TWISTED GROUP ALGEBRAS SATISFYING A

GENERALIZED POLYNOMIAL IDENTITY

D. S. PASSMAN

Dedicated to the memory of Professor A. I. Kostrikin

Abstract. If Kt[G] is a twisted group algebra satisfying a nondegeneratemultilinear generalized polynomial identity f(ζ1, ζ2, . . . , ζn) = 0, then we showthat G has certain normal subgroups of finite index which can be viewed asbeing almost central. For example, there exists H / G with |G : H|·|H ′|bounded by a fixed function of the support sizes of the nonzero Kt[G]-termsinvolved in f . Indeed, we obtain a more precise version of this result, withthe structure of H depending upon the specific twisting in the group algebra.We then go on to determine necessary and sufficient conditions for Kt[G] tosatisfy such an identity.

1. Introduction

Let Kt[G] denote a twisted group algebra of the group G over the field K. ThenKt[G] is a K-vector space with basis G, a copy of G. In particular, every elementα ∈ Kt[G] can be written uniquely as α =

∑x∈G kxx̄, a finite K-linear combination

of basis elements. Here kx ∈ K, and we use suppα = {x ∈ G | kx 6= 0} to denotethe support of α, a finite subset of G. In addition, Kt[G] is a K-algebra and itsmultiplication is defined distributively by

x̄ȳ = t(x, y)xy for all x, y ∈ G,

where t : G×G→ K• is a 2-cocycle, namely a function whose structure is preciselyequivalent to the associativity of the algebra. Without loss of generality, we canassume that 1̄ = 1 is the identity element of Kt[G]. Of course, Kt[G] = K[G] is theordinary group algebra of G over K when the twisting function t(x, y) is identicallyequal to 1.

Now let R be any K-algebra. Then a multilinear generalized polynomial over Ris a function f(ζ1, ζ2, . . . , ζn) of n noncommuting variables having the form

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j .

Here Symn is the symmetric group of degree n, and each αi,σ,j is an element of thering R. Obviously, f is linear in each ζi, and we say that R satisfies the generalizedpolynomial identity (g.p.i.) f if

f(r1, r2, . . . , rn) = 0 for all r1, r2, . . . , rn ∈ R.

2000 Mathematics Subject Classification. 16S34,16R50.The author’s research was supported in part by NSF Grant DMS-9820271.

1

2 D. S. PASSMAN

Unfortunately, this condition may say nothing about R since, for example, we couldhave all αi,σ,j = 0. Indeed, even if these terms are nonzero, the polynomial mightstill be of little interest. As with ordinary polynomial identities, we would like ageneralized identity to be indicative of some version of commutativity. To this end,write f =

∑σ∈Sym

n

fσ, where

fσ(ζ1, ζ2, . . . , ζn) =

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j .

Then, we say that f is nondegenerate if there exists some τ ∈ Symn with fτ notan identity for R. In other words, f is nondegerate if there exists τ ∈ Symn andsuitable elements s1, s2, . . . , sn ∈ R with fτ (s1, s2, . . . , sn) 6= 0. It follows that if Rsatisfies a nondegenerate multilinear polynomial identity, then at least two distinctpermutations must come into play nontrivially in f . By relabeling the variables,if necessary, we can always assume that f1 is the nontrivial component of f . Notethat an ordinary multilinear polynomial is a function of the form

g(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

ασζσ(1)ζσ(2) · · · ζσ(n)

with each ασ ∈ K, and g is clearly nondegenerate if and only if ατ 6= 0 for some τ .Group algebras satisfying an ordinary polynomial identity (p.i.) were classified

in [2], [9] and [6], while group algebras satisfying a nondegenerate generalized poly-nomial identity were handled in [7]. The latter paper also studied p.i. twisted groupalgebras. In recent years, there has been renewed interest in the g.p.i. result becauseit turns out to be a key ingredient in the solution of Hartley’s problem. Specifically,it was shown in [1] (and then in [3] for finite fields) that if G is periodic and if theunit group of K[G] satisfies a group identity, then K[G] must satisfy a polynomialidentity. Furthermore, [4] consided Hartley’s problem for twisted group algebrasand, in the course of that work, obtained basic properties of twisted group algebrassatisfying a g.p.i. In this paper, we sharpen the latter result and prove

Theorem 1.1. Let Kt[G] be a twisted group algebra of G over K which satisfiesthe nondegenerate multilinear generalized polynomial identity

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

with each αi,σ,j in Kt[G]. If

s =∑

σ∈Symn

aσ∑

j=1

n∏

i=1

| suppαi,σ,j |,

then G has a normal subgroup H with |G : H |·|H ′| bounded by a function of s.

Indeed, we obtain a more precise version of this theorem, with the structure of Hdepending upon the particular twisting in Kt[G]. Results of this nature are provedby a coset counting argument, discovered independently in [5] and [9], and knownas the Delta Method. As will be apparent, we push this method to the limit andbeyond in the following two sections. Specifically, Section 2 is concerned with linearidentities in twisted group algebras, namely identities of the form

α1x̄β1 + α2x̄β2 + · · · + αnx̄βn = 0 for all x ∈ T,

TWISTED GROUP ALGEBRAS 3

where T is a “large” subset of G. Section 3 then builds upon this to prove the maintheorems on generalized polynomial identities.

At this point, it is appropriate to recall a number of basic definitions, but in aslightly more general context. To start with, for any groups G ⊆ G, we let

D = DG(G) = {x ∈ G∣∣ |G : CG(x)|

4 D. S. PASSMAN

map. When G = G, we write θ, θt, θn, and θtn for φ, φ

t, φn, and φtn, respectively.

Additional notation will be introduced in the course of the proof.

2. Linear Identities

Let G be an arbitrary group and let T be a subset of G. The (right) index |G : T |is defined to be the smallest positive integer n such that

G = Tx1 ∪ Tx2 ∪ · · · ∪ Txn

for suitable x1, x2, . . . , xn ∈ G. Of course, if no such elements exist, then we set|G : T | = ∞. Note that this definition agrees with the usual index when T is asubgroup of G.

Lemma 2.1. Let S and T be subsets of the group G.

(i) If x ∈ G, then |G : xT | = |G : T |.(ii) Suppose S = Tx1 ∪ Tx2 ∪ · · · ∪ Txr with x1, x2, . . . , xr ∈ G. If |G : S| ≤ s,

then |G : T | ≤ rs.(iii) Assume that S = H1x1∪H2x2∪· · ·∪Hrxr where each Hixi is a coset of some

subgroup Hi of G. If G = S ∪ T and S 6= G, then |G : T | ≤ (r + 1)!.(iv) If G = H1x1 ∪H2x2 ∪ · · · ∪Hrxr where each Hixi is a coset of some subgroup

Hi of G, then |G : Hj | ≤ r for some subscript j.

Proof. Parts (i) and (ii) are obvious. Parts (iii) and (iv) follow from [8, Lemmas5.2.1 and 5.2.2]. �

Next, let Hom(G,K•) denote the dual group of G, namely the multiplicativegroup of all homomorphisms λ : G → K•, where K• \ {0}. The following resultgeneralizes Dedekind’s lemma on the linear independence of characters.

Lemma 2.2. Let K be a field and let λ1, λ2, . . . , λn ∈ Hom(G,K•) with λi 6= 1 forall i. Suppose that d0, d1, d2, . . . , dn ∈ K, and let T be a subset of G such that

d0 = d1λ1(g) + d2λ2(g) + · · · + dnλn(g) for all g ∈ G \ T.

Then either d0 = 0 or |G : T | ≤ 2n.

