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TWISTED GROUP RING ISOMORPHISM PROBLEM AND

INFINITE COHOMOLOGY GROUPS

LEO MARGOLIS AND OFIR SCHNABEL

Abstract. We continue our investigation of a variation of the group ring iso-morphism problem for twisted group algebras. Contrary to previous work, weinclude cohomology classes which do not contain any cocycle of finite order.This allows us to study the problem in particular over any field of charac-teristic 0. We prove that there are finite groups G and H which can not bedistinguished by their rational twisted group algebras, while G and H can beidentified by their semi-simple twisted group algebras over other fields. Thisis in contrast with the fact that the structural information on G which canobtained from all the semi-simple group algebras of G is already encoded inits rational group algebra.

We further show that for an odd prime p there are groups of order p4 whichcan not be distinguished by their twisted group algebras over F for any field F

of characteristic different from p. On the other hand we prove that the groupsconstructed by E. Dade, which have isomorphic group algebras over any field,can be distinguished by their rational twisted group algebras. We also answera question about sufficient conditions for the twisted group ring isomorphismproblem to hold over the complex numbers.

1. Introduction

In [MS18] we proposed a version of the celebrated group ring isomorphism prob-lem for twisted group rings, namely “the twisted group ring isomorphism problem”.

Recall that for a finite group G and a commutative ring R, the group ringisomorphism problem (GRIP) asks whether the ring structure of RG determines Gup to isomorphism. Roughly speaking the twisted group ring isomorphism problem(TGRIP) asks if the ring structure of all the twisted group rings of G over Rdetermines the group G. In a sense the (TGRIP) can also be understood as aquestion on how strongly the projective representation theory of a group over agiven ring R influences its structure, as was already shown in the ground layingwork of Schur [Sch07].

We denote by R∗ the unit group in a ring R. For a 2-cocycle α ∈ Z2(G,R∗) thetwisted group ring RαG of G over R with respect to α is the free R-module withbasis {ug}g∈G where the multiplication on the basis is defined via

uguh = α(g, h)ugh for all g, h ∈ G

2010 Mathematics Subject Classification. 16S35, 20C25, 20E99 .Key words and phrases. Twisted group algebras, Projective representations, Second cohomol-

ogy groups.This research has been supported by the Research Foundation Flanders (FWO - Vlaanderen).

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http://arxiv.org/abs/2011.10129v1

2 LEO MARGOLIS AND OFIR SCHNABEL

and any ug commutes with the elements of R. Owning to this definition RαG isautomatically associative. The ring structure of RαG depends only on the coho-mology class of α and not on the particular 2-cocycle. Notice that the ring R iscentral in the twisted group ring RαG and correspondingly the associated secondcohomology group is with respect to a trivial action of G on R∗. See [Kar85, Chap-ter 3] for details. We denote the second cohomology group of G with values in R∗

by H2(G,R∗), the cohomology class of α ∈ Z2(G,F ∗) by [α] and call G a groupbase of RαG.

Definition 1.1. Let R be a commutative ring and let G and H be finite groups.We say that G ∼R H if there exists a group isomorphism

ψ : H2(G,R∗) → H2(H,R∗)

such that for any [α] ∈ H2(G,R∗),

RαG ∼= Rψ(α)H.

The main problem we are interested in is the following.

The twisted group ring isomorphism problem (TGRIP). For a given com-mutative ring R, determine the ∼R-classes. Answer in particular, for which groupsG ∼R H implies G ∼= H .

In [MS18] we investigated (TGRIP) over the complex numbers and gave someresults for families of groups, e.g. abelian groups, p-groups, groups of central typeand groups of cardinality p4 and p2q2 for p, q primes. In [MS20] we continued ourinvestigation of the (TGRIP) considering also other fields, but under the conditionthat any cohomology class contains a cocycle of finite order. This covers, in partic-ular, twisted group algebras over the complex and real numbers and any finite field.In this situation in [MS20] we introduced for a group G a so-called Yamazaki cover,generalizing the Schur cover of G in the sense that any projective representation ofG is projectively equivalent to a linear representation of its Yamazaki cover. Thusa Yamazaki cover can be used to translate the problem of describing the Wedder-burn decomposition of a twisted group algebra of G to a problem of describing theWedderburn decomposition of a group algebra of the Yamazaki cover of G.

Such a translation can not be expected when we consider cohomology classeswhich do not contain a cocycle of finite order, as it is the case for instance for anabelian group over the rationals. In this case semi-simple twisted group algebrasmight contain Wedderburn components which are not a Wedderburn componentof any finite-dimensional group algebra over the given field and furthermore thecohomology groups which one has to investigate might be infinitely-generated. Inthis paper we develop tools to study the (TGRIP) over any field and apply thosetools to several examples.

