Group algebras with centrally metabelian unit groups

Publicacions Matem`atiques, Vol 40 (1996), 443–456. GROUP ALGEBRAS WITH CENTRALLY METABELIAN UNIT GROUPS Meena Sahai Abstract Given a field Kof characteristic p>2 and a finite group G, necessary and sufficient conditions for the unit group U(KG)of the group algebra KG to be centrally metabelian are obtained. It is observed that U(KG) is centrally metabelian if and only if KG is Lie centrally metabelian. 1. Introduction Let Gbe a finite group and let Kbe a field of characteristic p>0, p= 2. Necessary and sufficient conditions for the unit group U(KG)to be metabelian were recently obtained by Shalev [5]. In Char K=p≥5, it turns out that U(KG) is metabelian if and only if Gis abelian and in Char K=3,U(KG) is metabelian if and only if either Gis abelian or Gis central cyclic of order 3. The characterization of metabelian group algebras by Rosenberger and Levin [2] shows that for a finite group G and Ka field with Char K=2,U(KG) is metabelian if and only if the group algebra KG is Lie metabelian. Also, in this connection, we have an important result due to Sharma and Srivastava [6, Theorem 4.1], which is, δ2(U(R))−1⊆δ2(L(R))Rfor arbitrary rings R. This shows [6, Corollary 4.2] that the unit group of a Lie metabelian ring is a metabelian group. The aim, in this paper, is to find necessary and sufficient conditions for the unit group U(KG) to be centrally metabelian. Recall that a group G is centrally metabelian if the second derived term δ2(G) is contained in the centre ζ(G), that is, (δ2(G),G) = 1. Recently Sharma and Srivastava [6] and Sahai and Srivastava [4] have obtained necessary and sufficient conditions for the group algebra KG to be Lie centrally metabelian. Our investigations show that U(KG) is centrally metabelian as a group if and only if KG is Lie centrally metabelian, at least when Char K=p=2 and Gis a finite group. This is not true in general as Tasic’ [7] has given 444 M. Sahai example of a Lie centrally metabelian algebra of characteristic 2 whose unit group is not centrally metabelian. Our notations are standard. We use (x, y)=x−1y−1xy for group commutators and [x, y]=xy −yx for Lie commutators. We now start with our work. 2. Sufficient conditions Theorem 2.1. Let Kbe a field, Char K=p=2and let Gbe a group, finite or infinite. If KG is Lie centrally metabelian, then U(KG) is centrally metabelian. Proof: Suppose that KG is Lie centrally metabelian. By [4, Theorem B] either Gis abelian or Char K= 3 and G=C3.IfGis abelian, then clearly U(KG) is abelian. Assume that Char K= 3 and G=t, t3= 1. Since Gis normal in G, we see that (t−1)2=t2+t+1 is central in KG. Also if Gis central, then by [2], KG is Lie metabelian and by [6, Corollary 4.2], U(KG) is metabelian. This is also given in Shalev [5, Theorem B] for finite groups. So we are left with the case when G=t,t3= 1, Char K= 3 and tis not central in G.Now∆(G)KG =(t−1)KG. In this case, γ3(G)=G, δ(1)(KG)=∆(G)KG and δ(2)(KG)=∆(G)2KG =(t−1)2KG.We know by [6, Theorem 4.1], δ2(U(KG))−1⊆δ2(L(KG))KG ⊆δ(2)(KG). So δ2(U(KG)) ⊆1+(t−1)2KG. Let u∈δ2(U(KG)) and g∈G. Then u−1∈(t−1)2KG and we have (u, g)−1=u−1g−1[u−1,g]∈ KG[(t−1)2KG,KG]. Thus (u, g)−1∈(t−1)2∆(G)KG = 0, since (t−1)2is central in KG and ∆(G)3= 0. This shows that (u, g)=1 for every u∈δ2(U(KG)) and for every g∈Gand hence δ2(U(KG)) is contained in the centre of KG, as desired. 