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Lie Centrally Metabelian Group Rings in Characteristic 3
JOURNAL OF ALGEBRA ARTICLE NO.
180, 111]120 Ž1996.
0055
Lie Centrally Metabelian Group Rings in Characteristic 3 B. Kulshammer ¨ Department of Mathematics, Uni¨ ersity of Augsburg, D-86135 Augsburg, Germany
and R. K. Sharma Department of Mathematics, Indian Institute of Technology, Kharagpur (W. B.)-721302, India Communicated by Walter Feit Received January 12, 1995
We classify Lie centrally metabelian group algebras over fields of characteristic 3. Q 1996 Academic Press, Inc.
Let F be a field of characteristic p G 0, and let L be a Lie algebra over F. For subsets X, Y of L, we denote by w X, Y x the F-subspace of L spanned by all elements of the type w x, y x with x g X and y g Y. Then LX [ w L, L x is the derived Lie algebra of L, LY [ Ž LX .X is the derived Lie algebra of LX , and L is called centrally metabelian if w LY , L x s 0. For an associative, unitary algebra A over F, we denote by L Ž A. the Lie algebra which has the same underlying vector space as A and which satisfies w a, b x [ ab y ba for a, b g A. We call A Lie centrally metabelian, if L Ž A. is centrally metabelian. Let G be a group. For subsets X, Y of G, we denote by Ž X, Y . the subgroup of G generated by all elements of the form Ž x, y . [ xyxy1 yy1 with x g X and y g Y. Then GX [ Ž G, G . is the derived subgroup of G, GY [ Ž GX .X is the derived subgroup of GX , and G is called centrally metabelian if Ž GY , G . s 1. Moreover, the lower central series of G is defined by g 1Ž G . [ G and gnq1Ž G . [ ŽgnŽ G ., G . for every positive integer n. We are interested in the question of when the group ring FG of G over F is Lie centrally metabelian. For the case p s 0 it is known that FG is Lie centrally metabelian if and only if G is abelian. It is proved in w5x that 111 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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the same is true if p ) 3. This leaves the Žmore interesting . cases p g 2, 34 . If p s 3 and if FG is Lie centrally metabelian then it is also proved in w5x that GX is a finite 3-Engel 3-group of exponent at most 9 and therefore nilpotent of class at most 4. The main purpose of this paper is to solve the case p s 3 completely by showing: THEOREM 1. Let F be a field of characteristic 3, and G a group. Then the group algebra FG of G o¨ er F is Lie centrally metabelian if and only if < GX < F 3. Further, U Ž FG ., the group of units of FG, is centrally metabelian and its deri¨ ed subgroup is of exponent F 3. We also improve an earlier theorem for arbitrary Lie solvable rings of w4x as: THEOREM 2. Let R be a Lie sol¨ able ring of deri¨ ed length n. Then the ideal of R generated by all elements of the type www x, y x, w z, u xx, ¨ x; x, y, z, u, ¨ g R is nilpotent of index at most 29 Ž19.10 ny 3 y 1.. Further, if 2 is in¨ ertible in R, then the ideal of R generated by all elements of the type w x, y x s xy y yx is a nil ideal. We start by proving the easy half of Theorem 1. LEMMA 3. Let F be a field of characteristic 3, and let G be a group such that < GX < F 3. Then the group algebra FG of G o¨ er F is Lie centrally metabelian, and U Ž FG ., the group of units of FG, is centrally metabelian with its deri¨ ed subgroup of exponent at most 3. Proof. For any group H, let DŽ FH . denote the augmentation ideal of FH. Thus DŽ FH . is the F-span of all elements of the form h y 1 with h g H. The canonical epimorphism w : FG ª F w GrGX x has kernel DŽ FGX . FG; in particular, FGrDŽ FGX . FG ( F w GrGX x and thus, L Ž FG .X : DŽ FGX . FG and L Ž FG .Y : DŽ FGX . FGDŽ FGX . FG s DŽ FGX . 2 FG. We may assume that < GX < s 3, for otherwise the result is trivial. If g denotes a generator of GX then DŽ FGX . s Ž g y 1. FGX , DŽ FGX . FG s Ž g y 1. FG, and DŽ FGX . 2 FG s Ž g y 1. 2 FG, where Ž g y 1. 2 s g 2 q g q 1 is contained in the centre Z FG of FG. Thus X
2 2 2 3 Ž g y 1 . a, b s Ž g y 1 . w a, b x g Ž g y 1 . L Ž FG . : Ž g y 1 . FG s 0
for a, b g FG, so that FG is Lie centrally metabelian. Theorem 4.1 in w5x shows that Y
Y
2
U Ž FG . : 1 q L Ž FG . FG : 1 q Ž g y 1 . FG s 1 q Ž g 2 q g q 1 . FG. If x, y g G then Ž y, x . g GX s 1, g, g 2 4 and therefore Ž g 2 q g q 1.Ž y, x . s g 2 q g q 1. Thus
Ž g 2 q g q 1 . xy s Ž g 2 q g q 1 . Ž y, x . xy s Ž g 2 q g q 1 . yx s y Ž g 2 q g q 1 . x.
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This shows that Ž g 2 q g q 1. FG : Z FG; in particular, Ž U Ž FG .Y , U Ž FG .. s 1. Finally, U Ž FG .X : 1 q L Ž FG .X FG : 1 q DŽ FGX . FG s 1 q Ž g y 1. FG implies that X
3 3 Ž U Ž FG . . : Ž 1 q Ž g y 1 . FG . s 1 q Ž g y 1 . 3 FG s 1.
