Group algebras with symmetric units satisfying a group identity

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1 manuscripta math. 119, (2006) Springer-Verlag 2005 S.K. Sehgal A. Valenti Group algebras with symmetric units satisfying a group identity Received: 12 July 2005 / Revised version: 24 October 2005 Published online: 28 December 2005 Abstract. We study group algebras FGfor which the symmetric units under the natural involution: g = g 1 satisfy a group identity. For infinite fields F of characteristic 2, a classification of torsion groups G whose symmetric units U + (F G) satisfy a group identity was given in [3] by Giambruno-Sehgal-Valenti. We extend this work to non torsion groups. 1. Introduction Let F be a field, G a group and FGthe group algebra of G over F. Recall that FG has a natural involution induced by the map g g 1 for all g G. Let U(FG) denote the group of units of FG and U + (F G) ={u U(FG) u = u } the set of symmetric units of FG. The study of group algebras whose group of units satisfies a group identity has received significant contributions in the last few years. A conjecture of Hartley claimed that if G is a torsion group and the group of units of the group algebra FG over the field F satisfies a group identity then FG must satisfy a polynomial identity. This conjecture was completely solved in [2] and [3] over an infinite field and in [6] over a finite field and a complete characterization of such group algebras has been given in [9], [7]. Moreover under some mild hypothesis a classification of non torsion group G such that U(FG) satisfies a group identity has been given in [5]. In this paper we study group algebras FG such that the symmetric units satisfy a group identity, in the sense that there is a non trivial word w(x 1,...,x n ) in the free group generated by x 1,...,x n such that w(u 1,...,u n ) = 1 for all u 1,...,u n U + (F G). A study of group algebras whose symmetric units satisfy a group identity was started in [4] and a characterization was found for torsion groups. In this paper we treat the general case of non torsion groups G such that U + (F G) satisfies a group identity when F is an infinite field. A classification is given in Theorem 14. Research supported by NSERC of Canada and MIUR of Italy. S.K. Sehgal: Department of Mathematical and Statistical Sciences, University of Alberta, T6G 2G1 Edmonton, Canada. s.sehgal@ualberta.ca A. Valenti: Dipartimento di Metodi e Modelli Matematici, Università di Palermo, Viale delle Scienze, Palermo, Italy. avalenti@unipa.it DOI: /s

2 244 S.K. Sehgal, A. Valenti 2. Symmetric units of semiprime group algebras Throughout F will be an infinite field of char F = p 0,p 2,Gwill be a group and T = T (G) ={g G o(g) < } its torsion subset. In case char F = p>0, we define P ={g G g pt = 1, for some t 0} the set of p-elements and Q ={g T p o(g)}. For convenience, when p = 0 we set T = Q, P = 1. Also {0} is considered a nilpotent ideal. We recall the following result for semiprime rings proved in Theorem 2, Corollary 2 and Theorem 4 in [4]. Theorem 1. Let C be an infinite commutative domain of characteristic p 0 and let G be any group. If the group algebra CG is semiprime and U + (CG) satisfies a group identity, then 1) every subgroup of Q is normal in G; 2) Q is either abelian or a Hamiltonian 2-group; 3) every symmetric idempotent of CG is central. In the next two lemmas we assume that the group algebra FGis semiprime and U + (F G) satisfies a group identity. We start with the following Lemma 2. Suppose that G is generated by a finite group H and an element x of infinite order. If FH has no nilpotent elements then all idempotents of FH are central in FG. Proof. It is well known that the algebra FH, having no nilpotent elements, is semisimple and so isomorphic to a direct sum of division rings. Hence every idempotent of FH is a sum of primitive idempotents and it is enough to prove the lemma for primitive idempotents. Let e = e 2 be a primitive idempotent. If e = e, then by the previous theorem e is central. Hence we may assume that e 2 = e is not a symmetric idempotent. Since also e is a primitive idempotent, we must have ee = e e = 0. It follows that e + e is a symmetric idempotent and by the above is central in FG.Asa consequence we have e x + e x = (e + e ) x = e + e. Notice that by the previous theorem H is normal in G; hence e x and e x are still primitive idempotents of FH.Now,ife x = e, then e is central in FGand we are done. Hence we may assume that e x = e. This says that xe = e x and, by working with x 1 instead of x, wegetx 1 e = e x 1. From this we also have ex = xe and ex 1 = x 1 e. We now consider the elements s 1 = (x + x 1 )e, s 2 = (x + x 1 )e. It is easily checked that s 1 and s 2 are symmetric elements and s1 2 = s2 2 = 0. Moreover s 1 s 2 = (x +x 1 ) 2 e which implies that (s 1 s 2 ) N = (x +x 1 ) 2N e 0, for all N 1. But by [4, Lemma 2], the product of any square zero symmetric elements must be nilpotent and with this contradiction the proof is complete.

