TWISTED GROUP ALGEBRAS SATISFYING A
GENERALIZED POLYNOMIAL IDENTITY
D. S. PASSMAN
Dedicated to the memory of Professor A. I. Kostrikin
Abstract. If K t [G] is a twisted group algebra satisfying a nondegenerate
multilinear generalized polynomial identity f (ζ1 , ζ2 , . . . , ζn ) = 0, then we show
that G has certain normal subgroups of finite index which can be viewed as
being almost central. For example, there exists H / G with |G : H|·|H 0 |
bounded by a fixed function of the support sizes of the nonzero K t [G]-terms
involved in f . Indeed, we obtain a more precise version of this result, with
the structure of H depending upon the specific twisting in the group algebra.
We then go on to determine necessary and sufficient conditions for K t [G] to
satisfy such an identity.
1. Introduction
t
Let K [G] denote a twisted group algebra of the group G over the field K. Then
K t [G] is a K-vector space with basis G, aP
copy of G. In particular, every element
α ∈ K t [G] can be written uniquely as α = x∈G kx x̄, a finite K-linear combination
of basis elements. Here kx ∈ K, and we use supp α = {x ∈ G | kx 6= 0} to denote
the support of α, a finite subset of G. In addition, K t [G] is a K-algebra and its
multiplication is defined distributively by
for all x, y ∈ G,
x̄ȳ = t(x, y)xy
•
where t : G × G → K is a 2-cocycle, namely a function whose structure is precisely
equivalent to the associativity of the algebra. Without loss of generality, we can
assume that 1̄ = 1 is the identity element of K t [G]. Of course, K t [G] = K[G] is the
ordinary group algebra of G over K when the twisting function t(x, y) is identically
equal to 1.
Now let R be any K-algebra. Then a multilinear generalized polynomial over R
is a function f (ζ1 , ζ2 , . . . , ζn ) of n noncommuting variables having the form
f (ζ1 , ζ2 , . . . , ζn ) =
aσ
X X
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j .
σ∈Symn j=1
Here Symn is the symmetric group of degree n, and each αi,σ,j is an element of the
ring R. Obviously, f is linear in each ζi , and we say that R satisfies the generalized
polynomial 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 could
have all αi,σ,j = 0. Indeed, even if these terms are nonzero, the polynomial might
still be of little interest. As with ordinary polynomial identities, we would like a
generalizedPidentity to be indicative of some version of commutativity. To this end,
write f = σ∈Symn fσ , where
fσ (ζ1 , ζ2 , . . . , ζn ) =
aσ
X
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j .
j=1
Then, we say that f is nondegenerate if there exists some τ ∈ Symn with fτ not
an identity for R. In other words, f is nondegerate if there exists τ ∈ Symn and
suitable elements s1 , s2 , . . . , sn ∈ R with fτ (s1 , s2 , . . . , sn ) 6= 0. It follows that if R
satisfies a nondegenerate multilinear polynomial identity, then at least two distinct
permutations 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 . Note
that an ordinary multilinear polynomial is a function of the form
X
g(ζ1 , ζ2 , . . . , ζn ) =
ασ ζσ(1) ζσ(2) · · · ζσ(n)
σ∈Symn
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 polynomial identity were handled in [7]. The latter paper also studied p.i. twisted group
algebras. In recent years, there has been renewed interest in the g.p.i. result because
it 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 the
unit group of K[G] satisfies a group identity, then K[G] must satisfy a polynomial
identity. Furthermore, [4] consided Hartley’s problem for twisted group algebras
and, in the course of that work, obtained basic properties of twisted group algebras
satisfying a g.p.i. In this paper, we sharpen the latter result and prove
Theorem 1.1. Let K t [G] be a twisted group algebra of G over K which satisfies
the nondegenerate multilinear generalized polynomial identity
f (ζ1 , ζ2 , . . . , ζn ) =
aσ
X X
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
σ∈Symn j=1
with each αi,σ,j in K t [G]. If
s=
aσ Y
n
X X
| supp αi,σ,j |,
σ∈Symn j=1 i=1
then G has a normal subgroup H with |G : H|·|H 0 | bounded by a function of s.
Indeed, we obtain a more precise version of this theorem, with the structure of H
depending upon the particular twisting in K t [G]. Results of this nature are proved
by a coset counting argument, discovered independently in [5] and [9], and known
as the Delta Method. As will be apparent, we push this method to the limit and
beyond in the following two sections. Specifically, Section 2 is concerned with linear
identities in twisted group algebras, namely identities of the form
α1 x̄β1 + α2 x̄β2 + · · · + αn x̄βn = 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 main
theorems on generalized polynomial identities.
At this point, it is appropriate to recall a number of basic definitions, but in a
slightly more general context. To start with, for any groups G ⊆ G, we let
D = DG (G) = {x ∈ G |G : CG (x)| < ∞}.
Then D is a subgroup of G normalized by G, and it is the set of all elements of G
having a finite number of G-conjugates. Next, for each integer n ≥ 1, we define
Dn = Dn,G (G) = {x ∈ G |G : CG (x)| ≤ n}.
This time Dn is a subset of G normalized by G, and contained in the group D.
Clearly, Dn Dm ⊆ Dnm . If G = G, we usually write DG (G) = ∆(G) and Dn,G (G) =
∆n (G). The characteristic subgroup ∆(G) is the f.c. (finite conjugate) center of G.
Now suppose K t [G] is a twisted group algebra and, for each x ∈ G, let
CtG (x) = {y ∈ G | x̄ȳ = ȳx̄}
denote its twisted centralizer in G. Then it is easy to see that CtG (x) is a subgroup
of G contained in the ordinary centralizer CG (x), and that its index |G : CtG (x)|
measures the number of distinct conjugates x̄ȳ = ȳ −1 x̄ȳ ∈ K t [G] with y ∈ G. There
are also twisted versions of D and Dn given by
Dt = DtG (G) = {x ∈ G |G : CtG (x)| < ∞}
and
Dtn = Dtn,G (G) = {x ∈ G |G : CtG (x)| ≤ n}.
Then Dt is a subgroup of G, normalized by G, and contained in D(G). Furthermore,
Dtn is a subset of G, normalized by G, and contained in Dn (G). Of course, Dtn Dtm ⊆
Dtnm , and if G / G then D, Dt , Dn and Dtn are normal subsets of G. When G = G,
we write DtG (G) = ∆t (G) and Dtn,G (G) = ∆tn (G).
Finally, if S is any subset of G, let K t [S] denote the set of all elements of K t [G]
with support contained in S. For example, if S = H is a subgroup of G, then K t [H]
is the usual subalgebra of K t [G] associated with H. Now, for any such subset S,
there is a natural K-linear projection map πS : K t [G] → K t [S] ⊆ K t [G] given by
(
x̄, if x ∈ S, or
πS (x̄) =
0, if x ∈ G \ S.
Thus
πS :
X
kx x̄ 7→
x∈G
X
kx x̄,
x∈S
and it is easy to see that if S is normalized by G, then πS (ȳ −1 αȳ) = ȳ −1 πS (α)ȳ
for all α ∈ K t [G] and y ∈ G. For the most part, we are concerned with the special
subsets of G defined in the preceding paragraphs, and the corresponding projection
maps for these sets are denoted by
φ : K t [G] → K t [D],
φt : K t [G] → K t [Dt ]
and
φn : K t [G] → K t [Dn ],
φtn : K t [G] → K t [Dtn ].
Since D and Dt are subgroups of G, it is easy to see that φ : K t [G] → K t [D] is a
K [D]-bimodule homomorphism and that φt : K t [G] → K t [Dt ] is a K t [Dt ]-bimodule
t
4
D. S. PASSMAN
map. When G = G, we write θ, θ t , θn , and θnt for φ, φt , 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 = T x 1 ∪ T x2 ∪ · · · ∪ T x n
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 a
subgroup 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 = T x1 ∪ T x2 ∪ · · · ∪ T xr with x1 , x2 , . . . , xr ∈ G. If |G : S| ≤ s,
then |G : T | ≤ rs.
