Finite groups all of whose minimal subgroups are N E ^∗-subgroups

Abstract

Let G be a finite group. A subgroup H of G is called an NE-subgroup of G if it satisfies H ^G∩N _G (H)=H. A subgroup H of G is said to be a N E ^∗-subgroup of G if there exists a subnormal subgroup T of G such that G=H T and H∩T is a NE-subgroup of G. In this article, we investigate the structure of G under the assumption that subgroups of prime order are N E ^∗-subgroups of G. The finite groups, all of whose minimal subgroups of the generalized Fitting subgroup are N E ^∗-subgroups are classified.

1 Introduction

All groups considered will be finite. We use conventional notions and notation, as in Huppert [10]. Throughout this article, G stands for a finite group and π(G) denotes the set of primes dividing |G|. Notation and basic results in the theory of formations are taken mainly from Doerk and Hawkes [6].

Recall that a subgroup H of a group G is called c-supplemented (c-normal, weakly c-normal, respectively) in G if there exists a subgroup (normal subgroup, subnormal subgroup, respectively) K of G such that G=H K and H∩K≤H _G , where H _G =Core _G (H) is the largest normal subgroup of G contained in H (see [3; 15; 17]). Following Li [12], a subgroup R of G is called a NE-subgroup of G if R ^G∩N _G (R)=R. In the recent years, there has been much interest in investigating the influence of NE-subgroups of prime order and cyclic subgroups of order 4 on the structure of the groups. In [4], Bianchi et al. introduced the concept of a $\mathcal {H}$-subgroup and investigated the influence of $\mathcal {H}$-subgroups on the structure of a group G: a subgroup H of G is said to be a $\mathcal {H}$-subgroup of G if N _G (H)∩H ^g≤H for all g∈G. Asaad [1] described the groups, all of whose certain subgroups of prime power orders are $\mathcal {H}$-subgroups. Clearly an NE-subgroup is an $\mathcal {H}$-subgroup in G. The converse is not true in general (see [13]).

The aim of this paper is threefold. First, we introduce a new concept called N E ^∗-subgroup which covers properly the notion of NE-subgroup (see Definition 1.1 and Example 1.2 below). Our second aim is to characterize the structure of a group G with the requirement that certain subgroups of G possess the N E ^∗-property. We state our results in the broader context of formation theory and only consider the conditions on minimal subgroups of G (dropping the assumption that every cyclic subgroup of order 4 is an N E ^∗-subgroup). Our final aim is to investigate the structure of groups G with the property that all the cyclic subgroups of prime order or order 4 of G satisfy the N E ^∗-property. We first introduce the following concept:

DEFINITION 1.1

A subgroup H of a finite group G is said to be an N E ^∗-subgroup of G if there exists a subnormal subgroup T of G such that G=H T and H∩T is an NE-subgroup of G.

It is clear that NE-subgroups are N E ^∗-subgroups but the converse is not true in general.

Example 1.2.

G=S ₄, the symmetric group of degree 4, and L=A ₄, the alternating group of degree 4. Clearly, G=L⋊H, where H=〈(13)〉. Observe that H∩A ₄=1, this yields that H is an N E ^∗-subgroup of G by Definition 1.1. Now H ^(12)(34)=〈(24)〉≤N _G (H) and (12)(34)∉N _G (H) show that H is not an NE-subgroup of G.

Buckley [5] proved that a finite group of odd order, all of whose minimal subgroups are normal is supersolvable. We prove the following theorem which is an improvement of a recent result due to Asaad and Ramadan (see Theorem 1.1 of [2]). Hence, Q ₈ will denote the quaternion group of order 8 and a group G is called Q ₈-free if no quotient group of any subgroup of G is isomorphic to Q ₈. Throughout this paper, $\mathcal {U}$ will denote the class of all supersolvable groups. Clearly, $\mathcal {U}$ is a formation. The $\mathcal {U}$-hypercentre $Z_{\mathcal {U}}(G)$ of G is the product of all normal subgroups H of G such that each chief factor of G below H has prime order.

Theorem 1.3.