3 Group algebras with symmetric units satisfying a group identity 245 Lemma 3. Suppose that G is generated by a finite Hamiltonian 2-group H and an element x of infinite order. If FG is semiprime, then char F = 0 and every idempotent of FH is central in FG. Proof. Let K 8 = a,b a 4 = b 4 = 1, a 2 = b 2 = z, ab 1 = ba be the quaternion group of order 8. Then it is well known that FK 8 = 4F M 2 (F ) if and only if the equation X 2 + Y 2 = 1can be solved in F (see [10, p. 229]). Suppose that we have M 2 (F ) as a simple component of FK 8. A projection θ : FK 8 M 2 (F ) is given by θ(a) = [ ] α β, θ(b) = β α [ ] where α, β F are such that α 2 + β 2 = 1. Since every finite subgroup is normal in G, the automorphism of K 8 induced by x preserves all subgroups of K 8. Such an automorphism is inner, say it is conjugation by c. Replacing x by xc 1 we can assume that x centralizes K 8. We note that M 2 (F ) is spanned by I, θ(a), θ(b), and θ(ab), so these four matrices are a basis. Moreover, the latter three have trace zero. Thus they span the matrices of trace zero. Since sends each of these three elements to its negative it follows that matrices of trace zero are all skew symmetric (with respect to the involution induced by ). Now,(x x 1 )e 12 and (x x 1 )e 21 are both symmetric of square zero but their product (x x 1 ) 2 e 11 is not nilpotent, contradicting [4, Lemma 2 ]. Since the equation α 2 + β 2 = 1can be solved over any finite field we must have char F = 0 and FK 8 = F F F F D where D is a four dimensional division algebra. Let now e be an idempotent of FH. Since H is finite and, by the above, FH has no nilpotent elements then, by the previous lemma, e is central in FG. Theorem 4. Let F be an infinite field and G a non torsion group. Suppose FG is semiprime. If U + (F G) satisfies a group identity, then 1) if char F = p>2,t is an abelian p subgroup of G; 2) if char F = 0,T is either abelian or a Hamiltonian 2-group; 3) all idempotents of FT are central in FG; 4) if not all idempotents of FT are symmetric then G/T satisfies a group identity. Conversely, if G/T is a u.p. group and G satisfies 1)-4)then U + (F G) satisfies a GI. Proof. Suppose U + (F G) satisfies a group identity. In view of Theorem 1, Q is either abelian or a Hamiltonian 2-group. Hence in case char F = 0, by Lemma2 and Lemma 3 we obtain that 3) holds. Let us assume that char F = p>2, we shall prove that P = 1. Suppose by contradiction that P 1 and let g P,o(g) = p. Let h P, o(h) = p k,k 1. Then ĝ = 1 + g + +g o(g) 1,(h 1)(h 1 1) (F G) +. Since ĝ 2 = 0 and ((h 1)(h 1 1)) pk = 0, by [4, Lemma 3] it follows that ĝ(h 1)(h 1 1)ĝ = 0. This says that ĝhĝ + ĝh 1 ĝ = 0 and, as in the proof of [4, Theorem 5], it follows that g is normalized by P.

