On the Jacobson radical and unit groups of group algebras

Publicacions Matem`atiques, Vol 42 (1998), 339–346. ON THE JACOBSON RADICAL AND UNIT GROUPS OF GROUP ALGEBRAS Meena Sahai Abstract In this paper, we study the situation as to when the unit group U(KG) of a group algebra KG equals K∗G(1 + J(KG)), where Kis a field of characteristic p>0 and Gis a finite group. 1. Introduction Let Rbe any associative ring with identity 1 6= 0. Then Rmay be treated as a Lie ring under the Lie multiplication [x, y]=xy −yx, x,y∈R. The Lie ring thus obtained is denoted by L(R) and is called the associated Lie ring of R. The lower central chain {γn(L(R)) |n= 1,2,...}and the derived chain {δn(L(R)) |n=0,1,2,...}of L(R) are defined inductively as follows: γ1(L(R)) = δ0(L(R)) = L(R), γn+1(L(R))=[γ n (L(R)),L(R)], δn(L(R))=[δ n−1 (L(R)),δn−1(L(R))]. The Lie ring L(R) is solvable of length nif δn(L(R)) = (0) but δn−1(L(R)) 6= (0). Let J(R) denote the Jacobson radical of R. Then 1+J(R) is a normal subgroup of the unit group U(R) and we have the exact sequence of groups 1→1+J(R)→U(R)→U(R/J(R)) →1. Keywords. Solvable group, p-solvable group, locally finite group. 1991 Mathematics subject classifications: 16N20, 16S34, 16U60. 340 M. Sahai Thus U(R)/(1 + J(R)) ∼ =U(R/J(R)). If further 2 and 3 are invertible in Rand the associated Lie ring L(R) is solvable, then γ2(L(R))R= δ1(L(R))Ris a nil ideal of Rby Sharma and Srivastava [7, Theorem 2.4]. Since nil ideals are always contained in the Jacobson radical, we have, in this situation, γ2(L(R))R⊆J(R) and thus R/J(R) is commutative. Thus the commutator subgroup U(R)0of U(R) is contained in 1 +J(R). If J(R) is nilpotent as an ideal, then 1+J(R) is nilpotent as a group and so U(R) is solvable. In particular, in the above situation, if (J(R))2=0, then U(R) is metabelian. We wish to study, in this paper, some connections in the above direction when R=KG is the group algebra of the group Gover the field K, where Char K=p>0 and Gis finite. Throughout the paper, Zpdenotes the field with pelements. 2. Preliminaries Let KG be the group algebra of the group Gover the field K.We denote by ∆(G), the augmentation ideal of KG. Clearly 1 + J(KG) defines a normal subgroup of the unit group U(KG). Also there are the trivial units of the form kg,06=k∈K,g∈G,inU(KG). Our aim, in this paper, is to investigate situations where U(KG)=K ∗ G(1+J(KG)), K∗=K\{0}. Obviously U(KG) can not be smaller than this as the right hand side is always contained in U(KG). Almost in all the known cases the Jacobson radical J(KG) of a group algebra KG is a nil ideal; (see Passman [5, Chap. 8]), and at least, for sure, this is the case for the class of solvable, linear and locally finite groups. Suppose Char K=p,p>0 and J(KG) is nil. Then for any α∈J(KG), αpn= 0 for some n≥0 and thus (1 + α)pn=1+α p n=1. This shows that 1 + J(KG) is a normal p-subgroup of U(KG)ifJ(KG) is a nil ideal. We make the following observations. Lemma 2.1. Let Kbe a field with Char K=p>0and let Gbe a group. Then G∩{1+J(KG)}is a normal p-subgroup of G. Further if Gis locally finite, then Op(G)=G∩{1+J(KG)}. Proof: Clearly G∩{1+J(KG)}is a normal subgroup