The proof of the other half of Theorem 1 is more involved. We start by showing that the group algebra of a certain group of order 18 is not Lie centrally metabelian. LEMMA 4. Let F be a field of characteristic 3, let N be an elementary abelian group of order 9, let a be the automorphism of N in¨ erting all elements in N, and let G be the semidirect product of N and ² a:. Then the group algebra FG is not Lie centrally metabelian. Proof. We write N s ² b, c :. Then G s ² a, b, c : where b 3 s c 3 s 1, bc s cb, abay1 s by1 , and acay1 s cy1 . A simple calculation shows that w a y baby1 , a y cacy1 x f Z FG. Thus FG is not Lie centrally metabelian. We now prove a result on the derived Lie algebra of a tensor product of associative algebras. For simplicity of notation, we shall write AX , AY , . . . instead of L Ž A.X , L Ž A.Y , . . . in the following: PROPOSITION 5. field F. Then
Let A and B be two associati¨ e, unitary algebras o¨ er a
Ži. Ž A m B .X s AX m B q A m BX , Žii. w AX m B, AX m B x s AY m B q Ž AX . 2 m BX ; Žiii. Ž A m B .Y s AY m B q Ž AX . 2 m BX q w AX m B, A m BX x q A m Y B q AX m Ž BX . 2 . Proof. Ži. Since Ž A m B .X is spanned by elements of the form
w a1 m b1 , a2 m b 2 x s a1 a2 m b1 b 2 y a2 a1 m b 2 b1 s a1 a2 m b1 b 2 y a2 a1 m b1 b 2 q a2 a1 m b1 b 2 y a2 a1 m b 2 b1 s w a1 , a2 x m b1 b 2 q a2 a1 m w b1 , b 2 x with a1 , a2 g A, b1 , b 2 g B, we obtain Ž A m B .X : AX m B q A m BX . Conversely, AX m B is spanned by elements of the form
w a1 , a2 x m b s Ž a1 a2 y a2 a1 . m b s a1 a2 m b y a2 a1 m b s Ž a1 m 1 . Ž a2 m b . y Ž a2 m b . Ž a1 m 1 . s w a1 m 1, a2 m b x
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with a1 , a2 g A, b g B. Thus AX m B : Ž A m B .X and, similarly, A m BX : Ž A m B .X . Žii. As in the proof of Ži., w AX m B, AX m B x is spanned by elements of the form w a1 m b1 , a2 m b 2 x s w a1 , a2 x m b1 b 2 q a2 a1 m w b1 , b 2 x with a1 , a2 g AX , b1 , b 2 g B. Thus w AX m B, AX m B x : AY m B q Ž AX . 2 m BX . Conversely AY m B is spanned by elements of the form w a1 , a2 x m b s w a1 m 1, a2 m b x with a1 , a2 g AX , b g B. Thus AY m B : w AX m B, AX m B x. Also, Ž AX . 2 m BX is spanned by elements of the form a2 a1 m w b1 , b 2 x s w a1 m b1 , a2 m b 2 x y w a1 , a2 x m b1 b 2 with a1 , a2 g AX , b1 , b 2 g B, a1 m b1 , a2 m b 2 g AX m B, and w a1 , a2 x m b1 b 2 g AY m B : w AX m B, AX m B x. Thus Ž AX . 2 m BX : w AX m B, AX m B x. Žiii. An argument similar to the one in Žii. shows that
w A m BX , A m BX x s A m BY q AX m Ž BX . 2 . The result follows from this, Ži., and Žii.. In the following Sn will denote the symmetric group of degree n, for any positive integer n. We use Proposition 5 to show that the group rings of certain groups are not Lie centrally metabelian. LEMMA 6. Let F be a field of characteristic 3, let X be a group, and let G [ S3 = X. If the group ring FG is Lie centrally metabelian, then X is abelian. Proof. We identify FG with FS3 m FX as usual. By Proposition 5, Y
X 2
X
Ž FS3 m FX . = Ž Ž FS3 . . m Ž FX . . We write S3 s ² g, h: where g 3 s h2 s 1 and hghy1 s g 2 . Then g y g 2 s g y hghy1 g Ž FS3 .X and z [ Ž g y g 2 . 2 s 1 q g q g 2 g Z FS3 R 04 . If FG is Lie centrally metabelian then z m Ž FX .X is contained in Z Ž FS3 m FX . s Z FS3 m Z FX, so Ž FX .X : Z FX. It can be checked that this implies X to be abelian. We now embark on the proof of the more difficult half of Theorem 1. LEMMA 7. Let F be a field of characteristic 3, let G be a group, and suppose that the group ring FG of G o¨ er F is Lie centrally metabelian. Then < GX < F 3. Proof. Let G be a counterexample. We will often make use of the fact that the hypothesis of Lemma 7 carries over to the subgroups and factor groups of G. The results obtained in w5x show that GX is a finite 3-group
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and that Ž g, h. 3 s 1 for g, h g G; in particular, GX cannot be cyclic. This implies, by Burnside’s basis theorem, that GXrF Ž GX . is not cyclic; here F Ž H . denotes the Frattini subgroup of a group H. Thus GrFŽ GX . is also a counterexample. We may therefore replace G by GrFŽ GX . and assume that GX is elementary abelian of order 3 n ) 3. Let C [ CG Ž GX . denote the centralizer of GX in G. The action of G on X C into the automorphism group AŽ GX . of G by conjugation embeds GrC X X G . Note that AŽ G . is isomorphic to the general linear group G LnŽ3. of C is finite. degree n over the field F3 with 3 elements, in particular GrC C is abelian. We now Since GX is abelian, we also have GX : C , so that GrC C in more detail. ŽOnce the lemma is try to determine the structure of GrC < < C F 2.. proved we will know that GrC C contains an element aC C of prime order p ) 3. Suppose first that GrC C : acts completely reducibly by conjugation on By Maschke’s theorem, ² aC C :-modules GX . We write GX s M1 = M2 = ??? = Mr with irreducible ² aC X ² : C does not act trivially on G , at least one of the M1 , M2 , . . . , Mr . Since aC C :-module. We may assume that i s 1. Then < M1 < ) 3, Mi is a nontrivial ² aC and we set H [ ² a, GX :rM2 = M3 = ??? = Mr . If we denote the images of a and M1 in H by a and M1 , respectively, then H s ² a, M1 : s ² a: M1 where a induces an automorphism of order p on M1 ( M1. Since M1 is a nontrivial irreducible ² a:-module we get M1 s Ž M1 , ² a:. : H X Žcf. Satz III.13.4 in w2x.; in particular, H is also a counterexample. Thus we may replace G by H and therefore assume that G s GX² a: where GX is a C :-module. nontrivial irreducible ² aC p Since a centralizes both GX and a we must have a p g Z G, the centre C :-submodule of GX on which ² aC C : acts of G. Thus ² a p : l GX is an ² aC X p: p: ² ² trivially. Hence a l G s 1. But now Gr a is also a counterexample. We may therefore replace G by Gr² a p : and assume that G is the semidirect product of GX and ² a: where a has order p. However, in this situation, Theorem C in w3x implies that L Ž FG .Z / 0, contradicting our hypothesis, w L Ž FG .Y , L Ž FG .x s 0. C has order 2 k 3 l , with integers k, l. This contradiction means that GrC C contains an element of order 4. Then we argue Next, we assume that GrC in a similar way, this time choosing M1 to be an irreducible faithful ² aC C :-module. Thus < M1 < s 9. As before we are led to the situation where G s ² a: M1 is the semidirect product of M1 and ² a:, where ² a: has order 4. Thus G has order 36, and the subgroup H or order 18 in G is isomorphic to the group appearing in Lemma 4. Since FH is not Lie centrally metabelian by Lemma 4, this is a contradiction. C is abelian of order 2 k 3 l with This contradiction shows that GrC elementary abelian Sylow 2-subgroup; in particular, F3 is a splitting field C . Thus the GrC C-module GX has a composition series with compofor GrC sition factors of F3-dimension 1, i.e., of order 3. In particular, G has a