Let G be Q ₈ -free and let P be a nontrivial normal p-subgroup of G. If all minimal subgroups of P are NE ^∗ -subgroups of G, then $P \leq Z_{\mathcal {U}}(G)$ , where $\mathcal {U}$ is the formation of all supersolvable groups.

Theorem 1.3 may be false if we drop the first condition. The following example shows the necessity of the ‘ Q ₈-free’ hypothesis in Theorem 1.3.

Example 1.4.

Let G be the semidirect product of the quaternion group P of order 8 and the cyclic group 〈c〉 of order 9, where P=〈a,b|a ⁴=1,b ²=a ²,a ^b=a ⁻¹〉, which is isomorphic to Q ₈ and c acts on P or equivalently, c is the automorphism of order 3 of G given by a ^c=b,b ^c=a b. Thus G=P⋊〈c〉 is a group of order 2³⋅3². We obtain that there are only two minimal subgroups, i.e., 〈a ²〉 and 〈c ³〉 in G, and the centre of G is Z(G)=〈a ²〉×〈c ³〉 (see p. 292 of [14]). Thus all minimal subgroups of G are normal and hence are certainly N E ^∗-subgroups of G. Note that the chief series of G containing P is $1\lhd Z(P)\lhd P\lhd (P\times \langle c^{3}\rangle ) \lhd G$. This yields that $P \not \leq Z_{\mathcal {U}}(G)$.

Li showed that if every minimal subgroup of G is an NE-subgroup of G, then G is solvable (see Theorem 1(b) of [12]). The following theorem shows that this result remains true if, in Theorem 1 of [12], we consider N E ^∗-subgroups instead of NE-subgroups.

Theorem 1.5.

If all minimal subgroups of a group G are NE ^∗ -subgroups of G, then G is solvable.

Recall that a p-group G is called ultra-special if G ^′=Φ(G)=Z(G)=Ω₁(G). For any group G, F ^∗(G) denotes the generalized fitting subgroup: the set of all elements g of G which induce inner automorphisms on every chief factor of G. The following new characterizations of groups involve the requirement that certain minimal subgroups of F ^∗(H) possess the N E ^∗-property.

Theorem 1.6.

Let $\mathcal {F}$ be a saturated formation containing all supersolvable groups and let H be a normal subgroup of G such that $G/H\in \mathcal {F}$. If all minimal subgroups of F ^∗ (H) are NE ^∗ -subgroups of G, then either $G\in \mathcal {F}$ or G contains a minimal non-nilpotent subgroup K with the following properties:

• (1) K has a nontrivial normal Sylow 2-subgroup, say K ₂;

• (2) K ₂ ≤O ₂ (H),|K ₂ |=2 ^3s , and |Φ(K ₂)|=2^s , where s≥1;

• (3) K ₂ is ultra-special, that is, $K_{2}^{\prime } ={\Phi } (K_{2}) = Z(K_{2}) = {\Omega }_{1}(K_{2});$

• (4) If p is the odd prime dividing |K|, then p divides 2 ^s+1.

As an easy consequence of Theorem 1.6, we obtain the following result.

COROLLARY 1.7

Let $\mathcal {F}$ be a saturated formation containing all supersolvable groups and let H be a normal subgroup of a group G such that $G/H\in \mathcal {F}$. Let F ^∗(H) be of even order and let S be a Sylow 2-subgroup of F ^∗(H). Further, assume that every minimal subgroup of F ^∗(H) is an N E ^∗-subgroup of G. Then $G\in \mathcal {F}$ if one of the following conditions holds:

• (1) Ω₂(S)≤Z(S);

• (2) For all primes p dividing |G| and all s≥1, we have that p does not divide 2^s+1.

Theorems 1.5 and 1.6 are not true if the hypothesis of the N E ^∗-condition on minimal subgroups of G (respectively of F ^∗(H)) is replaced by just the condition on minimal subgroups of noncyclic Sylow subgroups of G (respectively of F ^∗(H)). For example, the group G:=S L(2,5) shows these facts: the only Sylow subgroups of G which are noncyclic are the Sylow 2-subgroups, which are quaternion groups. Then the only minimal subgroup under consideration would be the centre Z(G) of the group, which is normal. It is clear that Z(G) satisfies the N E ^∗-condition in G.

