Lie solvable group algebras of derived length three

Publicacions Matem`atiques, Vol 39 (1995), 233–240. LIE SOLVABLE GROUP ALGEBRAS OF DERIVED LENGTH THREE Meena Sahai Abstract Let Kbe a field of characteristic p>2 and let Gbe a group. Necessary and sufficient conditions are obtained so that the group algebra KG is strongly Lie solvable of derived length at most 3. It is also shown that these conditions are equivalent to KG Lie solvable of derived length 3 in characteristic p≥7. 1. Introduction Any associative ring Rgives rise to the associated Lie ring L(R) under the Lie multiplication [x, y]=xy −yx,x,y∈R. We define, inductively, [x1,x 2,... ,x n]=[[x1,x 2,... ,x n−1],x n]. An additive subgroup Vof R is called a Lie ideal of Rif [v,r]∈Vfor all v∈Vand r∈R. For any two Lie ideals Vand W, we denote by [V,W] to be the additive subgroup of Rgenerated by {[v,w]|v∈Vand w∈W}. We define the Lie derived series δ[n](L(R)) and the strong Lie derived series δ(n)(R), n≥0, by induction as follows: δ[0](L(R)) = δ(0)(R)=R, δ[n](L(R)) = [δ[n−1](L(R)),δ[n−1](L(R))], δ[n](R)=[δ(n−1)(R),δ(n−1)(R)]R. Ris Lie solvable of derived length nif δ[n](L(R)) = 0 but δ[n−1](L(R)) = 0. Similarly Ris strongly Lie solvable of derived length nif δ(n)(R)=0 but δ(n−1)(R)= 0. Lie solvable/strongly Lie solvable rings of derived length 2 are called Lie metabelian/strongly Lie metabelian rings. Also Ris said to be Lie centrally metabelian if [δ[2](L(R)),R]=0. Let Kbe a field with Char K=pand let Gbe a group. It is known that the group algebra KG is Lie solvable if and only if Ghas a 2-abelian subgroup of index at most 2, when p= 2 and Gis p-abelian when p=2. 234 M. Sahai For p= 2 this is equivalent to KG strongly Lie solvable (see [4, Chapter V]). But the connection between the order of the derived subgroup Gof Gand the derived length of KG is not known as yet. In this direction Levin and Rosenberger [1] have characterized Lie metabelian group rings. Lie centrally metabelian group algebras have been studied by Sharma and Srivastava [6] and Sahai and Srivastava [3]. It is shown in [1] that the group ring RG of a group Gover a commutative ring R is Lie metabelian if and only if it is strongly Lie metabelian. In this paper, necessary and sufficient conditions for the group algebra KG, Char K=p≥3, to be strongly Lie solvable of derived length at most 3 have been obtained. It is shown that these conditions are equivalent to KG Lie solvable of derived length at most 3 for p≥7. 