Proof. We show, by induction on n, that if d0 6= 0, then |G : T | ≤ 2n. The casen = 0 is clear, so assume that n ≥ 1. Since λn 6= 1, we can choose x ∈ G withλn(x) 6= 1. Then

d0λn(x) =

n∑

1

diλi(g)λn(x) for all g ∈ G \ T

and

d0 =

n∑

1

diλi(gx) =

n∑

1

diλi(g)λi(x) for all g ∈ G \ Tx−1.

Thus, by subtracting, we obtain

d0(λn(x) − 1) =n−1∑

1

diλi(g)(λn(x) − λi(x)) for all g ∈ G \ (T ∪ Tx−1).

Note that d0(λn(x) − 1) 6= 0 so, by induction, T ∪ Tx−1 has index ≤ 2n−1 in G.Hence |G : T | ≤ 2n by Lemma 2.1(ii). �

TWISTED GROUP ALGEBRAS 5

Now let Kt[G] be a twisted group algebra of G over K, and let G be a subgroupof G. In the following five lemmas, we suppose that α1, α2, . . . , αn ∈ Kt[G] andβ1, β2, . . . , βn ∈ Kt[G] with A =

⋃i suppαi, B =

⋃i suppβi, |A| ≤ a and |B| ≤ b.

For convenience, we also assume that a ≥ 2. Recall that, if α ∈ K t[G] and x ∈ G,then αx̄ = x̄−1αx̄. The following is the usual coset counting argument associatedwith the Delta Method.

Lemma 2.3. If g ∈ G, then there exists a subset C of G which is the set-theoreticunion of at most ab right cosets of G-centralizers of elements of A, such that

g /∈ supp(αx̄1β1 + αx̄2β2 + · · · + α

x̄nβn) for all x ∈ G \ C.

Proof. Let u ∈ A and v ∈ B. If u is G-conjugate to gv−1, choose xu,v ∈ G withx−1u,vuxu,v = gv

−1. We define C to be the union of the cosets CG(u)xu,v for all

appropriate u and v. Now suppose x ∈ G and g ∈ supp(αx̄1β1 +αx̄2β2 + · · ·+α

x̄nβn).

Then there must exist some u ∈ A and v ∈ B with g = uxv. Thus ux = gv−1, soxu,v is defined, and x

−1ux = gv−1 = x−1u,vuxu,v . It follows that xx−1u,v ∈ CG(u), so

x ∈ CG(u)xu,v ⊆ C, as required. �

We continue with the preceding notation. In addition, for the next four lemmas,we assume that T is a subset of G with

α1x̄β1 + α2x̄β2 + · · · + αnx̄βn = 0 for all x ∈ G \ T.

For convenience, we define the integer valued function f , depending upon a andb, by f(k) = ab2a−1ka for all integers k ≥ 1. It is clear that f(k) ≥ k provideda, b ≥ 1. At this point, the notation of the preceding section comes into play.

Lemma 2.4. Let k ≥ 1 be an integer and assume that A ∩ Df(k) = A ∩ Dk andthat A ∩ Dtf(k) = A ∩ D

tk. Then either |G : T | ≤ k

a(ab2a−1 + 2)! or

φtk(α1)β1 + φtk(α2)β2 + · · · + φ

tk(αn)βn = 0.

Proof. We will show that if φtk(α1)β1 +φtk(α2)β2 + · · ·+φ

tk(αn)βn 6= 0, then |G : T |

is suitably bounded. To this end, write A = R ∪ S ∪ Z where R = A ∩ Dtk,S = (A ∩ Dk) \R and Z = A \ (R ∪ S). Then R, S and Z are disjoint and we canwrite each αi uniquely as αi = ρi − σi + ζi where supp ρi ⊆ R, suppσi ⊆ S andsupp ζi ⊆ Z. It is clear that ρi = φtk(αi). In particular, |R| ≥ 1.

SetW =

⋂

u∈R

CtG(u) ∩

⋂

u∈S

CG(u).

Then, by definition, each of the intersecting subgroups here has index at most k, so|R ∪ S| ≤ |A| = a implies that |G : W | ≤ ka. Now let g ∈ G be a fixed element inthe support of φtk(α1)β1 +φ

tk(α2)β2 + · · ·+φ

tk(αn)βn 6= 0, and let 0 6= d ∈ K be the

coefficient of ḡ. For convenience, set T ′ = T ∩W . We will bound |G : T ′| ≥ |G : T |.Let x ∈ W \ T ′. Then x ∈ G \ T , so

0 = x̄−1∑

i

αix̄βi =∑

i

αx̄i βi =∑

i

ρx̄i βi −∑

i

σx̄i βi +∑

i

ζ x̄i βi

Note that W ⊆ CtG(R), by definition of W , so ρx̄i = ρi. Furthermore, Lemma 2.3

implies that there exists a subset C of W , that is a union of at most ab right cosetsof W -centralizers of elements of Z, with

g /∈ supp∑

i

ζ x̄i βi for all x ∈W \ C.

6 D. S. PASSMAN

Thus, for all x ∈W \ (T ′ ∪ C), the coefficient of ḡ in∑

i

σx̄i βi =∑

i

ρiβi +∑

i

ζ x̄i βi

is precisely equal to d 6= 0.Now if u ∈ suppσi ⊆ S, then W centralizes u, by the definition of W . In

particular, for all x ∈W , ūx̄ = λu(x)ū where λu : W → K• is a linear character. It

follows that the coefficient of ḡ in∑

i σx̄i βi is a fixed K-linear combination of the

various λu(x). In other words, there exist du ∈ K with

0 6= d =∑

u∈S

duλu(x) for all x ∈W \ (T′ ∪ C).

We note that λu 6= 1 for all u ∈ S. Indeed, if λu = 1, then CtG(u) ⊇ W . Hence|G : CtG(u)| ≤ |G : W | ≤ k

a ≤ f(k) and u ∈ A ∩ Dtf(k) = A ∩ Dtk = R, a

contradiction. Thus, since |S| ≤ |A| − |R| ≤ a − 1, we can apply Lemma 2.2 toconclude that |W : T ′ ∪ C| ≤ 2a−1.

Let Y be a subset of W of size ≤ 2a−1 with W = (T ′ ∪ C)Y = T ′Y ∪ CY . Weclaim that CY 6= W . Indeed, since C is a union of ≤ ab cosets of W -centralizers ofelements of Z, the equality CY = W along with Lemma 2.1(iv) would imply that|W : CW (u)| ≤ ab2

a−1 for some u ∈ Z. But then

|G : CG(u)| ≤ |G : W ||W : CW (u)| ≤ ab2a−1ka = f(k),

so u ∈ A∩Df(k) = A∩Dk, by hypothesis. In particular, u ∈ R∪S, a contradiction.Thus, CY 6= W , so W = T ′Y ∪ CY and Lemma 2.1(iii) yield the inequality|W : T ′Y | ≤ (ab2a−1 + 1)! and hence |W : T ′| ≤ 2a−1(ab2a−1 + 1)! ≤ (ab2a−1 + 2)!since |Y | ≤ 2a−1. Furthermore, by Lemma 2.1(ii) again,

|G : T | ≤ |G : T ′| ≤ |G : W ||W : T ′| ≤ ka(ab2a−1 + 2)!

and |G : T | is suitably bounded. �

As an immediate consequence, we obtain

Lemma 2.5. Let k ≥ 1 be an integer and assume that A ∩ Df(k) = A ∩ Dk andA ∩ Dt

f(k) = A ∩ Dtk. Then either |G : T | ≤ k

a(ab2a−1 + 2)! or

φtk(α1)x̄β1 + φtk(α2)x̄β2 + · · · + φ

tk(αn)x̄βn = 0 for all x ∈ G.