Using Theorems 1.5 and 1.6, we can derive the following results.

Theorem 1.8.

Let $\mathcal {F}$ be a saturated formation containing all supersolvable groups and let H be a normal subgroup of G such that $G/H\in \mathcal {F}$ . If all minimal subgroups and cyclic subgroups of order 4 of F ^∗ (H) are NE ^∗ -subgroups of G, then $G\in \mathcal {F}$.

Theorem 1.9.

Assume that every minimal subgroup of a group G is an NE ^∗ -subgroup of G. Then either G is supersolvable or G is solvable, $G_{2^{\prime }}$ is supersolvable and $G_{2^{\prime }}/C_{G_{2^{\prime }}}(O_{2}(G))$ is abelian for any Hall 2 ^′ -subgroup $G_{2^{\prime }}$ of G.

Theorem 1.8 is an improvement of Theorem 4.2 of Li in [13]. We do not know whether can be extended by considering minimal subgroups and cyclic subgroups of order 4 of only noncyclic Sylow subgroups of F ^∗(H).

It is natural to ask the question: are there differences between both N E ^∗-subgroups and weakly c-normal subgroups? For any subgroup H of a finite group G, it follows that every weakly c-normal subgroup is an N E ^∗-subgroup. Here we give an explanation. Let H be a weakly c-normal subgroup of G. Then there exists a subnormal subgroup K of G such that G=H K and H∩K≤H _G . Let K ₁=H _G K, applying Wielandt’s results (see [16]), we have K ₁=〈H _G ,K〉 is a subnormal subgroup of G. Since G=H K ₁ and H∩K ₁=H _G (H∩K)=H _G , so H∩K ₁ is normal in G. Thus H is an N E ^∗-subgroup of G. In general, if R is an N E ^∗-subgroup of G, R is not necessary to be weakly c-normal in G (see Example 1.10 below). But if H is an N E ^∗-subgroup contained in a normal nilpotent subgroup K of G, then it is true that H is weakly c-normal in G (see Lemma 2.2).

Example 1.10.

Let G=A ₅ and H=A ₄, the alternating subgroups with degree 5 and 4, respectively. Then G=H G and H ^G∩N _G (H)=H. Thus H is a N E ^∗-subgroup of G but not weakly c-normal in G.

Example 1.11.

Let G=A ₅, the alternating group of degree 5, and H a Sylow 5-subgroup. Noting that G is a nonabelian simple group, we get that H ^G∩N _G (H)=N _G (H) of order 10. Hence H is neither an N E ^∗-subgroup nor a weakly c-normal subgroup of G.

2 Preliminaries

In this section, we state some lemmas which are useful.

Lemma 2.1.

Let K and H be subgroups of a group G.

• (1) If H≤K and H is an NE-subgroup of G, then H is an NE-subgroup of K.

• (2) If H is a subnormal subgroup of K and H is an NE-subgroup of G, then H is normal in K.

Proof.

See Lemmas 1 and 4 of [12]. □

Lemma 2.2.

Let K and H be subgroups of a group G.

• (1) If H≤K and H is a NE ^∗ -subgroup of G, then H is a NE ^∗ -subgroup of K.

• (2) If H is a NE ^∗ -subgroup that is contained in a normal nilpotent subgroup K, then H is weakly c-normal in G.

Proof.

• (1) Since H is an N E ^∗-subgroup of G, there exists a subnormal subgroup L of G such that G=H L and H∩L is an NE-subgroup of G. It follows that K=K∩H L=H(K∩L) and K∩L is subnormal in K (see [16]). This implies that H∩L is an NE-subgroup of K by Lemma 2.1(1). So claim (1) holds.

• (2) Clearly H is subnormal in G. Since H is an N E ^∗-subgroup of G, there exists a subnormal subgroup L of G such that G=H L and H∩L is a NE-subgroup of G. It follows that the intersection H∩L is a subnormal subgroup of G. Therefore it follows immediately from Lemma 2.1(2) that H∩L is normal in G. Thus H∩L≤H _G holds.

