On a question of Beidleman and Robinson

Document downloaded from: This paper must be cited as: The final publication is available at Copyright Additional Information http://hdl.handle.net/10251/18474 Ballester Bolinches, A.; Esteban Romero, R. (2002). On a question of Beidleman and Robinson. Communications in Algebra. 12(30). doi:10.1081/AGB-120016008 http://www.tandfonline.com/10.1081/AGB-120016008 Taylor & Francis [EN] This is an Author's Original Manuscript of an article submitted for consideration in Communications in Algebra 30 (12):5757-5770 (2012) (copyright Taylor & Francis). Communications in Algebra is available online at http://tandfonline.com/10.1081/AGB120016008 This is an Author’s Original Manuscript of an article submitted for consideration in Communications in Algebra 30(12):5757–5770 (2002) (copyright Taylor & Francis). Communications in Algebra is available online at http://www.tandfonline.com/10.1081/AGB-120016008 DOI 10.1081/AGB-120016008 ON A QUESTION OF BEIDLEMAN AND ROBINSON∗ A. Ballester-Bolinches Departament d’` Algebra Universitat de Val`encia Dr. Moliner, 50 E-46100 Burjassot (Val`encia) Spain email: [email protected] R. Esteban-Romero Departament de Matem`atica Aplicada Universitat Polit`ecnica de Val`encia Cam´ı de Vera, s/n E-46022 Val`encia Spain email: [email protected]v.es Abstract In (1, Theorem A), Beidleman and Robinson proved that if a group satisfies the permutizer condition, it is soluble, its chief factors have order a prime number or 4 and Ginduces the full group of automorphisms in the chief factors of order 4. In this paper, we show that the converse of this theorem is false by showing some counterexamples. We also find some sufficient conditions for a group satisfying the converse of that theorem to satisfy the permutizer condition. ∗Supported by Proyecto PB97-0674-C02-02 and Proyecto PB97-0604 from DGICYT, Ministerio de Educaci´on y Ciencia 1 1 INTRODUCTION All groups considered in this paper are finite. Given a subgroup Hof a group G, the permutizer PG(H) of Hin Gis defined as the subgroup generated by all cyclic subgroups of Gthat permute with H. Thus H≤PG(H) and H6=PG(H) if and only if Hhxi=hxiHfor some x∈G\H. A group Gsuch that H6=PG(H) for every proper subgroup His said to satisfy the permutizer condition or to be a P-group. Beidleman and Robinson (1, Theorem A) proved the following result: Theorem 1. Let Gbe a finite group satisfying the permutizer condition. Then Gis soluble and each chief factor of Ghas order 4or a prime. In addition, if Fis a chief factor of order 4, then Ginduces the full group of automorphisms in F, i. e., G/CG(F)∼ =Σ3. In the same paper, the authors asked whether the converse is true. We show in this paper that the converse is not true and find sufficient conditions for a group satisfying the converse of that theorem to be a P-group. 