A main motivation, based on the results in [MS18, MS20] and continuing ourprevious work, is also to explore:

(a) The differences between the (TGRIP) and the (GRIP).(b) The differences between the (TGRIP) over fields where any cohomology

class contains a cocyle of finite order and the (TGRIP) over general fields.

The complex and rational numbers play a special role in the semi-simple linearrepresentation theory of finite groups, in the sense that if F is a field of characteristic

TWISTED GROUP RING ISOMORPHISM PROBLEM 3

not dividing the order of a finite group G, then

(1) FG ∼= FH ⇒ CG ∼= CH and QG ∼= QH ⇒ FG ∼= FH.

Cf. [MS20, p. 220] and [Pas77, Theorem 14.1.9] for proofs of these facts. Henceone could say that in the semi-simple case of the (GRIP) the complex numbers“know the least”, while the rationals “know the most”. In [MS20, Examples 3.1]we already observed that C is not the least knowing field for the (TGRIP) byexhibiting two non-isomorphic abelian groups G and H such that G ∼F H for acertain finite field F , while G ∼C H and G abelian implies G ∼= H [MS18, Lemma1.2]. The second part of (1) leads to:

Question 1.2. Are there non-isomorphic finite groups G and H and a field F withthe characteristic of F not dividing the order of G such that G ∼Q H but G 6∼F H?

Besides the absence of a Yamazaki cover, a major difficulty is that unlike in thework over C in [MS18], or over finite fields in [MS20], over Q or a general field F ,division algebras can appear in the Wedderburn decomposition of twisted groupalgebras, so here the Brauer group of the group base over F comes into play. Thisleads to deep number-theoretical questions which are in general hard to answer.

As for groups of order at most 16 the rational group algebra determines theisomorphism type of the group base, in the structural sense the smallest non-isomorphic groups which have isomorphic rational group algebras are the two non-abelian groups of order p3 for p an odd prime. By (1) these groups have isomorphicgroup algebras over any field of characteristic different from p. It turns out thatthe answer of the (TGRIP) in this case depends on a number theoretical condition.

Theorem 1. Let p be an odd prime, G and H the two non-abelian groups of orderp3 and F a field of characteristic different from p. Denote by ζ a primitive p-throot of unity in an extension of F .

Then G ∼F H if and only if F does not contain a primitive p-th root of unityand ζ is in the image of the norm map of the field extension F ( p

√λ, ζ)/F (ζ) for all

λ ∈ F ∗. In particular, G 6∼Q H.

The next natural candidates to answer Question 1.2 are given by a triple ofgroups of order 81 and two pairs of groups of order p4 for p a prime bigger than3. These groups have isomorphic group algebras over the rationals and we cancompletely solve the (TGRIP) for them.

Theorem 2. Let p be an odd prime and F a field of characteristic different fromp. Denote by Gi the i-th group of order p4 in the SmallGroupLibrary of GAP.

(1) Let p = 3. Then G8 ∼F G10. Moreover, G8 ∼F G9 if and only if Fcontains a primitive 3-rd root of unity.

(2) Let p > 3. Then G9 ∼F G10. Moreover, G7 ∼F G8 if and only if F doesnot contain a primitive p-th root of unity.

In particular Question 1.2 has a positive answer.

Theorem 2 exhibits for each odd prime p a pair of groups G and H of orderp4 such that G ∼Q H , but not G ∼F H for any field F of characteristic differentfrom p. Hence we obtain an answer to Question 1.2. Theorem 2 also exhibits apair of groups of order p4 which satisfy the relation of the (TGRIP) over any fieldof characteristic different from p. Note that if F is a field of characteristic p, then


FG 6∼= FH for G a group of order p4 and H a group not isomorphic to G [Pas77,pp. 671-675], hence in this case G 6∼F H .

Concerning the (TGRIP) over all fields simultaneously, including those of char-acteristic dividing the order of the group base, we study a twisted version of aquestion posed by Brauer: For finite groups G and H such that FG ∼= FH for anyfield F , is it necessary true that G and H are isomorphic? In [MS20, Problem 1.2]we suggested a twisted version of this question.

Problem 1.3. Are there non-isomorphic finite groups G and H such that G ∼F Hover any field F?

Dade [Dad71] constructed a family of pairs of groups G and H of order p3q6

which provided a counterexample for the problem of Brauer. Here p and q are anypair of primes satisfying q ≡ 1 mod p2. In [MS20, Theorem 1] we proved that if Gand H are the groups from Dade’s example and they have even order, then thereexists an infinite number of fields F such that G 6∼F H . Here we can remove thecondition on the order and show the following:

Theorem 3. For each pair of groups G and H from Dade’s example G 6∼Q H.

Therefore, we do not know the answer to Problem 1.3, but we know that the clas-sical counterexample for Brauer’s problem is not a counterexample to the twistedversion. See Remark 7.4 for other candidates for an answer to Problem 1.3.

