JOURNAL OF ALGEBRA ARTICLE NO.
184, 207]212 Ž1996.
0256
Zassenhaus Conjecture and Infinite Nilpotent Groups Zbigniew S. Marciniak* Institute of Mathematics, Warsaw Uni¨ ersity, ul. Banacha 2, 02-097 Warsaw, Poland
and Sudarshan K. Sehgal Department of Mathematical Sciences, Uni¨ ersity of Alberta, Edmonton, Alberta, T6G 2G1 Canada Communicated by Walter Feit Received October 16, 1995
Zassenhaus conjectured that any torsion unit in an integral group ring ZG of a finite group G is rationally conjugate to "g for some g g G. This has been confirmed for all finite nilpotent groups. We prove that such a result does not hold for infinite nilpotent groups. Q 1996 Academic Press, Inc.
INTRODUCTION There is a famous conjecture of Zassenhaus that a torsion unit in an integral group ring ZG of a finite group G is rationally conjugate to "g for some g g G. This conjecture has been confirmed by Weiss for finite nilpotent groups w9, 10x. So far no counterexample has been found. It has been proved that the Zassenhaus Conjecture holds for a finite group G if and only if G is a UT group. The group g is said to be a UT group if for every torsion unit in ZG there is a unique conjugacy class for which the associated Bass trace function does not vanish. For details see w2x and w8, Section 41x. *E-mail:
[email protected]. † E-mail:
[email protected]. Supported by Canadian NSERC Grant A-5300 and Polish Scientific Grant 2P30101007. 207 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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It is easy to see that if an arbitrary group G has the property that every torsion unit is conjugate to a trivial unit then G is a UT group. It is also known that arbitrary nilpotent groups w5x and some polycyclic-by-finite groups w6x have the UT property. Therefore it makes sense to ask about an extension of the Zassenhaus Conjecture for those groups. In fact, for some infinite groups it is possible to conjugate rationally all torsion units to trivial units; see w2, Prop. 12; 4x. The object of this paper is to show that the exact analog of the Zassenhaus Conjecture does not hold for all infinite nilpotent groups. In the proof we apply the main idea of w3x which relates idempotents in K w G = H x to projective KH-modules. It seems that the most we can expect for infinite groups is the possibility of stable conjugation of torsion units to trivial units. This has been done for all nilpotent groups w5, Theorem 7.2x. In fact, one can prove a stronger result: finite p-groups of normalized units can be simultaneously stably conjugated to group elements w5, Prop. 7.1x.
EXAMPLE Let s : Z [ Z ª Z [ Z be the automorphism defined by s Ž x, y . s Ž x q y, y .. The semi-direct product H s ŽZ [ Z. i s Z is a nilpotent group, as w H, H x s Z Ž H . s ²Ž1, 0.:. Choose an element t g H whose coset generates the quotient group HrZ [ Z. We may assume that conjugation by t coincides with s on Z [ Z - H. We will also need the elements x i s Ž i, 1. g Z [ Z - H for i s 0, 1, 2, 3. Note that x i s t i x 0 tyi . Let us define four elements of the group ring Z H:
a s 2 ty2 x 1 x 2 ,
g s wty2 x 0 x 2 q ty1 x 0 ,
b s 2 ty2 x 0 x 3 ,
d s 2 ty2 x 1 x 3 y ty1 x 2 ,
where x i s x i y 1 for i s 0, 1, 2, 3. Finally, let D 8 s ² a, b N a4 s 1, b 2 s 1, baby1 s ay1 : be the dihedral group. Our example will live in ZG, where G s H = D 8 . THEOREM. The element u s b q wŽ a y b . q Ž a q b . b q Žg q d . a q Žg y d . baxŽ a2 y 1. of the integral group ring ZG has the following properties: Ži. u 2 s 1, i.e., u is a torsion unit; Žii. u is not conjugate in KG to a group element for any field K of characteristic zero;
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0 . 0 . Žiii. the matrices Ž u0 y1 and Ž 0b y1 are conjugate in the linear group GL2 ŽQG ., i.e., u is stably conjugate to b g G in the sense of w5x.
Proof. We select a Wedderburn isomorphism f
Q D 8r² a2 : [ M2 Ž Q .