of G. Let 1 6= x∈G∩{1+J(KG)}. Then x−1∈J(KG) and ∆(hxi)=(x−1)Khxi⊆ J(Khxi). Thus J(Khxi)6= 0 and so hxiis finite. Also J(Khxi)⊇∆(hxi) is nilpotent, since Khxiis Artinian. Hence hxiis a finite p-group and G∩{1+J(KG)}is a normal p-subgroup. Jacobson radical and unit groups 341 If Gis locally finite, then Op(G) is a locally finite normal p-subgroup and so ∆(Op(G)) = J(KOp(G)) ⊆J(KG). Thus Op(G)⊆G∩{1+ J(KG)}and by the first part, we get G∩{1+J(KG)}=Op(G), as desired. This result easily yields Corollary 2.2. If Gis locally finite and Char K=p>0, then ∆(N)KG ⊆J(KG)for every normal p-subgroup Nof Gand equality holds if Nis a normal Sylow p-subgroup of G. It may be noted that ∆(G)=J(KG) for any locally finite p-group G if Char K=p>0 (Passman [5, Chap. 8]). 3. Main results Now we start our study of the problem: When is U(KG)=K ∗ G(1 + J(KG))? Proposition 3.1. Let Kbe a field with Char K=p>0and let Gbe a locally finite group having a normal Sylow p-subgroup P. Then U(KG)=K ∗ G(1+J(KG)) if and only if one of the following holds: (i) G=P; (ii) K=Z2and G/P ∼ =C3; (iii) K=Z3and G/P ∼ =C2. Proof: First suppose that U(KG)=K ∗ G(1 + J(KG)). By Corollary 2.2, J(KG)=∆(P)KG and KG/J(KG)∼ =KG/P. Further U(KG/J(KG)) ∼ =U(KG)/(1+J(KG))=K∗G(1+J(KG))/(1+J(KG)). So U(KG/J(KG)) ∼ =K∗G/(G∩{1+J(KG)}). Also U(KG/J(KG)) ∼ = U(KG/P). Since by Lemma 2.1, G∩{1+J(KG)}=Op(G)=P, we see that U(KG/P)=K ∗·G/P using the natural epimorphism U(KG)→U(KG/P). Thus the group algebra KG/P has only trivial units. So by Passman [5, Lemma 13.1.1], either G/P is trivial, that is, G=Por K=Z2and G/P ∼ =C3since G/P is a p0-group or K=Z3 and G/P ∼ =C2. Conversely if G=P, then J(KG)=∆(G) and we are through as U(KG)=K ∗ (1 + J(KG)). In the other two cases, the units of KG/P are trivial, J(KG)=∆(P)KG and G∩{1+J(KG)}=P. Hence clearly U(KG)=K ∗ G(1+J(KG)). 342 M. Sahai In fact, 16=G/P =G/(G∩{1+J(KG)})∼ =G(1+J(KG))/(1+J(KG)) and this is a subgroup of U(KG)/(1+J(KG)) ∼ =U(KG/J(KG)) = U(KG/∆(P)KG)∼ =U(KG/P). But U(Z2C3)=C 3and U(Z3C2)= ±C 2 , hence the result. Now we turn to finite groups. If Char K=p>0 and Ghas no pelements, then J(KG) = 0, so our problem U(KG)=K ∗ G(1+J(KG)) reduces to U(KG)=K ∗ G. This is the case of trivial units. So we assume that Gis finite, it has p-elements and hence J(KG)6= 0. Also if Gis a finite p-group or Ghas a normal Sylow p-subgroup, then Proposition 3.1 above gives the answer. Theorem 3.2. If Char K=p>0and Gis a finite solvable group having no normal Sylow p-subgroup, then U(KG)=K ∗ G(1+J(KG)) if and only if K=Z2and G/O2(G)∼ =S3. Proof: Suppose U(KG)=K ∗ G(1+J(KG)). Then U(KG) is solvable. Further G/Op(G) is not abelian, otherwise Sylow p-subgroup will be normal. By Passman’s Theorem (see Karpilovsky [4, Theorem 3.8.9] or Bateman [2, Theorem 5]), K=Z2or Z3. But K=Z3case gives that G/O3(G) is a 2-group, so Sylow 3-subgroup is normal. Hence we are left with only one case when K=Z2and G/O2(G)=Ahxi, where Ais an elementary abelian 3-group and xis an element of order 2 such that x−1ax =a−1for all a∈A. We wish to show that A=C3.Now U(KG/J(KG)) ∼ =U(KG) 1+J(KG)=K∗G(1+J(KG)) 1+J(KG) ∼ =K∗G K∗G∩(1+J(KG)) =G G∩(1+J(KG)) =G O2(G)=Ahxi. Here