While for a finite group G and a normal subgroup N there exists a naturalsurjective ring homomorphism FG → F (G/N), this is not always the case fortwisted group algebras. A crucial role in our proofs of Theorems 1-3 is played by atheorem, which is also of independent interest, and which allows to write a twistedgroups algebra FαG as a direct sum of twisted group algebras of G/N when N is acentral subgroup, α a cocycle inflated from a cocycle of G/N and the characteristicof F does not divide the order of N . See Theorem 5.3 for more details.

In view of a possible solution of Problem 1.3 it would be of interest to determineif an implication like the second part of (1) could be formulated for given groupsG and H and a finite number of fields, i.e.:

Problem 1.4. Does there exists a general procedure to determine some fieldsF1, . . . , Fn, depending on given finite groups G and H , such that G ∼Fi

H for1 ≤ i ≤ n implies G ∼F H for any field F?

We do not know an answer to this problem.We also answer one more question for complex twisted group algebras which

remained open in [MS18]. It is clear that if G ∼F H , for some field F and finitegroups G and H , then H2(G,F ∗) ∼= H2(H,F ∗) and the following condition holds:

F -bijectivity condition: There exists a set bijection ϕ : H2(G,F ∗) → H2(H,F ∗)such that FαG ∼= Fϕ(α)H for any [α] ∈ H2(G,F ∗).

For F being the complex numbers the F -bijectivity condition in particular im-plies, that FG ∼= FH . Write H2(G,C∗) = M(G). Neither the condition M(G) ∼=M(H) nor the C-bijectivity condition by itself is sufficient to imply G ∼C H [MS18,Theorem 1.6]. Since introducing the (TGRIP) we tried to answer the followingquestions.

Question 1.5. Let G and H be finite groups.


(1) DoM(G) ∼=M(H) and the C-bijectivity condition together implyG ∼C H?(2) If the answer to the above is negative what additional conditions will imply

G ∼C H?

We answer this question in the following way.

Theorem 4. There exist finite groups G and H of order 288 such that M(G) ∼=M(H) and the C-bijectivity condition holds for G and H, but G 6∼C H.

However, if M(G) ∼= M(H) and the C-bijectivity condition holds for G and H,and additionally M(G) is cyclic or isomorphic to an elementary abelian group oforder 4 or 9, then G ∼C H.

A theorem, which is important on its own, which is crucial for the proof ofTheorem 4 is Theorem 3.1 stating that

CαG ∼= Cαr

G⇔ r is coprime with the order of [α].

The paper is organized as follows. In §2 we recall definitions and results we willuse throughout the paper and prove some auxiliary results which will be used later.In §3 for a group G and r ∈ N we prove that , CαG ∼= Cα

r

G if and only if r is primeto the order of [α] in M(G). In §4 we prove Theorem 4. More specifically, in §4.1we prove the second part of Theorem 4 and in §4.2 we prove the first part. In §5we prove how under certain conditions a twisted group algebra can be decomposedas a direct sum of twisted group algebra of a quotient group. This generalize aknown results for group algebras [Kar85, Theorem 3.2.9]. The main results of §6are Theorem 1 and 2, but we also include the description of twisted group algebrasfor elementary abelian p-groups of rank 2. The main result of §7 is Theorem 3.

2. Preliminaries

Throughout the whole paper F denotes a field and its unit group is denoted byF ∗. By G we always denote a finite group. All the groups which will appear asthe basis of a group algebra or twisted group algebra will be assumed to be finite.For a twisted group algebra FαG we denote by {ug}g∈G the basis correspondingto the elements of G on which multiplication is explained by the cocycle α. Allcohomology groups are taken with respect to a trivial module structure. We oftendefine a cohomology class [α] ∈ H2(G,F ∗) by giving the multiplication relations inFαG where we frequently omit the relations in the basis {ug}g∈G of FαG whichremain unchanged from FG.

We write Cn for the cyclic group of order n. MoreoverMn(F ) denotes the n×n-matrix ring over F . When A is an F -algebra, x ∈ A and u a unit in A we writexu = u−1xu.

2.1. The second cohomology group of a finite group and projective rep-resentations. We start by giving some basic results on the second cohomologygroup which we will need in the sequel. The results here are classical and containedin [Kar85]. We will use without additional explanation the known structure of thesecond cohomology group of abelian groups (see [Sch07] for the complex case ande.g. [Yam64, Corollary in §2.2] for the general case).

The second cohomology group of a group G over the complex numbers is denotedby M(G) and is called the Schur multiplier. For an abelian group A we denote byZ2(G,A) the 2-cocycles of G with values in A. For a cocycle α ∈ Z2(G,A) we


denote by [α] the cohomology class of α in H2(G,A). Moreover, α is symmetric ifα(x, y) = α(y, x) for all x, y ∈ G. If G is also abelian we set

Ext(G,A) = {[α] ∈ H2(G,A) | α is symmetric}.