(iii) Assume that S = H1 x1 ∪ H2 x2 ∪ · · · ∪ Hr xr where each Hi xi 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 = H1 x1 ∪ H2 x2 ∪ · · · ∪ Hr xr where each Hi xi 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, Lemmas
5.2.1 and 5.2.2].
Next, let Hom(G, K • ) denote the dual group of G, namely the multiplicative
group of all homomorphisms λ : G → K • , where K • \ {0}. The following result
generalizes 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 for
all i. Suppose that d0 , d1 , d2 , . . . , dn ∈ K, and let T be a subset of G such that
d0 = d1 λ1 (g) + d2 λ2 (g) + · · · + dn λn (g)
for all g ∈ G \ T.
n
Then either d0 = 0 or |G : T | ≤ 2 .
Proof. We show, by induction on n, that if d0 6= 0, then |G : T | ≤ 2n . The case
n = 0 is clear, so assume that n ≥ 1. Since λn 6= 1, we can choose x ∈ G with
λn (x) 6= 1. Then
d0 λn (x) =
n
X
di λi (g)λn (x)
for all g ∈ G \ T
1
and
d0 =
n
X
di λi (gx) =
1
n
X
di λi (g)λi (x)
for all g ∈ G \ T x−1 .
1
Thus, by subtracting, we obtain
d0 (λn (x) − 1) =
n−1
X
di λi (g)(λn (x) − λi (x))
for all g ∈ G \ (T ∪ T x−1 ).
1
Note that d0 (λn (x) − 1) 6= 0 so, by induction, T ∪ T x−1 has index ≤ 2n−1 in G.
Hence |G : T | ≤ 2n by Lemma 2.1(ii).
TWISTED GROUP ALGEBRAS
5
Now let K t [G] be a twisted group algebra of G over K, and let G be a subgroup
t
of G. In the following five lemmas,
S we suppose that
S α1 , α2 , . . . , αn ∈ K [G] and
t
β1 , β2 , . . . , βn ∈ K [G] with A = i supp αi , B = i supp βi , |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−1 αx̄. The following is the usual coset counting argument associated
with the Delta Method.
Lemma 2.3. If g ∈ G, then there exists a subset C of G which is the set-theoretic
union of at most ab right cosets of G-centralizers of elements of A, such that
g∈
/ supp(αx̄1 β1 + αx̄2 β2 + · · · + αx̄n βn )
for all x ∈ G \ C.
Proof. Let u ∈ A and v ∈ B. If u is G-conjugate to gv −1 , choose xu,v ∈ G with
−1
x−1
. We define C to be the union of the cosets CG (u)xu,v for all
u,v uxu,v = gv
appropriate u and v. Now suppose x ∈ G and g ∈ supp(αx̄1 β1 + αx̄2 β2 + · · · + αx̄n βn ).
Then there must exist some u ∈ A and v ∈ B with g = ux v. Thus ux = gv −1 , so
−1
xu,v is defined, and x−1 ux = gv −1 = x−1
u,v uxu,v . It follows that xxu,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
α1 x̄β1 + α2 x̄β2 + · · · + αn x̄βn = 0
for all x ∈ G \ T.
For convenience, we define the integer valued function f , depending upon a and
b, by f (k) = ab2a−1 k a for all integers k ≥ 1. It is clear that f (k) ≥ k provided
a, 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 and
that A ∩ Dtf (k) = A ∩ Dtk . Then either |G : T | ≤ k a (ab2a−1 + 2)! or
φtk (α1 )β1 + φtk (α2 )β2 + · · · + φtk (αn )βn = 0.
Proof. We will show that if φtk (α1 )β1 + φtk (α2 )β2 + · · · + φtk (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 can
write each αi uniquely as αi = ρi − σi + ζi where supp ρi ⊆ R, supp σi ⊆ S and
supp ζi ⊆ Z. It is clear that ρi = φtk (αi ). In particular, |R| ≥ 1.
Set
\
\
W =
CtG (u) ∩
CG (u).
u∈R
u∈S
Then, by definition, each of the intersecting subgroups here has index at most k, so
|R ∪ S| ≤ |A| = a implies that |G : W | ≤ k a . Now let g ∈ G be a fixed element in
the support of φtk (α1 )β1 + φtk (α2 )β2 + · · · + φtk (αn )βn 6= 0, and let 0 6= d ∈ K be the
coefficient of ḡ. For convenience, set T 0 = T ∩ W . We will bound |G : T 0 | ≥ |G : T |.
Let x ∈ W \ T 0 . Then x ∈ G \ T , so
X
X
X
X
X
0 = x̄−1
αi x̄βi =
αx̄i βi =
ρx̄i βi −
σix̄ βi +
ζix̄ βi
i
i
i
i
i
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 cosets
of W -centralizers of elements of Z, with
X
g∈
/ supp
ζix̄ βi
for all x ∈ W \ C.
i
6
D. S. PASSMAN
Thus, for all x ∈ W \ (T 0 ∪ C), the coefficient of ḡ in
X
X
X
ζix̄ βi
ρ i βi +
σix̄ βi =
i
i
i
is precisely equal to d 6= 0.
Now if u ∈ supp σi ⊆ S, then W centralizes u, by the definition of W . In
particular, for all x ∈ W , ūx̄ = λu (x)ū
λu : W → K • is a linear character. It
P where
x̄
follows that the coefficient of ḡ in i σi βi is a fixed K-linear combination of the
various λu (x). In other words, there exist du ∈ K with
X
0 6= d =
du λu (x)
for all x ∈ W \ (T 0 ∪ C).
u∈S
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 to
conclude that |W : T 0 ∪ C| ≤ 2a−1 .
Let Y be a subset of W of size ≤ 2a−1 with W = (T 0 ∪ C)Y = T 0 Y ∪ CY . We
claim that CY 6= W . Indeed, since C is a union of ≤ ab cosets of W -centralizers of
elements of Z, the equality CY = W along with Lemma 2.1(iv) would imply that
|W : CW (u)| ≤ ab2a−1 for some u ∈ Z. But then
|G : CG (u)| ≤ |G : W ||W : CW (u)| ≤ ab2a−1 k a = 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 0 Y ∪ CY and Lemma 2.1(iii) yield the inequality
|W : T 0 Y | ≤ (ab2a−1 + 1)! and hence |W : T 0 | ≤ 2a−1 (ab2a−1 + 1)! ≤ (ab2a−1 + 2)!
since |Y | ≤ 2a−1 . Furthermore, by Lemma 2.1(ii) again,
|G : T | ≤ |G : T 0 | ≤ |G : W ||W : T 0 | ≤ k a (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 and
A ∩ Dtf (k) = A ∩ Dtk . Then either |G : T | ≤ k a (ab2a−1 + 2)! or
φtk (α1 )x̄β1 + φtk (α2 )x̄β2 + · · · + φtk (αn )x̄βn = 0
for all x ∈ G.
