On second minimal subgroups of Sylow subgroups of finite groups

This paper has been published in Journal of Algebra, 342(2):134–146 (2011). Copyright 2011 by Elsevier. The final publication is available at www.sciencedirect.com. http://dx.doi.org/10.1016/j.jalgebra.2011.06.016 http://www.sciencedirect.com/science/article/pii/S0021869311003553 On second minimal subgroups of Sylow subgroups of finite groups A. Ballester-Bolinches∗R. Esteban-Romero† Yangming Li‡ Abstract A subgroup Hof a finite group Gis a partial CAP-subgroup of Gif there is a chief series of Gsuch that Heither covers or avoids its chief factors. Partial cover and avoidance property has turned out to be very useful to clear up the group structure. In this paper, finite groups in which the second minimal subgroups of their Sylow p-subgroups, pa fixed prime, are partial CAP-subgroups are completely classified. Keywords: finite group, partial CAP-subgroup, second minimal subgroup, supersoluble group. Mathematics Subject Classification (2000): 20D10, 20D20. 1 Introduction All groups considered in this paper are finite. Arguably, the study of subgroup embedding properties has been one of the most efficient methods to clear up the structure of the groups. In particular, the embedding properties of 2-maximal and 2-minimal subgroups tend to give additional information about the group ([4, 17, 23, 27]). During the past four decades, the subgroup property known as the cover-avoidance property has gained more and more currency, first in the context of soluble groups ([8–10,12,24,25] and [2, Chapter 4]), and more recently as a way of describing certain classes of soluble and supersoluble groups and their local versions ([3,7,11,13–16,20–22,26]). ∗Departament d’Àlgebra, Universitat de València, Dr. Moliner, 50, E-46100 Burjassot, València, Spain, email: [email protected] †Institut Universitari de Matemàtica Pura i Aplicada, Universitat Politècnica de València, Camí de Vera, s/n, E-46022 València, Spain, email: [email protected] ‡Department of Mathematics, Guangdong University of Education, Guangzhou, 510310, People’s Republic of China; Departament d’Àlgebra, Universitat de València, email: [email protected] 1 Let Abe a subgroup of a group Gand H/K a section of G. We say that Acovers H/K if HA =KA and Aavoids H/K if A∩H=A∩K. If Aeither covers or avoids every chief factor of G, then we say that Ahas the cover and avoidance property in Gor Ais a CAP-subgroup of G. Unfortunately the cover and avoidance property is not hereditary in intermediate subgroups, that is, if Ais a CAP-subgroup of Gand Ais contained in a subgroup Bof G, it does not follow in general that Ahas the cover and avoidance property in B(see [4, Example 3]). The failure of the cover and avoidance property to hold in intermediate subgroups leads to the following weaker property, which is persistent in subgroups and is also extremely useful in the structural study of the groups: Definition 1.1. A subgroup Aof a group Gis called a partial CAP-subgroup of Gif there exists a chief series ΓAof Gsuch that Aeither covers or avoids each factor of ΓA(see [11,21] for alternative terminologies). Clearly, every CAP-subgroup is a partial CAP-subgroup, but the converse does not hold ([4, Example 3]). In [4], the authors considered the effect of imposing the partial cover and avoidance property to the second maximal subgroups of the Sylow p-subgroups, pa fixed prime. In the present paper the emphasis is on second minimal subgroups, and we consider what might be considered an opposite extreme, where the second minimal subgroups (2-minimal subgroups for short) of the Sylow p-subgroups are partial CAPsubgroups. In one result, we characterise the groups with this property, identifying a remarkable analogy