An important tool to understand H2(G,F ∗) is the following exact sequence.

Theorem 2.1. [Kar93, Theorem 11.5.2] There is the following split short exactsequence

1 → Ext(G/G′, F ∗) → H2(G,F ∗) → Hom(M(G), F ∗) → 1.

Let n1, . . . , nr be natural numbers. Recall that [Kar85, Corollary 2.3.17]

(2) Ext(Πri=1Cnr, F ∗) ∼= Πri=1 Ext(Cnr

, F ∗).

Therefore, in order to understand Ext(G/G′, F ∗) it is sufficient to understand thedescription of Ext(Cn, F

∗) ∼= H2(Cn, F∗). This is well known [Kar85, Theorem

1.3.1]:

(3) Ext(Cn, F∗) ∼= H2(Cn, F

∗) ∼= F ∗/(F ∗)n.

The map from Ext(G/G′, F ∗) to H2(G,F ∗) in Theorem 2.1 is known as theinflation map and we denote it as inf. More general for a normal subgroup N of Gand α ∈ Z2(G/N,F ∗) one has inf(α)(x, y) = α(xN, yN) for all x, y ∈ G and thismap implies a homomorphism from H2(G/N,F ∗) to H2(G,F ∗).

If V is an F -vector space, a map η : G → GL(V ), such that the compositionof η with the natural projection from GL(V ) to PGL(V ) is a group homomor-phism is called a projective representation of G. The projective representation η isirreducible, if V admits no proper G-invariant subspace. Two projective represen-tations η1 : G → GL(V1) and η2 : G → GL(V2) are called projectively equivalentif there is a map µ : G → F ∗ satisfying µ(1) = 1 and a vector space isomorphismf : V1 → V2 such that

η1(g) = µ(g)f−1η2(g)f

for every g ∈ G.If we define now a map α : G×G→ F by

α(g1, g2) = η(g1)η(g2)η(g1g2)−1,

then α lies in Z2(G,F ∗) and we refer to η as an α-representation of G. For a fixedα ∈ Z2(G,F ∗), the set of projective equivalence classes of α-representations of G isdenoted by Proj(G,α). As in the ordinary case, there is a natural correspondencebetween projective representations of G over F with an associated cohomologyclass [α] and FαG-modules. In particular, if the characteristic of F does not dividethe order of G, then FαG is the direct sum of the simple FαG-modules whichcorrespond to the projective equivalence classes of α-representations. See [Kar85,Section 3.1] for more details.

We now record some facts on the Schur multiplier. We first recall that for anyfinite group G there exists a Schur cover SG, which is also a finite group, such thatCSG ∼= ⊕[α]∈H2(G,C∗)C

αG.Recall that a group action of a group T on a group N induces an action of T on

M(N).


Lemma 2.2. [Kar87, Section 2.2] Let N and T be subgroups of G and assumeG = N ⋊ T . Then M(N)T is a subgroup of M(G). Moreover, if N and T are ofcoprime order, then

M(G) =M(T )×M(N)T .

Here M(N)T consists precisely of those elements in M(N) which are invariantunder the T -action.

Set

cG,α = gcd{dim(η) | η ∈ Proj(G,α)},and denote, for a prime p, the biggest power of p dividing cG,α by (cG,α)p.

Proposition 2.3. [Hig88, Lemma 1 and Proposition 1] With the above notation

(1) ◦([α]) divides cG,α.(2) If a prime p divides cG,α, then p divides ◦([α]).(3) For a Sylow p-subgroup Gp of G

(cG,α)p = cGp,α

where on the right side α is the restriction of α to Gp.

Another important result when working over the complex numbers is sometimesreferred to as “Brauer’s Trick”.

Proposition 2.4. [Kar85, Theorem 2.3.2] Let α ∈ Z2(G,C∗). Then there exists acocycle α′ ∈ Z2(G,C∗) which is cohomologous to α such that α′(g, h) is a root ofunity of order dividing the order of [α] for any g, h ∈ G.

2.2. Cyclic algebras and symbol algebras. We will need some basic facts aboutsome well-known classes of central simple algebras. We fix here our notation forthese algebras and cite the results on them which will be useful for our investiga-tions. We refer to [Pie82, Chapter 15] and [BO13, Chapter VII] for the facts oncyclic algebras. We also prove a lemma of number theoretical nature which will beimportant later to disprove isomorphisms between twisted group algebras.

Definition 2.5. Let K/F be a finite cyclic field extension of degree n with Galoisgroup generated by an element δ and let λ ∈ F ∗. Let A be the K-algebra whichcontains an element a such that as a K-vector space we have A = ⊕n−1

i=0 aiK and

for each µ ∈ K we have µa = aδ(µ) and an = λ. Then A is called a cyclic algebrawhich we denote by (λ,K/F, δ).