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normal subgroup N contained in GX with < GX : N < s 9. Thus we can replace G by GrN and therefore assume that < GX < s 9. In this situation the action of G on GX , by conjugation, embeds the C into the subgroup B of G L 2 Ž3. consisting of all abelian group GrC elements of the shape
ž
* 0
* . *
/
C contains an Note that B is isomorphic to S2 = S3 . We assume that GrC C whose image in B is y1. Arguing as before, we see that we element aC may assume that G s ² a, GX : s ² a: GX is the semidirect product of GX and ² a: where a has order 2. Thus G is the group appearing in Lemma 4, C which is impossible. This contradiction shows that the abelian group GrC C < F 3. is isomorphic to a subgroup of Br² y 1: ( S3 , in particular, < GrC Here we stop our first line of attack and prepare for the second. We can write GX s ²Ž a, b ., Ž c, d .: and set H [ ² a, b, c, d :. Thus H X s X G , so H is also a counterexample. Hence we may replace G by H and therefore assume that G s ² a, b, c, d :; in particular, G is finitely generated. It follows that the first cohomology group H 1 Ž GrGX , GX . is finite, a fact we are going to exploit in a moment. Each element in C acts trivially on both GrGX and GX . So the method of Satz I.17.1 in w2x yields a homomorphism C ª H 1 Ž GrG X , G X . with kernel G X Z G. Since H 1 Ž GrGX , GX . is a finite 3-group, CrGX Z G is a finite 3-group as well. Z G is a Moreover, < GX Z G : Z G < s < GX : GX l Z G < divides 9. Thus CrZ finite 3-group; in particular, Z G is a subgroup of finite index in the finitely generated group G. This implies by Satz I.19.10 in w2x that Z G is a finitely generated abelian group. Suppose that Z G contains an element z such that ² z : l GX s 1. Then Gr² z : is also a counterexample, so we can replace G by Gr² z :. This applies, in particular, whenever z is an element of infinite order in Z G. Thus we may assume that Z G is finite. By a similar argument, we may even assume that Z G is a finite 3-group. Now C is a finite 3-group, and < G : C < F 3; in particular, G is a finite group of order 3 m or 2.3 m for some integer m. We may also assume that our counterexample has least possible order. Then every proper subgroup H of G satisfies < H X < F 3. Let us first consider the case when < G < s 2.3 m . We denote by P a Sylow 3-subgroup of G and by a an element of order 2 in G, so that G s ² a: P is a semidirect product. If P is abelian then P s CP Ž a. = Ž P, ² a:., by Satz III.13.4 in w2x. Since GrŽ P, ² a:. is abelian we see that Ž P, ² a:. s GX . But CP Ž a. is isomorphic to the group of order 18 appearing in Lemma now GrC 4, a contradiction. Next we consider the case when P is non abelian, i.e., < P X < s 3. In this case, P [ PrPX s CP Ž a. = Ž P, ² a:. by Satz III.13.4 in w2x.
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Let G [ GrPX . Since GrŽ P, ² a:. is abelian, we conclude that Ž P, ² a:. s GXs GXrP X ; in particular, <Ž P, ² a:.< s 3. Now H [ ² a, GX : is a group of order 18, and a acts non-trivially on GXrP X . If a acts also non-trivially on P X , H is isomorphic to the group appearing in Lemma 4, which is impossible. Thus a acts trivially on P X , and H s P X = S for a subgroup S of H isomorphic to S3 . We write CP Ž a. s XrP X . Then a acts trivially on P X and on XrP X , so a acts trivially on X Žcf. Theorem 5.3.2 in w1x.. Thus X s CP Ž a.. Any other involution aX in G is conjugate to a by an element in P. Hence CP Ž aX . s CP Ž a. s X. But S is generated by its involutions, so X : CP Ž S .. This means that G s X = S, where < X X < s 3. However, this is impossible by Lemma 6. We are thus left with the case when G is a finite 3-group. Recall that we may write GX s ²Ž a, b ., Ž c, d .: and G s ² a, b, c, d :, so that < GrFŽ G .< F 3 4 . We assume first that < GrFŽ G .< s 3 4 . ŽThe following argument is due X X to Zhang.. Then ² a, b, c : / G, so ² a, b, c : s ²Ž a, b .:. Similarly, ² a, b, d : X X s ²Ž a, b .:, ² a, c, d : s ²Ž c, d .:, and ² b, c, d : s ²Ž c, d .:. Hence Ž a, c . g ²Ž a, b .: l ²Ž c, d .: s 1. Similarly, 1 s Ž a, d . s Ž b, c . s Ž b, d .. But now <² a, bc, d :X < s 9, so G s ² a, bc, d : and < GrFŽ G .< F 3 3 , a contradiction. C< Suppose next that < GrFŽ G .< s 3 2 , and write G s ² a, b :. Since < GrC X Ž . F 3 we may assume that b g C . Then G is generated by c [ a, b and g 3 Ž G .. And g 3 Ž G . is generated by d [ Ž a, c . since Ž b, c . s 1. It is now easy Žbut lengthy. to verify that ww ab y ba, ac y cax, b x / 0, so we get a contradiction in this case. It remains to deal with the case < GrFŽ G .< s 3 3. If x, y g G, then ² x, y : / G, so there exists a maximal subgroup M of G containing ² x, y :. Thus < M X < F 3, so Ž x, y . g M X : Z G; in particular, G is nilpotent of class 2. We write G s ² a, b, c :. Then GX s ²Ž a, b ., Ž a, c ., Ž b, c .:. Since < GX < F 9, we may assume that GX s ²Ž a, b ., Ž a, c .:, and we can write Ž b, c . s Ž a, b . i Ž a, c . j with integers i, j. Then we set ˜ b [ ayj b, ˜ c [ a i c, so that ˜ ˜ ² : Ž . Ž . Ž . Ž . G s a, b, ˜ c , a, b s a, b , a, ˜ c s a, c , and
Ž ˜b , ˜c . s Ž ayj b, ai c . s Ž a, c . yj Ž b, a. i Ž b, c . s 1. Thus we can replace b by ˜ b and c by ˜ c to obtain Ž b, c . s 1. An easy Žbut lengthy. calculation shows that ww ab y ba, ac y cax, b x / 0 again. This means that we have reached our final contradiction, and hence Lemma 7 is proved. Theorem 1 now follows from Lemmas 3 and 7. It follows from Lemma 3 that if F is a field of characteristic 3 and G is a group with < GX < F 3 then the group of units U Ž FG . of FG is centrally metabelian. This may be the best possible result in this direction. We give the following example in order to substantiate this fact.