between the partial cover and avoidance property of the subgroups of index p2and the partial cover and avoidance property of the subgroups of order p2. Main theorem. Let pbe a prime number, let Gbe a group, and let G+= G/Op0(G). Then every subgroup of Gof order p2is a partial CAP-subgroup of Gif and only if one of the following statements holds: 1. the order of the Sylow p-subgroups of Gis at most p; 2. Gis a p-supersoluble group; 3. Φ(G+)=1and, if Pis a Sylow p-subgroup of G,P+= Soc(G+) = V1× · · · × Vr, where V1, . . . , Vrare minimal normal subgroups of G+which are G+-isomorphic to a 2-dimensional irreducible G+-module Vover the Galois field GF(p). Furthermore, Vis not an absolutely irreducible G+-module when r > 1. It seems desirable now to give an example of a group satisfying condition 3 of our main theorem. Existence of such groups was already shown in [1]. For the sake of completeness, we reproduce here the example of that paper. 2 Example 1.2. Consider an elementary abelian group H=ha, b |a5=b5= 1, ab =bai of order 25 and let αbe an automorphism of Hof order 3satisfying that aα=b,bα=a−1b−1. Let H1=H,H2=ha0, b0ibe a copy of H1and G= [H1×H2]hαi. For any subgroup Aof Gof order 25, there exists a minimal normal subgroup Nsuch that A∩N= 1. Then Acovers or avoids the factors of the chief series of G 1< N < AN < G. In other words, Ais a partial CAP-subgroup of G. However, Gis not 5supersoluble. Note that His not an absolutely irreducible G-module over the Galois field of 5elements. This example also shows that a group in which the second minimal subgroups of the Sylow subgroups are partial CAP-subgroups is not supersoluble in general. The best we are able to say is the following: Corollary 1.3. A group in which the second minimal subgroups of the Sylow subgroups are partial CAP-subgroups is soluble. 2 Preliminaries We begin with some preparatory lemmas before coming to the main result of the paper. The main basic properties of partial CAP-subgroups are listed in the following result appeared in [11]. They are particularly useful when induction arguments are applied. Lemma 2.1. Let Sbe a partial CAP-subgroup of a group G. 1. If S≤K≤G, then Sis a partial CAP-subgroup of K. 2. If N≤Sand NEG, then S/N is a partial CAP-subgroup of G/N. 3. If NEGand (|S|,|N|)=1, then SN/N is a partial CAP-subgroup of G/N. The information given in the following lemma comes in extremely useful when studying the partial cover and avoidance property. Lemma 2.2 ([1, Lemma 2.2]).Let Hbe a partial CAP-subgroup of a group G. Suppose that Qis a normal subgroup of Gsuch that His contained in Q. Then there exists a chief series ΩHof Gpassing through Qsuch that H either covers or avoids each chief factor in ΩH. 3 Let rbe a positive integer and let Hbe a subgroup of G. Then His called an r-minimal (respectively r-maximal) subgroup of Gif there exists a subgroup chain 1 = H0< H1<· · · < Hr=H(respectively H=H0< H1<· · · < Hr=G) such that Hiis a maximal subgroup of Hi+1 for all 0≤i≤r−1. In the present paper we investigate the effect of imposing the partial cover and avoidance property on the 2-minimal subgroups of the Sylow subgroups, and once more we get a sense of why the partial cover and avoidance property has such bearing in the study of soluble groups. In fact, we use a local approach and characterise the groups Genjoying the following property: (†)Every 2-minimal subgroup of every Sylow p-subgroup of Gis a partial CAP-subgroup of G,pa fixed prime. In the following pwill be a fixed prime. Since 2-minimal subgroups of p-groups have order p2, every group with Sylow p-subgroups of order psatisfies property (†). All p-supersoluble groups, or p-soluble groups whose p-chief factors have order p, also satisfy (†). Therefore we must think