Proof. We suppose that |G : T | > k a (ab2a−1 + 2)!. Let g ∈ G be arbitrary and note
that
α1 x̄ḡβ1 + α2 x̄ḡβ2 + · · · + αn x̄ḡβn = 0
for all x ∈ G \ T g −1,
S
since x̄ḡ = d xg for some 0 6= d ∈ K. Furthermore, if B 0 = i supp ḡβi , then
B 0 = gB, so |B 0 | = |B| = b. Thus, since |G : T g −1| = |G : T | is suitably large,
Lemma 2.4 implies that
φtk (α1 )ḡβ1 + φtk (α2 )ḡβ2 + · · · + φtk (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) ≤ k r . Indeed, if
m = max{a, b}, then 2m+1 ≥ m2 , so k ≥ 4 yields k m ≥ 22m ≥ m2 2m−1 ≥ ab2a−1
and hence k r ≥ k m+a ≥ ab2a−1 k a = f (k). In the next lemma, we use a pigeon-hole
argument to show that
TWISTED GROUP ALGEBRAS
Lemma 2.6. Either |G : T | ≤ (1/r)3 ·4r
r+1
7
r
! or there exists an integer k ≤ 4r with
φtk (α1 )x̄β1 + φtk (α2 )x̄β2 + · · · + φtk (αn )x̄βn = 0
for all x ∈ G.
i
Proof. Let si = 4r 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 can
be 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 with
r
A∩(Dtsi+1 \Dtsi ) = ∅ = A∩(Dsi+1 \Dsi ). If k = si , then 4 ≤ k ≤ sr = 4r and, as we
observed, f (k) ≤ k r = sri = si+1 . Thus A ∩ Dtf (k) = A ∩ Dtk and A ∩ Df (k) = A ∩ Dk ,
so Lemma 2.5 implies that either
φtk (α1 )x̄β1 + φtk (α2 )x̄β2 + · · · + φtk (αn )x̄βn = 0
for all x ∈ G
or
|G : T | ≤ k a (ab2a−1 + 2)! ≤ (1/8ab)3 ·(ab2a−1 k a )!
≤ (1/r)3 ·f (k)! ≤ (1/r)3 ·sr+1 ! = (1/r)3 ·4r
r+1
!
2
a
since k ≥ 4 = 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 | ≤ 4r
t
r+1
! or
t
φ (α1 )x̄β1 + φ (α2 )x̄β2 + · · · + φt (αn )x̄βn = 0
for all x ∈ G.
r+1
Proof. Assume that |G : T | > 4r !. We proceed by induction on |A ∩ Dt | to show
that φt (α1 )x̄β1 + φt (α2 )x̄β2 + · · · + φt (αn )x̄βn = 0 for all x ∈ G. The result is, of
course, clear if A ∩ Dt = ∅. So suppose g ∈ A ∩ Dt and observe that
α01 x̄β1 + α02 x̄β2 + · · · + α0n x̄βn = 0
for all x ∈ G \ T,
S
−1
0
0
0
where αi = ḡ αi . Furthermore, if A = i supp αi , then A0 = g −1 A and A0 ∩ Dt =
g −1 (A ∩ Dt ), since g ∈ Dt . Thus the corresponding sets have the same size, and
1 ∈ A0 ∩ Dt . By the preceding lemma and the assumption on |G : T |, there exists
an integer k with
φtk (α01 )x̄β1 + φtk (α02 )x̄β2 + · · · + φtk (α0n )x̄βn = 0
Hence,
setting α00i = α0i − φtk (α0i ), we
α001 x̄β1 + α002 x̄β2 + · · · +
S
S
for all x ∈ G.
have
α00n x̄βn = 0
for all x ∈ G \ T.
Note that A00 = i supp α00i ⊆ i supp α0i = A0 and that 1 ∈ A0 ∩ Dt but 1 ∈
/
A ∩ Dt since 1 ∈ A0 ∩ Dtk . Thus |A00 ∩ Dt | < |A0 ∩ Dt | and, by induction,
00
φt (α001 )x̄β1 + φt (α002 )x̄β2 + · · · + φt (α00n )x̄βn = 0
(α00i )
for all x ∈ G.
Finally, note that φ
=φ
− φtk (α0i )) = φt (α0i ) − φtk (α0i ),
φt (α01 )x̄β1 + φt (α02 )x̄β2 + · · · + φt (α0n )x̄βn
= φtk (α01 )x̄β1 + φtk (α02 )x̄β2 + · · · + φtk (α0n )x̄βn = 0
t
t
(α0i
so
for all x ∈ G.
Thus, since φt (α0i ) = φt (ḡ −1 αi ) = ḡ −1 φt (αi ), we conclude that
φt (α1 )x̄β1 + φt (α2 )x̄β2 + · · · + φt (αn )x̄βn = 0
as required.
for all x ∈ G,
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 the
twisted group ring of support size ≤ 1. Alternately, we could state the result in
terms of the various truncations of the elements γj . Recall that a truncation of
an element in a twisted group ring is a partial sum of the natural expression of
that element as a K-linear combination of members of the basis G. Thus, for
example, φt (α) and α − φt (α) are both truncations of α ∈ K t [G]. Note also that
any truncation of a singleton is either itself or 0.
Proposition 2.8. Let K t [G] be a twisted group algebra and let αi , βi , γj , δj ∈ K t [G]
with i = 1, 2, . . . , n and j = 1, 2, . . . , m. Assume that each γj is a singleton. Let G
be a subgroup of G, let T be a subset of G, and suppose that
n
m
X
X
αi x̄βi +
γj x̄δj = 0
for all x ∈ G \ T.
Write A =
S
i=1
i
j=1
supp αi , B =
S
i
supp βi , C =
S
j
supp γj , D =
S
j
supp δj , and let
r ≥ 2 max{|A ∪ C|, |B ∪ D|} with r ≥ 4. Then either |G : T | ≤ (1/r)3 ·4r
n
m
X
X
φt (αi )x̄βi +
ψ(γj )x̄δj = 0
for all x ∈ G
i=1
r+1
! or
j=1
where ψ(γj ) is equal to 0 or to ȳφtk (ȳ −1 γj ) for some y ∈ A ∩ Dt and some integer
r
k ≤ 4r . In particular, if γj = 0, then ψ(γj ) = 0.
r+1
Proof. AssumePthat |G : T | >P
(1/r)3 ·4r !. We proceed by induction on |A ∩ Dt |
t
to show that i φ (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 and
observe that
n
m
X
X
α0i x̄βi +
γj0 x̄δj = 0
for all x ∈ G \ T,
i=1
j=1
S
where α0i = ḡ −1 αi and γj0 = ḡ −1 γj . Furthermore, if A0 = i supp α0i and C 0 =
S
0
−1
0
A, C 0 = g −1 C and A0 ∩ Dt = g −1 (A ∩ Dt ), since g ∈ Dt .
j supp γj , then A = g
Thus the corresponding sets have the same size, and 1 ∈ A0 ∩ Dt . By Lemma 2.6
r
and the assumption on |G : T |, there exists an integer k ≤ 4r with
n
m
X
X
(∗)
φtk (α0i )x̄βi +
φtk (γj0 )x̄δj = 0
for all x ∈ G.
i=1
Hence, setting
α00i
=
j=1
α0i
n
X
i=1
−
φtk (α0i )
α00i x̄βi +
and γj00 = γj0 − φtk (γj0 ), we have
m
X
γj00 x̄δj = 0
j=1
for all x ∈ G \ T.
S
S
Note that A00 = i supp α00i ⊆ i supp α0i =SA0 and that S
1 ∈ A0 ∩ Dt but 1 ∈
/
00
t
0
t
00
00
A ∩ D since 1 ∈ A ∩ Dk . Furthermore, C = j supp γj ⊆ j supp γj0 = C 0 , and
γj00 = γj0 − φtk (γj0 ) is a truncation of γj0 and hence it is also a singleton. Thus, since
|A00 ∩ Dt | < |A0 ∩ Dt |, induction implies that
n
m
X
X
(∗∗)
φt (α00i )x̄βi +
ψ(γj00 )x̄δj = 0
for all x ∈ G
i=1
j=1
TWISTED GROUP ALGEBRAS
9
and for appropriate ψ(γj00 ). Next, φt (α00i ) = φt (α0i − φtk (α0i )) = φt (α0i ) − φtk (α0i ), so
φt (α0i ) = φt (α00i ) + φtk (α0i ) and, by adding (∗) and (∗∗), we obtain
n
X
φt (α0i )x̄βi +
i=1
m
X
[φtk (γj0 ) + ψ(γj00 )]x̄δj = 0
for all x ∈ G.
j=1
Thus, since φt (α0i ) = φt (ḡ −1 αi ) = ḡ −1 φt (αi ), we conclude that
n
X
φt (αi )x̄βi +
i=1
m
X
ψ(γj )x̄δj = 0
for all x ∈ G,
j=1
where we set ψ(γj ) = ḡφtk (γj0 ) + ḡψ(γj00 ) = ḡφtk (ḡ −1 γj ) + ḡψ(γj00 ). It remains to
better understand ψ(γj ).