4 246 S.K. Sehgal, A. Valenti Let now h T be a p -element. By [4, Theorem 4], h is normal in G, hence H = g, h, the subgroup generated by g and h, is finite. But then by [4, Theorem 3] the p-elements in H form a normal subgroup. It follows that (g, h) = 1, being a p-element in H. Hence we have proved that g is normalized by Q and by P, for any g P,o(g) = p. Let K be the subgroup of G generated by all elements of order p. These elements commute as each of them generates a normal subgroup of P of order p. Thus K is abelian. Let now x P,o(x) = p and y G \ T. We claim that x has only a finite number of conjugates under y. Consider K as a vector space V over GF (p) and y as an automorphism. Then V is a module for the ring R = GF (p)[y,y 1 ]. For x = v V let I be the annihilator of v in R. Then vr R/I. If I 0 then R/I is a finitely generated GF (p) module. Hence vr is finite and v has only a finite number of conjugates under y. Otherwise I = 0 and vr R is free of rank 1. Thus if x did have infinitely many conjugates under y then it follows from the freeness that all groups x yi direct sum. Let X = x. Then α = X(y + y 1 )X is a symmetric square-zero element and X y2 (1 x y2 ) = 0. It follows by [4, Lemma 3] that X y2 X(y + y 1 )X(1 x y2 ) = 0. Since y and y 1 belong to different cosets of K G, wehave and therefore X y2 XyX(1 x y2 ) = 0 X y2 XX y (1 x y3 ) = 0. This is a contradiction to the direct summing of x yi. Thus x has only a finite number of conjugates under y as claimed. Let H = x,y and let x H be the normal closure in H of the subgroup generated by x. Since x H is generated by elements of order p, x H is abelian. Hence, by the above, x H is a finite p-group. It follows that (H, x H ), the augmentation ideal of x H in H, is a nilpotent ideal of FH. The outcome of this is that, for any x P,o(x) = p and for every y G \ T,the element (1 x)y + y 1 (1 x 1 ) lying in (H, x H ) is nilpotent. Since FGis semiprime, x is not normal in G. Hence there exists x 1 P, x 1 x and o(x 1 ) = p. Since X 1 = x 1,(1 x)y+y 1 (1 x 1 ) (F G) + and X 2 1 = 0, by [4, Lemma 3], we get that X 1 ((1 x)y + y 1 (1 x 1 ))X 1 = 0. This implies that X 1 (1 x)x y 1 y = 0 which says that x y 1 x,x 1 = x x 1. By exchanging the role of x and x 1 we also get that x y x,x 1. Hence x,x 1 is a normal finite p-subgroup of G,

5 Group algebras with symmetric units satisfying a group identity 247 contrary to the semiprimeness of FG.Thus P = 1 and 1) holds. By Lemma 2 we obtain that 3) holds. Now, we shall prove 4). If T = K 8 E is a Hamiltonian 2-group, then all idempotents of F(K 8 E), are symmetric and there is nothing to prove. Hence we may assume that T is abelian. Let e = e 1 be a nonsymmetric primitive idempotent of FT.We may clearly assume that T is finite. We can write 1 = e 1 + e 1 + e 2 + as a sum of primitive idempotents of FT. Then, since all idempotents of FT are central in FG, the element u = ge 1 +g 1 e 1 +e 2 + is a symmetric unit, for any g G. Moreover, if U + (F G) satisfies w(u 1,...,u r ) = 1 then an easy calculation shows that w(g 1,...,g r )e 1 = e 1 for g i G. Since e 1 FT, it follows that G/T satisfies w(x 1,...,x r ) = 1 and 4) is proved. Conversely, suppose that G satisfies 1), 2), 3) and all idempotents of FT are symmetric. Then we know by [5, p. 503] that we can write an element u U(FG) as u = (u 1 g 1,u 2 g 2,...,u r g r ) where g i X a transversal of G/T, u i D i where T 1 is a finite subgroup of T and FT 1 = D 1 D r a direct sum of division rings. Then if u U + (F G) we have u = (u 1 g 1,...,u r g r ) = (g 1 1 u 1,...,g 1 r u r ) = u. It follows that gi 2 = u i u g i i D i FT.We conclude that gi 2 T so g i has finite order and hence g i T.Thus u FT.But U + (F T ) is abelian and we are done. In