We recall that two simple F -algebras A and B are called Brauer equivalent ifand only if there exists a division F -algebra D and integers n and m such thatA ∼= Mn(D) and B ∼= Mm(D). We write A ≡ B to express Brauer equivalence.Moreover, we write A ≡ 1, if A is isomorphic to a matrix ring over F .

Proposition 2.6. [BO13, Proposition VII.1.9] The algebra (λ,K/F, δ) is a centralsimple F -algebra. If L is a field such that F ⊆ L ⊆ K, then (λ,K/F, δ) ⊗F L is

Brauer equivalent to (λ,K/L, δ) where δ = δ[L:F ].

An especially nice form of cyclic algebras is the following.

Definition 2.7. [Row08, Remark 24.3] Let n be a natural number and F a fieldcontaining an n-th primitive root of unity ζ. For λ, µ ∈ F ∗ define an F -algebraA which is generated by two elements x and y and the relations xn = λ, yn = µ


and yx = ζxy. Then this algebra is called a symbol algebra and is denoted by(λ, µ;F ; ζ)n. When the field F and the element ζ is clear from the context wesimply write (λ, µ)n.

Note that the algebra (λ, µ;F ; ζ)n can also be viewed as a twisted group algebraof G ∼= Cn × Cn = 〈g〉 × 〈h〉 over F . Namely, there is a cohomology class [α] ∈H2(G,F ∗) determined by the following relations in FαG,

(ug)n = λ, (uh)

n = µ, uhug = ζuguh.

It is easy to see that FαG ∼= (λ, µ;F ; ζ)n.

Notation: LetK/F be a finite field extension and denote by Nr(K/F ) the normmap of this extension. We write a ∈ Nr(K/F ), if a is in the image of Nr(K/F ).

Lemma 2.8. [Row08, Proposition 24.48], [BO13, Proposition VII.1.9]Let F be a field containing a primitive n-th root of unity ζ and set (λ, µ)n =(λ, µ;F ; ζ)n for any λ, µ ∈ F ∗. The following hold.

(1) (λ, 1)n ≡ 1.

(2) (λ, µ)n ≡ 1 if and only if µ ∈ Nr(F ( n√λ)/F ).

(3) (λ, µ1)n ⊗ (λ, µ2)n ≡ (λ, µ1µ2)n.(4) (λ, µ)n ∼= (µ, λ−1)n.

(5) (λ, µ1) ∼= (λ, µ2) if and only if µ1µ−12 ∈ Nr(F ( n

√λ)/F ).

The preceding lemma shows that to decide on isomorphisms of symbol algebrasone has to understand the images of the norm map of field extensions. Though ingeneral this is not an easy task, at least for some situations important to us we canachieve this. The proof of the following lemma is essentially due to Danny Neftin.

Lemma 2.9. Let p be an odd prime, ζ a primitive complex p-th root of unity andset F = Q(ζ). Then here exists λ ∈ Q∗ such that ζ /∈ Nr(F ( p

√λ)/F ).

Proof. For a fixed λ set z = p√λ. Let q be a prime such that q 6≡ 1 mod p, which

exists by Dirichlet’s theorem. Denote by ν the q-adic evaluation of F and by Fν thecompletion of F with respect to ν. Note that by our choice of q the field Fν doesnot contain a root of unity of order p2. By the Hasse Norm Theorem [Pie82, Section18.4] to prove that ζ /∈ Nr(F (z)/F ) it is sufficient to show that ζ /∈ Nr(F (z)ν/Fν).By a theorem of Albert [Alb37, Chapter 9, Theorem 11] this is equivalent to thecondition that there exists a field K such that Fν ⊆ F (z)ν ⊆ K and K/Fν is acyclic extension of degree p2.

Assume now that λ is a uniformizing element of Fν . Then the extension Fν(z)/Fνis totally ramified, i.e. the ramification index equals p. Because there are exactly proots of unity in Fν with order a power of p, it then follows from [HS11, Proposition4.1] that there is no field K such that Fν ⊆ F (z)ν ⊆ K and K/Fν is a cyclicextension of degree p2. �

2.3. Partial isomorphisms of second cohomology groups. We record herea fact which restricts the possible isomorphisms of the second cohomology groupswhich can come up in the (TGRIP). This fact will be particularly useful to us laterwhen H2(G,F ∗) will be infinitely generated. First recall the following result oncommutative components of twisted group algebras [MS20, Proposition 2.5].


Proposition 2.10. Let F be a field such that char(F ) does not divide the orderof G and let [α] ∈ H2(G,F ∗). Then FαG admits a commutative Wedderburncomponent if and only if [α] is in the image of the inflation map from Ext(G/G′, F ∗)to H2(G,F ∗).

Proposition 2.11. Let G and H be finite groups such that G ∼F H. Assumemoreover that the characteristic of F does not divide the order of G.Then Ext(G/G′, F ∗) ∼= Ext(H/H ′, F ∗) and Hom(M(G), F ∗) ∼= Hom(M(H), F ∗).