3. Necessary conditions We have seen in the previous section that for arbitrary groups G,KG Lie centrally metabelian implies either Gis abelian or Char K= 3 and G=C3and this, in turn, implies that the unit group U(KG) is centrally metabelian. For finite groups, now we assume that U(KG) is centrally metabelian and establish the converse. We first make the following observation: Lemma 3.1. GL2(Z3)is not centrally metabelian. Group algebras 445 Proof: Let A=−11 0−1,B=1−1 −1−1in SL2(Z3)=GL2(Z3). Then A−1B−1 AB =11 1−1belongs to GL2(Z3), however, 11 1−1 is not in the centre of GL2(Z3). Lemma 3.2. Let Gbe a finite group and let Char K=p=2such that the unit group U(KG)is centrally metabelian. Then G/Op(G)is abelian. Proof: We have the exact sequence of groups 1→1+J(KG)→U(KG)→U(KG/J(KG)) →1. Now KG/J(KG)∼ =m i=1 Mni(Di) and so U(KG/J(KG)) ∼ =m i=1 GLni(Di). But U(KG) is centrally metabelian implies U(KG/J(KG)) is centrally metabelian. Thus GLni(Di) is centrally metabelian for all iand therefore all Di’s are fields and in view of Lemma 3.1, ni= 1 for all i. This is because GLn(D) is solvable, n=1, Char D= 2, implies n=2,D=Z3and thus GLn(D)=GL2(Z3) but by Lemma 3.1, GL2(Z3) is not centrally metabelian. Thus U(KG/J(KG)) is a direct product of multiplicative groups of fields and hence abelian. But then U(KG)/{1+J(KG)}is abelian and U(KG)⊆1+J(KG). We get G⊆G∩{1+J(KG)}=Op(G) and therefore, G/Op(G)is abelian, as desired. Corollary 3.3. Let Char K=p=2and let Gbe a finite group such that Op(G)=1and U(KG)is centrally metabelian. Then Gmust be abelian. Corollary 3.4. Let Char K=p=2and let Gbe a finite group such that U(KG)is solvable. Then G=PH, a split extension of a p-group Pby a p-group H. Proof: Since U(KG) is solvable, either G/Op(G) is abelian or p=3 and G/O3(G) is a 2-group, see [3]. In either case, Sylow p-subgroup of Gis normal in G. Let it be P.Now|P|and |G:P|are relatively prime, hence by Schur-Zassenhaus Theorem G=PH, with desired properties. Lemma 3.5. Let Gbe a finite p-group, p≥5and let Kbe a field with Char K=psuch that U(KG)is centrally metabelian. Then Gis abelian. 446 M. Sahai Proof: If not, let Gbe a counter example of least order. Then G= x, y,z=(x, y)=1,G=z,zp= 1 and zcentral. Let u1=(1+x, y) and u2=(1+y,x), then using centrality of z,we get (u1,u 2)−1= u−1 1u−1 2[u1−1,u 2−1] =u−1 1u−1 2[(1 + x)−1y−1[1 + x, y],(1 + y)−1x−1[1 + y,x]] =u−1 1u−1 2[(1+x)−1 y−1yx,(1+y)−1x−1xy]((x, y)−1)((y,x)−1) =−u−1 1u−1 2[(1+x)−1(1+x−1),(1+y)−1(1+y−1)](z−1)2z−1 =−u−1 1u−1 2[(1 + x)−1,(1 + y)−1](z−1)2z−1 =−u−1 1u−1 2(1 + x)−1(1 + y)−1[1 + x, 1+y] (1 + y)−1(1 + x)−1(z−1)2z−1 =−u−1 1u−1 2(1+x)−1(1+y)−1yx(1+y)−1(1+x)−1(z−1)3z−1 =−u−1 1u−1 2γ(z−1)3z−1, where γ=(1+x)−1(1 + y)−1yx(1 + y)−1(1 + x)−1. Since (u1,u 2) is central in KG,so 0=[(u1,u 2)−1,x] =−[u−1 1u−1 2γ,x](z−1)3z−1 =−{u−1 1[u−1 2,x]+[u−1 1,x]u−1 2}γ(z−1)3z−1−u−1 1u−1 2[γ,x](z−1)3z−1 . It is not difficult to see that both [u−1 2,x] and [u−1 1,x] belong to KG(z−1)2. Now multiplying by (z−1)p−5and using (z−1)p=0, given p≥5, we get [γ,x](z−1)p−2= 0. With routine calculations, [γ,x] = (1 + x)−1[(1 + y)−1yx(1 + y)−1,x](1 + x)−1 =(1+x)−1(1 + y)−1[y,x]x(1 + y)−1+yx[(1 + y)−1,x](1 + x)−1 +(1+x)−1[(1 + y)−1,x]yx(1 + y)−1(1 + x)−1. Using [(1 + y)−1,x]=−(1 + y)−1[1 + y,x](1 + y)−1=(1+y)−1yx(1 + y)−1(z−1), we get [γ,x] = (1 + x)−1(1 + y)−1{−yx2(1 + y)−1+yx(1 + y)−1yx(1 + y)−1} (1 + x)−1(z−1) +(1+x)−1(1 + y)−1yx(1 + y)−1yx(1 + y)−1(1 + x)−1(z−1) =(1+x)−1(1 + y)−1yx{−x+ 2(1 + y)−1yx} (1 + y)−1(1 + x)−1(z−1) =(1+x)−1(1 + y)−1yx(1 + y)−1{−(1 + y)x+2yx} (1 + y)−1(1 + x)−1(z−1) =−(1+x)−1(1+y)−1yx(1+y)−1(y−1)x(1 + y)−1(1 + x)−1(z−1). Group algebras 447 Now [γ,x](z−1)p−2= 0 implies (y−1)(z−1)p−1=0. Soy∈zand yis central. But then z=(x, y) = 1, a contradiction. We now apply this lemma to settle the case when Char K=p≥5 and Gis an arbitrary finite group. Theorem 3.6. Let Char K=p≥5and let Gbe any finite group such that U(KG)is centrally metabelian. Then Gis abelian. Proof: By Corollary 3.4, G=PH, a split extension of a p-group P by a p-group H. By Corollary 3.3, His abelian and by Lemma 3.5, P is abelian. Suppose, if possible, Gis non-abelian. Then (P,h)= 1 for some 1 =h∈H. Since hinduces a p-automorphism on P,by[1, Theorem 5.3.6], (P, h, h)=(P,h). Let L=(P, h),h. Then L=(P, h, h)= (P,h)= 1. The Jacobson radical J=J(KL) = ∆((P,h))KL. Since 1+J⊆U(KL), (1 + J, h)⊆U(KL)and ((1 + J, h),(P,h)) ⊆U(KG) which is central in U(KG). Let x, y, z ∈(P,h). Put a=1−x, then a∈J. Let u1=(1−ha, h). Then u1=(1−ha)−1(1 −ha)h ={1+ha +(ha)2+(ha)3+···}(1 −hah) =1+ha −hah+(ha)2−(ha)hah +(ha)3−(ha)2hah+(ha)4−(ha)3hah+··· ≡1+h(a−ah)+h2ah(a−ah)+h3ah2ah(a−ah) (mod J4). Now, since Pis abelian, working modulo J4,wehave u2=(u1,y) =1+u−1 1(uy 1−u1) ≡1+u−1 1{(hy−h)(a−ah)+(h2y−h2)ah(a−ah) +(h3y−h3)ah2ah(a−ah)} ≡1+u−1 1h{(h, y)−1+h((h2,y)−1)ah+h2((h3,y)−1)ah2 ah}(a−ah) ≡1+u−1 1h{(h, y)−1+h((h2,y)−1)ah}(a−ah),(mod J4), since (h3,y)−1∈J.Nowu2is central. So we have working modulo J4 0=[u2−1,z] ≡[u−1 1h{(h, y)−1+h((h2,y)−1)ah}(a−ah),z] ≡[u−1 1,z]h{(h, y)−1+h((h2,y)−1)ah}(a−ah) +u−1 1[h((h, y)−1) + h2((h2,y)−1)ah,z](a−ah) ≡u−1 1{[h, z]((h, y)−1)+[h2,z]((h2,y)−1)ah}(xh−x), 448 M. Sahai since [u−1 1,z]=−u−1 1[u1,z]u−1 1∈J2. Thus we get ((x, h)−1)((y,h)− 1)((z,h)−1) ∈J4for all x, y, z ∈(P, h)=(P, h, h). So J3⊆J4and J3= 0, since Jis nilpotent. Thus (∆((P,h)))3= 0 and ((P,h)−1)3= 0. Now Char K=p≥5 and (P,h)isap-group implies (P,h)=1, a contradiction to our assumption that (P,h)=1. ThusGmust be abelian. Remark 3.7. The entire proof of Theorem 3.6 goes through upto (∆((P,h)))3= 