case G satisfies 1), 2), 3) and not all idempotents of FT are symmetric, then T is abelian and by 4) G/T satisfies a group identity, say, w(x 1,...,x k ) = 1. Then for v 1,...,v k U(F G), w(v 1,...,v k ) U(FT)which is abelian. It follows that U(FG) satisfies the group identity (w(t 1,...,t k ), w(v 1,...,v k )) = 1. Lemma 5. Suppose that N is a nil ideal of FG such that N = N and p 2. Then all symmetric units of FG/N lift to symmetric units of FG. Proof. Let us be given a symmetric unit γ F G/N. Lift it to γ U(F G). Then γ = γ(1 + η) with η N. Let µ = 1 2 (γ + γ ) = γ(1 + η/2). Then µ = µ, µ = γ and µ is a unit, proving the lemma. Let N = N(FG) be the sum of all the nilpotent ideals of the group algebra FG. We shall next extend the previous theorem in case N is a nilpotent ideal. Recall that if φ is the FC-subgroup of G then φ + = T φ and φ p = P φ. Also, P = 1 and Q = T if p = 0. We have Theorem 6. Let G be a non torsion group and F an infinite field with charf 2. Suppose that N(FG)is a nilpotent ideal. If U + (F G) satisfies a group identity then

6 248 S.K. Sehgal, A. Valenti 1) T is a subgroup of G; 2) P is a finite subgroup of G; 3) T/P is either an abelian p -group or a Hamiltonian 2-group with p = 0; 4) every idempotent of F(T/P)is central. 5) If not all idempotents of F(T/P) are symmetric then G/T satisfies a group identity. Conversely, if G/T is a u.p. group and G satisfies 1) 5)then U + (F G) satisfies a GI. Proof. Since N is nilpotent, by [8, Theorem ], φ p = φ p (G) is a finite p-group. Also φ p (G/φ p ) = 1 and the FC-subgroup of G/φ p has no p-elements. Then, by [8, Theorem and Theorem ], F(G/φ p ) is semiprime. Also, (G, φ p (G)) nilpotent implies that U + (F (G/φ p )) satisfies a GI. By the previous theorem T(G/φ p ) = T/φ p is either an abelian p -subgroup or a Hamiltonian 2-group and p = 0. In any case, T is a subgroup of G and P = φ p is finite. Since F(G/φ p ) is semiprime, 4) and 5) hold by Theorem 4. The converse follows from Theorem 4 and Lemma Polynomial identities Our aim is to determine conditions on FG so that a group identity on U + (F G) forces FG to satisfy a polynomial identity (PI). We assume in this section that p>2. In case N = N(FG) is not a nilpotent ideal then, as we see below, a group identity on the symmetric units of FGforces FGto satisfy a PI. Lemma 7. Let U + (F G) satisfy a group identity. If N(FG) is a nil not nilpotent ideal then FGsatisfies a PI. Proof. By [4, Remark 2], since N is a nil subring invariant under, we get that N satisfies a PI. Since N is nil not nilpotent, by the argument in [3, Theorem 2] we get that FG satisfies a GPI. Then, by [8, Theorem ], [G : φ] < and φ <. Since [G : φ] < in order to prove that FGsatisfies a PI, it is enough to show that Fφ satisfies a PI. Since φ + is a torsion group and U + (F G) is GI, by [4, Theorem 3], φ p + = φ+ P is a subgroup. This group is infinite as N(FG) is not nilpotent [8, Theorem ]. If G is factored by a finite p-group then N(FḠ) is still nilpotent and φ p + (Ḡ) is infinite. Also, since (φ+ p ) φ is finite, then (φ p + ) is a finite p-group. By factoring by (φ p + ) we may assume that φ p + is abelian. Now, since φ is a finite group, again by [4, Theorem 3], we get that φ p is a subgroup. Hence by factoring by φ p, we may assume that φ is a finite p -group. We have (φ, φ p + ) φ φ p + = 1. Thus φ+ p ζ(φ), the center of φ. Let P 1 = φ p +. If P 1 is finite then by [8, Theorem ], we get that N(FG) is nilpotent contrary to our assumption. Thus P 1 is infinite. Now, by [4, Remark 2], N(Fφ) satisfies a polynomial identity. Let f(x 1,...,x n ) be such a PI which we may assume to be multilinear. As in the proof of [5, Proposition 4.4], it follows that Fφ satisfies f and we are done.