Proof. Let ϕ : H2(G,F ∗) → H2(H,F ∗) be an isomorphism realizing the relationG ∼F H . Recall that

(4) H2(G,F ∗) ∼= Ext(G/G′, F ∗)×Hom(M(G), F ∗)

and Ext(G/G′, F ∗) can be identified with a subgroup of H2(G,F ∗) via the inflationmap. For [α] ∈ H2(G,F ∗) we know that FαG admits an commutative simpleWedderburn component if and only if [α] ∈ Ext(G/G′, F ∗) by Proposition 2.10. So

[α] ∈ Ext(G/G′, F ∗) ⇔ FαG admits a commutative component

⇔ Fϕ(α)H admits a commutative component ⇔ ϕ([α]) ∈ Ext(H/H ′, F ∗).

This implies that ϕ restricts to an isomorphism between the groups Ext(G/G′, F ∗)and Ext(H/H ′, F ∗). In view of (4) it then also induces an isomorphism betweenHom(M(G), F ∗) and Hom(M(H), F ∗). �

2.4. Other useful results to determine twisted group algebras. For theconvenience of the reader we include several more known results which we will uselater to determine isomorphism types of twisted group algebras.

Proposition 2.12. [Kar85, Proposition 3.3.8] Let α ∈ Z2(G,C∗) be of order n andlet Gα = 〈α(x, y)g | x, y, g ∈ G〉. Then

CGα ∼=n−1⊕

i=0

Cαi

G.

Lemma 2.13. [Pas77, Lemma 6.1.6] In a ring R let 1 = e1 + . . . + en be a de-composition into pairwise orthogonal idempotents. Assume R∗ permutes the set{e1, . . . , en} transitively by conjugation. Then R ∼=Mn(e1Re1).

The following two results are due to Ito.

Theorem 2.14. [Isa06, (12.34) Corollary] Let G be solvable. Then G has a normalabellian Sylow p-subgroup if and only if p does not divide the degree of any complexirreducible character of G.

Theorem 2.15. [Isa06, (6.15) Theorem] Let A be a normal abelian subgroup of G.Then the degree of any irreducible complex character of G divides [G : A].

3. Powering of elements in the Schur multiplier

The following theorem will be an important tool in the proof of Theorem 4. Theproof presented here is due to Y. Ginosar. Recall that for [α] ∈ H2(G,C∗) wedenote by cG,α the greatest common divisor of the dimensions of the irreducibleα-representations.


Theorem 3.1. Let G be a finite group, let [α] ∈M(G) be of order n and let r be anatural number. If r is coprime with n, then CαG ∼= Cα

r

G and if r is not coprimewith n, then cG,α > cG,αr . In particular, CαG ∼= Cα

r

G if and only if gcd(n, r) = 1.

Proof. Let ζ be a primitive n-th root of unity in C. Assume first gcd(n, r) = 1.Using Proposition 2.4 we may assume that the values of α in C∗ are all powers ofζ. Since gcd(n, r) = 1, the following is an isomorphism

ψ : Q(ζ) → Q(ζ)ζ 7→ ζr.

Let

CαG = spanC{ug}g∈G and Cαr

G = spanC{vg}g∈G.Then ψ can be extended to an automorphism of C which can be extended as follows

ψ : CαG → Cαr

G∑

g∈G agug 7→∑

g∈G ψ(ag)vg.

In order to show that ψ is a ring isomorphism it is enough to check that

ψ(uguh) = ψ(α(g, h)ugh) = ψ(α(g, h))vgh = αr(g, h)vgh = vgvh = ψ(ug)ψ(uh).

Next, we assume that gcd(n, r) = d where d 6= 1. Let s = rd . Then, r = sd and

gcd(s, n) = 1. Consequently

Cαr

G = C(αd)sG ∼= Cαd

G,

where the right isomorphism derives from the fact that the order of [α]d is primeto s. Hence we may assume that r is a divisor of n.

We will reduce the problem to r being a prime divisor of n. Let p be a primedivisor of r. Denote by Gp a Sylow p-subgroup of G. By Proposition 2.3 in orderto show that cG,α > cG,αr it is sufficient to show that cGp,α is greater than cGp,αr .Note that since the p-part of M(G) is embedded inside M(Gp) [Kar85, Corollary2.3.24 and its proof] we know that [α] is not trivial as an element ofM(Gp). Henceit will be sufficient to assume that order of [α] ∈ M(Gp) is a power of p. In orderto finish the proof we will show that cGp,α is greater than cGp,αp .

