − 1 ) / 2 is square-free or r 2 for an odd prime r .

Proof

By Lemma 2.7, we have that G is isomorphic to one of the following simple groups.

1. PSL ( 3 , 3 ) ;

2. The Suzuki group Sz ( 2 q ) , where q is an odd prime;

3. PSL ( 2 , p ) , where p is a prime with p > 3 and p 2 ≢ 1 ( mod 5 ) ;

4. PSL ( 2 , 2 q ) , where q is a prime;

5. PSL ( 2 , 3 q ) , where q is an odd prime.

Case 1 G ≇ PSL ( 3 , 3 ) .

Suppose G ≅ PSL ( 3 , 3 ) . Then G contains a maximal subgroup which is isomorphic to the Hesse group A of order 2 4 3 3 . By hypothesis, A is either an MSN-group or a minimal non-MSN-group. Clearly, all Sylow subgroups of A are non-cyclic, so at least one of the Sylow subgroups of A is normal in A by Lemma 2.3 and a result in [2, Theorem 3.2]. However, the Sylow subgroups of A are all non-normal. Hence G ≇ PSL ( 3 , 3 ) .

Case 2 G ≅ Sz ( 2 3 ) .

Suppose G ≅ Sz ( 2 q ) . Then by [11, Theorem 9], G has maximal subgroups: the Frobenius group M with a cyclic complement H of order 2 q − 1 and kernel K of order 2 2 q ; the dihedral group D of order 2 ( 2 q − 1 ) and the Frobenius group F of order 4 ( 2 q ± 2 q + 1 2 + 1 ) . Clearly, the dihedral group D of order 2 ( 2 q − 1 ) is an MSN-group. The Frobenius group M is either an MSN-group or a minimal non-MSN-group by hypothesis. If M is a minimal non-MSN-group, then 2 q − 1 is p l by Lemma 2.5 for an odd prime p and a positive integer l with l > 1 . Since K is non-abelian, then Φ ( K ) H is an MSN-group with Φ ( H ) ≠ 1 , and so Φ ( K ) Φ ( H ) = Φ ( K ) × Φ ( H ) by Remark 2.4, which contradicts the fact that M is a Frobenius group. Thus, M must be an MSN-group, and we have that 2 q − 1 is square-free. Otherwise, by Remark 2.4, there exists a non-trivial maximal subgroup H 1 of the Sylow subgroups of H such that K H 1 = K × H 1 , which contradicts the fact M is a Frobenius group. By hypothesis, the Frobenius group F is either an MSN-group or a minimal non-MSN-group. Clearly, the Sylow 2-subgroup of F is neither cyclic nor normal, so F is a minimal non-MSN-group. By the similar arguments as given earlier, 2 q ± 2 q + 1 2 + 1 is r t for an odd prime r and a positive integer t . If t > 1 , then there exists a non-trivial subgroup F 1 of F such that the Sylow 2-subgroup of F 1 is neither cyclic nor normal, which contradicts that F is a minimal non-MSN-group. Therefore, 2 q ± 2 q + 1 2 + 1 are odd primes. By the computation, we easily have that at least one of 2 q + 2 q + 1 2 + 1 and 2 q − 2 q + 1 2 + 1 is a multiple of 5 for any odd prime q with q ≥ 5 , a contradiction. Thus, only odd prime 3 satisfies all of the aforementioned conditions, and so G is of type (II).

Case 3 G ≅ A 5 or G ≅ PSL ( 2 , p ) , where p is an odd prime with p > 5 , 5 ∤ p 2 − 1 , only one of ( p + 1 ) / 4 and ( p − 1 ) / 4 is an odd prime if p ≥ 13 , and ( p − 1 ) / 2 is square-free or r 2 for an odd prime r if ( p + 1 ) / 4 is an odd prime.

Suppose G ≅ PSL ( 2 , p ) . By [12, Corollary 2.2], G has maximal subgroups: the alternating group A 4 of degree 4 when p ≡ ± 3 ( mod 8 ) ; the symmetric group S 4 of degree 4 when p 2 ≡ 1 ( mod 16 ) ; the dihedral groups of order p ± 1 ; and the Frobenius group M with a cyclic complement H of order ( p − 1 ) / 2 and kernel K of order p . G ≅ A 5 if p = 5 , so G is of type (I). Obviously, the prime 7 satisfies the condition, so we only consider the case when p ≥ 13 . Furthermore, 4 must divide the order of either D p − 1 or D p + 1 if p ≥ 13 .