7 Group algebras with symmetric units satisfying a group identity 249 Lemma 8. Let U + (F G) satisfy a group identity. If I = I is a nil ideal of FGthen I satisfies the - polynomial identities [x 1 +x 1,x 2] n 0 and ([x 1 +x 1,x 2]x 3 ) n 0 for some n 0. Proof. Let F {x 1,x1,x 2,x2 } be the free algebra with involution on x 1 and x 2 and consider the algebra F {x 1,x1,x 2,x2 }[[t]] of formal series in the variable t. The elements (1 + x 1 t)(1 + x1 t)and (1 + x 2t)(1 + x2 t)generate a free group in U(R), the unit group of R. Hence, if w = w(x 1,x 2 ) is the group identity satisfied by U + (F G), w((1 + x 1 t)(1 + x1 t),(1 + x 2t)(1 + x2 t)) 1. We can write w((1 + x 1 t)(1 + x 1 t),(1 + x 2t)(1 + x 2 t) = 1 + i 0 f i (x 1,x 1,x 2,x 2 )ti (1) where f i (x 1,x1,x 2,x2 ) are homogeneous polynomials in x 1,x1,x 2,x2. Moreover by evaluating x i into nilpotent elements of I and t to λ F, as in the proof of [5, Lemma 2.2] or [4, Remark 2], we get that for all i 0,f i (x 1,x1,x 2,x2 ) is a *-PI for I. We first claim that there exists f l (x 1,x1,x 2,x2 ) which is not a *-PI for M 2(F ), the algebra of 2 2 matrices over F with transpose involution. In fact, suppose to the contrary that all f i vanish in M 2 (F ). Then if we evaluate x 1 = x2 = a where a M 2 (F ) is such that a 2 = 0, then as in the proof of [4, Lemma 2], we get that there exists f io such that f io (a, a,a,a)= αa ɛ (a a) l (a ) µ where ɛ, µ {0, 1},l 1 and α F,α 0. But then f io (e 12,e 21,e 21,e 12 ) = αe12 ɛ el 22 eµ 21 0 a contradiction. We next claim that there exists f m (x 1,x1,x 2,x2 ) which is not a *-PI for M 4(F ), the algebra of 4 4 matrices over F with symplectic involution. In fact, as above we get down to the evaluation of a polynomial f io and we get f io (e 12 + e 34,e 43 + e 21,e 43 + e 21,e 12 + e 34 ) = α(e 12 + e 34 ) ɛ (e 44 + e 22 ) l (e 43 + e 21 ) µ 0. The claim is proved also in this case. We finally claim that there exist f n (x 1,x1,x 2,x2 ) such that the sum of all monomials of f n where no * appears is not a polynomial identity for M 2 (F ). In order to prove the claim we first fix the notation: for a polynomial f i (x 1,x1, x 2,x2 ), we let f i (x 1,x 2 ) be the sum of all monomials where only x 1 and x 2 appear (and no x1 and x 2 ). So, suppose to the contrary that for all i 0, f i (x 1,x 2 ) isapi for M 2 (F ). Then in (1) we evaluate x 1 = a,x 2 = b, where a,b I,a 2 = b 2 = 0 and t = λ F. We get w((1 + aλ)(1 + a λ), (1 + bλ)(1 + b λ)) = ((1 + aλ)(1 + a λ)) i 1 ((1 + bλ)(1 + b λ)) j1 =1 and there exists i 0 such that f i0 (a, b) = αa ɛ (ba) l b µ = 0 with α 2Z and ɛ, µ {0, 1}. But f i0 (e 12,e 21 ) = αe12 ɛ (e 22) l e µ 21 0, a contradiction.this proves the claim.