about groups whose order is divisible by p2which are not p-supersoluble. An interesting special case is when the Sylow p-subgroups of Ghave order p2. In this case, the structure of Gis quite restricted as the following lemma shows. Lemma 2.3. Let Gbe a group whose Sylow p-subgroups have order p2. Suppose that Gsatisfies property (†). Then Gis p-soluble and either Gis psupersoluble or Soc(G/Op0(G)) = POp0(G)/Op0(G)is an elementary abelian group of order p2for each Sylow p-subgroup Pof G. Proof. We can assume without loss of generality that Op0(G)=1. Suppose that Gis not p-soluble. Then every minimal normal subgroup of Gis nonabelian and its order is divisible by pby [4, Theorem 7]. It is clear then that a Sylow p-subgroup Pof Gneither covers nor avoids any minimal normal subgroup of G, a contradiction which shows that Gis p-soluble. In that case, S= Soc(G)is a minimal normal subgroup of Gcontained in Pby [4, Theorem 7]. Consequently, either Sis of order pand Gis p-supersoluble or S=Pis the Sylow p-subgroup of G. The next lemmas will be applied to the consideration of groups satisfying property (†). Lemma 2.4 ([5, Proposition 1]).Let Fbe a saturated formation. Assume that Gis group such that Gdoes not belong to Fand there exists a maximal 4 subgroup Mof Gsuch that M∈Fand G=MF(G). Then GF/(GF)0is a chief factor of G,GFis a p-group for some prime p,GFhas exponent pif p > 2and exponent at most 4 if p= 2. Moreover, either GFis elementary abelian or (GF)0= Z(GF) = Φ(GF). As an important deduction we have the Lemma 2.5 ([5, Theorem 6]).Let Fbe a saturated formation and Ga group with a normal subgroup Ksuch that G/K ∈F. If for some prime p, every subgroup of order pof Kis contained in the F-hypercentre ZF(G)of G, then G/Op0(K)∈F. There are some places where we use a known criterion for a normal psubgroup to be contained in the hypercentre. For convenience, this is stated here as: Lemma 2.6. Suppose that Pis a normal p-subgroup of G. Then P≤Z∞(G) if and only if Op(G)≤CG(P). 3 Main results In this section we analyse the structure of the groups satisfying property (†), and prepare the way for the proof of the main result. We begin with a theorem about the minimal normal subgroups of the groups satisfying property (†). Theorem 3.1. Let Gbe a group satisfying property (†)whose order is divisible by p2. Then every minimal normal subgroup of Gis either a p0-group or a p-group. The minimal normal p-subgroups of Gare of the same order, and it is at most p2. Proof. Let Nbe a minimal normal subgroup of G, and suppose that Nis not a p0-group. Let 16=Npbe a Sylow p-subgroup of N, and let Qbe a subgroup of Gof order p2such that Q∩Np6= 1. We consider a chief series of G (Γ) : 1 = G0<· · · < Gi<· · · < Gj< Gj+1 <· · · < Gm=G such that Qeither covers of avoids each chief factor of Gin (Γ). Then there exists an index i∈ {1, . . . , m}such that N∩Gi−1= 1 and Gi=Gi−1N. In that case, Gi/Gi−1is G-isomorphic to N. Suppose that Qavoids Gi+1/Gi, then Q∩Np≤Q∩Gi∩N=Q∩Gi−1∩N= 1, against the choice of Q. Consequently, Qcovers Gi/Gi−1. Then Gi/Gi−1is of order at most p2and Nis a p-group of order at most p2. 5 Next we prove that all minimal normal p-subgroups of Gare of the same order. Suppose, arguing by contradiction, that Ghas two minimal normal p-subgroups, N1and N2say, such that |N1|=pand |N2|=p2. Let Hbe a subgroup of N2of order p. Then N1His a subgroup of Gof order p2which is a partial CAP-subgroup of G. By Lemma 2.2, there exists a chief series of G, (∆) : 1 ≤N3≤N1N2≤ · · · ≤ G, passing through N1N2such that N1Heither covers or avoids each chief factor of Gin (∆). In addition, the order of N3is por p2. Assume that N3is of order p. If N1H∩N3= 1, then N1HN3=N1N2and N2=H(N2∩N1N3). It means that either N2=Hor N2=HN1N3. This contradiction