Proof. We suppose that |G : T | > ka(ab2a−1 +2)!. Let g ∈ G be arbitrary and notethat

α1x̄ḡβ1 + α2x̄ḡβ2 + · · · + αnx̄ḡβn = 0 for all x ∈ G \ Tg−1,

since x̄ḡ = d xg for some 0 6= d ∈ K. Furthermore, if B ′ =⋃

i supp ḡβi, thenB′ = gB, so |B′| = |B| = b. Thus, since |G : Tg−1| = |G : T | is suitably large,Lemma 2.4 implies that

φtk(α1)ḡβ1 + φtk(α2)ḡβ2 + · · · + φ

tk(αn)ḡβn = 0,

and the result follows since this holds for all g ∈ G. �

Now let r = 2 max{a, b} and note that if k ≥ 4 then f(k) ≤ kr. Indeed, ifm = max{a, b}, then 2m+1 ≥ m2, so k ≥ 4 yields km ≥ 22m ≥ m22m−1 ≥ ab2a−1

and hence kr ≥ km+a ≥ ab2a−1ka = f(k). In the next lemma, we use a pigeon-holeargument to show that

TWISTED GROUP ALGEBRAS 7

Lemma 2.6. Either |G : T | ≤ (1/r)3·4rr+1

! or there exists an integer k ≤ 4rr

with

φtk(α1)x̄β1 + φtk(α2)x̄β2 + · · · + φ

tk(αn)x̄βn = 0 for all x ∈ G.

Proof. Let si = 4ri for i ≥ 0 and note that 4 = s0 ≤ s1 ≤ s2 ≤ · · · . Hence the

subsets A∩ (Dtsi+1 \Dtsi

) are disjoint, and consequently at most a = |A| of these canbe nonempty. Similarly, at most a of the subsets A∩(Dsi+1 \Dsi) can be nonempty.Since r ≥ 2a, it follows that there exists a subscript i with 0 ≤ i ≤ r and withA∩(Dtsi+1 \D

tsi

) = ∅ = A∩(Dsi+1 \Dsi). If k = si, then 4 ≤ k ≤ sr = 4rr and, as we

observed, f(k) ≤ kr = sri = si+1. Thus A∩Dtf(k) = A∩D

tk and A∩Df(k) = A∩Dk,

so Lemma 2.5 implies that either

φtk(α1)x̄β1 + φtk(α2)x̄β2 + · · · + φ

tk(αn)x̄βn = 0 for all x ∈ G

or

|G : T | ≤ ka(ab2a−1 + 2)! ≤ (1/8ab)3·(ab2a−1ka)!

≤ (1/r)3·f(k)! ≤ (1/r)3·sr+1! = (1/r)3·4r

r+1

!

since ka ≥ 42 = 16, and the lemma is proved. �

The next result is actually a special case of the more important Proposition 2.8.It is certainly simpler to state.

Lemma 2.7. Either |G : T | ≤ 4rr+1

! or

φt(α1)x̄β1 + φt(α2)x̄β2 + · · · + φ

t(αn)x̄βn = 0 for all x ∈ G.

Proof. Assume that |G : T | > 4rr+1

!. We proceed by induction on |A∩Dt| to showthat φt(α1)x̄β1 + φ

t(α2)x̄β2 + · · · + φt(αn)x̄βn = 0 for all x ∈ G. The result is, ofcourse, clear if A ∩ Dt = ∅. So suppose g ∈ A ∩ Dt and observe that

α′1x̄β1 + α′2x̄β2 + · · · + α

′nx̄βn = 0 for all x ∈ G \ T,

where α′i = ḡ−1αi. Furthermore, if A

′ =⋃

i suppα′i, then A

′ = g−1A and A′∩Dt =g−1(A ∩ Dt), since g ∈ Dt. Thus the corresponding sets have the same size, and1 ∈ A′ ∩ Dt. By the preceding lemma and the assumption on |G : T |, there existsan integer k with

φtk(α′1)x̄β1 + φ

tk(α

′2)x̄β2 + · · · + φ

tk(α

′n)x̄βn = 0 for all x ∈ G.

Hence, setting α′′i = α′i − φ

tk(α

′i), we have

α′′1 x̄β1 + α′′2 x̄β2 + · · · + α

′′nx̄βn = 0 for all x ∈ G \ T.

Note that A′′ =⋃

i suppα′′i ⊆

⋃i suppα

′i = A

′ and that 1 ∈ A′ ∩ Dt but 1 /∈A′′ ∩ Dt since 1 ∈ A′ ∩ Dtk. Thus |A

′′ ∩ Dt| < |A′ ∩ Dt| and, by induction,

φt(α′′1 )x̄β1 + φt(α′′2 )x̄β2 + · · · + φ

t(α′′n)x̄βn = 0 for all x ∈ G.

Finally, note that φt(α′′i ) = φt(α′i − φ

tk(α

′i)) = φ

t(α′i) − φtk(α

′i), so

φt(α′1)x̄β1 + φt(α′2)x̄β2 + · · · + φ

t(α′n)x̄βn

= φtk(α′1)x̄β1 + φ

tk(α

′2)x̄β2 + · · · + φ

tk(α

′n)x̄βn = 0 for all x ∈ G.

Thus, since φt(α′i) = φt(ḡ−1αi) = ḡ

−1φt(αi), we conclude that

φt(α1)x̄β1 + φt(α2)x̄β2 + · · · + φ

t(αn)x̄βn = 0 for all x ∈ G,

as required. �

8 D. S. PASSMAN

The following is the key result needed to handle generalized polynomial identities.We state it in a somewhat cleaner form using singletons, that is elements of thetwisted group ring of support size ≤ 1. Alternately, we could state the result interms of the various truncations of the elements γj . Recall that a truncation ofan element in a twisted group ring is a partial sum of the natural expression ofthat element as a K-linear combination of members of the basis G. Thus, forexample, φt(α) and α − φt(α) are both truncations of α ∈ Kt[G]. Note also thatany truncation of a singleton is either itself or 0.

Proposition 2.8. Let Kt[G] be a twisted group algebra and let αi, βi, γj , δj ∈ Kt[G]with i = 1, 2, . . . , n and j = 1, 2, . . . ,m. Assume that each γj is a singleton. Let Gbe a subgroup of G, let T be a subset of G, and suppose that

n∑

i=1

αix̄βi +

m∑

j=1

γj x̄δj = 0 for all x ∈ G \ T.

Write A =⋃

i suppαi, B =⋃

i suppβi, C =⋃

j supp γj , D =⋃

j supp δj , and let

r ≥ 2 max{|A ∪ C|, |B ∪D|} with r ≥ 4. Then either |G : T | ≤ (1/r)3·4rr+1

! orn∑

i=1

φt(αi)x̄βi +

m∑

j=1

ψ(γj)x̄δj = 0 for all x ∈ G

where ψ(γj) is equal to 0 or to ȳφtk(ȳ

−1γj) for some y ∈ A ∩ Dt and some integerk ≤ 4r

r

. In particular, if γj = 0, then ψ(γj) = 0.

Proof. Assume that |G : T | > (1/r)3·4rr+1

!. We proceed by induction on |A ∩ Dt|to show that

∑i φ

t(αi)x̄βi +∑

j ψ(γj)x̄δj = 0 for all x ∈ G and for appropriate

ψ(γj). The result is, of course, clear if A ∩ Dt = ∅. So suppose g ∈ A ∩ Dt andobserve that

n∑

i=1

α′ix̄βi +

m∑

j=1

γ′j x̄δj = 0 for all x ∈ G \ T,

where α′i = ḡ−1αi and γ

′j = ḡ

−1γj . Furthermore, if A′ =

⋃i suppα

′i and C

′ =⋃j supp γ

′j , then A

′ = g−1A, C ′ = g−1C and A′ ∩ Dt = g−1(A ∩ Dt), since g ∈ Dt.

Thus the corresponding sets have the same size, and 1 ∈ A′ ∩ Dt. By Lemma 2.6and the assumption on |G : T |, there exists an integer k ≤ 4r

r

with

(∗)n∑

i=1

φtk(α′i)x̄βi +

m∑

j=1

φtk(γ′j)x̄δj = 0 for all x ∈ G.