2. Results and Proofs Throughout this section Kdenotes a field with Char K=p≥3 and Gdenotes a group. It is well known and easy to see that if Mand Nare normal subgroups of G, then [∆(M)KG,∆(N)KG]KG = ∆((M,N))KG +∆(M)∆(N)∆(G)KG + ∆((M,G))∆(N)KG +∆(M)∆((N,G))KG. In particular, if we take M=N=G, then δ(2)(KG) = [∆(G)KG,∆(G)KG]KG =∆(G)KG +∆(G)3KG +∆(γ3(G))∆(G)KG +∆(G)∆(γ3(G))KG. Remark 2.1. If Gis central, then the above equation gives δ(2)(KG)=∆(G)3KG. Further, if γ3(G)=G, then δ(2)(KG)= ∆(G)2KG. If |G|=pn, Char K=pand t(G) denotes the nilpotency index of the augmentation ideal ∆(G), then it is known that n(p−1)+1 ≤t(G)≤pn with equality on the left/right hand side if and only if Gis elementary abelian/cyclic (see [2]). This will be used for t(G), as Gis a finite p-group if KG is Lie solvable. For any element x∈Gwe denote ˆx= 1+x+x2+···+xn−1where order of xis n. We start with the following straightforward observation. Lie Solvable Group Algebras 235 Lemma 2.2. For all n≥1,∆(G)2n−1KG ⊆δ(n)(KG)⊆ ∆(G)2n−1KG. Proof: The right hand side inclusion is immediate by induction on n. Since δ(1)(KG)=∆(G)KG and the identity δ1δ2[g1,g 2]=[δ1g1,δ 2g2]−[δ1,δ 2g2]g1−[δ1g1,δ 2]g2+[δ1,δ 2]g1g2 is true for all δ1,δ2∈∆(G)2n−1and g1,g2∈G,wehave ∆(G)2n+1−1KG =∆(G)2n−1∆(G)2n−1∆(G)KG =∆(G)2n−1∆(G)2n−1[KG,KG]KG ⊆[∆(G)2n−1KG,∆(G)2n−1KG]KG ⊆[δ(n)(KG),δ(n)(KG)]KG =δ(n+1)(KG). This proves the left hand side inclusion by induction on n. Theorem 2.3. Let Kbe a field of characteristic p=2and let Gbe a group. Then δ(3)(KG)=0if and only if one of the following holds: (i) Gis abelian. (ii) p=7,G=C7and γ3(G)=1. (iii) p=5,G=C5and either γ3(G)=1or γn(G)=Gfor all n≥3 with xg=x−1for all x∈Gand for all g/∈CG(G). (iv) p=3,Gis a group of one of the following types: (a) G=C3. (b) G=C3×C3and either γ3(G)=1or γ3(G)=C3,γ4(G)=1 or γn(G)=G, for all n≥3with xg=x−1for all x∈G and for all g/∈CG(G). (c) G=C3×C3×C3,γ3(G)=1. Proof: Suppose that δ(3)(KG) = 0. Since Char K=2,Gis a finite p-group. Let |G|=pn. By Lemma 2.2, ∆(G)7KG ⊆δ(3)(KG)⊆ ∆(G)4KG.Thus∆(G)7= 0. This in turn implies t(G)≤7. By the discussion following just after Remark 2.1, we conclude that (i) p≥11 implies n= 0. In this case G= 1 and thus Gis abelian. (ii) p= 7 and Gis non-abelian implies n= 1 and G=C7. (iii) p= 5 and Gis non-abelian implies n= 1 and G=C5. (iv) p= 3 and Gis non-abelian implies Gis C3or C3×C3or C3× C3×C3. 236 M. Sahai If Gis abelian, we are through. So we discuss each non-abelian case separately. Case (ii). p= 7. In this case G=C7. We shall show that Gis central, i.e., γ3(G) = 1. If not then γ3(G)=Gand by Remark 2.1, δ(2)(KG)=∆(G)2KG. Let G=x. Then (x−1)6=ˆx. Now for any g∈G,wehave 0=[(x−1)2,(x−1)2g−1] =(x−1)3[x, g−1]+(x−1)2[x, g−1](x−1) =−x{(x−1)3((x, g)−1)+(x−1)2((x, g)−1)(xg−1)}g−1. If (x, g)=xk,1≤k≤5, then