The lattice of Fitting classes which are right extensible by soluble groups

Publ. Mat. 49 (2005), 351–362 THE LATTICE OF FITTING CLASSES WHICH ARE RIGHT EXTENSIBLE BY SOLUBLE GROUPS M. J. Iranzo∗, Julio P. Lafuente†and F. P´ erez-Monasor∗ In memory of Professor Klaus Doerk Abstract In this paper we study the set of Fitting classes which are right extensible by soluble groups ordered by the inclusion relation. The consideration of the associated lattices gives rise to new Fitting classes and it allows to obtain some injectivity criteria for general Fitting classes. All considered groups are assumed to be finite. Set T={T;Tis a Fitting class such that T=TS}, where Sis the class of soluble groups. The classes in Thave a particular interest in the Theory of Fitting classes of finite groups which are not necessarily soluble. For instance, in a previous paper the authors show that, if T∈ T , then each intermediate Fitting class Fbetween T∗ and hT,b(T)iis injective (that is each group has F-injectors) —in particular, each Fitting class in Locksec(S) is injective, in accordance with a wider conjecture of Shemetkov [4, 11.117]—, being normal Fitting classes precisely those belonging to Locksec(hT,b(T)i). In this paper we study the set Tordered by the inclusion relation. The consideration of the lattices associated to Tgives rise to new Fitting classes, as the complement in XNof the class of generalized nilpotent groups and the complement in TNof the class of nilpotent-constrained groups. The notations and terminology are referred to [1]. We denote by Ethe class of all (finite) groups. Recall that a class of groups Fis a Fitting class if (i) NEG∈Fimplies N∈F, and (ii) N, M EG∈E, 2000 Mathematics Subject Classification. 20D10. Key words. Finite group, Fitting class, lattice, preboundary. ∗Supported by MTM2004-06067-C02-01. †Supported by MTM2004-08219-C02-01. 352 M. J. Iranzo, J. P. Lafuente, F. P´ erez-Monasor N, M ∈Fimplies NM ∈F. If Fis a class of groups, the product GF of all normal subgroups of G∈Ewhich belong to Fis called the radical of Gwith respect to F. The Fitting product of two Fitting classes F and Gis FG= (G∈E;G/GF∈G). The product of two classes of groups Fand Gis the class FG of the groups G∈Ewhich have a normal subgroup Nsuch that N∈Fand G/N ∈G. For a set of groups X, hXidenotes the smallest Fitting class containing X. The boundary of the Fitting class Fis the set b(F) of the groups Xsuch that X6∈ F and Nsn G,N6=Gimplies N∈F(where Nsn Gmeans that Nis a subnormal subgroup of G). Each group X∈b(F) is single-headed (that is Xhas a unique maximal normal subgroup, denoted Cos(X)) and b(F) is subnormally independent (that is Xsn Y∈b(F), X6=Yimplies X6∈ b(F)). A group G∈Eis perfect if G=G0.b(F) is the set of all perfect groups X∈b(F). Set Fb=hCos(X); X∈b(F)i,Fm=hb(F)i and Fs=hG∈F;G0=Gi. If Xis a class of groups, D0Xdenotes the class of all groups which are a direct product of elements of X. If F⊆X are Fitting classes, we set XF= (G∈X;G/GF∈D0Σ), where Σ is the class of all nonabelian simple groups. A class mof groups is a preboundary if it is subnormally independent and