2 A COUPLE OF EXAMPLES In this section we present a couple of examples to show that the converse of Theorem 1 is false. Example 1.Let Vbe an irreducible and faithful hbi-module over the field of 2 elements, where hbiis a cyclic group of order 3, such that the corresponding semidirect product is isomorphic to A4, the alternating group of degree 4. Let A1and A2be two copies of V, and consider A=A1×A2. It is clear that Ais a faithful hbi-module. Denote B= [A]hbithe corresponding semidirect product. Then we can choose generators c,dof A1and e,fof A2such that cb=cd,db=c,eb=ef and fb=e. The group Bhas an automorphism a of order 2 such that ba=b2,ca=ef,da=f,ea=cd and fa=d, so we can consider the semidirect product G= [B]hai. The normal series 1< N =hcef, dei< A < B < G is a chief series of G. Hence the chief factors of Ghave order 2, 3 or 4. On the other hand, CG(N) = A,CG(A/N) = A,CG(B/A) = Band CG(G/B) = G. Consequently, Gsatisfies the converse of Theorem 1. Let H= [A2]hbi. We show that PG(H) = H. Suppose there exists an element x∈Gof order 3 such that Hhxiis a subgroup of G. Then there 2 exists a Sylow 3-subgroup H3of Hsuch that H3hxiis a Sylow 3-subgroup of Hhxi. Since Sylow 3-subgroups of Ghave order 3, it follows that H3=hxi and x∈H. Suppose now that there exists a 2-element y∈Gsuch that Hhyiis a subgroup of G. It is clear that in this case A2hyiis a Sylow 2-subgroup of Hhyi. If y∈A, then A2hyiis an elementary abelian 2-group normalized by b. Hence, if A26=A2hyi, there exists an element z∈A2hyisuch that A2hyi=A2× hziand b∈CG(hzi), a contradiction. Therefore A2hyi=A2 and y∈H. Assume that y=y1afor some y1∈A. Notice that y2∈Aand so A∩A2hyi=A2hy2i. Thus A2hy2i ≤ Ais a normal subgroup of A2hyi. In particular, A1=Ay 2=Aa 2≤A2hy2i, a contradiction. Finally, assume that there exists an element g∈Gsuch that Hhgiis a subgroup of G. Then hgi=hg1i × hg2i, where |hg1i| is a 2-number and |hg2i| ∈ {1,3}. We can find a Sylow 3-subgroup H3of Hsuch that H3hg2iis a Sylow 3-subgroup of Hhgi. Therefore H3=hg2iand g2∈Hbecause Sylow 3-subgroups of Ghave order 3. Hence Hhgi=Hhg2iand so g2∈Hby the above case. Consequently g∈Has we want to prove. Our next example is quite surprising bearing in mind the results of the next section. Theorem 2. G= Σ4×Σ4×Σ4×Σ4is not a P-group. Proof. It is enough to find a proper subgroup Hof Gsuch that PG(H) = H. Denote by Gi, 1 ≤i≤4, the factors of Gisomorphic to Σ4and by Ni= hai, biithe unique minimal normal subgroup of Gi. Let gibe an element of order 3 in Gisuch that Li=Nihgiiis isomorphic to A4. We consider the set H=H1hgi, where H1=ha1a3a4, ag1 1ag3 3ag4 4, a2a3ag4 4, ag2 2ag3 3ag2 4 4i and g=g1g2g3g4. Since ag2 i i=aiagi ifor all i, it follows that g∈NG(H1). Hence His a subgroup of G. We prove that PG(H) = H. First of all, His not permutable with any subgroup of order 3 which is not contained in H. Assume not and let q∈G\Hbe an element of order 3 such that Hhqi ≤ G. Then H1is a Sylow 2-subgroup of Hhqiwhich is normalized by q. Consequently, aq,bq∈H1, where a=a1a3a4and b=a2a3ag4 4. Notice that H1=ha, agi × hb, bgi. Since the component of aqin G2is trivial and the component of bqin G1is also trivial, it follows that aq∈ ha, agiand bq∈ hb, bgi. Hence either aq=a,aq=agor aq=aag, and either bq=b, bq=bgor bq=bbg. If q=q1q2q3q4, with qi∈Gi, 1 ≤i≤4, then one of the following cases holds: 3 Case (a) aq1 1=a1,aq2 2=a2,aq3 3=a3,aq4 4=a4,ag4q4 4=ag4 4. Case (b) aq1 1=ag1 1,aq2 2=ag2 2,aq3 3=ag3 3,aq4 4=ag4 4,ag4q4 4=a4ag4 4. Case (c) aq1 1=a1ag1 1,aq2 2=a2ag2 2,aq3 3=a3ag3 3,aq4 4=a4ag4 4,ag4q4 4=a4. The case (a) is