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EXAMPLE 8. Let F be a field of characteristic 3 and S3 be the symmetric group on 3 symbols, then the group of units U Ž FS3 . of FS3 is centrally metabelian but not metabelian. Proof. Let A s FS3 and U s U Ž A ., then U is centrally metabelian by Lemma 3. We now show that U is not metabelian. Let S3 s ² g, h : g 3 s h2 s 1, hghy1 s g 2 : and J s DŽ FSX3 . FS3 . The elements e s y1 y h and f s 1 y e are two orthogonal idempotents in A such that e q f s 1, and hence the two-sided Peirce decomposition of A can be written as A s e Ae [ e A f [ f Ae [ f A f . Also, for some suitable a, b g A we can write e Ae s Fe q Fab,
e A f s Fa,
f Ae s Fb,
f A f s Ff q Fba
and J s Fab q Fa q Fb q Fba. Now, J 2 is central in A. For any a , b g F, the following relations hold in A modulo J 2
Ž Ž e y f . , Ž1 q a a q b b. . s Ž e y f . Ž1 q a a q b b. Ž e y f .
y1
Ž1 q a a q b b.
y1
s Ž e y f . Ž1 q a a q b b. Ž e y f . Ž1 y a a y b b. s Ž e q a a y f y b b. Ž e y a a y f q b b. s e y aa y aa q f y bb y bb s 1 q a a q b b. This implies U X q J 2 = 1 q Fa q Fb q J 2 s 1 q J = U X q J 2 . In particular, U X q J 2 s 1 q J. If U is metabelian, then w U X , U X x s 0 and since J 2 is central in A, this implies that w U X q J 2 , U X q J 2 x s 0. This further implies that w J, J x s w1 q J, 1 q J x s 0. Hence J must be commutative. But Ž g y 1., Ž g y 1. h g J are such that
Ž g y 1. , Ž g y 1. h 2
2
s Ž g y 1. h y Ž g y 1. h Ž g y 1. s Ž g y 1. h y Ž g y 1. Ž g 2 y 1. h 2
2
2
s Ž g y 1. h y Ž g y 1. Ž g q 1. h s Ž g y 1. Ž 1 y g y 1. h 2
s y Ž g y 1 . gh / 0. Hence U is not metabelian.
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We now concentrate on Theorem 2. An ideal U of the associated Lie ring L Ž R . of the ring R is called a Lie ideal of R. Thus a Lie ideal U of R is an additive subgroup of R such that wU, R x : U. Let R be a Lie solvable ring of length n G 3, c s 29 Ž19.10 ny 3 y 1., J be the ideal of R generated by all elements of the type www x, y x, w z, u xx, ¨ x; x, y, z, u, ¨ g R, and I be the derived ideal of R generated by all elements of the type w a , b x; a , b g R. It is proved in w4x that if 3 is invertible in R then J is nilpotent of index at most c and when 2 and 3 both are invertible in R then I is a nil ideal in R. Theorem 2 proves this result without the restriction of invertibility of 3. However, the invertibility of 2 in the context cannot be dropped. We first prove the following: LEMMA 9.
Let U be a Lie ideal of R, then Žg 3 ŽU .. 2 : d Ž2. ŽU . R.
Proof. The following identity holds true for any elements x 1 , x 2 , x 3 , y 1 , y 2 , y 3 g R: 2 w x 1 , x 2 , x 3 xw y 1 , y 2 , y 3 x s w x1 , x 2 x , w y 3 y 2 , y1 , x 3 x q w x1 , x 2 x , w y 3 x 3 , y 2 , y1 x q w x1 , x 2 x , w y 3 y1 , x 3 , y 2 x y w y1 , y 2 x , w x 3 x1 , x 2 , y 3 x y w y1 , y 2 x , w x 3 x 2 , y 3 , x1 x y w y1 , y 2 x , w x 3 y 3 , x1 , x 2 x q w y 3 , x 3 x , w x 2 x1 , y1 , y 2 x q w y 3 , x 3 x , w x 2 y1 , y 2 , x1 x q w y 3 , x 3 x , w x 2 y 2 , x1 , y1 x y w x1 , x 2 x , w y 3 , y1 , x 3 y 2 x q w y1 , y 2 x , w x 3 , x 2 , y 3 x1 x y w y 3 , x 3 x , w x 2 , y 2 , x1 y1 x q w x1 , x 2 , x 3 x , w y 3 , y 2 , y1 x q w y1 , y 2 , y 3 x , w x 2 , x 3 , x1 x q w y 3 , x 3 , y 2 x , w x 2 , x1 , y1 x q w y 3 , x 3 , x1 x , w x 2 , y1 , y 2 x y w x1 , x 2 x , w y 3 , x 3 , y 2 x y1 q w y1 , y 2 x , w x 3 , y 3 , x1 x x 2 y w y 3 , x 3 x , w x 2 , x1 , y1 x y 2 y w x1 , x 2 x , w y 3 , y 2 , y1 x x 3 q x 3 w x1 , x 2 x , w y 3 , y1 , y 2 x q w y1 , y 2 x , w x 3 , x1 , x 2 x y 3 y y 3 w y1 , y 2 x , w x 3 , x 2 , x1 x y w y 3 , x 3 x , w x 2 , y1 , y 2 x x1 q x1 w y 3 , x 3 x , w x 2 , y 2 , y1 x . Now by letting the elements x 1 , x 2 , x 3 , y 1 , y 2 , y 3 g U in the above identity and using the fact that RUR s UR s RU for any Lie ideal U of R, we get that 2Žg 3 ŽU .. 2 : d Ž2. ŽU . R. Also, it follows from the Proof of Lemma 1.7 of w4x that 3Žg 3 ŽU .. 2 : d Ž2. ŽU . R. On combining the two, we get Lemma 9.
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The proof of Theorem 2 follows on the same lines as those in w4x by replacing Lemma 1.7 in w4x by the above Lemma 9. Finally, since for a Lie solvable ring R, the ideal J is nilpotent Žof index at most c ., therefore U Ž RrJ . s Ž U Ž R . q J .rJ, where U Ž A. denotes the group of units for a ring A. In fact, for any positive integer n, we have d Ž n. Ž U Ž RrJ .. s Ž d Ž n. Ž U Ž R .. q J .rJ. Hence U Ž R . is solvable if and only if U Ž RrJ . is solvable. But RrJ is a Lie centrally metabelian ring. Obviously if one knows the solvability length of the group of units of a Lie centrally metabelian ring one then knows a bound for the solvability length of the group of units of an arbitrary Lie solvable ring. Thus their study becomes important.