To this end, suppose first that ḡφtk (γj0 ) = ḡφtk (ḡ −1 γj ) 6= 0. Then since γj0 =
−1
ḡ γj is a singleton, it follows that γj00 = γj0 − φtk (γj0 ) = 0. Hence ψ(γj00 ) = 0 and
ψ(γj ) = ḡφtk (ḡ −1 γj ), as required. On the other hand, if ḡφtk (γj0 ) = ḡφtk (ḡ −1 γj ) = 0,
then γj00 = γj0 = ḡ −1 γj is a singleton and ψ(γj ) = ḡψ(γj00 ) is either 0 or equal to
r
ḡȳθ`t (ȳ −1 γj00 ) = ḡȳθ`t (ȳ −1 ḡ −1 γj ) for some y ∈ A00 ∩ Dt and ` ≤ 4r . Note that, in
the latter case, gy ∈ Dt and y ∈ A00 ⊆ A0 = g −1 A so gy ∈ A. Thus gy ∈ A ∩ Dt
and, since gy = dḡȳ for some 0 6= d ∈ K, we conclude that ψ(γj ) = gyθ`t (gy −1 γj )
also has the appropriate form.
3. Generalized Polynomial Identities
We now move on to consider multilinear generalized polynomial identities. Since
these identities are, in some sense, combinations of linear identities, it is not surprising 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 K t [G] is given and that G is a normal subgroup
of G. As we observed, the latter implies that D, Dt , Dk and Dtk are all normal
subsets of G. Let
aσ
X X
f (ζ1 , ζ2 , . . . , ζn ) =
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
σ∈Symn j=1
be a multilinear generalized polynomial with each αi,σ,j in K t [G], and assume
throughout that K t [G] satisfies f , so that
f (γ1 , γ2 , . . . , γn ) = 0
for all γ1 , γ2 , . . . , γn ∈ K t [G].
As usual, for each σ ∈ Symn , we write
fσ (ζ1 , ζ2 , . . . , ζn ) =
aσ
X
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
j=1
for the σ-component of f . We start with
Lemma 3.1. Let K t [G], G, and f be as above. Assume further that each αi,σ,j is
a singleton and that
a1
X
j=1
φt (α0,1,j )φt (α1,1,j ) · · · φt (αn,1,j ) 6= 0.
10
D. S. PASSMAN
P
r
If r ≥ 2 σ aσ , twice the number of monomials which occur in f , and if k = 4r ,
r+1
then |G : ∆tk2 | ≤ (n/r)·4r !.
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 generalized
multilinear polynomial fi of degree n − i by
fi (ζi+1 , ζi+2 , . . . , ζn ) =
aσ
XX
βi,σ,j ζσ(i+1) αi+1,σ,j · · · αn−1,σ,j ζσ(n) αn,σ,j
σ∈Si j=1
where
βi,σ,j = φt (α0,σ,j )φt (α1,σ,j ) · · · φt (αi−1,σ,j )αi,σ,j .
Then f0 = f and
fn =
a1
X
βn,1,j =
a1
X
φt (α0,1,j )φt (α1,1,j ) · · · φt (αn−1,1,j )αn,1,j
j=1
j=1
t
is a nonzero element of K [G] since
t
φ (fn ) =
a1
X
φt (α0,1,j )φt (α1,1,j ) · · · φt (αn−1,1,j )φt (αn,1,j ) 6= 0.
j=1
Next, set M00 = ∅ and, for i = 1, 2, . . . , n, let M0i to be the set of all monomials
µ of the form
−1
µ = βi−1,τ,j
0 βi−1,σ,j ζσ(i) αi,σ,j ζσ(i+1) αi+1,σ,j · · · α`−2,σ,j ζσ(`−1) α`−1,σ,j
with σ ∈ Si−1 \ Si , σ(`) = i, and τ ∈ Si . Furthermore, we must have βi−1,τ,j 0 6= 0
and, since this element is a singleton, it is invertible in K t [G]. Note that any such
µ involves only variables from the set {ζi+1 , ζi+2 , . . . , ζn } and that at least one
variable occurs since σ ∈ Si−1 \ Si implies that σ(i) 6= i. Furthermore, since µ is
uniquely determined by the
j} and {τ, j 0 }, we see that |M0i | ≤ r2 and that
Snpairs {σ,
0
0
Mn = ∅. Now let M = i=0 Mi 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| ≤ nr 2 ,
Mi ⊇ M0i , and M = M0 ⊇ M1 ⊇ · · · ⊇ Mn = ∅.
r
r+1
Set k = 4r . Our goal is to show that |G : ∆tk2 | ≤ (n/r)·4r !. Suppose, by
way of contradiction, that this is not the case. Then we will prove by induction on
i = 0, 1, . . . , n that, for all xi+1 , xi+2 , . . . , xn ∈ G, either fi (x̄i+1 , x̄i+2 , . . . , x̄n ) = 0
or there exists a monomial µ in Mi with supp µ(x̄i+1 , x̄i+2 , . . . , x̄n ) ∩ Dtk 6= ∅. This
is, of course, clear for i = 0 since f0 = f is a generalized polynomial identity
satisfied by K t [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 ith
variable. Let µ ∈ Mi . If supp µ(x̄i+1 , x̄i+2 , . . . , x̄n ) ∩ Dtk 6= ∅, we are done. Thus
we may suppose that supp µ(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 . Write
µ = µ0 ζi µ00 where µ0 and µ00 are monomials in the variables ζi+1 , ζi+2 , . . . , ζn . Since
Dtk is a normal subset of G and since each αi,σ,j is a singleton, it follows that
supp µ(x̄, x̄i+1 , . . . , x̄n ) ∩ Dtk 6= ∅ if and only if
x ∈ (h0 )−1 Dtk (h00 )−1 = Dtk (h0 )−1 (h00 )−1 = Dtk yµ
TWISTED GROUP ALGEBRAS
11
where supp µ0 (x̄i+1 , x̄i+2 , . . . , x̄n ) = {h0 }, supp µ00 (x̄i+1 , x̄i+2 , . . . , x̄n ) = {h00 }, and
yµ = (h0 )−1 (h00 )−1 . In particular, if x ∈ G \ T where
[
T =
G ∩ Dtk yµ ,
µ∈Ni−1
then we have supp µ(x̄, x̄i+1 , . . . , x̄n ) ∩ Dtk = ∅ for all µ ∈ Ni−1 = Mi−1 \ Mi and
hence for all µ ∈ Mi−1 .