consists of single-headed groups. Denote by Pthe set of all preboundaries of perfect groups. If m∈ P and G∈E, set bm(G) = {Xsn G;X∈m}. The map b: T → P is a bijection whose inverse h: P → T is given by h(m) = (G∈E; bm(G) = ∅) if m∈ P [1, XI 4.4]. We begin with Theorem 1. (T,⊆)with the operations T∧R=T∩R,T∨R=hT,RiS, if T,R∈ T , is a complete, distributive and atomic lattice. Thas no coatoms. Proof: It is immediate that (T∩R)S=T∩R, and therefore (T,⊆) is a complete lattice. Denote by ∨and ∧the corresponding lattice operations. We have that, if T,R∈ T , then T∧R=T∩R. And immediatly T∨R=hT,RiS. Let us see that the lattice is distributive. We must show that, if T,R,K∈ T , then T∩FS =HS, where F=hR,Kiand H=hT∩R,T∩Ki. As T∩FS = (T∩F)S, this is equivalent to (T∩F)S=HS. Lattices of Fitting Classes 353 Clearly H⊆T∩F. Let us assume that HS 6= (T∩F)S. Then there exists X∈(T∩F)S∩b(HS). Xis a perfect and single-headed group, hence X∈T∩F. As F=hR,Ki, X∈SnR∪SnKby [1, XI 4.14]. Therefore X∈(T∩R)∪(T∩K)⊆H, a contradiction. The lattice is atomic as a particular case of Proposition 4 and has no coatoms as a consequence of Proposition 7. Remarks 2.i) If T,R∈ T and T6=E6=R, then hT,RiS6=E. In particular, the unique elements of Twhich have a complement are S and E. Suppose, to have a contradiction, that hT,RiS=E. Let X∈b(T), Y∈b(R) and take S∈Σ. Then the regular wreath product W= (X×Y)oSis perfect and single-headed, by [1, A 18.8]. By hypothesis, W∈ hT,Ri, hence either W∈Tor W∈R, by [1, XI 4.14], against X6∈ Tand Y6∈ R. ii) Let T,R∈ T ,T6⊆ R6⊆ T. Then hT,Riis strictly contained in T∨R. To see it, consider H= (G∈E;G/(GTGR)∈N). By [1, IX 2.1], His a Fitting class which contains both Tand R. Let us assume that G∈ T is such that T,R⊆G. Let G∈H. Then GTGR∈G and G/(GTGR)∈N. Thus G∈GS =G. Therefore H⊆T∨R. Let us take R∈R∩b(T) and T∈T∩b(R) and form the regular wreath product G= (T×R)oS3. By [1, X 1.15], Tand Rare Lockett classes. Then, by [1, X 2.1], G/(GTGR)∼ =S36∈ N. Therefore G6∈ H. But Ghas a subgroup Hsuch that H/(HTHR)∼ =C3∈Nand G/H ∼ = C2, that is G∈HS. In particular G∈(T∨R)\ hT,Ri. If Xand Yare classes of groups, we set X·Y= (G∈E;G=GXGY). Notice that if Xand Yare Fitting classes, not necessarily X·Yis a Fitting class (see [1, Remark (b) after IX 2.1]). If m∈ P, we write Em(G) = hX;X∈bm(G)i. Lemma 3. Let Fbe a Fitting class and n⊆b(F). Then we have: hF,ni=F· hni= (G∈E;G=GFEn(G)). 354 M. J. Iranzo, J. P. Lafuente, F. P´ erez-Monasor Proof: Let G∈E. If X∈bn(G), then Cos(X) = XF. Set X= (G∈E;G=GFEn(G)). Let us see that it is a Fitting class. If G∈X, then G/GF∼ =En(G)/En(G)F∈D0Σ [2, Lemma 6]. Let NEG. Then bn(N)⊆bn(G). Thus, if bn(N) = {X1,...,Xr}, then NGF/GF=X1GF/GF× · · · × XrGF/GF, hence N=N∩NGF=N∩X1. . . XrGF=N∩En(N)GF= En(N)NF∈X. On the other side, if N, M EG,G=NM,N, M ∈X, then G=NM =NFEn(N)MFEn(M)≤GFEn(G), hence G∈X. Therefore Xis a Fitting class. Obviously X⊆ hF,ni, hence X=hF,ni. Set L=hni. As En(G)≤GL, we have GFEn(G)≤GFGLfor each G∈E. As L=hni ⊆ hF,ni=X, we have GFGL≤GFEn(G)≤GFGL, hence GFEn(G) = GFGL. Proposition 4. If E6=T∈ T , then [T,→) = {H∈ T ;T⊆H} is atomic. Moreover there