Hence, setting α′′i = α′i − φ

tk(α

′i) and γ

′′j = γ

′j − φ

tk(γ

′j), we have

n∑

i=1

α′′i x̄βi +m∑

j=1

γ′′j x̄δj = 0 for all x ∈ G \ T.

Note that A′′ =⋃

i suppα′′i ⊆

⋃i suppα

′i = A

′ and that 1 ∈ A′ ∩ Dt but 1 /∈A′′ ∩ Dt since 1 ∈ A′ ∩ Dtk. Furthermore, C

′′ =⋃

j supp γ′′j ⊆

⋃j supp γ

′j = C

′, and

γ′′j = γ′j − φ

tk(γ

′j) is a truncation of γ

′j and hence it is also a singleton. Thus, since

|A′′ ∩ Dt| < |A′ ∩ Dt|, induction implies that

(∗∗)n∑

i=1

φt(α′′i )x̄βi +

m∑

j=1

ψ(γ′′j )x̄δj = 0 for all x ∈ G

TWISTED GROUP ALGEBRAS 9

and for appropriate ψ(γ ′′j ). Next, φt(α′′i ) = φ

t(α′i − φtk(α

′i)) = φ

t(α′i) − φtk(α

′i), so

φt(α′i) = φt(α′′i ) + φ

tk(α

′i) and, by adding (∗) and (∗∗), we obtain

n∑

i=1

φt(α′i)x̄βi +

m∑

j=1

[φtk(γ′j) + ψ(γ

′′j )]x̄δj = 0 for all x ∈ G.

Thus, since φt(α′i) = φt(ḡ−1αi) = ḡ

−1φt(αi), we conclude that

n∑

i=1

φt(αi)x̄βi +

m∑

j=1

ψ(γj)x̄δj = 0 for all x ∈ G,

where we set ψ(γj) = ḡφtk(γ

′j) + ḡψ(γ

′′j ) = ḡφ

tk(ḡ

−1γj) + ḡψ(γ′′j ). It remains to

better understand ψ(γj).To this end, suppose first that ḡφtk(γ

′j) = ḡφ

tk(ḡ

−1γj) 6= 0. Then since γ′j =

ḡ−1γj is a singleton, it follows that γ′′j = γ

′j − φ

tk(γ

′j) = 0. Hence ψ(γ

′′j ) = 0 and

ψ(γj) = ḡφtk(ḡ

−1γj), as required. On the other hand, if ḡφtk(γ

′j) = ḡφ

tk(ḡ

−1γj) = 0,

then γ′′j = γ′j = ḡ

−1γj is a singleton and ψ(γj) = ḡψ(γ′′j ) is either 0 or equal to

ḡȳθt`(ȳ−1γ′′j ) = ḡȳθ

t`(ȳ

−1ḡ−1γj) for some y ∈ A′′ ∩ Dt and ` ≤ 4rr

. Note that, in

the latter case, gy ∈ Dt and y ∈ A′′ ⊆ A′ = g−1A so gy ∈ A. Thus gy ∈ A ∩ Dt

and, since gy = dḡȳ for some 0 6= d ∈ K, we conclude that ψ(γj) = gyθt`(gy−1γj)

also has the appropriate form. �

3. Generalized Polynomial Identities

We now move on to consider multilinear generalized polynomial identities. Sincethese identities are, in some sense, combinations of linear identities, it is not sur-prising that the work of the previous section should come into play. Fortunately,the arguments here are somewhat less computational.

In this section, we assume that Kt[G] is given and that G is a normal subgroupof G. As we observed, the latter implies that D, Dt, Dk and D

tk are all normal

subsets of G. Let

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

be a multilinear generalized polynomial with each αi,σ,j in Kt[G], and assume

throughout that Kt[G] satisfies f , so that

f(γ1, γ2, . . . , γn) = 0 for all γ1, γ2, . . . , γn ∈ Kt[G].

As usual, for each σ ∈ Symn, we write

fσ(ζ1, ζ2, . . . , ζn) =

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

for the σ-component of f . We start with

Lemma 3.1. Let Kt[G], G, and f be as above. Assume further that each αi,σ,j isa singleton and that

a1∑

j=1

φt(α0,1,j)φt(α1,1,j) · · ·φ

t(αn,1,j) 6= 0.

10 D. S. PASSMAN

If r ≥ 2∑

σ aσ, twice the number of monomials which occur in f , and if k = 4rr ,

then |G : ∆tk2| ≤ (n/r)·4r

r+1

!.

Proof. For each i = 0, 1, . . . , n, define Si ⊆ Symn by

Si = {σ ∈ Symn | σ(1) = 1, σ(2) = 2, . . . , σ(i) = i}.

Then S0 = Symn, Sn = {1} and Si ∼= Symn−i. Using this, we define the generalizedmultilinear polynomial fi of degree n− i by

fi(ζi+1, ζi+2, . . . , ζn) =∑

σ∈Si

aσ∑

j=1

βi,σ,jζσ(i+1)αi+1,σ,j · · ·αn−1,σ,jζσ(n)αn,σ,j

where

βi,σ,j = φt(α0,σ,j)φ

t(α1,σ,j) · · ·φt(αi−1,σ,j)αi,σ,j .

Then f0 = f and

fn =

a1∑

j=1

βn,1,j =

a1∑

j=1

φt(α0,1,j)φt(α1,1,j) · · ·φ

t(αn−1,1,j)αn,1,j

is a nonzero element of Kt[G] since

φt(fn) =

a1∑

j=1

φt(α0,1,j)φt(α1,1,j) · · ·φ

t(αn−1,1,j)φt(αn,1,j) 6= 0.

Next, set M′0 = ∅ and, for i = 1, 2, . . . , n, let M′i to be the set of all monomials

µ of the form

µ = β−1i−1,τ,j′βi−1,σ,jζσ(i)αi,σ,jζσ(i+1)αi+1,σ,j · · ·α`−2,σ,jζσ(`−1)α`−1,σ,j

with σ ∈ Si−1 \ Si, σ(`) = i, and τ ∈ Si. Furthermore, we must have βi−1,τ,j′ 6= 0and, since this element is a singleton, it is invertible in Kt[G]. Note that any suchµ involves only variables from the set {ζi+1, ζi+2, . . . , ζn} and that at least onevariable occurs since σ ∈ Si−1 \ Si implies that σ(i) 6= i. Furthermore, since µ isuniquely determined by the pairs {σ, j} and {τ, j ′}, we see that |M′i| ≤ r

2 and thatM′n = ∅. Now let M =

⋃ni=0 M

′i and let Mi denote the set of monomials in M

which involve only variables in the set {ζi+1, ζi+2, . . . , ζn}. Then clearly |M| ≤ nr2,Mi ⊇ M′i, and M = M0 ⊇ M1 ⊇ · · · ⊇ Mn = ∅.

Set k = 4rr

. Our goal is to show that |G : ∆tk2| ≤ (n/r)·4r

r+1

!. Suppose, byway of contradiction, that this is not the case. Then we will prove by induction oni = 0, 1, . . . , n that, for all xi+1, xi+2, . . . , xn ∈ G, either fi(x̄i+1, x̄i+2, . . . , x̄n) = 0or there exists a monomial µ in Mi with suppµ(x̄i+1, x̄i+2, . . . , x̄n)∩Dtk 6= ∅. Thisis, of course, clear for i = 0 since f0 = f is a generalized polynomial identitysatisfied by Kt[G]. Assume now that 1 ≤ i ≤ n and that the result holds for i− 1.Fix the elements xi+1, xi+2, . . . , xn ∈ G and let x ∈ G play the role of the ithvariable. Let µ ∈ Mi. If suppµ(x̄i+1, x̄i+2, . . . , x̄n) ∩ Dtk 6= ∅, we are done. Thuswe may suppose that suppµ(x̄i+1, x̄i+2, . . . , x̄n) ∩ Dtk = ∅ for all µ ∈ Mi.