Since p is a divisor of the order of [α], the dimension of any element in Proj(Gp, α)is divisible by p, cf. Proposition 2.3. Therefore the dimension of any irreducibleα-representation can be written as pj such that j ≥ 1 [Hig88, Main Theorem].Hence cGp,α is a p-power and there is an irreducible α-representation of dimensioncGp,α. Let η ∈ Proj(Gp, α) be such a representation. Now, denote by γ the p-exterior power of η with itself, cf. [Rot10, Section 9.8] for the definition. Then,γ ∈ Proj(Gp, α

p) [GS14, Section 2.2], and additionally, by [Rot10, Theorem 9.140],

dim(γ) =

(

cGp,α

p

)

,

and hence dim(γ) is divisible bycGp,α

p . Thus cGp,α is greater than cGp,αp . �

Remark 3.2. It would be interesting to know if a result like Theorem 3.1 holdsfor all fields, i.e. whether FαG ∼= Fα

r

G if and only if r is coprime with the orderof [α] ∈ H2(G,F ∗). We are not aware of a counterexample, but it is also clear thatthe proof given above can not be directly translated to the general situation.


4. Proof of Theorem 4

In this section we will prove Theorem 4 and give some additional results con-cerning sufficient conditions for G ∼F H . We split this section into two parts. Inthe first part we give some positive results for Question 1.5 (2) and in the secondpart we give a negative answer to Question 1.5 (1).

4.1. Positive results for Question 1.5 (2). For a group G, denote by SG aSchur cover of G. Then, the following is an immidate corollary of Theorem 3.1.

Corollary 4.1. Let G and H be groups such that CG ∼= CH, CSG ∼= CSH andM(G) ∼=M(H) ∼= Cp. Then G ∼C H.

Proof. Let [α] and [β] be generators of M(G) and M(H) respectively. Then, sinceCG ∼= CH , we have

p−1⊕

i=1

Cαi

G ∼=p−1⊕

i=1

Cβi

H.

The result now follows from Theorem 3.1 since Cαi

G ∼= Cαj

G and Cβi

G ∼= Cβj

Gfor any 1 ≤ i, j ≤ p− 1. �

The following is an elementary observation.

Lemma 4.2. Let F be a field and let G and H be groups satisfying the F -bijectivitycondition and in addition H2(G,F ∗) ∼= H2(H,F ∗) ∼= C2 × C2. Then G ∼F H.

Proof. The claim follows immediately from the fact that any set bijection fromC2 × C2 to itself fixing the trivial element is a group automorphism. �

We will prove a similar lemma for H2(G,F ∗) ∼= H2(H,F ∗) ∼= C3 ×C3, but onlyfor F being the complex numbers.

Lemma 4.3. Let G and H be groups satisfying the C-bijectivivty condition and inaddition M(G) ∼=M(H) ∼= C3 × C3. Then G ∼C H.

Proof. DenoteM(G) = 〈α〉 × 〈β〉, M(H) = 〈α〉 × 〈β〉.

By Theorem 3.1

CαG ∼= Cα2

G, CβG ∼= Cβ2

G, CαβG ∼= Cα2β2

G, Cα2βG ∼= Cαβ

2

G,

and similarly

CαH ∼= Cα2

H, CβH ∼= Cβ2

H, CαβH ∼= Cα2β2

H, Cα2βH ∼= Cαβ

2

H.

So, the map ψ providing the C-bijectivity is a permutation on these (at most) fourisomorphism types. The action of Aut(C3×C3) on the non-trivial cyclic subgroupsof C3 × C3 is given by Aut(C3 × C3)/Z(Aut(C3 × C3)) ∼= PGL(2, 3) ∼= S4. Henceany permutation of the non-trivial cyclic subgroups of C3 × C3 can be realized byan automorphism which implies that any permutation of the isomorphism types ofthe twisted group algebras can be realized by an isomorphism between the Schurmultipliers. �

We do not have a similar proof for the case that the Schur multiplier is Cp ×Cpfor p ≥ 5. Clearly the proofs above do not work. However, we do neither have acounterexample in this case.

Maybe the most interesting positive result on Question 1.5 is the following.


Theorem 4.4. Let G and H be groups satisfying the C-bijectivity condition and inaddition M(G) ∼=M(H) ∼= Cn. Then G ∼C H.

Proof. Denote the map which corresponds to the C-bijectivity by ψ : M(G) →M(H). Let [α] be a generator of M(G). We will prove that ψ([α]) = [β] is agenerator ofM(H). Assume [β] is of order k, a divisor of n. Owing to Theorem 3.1,ϕ(n) = ϕ(k) where ϕ denotes Euler’s totient function. Since k is a divisor of n weget that either k = n or n = 2k where k is an odd integer. Assume the latter. Then

there exists a generator [γ] ∈M(H) such that [β] = [γ]2, and additionally Cα2

G ∼=CγH . However by Theorem 3.1 (cG,β)2 is strictly smaller than (cG,γ)2 which isstrictly smaller than (cG,α)2 which contradicts CαG ∼= CβH . We proved that ψsends a generator ofM(G) to a generator ofM(H). From here it is straightforwardto prove G ∼C H using Theorem 3.1. �

4.2. Negative answer for Question 1.5 (1). Throughout this section G and Hwill be the following groups, which we introduce with abuse of notation (we usethe same names for generators of the groups). The identities of these groups inthe SmallGroupLibrary of GAP [GAP19] are [288, 484] and [288, 603], but wewill only employ theoretical arguments. The groups G and H are generated byelements a, b, c, d, e such that

a3 = b3 = c4 = d4 = e2 = 1.