REFERENCES 1. D. Gorenstein, ‘‘Finite Groups,’’ Harper and Row, New York, 1968. 2. B. Huppert, ‘‘Endliche Gruppen,’’ Vol. I, Springer-Verlag, Berlin, 1967. 3. A. Shalev, The derived length of Lie solvable group rings, I, J. Pure Appl. Algebra 78 Ž1992., 291]300. 4. R. K. Sharma and J. B. Srivastava, Lie solvable group rings, Proc. Amer. Math. Soc. 94 Ž1985., 1]8. 5. R. K. Sharma and J. B. Srivastava, Lie centrally metabelian group rings, J. Algebra 151 Ž1992., 476]486.
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Lie Centrally Metabelian Group Rings in Characteristic 3 B. Kulshammer ¨ Department of Mathematics, Uni¨ ersity of Augsburg, D-86135 Augsburg, Germany
and R. K. Sharma Department of Mathematics, Indian Institute of Technology, Kharagpur (W. B.)-721302, India Communicated by Walter Feit Received January 12, 1995
We classify Lie centrally metabelian group algebras over fields of characteristic 3. Q 1996 Academic Press, Inc.
Let F be a field of characteristic p G 0, and let L be a Lie algebra over F. For subsets X, Y of L, we denote by w X, Y x the F-subspace of L spanned by all elements of the type w x, y x with x g X and y g Y. Then LX [ w L, L x is the derived Lie algebra of L, LY [ Ž LX .X is the derived Lie algebra of LX , and L is called centrally metabelian if w LY , L x s 0. For an associative, unitary algebra A over F, we denote by L Ž A. the Lie algebra which has the same underlying vector space as A and which satisfies w a, b x [ ab y ba for a, b g A. We call A Lie centrally metabelian, if L Ž A. is centrally metabelian. Let G be a group. For subsets X, Y of G, we denote by Ž X, Y . the subgroup of G generated by all elements of the form Ž x, y . [ xyxy1 yy1 with x g X and y g Y. Then GX [ Ž G, G . is the derived subgroup of G, GY [ Ž GX .X is the derived subgroup of GX , and G is called centrally metabelian if Ž GY , G . s 1. Moreover, the lower central series of G is defined by g 1Ž G . [ G and gnq1Ž G . [ ŽgnŽ G ., G . for every positive integer n. We are interested in the question of when the group ring FG of G over F is Lie centrally metabelian. For the case p s 0 it is known that FG is Lie centrally metabelian if and only if G is abelian. It is proved in w5x that 111 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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the same is true if p ) 3. This leaves the Žmore interesting . cases p g 2, 34 . If p s 3 and if FG is Lie centrally metabelian then it is also proved in w5x that GX is a finite 3-Engel 3-group of exponent at most 9 and therefore nilpotent of class at most 4. The main purpose of this paper is to solve the case p s 3 completely by showing: THEOREM 1. Let F be a field of characteristic 3, and G a group. Then the group algebra FG of G o¨ er F is Lie centrally metabelian if and only if < GX < F 3. Further, U Ž FG ., the group of units of FG, is centrally metabelian and its deri¨ ed subgroup is of exponent F 3. We also improve an earlier theorem for arbitrary Lie solvable rings of w4x as: THEOREM 2. Let R be a Lie sol¨ able ring of deri¨ ed length n. Then the ideal of R generated by all elements of the type www x, y x, w z, u xx, ¨ x; x, y, z, u, ¨ g R is nilpotent of index at most 29 Ž19.10 ny 3 y 1.. Further, if 2 is in¨ ertible in R, then the ideal of R generated by all elements of the type w x, y x s xy y yx is a nil ideal. We start by proving the easy half of Theorem 1. LEMMA 3. Let F be a field of characteristic 3, and let G be a group such that < GX < F 3. Then the group algebra FG of G o¨ er F is Lie centrally metabelian, and U Ž FG ., the group of units of FG, is centrally metabelian with its deri¨ ed subgroup of exponent at most 3. Proof. For any group H, let DŽ FH . denote the augmentation ideal of FH. Thus DŽ FH . is the F-span of all elements of the form h y 1 with h g H. The canonical epimorphism w : FG ª F w GrGX x has kernel DŽ FGX . FG; in particular, FGrDŽ FGX . FG ( F w GrGX x and thus, L Ž FG .X : DŽ FGX . FG and L Ž FG .Y : DŽ FGX . FGDŽ FGX . FG s DŽ FGX . 2 FG. We may assume that < GX < s 3, for otherwise the result is trivial. If g denotes a generator of GX then DŽ FGX . s Ž g y 1. FGX , DŽ FGX . FG s Ž g y 1. FG, and DŽ FGX . 2 FG s Ž g y 1. 2 FG, where Ž g y 1. 2 s g 2 q g q 1 is contained in the centre Z FG of FG. Thus X
2 2 2 3 Ž g y 1 . a, b s Ž g y 1 . w a, b x g Ž g y 1 . L Ž FG . : Ž g y 1 . FG s 0
for a, b g FG, so that FG is Lie centrally metabelian. Theorem 4.1 in w5x shows that Y
Y
2
U Ž FG . : 1 q L Ž FG . FG : 1 q Ž g y 1 . FG s 1 q Ž g 2 q g q 1 . FG. If x, y g G then Ž y, x . g GX s 1, g, g 2 4 and therefore Ž g 2 q g q 1.Ž y, x . s g 2 q g q 1. Thus
Ž g 2 q g q 1 . xy s Ž g 2 q g q 1 . Ž y, x . xy s Ž g 2 q g q 1 . yx s y Ž g 2 q g q 1 . x.
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This shows that Ž g 2 q g q 1. FG : Z FG; in particular, Ž U Ž FG .Y , U Ž FG .. s 1. Finally, U Ž FG .X : 1 q L Ž FG .X FG : 1 q DŽ FGX . FG s 1 q Ž g y 1. FG implies that X
3 3 Ž U Ž FG . . : Ž 1 q Ž g y 1 . FG . s 1 q Ž g y 1 . 3 FG s 1.