impossible because the Sylow 3-subgroups of Giact fixedpoint freely on Nifor all i. Suppose that (b) holds. Then qig−1 i∈CLi(ai) = Nifor all i. Therefore qg−1∈Hhqi ∩ Soc(G) = H1and q∈H1hqi=H, a contradiction. The case (c) is analogous. Assume that His permutable with a subgroup hsicontained in Soc(G). Then, if s /∈H1, we have that H1× hsiis a Sylow 2-subgroup of Hhsiwhich is normalized by g. This means that there exists a subgroup hzi ≤ H1× hsi such that g∈CG(hzi), a contradiction. Hence s∈H. If His permutable with a subgroup hxiof order 2, with x∈G\V, where V=O{2,3}(G), then |Hhxi:H|= 2 and so Hmust be normalized by x. In particular, x∈NG(H1). Arguing as above, we have that ax∈ {a, ag, aag} and bx∈ {b, bg, bbg}. Consequently one of the following cases holds if x= x1x2x3x4with xi∈Gi: Case 1 ax1 1=a1,ax2 2=a2,ax3 3=a3,ax4 4=a4,ag4x4 4=ag4 4. Case 2 ax1 1=ag1 1,ax2 2=ag2 2,ax3 3=ag3 3,ax4 4=ag4 4,ag4x4 4=a4ag4 4. Case 3 ax1 1=a1ag1 1,ax2 2=a2ag2 2,ax3 3=a3ag3 3,ax4 4=a4ag4 4,ag4x4 4=a4. Suppose that either Case 2 or Case 3 holds. Then the automorphism induced by x4on N4has order 3, a contradiction. Consequently, Case 1 must hold. In this case, x4centralizes N4. Hence x4= 1. Since (a1a3a4)gx = ag1x1 1ag3x3 3ag4 4is an element of H1, we have that ag1x1 1ag3x3 3ag4 4=ag1 1ag3 3ag4 4. Hence x1centralizes N1and x3centralizes N3. This implies that x1=x3= 1. The same argument applied to (a2a3ag4 4)gx shows that x2= 1. Therefore x= 1, a contradiction. Assume now that His permutable with a cyclic subgroup hxiof order 4 which is not contained in H. It is clear that x2∈Soc(G). If x2/∈H, then H∩ hxi= 1 and |Hhxi:H|= 4. Notice that (x2)gand (x2)g2are not in H. Therefore Hhxiis equal to the disjoint union Hhxi=H∪Hx2∪H(x2)g∪ H(x2)g2. This is a contradiction, because xdoes not belong to this union. Consequently, x2∈H∩Soc(G) and |Hhxi:H|= 2. This implies that x normalizes Hand so xalso normalizes H1. Arguing as in the above paragraph, if x=x1x2x3x4, we have that either ax4 4=a4and ag4x4 4=ag4 4, or ax4 4=ag4 4and ax4g4 4=a4ag4 4, or ax4 4=a4ag4 4 and ag4x4 4=a4. In the first case, x4centralizes N4, and in the second and 4 third cases, the automorphism induced by x4on N4is of order 3. The latter possibility gives a contradiction. Hence the first case holds and then ax1 1=a1, ax2 2=a2and ax3 3=a3. Moreover, x4∈Soc(G). Arguing as above with the elements (a1a3a4)gx and (a2a3ag4 4)gx, it follows that x∈Soc(G). This is a contradiction because Soc(G) has no elements of order 4. We would like to mention at this point that what is proved in the above cases is that if xis either an element of order 2 in G\Vor xis an element of order 4, then hxidoes not permute with H1. Finally, suppose that hxi=hxi2×hxi3is a subgroup of Gsuch that Hhxi is a subgroup of G. Then H1hxi2is a Sylow 2-subgroup of Hhxi. Suppose that hxi2is not contained in H. By the above remark, hxi2is a subgroup of order 2 contained in Soc(G). Therefore Hhxi ≤ L1×L2×L3×L4, and then g∈NG(H1hxi2). This means that there exists an element y∈H1hxi2such that H1hxi2=H1× hyiand g∈CG(hyi), a contradiction. Thus hxi2≤H and Hhxi=Hhxi3. Bearing in mind the first case, it follows that hxi3≤H and hxi ≤ H. Consequently, PG(H) = Hand the theorem is proved. 