0 in Char K=3 also if we assume that Pis abelian. We, now, turn to Char K=3. Lemma 3.8. Let Char K=p=3and let Gbe a finite group of odd order such that U(KG)is centrally metabelian. Then G=PH,Pa p-group, Han abelian p-group. Further G=P. Proof: By Corollary 3.3 and 3.4, G=PH,Pap-group, Han abelian p-group. Assume, further, that Pis abelian. Then G=(P,H). If G= 1, choose x∈P,h∈Hsuch that (x, h, h)= 1 which is possible because (P, h, h)=(P,h)= 1 for some h∈H. By Remark 3.7, (∆((P,h)))3=0. Now (P,h)isap-group, p=3,so(P, h) is cyclic of order 3. Then (P,h)= (x, h). It is easy to see that (x, h)h=(x, h)(x, h, h)∈(P,h), hence (x, h)h=(x, h)or(x, h)−1.If(x, h)h=(x, h)−1, then (x, h)h2=(x, h) implying (x, h)h=(x, h), because order of his odd. Thus (x, h, h)=1, a contradiction. Hence G=(P,H) = 1 and Gis abelian. So G=P. Now let Pbe non-abelian. By applying the above case to the group G/P, we get (P,H)≤Pand so G=Pin this case also. Next result is for finite 3-groups. Proposition 3.9. Suppose that Char K=3and Pis a finite 3-group such that U(KP)is centrally metabelian. Then either Pis abelian or P=C3. Proof: If not, let Gbe a minimal counter example. Then |G|= 9 and we have the following three cases. Case (i): Gis central cyclic of order 9. Let G=z,,z=(x, y), x, y ∈G,z9= 1. Exactly as in the proof of Lemma 3.5, G=x, y,z=(x, y)= 1, and we conclude that (y−1)(z− 1)8=0. Thusy∈z⊆ζ(G) and so (x, y) = 1, a contradiction. Hence this case will not arise. Case (ii): Gis central and G=C3×C3. Clearly ∆(G)5= 0. Since Gis not cyclic, there exist elements x, y1,y 2∈Gsuch that z1=(x, y1)= 1 and z2=(x, y2)/∈z1, see Group algebras 449 [4, proof of Theorem B]. Let u1=(1+x, y1), u2=(1+y2,x). Then exactly as in the proof of Lemma 3.5, we get (u1,u 2)−1=−u−1 1u−1 2γ(z1−1)(z2−1)2z−1 2, where γ=(1+x)−1(1 + y2)−1y2x(1 + y2)−1(1 + x)−1. Hence 0=[(u1,u 2)−1,y 1] =−[u−1 1u−1 2γ,y1](z1−1)(z2−1)2z−1 2 =−{[u−1 1u−1 2,y 1]γ+u−1 1u−1 2[γ,y1]}(z1−1)(z2−1)2z−1 2. We get [γ,y1](z1−1)(z2−1)2= 0, first term above being 0 because [u−1 1u−1 2,y 1]∈∆(G)2KG and ∆(G)5=0. Now−[γ−1,y 1]= γ−1[γ,y1]γ−1and z1,z2are central. So [γ−1,y 1](z1−1)(z2−1)2=0. Now γ−1=(1+x)(1 + y2)(y2x)−1(1 + y2)(1 + x) =(1+x)(1 + y2)z2y−1 2x−1(1 + y2)(1 + x) =(1+x)(1 + y−1 2)(z2+yx 2z2)(1 + x−1) =(1+x)(1 + y−1 2)(z2+y2)(1 + x−1). Since G=C3×C3, let (y1,y 2)=zi 1zj 2for some 0 ≤i,j≤2. Now using (zi 1−1)(z1−1) = i(z1−1)2,z2(z2−1)2=(z2−1)2, and expanding [γ−1,y 1] in the usual way, we get {y1x(1 + y−1 2)(1 + y2)(1 + x−1)+i(1 + x)y1y−1 2(1 + y2)(1 + x−1) −i(1 + x)(1 + y−1 2)y2y1(1 + x−1) −(1 + x)(1 + y−1 2)(1 + y2)x−1y1}(z1−1)2(z2−1)2=0. Since [α, β]∈∆(G)KG for all α, β ∈KG and ∆(G)5= 0, on combining first term with last term and second term with third term, we get, using [α, β](z1−1)2(z2−1)2= 0, that 0={(y1x−x−1y1)(1 + y−1 2)(1 + y2) +i(1 + x)(y1y−1 2−y2y1)(1 + x−1)}(z1−1)2(z2−1)2 ={y1x−1(x2−1)(1 + y2)2y−1 2 +i(1 + x)2(1 −y2 2)y−1 2y1x−1}(z1−1)2(z2−1)2 =y1x−1(1 + x){(x−1)(y2+1) +i(1 + x)(1 −y2)}(1 + y2)y−1 2(z1−1)2(z2−1)2. 