8 250 S.K. Sehgal, A. Valenti Set B = F {x 1,x1,x 2,x2 }/ f l,f m,f n T where f l,f m,f n T is the T -ideal of the free algebra F {x 1,x1,x 2,x2 } generated by f l,f m,f n. Thus B is a finitely generated PI-algebra with involution satisfying the -PIs f l,f m,f n. By Amitsur s theorem [1] then B satisfies an ordinary identity. Let P be a prime ideal of B and suppose first that P P. Then B/P has an induced involution and f l,f m,f n are *-PIs for B/P. Since B/P is prime, it satisfies all the *-PIs of n n matrices with either transpose or symplectic involution for some n 1 (see [11, Corollary ]). Since f l and f m do not vanish on M 2 (F ) with =transpose or on M 4 (F ) with =symplectic respectively, it follows that B/P satisfies all *-PIs of M 2 (F ), symplectic. Thus [x 1 + x1,x 2]is a *-PI for B/P. Suppose now that P P. Consider the ideal (P + P )/P of B/P. We can choose representatives of the elements of (P + P )/P to be in P. It follows that when evaluating f n on (P + P )/P we are actually evaluating x1 = x 2 = 0. Hence it follows that (P + P )/P satisfies the identity f n (x 1,x 2 ) obtained from f n by summing all monomials not involving x1 and x 2. Since f n (x 1,x 2 ) is not a PI for M 2 (F ), it follows that (P + P )/P, must be commutative. But B/P is a prime ring, hence also B/P is commutative and [x 1,x 2 ]isapiforb/p. In conclusion we have that for all a,b B,[a + a,b] Pprime P = L, where L is the lower nil radical of B. Since B is PI and finitely generated, by [12, Th ], L is nilpotent. Thus [x 1 + x1,x 2] k 0 and ([x 1 + x1,x 2]x 3 ) k 0 are *-PI for B for some k 1. It follows that they are also -PIs for I and the proof is complete. Lemma 9. If U + (F G) satisfies a group identity then P and T are subgroups of G and (P ) is locally nilpotent. Proof. If N(FG) is nilpotent, then by Theorem 6, we know that P and T are subgroups and P is a finite group and we get the desired conclusion. Hence we may assume that N(FG) is nil not nilpotent. But then by Lemma 7, FG is PI. Let g 1,...,g n P and let H = g 1,...,g n. Then FH is a finitely generated PI-algebra. By [12, Theorem ] it follows that J(FH), the Jacobson radical of FH is nilpotent. Thus N(FH) is nilpotent and U + (F G) is GI. By Theorem 6 then the p-elements of H form a finite subgroup. Hence H is finite and so, P is locally finite and (P ) is locally nilpotent. Since G/P has no p-elements, F(G/P) is semiprime and, by Theorem 4, T(G/P)is an abelian p -group. It follows that T = T (G) is a subgroup of G. Lemma 10. Suppose that N(FG) is nil not nilpotent and U + (F G) satisfies a GI. Then (G, P ) is a locally nilpotent ideal satisfying the polynomial identities [x 1 +x 1,x 2] pt 0 and ([x 1 +x 1,x 2]x 3 ) pt 0 for some t 0. Moreover (G, P ) is nil of bounded exponent if and only if P is of bounded exponent. Proof. As in the proof of [5, Lemma 5.1] it follows that (G, P ) is a locally nilpotent ideal and that (G, P ) is nil of bounded exponent if P is of bounded exponent.

9 Group algebras with symmetric units satisfying a group identity 251 Since (G, P ) is -invariant, by Lemma 8, (G, P ) satisfies the polynomial identities [x 1 + x1,x 2] pt 0 and ([x 1 + x1,x 2]x 3 ) pt 0 for some t 0. The converse of the second statement is trivial. 4. Classification theorems In this section we will give the characterization of groups G such that U + (F G) satisfies a group identity when char F 2. Lemma 11. Let N(FG) be nil not nilpotent and suppose that U + (F G) satisfies a group identity. If A is a normal abelian subgroup of G of finite index, then (G, A P) is of bounded p-power exponent. Proof. Let a A P and g G. Then (a 1)g + g 1 (a 1 1) (G, A P) + and (a 1)g (G, A P).Moreover, [(a 1)g + g 1 (a 1 1), (a 1)g] = [g 1 (a 1 1), (a 1)g] = g 1 (a 1 1)(a 1)g (a 1)(a 1 1) = ((a 1 ) g 1)(a g 1) (a 1)((a 1 1) = a + a 1 a g (a g ) 1. Since [x+x,y] pt is a *-PI for (G, A P),we get a pt +a pt (a pt ) g (a pt ) g = 0 and so a pt + a pt = (a pt ) g + (a pt ) g. If a pt = (a pt ) g then (g, a) pt = 1 and we are done. Otherwise a pt = (a pt ) g which says that ga pt = a pt g and we consider this case below. Let a pt = b. Then gb = b 1 g and [(b 1)g + g 1 (b 1 1), g(b 1)] = (b 1) 2 g 2 (b 1 1) 2 g 2 + (b 1) 2 (b 1 1) 2 = ((b 1) 2 (b 1 1) 2 )(g 2 + 1) = ((b 1) 2 b 2 (b 1) 2 )(g 2 + 1) = (1 b 2 )(b 1) 2 (g 2 + 1) = b 2 (b 2 1)(b 1) 2 (g 2 + 1) = b 2 (b + 1)(b 1) 3 (g 2 + 1). Since (G, A P)satisfies the *-PI [x + x,y] pt = 0 and g 2 commutes with b, we get b 2pt (b + 1) pt (b 1) pt+1 (g 2 + 1) pt = 0. Since b is a p-element and charf 2,(b+ 1) is not a zero divisor; hence we get (b pt+1 1)(g 2pt + 1) = 0. So either b pt+1 = 1org 2pt = b pt+1 and b pt+1 g 2pt = 1. Thus b pt+1 = 1 in any case since b P. Since A is normal in G, weget that (g, a) pt p t+1 = 1. We have proved that (G, A P) is of bounded p-power exponent.