Now let µ ∈ Mi−1 \ Mi = Ni−1 so that µ involves the variable ζi. Writeµ = µ′ζiµ

′′ where µ′ and µ′′ are monomials in the variables ζi+1, ζi+2, . . . , ζn. SinceDtk is a normal subset of G and since each αi,σ,j is a singleton, it follows thatsuppµ(x̄, x̄i+1, . . . , x̄n) ∩ D

tk 6= ∅ if and only if

x ∈ (h′)−1Dtk(h′′)−1 = Dtk(h

′)−1(h′′)−1 = Dtkyµ

TWISTED GROUP ALGEBRAS 11

where suppµ′(x̄i+1, x̄i+2, . . . , x̄n) = {h′}, suppµ′′(x̄i+1, x̄i+2, . . . , x̄n) = {h′′}, andyµ = (h

′)−1(h′′)−1. In particular, if x ∈ G \ T where

T =⋃

µ∈Ni−1

G ∩ Dtkyµ,

then we have suppµ(x̄, x̄i+1, . . . , x̄n) ∩ Dtk = ∅ for all µ ∈ Ni−1 = Mi−1 \Mi andhence for all µ ∈ Mi−1.

By induction, we conclude that

0 = fi−1(x̄, x̄i+1, . . . , x̄n)

=∑

σ∈Si−1

aσ∑

j=1

βi−1,σ,j x̄σ(i)αi,σ,j · · ·αn−1,σ,j x̄σ(n)αn,σ,j

for all x = xi ∈ G \ T

and we study this linear identity using Proposition 2.8. Specifically, we let the{α, β}-terms correspond to those σ ∈ Si and the {γ, δ}-terms correspond to thoseσ ∈ Si−1 \ Si. Since all the summands in fi−1 are singletons, it is clear, in thenotation of the above mentioned proposition, that r ≥ 2 max{|A ∪ C|, |B ∪ D|}.Furthermore, since f0 must contain at least two summands, it follows that r ≥ 4.Thus, the hypotheses of Proposition 2.8 are satisfied, and there are two possibleconclusions.

Suppose first that |G : T | ≤ (1/r)3·4rr+1

!, and note that T is a union of |Ni−1| ≤|M| ≤ nr2 subsets of the form G ∩ Dtkyµ with yµ ∈ G. Furthermore, if g, h ∈G ∩ Dtkyµ, then gh

−1 ∈ G ∩ (Dtk)2 ⊆ G ∩ Dtk2 = ∆

tk2 . Thus G ∩ D

tkyµ is contained

in a right G-translate of ∆tk2

, and it follows from Lemma 2.1(ii) that |G : ∆tk| ≤

nr2·(1/r)3·4rr+1

! = (n/r)·4rr+1

!, contrary to our assumption. Consequently, thesecond possibility must occur and we consider that conclusion with x = 1.

Observe that if σ ∈ Si−1, then σ ∈ Si if and only if σ(i) = i and hence if andonly if

βi−1,σ,j x̄σ(i)αi,σ,j · · ·αn−1,σ,j x̄σ(n)αn,σ,j

= βi−1,σ,j x̄αi,σ,j · · ·αn−1,σ,j x̄σ(n)αn,σ,j

starts with βi−1,σ,j x̄αi,σ,j . Thus, since

φt(βi−1,σ,j)αi,σ,j = φt(φt(α0,σ,j)φ

t(α1,σ,j) · · ·φt(αi−2,σ,j)αi−1,σ,j

)αi,σ,j

= φt(α0,σ,j)φt(α1,σ,j) · · ·φ

t(αi−2,σ,j)φt(αi−1,σ,j)αi,σ,j

= βi,σ,j ,

it follows that the {α, β}-terms (with x = 1) in the conclusion of Proposition 2.8sum to fi(x̄i+1, x̄i+2, . . . , x̄n).

Next, note that a typical γ term looks like

γ = βi−1,σ,j x̄σ(i)αi,σ,j x̄σ(i+1)αi+1,σ,j · · ·α`−2,σ,j x̄σ(`−1)α`−1,σ,j

with σ(`) = i and with ` 6= i since σ ∈ Si−1 \ Si. Futhermore, if y ∈ A ∩ Dt, thencertainly ȳ = dβi−1,τ,j′ for some τ ∈ Si, 1 ≤ j′ ≤ aτ and 0 6= d ∈ K. Thus ψ(γ) iseither equal to 0 or to a suitable

ȳφtk′ (ȳ−1γ) = βi−1,τ,j′φ

tk′

(β−1i−1,τ,j′γ

)= βi−1,τ,j′φ

tk′

(µ(x̄i+1, x̄i+2, . . . , x̄n)

)

12 D. S. PASSMAN

with k′ ≤ k and with

µ = β−1i−1,τ,j′γ

= β−1i−1,τ,j′βi−1,σ,j x̄σ(i)αi,σ,j x̄σ(i+1)αi+1,σ,j · · ·α`−2,σ,j x̄σ(`−1)α`−1,σ,j

∈ M′i ⊆ Mi.

But, by assumption, we have suppµ(x̄i+1, x̄i+2, . . . , x̄n)∩Dtk = ∅ and consequentlyφtk′

(µ(x̄i+1, x̄i+2, . . . , x̄n)

)= 0. Thus, again ψ(γ) = 0, and it follows that the

{γ, δ}-terms contribute nothing to the sum in the conclusion of Proposition 2.8. Inother words, that result yields fi(x̄i+1, x̄i+2, . . . , x̄n) = 0, and the inductive step isproved.

In particular, when i = n, the inductive result implies that either fn = 0 orthere exists µ ∈ Mn with suppµ ∩ D

tk 6= ∅. However, fn is known to be a nonzero

constant and Mn = ∅. Hence we have a contradiction and therefore it follows that|G : ∆tk| ≤ (n/r)·4

rr+1 !, as required. �

Next, we isolate the procedure of [7, Lemma 2.1] which really only uses the factthat G and Dt are normal in G.

Lemma 3.2. If f1(1, 1, . . . , 1) 6= 0, then there exist z1, z2, . . . , zn ∈ G so that

f̃(ζ1, ζ2, . . . , ζn) = f(z̄−11 ζ1z̄1, z̄

−12 ζ2z̄2, . . . , z̄

−1n ζnz̄n)

=∑

σ∈Symn

aσ∑

j=1

α̃0,σ,jζσ(1)α̃1,σ,jζσ(2) · · · α̃n−1,σ,jζσ(n)α̃n,σ,j

is a multilinear generalized polynomial identity for K t[G] with

a1∑

j=1

φt(α̃0,1,j)φt(α̃1,1,j) · · ·φ

t(α̃n,1,j) 6= 0.

Proof. Let Y be a right transversal for Dt in G. Then we can write each αi,1,j asαi,1,j =

∑s βi,j,sȳs with βi,j,s ∈ K

t[Dt] and ys ∈ Y . Since

0 6= f1(1, 1, . . . , 1) =a1∑

j=1

α0,1,jα1,1,j · · ·αn,1,j

and since each αi,1,j =∑

s βi,j,sȳs, there exist ys0 , ys1 , . . . , ysn ∈ Y with

0 6=a1∑

j=1

β0,j,s0 ȳs0β1,j,s1 ȳs1 · · ·βn,j,sn ȳsn .

In particular, if we set z̄i = ȳs0 ȳs1 · · · ȳsi−1 and z̄0 = 1, then

0 6=a1∑

j=1

(β0,j,s0)z̄−1

0 (β1,j,s1)z̄−1

1 · · · (βn,j,sn)z̄−1

n .