□

Lemma 2.3.

Let S be a nontrivial 2-group and let H be a nontrivial group of automorphisms of S fixing the involutions of S. If H is cyclic of odd order and H acts irreducibly on S/Φ(S), then |S|=2^3s,|Φ(S)|=2^s, where s≥1,S is ultra-special and |H| divides 2 ^s+1.

Proof.

See Theorems 1.3 and 2.2 of [9]. □

Lemma 2.4.

Let S be a nontrivial 2-group and let H be a nontrivial group of automorphisms of S fixing the involutions of S. If 2 does not divide |H|, then H is abelian.

Proof.

See Theorem 4.4 of [9]. □

3 Proofs

Proof of Theorem 1.3.

We prove the theorem by induction on |G|+|P|. Suppose, first, that p>2. Then by Lemma 2.2(2), the condition that every minimal subgroup of P is N E ^∗-subgroup of G implies that every minimal subgroup of P is weakly c-normal in G. In particular, every minimal subgroup of P is c-supplemented in G. Hence we conclude that $P \leq Z_{\mathcal {U}}(G)$ by Theorem 1.1 of [2]. Thus we may assume that p=2. If every minimal subgroup H of P is normal in G, then H Q=H×Q for any Sylow q-subgroup Q of G, where q is an odd prime. This implies that Ω₁(P)≤C _G (Q). Since G is Q ₈-free, by Lemma 2.15 of [7], we obtain that Q≤C _G (P), yielding that G/C _G (P) is a p-group. This means that $P \leq Z_{\mathcal {U}}(G)$, as claimed. Then we may assume that P has a minimal subgroup H such that H is not normal in G, which implies that H is a N E ^∗-subgroup of G. It follows that there exists a subnormal subgroup K of G such that G=H K and H∩K is a NE-subgroup of G. Assume H∩K≠1, G=K and so H is a NE-subgroup and, of course, a $\mathcal {H}$-subgroup. Since H is a subnormal subgroup of G, it follows that $H \lhd G$ by Lemma 2.1, a contradiction. Hence we conclude that H∩K must be 1. Let L=P∩K. Since K is a maximal subgroup of G, we conclude that the subnormal subgroup K of G is normal in G. Thus $L=P \cap K\lhd G$. Because H≤P, Dedekind’s law implies P=H L. By our hypothesis, every minimal subgroup of L is a N E ^∗-subgroup of G. Therefore, $L \leq Z_{\mathcal {U}}(G)$ by induction. Observe that if P/L is normal in G/L with order p then $P/L \leq Z_{\mathcal {U}}(G/L)$. So $L\leq Z_{\mathcal {U}}(G)$, yielding $Z_{\mathcal {U}}(G/L) = Z_{\mathcal {U}}(G)/L$, which implies that $P\leq Z_{\mathcal {U}}(G)$ and the proof is complete. □

Proof of Theorem 1.5.

Assume that the theorem is false and let G be a counterexample of minimal order. Then:

• (1) Every proper subgroup of G is solvable. Let T be a proper subgroup of G. By Lemma 2.2(1), every minimal subgroup of T is a N E ^∗-subgroup of T, and so T satisfies the hypothesis of G. The minimal choice of G yields that T is solvable.

• (2) G/Φ(G) is a minimal simple group. By (1), G has a nontrivial maximal normal solvable subgroup, say M. Clearly, Φ(G) is a subgroup of M. We shall show that Φ(G)=M. Because otherwise we have M≦̸Φ(G), and we can conclude that there exists a maximal subgroup N of G such that G=M N and consequently G is solvable by (1), a contradiction. Thus the subgroup Φ(G) is the unique maximal subgroup of G, and so G/Φ(G) is a minimal simple group.