By induction, we conclude that
0 = fi−1 (x̄, x̄i+1 , . . . , x̄n )
=
aσ
X X
βi−1,σ,j x̄σ(i) αi,σ,j · · · αn−1,σ,j x̄σ(n) αn,σ,j
σ∈Si−1 j=1
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 those
σ ∈ Si−1 \ Si . Since all the summands in fi−1 are singletons, it is clear, in the
notation 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 possible
conclusions.
r+1
Suppose first that |G : T | ≤ (1/r)3 ·4r !, and note that T is a union of |Ni−1 | ≤
|M| ≤ nr2 subsets of the form G ∩ Dtk yµ with yµ ∈ G. Furthermore, if g, h ∈
G ∩ Dtk yµ , then gh−1 ∈ G ∩ (Dtk )2 ⊆ G ∩ Dtk2 = ∆tk2 . Thus G ∩ Dtk yµ is contained
in a right G-translate of ∆tk2 , and it follows from Lemma 2.1(ii) that |G : ∆tk | ≤
r+1
r+1
nr2 ·(1/r)3 ·4r ! = (n/r)·4r !, contrary to our assumption. Consequently, the
second 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 and
only 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 )φt (α1,σ,j ) · · · φt (αi−2,σ,j )αi−1,σ,j αi,σ,j
= φt (α0,σ,j )φt (α1,σ,j ) · · · φt (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 2.8
sum 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 , then
certainly ȳ = dβi−1,τ,j 0 for some τ ∈ Si , 1 ≤ j 0 ≤ aτ and 0 6= d ∈ K. Thus ψ(γ) is
either equal to 0 or to a suitable
−1
ȳφtk0 (ȳ −1 γ) = βi−1,τ,j 0 φtk0 βi−1,τ,j
= βi−1,τ,j 0 φtk0 µ(x̄i+1 , x̄i+2 , . . . , x̄n )
0γ
12
D. S. PASSMAN
with k 0 ≤ k and with
−1
µ = βi−1,τ,j
0γ
−1
= βi−1,τ,j
0 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
∈ M0i ⊆ Mi .
But, by assumption, we have
supp µ(x̄i+1 , x̄i+2 , . . . , x̄n ) ∩ Dtk = ∅ and consequently
t
φk0 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. In
other words, that result yields fi (x̄i+1 , x̄i+2 , . . . , x̄n ) = 0, and the inductive step is
proved.
In particular, when i = n, the inductive result implies that either fn = 0 or
there exists µ ∈ Mn with supp µ ∩ Dtk 6= ∅. However, fn is known to be a nonzero
constant and Mn = ∅. Hence we have a contradiction and therefore it follows that
r+1
|G : ∆tk | ≤ (n/r)·4r !, as required.
Next, we isolate the procedure of [7, Lemma 2.1] which really only uses the fact
that 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
˜ 1 , ζ2 , . . . , ζn ) = f (z̄ −1 ζ1 z̄1 , z̄ −1 ζ2 z̄2 , . . . , z̄ −1 ζn z̄n )
f(ζ
n
1
2
aσ
X X
=
α̃0,σ,j ζσ(1) α̃1,σ,j ζσ(2) · · · α̃n−1,σ,j ζσ(n) α̃n,σ,j
σ∈Symn j=1
is a multilinear generalized polynomial identity for K t [G] with
a1
X
φt (α̃0,1,j )φt (α̃1,1,j ) · · · φt (α̃n,1,j ) 6= 0.
j=1
Proof. Let
for Dt in G. Then we can write each αi,1,j as
P Y be a right transversal
t t
αi,1,j = s βi,j,s ȳs with βi,j,s ∈ K [D ] and ys ∈ Y . Since
0 6= f1 (1, 1, . . . , 1) =
and since each αi,1,j =
P
0 6=
a1
X
α0,1,j α1,1,j · · · αn,1,j
j=1
s
βi,j,s ȳs , there exist ys0 , ys1 , . . . , ysn ∈ Y with
a1
X
β0,j,s0 ȳs0 β1,j,s1 ȳs1 · · · βn,j,sn ȳsn .
j=1
In particular, if we set z̄i = ȳs0 ȳs1 · · · ȳsi−1 and z̄0 = 1, then
0 6=
a1
X
−1
−1
−1
(β0,j,s0 )z̄0 (β1,j,s1 )z̄1 · · · (βn,j,sn )z̄n .
j=1
Now βi,j,si = φ
t
(αi,1,j ȳs−1
)
i
−1
(βi,j,si )z̄i
so, since Dt / G,
−1
).
= φt (z̄i αi,1,j ȳs−1
z̄i−1 ) = φt (z̄i αi,1,j z̄i+1
i
It therefore follows that if we replace ζi in f by z̄i−1 ζi z̄i and if we multiply f on the
−1
right by z̄n+1
, then this new multilinear generalized polynomial f˜ has the required
TWISTED GROUP ALGEBRAS
13
property concerning φt . Furthermore, since G is normal in G, f˜ is also an identity
for K t [G].
With this, we now quickly prove our main result.
Theorem 3.3. Let K t [G] be a twisted group algebra of G over K and let G be
a normal subgroup of G. Suppose that K t [G] satisfies the multilinear generalized
polynomial identity
aσ
X X
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
f (ζ1 , ζ2 , . . . , ζn ) =
σ∈Symn j=1
with each αi,σ,j in K t [G], but that there exists τ ∈ Symn with fτ not an identity
for K t [G]. If
aσ Y
n
X X
r=2
| supp αi,σ,j |,
σ∈Symn j=1 i=0
then |G :
∆tk2 |
≤4
r r+1
r
! where k = 4r .
Proof. We modify the identity f so that Lemma 3.1 can apply. To start with, write
each αi,σ,j in terms of the K-basis G and apply the distributive law to each product
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j .
Q
By doing this, we obtain ni=0 | supp αi,σ,j | similar products but with parameters
which are all singletons. In other words, we can now assume that each αi,σ,j is a
singleton and that r/2 is equal to the total number of terms that occur. Indeed, if
we choose to combine terms which are K-multiples of each other, then r/2 will be
greater than or equal to this total number of terms.
Next, we know that fτ is not an identity for K t [G] and, by suitably permuting
the variables, we can clearly assume that τ = 1. Furthermore, since f1 is multilinear and since G is a K-basis for K t [G], there exist g1 , g2 , . . . , gn ∈ G with
f1 (ḡ1 , ḡ2 , . . . , ḡn ) 6= 0. Thus, by replacing each ζi by ζi ḡi , which does not effect the
number of terms in f or the fact that each αi,σ,j is a singleton, we can assume that
f1 (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 the
collection of all such iσ and jσ , and note that |U | ≤ 2·(r/2) = r since f has at most
r/2 terms. Suppose f 0 is the polynomial obtained from f by setting ζi = 1 for all
i∈
/ U . Then f 0 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 0 = |U | ≤ r.
Furthermore, since U contains all iσ and jσ , we see that, if the ζi -terms in
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
are not in their natural order, then the truncated set of ζi ’s which remain in f 0 are
also not in their natural order. In other words, f10 (1, 1, . . . , 1) = f1 (1, 1, . . . , 1) 6= 0
and f 0 is a nondegenerate multilinear identity for K t [G]. Since deg f 0 ≤ r, this now
allows us to assume that n = deg f ≤ r.
Finally, we apply Lemma 3.2 to f , and note that this does not change the degree
of 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, and
since 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 K t [G] be a twisted group algebra of G over K and let G be
a normal subgroup of G. Suppose that K t [G] satisfies the multilinear generalized
polynomial identity
f (ζ1 , ζ2 , . . . , ζn ) =
aσ
X X
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
σ∈Symn j=1
with each αi,σ,j in K t [G], but that there exists τ ∈ Symn with fτ not an identity
for K t [G]. If
aσ Y
n
X X
| supp αi,σ,j |,
r=2
σ∈Symn j=1 i=0
then
r+1
(i) |G : ∆| ≤ 4r ! and the commutator subgroup of ∆ is finite.
r+1
(ii) |G : ∆t | ≤ 4r ! and the commutator subgroup of ∆t is finite.
r+1
(iii) G has a normal subgroup H ⊆ ∆t with |G : H| ≤ 4r ! and with |H 0 | bounded
by a function of r. Furthermore, H = ∆th (H) for some integer h depending
upon r.
r
r+1
Proof. If k = 4r , then the previous theorem implies that |G : ∆tk2 | ≤ 4r ! = `. In
particular, since ∆ ⊇ ∆t ⊇ ∆tk2 , it follows that ` also bounds |G : ∆| and |G : ∆t |.
Furthermore, since ∆k2 ⊇ ∆tk2 , we have |G : ∆k2 | < ∞ and [7, Lemma 2.3] implies
that the commutator subgroup of ∆ is finite. Clearly, the same is also true of the
commutator subgroup of ∆t ⊆ ∆.