3 QP-GROUPS We say that a group Gis a QP-group (or Gsatisfies the property QP) if Gis soluble, each chief factor of Ghas order 4 or a prime, and if A/B is a chief factor of Gof order 4, then Ginduces the full group of automorphisms in A/B, i. e., G/CG(A/B)∼ =Σ3. Recall that in any group G, there is a unique maximum normal supersolubly embedded subgroup, denoted here by ZU(G). It is known that there is a G-invariant series in ZU(G) with cyclic factors, while G/ZU(G) has no nontrivial normal cyclic subgroups. ZU(G) is the U-hypercentre of G, where Uis the formation of all supersoluble groups (see (2, Section IV.6)). Beidleman and Robinson proved in (1, 3.1) the following result: Lemma 1. A group Gis a P-group if and only if G/ZU(G)is a P-group. In order to prove that a QP-group is a P-group, one often can assume that the U-hypercentre is trivial. Hence the following result applies. Lemma 2. Let Gbe a QP-group such that ZU(G) = 1. Then: 1. O20(G) = 1 and Gis a {2,3}-group. 2. G/F(G)is isomorphic to a subgroup of a direct product of Σ3and C2. Consequently, G/F(G)is an extension of a 3-group V/F(G)by a 2group G/V , where V=O{2,3}(G). 5 3. The supersoluble normalizers of Gare exactly the normalizers of the Sylow 3-subgroups of G. Proof. 1. Since every chief factor of Gof odd order is cyclic, it follows that O20(G) is supersolubly embedded in G. Hence O20(G)≤ZU(G) = 1. Applying (1, (3.5)), Gis a {2,3}-group. 2. follows from the fact that F(G) is the intersection of the centralizers of the chief factors of G. 3. Let Dbe a supersoluble normalizer of G(for properties of normalizers see, for example, (2, Chapter V)). Then there exists a Hall system Σ = {1, G2, G3, G}of G, where G2is a Sylow 2-subgroup of Gand G3 is a Sylow 3-subgroup of G, such that Dis the supersoluble normalizer associated to Σ. For each prime p, we denote f(p) the formation of all abelian groups whose exponent divides p−1 and v(p) = Gf(p), the f(p)-residual of G. It is known ((2, IV.3.4)) that fis an integrated local definition of U. Then, according to (2, V.1.1), D=\ p∈{2,3} NGGp0∩v(p)=NGG20∩v(2)∩NGG30∩v(3). Since f(2) = 1, it follows that v(2) = G. Moreover, v(3) is contained in Vbecause G/V is an elementary abelian 2-group. On the other hand, G2∩V=F(G). Hence G30∩v(3) = G2∩v(3) ≤F(G)∩v(3) ≤ G2∩v(3) and NGG30∩v(3)=NGF(G)∩v(3)=G. This implies that there exists a Sylow 3-subgroup G3of Gsuch that D=NG(G3). The result now follows from the fact that the supersoluble normalizers are conjugate (see (2, V.2.3)). The following results turn out to be crucial in the proof of our main theorems. Lemma 3. Let Gbe a QP-group contained in a direct product S1×· · ·×Srof rcopies of Σ4and containing the corresponding direct product A1×· · ·×Arof rcopies of A4. Assume that His a subgroup of Gsuch that PG(H) = Hand let H2and H3be, respectively, a Sylow 2-subgroup and a Sylow 3-subgroup of H. If Niis the minimal normal subgroup of Gcontained in Ai, then: 1. If Ni∩H6= 1, then Ai≤H. 2. If Ni∩H= 1, then H2≤CG(Ni)and H36≤ CG(Ni). 6 Proof. 