is a bijective correspondence between the set of atoms of [T,→)and b(T). Proof: Let us see that H∈ T is an atom in [T,→) if and only if H= hT, XiS, where X∈b(T). Assume that His an atom in [T,→). Let X∈H∩b(T). Then T⊂ hT, XiS⊆H hence H=hT, XiS. Assume conversely that X∈b(T), H∈ T and T⊂H⊆ hT, XiS. Take Y∈b(T)∩H. As Yis perfect, then Y∈ hT, Xi. Then Y∈SnX, by [1, XI 4.14], hence Y∼ =Xbecause Cos(X)∈T. Therefore H= hT, XiS. Now if T⊂H∈ T , let X∈b(T)∩H. Then hT, XiS⊆H, hence [T,→) is atomic. Finally we can show as above that if X, Y ∈b(T), then hT, XiS= hT, Y iSif and only if X∼ =Y. Lattices of Fitting Classes 355 The classes T∈ T for which b(T) is maximal in (P,⊆), where Pis the set of all preboundaries of perfect groups, are precisely those with Tb=T∗[2, Proposition 13]. We have Proposition 5. Let E6=T∈ T . If His an atom in [T,→), then Hb=H∗. Proof: By Proposition 4, H=hT, XiS, where X∈b(T). Now if Y∈H is perfect and single-headed, then either Y∈Tor Y∼ =X, by [1, XI 4.14], hence Hs=hTs, Xi. On the other hand, if S∈Σ, then XoS∈b(H), because Xis perfect and single-headed by [1, A 18.8 (d)], hence X∈Hb. We proceed now to show that Ts⊆Hb. Let Y∈Tperfect and single-headed and take G=X×Y. Let S∈Σ and W=GoS.Wis perfect and single-headed again by [1, A 18.8 (d)]. Assume that W∈H. Then W∈ hT, Xi, hence either W∈Tor Wsn X, by [1, XI 4.14], a contradiction. As Cos(W)∈H, we have that W∈b(H). Therefore Cos(W)∈Hb, hence Y∈Hb. And Ts⊆Hb. Finally H∗=hHs,Hbi=hTs,X,Hbi=Hb, by [3, 1.1]. Remark 6.As a consequence of Proposition 5 we have that {T∈ T ;Tb= T∗}is not a sublattice of T. Take any T∈ T such that Tb6=T∗ and T6=E(for instance the class Cof N-constrained groups, see [2, Example (iii)]) and two different atoms H,K∈[T,→). Then Hb=H∗ and Kb=K∗, by Proposition 5, but (H∩K)b=Tb6=T∗= (H∩K)∗. Proposition 7. Let S6=H∈ T . Then the set of coatoms of (←,H] = {G∈ T ;G⊆H} is in bijective correspondence with the maximal elements of psh(H) = (X∈H;Xis perfect and single-headed) ordered by the relation “to be subnormally embedded”. Proof: Let us assume that Tis a coatom in (←,H]. Then, if X∈H∩ b(T), we have H=hT, XiS. If Yis a perfect and single-headed group in Hand Xsn Y, then X∼ =Y, by [1, XI 4.14], and Xis maximal. Further if Z∈H∩b(T), then hT, XiS=H=hT, ZiS. Therefore X∈ hT, ZiS, hence Xsn Zand X∼ =Zby the maximality of X. Thus H∩b(T) = (X). 356 M. J. Iranzo, J. P. Lafuente, F. P´ erez-Monasor Assume now that Xis a maximal element in psh(H). Consider m= (b(H)\(Y;Xsn Y)) ∪(X). Clearly m∈ P. Let T= h(m). Let us see that H=hT, XiS. Suppose that there exists Y∈T∩b(H). As b(T) = m, it follows that Xsn Y∈ b(H), hence X∈T, a contradiction. Therefore T⊆H, hence hT, XiS⊆ H. Suppose Y∈H∩b(hT, XiS). Yis a perfect and single-headed group; by the maximality of Xwe have that Xis not subnormally embedded in Y. Therefore, if Zsn Y, then Z6∈ m. Thus Y∈h(m) = T, a contradiction. So H=hT, XiSand Tis a coatom in (←,H]. Corollary 8. If T∈ T is such that Tb6=T∗, then each perfect and single-headed group in Tis strictly subnormally embedded in another perfect and single-headed group in T. Proof: Assume the