Finally, we let H be the subgroup of G generated by ∆tk2 . Then H / G, H ⊆ ∆t ,
`
and |G : H| ≤ |G : ∆tk2 | ≤ `. Since H = (∆tk2 )4 , by [8, Lemma 5.2.3], it follows
4`
that H = ∆th (H) where h = (k 2 ) . In particular, we also have H = ∆h (H), so [8,
Theorem 5.2.9], a result of Wiegold, implies that |H 0 | is bounded by a function of
h and hence by a function of r.
Theorem 1.1 is, of course, an immediate consequence of part (iii) above and, in
some sense, it is the most that we can conclude about the group-theoretic structure
of G. Indeed, it was shown in [7] that an ordinary group algebra K[G] satisfies
a nondegenerate multilinear polynomial identity if and only if G has a normal
subgroup H of finite index with H 0 finite. But this structure does not guarantee a
g.p.i. in the case of twisted group algebras as can be seen in the following
Example 3.5. Let p be a prime and let K be a field containing a nonidentity
pth root of unity. If G is a countably infinite elementary abelian p-group, then
there exists a twisted group algebra K t [G] which does not satisfy a nondegenerate
multilinear generalized polynomial identity, but does have G = ∆tp (G).
Proof. Let ε 6= 1 be the given pth root of unity in K, and let x1 , x2 , . . . , y1 , y2 , . . . be
a basis for G. Define R to be the K-algebra whose generators x̄1 , x̄2 , . . . , ȳ1 , ȳ2 , . . .
satisfy the relations x̄pi = 1 = ȳip , x̄i x̄j = x̄j x̄i , ȳi ȳj = ȳj ȳi , and
(
ȳj x̄i , if i 6= j, or
x̄i ȳj =
εȳj x̄i , if i = j.
TWISTED GROUP ALGEBRAS
15
Then it is easy to see that R = K t [G] is a twisted group algebra of G. In fact, if
g ∈ G, then the possible G-conjugates of ḡ are ḡ, εḡ, . . . , εp−1 ḡ, so g ∈ ∆tp (G), and
hence G = ∆tp (G).
Note that the subalgebra of K t [G] generated by x̄i , ȳi is isomorphic to the matrix
ring Mp (K), since this corresponds to a nonlinear irreducible representation of a
nonabelian group of order p3 . Thus, the subalgebra of K t [G] generated by k pairs
{x̄i , ȳi } is isomorphic to Mp (K)⊗k = Mpk (K).
Finally, suppose that K t [G] satisfies the nondegenerate g.p.i.
X
fσ (ζ1 , ζ2 , . . . , ζn )
f (ζ1 , ζ2 , . . . , ζn ) =
σ∈Symn
t
and let s1 , s2 , . . . , sn ∈ K [G] with f1 (s1 , s2 , . . . , sn ) 6= 0. We may assume that all
terms in f (ζ1 , ζ2 , . . . , ζn ) and all s1 , s2 , . . . , sn are contained in K t [H], where H =
hx1 , y1 , x2 , y2 , . . . , xr , yr i. Now let W = hxr+1 , yr+1 , xr+2 , yr+2 , . . . , xr+k , yr+k i so
that K t [W ] ∼
= Mpk (K). Furthermore, since K t [W ] and K t [H] commute elementwise, we see that K t [W ] satisfies the identity
X
f (s1 ζ1 , s2 ζ2 , . . . , sn ζn ) =
fσ (s1 , s2 , . . . , sn )ζσ(1) ζσ(2) · · · ζσ(n) .
σ∈Symn
In particular, if k is chosen so that pk > n, then we can take ζ1 , ζ2 , . . . , ζn to be the
matrix units ζi = ei,i+1 and conclude that
f (s1 e1,2 , s2 e2,3 , . . . , sn en,n+1 ) = f1 (s1 , s2 , . . . , sn )e1,n+1
since all products of the matrix units other than e1,2 e2,3 · · · en,n+1 = e1,n+1 are
zero. But 0 6= f1 (s1 , s2 , . . . , sn ) ∈ K t [H], and 0 6= e1,n+1 ∈ K t [W ], so we have a
contradiction since K t [H × W ] = K t [H] ⊗ K t [W ].
In the preceding example, if H is any subgroup of G of finite index, then it is
easy to see that K t [H] is noncommutative. On the other hand, G is an abelian
group, so the ordinary group algebra K[H] is certainly commutative. In particular,
the two algebras K t [H] and K[H] are not K-isomorphic, nor can they become
isomorphic under any field extension of K. As we will see in the next section,
it is precisely this condition which keeps K t [G] from satisfying a nondegenerate
multilinear polynomial identity.
4. Stable Untwisting
We say that the twisted group algebra K t [G] is untwisted if there exists a function
G → K • , given by x 7→ dx , such that the elements x̃ = dx x̄ with x ∈ G form a
group basis for K t [G]. Thus x̃ỹ = x
fy for all x, y ∈ G, and K t [G] ∼
= K[G] via this
diagonal change of basis. In addition, we say that K t [G] is stably untwisted if there
exists a field F ⊇ K such that F t [G] = F ⊗K K t [G] is untwisted. The standard
untwisting criterion is
Lemma 4.1. K t [G] is untwisted if and only if there exists an algebra homomorphism ρ : K t [G] → K. Furthermore, K t [G] is stably untwisted if and only if there
exists an algebra homomorphism ρ : K t [G] → R with R a commutative algebra.
Proof. If K t [G] ∼
= K[G], then there exists a homomorphism ρ : K t [G] → K corresponding to the augmentation map of K[G]. Conversely, if ρ exists, we define
x̃ = ρ(x)−1 x̄ for all x ∈ G. Then x̃ is the unique element in K x̄ with ρ(x̃) = 1, and
16
D. S. PASSMAN
e = {x̃ | x ∈ G} is a group basis for
consequently x̃ỹ = x
fy for all x, y ∈ G. Thus G
t
t
∼
K [G], and K [G] = K[G].
If K t [G] is stably untwisted with associated field F ⊇ K, then by the above we
have an algebra homomorphism ρ : K t [G] → F t [G] → F . Conversely, if ρ exists, let
F be an algebraically closed field extension of K with |F | > |G|, an inequality of
cardinal numbers. Then ρ extends to an F -algebra map ρ : F t [G] = F ⊗K K t [G] →
F ⊗K R → L where the field L is equal to F ⊗ R modulo a maximal ideal. Since
dimF L ≤ dimF F t [G] = |G| < |F |, [8, Lemma 7.1.2(iii)] implies that L = F . Thus
F t [G] is untwisted and K t [G] is stably untwisted.
As an application of the above, suppose a ∈ K does not have an nth root in the
field, and let G = hxi be cyclic of order n. Then K t [G] = K[x̄ | x̄n = a] cannot be
untwisted over K, but it can be untwisted once we adjoin an nth root of a to the
field. Next, we show
Lemma 4.2. Let K t [G] be given and let H be a normal subgroup of G of finite
index with |H 0 | < ∞ and with K t [H] stably untwisted. Then K t [G] satisfies a
nondegenerate multilinear generalized polynomial identity.
Proof. By assumption, there exists a field F ⊇ K with F ⊗K K t [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 the
smallest ideal of F t [H] with commutative quotient. Viewed in F [H], we see that
I ⊆ ω(F [H 0 ])·F [H], the kernel of the natural map F [H] → F [H/H 0 ], and hence,
since H 0 is finite, we have αI = 0 where α 6= 0 is the sum in F [H 0 ] of all elements
of H 0 . In other words, there exists 0 6= α ∈ F t [H 0 ] ⊆ F t [H] with αI = 0.