• (3) Φ(G) is a 2-group. By (2), applying Thompson’s classification of minimal simple groups, we obtain that G/Φ(G) is isomorphic to one of the following groups:

  • (i) P S L(3,3);

  • (ii) the Suzuki group S _z (2^q), where q is an odd prime;

  • (iii) P S L(2,p), where p is an odd prime with p ²≡1(mod 5);

  • (iv) P S L(2,2^q), where q is a prime;

  • (v) P S L(2,3^q), where q is an odd prime.

  Using this result, we shall show that Φ(G) is a 2-group. Let K be the 2-complement of Φ(G), then $K\lhd G$ and K is nilpotent. We wish to show, first, that K≤Z(G). Let p be a prime dividing |K| and let P∈ Syl _p (K). It is clear that P is normal in G. By hypothesis, every subgroup L of order p in P is a N E ^∗-subgroup of G. It follows that there exists a subnormal subgroup T of G such that G=L T and L∩T is a NE-subgroup of G. If L∩T=1, then G has a subgroup T of index p. Since T is a maximal subgroup of G and T is a subnormal subgroup of G, we have $T\lhd G$. Thus G is solvable by (1), a contradiction. So L∩T=L and so T=G. This implies that L is a NE-subgroup of G, and is normal in G by Lemma 2.1(2). Assume that L≦̸Z(G). Then C _G (L) is a proper subgroup and so C _G (L)≤Φ(G) by simplicity of G/Φ(G). This implies that G/C _G (L) is cyclic and so G is solvable, a contradiction. Consequently, each subgroup of P of order p lies in the centre Z(G). Consider the group D=S P, where S is a Sylow 2-subgroup of G. It follows from It$\hat {\mathrm {o}}^{\prime }$s lemma (see Chapter IV, Satz 5.5 of [10]) that D is p-nilpotent, and hence, D is nilpotent. Thus $S\leq C_{G}(P)\lhd G$. Applying the simplicity of G/Φ(G) again, we can conclude that P≤Z(G). Next, we denote by S ₀ a Sylow 2-subgroup of Φ(G), and consider the group $\overline {G} = G/S_{0}$. Since K≤Z(G), we have that G/Z(G)≅G/Φ(G) and $\overline {G}$ is a quasisimple group with the centre of odd order. By checking the table on Schur multipliers of the known simple groups (see p. 302 of [8]), we can conclude that the Schur multiplier of each of the minimal simple groups is a 2-group. It follows that $Z(\overline {G})$ must be 1, and therefore Φ(G) is a 2-group.

• (4) Let R be a Sylow r-subgroup of G, where r>2. Then there exists a subgroup L of order r such that L is not normal in G. Obviously, C _G (Ω₁(R))<G, because otherwise we would have Ω₁(R)≤Z(G) and so G would be r-nilpotent by Chapter IV, Satz 5.5 of [10]. It follows that G is solvable by (1), a contradiction. Then Ω₁(R) is solvable by (1). Assume that every minimal subgroup of R is normal in G. Then we can conclude that Ω₁(R) is an elementary abelian normal subgroup of G and every chief factor of G which lies below Ω₁(R) is cyclic of order r, which implies that ${\Omega }_{1}(R)\leq Z_{\mathcal {U}}(G)$. It follows that G/C _G (Ω₁(R)) is supersolvable by Chapter IV, Theorem 6.10 of [6] and so G is solvable, a contradiction. Thus there exists a subgroup L of order r such that L is not normal in G.

• (5) Let $\overline {G}=G/{\Phi } (G)$. Then 3 does not divide $|\overline {G}|$. Assume that 3 divides $|\overline {G}|$. Then G has a subgroup L of order 3 such that L is not normal in G by (4). By hypothesis, L is a N E ^∗-subgroup of G. It follows that there exists a subnormal subgroup T of G such that G=L T and L∩T is a NE-subgroup of G. If L∩T=1, consequently G has a subgroup T of index 3. Since T is a maximal subgroup of G, we conclude that the subnormal subgroup T of G is normal in G. Thus G is solvable by (1), a contradiction. So assume for the rest of this paragraph that L≤T. Clearly L is a NE-subgroup of G, which implies that L ^G∩N _G (L)=L and so L is a Sylow subgroup of L ^G. By the Frattini argument, we can conclude that G=N _G (L)L ^G. Moreover L ^G is a Frobenius group with complement L. Let N be the kernel of L ^G. Then N is nilpotent and so normal in G. Hence G=N _G (L)N, which means that G is solvable, a contradiction. Thus 3 does not divide $|\overline {G}|$.