we get (x−1)4(1 + x+x2+···+xk−1)(2 + x+x2+···+xk)=0. Multiplying by (x−1)2, we have k(k+2)ˆx=0. Thusk(k+2)=0in K.Sok=5. Nowk= 5 is not possible because otherwise (x−1)4(1 + x+x2+x3+x4)(2 + x+x2+x3+x4+x5)=0, which gives (x−1)5(1 + x+x2+x3+x4)(5+4x+3x2+2x3+x4)=0. Multiplying by (x−1) we get 75ˆx= 0 which is not true. Thus k=0 and (x, g) = 1 for all g∈G. Hence Gis central. Case (iii). p= 5. In this case G=C5. Let G=x.IfGis not central, then γ3(G)=G, and δ(2)(KG)=∆(G)2KG. Let (x, g)=xk for some g∈G,1≤k≤3. If we proceed exactly as in the previous case, we get k(k+2)=0inK. This gives k=3. Thus(x, g)=1orx3, i.e., if g/∈CG(G), then xg=x−1, as desired. Case (iv). p=3. IfG=C3, we are through. If G=C3×C3 then t(G) = 5 and γ3(G)=1orC3or G.Ifγ3(G) = 1, we are through. Consider the case when γ3(G)=C3=z. Let y∈Gsuch that y/∈γ3(G). Since ∆(γ3(G))∆(G)KG ⊆δ(2)(KG), we have that for all g∈G 0=[(z−1)(y−1)g−1,(z−1)(y−1)] =(z−1)2(y−1)[g−1,y]+(z−1)(y−1)[g−1,z](y−1) =(z−1)2(y−1)y(( y,g)−1)g−1+(z−1)(y−1)z(( z,g)−1)(yg−1)g−1 . Lie Solvable Group Algebras 237 First term is zero because (y,g)∈γ3(G) and ∆(γ3(G))3=0. Thus (z−1)(y−1)(zk−1)(yg−1) = 0, where (z,g)=zk. This implies that if k= 0, then (z−1)2(y−1)2= 0, because yg=y(y,g) and (y,g)∈γ3(G). But this is a contradiction to the fact that y/∈γ3(G). Hence k= 0. This shows that γ3(G) is central, i.e., γ4(G)=1. If γ3(G)=G, then δ(2)(KG)=∆(G)2KG. Let G=x×yand let g∈Gsuch that g/∈CG(G). Now [(x−1)2,(x−1)2g−1] = 0. Using the fact that (x−1)3= 0 and expanding we get (x−1)2(xg−1)2=0. This implies xg∈x. Similarly yg∈y. Suppose that xg=x,so xg=x−1.Ifyg=y, then [(x−1)(y−1)g−1,(x−1)(y−1)] = 0 gives that (x−1)2(y−1)2=0.Soy∈xwhich is not possible. Hence yg=y and we must have yg=y−1.Thusifg/∈CG(G), then xg=x−1and yg=y−1which proves that ug=u−1for every u∈G. Now if G=C3×C3×C3, then t(G) = 7. Also ∆(G)3(KG)⊆ δ(2)(KG) and ∆(γ3(G))∆(G)KG ⊆δ(2)(KG). We wish to prove that Gis central, i.e., γ3(G) = 1. Suppose, if possible γ3(G)= 1. Let 1=x∈γ3(G). Choose y,z∈Gsuch that G=x×y×z. As before for any g∈G,[(x−1)2g−1,(x−1)2] = 0, implies (x−1)2(xg−1)2= 0. This implies xg∈x, i.e., xg=xor x−1. Next observe that for any u,v∈G, we have [u−1,v]={vu−1−(v−1)}u−1.Wehave [(y−1)(z−1)2g−1,(x−1)(z−1)] = 0 and so (y−1)(z−1)2({(xg−1) − (x−1)}(zg−1) + (x−1){(zg−1) −(z−1)})=0. Letzg=xryszt. If xg=x, we get (y−1)(x−1)(xrys−1)ˆz= 0. This is possible only if xrys= 1, because x,y,zare independent. If xg=x−1, we get, after simplification that (y−1)(x−1−1)(xrys−1)ˆz= 0 and hence again xrys=1. Thuszg∈z. That is, zg=zor z−1for all g∈G. Further for any g∈G,[(y−1)2(z−1)g−1,(x−1)(z−1)] = 0 implies (y−1)2(z−1)({(xg−1)−(x−1)}(zg−1)+(x−1){(zg−1)−(z−1)})=0. If zg=zand xg=x−1, then we get (y−1)2(z−1)2(x−1) = 0, which is impossible since x,y,zare independent. Thus zg=zimplies xg=x. If zg=z−1and xg=x, we again get (y−1)2(z−1)2(x−1) = 0, which is not possible as before. Finally if zg=z−1and xg=x−1, we get (y−1)2(z−1)2(x−1){x(z+1)+z}= 0, which is again not possible because x,y,zare independent. Thus for any g∈G, we have zg=z and xg=x. Similarly for any g∈G, we must have yg=y.ThusGis central. Now we prove the converse. If Gis abelian, then clearly δ(3)(KG)=0. If Char K= 7 and G=C7with γ3(G) = 1, then by Remark 2.1 δ(2)(KG)=∆(G)3KG.Thus δ(3)(KG) = [∆(G)3KG,∆(G)3KG]KG ⊆∆(G)6[KG,KG]KG ⊆∆(G)7KG =0. 238 M. Sahai If Char K= 5 and G=C5=x, say, then t(G) = 5. First let γ3(G) = 1. Then as above δ(3)(KG)⊆∆(G)7KG = 0. Now let γ3(G)= Gwith the condition that if g/∈CG(G), then xg=x−1. Clearly ∆(G)KG =(x−1)KG and by Remark 2.1, δ(2)(KG)=∆(G)2KG. Thus δ(3)(KG)=[(x−1)2KG,(x−1)2KG]KG. Let g1,g2∈G, then [(x−1)2g1,(x−1)2g2] =(x−1)4[g1,g 2]+(x−1)3[g1,x]g2+(x−1)2[g1,x](x−1)g2 +(x−1)3[x, g2]g1+(x−1)2[x, g2](x−1)g1 =(x−1)4((g−1 1,g−1 2)−1)g2g1+(x−1)3((g−1 1,x −1)−1)xg1g2 +(x−1)2((g−1 1,x −1)−1)x(xg−1 1−1)g1g2 +(x−1)3((x−1,g−1 2)−1)xg−1 2g2g1 +(x−1)2((x−1,g−1 2)−1)xg−1 2(xg−1 2−1)g2g1 =(x−1)2((g−1 1,x −1)−1)x(x−1)+(xg−1 1−1)g1g2 +(x−1)2((x−1,g−1 2)−1)xg−1 2(x−1)+(xg−1 2−1)g2g1 =(x−1)4((g−1 1,x −1)−1)g1g2+(x−1)4((x−1,g−1 2)−1)g2g1 assuming that xg1=x−1,xg2=x−1, other cases give 0. The expression on the right hand side is 0 as t(G) = 5. Thus δ(3)(KG)=0. If Char K= 3 and G=C3then t(G) = 3 and so δ(3)(KG)⊆ ∆(G)4KG =0. Assume that G=C3×C3. Then t(G) = 5. Now if γ3(G) = 1 then δ(2)(KG)=∆(G)3KG and δ(3)(KG)⊆∆(G)7KG =0. Ifγ3(G)=C3 and γ4(G) = 1, then δ(2)(KG)=∆(G)3KG +∆(γ3(G))∆(G)KG. δ(3)(KG) = 0 because t(G)=5andt(γ3(G))=3. Ifγ3(G)=Gand ug=u−1for every u∈G,g/∈CG(G), assume that G=x×y. Then ∆(G)KG =(x−1)KG +(y−1)KG and ∆(G)2KG =(x− 1)2KG+(y−1)2KG+(x−1)(y−1)KG. Also δ(2)(KG)=∆(G)2KG. For any g/∈CG(G) and u,v∈G, we have [g,(u−1)(v−1)]=(u−1)[g,v]+[g,u](v−1) =(u−1)((g−1,v−1)−1)vg+((g−1,u −1)−1)ug(v−1) =(u−1)(v−1)vg +(u−1)u(vg−1−1)g =(u−1)(v−1)(v+uv +u)g∈∆(G)3KG. Lie Solvable Group Algebras 239 Using this and above we see that δ(3)(KG)⊆∆(G)5KG =0. The case when G=C3×C3×C3and γ3(G) = 1 follows easily as t(G)=7andδ(2)(KG)=∆(G)3KG. Example 2.4. By above Theorem δ(3)(KD10) = 0, if Char K=5 where D10 denotes the Dihedral group of order 10. Thus in this case G=C5need not be central and Gneed not even be nilpotent. Also if G is the semidirect product of C3×C3by C2induced by the automorphism sending every element of C3×C3to its