K=Z2,soifKG/J(KG)∼ =Qr i=1 Mni(Di), by Bateman [2, Theorem 5], U(KG/J(KG)) ∼ =Qs i=0 K∗ i×Qt j=1 GL2(Z2), where Ki are finite fields of characteristic 2 and second term is a direct product of t-copies of GL2(Z2)∼ =S3. Also |U(KG/J(KG))|=|G/O2(G)|= |A||hxi| =3 m·2 where A=C3×C3× ··· × C 3(m-copies). Thus clearly t= 1. Also |Ki|=2 n ifor some ni,so|K ∗ i |=2 n i−1 for i=0,1,2,... ,s.Thusn i= 2 for every i. We show that s= 0 and U(KG/J(KG)) ∼ =G/O2(G)∼ =GL2(Z2)∼ =S3. Jacobson radical and unit groups 343 Suppose |A|=3 mand m>1. Then there exist a,b∈Asuch that hai×hbi⊆A,a 3=b 3=1,x −1 ax =a−1,x−1bx =b−1.We have Ahxi=G/O2(G)∼ =Qs i=0 K∗ i×GL2(Z2) and denote by φthe isomorphism. Then φ(a)=( Qs i=0 ki,g 1),φ(b)=( Qs i=0 k0 i,g 2), a,bnoncentral implies g16=1,g 26= 1. Also a3=b3= 1 gives g3 1=g3 2=1. In GL2(Z2)∼ =S3, either g1=g2or g2=g−1 1=g2 1.Ifg 1=g 2 , then φ(a2b) is central and so a2bis central. But x−1a2bx =(a 2 b) −1 ,so(a 2 b) −1=a 2 b and we get a=b.Ifg 2=g −1 1, then φ(ab) is central, so ab is central and x−1abx =(ab)−1=ab.Thusa=b −1 . In both cases we get a contradiction, since hai∩hbi=1. ThusA=hai=C 3and G/O2(G)= GL2(Z2)∼ =S3, as desired. Conversely, let K=Z2and G/O2(G)∼ =S3.By[6, 6.2, p. 215] ¯¯¯¯ U(Z2G) 1+∆(O 2 (G))Z2G¯¯¯¯ =|U(Z2G/O2(G))|=|U(Z2S3)|=12. Also U(Z2G) 1+J(Z 2 G)∼ =U(Z 2 G)/{1+∆(O 2 (G))Z2G} {1+J(Z 2 G)}/{1+∆(O 2 (G))Z2G} and so ¯¯¯¯ U(Z2G) 1+J(Z 2 G)¯ ¯ ¯ ¯ =12 |{1+J(Z 2 G)}/{1+∆(O 2 (G))Z2G}|. Since the Sylow 2-subgroups are not normal, G/O2(G) contains 2-elements and J(Z2G)⊃∆(O2(G))Z2G. Further U(Z2G) 1+J(Z 2 G)∼ =UµZ 2 G J(Z 2 G)¶ =GL2(Z2)× s Y i=0 K∗ i,K i =2 n i,K ∗ =K\{0} since U(Z2G) is solvable and U(Z2G/J(Z2G)) is non-abelian, otherwise G0⊆G∩{1+J(Z 2 G)}=O 2 (G) implies that a Sylow 2-subgroup is normal. All this forces |(1 + J(Z2G))/{1+∆(O 2 (G))Z2G}| =2 and U(Z2G) 1+J(Z2G)∼ =GL2(Z2)∼ =S3∼ =G/O2(G)= G G∩(1+J(Z2G)) .Thus U(Z 2 G)=G(1+J(Z2G)), as desired. In general if Gis a finite group and Kis a field with Char K=p such that U(KG)=K ∗ G(1 + J(KG)), then U(KG)n⊆ζ(U(KG)), the center of U(KG), for some fixed n. This can be seen as follows. Since J(KG) is nilpotent, we have J(KG)pl= 0 for some fixed l. Now let u∈U(KG), then u=kg(1+α) for some k∈K∗,g∈G,α∈J(KG). 344 M. Sahai It is easy to see that for all m,wehave u m=k m g m (1+αgm−1)(1 + αgm−2)...(1+αg)(1 + α). Thus if n0=|G|, then un0=kn0(1 + β), for some β∈J(KG). Furthermore un0pl=kn0pland thus if n=n0pl, then unis central. Thus U(KG)n⊆ζ(U(KG)) and we can use Coelho [3, Lemma 1.1]. Let A={g∈G|gis a p0-element}.IfAconsists of central elements alone, then Ais a normal subgroup of Gand G=AP for any Sylow p-subgroup Pof G. Clearly then PCGand Proposition 3.1 handles the situation U(KG)=K ∗ G(1+J(KG)). We wish to tackle, now, the case when Ghas a non-central p0-element. By Coelho [3, Lemma 1.1] and the above discussion we must have that Kis a finite field. Lemma 3.3. Let Gbe a finite group and let Char K=p>0such that U(KG)=K ∗ G(1 + J(KG)). Then U(K¯ G)=K ∗¯ G(1 + J(K¯ G)), where ¯ G=G/Op(G). Proof: Since ∆(Op(G))KG ⊆J(KG), U(KG/J(KG)) ∼ =U(K¯ G/J(K¯ G)). Therefore, U(KG) 1+J(KG)=K∗G(1+J(KG)) 1+J(KG)∼ =K∗G G∩(1+J(KG)) =K∗G