shows that N3is a subgroup of N1H. In particular, N1Hcannot cover N1N2/N3. Hence N1H=N1N2∩N1H=N1H∩N3=N3, a contradiction which shows that N3must be of order p2. Since N1N2is of order p3and N1His of order p2, it follows that N1Hcovers N3. Thus N1H=N3, which contradicts the fact that N1and N3are two different minimal normal subgroups of G. This proves the result. We now touch the question of the p-length of p-soluble groups satisfying property (†). We prove that these groups belong to the saturated formation Fof all p-soluble groups whose p-length is at most one. Theorem 3.2. Let Gbe a p-soluble group satisfying property (†). Then the p-length of Gis at most 1. Proof. We will obtain a contradiction by supposing that the result is false and choosing a counterexample Gof least order. For the ease of reading, we break the argument into separately-stated steps. 1. Op0(G) = 1. Assume that Op0(G)6= 1. By Lemma 2.1 the hypothesis holds in the group G/Op0(G). The minimal choice of Gimplies that G/Op0(G)belongs to F. Hence Gis an F-group, against the choice of G. Thus Op0(G) = 1. 2. Any proper subgroup of Gbelongs to F. Let Hbe a proper subgroup of G. If a Sylow p-subgroup Hpof His of order at most p, we have that H∈Fby [3, Lemma 3.1]. Assume that p2divides |Hp|and let Lbe a subgroup of Hof order p2. Then Lis a partial CAP-subgroup of Hby Lemma 2.1, and so Hsatisfies property (†). The minimality of Gyields H∈F. This confirms Step 2. 6 Let GFdenote the F-residual of G, that is, the smallest normal subgroup of Gwith quotient in F. 3. There exists a maximal subgroup Mof Gsuch that M∈Fand G= MGF. Moreover, GF/Φ(GF)is a chief factor of G, and the exponent of GFis por at most 4if p= 2. Since Gis not an F-group and Fis saturated, it follows that G/Φ(G) does not belong to F. Let N/Φ(G)a non-trivial normal subgroup of G/Φ(G). Then N/Φ(G)is supplemented in G/Φ(G). By Step 2, G/N belongs to F. Therefore, since Fis a formation, G/Φ(G)has a unique minimal normal subgroup, T/Φ(G)say. Moreover, T/Φ(G)is not a p0-group. Since Gis p-soluble, it follows that T/Φ(G)is an abelian p-chief factor of Gwhich is complemented in Gby a maximal subgroup Mof G. Then G=MF(G)and T=GFΦ(G). Step 2 implies that M∈Fand so GF/Φ(GF)is a chief factor of G, and the exponent of GF is por at most 4if p= 2 by Lemma 2.4. This proves our claim. 4. Φ(GF) = 1 and |GF|=p2. Suppose that GF/Φ(GF)has order p. Since GF/Φ(GF)is G-isomorphic to Soc(G/MG), it follows that G/MGis in F. This contradiction shows that GF/Φ(GF)has order greater than p. Let Hbe a subgroup of GF of order p2such that H6≤ Φ(GF). Then His a partial CAP-subgroup of Gand, by Lemma 2.2, there exists a chief series of G, (Γ1) : 1 = G0≤G1≤ · · · ≤ K≤GF≤ · · · ≤ Gn=G, passing through GFsuch that Heither covers or avoids each G-chief factor in (Γ1). Since GF/Φ(GF)is a chief factor of Gby Step 3, it follows that either KΦ(GF) = Φ(GF)or KΦ(GF) = GF. If KΦ(GF) = GF, then K=GF, contrary to assumption. Thus K≤Φ(GF). Since GF/K is a chief factor of G, we have K= Φ(GF)and so GF/Φ(GF)is a chief factor of Gin (Γ). It therefore follows that Heither covers or avoids GF/Φ(GF). If Havoids GF/Φ(GF), then H=H∩GF≤Φ(GF), contrary to the choice of H. Hence we have that Hcovers GF/Φ(GF). Consequently GF=HΦ(GF) = H,GFis of order p2and Φ(GF) = 1. 5. Gis not a primitive group. In particular MG6= 1. Suppose, arguing by contradiction, that Gis primitive. Then GFis the unique minimal normal subgroup of G. Since Ghas p-length greater than 1and GFis a p-group, it follows that pdivides |M|. Then we can choose an element a∈GFand an element b∈Msuch that ha, biis a 7 subgroup of Gof order p2. Obviously ha, bineither covers nor avoids GF, a contradiction which proves Step 5. 