Next, let |G : H| = n, write M = Mn (F t [H]) and consider the natural homomorphism Mn (F t [H]) → Mn (F t [H]/I). Since F t [H]/I is commutative, the AmitsurLevitzki theorem (see [8, Theorem 5.1.9]) implies that Mn (F t [H]/I) satisfies the
standard identity
X
sm (ζ1 , ζ2 , . . . , ζm ) =
(−1)σ ζσ(1) ζσ(2) · · · ζσ(m)
σ∈Symm
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 (F t [H]) as in [8, Lemma 5.1.10]. Then under this
embedding, α ∈ F t [H] maps to the diagonal matrix
−1
−1
diag(x̄1 αx̄−1
1 , x̄2 αx̄2 , . . . , x̄n αx̄n ),
annihilates I since α annihilates I and since conjugation
and note that each x̄i αx̄−1
i
by x̄−1
is
an
automorphism
of F t [H] stabilizing the characteristic ideal I. With
i
t
this, we conclude that αsm (F [G], F t [G], . . . , F t [G]) maps to 0 under the embedding
into M , and hence we see that F t [G] satisfies the multilinear generalized polynomial
identity
X
αsm (ζ1 , ζ2 , . . . , ζm ) =
(−1)σ αζσ(1) ζσ(2) · · · ζσ(m) .
σ∈Symm
t
t
Finally, since F [G] is free over K [G], we conclude that K t [G] satisfies the multilinear generalized polynomial identity
X
f (ζ1 , ζ2 , . . . , ζm ) = βsm (ζ1 , ζ2 , . . . , ζm ) =
(−1)σ βζσ(1) ζσ(2) · · · ζσ(m)
σ∈Symm
TWISTED GROUP ALGEBRAS
17
where 0 6= β ∈ K[H 0 ] occurs in the representation of α ∈ F [H 0 ] = F ⊗K K[H 0 ]. Of
course, 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 general
context of G ⊆ G. To start with, we need the following standard fact.
Lemma 4.3. Let K t [G] be a twisted group algebra of the abelian group G and
suppose that K t [G] satisfies an ordinary polynomial identity of degree n. Then G
has a subgroup Z of index ≤ (n/2)2 with K t [Z] central in K t [G].
Proof. Let F be an algebraically closed field extension of K with |F | > |G|, and
observe that F t [G] = F ⊗K K t [G] also satisfies an ordinary polynomial identity of
degree n. In particular, if ξ : F t [G] → R is any irreducible representation, then R
is a primitive p.i. algebra, so Kaplansky’s theorem (see [8, Theorem 5.3.4]) implies
that 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 K t [G]}. Then Z is a subgroup of
G and K t [Z] is central in K t [G].
Now define χ : G → F so that, for each g ∈ G, χ(g) is the trace of the m × m
matrix 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 there
exists h ∈ G so that ḡ and h̄ do not commute. In particular, since G is abelian,
we have h̄−1 ḡh̄ = kḡ for some 1 6= k ∈ K • , and then ξ(h̄)−1 ξ(ḡ)ξ(h̄) = k·ξ(ḡ). But
similar 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 matrices
with nonzero trace.
Finally, if {g1 , g2 , . . .} is a transversal for Z in G, then we claim
P that the set
{ξ(ḡ1 ), ξ(ḡ2 ), . . .} is F -linearly independent in Mm (F ). Indeed, if sj=1 fj ξ(ḡj ) = 0
P
is an F -linear relation, then for all i we have sj=1 fj ξ(ḡj )ξ(ḡi )−1 = 0. Moreover,
gj,i = gj gi−1 ∈ Z if and only if i = j, and in all cases ḡj ḡi−1 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 in
F . 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 K t [G] is a commutative twisted group algebra of the finite
group G over the algebraically closed field K. If Ω is a nonempty, nonzero subset
of K t [G], then there exists γ ∈ K t [G] with 0 6= γΩ ⊆ I, a 1-dimensional ideal.
Proof. Since K is algebraically closed, and K t [G] is commutative and finite dimensional, it follows from Lemma 4.1 that K t [G] = K[G] is an ordinary group algebra
of the finite abelian group G. Let char K = p ≥ 0 and write G = P × H where P
is the Sylow p-subgroup of G and H is the p-complement. Of course, if p = 0, then
G = H and P = 1.
P
Since K[H] is semiprimitive, write K[H] = t1 ej K, where {ej | j = 1, . . . , t}
is the set of primitive idempotents of this group algebra. Certainly, ei Ω 6= 0 for
some i, and ei Ω ⊆ ei K[P ] ∼
= K[P ] since K[G] ∼
= K[H] ⊗K K[P ]. Finally, let
ω(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 ])m ei Ω 6= 0, and we can
choose γ ∈ ω(K[P ])m ei with γΩ 6= 0. But then γΩ is a subset of ei K[P ] ∼
= K[P ]
18
D. S. PASSMAN
annihilated by the augmentation ideal, so γΩ ⊆ Kei P̂ = I, where P̂ is the sum of
the 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 different
irreducible representation. On the other hand, if A = K[x, y | (x, y)2 = 0], then it
is clear that there is no multiple of Ω = {x, y} which is both nonzero and contained
in a 1-dimensional subspace. Now we return to infinite groups.
Lemma 4.5. Let K t [G] be given and let G be a normal subgroup of G. Suppose
that K t [G] satisfies the multilinear generalized polynomial identity
aσ
X X
f (ζ1 , ζ2 , . . . , ζn ) =
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
σ∈Symn j=1
with each αi,σ,j in K t [G], and assume that there exists τ ∈ Symn with fτ not an
identity for K t [G]. Then G has a subgroup H of finite index with K t [H] satisfying
an identity of the form
X
ασ ζσ(1) ζσ(2) · · · ζσ(n)
f˜(ζ1 , ζ2 , . . . , ζn ) =
σ∈Symn
with ασ ∈ K [G] for all σ ∈ Symn and with α1 6= 0. Furthermore, |H 0 | < ∞ and
K t [H 0 ] is central in K t [H].
t
Proof. As in the proof of Theorem 3.3, we first modify f so that f1 (1, 1, . . . , 1) 6= 0
and that all αi,σ,j are singletons. Then we apply Lemma 3.2 to obtain a new identity
aσ
X X
f (ζ1 , ζ2 , . . . , ζn ) =
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
σ∈Symn j=1
with
a1
X
φt (α0,1,j )φt (α1,1,j ) · · · φt (αn,1,j ) 6= 0,
j=1
while still preserving the fact that each αi,σ,j is a singleton. Let H = ∆t (G) so
that |G : H| < ∞ and |H 0 | < ∞ by Corollary 3.4(ii). Since |G : H| < ∞, it is clear
that DtG (G) = DtG (H), and hence φt above maps K t [G] to K t [DtG (H)].
Fix h2 , h3 , . . . , hn ∈ H and consider the linear identity
0 = f (h̄1 , h̄2 , . . . , h̄n )
X
=
α0,σ,j h̄σ(1) α1,σ,j h̄σ(2) · · · αn−1,σ,j h̄σ(n) αn,σ,j
σ,j
=
X
βσ,j h̄1 γσ,j
for all h1 ∈ H.
σ,j
P
Then, by Lemma 2.7 with T = ∅, we have 0 = σ,j φt (βσ,j )h̄1 γσ,j for all h1 ∈ H
and, by applying φt to this expression, we obtain
X
0=
φt (βσ,j )h̄1 φt (γσ,j )
for all h1 ∈ H,
σ,j
t
since φ (βσ,j )h̄1 ∈ K[DtG (H)]. Note that each αi,σ,j is a singleton, so it follows that
each βσ,j and γσ,j is also a singleton. Hence φt (βσ,j ) = 0 or βσ,j , and φt (γσ,j ) = 0
TWISTED GROUP ALGEBRAS
19
or γσ,j . Furthermore, since H ⊆ DtG (H) = DtG (G) / G, the question of whether
supp βσ,j ∈ DtG (H) or not does not depend upon the elements h̄2 , h̄3 , . . . , h̄n that
occur in the product expression for βσ,j , and similarly this is also true for γσ,j .