Op(G)∼ =U(K¯ G) 1+J(K¯ G). This clearly shows that U(K¯ G)=K ∗¯ G(1+J(K¯ G)). When p0-elements are not central, Aneed not form a subgroup. Even when Aforms a subgroup, Sylow p-subgroup need not be normal. However, we have the following. Theorem 3.4. Let Gbe a finite group such that Aforms a non-central subgroup and Char K=P>0.IfU(KG)=K ∗ G(1+J(KG)) then G is solvable and Kis finite. Jacobson radical and unit groups 345 Proof: Since U(KG)=K ∗ G(1+J(KG)) and Gis finite, Kis a finite field. Hence in the decomposition KG/J(KG)∼ =Qr i=1 Mni(Di), each Di=Kiis a field, being finite division rings. Thus U(KG/J(KG)) ∼ = Qr i=1 GLni(Ki), Kifinite, Char Ki=p.If¯ G=G/Op(G) is solvable, then clearly Gis solvable. In view of Lemma 3.3, we can assume that Op(G)=1. Now UµKG J(KG)¶∼ =U(KG) 1+J(KG)∼ =K∗G G∩{1+J(KG)}=K∗G. Let Aidenote the set of p0-elements of GLni(Ki) for all i=1,2,... ,r. Clearly, Aiis a subgroup of GLni(Ki) for all i=1,2,... ,r. Also Aiis non-central in GLni(Ki), if ni>1. Therefore, ni= 1 or 2 and Ki∼ =K if ni= 2, where K=Z2or Z3(see Artin [1, p. 165]). Since both GL2(Z2) and GL2(Z3) are solvable, U(KG/J(KG)) is solvable and so G≤U(KG) is solvable, as desired. We now discuss finite p-solvable groups: Let Kbe a field with Char K=p>0 and Ga finite group such that U(KG)isp-solvable. Then U(ZpG)isp-solvable and hence U(ZpG/J(ZpG)) is p-solvable. But U(ZpG/J(ZpG)) = Qr i=1 GLni(Di), so each Diis a field, being a finite division ring. Thus for each i, GLni(Di)=GLni(GF(qi)), qi=pniand p-solvabiblity forces each ni=1orn i=2,q i=p,p= 2 or 3. But GL2(Z2) and GL2(Z3) are solvable. Thus U(ZpG/J(ZpG)) is solvable and therefore, U(ZpG) is solvable. This gives that Gis solvable. Thus U(KG)isp-solvable implies Gis solvable. In particular, we have Theorem 3.5. If Char K=p>0and Gis a p-solvable group such that U(KG)=K ∗ G(1+J(KG)), then Gis solvable. Proof: Clearly U(KG)isp-solvable. Rest follows from the above discussion. 4. Conclusion We have covered most of the cases for finite groups except for finite groups which are not p-solvable, in which the p0-elements are non-central and do not form a subgroup. This problem is still open. Some preliminary results have been obtained in this direction by the author and will be taken up separately in a subsequent paper. 346 M. Sahai References 1. E. Artin,“Geometric Algebra,” Interscience, New York, 1957. 2. J. M. Bateman, On the solvability of unit groups of group algebras, Trans. Amer. Math. Soc. 157 (1971), 73–86. 3. S. P. Coelho, Group rings with units of bounded exponent over the center, Canad. J. Math. 34 (1982), 1349–1364. 4. G. Karpilovsky,“Unit Groups of Group Rings,” Wiley Interscience, New York, 1989. 5. D. S. Passman,“The Algebraic Structure of Group Rings,” Wiley Interscience, New York, 1977. 6. S. K. Sehgal,“Topics in Groups Rings,” Marcel Dekker, New York, 1978. 7. R. K. Sharma and J. B. Srivastava, Lie solvable rings, Proc. Amer. Math. Soc. 94 (1985), 1–8. Department of Mathematics and Astronomy Lucknow University Lucknow - 226 007 INDIA e-mail: [email protected] Primera versi´o rebuda el 8 d’abril de 1997, darrera versi´o rebuda el 16 de febrer de 1998