Now βi,j,si = φt(αi,1,j ȳ

−1si

) so, since Dt /G,

(βi,j,si )z̄−1

i = φt(z̄iαi,1,j ȳ−1siz̄−1i ) = φ

t(z̄iαi,1,j z̄−1i+1).

It therefore follows that if we replace ζi in f by z̄−1i ζiz̄i and if we multiply f on the

right by z̄−1n+1, then this new multilinear generalized polynomial f̃ has the required

TWISTED GROUP ALGEBRAS 13

property concerning φt. Furthermore, since G is normal in G, f̃ is also an identityfor Kt[G]. �

With this, we now quickly prove our main result.

Theorem 3.3. Let Kt[G] be a twisted group algebra of G over K and let G bea normal subgroup of G. Suppose that Kt[G] satisfies the multilinear generalizedpolynomial identity

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

with each αi,σ,j in Kt[G], but that there exists τ ∈ Symn with fτ not an identity

for Kt[G]. If

r = 2∑

σ∈Symn

aσ∑

j=1

n∏

i=0

| suppαi,σ,j |,

then |G : ∆tk2 | ≤ 4rr+1 ! where k = 4r

r

.

Proof. We modify the identity f so that Lemma 3.1 can apply. To start with, writeeach αi,σ,j in terms of the K-basis G and apply the distributive law to each product

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j .

By doing this, we obtain∏n

i=0 | suppαi,σ,j | similar products but with parameterswhich are all singletons. In other words, we can now assume that each αi,σ,j is asingleton and that r/2 is equal to the total number of terms that occur. Indeed, ifwe choose to combine terms which are K-multiples of each other, then r/2 will begreater than or equal to this total number of terms.

Next, we know that fτ is not an identity for Kt[G] and, by suitably permuting

the variables, we can clearly assume that τ = 1. Furthermore, since f1 is mul-tilinear and since G is a K-basis for Kt[G], there exist g1, g2, . . . , gn ∈ G withf1(ḡ1, ḡ2, . . . , ḡn) 6= 0. Thus, by replacing each ζi by ζiḡi, which does not effect thenumber of terms in f or the fact that each αi,σ,j is a singleton, we can assume thatf1(1, 1, . . . , 1) 6= 0. Of course, this new f is still an identity for K

t[G].Now for each 1 6= σ ∈ Symn for which a σ-term occurs in f , choose iσ , jσ ∈

{1, 2, . . . , n} such that iσ < jσ but σ(iσ) > σ(jσ). Let U ⊆ {1, 2, . . . , n} be thecollection of all such iσ and jσ , and note that |U | ≤ 2·(r/2) = r since f has at mostr/2 terms. Suppose f ′ is the polynomial obtained from f by setting ζi = 1 for alli /∈ U . Then f ′ is certainly a multilinear generalized polynomial identity for K t[G],all its parameters are singletons, it has at most r/2 terms, and deg f ′ = |U | ≤ r.Furthermore, since U contains all iσ and jσ , we see that, if the ζi-terms in

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

are not in their natural order, then the truncated set of ζi’s which remain in f′ are

also not in their natural order. In other words, f ′1(1, 1, . . . , 1) = f1(1, 1, . . . , 1) 6= 0and f ′ is a nondegenerate multilinear identity for Kt[G]. Since deg f ′ ≤ r, this nowallows us to assume that n = deg f ≤ r.

Finally, we apply Lemma 3.2 to f , and note that this does not change the degreeof f , the number of terms it comprises, or the fact that each αi,σ,j is a singleton.With this last modification, all the hypotheses of Lemma 3.1 are now satisfied, andsince n ≤ r, Lemma 3.1 yields the result. �

14 D. S. PASSMAN

The remainder of the argument is entirely group theoretic.

Corollary 3.4. Let Kt[G] be a twisted group algebra of G over K and let G bea normal subgroup of G. Suppose that Kt[G] satisfies the multilinear generalizedpolynomial identity

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

with each αi,σ,j in Kt[G], but that there exists τ ∈ Symn with fτ not an identity

for Kt[G]. If

r = 2∑

σ∈Symn

aσ∑

j=1

n∏

i=0

| suppαi,σ,j |,

then

(i) |G : ∆| ≤ 4rr+1

! and the commutator subgroup of ∆ is finite.

(ii) |G : ∆t| ≤ 4rr+1

! and the commutator subgroup of ∆t is finite.

(iii) G has a normal subgroup H ⊆ ∆t with |G : H | ≤ 4rr+1

! and with |H ′| boundedby a function of r. Furthermore, H = ∆th(H) for some integer h dependingupon r.

Proof. If k = 4rr

, then the previous theorem implies that |G : ∆tk2 | ≤ 4rr+1 ! = `. In

particular, since ∆ ⊇ ∆t ⊇ ∆tk2

, it follows that ` also bounds |G : ∆| and |G : ∆t|.Furthermore, since ∆k2 ⊇ ∆

tk2 , we have |G : ∆k2 |

TWISTED GROUP ALGEBRAS 15

Then it is easy to see that R = Kt[G] is a twisted group algebra of G. In fact, ifg ∈ G, then the possible G-conjugates of ḡ are ḡ, εḡ, . . . , εp−1ḡ, so g ∈ ∆tp(G), andhence G = ∆tp(G).

Note that the subalgebra of Kt[G] generated by x̄i, ȳi is isomorphic to the matrixring Mp(K), since this corresponds to a nonlinear irreducible representation of anonabelian group of order p3. Thus, the subalgebra of Kt[G] generated by k pairs{x̄i, ȳi} is isomorphic to Mp(K)⊗k = Mpk (K).

Finally, suppose that Kt[G] satisfies the nondegenerate g.p.i.

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

fσ(ζ1, ζ2, . . . , ζn)

and let s1, s2, . . . , sn ∈ Kt[G] with f1(s1, s2, . . . , sn) 6= 0. We may assume that allterms in f(ζ1, ζ2, . . . , ζn) and all s1, s2, . . . , sn are contained in K

t[H ], where H =〈x1, y1, x2, y2, . . . , xr , yr〉. Now let W = 〈xr+1, yr+1, xr+2, yr+2, . . . , xr+k, yr+k〉 sothat Kt[W ] ∼= Mpk (K). Furthermore, since K

t[W ] and Kt[H ] commute element-wise, we see that Kt[W ] satisfies the identity

f(s1ζ1, s2ζ2, . . . , snζn) =∑

σ∈Symn

fσ(s1, s2, . . . , sn)ζσ(1)ζσ(2) · · · ζσ(n).

In particular, if k is chosen so that pk > n, then we can take ζ1, ζ2, . . . , ζn to be thematrix units ζi = ei,i+1 and conclude that

f(s1e1,2, s2e2,3, . . . , snen,n+1) = f1(s1, s2, . . . , sn)e1,n+1

since all products of the matrix units other than e1,2e2,3 · · · en,n+1 = e1,n+1 arezero. But 0 6= f1(s1, s2, . . . , sn) ∈ Kt[H ], and 0 6= e1,n+1 ∈ Kt[W ], so we have acontradiction since Kt[H ×W ] = Kt[H ] ⊗Kt[W ]. �

In the preceding example, if H is any subgroup of G of finite index, then it iseasy to see that Kt[H ] is noncommutative. On the other hand, G is an abeliangroup, so the ordinary group algebra K[H ] is certainly commutative. In particular,the two algebras Kt[H ] and K[H ] are not K-isomorphic, nor can they becomeisomorphic under any field extension of K. As we will see in the next section,it is precisely this condition which keeps Kt[G] from satisfying a nondegeneratemultilinear polynomial identity.