P0
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
X0
βσ,j h̄1 γσ,j
0=
σ,j
is a multilinear identity that holds for all h1 , h2 , . . . , hn ∈ H.
We continue this process, applying the above truncation procedure to the variables h2 , h3 , . . . , hn in turn, thereby obtaining a multilinear generalized polynomial
X00
f 00 (ζ1 , ζ2 , . . . , ζn ) =
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
σ,j
P
satisfied by K t [H]. Again 00σ,j denotes the partial sum of all those terms that
survive this process. We note that if
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
survives, then when the truncation procedure was applied to h̄σ(1) we must have
had supp α0,σ,j ∈ DtG (H). Furthermore, when the procedure was applied to h̄σ(2) ,
we must have had supp α0,σ,j α1,σ,j ∈ DtG (H), and hence supp α1,σ,j ∈ DtG (H).
Continuing in this manner, we see that
f 00 (ζ1 , ζ2 , . . . , ζn )
X
=
φt (α0,σ,j )ζσ(1) φt (α1,σ,j )ζσ(2) · · · φt (αn−1,σ,j )ζσ(n) φt (αn,σ,j ),
σ,j
where the αn,σ,j term comes from a consideration of the truncation procedure applied to h̄σ(n) .
S
If A = i,σ,j supp φt (αi,σ,j ), then A is a finite subset of DtG (H), and hence
T
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 that
K t [C] satisfies
X
ασ ζσ(1) ζσ(2) · · · ζσ(n)
σ∈Symn
where
ασ =
X
φt (α0,σ,j )φt (α1,σ,j ) · · · φt (αn−1,σ,j )φt (αn,σ,j ).
j
In particular, we know that α1 6= 0.
Finally, C ⊆ H, so C 0 ⊆ H 0 and |C 0 | < ∞. Thus, since C ⊆ H = ∆t (H), we
see that L = CtC (C 0 ) is a subgroup of C of finite index. Moreover, L0 ⊆ C 0 , so
L = CtL (L0 ) and K t [L0 ] is central in K t [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 K t [G] be a twisted group algebra of G over K and let G be
a normal subgroup of G. Suppose that K t [G] satisfies the multilinear generalized
polynomial identity
aσ
X X
f (ζ1 , ζ2 , . . . , ζn ) =
α0,σ,j ζσ(1) α1,σ,j ζσ(2) · · · αn−1,σ,j ζσ(n) αn,σ,j
σ∈Symn j=1
20
D. S. PASSMAN
with each αi,σ,j in K t [G], but that there exists τ ∈ Symn with fτ not an identity
for K t [G]. Then G has a normal subgroup H of finite index with |H 0 | < ∞ and
with K t [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 0 | < ∞ and with K t [H] stably
untwisted. If C is the core of H in G, then we know that C / G, C ⊆ H and
|G : C| < ∞. Furthermore, C 0 ⊆ H 0 so |C 0 | < ∞, and K t [C] ⊆ K t [H] so K t [C]
is stably untwisted, because untwistings involve a diagonal change of basis. In
view of all of this, it suffices to prove the result without the normality conclusion.
In particular, by the previous lemma, we can now assume that K t [G] satisfies an
identity of the form
X
f˜(ζ1 , ζ2 , . . . , ζn ) =
ασ ζσ(1) ζσ(2) · · · ζσ(n)
σ∈Symn
with ασ ∈ K t [G] for all σ ∈ Symn and with α1 6= 0. Furthermore, we can suppose
that |G0 | < ∞ and that K t [G0 ] is central in K t [G].
˜ 1 , ζ2 , . . . , ζn ). To start with, by multiplying
We now procede to further modify f(ζ
˜
f on the left by ḡ for some g ∈ G, we can assume that 1 ∈ supp α1 . Next, if
x1 , x2 , . . . , xn ∈ G, write x = x1 x2 · · · xn and note that x̄σ(1) x̄σ(2) · · · x̄σ(n) = βσ x̄
for some βσ ∈ K t [G0 ] depending upon x1 , x2 , . . . , xn . Thus
X
α σ βσ
0 = f˜(x̄1 , x̄2 , . . . , x̄n )x̄−1 =
σ∈Symn
and
0=
X
πG0 (ασ βσ ) =
X
πG0 (ασ )βσ
σ∈Symn
σ∈Symn
where πG0 : K t [G] → K t [G0 ] is the natural projection. In particular,
X
X
πG0 (ασ )x̄σ(1) x̄σ(2) · · · x̄σ(n) ,
0=
πG0 (ασ )βσ x̄ =
σ∈Symn
σ∈Symn
so we see that K t [G] satisfies
X
πG0 (ασ )ζσ(1) ζσ(2) · · · ζσ(n) .
σ∈Symn
Since 1 ∈ supp α1 implies that πG0 (α1 ) 6= 0, we can now assume that ασ ∈ K t [G0 ]
for all σ ∈ Symn . Note that Ω = {ασ | σ ∈ Symn } is a nonzero, nonempty subset of
the finite-dimensional commutative K-algebra K t [G0 ] with K algebraically closed.
Thus, by Lemma 4.4, there exists γ ∈ K t [G0 ] with 0 6= γΩ ⊆ I, a 1-dimensional
˜ 1 , ζ2 , . . . , ζn ), so by permuting the
ideal of K t [G0 ]. But K t [G] also satisfies γ f(ζ
variables if necessary, we can now assume that all ασ ∈ I and that α1 6= 0. In
particular, we can write each ασ = α1 kσ for some kσ ∈ K, and consequently K t [G]
satisfies
X
kσ ζσ(1) ζσ(2) · · · ζσ(n)
α1
σ∈Symn
with k1 = 1.
Finally, let J = annK t [G0 ] α1 = annK t [G0 ] I, so that J is a maximal ideal of K t [G0 ]
with K t [G0 ] = K + J. Since K t [G] is free over K t [G0 ], we then have annK t [G] α1 =
TWISTED GROUP ALGEBRAS
21
J·K t [G]. By the above, we first see that K t [G]/(annK t [G] α1 ) satisfies the ordinary
polynomial identity
X
kσ ζσ(1) ζσ(2) · · · ζσ(n) ,
σ∈Symn
and second that
K t [G]/(annK t [G] α1 ) = K t [G]/(J·K t [G]) ∼
= K t [G/G0 ],
where the latter is a naturally obtained twisted group algebra of G/G0 . But then
K t [G/G0 ] satisfies an ordinary polynomial identity, so Lemma 4.3 implies that
the abelian group G/G0 has a subgroup Z/G0 of finite index with K t [Z/G0 ] =
K t [Z]/(J·K t [Z]) commutative. In particular, K t [Z] has a commutative homomorphic image, and therefore it is stably untwisted by Lemma 4.1. Since Z 0 ⊆ G0 , we
have |Z 0 | < ∞, and the result follows.
As an immediate consequence of the above and Lemma 4.2, we obtain
Corollary 4.7. The twisted group algebra K t [G] satisfies a nondegenerate multilinear generalized polynomial identity if and only if G has a normal subgroup H of
finite index with |H 0 | < ∞ and with K t [H] stably untwisted.
Finally, since untwisting and stable untwisting are inherited by subgroups, the
preceding result yields a roundabout proof of the possibly surprising
Corollary 4.8. If K t [G] satisfies a nondegenerate multilinear generalized polynomial identity and if H is a subgroup of G, then K t [H] also satisfies a nondegenerate
multilinear generalized polynomial identity.
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identity, Proc. AMS 125 (1997), 629–634.
[2] I. M. Isaacs and D. S. Passman, Groups with representations of bounded degree, Canad. J.
Math. 16 (1964), 299-309.
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Mathematics Department, University of Wisconsin, Madison, Wisconsin 53706
E-mail address: [email protected]