1. Suppose that Ni∩H6= 1 and let 1 6=ai∈Ni∩H. Since Ni is an elementary abelian 2-group of order 4, we can find an element 16=bi∈Nisuch that Ni=hai, bii. Then Hhbii=HNiand so bi∈PG(H) = H. Hence Niis contained in H. Let now hgiibe a Sylow 3-subgroup of Ai. Then Ai=Nihgiiand so HAi=Hhgii. This means that gi∈PG(H) = H. Consequently Aiis contained in H. 2. Suppose that Ni∩H= 1. We prove that H2centralizes Ni. Assume that this is not true. We distinguish two cases: (a) There exists an element h∈H\CG(Ni) such that the component hiof Hin Sihas order 2. If hhas a component hj,j6=i, of order 3, then h3is an element of H\CG(Ni) whose component in Si has order 2. Hence without loss of generality we may assume that o(h) = 2 or 4. Let ai∈Ni\{1}such that ah i6=ai. Then haiis an element of Gof order 4 (notice that (hai)2=h2ah iai6= 1). In this case, Hhhaii= HNiby order considerations. Consequently hai∈PG(H) = H and so ai∈Ni∩H= 1, a contradiction. (b) No element of H\CG(Ni) has its component in Siof order 2. We may assume that there exists h∈H2\CG(Ni). Suppose that H3 does not centralize Niand let g∈H3\CG(Ni). We know that o(hi) = 4 and so either hg or hg2has its component in Siof order 2. Moreover hg ∈H\CG(Ni) and hg2∈H\CG(Ni) because hi/∈Ai, a contradiction. Therefore H3≤CG(Ni) = S1× · · · × Ni× · · · × Srand so the projection of H3in Siis 1. Assume that the projection of H2in Siis a Sylow 2-subgroup of Si. Then there exists an element x∈ H\CG(Ni) such that o(xi) = 2. This is impossible by (2a). Hence the projection of H2in Siis just hhii. In particular hh2 iipermutes with H. Thus h2 i∈H∩Ni= 1, a contradiction. Consequently H2centralizes Ni. If H3≤CG(Ni), then Ni≤H, a contradiction. Hence H3cannot centralize Ni. Lemma 4. Let Gbe a QP-group such that ZU(G)=1. Assume that G has rchief factors of order 4in a given chief series of G. Suppose that Soc(G) = N1× · · · × Nr, where Niis a minimal normal subgroup of Gof order 4for all i. Then the order of a Sylow 3-subgroup of Gis at most 3r. 7 Suppose that G3=hg1g2, g3i. There exist elements hi∈Si\Ai,i∈ {1,2,3}, of order 2, such that ghi i=g−1 iand h1h2h3∈G. If h1∈G, then (g1g2)h1=g−1 1g2, whence g2∈G, a contradiction. Consequently h1/∈Gand, analogously, h2/∈G. Let ai∈Ni\ {1}such that ahi i=agi i,i∈ {1,2,3}. If h3∈G, then we consider the subgroup H=ha1ag2 2, ag1 1ag2 2 2, a3, ag3 3, g1g2, g3, h3i, and if h3/∈G, we take the subgroup H=ha1ag2 2, ag1 1ag2 2 2, a3, ag3 3, g1g2, g3i. Like in Example 1 we have that His a proper subgroup of Gsuch that PG(H) = H. Suppose that G3=hg1g2, g1g3i. There exist elements hi∈Si\Ai,i∈ {1,2,3}, of order 2, such that ghi i=g−1 iand h1h2h3∈G. If h1∈G, then (g1g2)h1=g−1 1g2∈G, whence g1∈G, a contradiction. Consequently, h1/∈G. Analogously, h2/∈Gand h3/∈G. Consider an element ai∈Si\Ai such that ahi i=agi ifor 1 ≤i≤3. Notice that (g1g2)2(g1g3)2=g1g2 2g2 3. The subgroup H=ha1a2a3, ag1 1ag2 2 2ag2 3 3, g1g2 2g2 3iis a proper subgroup of Gsuch that PG(H) = H. REFERENCES [1] Beidleman, James C.; Robinson, Derek J. S. On finite groups satisfying the permutizer condition. J. Algebra 1997,191, 686–703. [2] Doerk, Klaus; Hawkes, Trevor. Finite Soluble Groups; Number 4 in De Gruyter Expositions in Mathematics; Walter de Gruyter: Berlin, New York, 1992. [3] The GAP Group, Aachen, St Andrews. GAP – Groups, Algorithms, and Programming, Version 4.1, 1999. (http://www-gap.dcs.st-and.ac.uk/~gap). 14