4. Stable Untwisting

We say that the twisted group algebraKt[G] is untwisted if there exists a functionG → K•, given by x 7→ dx, such that the elements x̃ = dxx̄ with x ∈ G form agroup basis for Kt[G]. Thus x̃ỹ = x̃y for all x, y ∈ G, and Kt[G] ∼= K[G] via thisdiagonal change of basis. In addition, we say that K t[G] is stably untwisted if thereexists a field F ⊇ K such that F t[G] = F ⊗K Kt[G] is untwisted. The standarduntwisting criterion is

Lemma 4.1. Kt[G] is untwisted if and only if there exists an algebra homomor-phism ρ : Kt[G] → K. Furthermore, Kt[G] is stably untwisted if and only if thereexists an algebra homomorphism ρ : Kt[G] → R with R a commutative algebra.

Proof. If Kt[G] ∼= K[G], then there exists a homomorphism ρ : Kt[G] → K cor-responding to the augmentation map of K[G]. Conversely, if ρ exists, we definex̃ = ρ(x)−1x̄ for all x ∈ G. Then x̃ is the unique element in Kx̄ with ρ(x̃) = 1, and

16 D. S. PASSMAN

consequently x̃ỹ = x̃y for all x, y ∈ G. Thus G̃ = {x̃ | x ∈ G} is a group basis forKt[G], and Kt[G] ∼= K[G].

If Kt[G] is stably untwisted with associated field F ⊇ K, then by the above wehave an algebra homomorphism ρ : Kt[G] → F t[G] → F . Conversely, if ρ exists, letF be an algebraically closed field extension of K with |F | > |G|, an inequality ofcardinal numbers. Then ρ extends to an F -algebra map ρ : F t[G] = F ⊗K Kt[G] →F ⊗K R → L where the field L is equal to F ⊗ R modulo a maximal ideal. SincedimF L ≤ dimF F t[G] = |G| < |F |, [8, Lemma 7.1.2(iii)] implies that L = F . ThusF t[G] is untwisted and Kt[G] is stably untwisted. �

As an application of the above, suppose a ∈ K does not have an nth root in thefield, and let G = 〈x〉 be cyclic of order n. Then Kt[G] = K[x̄ | x̄n = a] cannot beuntwisted over K, but it can be untwisted once we adjoin an nth root of a to thefield. Next, we show

Lemma 4.2. Let Kt[G] be given and let H be a normal subgroup of G of finiteindex with |H ′| < ∞ and with Kt[H ] stably untwisted. Then Kt[G] satisfies anondegenerate multilinear generalized polynomial identity.

Proof. By assumption, there exists a field F ⊇ K with F ⊗K Kt[H ] = F t[H ]untwisted, so that F t[H ] ∼= F [H ]. Define I to be the characteristic ideal of F t[H ]generated by all Lie products [γ, δ] = γδ − δγ with γ, δ ∈ F t[H ]. Thus I is thesmallest ideal of F t[H ] with commutative quotient. Viewed in F [H ], we see thatI ⊆ ω(F [H ′])·F [H ], the kernel of the natural map F [H ] → F [H/H ′], and hence,since H ′ is finite, we have αI = 0 where α 6= 0 is the sum in F [H ′] of all elementsof H ′. In other words, there exists 0 6= α ∈ F t[H ′] ⊆ F t[H ] with αI = 0.

Next, let |G : H | = n, write M = Mn(F t[H ]) and consider the natural homomor-phism Mn(F

t[H ]) → Mn(F t[H ]/I). Since F t[H ]/I is commutative, the Amitsur-Levitzki theorem (see [8, Theorem 5.1.9]) implies that Mn(F

t[H ]/I) satisfies thestandard identity

sm(ζ1, ζ2, . . . , ζm) =∑

σ∈Symm

(−1)σζσ(1)ζσ(2) · · · ζσ(m)

with m = 2n. In particular, we have sm(M,M, . . . ,M) ⊆ Mn(I).Let {x1, x2, . . . , xn} be a transversal for H in G, and use it to construct an

embedding of F t[G] into M = Mn(Ft[H ]) as in [8, Lemma 5.1.10]. Then under this

embedding, α ∈ F t[H ] maps to the diagonal matrix

diag(x̄1αx̄−11 , x̄2αx̄

−12 , . . . , x̄nαx̄

−1n ),

and note that each x̄iαx̄−1i annihilates I since α annihilates I and since conjugation

by x̄−1i is an automorphism of Ft[H ] stabilizing the characteristic ideal I . With

this, we conclude that αsm(Ft[G], F t[G], . . . , F t[G]) maps to 0 under the embedding

intoM , and hence we see that F t[G] satisfies the multilinear generalized polynomialidentity

αsm(ζ1, ζ2, . . . , ζm) =∑

σ∈Symm

(−1)σαζσ(1)ζσ(2) · · · ζσ(m).

Finally, since F t[G] is free over Kt[G], we conclude that Kt[G] satisfies the multi-linear generalized polynomial identity

f(ζ1, ζ2, . . . , ζm) = βsm(ζ1, ζ2, . . . , ζm) =∑

σ∈Symm

(−1)σβζσ(1)ζσ(2) · · · ζσ(m)

TWISTED GROUP ALGEBRAS 17

where 0 6= β ∈ K[H ′] occurs in the representation of α ∈ F [H ′] = F ⊗K K[H ′]. Ofcourse, f is nondegenerate since f1(1, 1, . . . , 1) = β 6= 0. �

As we will see, there is a converse to Lemma 4.2 which holds in the more generalcontext of G ⊆ G. To start with, we need the following standard fact.

Lemma 4.3. Let Kt[G] be a twisted group algebra of the abelian group G andsuppose that Kt[G] satisfies an ordinary polynomial identity of degree n. Then Ghas a subgroup Z of index ≤ (n/2)2 with Kt[Z] central in Kt[G].

Proof. Let F be an algebraically closed field extension of K with |F | > |G|, andobserve that F t[G] = F ⊗K K

t[G] also satisfies an ordinary polynomial identity ofdegree n. In particular, if ξ : F t[G] → R is any irreducible representation, then Ris a primitive p.i. algebra, so Kaplansky’s theorem (see [8, Theorem 5.3.4]) impliesthat R ∼= Mm(D) for some F -division algebra D and some integer m ≤ n/2.Furthermore, dimF D ≤ dimF F t[G] = |G| < |F | and hence, by [8, Lemma 7.1.2(iii)]again, D = F . Let Z = {g ∈ G | ḡ is central in Kt[G]}. Then Z is a subgroup ofG and Kt[Z] is central in Kt[G].

Now define χ : G → F so that, for each g ∈ G, χ(g) is the trace of the m ×mmatrix ξ(ḡ). If g ∈ Z, then ξ(ḡ) is central in Mm(F ) and hence it is a scalar matrix.Thus ξ(ḡ) ∈ F and χ(g) = m·ξ(ḡ). On the other hand, if g ∈ G \ Z, then thereexists h ∈ G so that ḡ and h̄ do not commute. In particular, since G is abelian,we have h̄−1ḡh̄ = kḡ for some 1 6= k ∈ K•, and then ξ(h̄)−1ξ(ḡ)ξ(h̄) = k·ξ(ḡ). Butsimilar matrices have the same trace, so χ(g) = kχ(g) and χ(g) = 0 since k 6= 1.It follows that m 6= 0 in F , since ξ(G) spans Mm(F ) and since there exist matriceswith nonzero trace.

Finally, if {g1, g2, . . .} is a transversal for Z in G, then we claim that the set{ξ(ḡ1), ξ(ḡ2), . . .} is F -linearly independent in Mm(F ). Indeed, if

∑sj=1 fjξ(ḡj) = 0

is an F -linear relation, then for all i we have∑s

j=1 fjξ(ḡj)ξ(ḡi)−1 = 0. Moreover,

gj,i = gjg−1i ∈ Z if and only if i = j, and in all cases ḡj ḡ

−1i is a scalar multiple of ḡj,i.