contrary. Then psh(T) has maximal elements for the relation “to be subnormally embedded”, hence there exists a coatom K in (←,T], by Proposition 7. But then Tis an atom in [K,→), hence Tb=T∗by Proposition 5, against the hypothesis. We know that T∈ T if and only if there exist Fitting classes Fand X such that F⊆X⊆XFwith T=T(F,X), where T(F,X) = (G∈E;GX∈ F) [2, Theorem 8]. We set B(X) for the boolean of X. Lemma 9. If Fis a Fitting class, we write XF={X;Xis a Fitting class such that F⊆X=XF}. Then i) The map f : B(b(F)) → XF given by f(m) = hF,miif m⊆b(F), is an isomorphism of inclusionordered sets. ii) (XF,⊆)with the operations X∧Y=X∩Y,X∨Y=X·Y, if X,Y∈ XF, is a complete, distributive, complemented and atomic lattice. Proof: i) Let us see that, if Fis a Fitting class and F⊆X, then X∈ XF if and only if X=hF,X∩b(F)i. Set m=X∩b(F). If X=hF,mi, then G∈Ximplies G=GFEm(G), by Lemma 3, hence G/GF∼ =Em(G)/Em(G)F∈D0Σ. Therefore X=XF. Assume conversely that X∈ XF. Obviously hF,mi ⊆ X. Let G∈X. As X=XF, we have G/GF∈D0Σ. We may assume that G6∈ F. Lattices of Fitting Classes 357 Let N/GFbe a minimal normal subgroup of G/GF. Take Xsn Ga minimal supplement of GFin N. Then Xis perfect and single-headed and Cos(X) = XF, hence X∈b(F)∩X=m. It follows that G= GFEm(G)∈ hF,mi, and X⊆ hF,mi. Therefore f is a well defined bijection between the boolean of b(F) and XF. Let m,n⊆b(F). If m⊆n, then obviously f(m)⊆f(n). Finally, assume that hF,mi ⊆ hF,ni. Let X∈m. Then X∈ hF,ni, hence X∈n by [1, XI 4.14], as nis subnormally independent. ii) It suffices to show that, if m,n⊆b(F), X=hF,miand Y=hF,ni, then X·Y=hF,m∪ni. Now G∈X·Y⇐⇒ G=GXGY⇐⇒ G=GFEm(G) En(G) (by Lemma 3) ⇐⇒ G∈ hF,m∪ni, as obviously Em(G) En(G) = Em∪n(G). Theorem 10. Let Fbe a Fitting class and set TF={T∈ T ;T=T(F,X),X∈ XF}. Then we have i) If T∈ T , then T∈ TFif and only if b(T)⊆b(F). ii) The map τF:XF→ TF given by τF(X) = T(F,X)if X∈ XFis a lattice antiisomorphism. iii) TFis a complete, distributive, complemented and atomic lattice. Proof: i) By [1, XI 4.7], b(T)⊆b(F) is equivalent to Tb⊆F⊆T. By [2, Proposition 3], if T∈ TF, then Tb⊆F⊆T. Conversely, if T∈ T and Tb⊆F⊆T, set X= (G∈E;G=GFGM), where M=Tm. Then X∈ XFand T=T(F,X) by [3, 1.3, 1.4], hence T∈ TF. ii) Consider the composition B(b(F)) f −→ XF τF −→ TF, where f(m) = hF,mi, if m⊆b(F). Then τFf(m) = T(F,hF,mi) = (G∈E;GFEm(G)∈F) = (G∈E; Em(G)∈F) = (G∈E;Sn(G)∩m=∅) = h(m), hence τFf is the restriction of h to B(b(F)). Moreover if T∈ TF, taking m= b(T), we have that h(m) = Tand, by i), m⊆b(F). Therefore τFf is a bijection between B(b(F)) and TF. 358 M. J. Iranzo, J. P. Lafuente, F. P´ erez-Monasor Let T,R∈ TF. Then, by i), Tb⊆F⊆R, and, by [1, XI 4.7], T⊆Rif and only if b(R)⊆b(T). Therefore τFf is an antiisomorphism of ordered sets. Now, by Lemma 9, we have that fis an isomorphism, hence τFis also an antiisomorphism. iii) That follows now immediately from Lemma 9. Corollary 11. Let T,R∈ T . Then the following assertions are equivalent i) There exists a Fitting class Fsuch that T,R∈ TF. ii) b(T)∪b(R)is a preboundary. iii) b(T)∪b(R) = b(T∩R). Proof: i) =⇒iii) by Theorem 10. Of course iii) =⇒ii). If