450 M. Sahai We have {(x−1)(y2+1)+i(x+ 1)(1 −y2)}(z1−1)2(z2−1)2=0. Itis not difficult to see that this is not possible for any i=0,1,2. Case (iii): Gis not central in G. Gis nilpotent, |G|=9,γ3(G)= 1 implies γ3(G)=C3and γ4(G)=1. Choose w∈G,x∈Gsuch that z=(x, w)= 1. Then z∈ζ(G), z3= 1 and (1+x, w, w) is central in KG. Also (x, G)γ3(G). For otherwise (x, G) will be in ζ(G) and then (x, g−1,h)(g,h−1,x)(h, x−1,g)=1, implies (g,h−1,x) = 1 for all g,h ∈G.So(G,x) = 1 and z=(x, w)=1. Choose y∈Gsuch that (x, y)/∈γ3(G). Let u=(1+x, w), then (1 + x, w, w)=1+u−1w−1[u−1,w] =1+u−1w−1[(1 + x)−1w−1[1 + x, w],w] =1+u−1w−1[(1 + x)−1w−1wx(z−1),w] =1+u−1w−1(1 + x)−1[x, w](1 + x)−1(z−1) =1+u−1w−1(1 + x)−1wx(1 + x)−1(z−1)2 =1+u−1(1 + xz)−1(1 + x−1)−1(z−1)2. Now [(u, w),y] = 0 implies [(1+xz)−1(1+x−1)−1,y](z−1)2= 0, because [u−1,y]∈(z−1)KG and (z−1)3= 0. Solving this further, we have 0 = [(1 + x−1)(1 + xz),y](z−1)2 =[x−1+xz, y](z−1)2 =[x−1+x, y](z−1)2. Hence 0={−x−1[x, y]x−1+[x, y]}(z−1)2 ={−x−1yx((x, y)−1)x−1+yx((x, y)−1)}(z−1)2 ={−x−1y((x, y)x−1−1) + yx((x, y)−1)}(z−1)2 ={−x−1y((x, y)(x, y, x−1)−1) + yx((x, y)−1)}(z−1)2 ={−x−1y+yx}((x, y)−1)(z−1)2, since (x, y, x−1)∈γ3(G)=z. This gives yx{x−1y−1x−1y−1}((x, y)− 1)(z−1)2= 0 and so {(y,x)xx−2−1}((x, y)−1)(z−1)2= 0. Since (x, y)/∈γ3(G), it follows that (y,x)xx−2is in Gand so x−2∈G. But then x∈Gas order of xis odd. This is a contradiction as (x, y)/∈γ3(G). Thus we have a contradiction in all the three cases, so either Pis abelian or P=C3. Group algebras 451 Corollary 3.10. Let Char K=3and let Gbe a finite group of odd order such that U(KG)is centrally metabelian. Then either Gis abelian or Gis cyclic of order 3. Proof: It can be deduced easily from Lemma 3.8 and Proposition 3.9. Now we shall study the case when Gis a group of even order. Lemma 3.11. Let G=Ph,Pa finite 3-group, o(h)is even and coprime to 3and let Char K=3, such that U(KG)is centrally metabelian. Then either G=1or G=C3. Proof: If (P,h)⊆P, then G=Pand by Proposition 3.9, G=1or C3. So we are through. Assume that (P,h)P. Then z=(x, h)/∈P for some x∈P. First suppose that Pis abelian. Consider the group L=(P,h),h. By Remark 3.7, (∆((P, h)))3= 0 and hence (P, h) is cyclic of order 3. So G=(P,h)=C3, since P=1. Now let Pbe non-abelian. Then P=C3=t, say. Applying the above case to G/P, we have G/P =(P, h)P/P∼ =C3. Then G/P=zP, since z/∈P.Thusz3∈P. This gives that |G|=9 and hence Gis abelian. Again take L=(P,h),h, then L=(P, h, h)=(P, h)=C3, since U(KL) is centrally metabelian, (P,h) is abelian and we can apply Remark 3.7. So (P,h)=z,z3= 1. Also (z,h)∈(P,h), (z,h)=1 and so (z,h)=zand zh=z−1. Clearly G=(P, h)P=z×t, z3=t3= 1 and ∆(G)5= 0. Since Pis nilpotent, t∈ζ(P). Further, (t, h)∈(P,h)∩P= 1 and tis central in G. Case (i): (z,P)=1. There exists y∈Pwith 1 =(z,y)∈P=t. So we may take (z,y)=t. Let a=1−z∈∆(P). Then 1−ha is a unit. Let u1=(1+z,y) and u2=(1−ha, h). Then u2=(1−ha)−1(1 −ha)h ={1+ha +(ha)2}(1 −hah) =1+h(z−1)z+h2(z−1)2,