6
f : Q D8
as follows. For any z g Q D 8 we set f Ž z . s Ž f 1Ž z ., f 2 Ž z .. where f 1 is induced by the natural epimorphism D 8 ª D 8r² a2 : and f 2 is the linear representation
f 2 Ž a. s
ž
0 1
y1 , 0
f2 Ž b. s
/
ž
1 0
0 . y1
/
Let K be any field of characteristic zero and let R denote the group algebra KH. Then we have an isomorphism f
K H = D 8r² a2 : [ M2 Ž R . .
6
f s id R m f : KG f R m Q D 8
It is clear that f Ž u. s Ž b, U . where b is the coset of b in H = D 8r² a2 : and U is some matrix from M2 Ž R .. Now our task translates as follows. To prove Ži. it is enough to show that 2 U s I, the identity matrix. Because all elements of finite order in G belong to D 8 , their images in M2 Ž R . lie in the subring M2 ŽQ. of rational matrices. Therefore, to prove Žii. it is sufficient to show that U is not conjugate in M2 Ž R . to any rational matrix. Finally, to prove Žiii. we must only show that U [ y I is conjugate to f 2 Ž b . [ y I in M4 ŽQ H .. A straightforward calculation shows that U s f 2 Ž u. s
ž
1 y 4a y4d
4g . y1 q 4b
/
From now on it will be easier to work with the matrix E s Ž I q U . r2 s
ž
1 y 2a y2 d
2g g M2 Ž Z H . . 2b
/
Let us substitute yi s 2x i . Then Es
s
ž ž
1 y ty2 y1 y2
ty2 y0 y2 q ty1 y0
ty1 y2 y ty2 y1 y3
ty2 y0 y3
Ž y0 q t . ty2 Ž t y y1 . y1 ty2 Ž t y y1 .
Ž y0 q t . ty2 y0
/
y1 ty2 y0
where V s Žy0 q t, y1 . and W s Ž ty2 Ž t y y1 ., ty2 y0 ..
/
sVT?W
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Observe now that W ? V T s ty2 Ž t y y1 .Žy0 q t . q ty2 y0 y1 s 1. It follows that E 2 s Ž V T W . Ž V T W . s V T Ž WV T . W s E, i.e., E is an idempotent matrix. Therefore U 2 s Ž2 E y I . 2 s 4 E 2 y 4 E q I s I and part Ži. of our theorem is proved. For the proof of part Žii. consider P s kerŽ?E .: R [ R ª R [ R. Clearly P is a left projective R-submodule of R [ R generated by the rows of the matrix I y E. To simplify things, we shall identify P with a left ideal in R. To this end, we show that q1: R [ R ª R, the projection on the first coordinate, is injective on P. For suppose Ž0, c . g P l kerŽ q1 .. Then c ? y1 s Ž0, c . ? V T s Ž0, c . ? E ? V T s Ž0, 0. ? V T s 0. As the algebra R has no zero divisors Žsee w7, p. 592x. and y1 / 0 it follows that c s 0. The left ideal J s q1Ž P . is generated by the elements of the first column of I y E, i.e., by the set ty2 y1 y2 , ty1 y2 y ty2 y1 y34 . Hence P f J s R ? y1 y 2 q R ? Ž t y 2 y y1 y 3 . . Recall that to prove part Žii. it suffices to show that the matrix U, or equivalently E, is not conjugate in M2 Ž R . to a rational matrix. So let us suppose that it is the case: E ; F g M2 ŽQ.. Clearly F is also an idempotent matrix. Moreover, we have trŽTrŽ F .. s Ž tr TrŽ E .. s 1 where Tr denotes the matrix trace and tr catches the coefficient at 1 g H. Therefore F ; diagŽ1, 0. and we may assume that F is already in this form. Let y g M2 Ž R . be such that E s Y FYy1 . Then Y maps isomorphically kerŽ E . s P onto kerŽ F . s 0 [ R and hence P f R. Because R is a domain it follows that J is a left principal ideal of R. Note that R s Ss w t, ty1 x, where S s K wZ [ Zx. In particular, each element of R can be written uniquely as Ýt i si for some si g S. Suppose now that the ideal J is principal, i.e., Ry1 