6. The final contradiction. By Step 5, MG6= 1. Let Nbe a minimal normal subgroup of G contained in MG. Then Nis a p-group by Step 1, and N∩GF= 1. Let us choose an element a∈GFand an element b∈Nsuch that ha, biis a group of order p2. By hypothesis, ha, biis a partial CAP-subgroup of G. Applying Lemma 2.2, there exists a chief series of G, (Γ2) : 1 = G0≤K≤GFN≤ · · · ≤ Gn=G, passing through GFNsuch that ha, bieither covers or avoids each chief factor of Gin (Γ2). It is clear that ha, bineither covers nor avoids GF. Hence K6=GF. Then Kis an F-central chief factor of G. Since Mis an F-normaliser of Gby [2, Theorem 4.2.17], we can apply [2, Theorem 4.2.4] to conclude that K≤M. Assume that ha, biavoids K. Then ha, bimust cover GFN/K, and so GFN≤ ha, biK. We therefore have that GF=GF∩(ha, biK) = hai(GF∩ hbiK) = hai, contrary to Step 4. Hence ha, bimust cover K. Thus K=M∩ ha, bi=hbi. Now ha, bieither covers or avoids GFN/K. If ha, bicovers GFN/K, then GFN≤ ha, biK=ha, bi. Then GFis of order p, against Step 4. Thus ha, biavoids GFN/K. This gives the final contradiction ha, bi= GFN∩ ha, bi=ha, bi ∩ K=K. Our next result shows that a group satisfying property (†)whose order is divisible by p2must be p-soluble. Theorem 3.3. Let Gbe a group satisfying property (†). Then either the Sylow p-subgroups of Gare of order por Gis a p-soluble group. Proof. Suppose the result false, and let the group Gprovide a counterexample of least possible order. Then p2divides the order of G. According to Lemma 2.1, the property of Gis inherited by G∗=G/Op0(G). Hence the minimality of Gimplies that Op0(G)=1. We reach a contradiction after the following steps. 1. If Kis a proper subgroup of Gand p2divides |K|, then Kis p-soluble. If, in addition, Kis normal in G, then a Sylow p-subgroup of Kis also normal in G. 8 it follows that Nis G-isomorphic to N2. Hence N1is G-isomorphic to N2. This contradiction, together with [2, 1.2.36], allow us to conclude that all non-cyclic p-chief factors of Gare G-isomorphic and of order p2, and they are all complemented in G. The theorem holds in this case. 2. Φ(G)6= 1. Consider the following two chief series of G: (γ1) : 1 ≤ · · · ≤ Φ(G)≤N1≤N1N2≤ · · · ≤ G, (γ2) : 1 ≤ · · · ≤ Φ(G)≤N2≤N1N2≤ · · · ≤ G. Intersecting the series (γi)term-by-term with Mi,i= 1,2, and deleting repetitions, we get the chief series of M1and M2, respectively: (γ1)∩M1: 1 ≤ · · · ≤ Φ(G) = N1∩M1≤N2=N1N2∩M1 ≤ · · · ≤ M1, (γ2)∩M2: 1 ≤ · · · ≤ Φ(G) = N2∩M2≤N1=N1N2∩M2 ≤ · · · ≤ M2. Now we intersect these two chief series with Xand delete repetitions. (γ1)∩X= (γ2)∩X: 1 ≤ · · · ≤ Φ(G) = N1∩X=N2∩X ≤ · · · ≤ X. It is clear that (γ1)∩X= (γ2)∩Xis a chief series of Xand the X-chief factors of this series below Φ(G)are Mi-chief factors for all i= 1,2. We know that either Miis p-supersoluble or all non-cyclic p-chief factors of Miare Mi-isomorphic and have order p2, and every complemented Michief factor of Miis non-cyclic, i= 1,2. Hence the orders of N1/Φ(G) and N2/Φ(G)are either por p2. Suppose that |N1/Φ(G)|=p2and |N2/Φ(G)|=p. Since N2/Φ(G)is a cyclic complemented p-chief factor of M1, it follows that M1is p-supersoluble. Therefore every p-chief factor of M1in (γ1)∩M1is of order p. Hence every G-chief factor below Φ(G)in (γ1)is of order p. On the other hand, N1/Φ(G)is a complemented chief factor of M2of order p2. Hence M2is not psupersoluble and so every non-cyclic chief factor of M2is of order p2. But every chief factor of M2below Φ(G)is of order p. Consequently, N1/Φ(G)is the unique complemented chief factor of M2in the chief series (γ2)∩M2of M2. It follows that Φ(G)≤Φ(M2). Since Φ(M2) is a nilpotent group, we obtain by order considerations that Φ(G) = 15 OpΦ(M2). However, the same arguments of the proof for r= 1 show now that Φ(G) = 1, contrary to supposition. Therefore N1/Φ(G)and N2/Φ(G)have the same