b(T)∪b(R) is a preboundary, as it consists of perfect groups, then, by [1, XI 4.3], H= h(b(T)∪b(R)) is a Fitting class (incidentally belonging to T) such that b(H) = b(T)∪ b(R). In particular b(T)⊆b(H) and b(R)⊆b(H), hence T,R∈ TH, by Theorem 10, and we have finally that ii) =⇒i). Remarks 12.Let Fbe a Fitting class. i) Obviously max(TF) = E. On the other hand, min(TF) = h(b(F)) by Theorem 10. And it is immediate that h(b(F)) = (G; Socn(G/GF) = 1). ii) TFis not a sublattice of T. If T,R∈ TF, then b(T∩R) = b(T)∪ b(R), by Corollary 11, hence the infimum of two elements in TFcoincides with that in T. But the question changes for the supremum: it suffices to consider that TFis complemented and Remark 2 (i). In fact, for all T,R∈ T , b(T)∩b(R)⊆b(hT,RiS), with equality if and only if b(T)⊆b(R) or b(R)⊆b(T). To see it, let us assume first that b(hT,RiS)⊆b(T)∩b(R) and suppose moreover that T6⊆ R6⊆ T. Take X∈R∩b(T), Y∈T∩b(R) and Was in Remark 2 (i); then W∈b(hT,RiS)\(b(T)∩b(R)), against the hypothesis. We may assume therefore that T⊆R. Then hT,RiS= R, hence b(R)⊆b(T)∩b(R) and so b(R)⊆b(T). Let us assume now that b(R)⊆b(T). Then T⊆Rby [1, XI 4.4], hence b(hT,RiS) = b(R)⊆b(T)∩b(R). iii) If b(F) = m∪nand m∩n=∅, then T= h(m) and R= h(n) are mutual complements in TF,Tm=hmiand Rm=hni. Obviously Lattices of Fitting Classes 359 m⊆Rsand n⊆Ts. In fact, hmi ⊂ Rsand hni ⊂ Ts. For if X∈m, take any S∈Σ and suppose that XoS6∈ R; then XoS∈b(R) = n, as XoS is perfect and single-headed with cosocle in R, by [1, A 18.8], which contradicts that m∪nis subnormally independent; therefore XoS∈Rs; but XoS6∈ hmiby [1, XI 4.14] because mis subnormally independent. Analogously hni ⊂ Ts. In particular, Tm⊂R∗and Rm⊂T∗, by [1, XI 4.15]. iv) By Theorem 10 (i) and [2, Proposition 13], there exists T∈ TF such that Tb=T∗if and only if b(F) is maximal in (P,⊆). In this case such a Tis unique, namely T= h(b(F)) = FS =F∗; moreover, F∈Locksec(T), hence Fis injective, by [3, 3.4]. v) If Fand Gare two Fitting classes, then TF=TGif and only if b(F) = b(G), by Theorem 10 (i). Therefore TF=Th(b(F)). In particular, by [2, Proposition 13], if T∈ T satisfies Tb=T∗, then there is a unique E ∈ {TF;Fis a Fitting class}such that T∈ E, namely E=TT. In general, given T∈ T , the set of lattices TFaccording to Theorem 10 for which T∈ TFis {TH;H∈ T ,Tb⊆H⊆T} by [1, XI 4.7]. Observe that if T∈ TH,H∈ T , then the complement of Tin THis R= h(b(H)\b(T)) = h(b(H)∩T). We have T∩R= min(TH) = H. On the other hand if Gis a Fitting class such that T∩G⊆H, then G∩b(R) = G∩b(H)∩T⊆H∩b(H) = ∅, hence G⊆R. Therefore Ris the greatest Fitting class Gsuch that T∩G⊆H. If H,K∈ T ,Tb⊆H,K⊆Tand R, resp. L, is the complement of T in TH, resp. TK, then H⊆Kif and only if R⊆L. Indeed if H⊆K, then R∩T=H⊆K, hence R⊆L. And if R⊆L, then H∩b(K) = T∩R∩b(K)⊆T∩L∩b(K) = K∩b(K) = ∅, hence H⊆K. Example 13. Let Nbe the class of nilpotent groups and consider the class C= (G; F(G) = F*(G)) of N-constrained groups. Then C=T(N, e N), where e N= (G;G= F*(G)) is the class of generalized nilpotent groups (see [1, IX 4.14]). We noted in [2, Example (iii)] that Cb=Nand Cm= e N.