y2 q RŽ t y2 y y1 y3 . s R j for some j g R. To reach the contradiction we use Artamonov’s argument from w1, Lemma 2x. As y1 y2 g R j , it must be of the form r ? j for some r g R. But y1 y2 belongs to S and so both r and j must be monomials of the form t k sk . Therefore, without loss of generality, we may assume that j g S. Also t y2 y y1 y3 g R j , so for some Ýt i si g R we have t y2 y y1 y3 s Ýt i Ž si j . with all si j g Sj . From the uniqueness of polynomial coefficients it follows that both y2 and y1 y3 belong to the ideal Sj . Hence S y2 q S y1 y3 : Sj . On the other hand, j g J and hence j s r 9y1 y2 q r 0 Ž t y2 y y1 y3 . for some polynomials r 9 s Ýt i sXi and r 0 s Ýt i sYi . Comparing free terms on
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both sides, we obtain
j s sX0 ? y1 y2 q sy1 Ž sYy1 . y2 y sY0 y1 y3 : S y2 q S y1 y3 . In consequence, S y 2 q S y1 y 3 s S j . Recall that S is a K-algebra of Laurent polynomials in two variables and hence it is a unique factorization domain. As all elements yi lie in the augmentation ideal, j is not invertible in S. Let p be a prime factor of j . From y2 g Sj and y1 y3 g Sj it follows that p N y2 and either p N y1 or p N y3 . In any case we have the following: for two linearly independent elements z1 , z 2 g Z [ Z, the group ring elements z1 y 1, z 2 y 1 have a common prime factor. We can choose a free basis ¨ , w4 for Z [ Z so that z1 s ¨ m and z 2 s w n. Then the polynomials ¨ m y 1 and w n y 1 will have a common prime factor in the ring K w ¨ , ¨ y1 , w, wy1 x, a contradiction. This completes the proof of part Žii.. Finally, to prove part Žiii. it is enough to show that the matrices E9 s
ž
E 0
0 , 0
/
F9 s
ž
1 2
Ž I q f2 Ž b. . 0
0 s diag Ž 1, 0, 0, 0 . 0
/
are conjugate in M4ŽQ H .. Let us write T s Q H for short. Recall that E s V T ? W and W ? V T s 1. Therefore imŽ?E . f T. Let P s kerŽ?E . ; T [ T. Then we have an isomorphism m : T [ T f kerŽ E . [ imŽ E . s P [ T. We have two similar decompositions T 4 f im Ž E9 . [ ker Ž E9 . s T [ Ž P [ T 2 . , T 4 f im Ž F9 . [ ker Ž F9 . s T [ T 3 . Let Y g M4Ž T . be the matrix of the isomorphism id[ m y1 [id
T [ T 2 [ T.
6
T[ ŽP[T. [T
It is easy to check that for any ¨ g T 4 holds ¨ ? E9Y s ¨ ? Y F9, i.e., E9 s Y F9Yy1 , as desired.
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3. D. R. Farkas and Z. S. Marciniak, Idempotents in group rings: A surprise, J. Algebra 81, No. 1 Ž1983., 266]267. 4. F. Levin and S. K. Sahgal, Units in the integral group ring of the infinite dihedral group, in ‘‘Proceedings of the AMS Special Session, Series in Algebra’’ Vol. 1, pp. 57]68, World Scientific, Singapore 1993. 5. Z. S. Marciniak and S. K. Sehgal, Finite matrix groups over nilpotent group rings, J. Algebra, 181 Ž1996., 565]583. 6. Z. S. Marciniak and S. K. Sehgal, Hirsch induction and torsion units in group rings, Arch. Math. 64 Ž1995., 374]384. 7. D. S. Passman, ‘‘Algebraic Structure of Group Rings,’’ Interscience, New York, 1977. 8. S. K. Sehgal, ‘‘Units in Integral Group Rings,’’ Longman’s, Essex, 1993. 9. A. Weiss, Rigidity of p-adic torsion, Ann. of Math. 127 Ž1988., 317]332. 10. A. Weiss, Torsion units in integral group rings, J. Reine Angew. Math. 415 Ž1991., 175]187.