Thus, by taking traces, we get 0 = fi·χ(1) = fi·m, and hence fi = 0 since m 6= 0 inF . We conclude from this linear independence that |G : Z| ≤ dimF Mm(F ) = m2,and since m ≤ n/2, the lemma is proved. �

Next, we need

Lemma 4.4. Suppose Kt[G] is a commutative twisted group algebra of the finitegroup G over the algebraically closed field K. If Ω is a nonempty, nonzero subsetof Kt[G], then there exists γ ∈ Kt[G] with 0 6= γΩ ⊆ I, a 1-dimensional ideal.

Proof. Since K is algebraically closed, and Kt[G] is commutative and finite dimen-sional, it follows from Lemma 4.1 that Kt[G] = K[G] is an ordinary group algebraof the finite abelian group G. Let charK = p ≥ 0 and write G = P ×H where Pis the Sylow p-subgroup of G and H is the p-complement. Of course, if p = 0, thenG = H and P = 1.

Since K[H ] is semiprimitive, write K[H ] =∑t

1 ejK, where {ej | j = 1, . . . , t}is the set of primitive idempotents of this group algebra. Certainly, eiΩ 6= 0 forsome i, and eiΩ ⊆ eiK[P ] ∼= K[P ] since K[G] ∼= K[H ] ⊗K K[P ]. Finally, letω(K[P ]) be the augmentation ideal of K[P ], and note that this ideal is nilpotent.Thus there exists an integer m ≥ 0 maximal with ω(K[P ])meiΩ 6= 0, and we canchoose γ ∈ ω(K[P ])mei with γΩ 6= 0. But then γΩ is a subset of eiK[P ] ∼= K[P ]

18 D. S. PASSMAN

annihilated by the augmentation ideal, so γΩ ⊆ KeiP̂ = I , where P̂ is the sum ofthe elements of P in K[G]. �

The above result actually relies upon the fact that K[G] is a Frobenius algebra,so that each minimal ideal of this commutative algebra corresponds to a differentirreducible representation. On the other hand, if A = K[x, y | (x, y)2 = 0], then itis clear that there is no multiple of Ω = {x, y} which is both nonzero and containedin a 1-dimensional subspace. Now we return to infinite groups.

Lemma 4.5. Let Kt[G] be given and let G be a normal subgroup of G. Supposethat Kt[G] satisfies the multilinear generalized polynomial identity

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

with each αi,σ,j in Kt[G], and assume that there exists τ ∈ Symn with fτ not an

identity for Kt[G]. Then G has a subgroup H of finite index with K t[H ] satisfyingan identity of the form

f̃(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

ασζσ(1)ζσ(2) · · · ζσ(n)

with ασ ∈ Kt[G] for all σ ∈ Symn and with α1 6= 0. Furthermore, |H′| < ∞ and

Kt[H ′] is central in Kt[H ].

Proof. As in the proof of Theorem 3.3, we first modify f so that f1(1, 1, . . . , 1) 6= 0and that all αi,σ,j are singletons. Then we apply Lemma 3.2 to obtain a new identity

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

witha1∑

j=1

φt(α0,1,j)φt(α1,1,j) · · ·φ

t(αn,1,j) 6= 0,

while still preserving the fact that each αi,σ,j is a singleton. Let H = ∆t(G) so

that |G : H |

TWISTED GROUP ALGEBRAS 19

or γσ,j . Furthermore, since H ⊆ DtG(H) = DtG

(G) / G, the question of whethersuppβσ,j ∈ DtG(H) or not does not depend upon the elements h̄2, h̄3, . . . , h̄n thatoccur in the product expression for βσ,j , and similarly this is also true for γσ,j .

Thus, if∑′

σ,j denotes the partial sum of all those terms with φt(βσ,j) = βσ,j and

φt(γσ,j) = γσ,j , then we see that

0 =∑′

σ,jβσ,jh̄1γσ,j

is a multilinear identity that holds for all h1, h2, . . . , hn ∈ H .We continue this process, applying the above truncation procedure to the vari-

ables h2, h3, . . . , hn in turn, thereby obtaining a multilinear generalized polynomial

f ′′(ζ1, ζ2, . . . , ζn) =∑′′

σ,jα0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

satisfied by Kt[H ]. Again∑′′

σ,j denotes the partial sum of all those terms thatsurvive this process. We note that if

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

survives, then when the truncation procedure was applied to h̄σ(1) we must have

had suppα0,σ,j ∈ DtG(H). Furthermore, when the procedure was applied to h̄σ(2),we must have had suppα0,σ,jα1,σ,j ∈ DtG(H), and hence suppα1,σ,j ∈ D

tG

(H).Continuing in this manner, we see that

f ′′(ζ1, ζ2, . . . , ζn)

=∑

σ,jφt(α0,σ,j)ζσ(1)φ

t(α1,σ,j)ζσ(2) · · ·φt(αn−1,σ,j)ζσ(n)φ

t(αn,σ,j),

where the αn,σ,j term comes from a consideration of the truncation procedure ap-plied to h̄σ(n).

If A =⋃

i,σ,j suppφt(αi,σ,j), then A is a finite subset of D

tG

(H), and hence

C =⋂

a∈A CtH(ā) is a subgroup of finite index in H . Furthermore, K

t[C] satisfies

the above identity and it centralizes all the factors φt(αi,σ,j). Thus we see thatKt[C] satisfies ∑

σ∈Symn

ασζσ(1)ζσ(2) · · · ζσ(n)

whereασ =

∑

j

φt(α0,σ,j)φt(α1,σ,j) · · ·φ

t(αn−1,σ,j)φt(αn,σ,j).

In particular, we know that α1 6= 0.Finally, C ⊆ H , so C ′ ⊆ H ′ and |C ′| < ∞. Thus, since C ⊆ H = ∆t(H), we

see that L = CtC(C′) is a subgroup of C of finite index. Moreover, L′ ⊆ C ′, so

L = CtL(L′) and Kt[L′] is central in Kt[L]. With this, the result follows using L as

the appropriate subgroup of G. �

We can now obtain the promised converse to Lemma 4.2.

Theorem 4.6. Let Kt[G] be a twisted group algebra of G over K and let G bea normal subgroup of G. Suppose that Kt[G] satisfies the multilinear generalizedpolynomial identity

f(ζ1, ζ2, . . . , ζn) =∑

σ∈Symn

aσ∑

j=1

α0,σ,jζσ(1)α1,σ,jζσ(2) · · ·αn−1,σ,jζσ(n)αn,σ,j

20 D. S. PASSMAN

with each αi,σ,j in Kt[G], but that there exists τ ∈ Symn with fτ not an identity

for Kt[G]. Then G has a normal subgroup H of finite index with |H ′| < ∞ andwith Kt[H ] stably untwisted.

Proof. Without loss of generality, we can assume that K is algebraically closed.Suppose G has a subgroup H of finite index with |H ′|

TWISTED GROUP ALGEBRAS 21

J ·Kt[G]. By the above, we first see that Kt[G]/(annKt[G] α1) satisfies the ordinarypolynomial identity ∑

σ∈Symn

kσζσ(1)ζσ(2) · · · ζσ(n),

and second that

Kt[G]/(annKt[G] α1) = Kt[G]/(J ·Kt[G]) ∼= Kt[G/G′],

where the latter is a naturally obtained twisted group algebra of G/G′. But thenKt[G/G′] satisfies an ordinary polynomial identity, so Lemma 4.3 implies thatthe abelian group G/G′ has a subgroup Z/G′ of finite index with Kt[Z/G′] =Kt[Z]/(J ·Kt[Z]) commutative. In particular, Kt[Z] has a commutative homomor-phic image, and therefore it is stably untwisted by Lemma 4.1. Since Z ′ ⊆ G′, wehave |Z ′|