inverse, then δ(3)(KG) = 0 where Char K= 3. Thus in Char K= 3 also Gneed not be nilpotent. Corollary 2.5. Let Kbe a field with Char K=p≥7and let Gbe a group. Then the following are equivalent: (i) δ(3)(KG)=0, (ii) δ[3](L(KG)) = 0. Proof: Clearly if δ(3)(KG) = 0, then δ[3](L(KG)) = 0. Suppose that δ[3](L(KG)) = 0. Let x,y∈G. Then by [6, Lemma 2.4(iii)] 2((x, y, y)−1)3∈γ3(δ[1](L(KG))). Since by [5, Lemma 1.7] [γ3(δ[1](L(KG)))]2KG ⊆δ[3](L(KG))KG, we get that 4((x, y, y)−1)6= 0. Since Gis a p-group and p≥7, (x, y, y)=1and Gis 2-Engel. It is well known that for a 2-Engel group G,(G,G)3=1. Again because Gis a p-group, p≥7, we conclude that γ3(G) = 1. Now [[x, y][x, y, y],[x, y]]=[[x, y],[xy, y],[x, y]] is in γ3(δ[1](L(KG))). Therefore as above 0=[[x, y][x, y, y],[x, y]]2 =[yx((x, y)−1)[yx((x, y)−1),y],yx((x, y)−1)]2 =[yxy[x, y],yx]2((x, y)−1)6 =[yxy2x((x, y)−1),yx]2((x, y)−1)6 =[yxy2x, yx]2((x, y)−1)8 =(yxy2xyx)2((x, y)−1)10. This gives that ((x, y)−1)10 = 0. But Gis a p-group and hence for p≥11, (x, y) = 1 and for p=7,(x, y)7= 1. We conclude that for p≥11, Gis abelian and for p=7,G7= 1. Now let p= 7 and let x,y, u,v∈G, then [[x, y][x, y, y],[x, y]][[u, v][u, v, v],[u, v]] ∈[γ3(δ[1](L(KG)))]2 ⊆δ[3](L(KG)) = 0. 240 M. Sahai Simplifying as above we get ((x, y)−1)5((u, v)−1)5= 0. And so (u, v)∈ (x, y).ThusGis cyclic. Rest follows from Theorem 2.3. Next we give an example to illustrate that there are group algebras which are Lie solvable of length three but not strongly Lie solvable of length three. Example 2.6. Let K=Z2and G=S3, the symmetric group on three letters. Then Gis a cyclic group of order three. It can be easily verified that δ[2](L(KG)) ⊆KGand hence δ[3](L(KG)) = 0. But Z2S3 is not strongly Lie solvable. Acknowledgements. The author is thankful to the referee for valuable suggestions. References 1. F. Levin and G. Rosenberger, Lie metabelian group rings, Preprint no. 60, Ruhr-Universit¨at, Bochum, Dec. 1985. 2. K. Motose and Y. Ninomiya, On the nilpotency index of the radical of a group algebra, Hokkaido Math. J. 4(1975), 261–264. 3. M. Sahai and J. B. Srivastava, A note on Lie centrally metabelian group algebras, to appear in J. Algebra. 4. S. K. Sehgal,“Topics in Group Rings,” Marcel Dekker, 1978. 5. R. K. Sharma and J. B. Srivastava, Lie solvable rings, Proc. Amer. Math. Soc. 94 (1985), 1–8. 6. R. K. Sharma and J. B. Srivastava, Lie centrally metabelian group rings, J. Algebra 151(2) (1992), 476–486. Department of Mathematics Indian Institute of Technology New Delhi - 110 016 INDIA Primera versi´o rebuda el 6 de Juliol de 1994, darrera versi´o rebuda el 29 de Juny de 1995