The proof of the other half of Theorem 1 is more involved. We start by showing that the group algebra of a certain group of order 18 is not Lie centrally metabelian. LEMMA 4. Let F be a field of characteristic 3, let N be an elementary abelian group of order 9, let a be the automorphism of N in¨ erting all elements in N, and let G be the semidirect product of N and ² a:. Then the group algebra FG is not Lie centrally metabelian. Proof. We write N s ² b, c :. Then G s ² a, b, c : where b 3 s c 3 s 1, bc s cb, abay1 s by1 , and acay1 s cy1 . A simple calculation shows that w a y baby1 , a y cacy1 x f Z FG. Thus FG is not Lie centrally metabelian. We now prove a result on the derived Lie algebra of a tensor product of associative algebras. For simplicity of notation, we shall write AX , AY , . . . instead of L Ž A.X , L Ž A.Y , . . . in the following: PROPOSITION 5. field F. Then
Let A and B be two associati¨ e, unitary algebras o¨ er a
Ži. Ž A m B .X s AX m B q A m BX , Žii. w AX m B, AX m B x s AY m B q Ž AX . 2 m BX ; Žiii. Ž A m B .Y s AY m B q Ž AX . 2 m BX q w AX m B, A m BX x q A m Y B q AX m Ž BX . 2 . Proof. Ži. Since Ž A m B .X is spanned by elements of the form
w a1 m b1 , a2 m b 2 x s a1 a2 m b1 b 2 y a2 a1 m b 2 b1 s a1 a2 m b1 b 2 y a2 a1 m b1 b 2 q a2 a1 m b1 b 2 y a2 a1 m b 2 b1 s w a1 , a2 x m b1 b 2 q a2 a1 m w b1 , b 2 x with a1 , a2 g A, b1 , b 2 g B, we obtain Ž A m B .X : AX m B q A m BX . Conversely, AX m B is spanned by elements of the form
w a1 , a2 x m b s Ž a1 a2 y a2 a1 . m b s a1 a2 m b y a2 a1 m b s Ž a1 m 1 . Ž a2 m b . y Ž a2 m b . Ž a1 m 1 . s w a1 m 1, a2 m b x
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with a1 , a2 g A, b g B. Thus AX m B : Ž A m B .X and, similarly, A m BX : Ž A m B .X . Žii. As in the proof of Ži., w AX m B, AX m B x is spanned by elements of the form w a1 m b1 , a2 m b 2 x s w a1 , a2 x m b1 b 2 q a2 a1 m w b1 , b 2 x with a1 , a2 g AX , b1 , b 2 g B. Thus w AX m B, AX m B x : AY m B q Ž AX . 2 m BX . Conversely AY m B is spanned by elements of the form w a1 , a2 x m b s w a1 m 1, a2 m b x with a1 , a2 g AX , b g B. Thus AY m B : w AX m B, AX m B x. Also, Ž AX . 2 m BX is spanned by elements of the form a2 a1 m w b1 , b 2 x s w a1 m b1 , a2 m b 2 x y w a1 , a2 x m b1 b 2 with a1 , a2 g AX , b1 , b 2 g B, a1 m b1 , a2 m b 2 g AX m B, and w a1 , a2 x m b1 b 2 g AY m B : w AX m B, AX m B x. Thus Ž AX . 2 m BX : w AX m B, AX m B x. Žiii. An argument similar to the one in Žii. shows that
w A m BX , A m BX x s A m BY q AX m Ž BX . 2 . The result follows from this, Ži., and Žii.. In the following Sn will denote the symmetric group of degree n, for any positive integer n. We use Proposition 5 to show that the group rings of certain groups are not Lie centrally metabelian. LEMMA 6. Let F be a field of characteristic 3, let X be a group, and let G [ S3 = X. If the group ring FG is Lie centrally metabelian, then X is abelian. Proof. We identify FG with FS3 m FX as usual. By Proposition 5, Y
X 2
X
Ž FS3 m FX . = Ž Ž FS3 . . m Ž FX . . We write S3 s ² g, h: where g 3 s h2 s 1 and hghy1 s g 2 . Then g y g 2 s g y hghy1 g Ž FS3 .X and z [ Ž g y g 2 . 2 s 1 q g q g 2 g Z FS3 R 04 . If FG is Lie centrally metabelian then z m Ž FX .X is contained in Z Ž FS3 m FX . s Z FS3 m Z FX, so Ž FX .X : Z FX. It can be checked that this implies X to be abelian. We now embark on the proof of the more difficult half of Theorem 1. LEMMA 7. Let F be a field of characteristic 3, let G be a group, and suppose that the group ring FG of G o¨ er F is Lie centrally metabelian. Then < GX < F 3. Proof. Let G be a counterexample. We will often make use of the fact that the hypothesis of Lemma 7 carries over to the subgroups and factor groups of G. The results obtained in w5x show that GX is a finite 3-group
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and that Ž g, h. 3 s 1 for g, h g G; in particular, GX cannot be cyclic. This implies, by Burnside’s basis theorem, that GXrF Ž GX . is not cyclic; here F Ž H . denotes the Frattini subgroup of a group H. Thus GrFŽ GX . is also a counterexample. We may therefore replace G by GrFŽ GX . and assume that GX is elementary abelian of order 3 n ) 3. Let C [ CG Ž GX . denote the centralizer of GX in G. The action of G on X C into the automorphism group AŽ GX . of G by conjugation embeds GrC X X G . Note that AŽ G . is isomorphic to the general linear group G LnŽ3. of C is finite. degree n over the field F3 with 3 elements, in particular GrC C is abelian. We now Since GX is abelian, we also have GX : C , so that GrC C in more detail. ŽOnce the lemma is try to determine the structure of GrC < < C F 2.. proved we will know that GrC C contains an element aC C of prime order p ) 3. Suppose first that GrC C : acts completely reducibly by conjugation on By Maschke’s theorem, ² aC C :-modules GX . We write GX s M1 = M2 = ??? = Mr with irreducible ² aC X ² : C does not act trivially on G , at least one of the M1 , M2 , . . . , Mr . Since aC C :-module. We may assume that i s 1. Then < M1 < ) 3, Mi is a nontrivial ² aC and we set H [ ² a, GX :rM2 = M3 = ??? = Mr . If we denote the images of a and M1 in H by a and M1 , respectively, then H s ² a, M1 : s ² a: M1 where a induces an automorphism of order p on M1 ( M1. Since M1 is a nontrivial irreducible ² a:-module we get M1 s Ž M1 , ² a:. : H X Žcf. Satz III.13.4 in w2x.; in particular, H is also a counterexample. Thus we may replace G by H and therefore assume that G s GX² a: where GX is a C :-module. nontrivial irreducible ² aC p Since a centralizes both GX and a we must have a p g Z G, the centre C :-submodule of GX on which ² aC C : acts of G. Thus ² a p : l GX is an ² aC X p: p: ² ² trivially. Hence a l G s 1. But now Gr a is also a counterexample. We may therefore replace G by Gr² a p : and assume that G is the semidirect product of GX and ² a: where a has order p. However, in this situation, Theorem C in w3x implies that L Ž FG .Z / 0, contradicting our hypothesis, w L Ž FG .Y , L Ž FG .x s 0. C has order 2 k 3 l , with integers k, l. This contradiction means that GrC C contains an element of order 4. Then we argue Next, we assume that GrC in a similar way, this time choosing M1 to be an irreducible faithful ² aC C :-module. Thus < M1 < s 9. As before we are led to the situation where G s ² a: M1 is the semidirect product of M1 and ² a:, where ² a: has order 4. Thus G has order 36, and the subgroup H or order 18 in G is isomorphic to the group appearing in Lemma 4. Since FH is not Lie centrally metabelian by Lemma 4, this is a contradiction. C is abelian of order 2 k 3 l with This contradiction shows that GrC elementary abelian Sylow 2-subgroup; in particular, F3 is a splitting field C . Thus the GrC C-module GX has a composition series with compofor GrC sition factors of F3-dimension 1, i.e., of order 3. In particular, G has a