order. If |N1/Φ(G)|=|N2/Φ(G)|=p, then G is p-supersoluble, and the theorem holds. Assume that |N1/Φ(G)|=|N2/Φ(G)|=p2. Since N3−i/Φ(G)is a chief factor of Mi,Miis not p-supersoluble, i= 1,2. Then all noncyclic chief factors of Miare Mi-isomorphic and have order p2, and every complemented chief factor of Miis non-cyclic. Assume that X is p-supersoluble. Then every chief factor of Xbelow Φ(G)is cyclic. Certainly these chief factors are also chief factors of G. Since Mi has no cyclic complemented chief factors, exactly similar arguments to those used above show that Φ(G) = Φ(Mi)=1. This contradiction shows that Xis not p-supersoluble and so Xsatisfies the properties enunciated in the statement of the theorem. In particular, Ni/Φ(G) is Mi-isomorphic to an X-chief factor of the form Φ(G)/C, which is also a G-chief factor. It implies that N1/Φ(G)and N2/Φ(G)are Gisomorphic, and Gsatisfies properties 1and 2by [2, 1.2.36]. Suppose that r≥3. Denote G=N1M1=N2M2, where M1, M2are maximal subgroups of Gsuch that N1∩M1=N2∩M2= Φ(G). Consider the following two chief series of G: (δ1) : 1 ≤ · · · ≤ Φ(G)≤N1≤N1N2≤N1N2N3≤ · · · ≤ G, (δ2) : 1 ≤ · · · ≤ Φ(G)≤N2≤N1N2≤N1N2N3≤ · · · ≤ G. Intersecting the series (δi)term-by-term with Mi,i= 1,2, we get the series: (δ1)∩M1: 1 ≤ · · · ≤ Φ(G) = N1∩M1≤N2=N1N2∩M1 ≤N2N3=N1N2N3∩M1≤ · · · ≤ M1, (δ2)∩M2: 1 ≤ · · · ≤ Φ(G) = N2∩M2≤N1=N1N2∩M2 ≤N1N3=N1N2N3∩M2≤ · · · ≤ M2. By induction, we have that Miis p-supersoluble or all non-cyclic chief factors of Miare Mi-isomorphic and have order p2and every complemented chief factor of Miis non-cyclic, i= 1,2. We distinguish two possibilities: 1. M1or M2is p-supersoluble. Assume that M1is p-supersoluble. Then it follows that N2/Φ(G)and N3/Φ(G)have order p. If M2were not p-supersoluble, then it would follow that all complemented p-chief factors of M2are M2-isomorphic and of order p2. In that case, N3/Φ(G)would have order p2. This 16 contradiction yields that M2is p-supersoluble. In that case, every chief factor Ni/Φ(G)has order p,i= 1, . . . , r. In this case Gis psupersoluble, and the result holds. 2. Neither M1nor M2is p-supersoluble. Then all non-cyclic M1-chief factors are M1-isomorphic and have order p2and every complemented p-chief factor of M1is non-cyclic. In particular, N2/Φ(G),N3/Φ(G), ... , Nr/Φ(G)are M1-isomorphic (and so G-isomorphic), and have order p2. Since M2is not p-supersoluble, it follows that N1/Φ(G),N3/Φ(G), .. . , Nr/Φ(G)are M1-isomorphic (and so G-isomorphic). Consequently, N1/Φ(G),N2/Φ(G), ..., Nr/Φ(G) are G-isomorphic. Applying [2, 1.2.36], Gsatisfies the properties enunciated in the statement of the theorem. Therefore we conclude that the result as stated is true. The proof of Theorem 3.4 leads also to the following result: Corollary 3.5. Let Gbe a p-soluble group satisfying property (†)such that Op0(G) = 1. If Gis not p-supersoluble and F(G)/Φ(G)is a chief factor of G, then Φ(G)=1. As an interesting deduction we have the Corollary 3.6. Let Gbe a p-soluble group satisfying property (†). Assume that Op0(G)=1. Then either Gis p-supersoluble or Φ(G)=1and all complemented p-chief factors of Gare G-isomorphic and have order p2. Proof. Applying Theorem 3.3, it follows that F(G) = Op(G)is the unique Sylow p-subgroup of G. We shall proceed by induction on |G|. As usual, we write FG/Φ(G)=N1/Φ(G)× · · · × Nr/Φ(G), where Ni/Φ(G)is a minimal normal subgroup of G/Φ(G)for all i. According to [2, 1.2.36], every complemented p-chief factor is G-isomorphic to Ni/Φ(G)for some i. Certainly, we can assume that r≥2by Theorem 3.4 and Corollary 3.5. Suppose that Gis not p-supersoluble. According to Theorem 3.4, Ni/Φ(G) has order p2by Theorem 3.4. Let Mibe a maximal subgroup of Gsuch that G=NiMiand Ni∩Mi= Φ(G),i= 1,2, . . . , r. Since Nj/Φ(G)is a p-chief