10 252 S.K. Sehgal, A. Valenti We are indebted to Professor D.S. Passman for the proof of the next lemma. Lemma 12. Suppose that N(FG)is nil not nilpotent and U + (F G) satisfies a group identity. If P is of unbounded exponent then G is of bounded p-power exponent. Proof. We know, by Lemma 7, that FG satisfies a polynomial identity. Let A be a normal p-abelian subgroup of G of finite index. By taking the quotient with the finite p-subgroup A, we may assume that A is abelian. Also, since by the previous lemma, (G, A P)is of bounded p-power exponent, by factoring with (G, A P) we may assume that A P is central in G. Since P is of unbounded exponent and [P : A P ] <, then also A P is of unbounded exponent. Notice that P is central by-finite. Hence by a theorem of Schur [13, p. 39] P is a finite group. By taking the quotient with P we may assume that P is abelian. Take g, h G and t A P ζ(g). Then (t t 1 )(g g 1 ) (G, P ) + and the elements h(t t 1 ), (t t 1 ) p 2 g 1 h 1 (G, P ). It follows, by Lemma 10, that Hence then and so ([(t t 1 )(g g 1 ), h(t t 1 )]g 1 h 1 (t t 1 ) p 2 ) pn = 0. ((t t 1 ) pn+1 ([g g 1,h]g 1 h 1 ) pn = 0 (t 2 1) pn+1 ([g g 1,h]g 1 h 1 ) pn = 0 (t 2pn+1 1)([g g 1,h]g 1 h 1 ) pn = 0. Since there are infinitely many t 2pn+1, we get that ([g g 1,h]g 1 h 1 ) pn = 0. This say that the element (g, h) g 1 hg 1 h hg 2 h 1 is nilpotent of exponent p n. Also, replacing g with tg we obtain that (g, h) t 2 g 1 hg 1 h t 2 hg 2 h 1 is nilpotent of exponent p n, for all t A P. Hence implies that ((g, h) t 2 g 1 hg 1 h t 2 hg 2 h 1 ) pn = 0 (g, h) pn t 2pn (g 1 hg 1 h 1 ) pn 1 + t 2pn (hg 2 h 1 ) pn 0 mod[fg,fg]. Notice that 1 supp(α), for all α [FG,FG]. Hence one of the remaining terms above must be 1. But we can choose t in such a way that t 2pn (g 1 hg 1 h 1 ) pn 1 and t 2pn (hg 2 h 1 ) pn 1. Therefore (g, h) pn = 1, for all g, h G. Thus (g, h) P and is of bounded p-power exponent. Since P is abelian, G is of bounded p-power exponent.