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normal subgroup N contained in GX with < GX : N < s 9. Thus we can replace G by GrN and therefore assume that < GX < s 9. In this situation the action of G on GX , by conjugation, embeds the C into the subgroup B of G L 2 Ž3. consisting of all abelian group GrC elements of the shape
ž
* 0
* . *
/
C contains an Note that B is isomorphic to S2 = S3 . We assume that GrC C whose image in B is y1. Arguing as before, we see that we element aC may assume that G s ² a, GX : s ² a: GX is the semidirect product of GX and ² a: where a has order 2. Thus G is the group appearing in Lemma 4, C which is impossible. This contradiction shows that the abelian group GrC C < F 3. is isomorphic to a subgroup of Br² y 1: ( S3 , in particular, < GrC Here we stop our first line of attack and prepare for the second. We can write GX s ²Ž a, b ., Ž c, d .: and set H [ ² a, b, c, d :. Thus H X s X G , so H is also a counterexample. Hence we may replace G by H and therefore assume that G s ² a, b, c, d :; in particular, G is finitely generated. It follows that the first cohomology group H 1 Ž GrGX , GX . is finite, a fact we are going to exploit in a moment. Each element in C acts trivially on both GrGX and GX . So the method of Satz I.17.1 in w2x yields a homomorphism C ª H 1 Ž GrG X , G X . with kernel G X Z G. Since H 1 Ž GrGX , GX . is a finite 3-group, CrGX Z G is a finite 3-group as well. Z G is a Moreover, < GX Z G : Z G < s < GX : GX l Z G < divides 9. Thus CrZ finite 3-group; in particular, Z G is a subgroup of finite index in the finitely generated group G. This implies by Satz I.19.10 in w2x that Z G is a finitely generated abelian group. Suppose that Z G contains an element z such that ² z : l GX s 1. Then Gr² z : is also a counterexample, so we can replace G by Gr² z :. This applies, in particular, whenever z is an element of infinite order in Z G. Thus we may assume that Z G is finite. By a similar argument, we may even assume that Z G is a finite 3-group. Now C is a finite 3-group, and < G : C < F 3; in particular, G is a finite group of order 3 m or 2.3 m for some integer m. We may also assume that our counterexample has least possible order. Then every proper subgroup H of G satisfies < H X < F 3. Let us first consider the case when < G < s 2.3 m . We denote by P a Sylow 3-subgroup of G and by a an element of order 2 in G, so that G s ² a: P is a semidirect product. If P is abelian then P s CP Ž a. = Ž P, ² a:., by Satz III.13.4 in w2x. Since GrŽ P, ² a:. is abelian we see that Ž P, ² a:. s GX . But CP Ž a. is isomorphic to the group of order 18 appearing in Lemma now GrC 4, a contradiction. Next we consider the case when P is non abelian, i.e., < P X < s 3. In this case, P [ PrPX s CP Ž a. = Ž P, ² a:. by Satz III.13.4 in w2x.
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Let G [ GrPX . Since GrŽ P, ² a:. is abelian, we conclude that Ž P, ² a:. s GXs GXrP X ; in particular, <Ž P, ² a:.< s 3. Now H [ ² a, GX : is a group of order 18, and a acts non-trivially on GXrP X . If a acts also non-trivially on P X , H is isomorphic to the group appearing in Lemma 4, which is impossible. Thus a acts trivially on P X , and H s P X = S for a subgroup S of H isomorphic to S3 . We write CP Ž a. s XrP X . Then a acts trivially on P X and on XrP X , so a acts trivially on X Žcf. Theorem 5.3.2 in w1x.. Thus X s CP Ž a.. Any other involution aX in G is conjugate to a by an element in P. Hence CP Ž aX . s CP Ž a. s X. But S is generated by its involutions, so X : CP Ž S .. This means that G s X = S, where < X X < s 3. However, this is impossible by Lemma 6. We are thus left with the case when G is a finite 3-group. Recall that we may write GX s ²Ž a, b ., Ž c, d .: and G s ² a, b, c, d :, so that < GrFŽ G .< F 3 4 . We assume first that < GrFŽ G .< s 3 4 . ŽThe following argument is due X X to Zhang.. Then ² a, b, c : / G, so ² a, b, c : s ²Ž a, b .:. Similarly, ² a, b, d : X X s ²Ž a, b .:, ² a, c, d : s ²Ž c, d .:, and ² b, c, d : s ²Ž c, d .:. Hence Ž a, c . g ²Ž a, b .: l ²Ž c, d .: s 1. Similarly, 1 s Ž a, d . s Ž b, c . s Ž b, d .. But now <² a, bc, d :X < s 9, so G s ² a, bc, d : and < GrFŽ G .< F 3 3 , a contradiction. C< Suppose next that < GrFŽ G .< s 3 2 , and write G s ² a, b :. Since < GrC X Ž . F 3 we may assume that b g C . Then G is generated by c [ a, b and g 3 Ž G .. And g 3 Ž G . is generated by d [ Ž a, c . since Ž b, c . s 1. It is now easy Žbut lengthy. to verify that ww ab y ba, ac y cax, b x / 0, so we get a contradiction in this case. It remains to deal with the case < GrFŽ G .< s 3 3. If x, y g G, then ² x, y : / G, so there exists a maximal subgroup M of G containing ² x, y :. Thus < M X < F 3, so Ž x, y . g M X : Z G; in particular, G is nilpotent of class 2. We write G s ² a, b, c :. Then GX s ²Ž a, b ., Ž a, c ., Ž b, c .:. Since < GX < F 9, we may assume that GX s ²Ž a, b ., Ž a, c .:, and we can write Ž b, c . s Ž a, b . i Ž a, c . j with integers i, j. Then we set ˜ b [ ayj b, ˜ c [ a i c, so that ˜ ˜ ² : Ž . Ž . Ž . Ž . G s a, b, ˜ c , a, b s a, b , a, ˜ c s a, c , and
Ž ˜b , ˜c . s Ž ayj b, ai c . s Ž a, c . yj Ž b, a. i Ž b, c . s 1. Thus we can replace b by ˜ b and c by ˜ c to obtain Ž b, c . s 1. An easy Žbut lengthy. calculation shows that ww ab y ba, ac y cax, b x / 0 again. This means that we have reached our final contradiction, and hence Lemma 7 is proved. Theorem 1 now follows from Lemmas 3 and 7. It follows from Lemma 3 that if F is a field of characteristic 3 and G is a group with < GX < F 3 then the group of units U Ž FG . of FG is centrally metabelian. This may be the best possible result in this direction. We give the following example in order to substantiate this fact.