• (6) Completing the proof. By (1), (2), and (5), $\overline {G}$ is a minimal simple group, $(3, |\overline {G}|)=1$. It follows by [ citeyearCR10, II, Bemerkung 7.5], that $\overline {G}$ is isomorphic to the Suzuki group S _z (2^q), where q is odd. However |S _z (2^q)|≡0 (mod 5) by Chapter XI, Remarks 3.7(b) of [11] and so 5 divides $|\overline {G}|$. Therefore $\overline {G}$ has a subgroup of index 5 by a discussion similar to (5) above. This implies that $\overline {G}$ is isomorphic to a subgroup of S ₅, the symmetric group on five letters. Since 3 does not divide $|\overline {G}|$ by (5), it implies that |π(G)|=2 because Φ(G) is a 2-group by (3) and so G is solvable, a final contradiction.

□

Proof of Theorem 1.6.

Suppose that the result is false and let G be a counterexample of a minimal order. Then:

• (1) F ^∗(H)=F(H). By Lemma 2.2, every minimal subgroup of F ^∗(H) is a N E ^∗-subgroup of F ^∗(H). Then F ^∗(H) is solvable by Theorem 1.5. It follows that F ^∗(H)=F(H) by Chapter X, Theorem 13.13 of [11].

• (2) F(H) is of even order. Otherwise, F(H) is of odd order. Theorem 1.3 implies that $F(H)\leq Z_{\mathcal {U}}(G)$. Since $\mathcal {U}\subseteq \mathcal {F}$ and $\mathcal {U}$ and $\mathcal {F}$ are saturated formations, it follows that $Z_{\mathcal {U}}(G)\leq Z_{\mathcal {F}}(G)$ by Proposition 3.11 of [6]. Then we can conclude that $F(H)\leq Z_{\mathcal {F}}(G)$ and hence $G/C_{G}(F(H))\in {\mathcal {U}}$ by [6, IV, Theorem 6.10]. Moreover, since $G/H\in {\mathcal {U}}$ by our hypothesis, it follows that $G/C_{H}(F(H))\in {\mathcal {U}}$. By Chapter X, Theorem 13.12 of [11], we get that C _H (F ^∗(H))≤F(H) and C _H (F(H))≤F(H) since F ^∗(H)=F(H) by (1). Then $G/F(H)\in {\mathcal {F}}$ and since $F(H)\leq Z_{\mathcal {F}}(G)$, it follows that $G \in {\mathcal {F}}$, a contradiction. This proves (2).

• (3) There exists a Sylow subgroup P of G such that O ₂(H)P is not 2-nilpotent, where |O ₂(H)| and |P| are co-prime. If not, O ₂(H)≤Z _∞ (G), where Z _∞ (G) is the hypercentre of G. Since $Z_{\infty }(G)\leq Z_{\mathcal {U}}(G)$, it follows that $O_{2}(H)\leq Z_{\mathcal {U}}(G)$. Applying Theorem 1.3, we get that every Sylow subgroup of F(H) of odd order lies in $Z_{\mathcal {U}}(G)$ and hence $F(H)\leq Z_{\mathcal {U}}(G)$. By a discussion similar to Step (2), noting that $Z_{\mathcal {U}}(G)\leq Z_{\mathcal {F}}(G)$, it follows that $G \in {\mathcal {F}}$, a contradiction. Therfore there exists a Sylow subgroup P of G such that O ₂(H)P is not 2-nilpotent, where (|O ₂(H)|,|P|)=1.