Suppose p ≥ 13 and 4 ∣ ∣ D p − 1 ∣ . Clearly, the Sylow 2-subgroup of D p − 1 is non-cyclic. Since D p − 1 is not an MSN-group, we have that D p − 1 is a minimal non-MSN-group by hypothesis. Note 2 ∤ ( p − 1 ) / 4 . Otherwise, there exists a maximal subgroup M 1 of D p − 1 such that the Sylow 2-subgroup of M 1 is neither cyclic nor normal, which contradicts D p − 1 is a minimal non-MSN-group. Then by examining [2, Theorem 3.2], we have that ( p − 1 ) / 4 is an odd prime, say q . Then p + 1 = 2 ( 2 q + 1 ) , and so D p + 1 is an MSN-group, as desired. And ( p − 1 ) / 2 = 2 q , and hence, the Frobenius group M of order p ( p − 1 ) / 2 is an MSN-group, as desired. Suppose p ≥ 13 and 4 ∣ ∣ D p + 1 ∣ . By the similar arguments as given earlier, D p + 1 must be a minimal non-MSN-group and 2 ∤ ( p + 1 ) / 4 . Then by examining [2, Theorem 3.2], we have that ( p + 1 ) / 4 is an odd prime, say t . Then p − 1 = 2 ( 2 t − 1 ) , and so D p − 1 is an MSN-group, as desired. And ( p − 1 ) / 2 = 2 t − 1 is odd, using similar arguments as in Case 2, ( p − 1 ) / 2 is square-free if M is an MSN-group. If M is a minimal non-MSN-group, then ( p − 1 ) / 2 is r m by Lemma 2.5 for an odd prime r and a positive integer m with m > 1 . We have m = 2 , if not, there exists a maximal subgroup H 1 of H such that K H 1 is an MSN-group with Φ ( H 1 ) ≠ 1 , and so K Φ ( H 1 ) = K × Φ ( H 1 ) by Remark 2.4, which contradicts the fact M is a Frobenius group. Now we have that G is of type (III).

Case 4 G ≅ A 5 or G ≅ PSL ( 2 , 2 q ) , where q is an odd prime and 2 q − 1 is square-free or r 2 for an odd prime r .

If G ≅ PSL ( 2 , 2 q ) , then by [12, Corollary 2.2], G has maximal subgroups: the dihedral groups of order 2 ( 2 q ± 1 ) ; the Frobenius group M with a cyclic complement H of order 2 q − 1 and kernel K is elementary abelian of order 2 q ; and the alternating group A 4 of degree 4 when q = 2 . Clearly, G ≅ A 5 if q = 2 , so G is of type (I). Now we consider that q > 2 . By the similar arguments as in Case 2, 2 q − 1 is square-free if M is an MSN-group. If M is a minimal non-MSN-group, then 2 q − 1 is r m by Lemma 2.5 for an odd prime r and a positive integer m with m > 1 . We prove that there does not exist a non-trivial subgroup K 1 of K such that K 1 ⊴ M . If not, then K 1 H is an MSN-group, and so K 1 Φ ( H ) = K 1 × Φ ( H ) by Remark 2.4, which contradicts the fact M is a Frobenius group. Now we prove that m = 2 . If not, then by the similar arguments as the proof in Case 3, there exists a maximal subgroup H 1 of H with Φ ( H 1 ) ≠ 1 such that K Φ ( H 1 ) = K × Φ ( H 1 ) , which contradicts the fact M is a Frobenius group. Hence, G is of type (IV).

Case 5 G ≅ PSL ( 2 , 3 q ) , where q and ( 3 q + 1 ) / 4 are odd primes, ( 3 q − 1 ) / 2 is square-free or r 2 for an odd prime r .

If G ≅ PSL ( 2 , 3 q ) , then by [12, Corollary 2.2], G has maximal subgroups: the dihedral groups of order 3 q ± 1 ; the Frobenius group M with a cyclic complement H of order ( 3 q − 1 ) / 2 and kernel K is elementary abelian of order 3 q ; and the alternating group A 4 of degree 4. Clearly, 4 ∣ 3 q + 1 , and the dihedral group of order 3 q − 1 is an MSN-group, so we only consider the Frobenius group M and the dihedral group D of order 3 q + 1 . Using similar arguments as in Case 4, ( 3 q − 1 ) / 2 is square-free if M is an MSN-group, and ( 3 q − 1 ) / 2 is r 2 for an odd prime r if M is a minimal non-MSN-group. Using similar arguments as in Case 3, D must be a minimal non-MSN-group and 2 ∤ ( 3 q + 1 ) / 4 . By examining [2, Theorem 3.2], ( 3 q + 1 ) / 4 is an odd prime. So G is of type (V).□