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EXAMPLE 8. Let F be a field of characteristic 3 and S3 be the symmetric group on 3 symbols, then the group of units U Ž FS3 . of FS3 is centrally metabelian but not metabelian. Proof. Let A s FS3 and U s U Ž A ., then U is centrally metabelian by Lemma 3. We now show that U is not metabelian. Let S3 s ² g, h : g 3 s h2 s 1, hghy1 s g 2 : and J s DŽ FSX3 . FS3 . The elements e s y1 y h and f s 1 y e are two orthogonal idempotents in A such that e q f s 1, and hence the two-sided Peirce decomposition of A can be written as A s e Ae [ e A f [ f Ae [ f A f . Also, for some suitable a, b g A we can write e Ae s Fe q Fab,
e A f s Fa,
f Ae s Fb,
f A f s Ff q Fba
and J s Fab q Fa q Fb q Fba. Now, J 2 is central in A. For any a , b g F, the following relations hold in A modulo J 2
Ž Ž e y f . , Ž1 q a a q b b. . s Ž e y f . Ž1 q a a q b b. Ž e y f .
y1
Ž1 q a a q b b.
y1
s Ž e y f . Ž1 q a a q b b. Ž e y f . Ž1 y a a y b b. s Ž e q a a y f y b b. Ž e y a a y f q b b. s e y aa y aa q f y bb y bb s 1 q a a q b b. This implies U X q J 2 = 1 q Fa q Fb q J 2 s 1 q J = U X q J 2 . In particular, U X q J 2 s 1 q J. If U is metabelian, then w U X , U X x s 0 and since J 2 is central in A, this implies that w U X q J 2 , U X q J 2 x s 0. This further implies that w J, J x s w1 q J, 1 q J x s 0. Hence J must be commutative. But Ž g y 1., Ž g y 1. h g J are such that
Ž g y 1. , Ž g y 1. h 2
2
s Ž g y 1. h y Ž g y 1. h Ž g y 1. s Ž g y 1. h y Ž g y 1. Ž g 2 y 1. h 2
2
2
s Ž g y 1. h y Ž g y 1. Ž g q 1. h s Ž g y 1. Ž 1 y g y 1. h 2
s y Ž g y 1 . gh / 0. Hence U is not metabelian.
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We now concentrate on Theorem 2. An ideal U of the associated Lie ring L Ž R . of the ring R is called a Lie ideal of R. Thus a Lie ideal U of R is an additive subgroup of R such that wU, R x : U. Let R be a Lie solvable ring of length n G 3, c s 29 Ž19.10 ny 3 y 1., J be the ideal of R generated by all elements of the type www x, y x, w z, u xx, ¨ x; x, y, z, u, ¨ g R, and I be the derived ideal of R generated by all elements of the type w a , b x; a , b g R. It is proved in w4x that if 3 is invertible in R then J is nilpotent of index at most c and when 2 and 3 both are invertible in R then I is a nil ideal in R. Theorem 2 proves this result without the restriction of invertibility of 3. However, the invertibility of 2 in the context cannot be dropped. We first prove the following: LEMMA 9.
Let U be a Lie ideal of R, then Žg 3 ŽU .. 2 : d Ž2. ŽU . R.
Proof. The following identity holds true for any elements x 1 , x 2 , x 3 , y 1 , y 2 , y 3 g R: 2 w x 1 , x 2 , x 3 xw y 1 , y 2 , y 3 x s w x1 , x 2 x , w y 3 y 2 , y1 , x 3 x q w x1 , x 2 x , w y 3 x 3 , y 2 , y1 x q w x1 , x 2 x , w y 3 y1 , x 3 , y 2 x y w y1 , y 2 x , w x 3 x1 , x 2 , y 3 x y w y1 , y 2 x , w x 3 x 2 , y 3 , x1 x y w y1 , y 2 x , w x 3 y 3 , x1 , x 2 x q w y 3 , x 3 x , w x 2 x1 , y1 , y 2 x q w y 3 , x 3 x , w x 2 y1 , y 2 , x1 x q w y 3 , x 3 x , w x 2 y 2 , x1 , y1 x y w x1 , x 2 x , w y 3 , y1 , x 3 y 2 x q w y1 , y 2 x , w x 3 , x 2 , y 3 x1 x y w y 3 , x 3 x , w x 2 , y 2 , x1 y1 x q w x1 , x 2 , x 3 x , w y 3 , y 2 , y1 x q w y1 , y 2 , y 3 x , w x 2 , x 3 , x1 x q w y 3 , x 3 , y 2 x , w x 2 , x1 , y1 x q w y 3 , x 3 , x1 x , w x 2 , y1 , y 2 x y w x1 , x 2 x , w y 3 , x 3 , y 2 x y1 q w y1 , y 2 x , w x 3 , y 3 , x1 x x 2 y w y 3 , x 3 x , w x 2 , x1 , y1 x y 2 y w x1 , x 2 x , w y 3 , y 2 , y1 x x 3 q x 3 w x1 , x 2 x , w y 3 , y1 , y 2 x q w y1 , y 2 x , w x 3 , x1 , x 2 x y 3 y y 3 w y1 , y 2 x , w x 3 , x 2 , x1 x y w y 3 , x 3 x , w x 2 , y1 , y 2 x x1 q x1 w y 3 , x 3 x , w x 2 , y 2 , y1 x . Now by letting the elements x 1 , x 2 , x 3 , y 1 , y 2 , y 3 g U in the above identity and using the fact that RUR s UR s RU for any Lie ideal U of R, we get that 2Žg 3 ŽU .. 2 : d Ž2. ŽU . R. Also, it follows from the Proof of Lemma 1.7 of w4x that 3Žg 3 ŽU .. 2 : d Ž2. ŽU . R. On combining the two, we get Lemma 9.
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The proof of Theorem 2 follows on the same lines as those in w4x by replacing Lemma 1.7 in w4x by the above Lemma 9. Finally, since for a Lie solvable ring R, the ideal J is nilpotent Žof index at most c ., therefore U Ž RrJ . s Ž U Ž R . q J .rJ, where U Ž A. denotes the group of units for a ring A. In fact, for any positive integer n, we have d Ž n. Ž U Ž RrJ .. s Ž d Ž n. Ž U Ž R .. q J .rJ. Hence U Ž R . is solvable if and only if U Ž RrJ . is solvable. But RrJ is a Lie centrally metabelian ring. Obviously if one knows the solvability length of the group of units of a Lie centrally metabelian ring one then knows a bound for the solvability length of the group of units of an arbitrary Lie solvable ring. Thus their study becomes important.
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