factor of Mifor all j6=i, it follows that Miis not p-supersoluble. We observe that CMi(Nj/Φ(G)) is the Sylow p-subgroup of Mifor all j6=i. Therefore Op0(Mi)=1. Consequently ΦMi= 1 by induction. It therefore follows that Φ(Ni) = 1 for all iand Niis centralised by Njfor all j6= i. Then Niis elementary abelian and so it has the structure of G-module over the Galois field GF(p)as a natural way. Since Niis centralised by 17 F(G), Maschke’s theorem [6, A, 11.5] implies the complete reducibility of the representation space. In particular, there exists a minimal normal subgroup Kiof Gof order p2such that Ni= Φ(G)×Ki,i= 1,2, . . . , r. Consequently, G=N1· · · NrGp0=K1· · · KrGp0,F(G) = K1· · · Krand Φ(G) = 1, as required. Proof of the main theorem. We prove the necessity of the condition by induction on the order of G. Certainly, by Lemma 2.1, we may assume that Op0(G) = 1. Let Pbe a Sylow p-subgroup of G. It may be supposed that |P|is greater than p. Then, applying Theorem 3.3, Gis p-soluble and, by Theorem 3.2, the p-length of Gis at most 1. Hence P= Op(G) = F(G) is the Sylow p-subgroup of G. By Corollary 3.6, we have that either Gis p-supersoluble or Φ(G) = 1. Suppose that Gis not p-supersoluble. Then Φ(G)=1, and F(G)is elementary abelian and it can be regarded as a completely reducible G-module over the Galois field GF(p)by [6, A, 11.5]. This means that Pis expressible as a direct product of minimal normal subgroups of G, say P=V1× · · · × Vr, where Viis an irreducible G-module over GF(p) (i= 1, . . . , r). By Corollary 3.6, each |Vi|=p2,i= 1, . . . , r, and all of them are G-isomorphic. Now we consider the case that r > 1. Consider the submodule W=V1×V2. Write K= GF(p)and V=V1. Suppose that V is an absolutely irreducible G-module. Then E= EndKG(V) = Kand the G-endomorphisms of Vare exactly those defined by θt:V→V, given by vθt=vt, v ∈V, t ∈K. According to [6, B, 8.2], the irreducible submodules of Ware Vt={(v, vt) : v∈V}, for t∈K, and V2={(1, v) : v∈V}. On the other hand, V, as a vector K-space, has dimension 2. Let {a, b}be a basis of V. With the obvious notation, consider the subgroup A=h(a1b1,1),(1, a2b2)i of W. Then Ahas order p2and so Ais a partial CAP-subgroup of G. But A∩V2=h(1, a2b2)iand A∩Vt=h(a1b1,(a2b2)t)ifor any t∈K, that is, Aneither covers nor avoids any irreducible submodule of W, contrary to Lemma 2.2. Hence Vis not an absolutely G-module and the necessity of the condition holds. To prove the sufficiency, it may be assumed that Gsatisfies the condition 3and Op0(G)6= 1. Hence F(G) = Soc(G) = N1× · · · × Nr, where Ni∼ =GV is a G-irreducible module over GF(p)of dimension 2, for every i= 1, . . . , r. Moreover, F(G)is a Sylow p-subgroup of G. We may also assume that r≥2. Then Vis not an absolutely irreducible G-module. Let Q=ha, bibe a subgroup of Gof order p2. We prove that Qis a partial CAP-subgroup of Gby induction on the order of G. Obviously, we can suppose that Qis not a minimal normal subgroup of G. Let Nbe a minimal normal subgroup of G. Then Q∩Nis trivial or has order p. Assume that Q∩Nis of order pfor all minimal normal subgroups Nof G. Then two different minimal normal 18 subgroups produce different subgroups of order pof Q. Since the number of subgroups of order pof Qis exactly p+ 1, it follows that Ghas exactly p+ 1 minimal normal subgroups. However, according to [6, B, 8.2], Ghas at least pk+ 1 minimal normal subgroups, where pkis the number of elements in EndKG(V), and k≥2since Vis not an absolutely irreducible G-module. This contradiction implies that there exists a minimal normal subgroup Aof Gsuch that Q∩A= 1. Since the group G/A satisfies the conditions of the theorem, we have that QA/A is a partial CAP-subgroup of G/A. 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