11 Group algebras with symmetric units satisfying a group identity 253 Lemma 13. Suppose that U + (F G) satisfies a group identity. Then N is nilpotent if and only if P is finite. Proof. Suppose N is nilpotent. By Theorem 6, if U + (F G) satisfies a group identity then P is finite. Conversely if P is finite then (G, P ) = F G( P ) is nilpotent. Since G/P has no p-elements then F(G/P) has no nilpotent ideals and so N = (G, P ) is nilpotent. The next theorem gives a complete classification of groups G such that U + (F G) satisfies a group identity if F is infinite and G has an element of infinite order. For the sufficiency in c) we assume that G/T is a u.p. group. If G is torsion the problem has been solved in [4]. Theorem 14. Suppose that F is infinite, charf 2, and G has an element of infinite order. We have the following a) If U + (F G) satisfies a group identity then P is a subgroup. b) If P is of unbounded exponent and U + (F G) satisfies a group identity then 1) G contains a p-abelian subgroup of finite index. 2) G is of bounded p-power exponent. Conversely, if P is a subgroup and G satisfies 1) and 2) then U + (F G) satisfies a group identity. c) If P is of bounded exponent and U + (F G) satisfies a group identity then 0 )Pis finite or G has a p-abelian subgroup of finite index. 1 ) T (G/P ) is an abelian subgroup or a Hamiltonian 2-group with p = 0 and so T is a group. 2 ) Every idempotent of F(T/P)is central. 3 ) If not all idempotents of F(T/P)are symmetric then G/T satisfies a group identity. Conversely, if P is a subgroup of bounded exponent, G satisfies 0 ), 1 ), and 2 ) and G/T is a u.p. group then U + (F G) satisfies a group identity. Proof. a) It follows from Lemma 9. b) Suppose that U + (F G) satisfies a group identity. Then since P is infinite it follows by the last lemma that N is nil, not nilpotent. Hence 1) and 2) follow from Lemma 7 and Lemma 12 respectively. The converse is (b) of [5, Theorem 5.5]. c) Suppose that U + (F G) satisfies a group identity and that P is a group of bounded exponent. If P is not finite, by Lemma 7, FG satisfies a polynomial identity thus we have 0 ). Moreover, we conclude by Lemma 10 that (G, P ) is nil of bounded exponent. Clearly, if P is finite then (G, P ) is nilpotent. Thus, in any case, (G, P ) is nil of bounded exponent. So U(F(G/P)) is GI. Further, F(G/P) is semiprime as G has no p-elements. Then 1 ), 2 ) and 3 ) follow from Theorem 4. The converse follows from Theorem 4, Lemma 5 and Lemma 10.

12 254 S.K. Sehgal, A. Valenti Remark 15. In the converse part of c) we have assumed that G/T is a u.p. group. In fact, all we need is that the trivial unit conjecture holds for G/T, namely, for a crossed product D G/T where D is a division ring the unit group is trivial i.e. it equals D G/T. Remark 16. Comparing Theorem 14 and [5, Theorem 5.5] we see that we have proved if char F 2 and K 8 G then U + (F G) satisfies a GI U(FG) satisfies a GI, provided G mod torsion is a u.p. group. Acknowledgements. We would like to thank the referee for his suggestions to improve our presentation. References [1] Amitsur, S.A.: Identities in rings with involution. Israel J. Math 7, (1969) [2] Giambruno, A., Jespers, E., Valenti, A.: Group identities on units of rings. Arch. Math. 63, (1994) [3] Giambruno, A., Sehgal, S.K., Valenti, A.: Group algebras whose units satisfy a group identity. Proc. Am. Math. Soc. 125, (1997) [4] Giambruno, A., Sehgal, S.K., Valenti, A.: Symmetric units and group identities. Manuscripta Mathematica 96, (1998) [5] Giambruno, A., Sehgal, S.K., Valenti, A.: Group identities on units of group algebras. J. Algebra 226, (2000) [6] Liu, C.-H.: Group algebras with units satisfying a group identity. Proc. Am. Math. Soc. 127, (1999) [7] Liu, C.-H., Passman, D.S.: Group algebras with units satisfying a group identity II. Proc. Am. Math. Soc. 127, (1999) [8] Passman, D.S.: The Algebraic Structure of Group Rings. John Wiley, NewYork, 1977 [9] Passman, D.S.: Group algebras whose units satisfy a group identity II. Proc. Am. Math. Soc. 125, (1997) [10] Polcino Milies, C., Sehgal, S.K.: An Introduction to Group Rings. Kluwer, Dodrecht, 2002 [11] Rowen, L.H.: Polynomial Identities in Ring Theory. Academic Press, New York, 1980 [12] Rowen, L.H.: Ring Theory vol. II. Academic Press, New York, 1988 [13] Sehgal, S.K.: Topics in Group Rings. Marcel Dekker, New York, 1978