• (4) Completing the proof. By (3), it is clear that O ₂(H)P contains a minimal non-2-nilpotent subgroup, say K. By Chapter IV, Satz 5.4 of [10], we have that K is a minimal non-nilpotent subgroup of G. Applying Chapter III, Satz 5.2 of [10] we can conclude that K has a normal Sylow 2-subgroup K ₂ and a cyclic Sylow p-subgroup K _p , for a prime p≠2. Clearly K _p fixes the involutions of K ₂, because otherwise we get that K _p is normal in K, a contradiction. Moreover K _p acts irreducibly on K ₂/Φ(K ₂). It follows by Lemma 2.3 that |K ₂|=2^3s and Φ(|K ₂|)=2^s, where s≥1,K ₂ is ultra-special and $K_{p}/C_{K_{p}}(K_{2})$ divides 2^s+1. This is a final contradiction and the proof is complete.

□

Proof of Theorem 1.8.

Suppose that the result is false. By Theorem 1.6, there exists a minimal non-nilpotent subgroup K of G satisfying the properties (1), (2) and (3). For every cyclic subgroup L of K ₂ of order 4, since K ₂ is ultra-special, it follows that L≦̸Z(K ₂)=Ω₁(K ₂) and consequently K ₂≦̸C _K (L). If C _K (L) is normal in K, it follows that K _p is normal in K, where K _p is a Sylow p-subgroup of K and p>2, a contradiction. Thus C _K (L) is not normal in K and so L is not normal in K. We may conclude, by our assumptions, that L is a N E ^∗-subgroup of G and so L is a N E ^∗-subgroup of K. Then there exists a subnormal subgroup K ₁ of K such that K=L K ₁ and L∩K ₁ is a NE-subgroup of G. Since L is not normal and K is a minimal non-nilpotent group, it follows that if K ₁ is a proper subgroup of K, then the fact that K _p char K ₁ and $K_{1}\lhd K$ would imply that K _p is subnormal in K. Since K _p is a subnormal Hall-subgroup of K, K _p is normal in K, a contradiction. This means that K ₁=K and hence L is a NE-subgroup of G. Thus, by Theorem 4.2 of [13], we get that G belongs to $\mathcal {F}$. This is a final contradiction and the proof is complete. □

Proof of Theorem 1.9.

Theorem 1.5 immediately yields the solvability of G. Let $G_{2^{\prime }}$ be a Hall 2^′-subgroup of G. It follows by Lemma 2.2 and Theorem 1.6 that $G_{2^{\prime }}$ is supersolvable. Hence, if G has odd order, then G is supersolvable and we are done. Assume that 2 divides the order of G and that O ₂(G) is nontrivial. Then if $G_{2^{\prime }}$ centralizes the involutions of O ₂(G), then Lemma 2.4 implies that $G_{2^{\prime }}/C_{G_{2^{\prime }}}(O_{2}(G))$ is abelian. Hence we may assume that there exists an involution x∈O ₂(G) which is not centralized by G, which implies that 〈x〉 is not normal in G. Noting that 〈x〉 is a N E ^∗-subgroup of G, we deduce that G=〈x〉K for some subnormal subgroup K of G such that 〈x〉∩K is a NE-subgroup of G. If 〈x〉∩K=〈x〉 and so K=G. This implies that 〈x〉 is a NE-subgroup of G, and so normal in G by Lemma 2.1(2), a contradiction. Thus 〈x〉∩K must be 1, and so $K\lhd G$ since |G:K|=2. It follows that $G_{2^{\prime }}$ is a Hall 2-subgroup of K because $G_{2^{\prime }}\leq K$. It follows that $[O_{2}(G),G_{2^{\prime }}]\leq O_{2}(G)\cap K=O_{2}(K)$. If we argue by the induction on the order of G, we can deduce by inductive hypothesis that $[O_{2}(G),(G_{2^{\prime }})^{\prime }, (G_{2^{\prime }})^{\prime }]\leq [O_{2}(K),(G_{2^{\prime }})^{\prime }]=1$. This means that $[O_{2}(G),(G_{2^{\prime }})^{\prime }]=1$ by co-prime action. So $G_{2^{\prime }}/C_{G_{2^{\prime }}}(O_{2}(G))$ is abelian and the proof is complete. □