The Sylow 3-subgroup in both groups is 〈a〉 × 〈b〉 ∼= C3 × C3. A Sylow 2-subgroupin both groups is given by

(〈c〉 × 〈d〉)⋊ 〈e〉 where [c, e] = d2 and [d, e] = 1.

Both groups can be written as semidirect products of the form

(〈a〉 × 〈b〉)⋊ ((〈c〉 × 〈d〉)⋊ 〈e〉)where in both groups the following relations hold

[a, c] = 1, [a, d] = a, [b, d] = 1

and

in G : [a, e] = a, [b, c] = 1, [b, e] = b,

in H : [a, e] = 1, [b, c] = b, [b, e] = 1.

We will prove that G and H satisfy M(G) ∼=M(H) and the C-bijectivity condi-tion, but G 6∼C H . In both groups the commutator subgroup is

〈a〉 × 〈b〉 × 〈d2〉 ∼= C3 × C3 × C2.

We start with computing the Schur multipliers of these groups. Consider bothgroups as semi-direct products of the normal subgroup N = 〈a〉 × 〈b〉 and thesubgroup T = (〈c〉 × 〈d〉) ⋊ 〈e〉. We have M(C3 × C3) ∼= C3. So M(N) is cyclic oforder 3, a generator being determined by the relation [ua, ub] = ζ3 in the twistedgroup algebra, where ζ3 denotes a primitive 3-rd root of unity in C. However thiscohomology class is not invariant under the action of T . Indeed

[ud(a), ud(b)] = [ua2 , ub] = ζ23 6= [ua, ub].

Consequently, by Lemma 2.2

M(G) ∼=M(T ) ∼=M(H).


Now consider N1 = 〈c〉 × 〈d〉 ∼= C4 ×C4 and T1 = 〈e〉 ∼= C2. Then T = N1 ⋊ T1.Clearly M(N1) ∼= C4, a generator being determined by the relation [uc, ud] = ζ4 inthe twisted group algebra, where ζ4 denotes a primitive 4-th root of unity in C. It iseasy to check that this cohomology class is invariant under the T1-action. Anothernontrivial cohomology class of T is determined by [ud, ue] = −1 and it is easy toshow that these two non-trivial cohomology classes generate M(G) ∼=M(H).

Corollary 4.5. With the above notation

M(G) ∼=M(H) ∼= C2 × C4 = 〈α〉 × 〈β〉,where

α : [ud, ue] = −1, β : [uc, ud] = ζ4.

Notice that it is not surprising, by the results of Section 4.1, that in the neg-ative example we found to Question 1.5 (1) the groups admit a Schur multiplierisomorphic to C4 × C2.

We now start computing the Wedderburn decomposition of the twisted groupalgebras of these groups. We will start with the group algebras.

Lemma 4.6. With the above notation

CG ∼= CH ∼= 16C⊕ 28C2×2 ⊕ 10C4×4.

Proof. First, notice that by Theorem 2.14 the degrees of all the characters of Gand H are powers of 2. Notice that both groups G and H admit a normal abeliansubgroup of index 4. For G we can choose 〈a, b, c, d2〉 and for H we can choose〈a, b, c2, d2, e〉. Hence, it follows from Theorem 2.15 that the maximal degree of anirreducible character of G and H is smaller or equal to 4. Consequently, possibledimension of the irreducible representations of these groups are 1, 2 or 4. Next, since|G/G′| = 16 = |H/H ′| we have that CG and CH admit exactly 16 copies of thefield C in their Wedderburn decomposition. To finish the proof observe that boththese groups admit 54 conjugacy classes, and hence the dimension of the center ofCG and CH is 54. �

In order to compute the Wedderburn decomposition for the twisted group alge-bras we will use Proposition 2.12.

Lemma 4.7. With the above notation,

Cβ2

G ∼= 32C2×2 ⊕ 10C4×4 and Cβ2

H ∼= 16C2×2 ⊕ 14C4×4.

Proof. We will use Proposition 2.12. By the construction of Gω, we get that

Gβ2∼=

(

a

C3 ×b

C3

)

⋊

(

(f

C2 ×c

C4)⋊d

C4)⋊e

C2

)

,

where f is central and [c, d] = f and all the other relations are the same as in G.Similarly,

Hβ2∼=

(

a

C3 ×b

C3

)

⋊

(

(f

C2 ×c

C4)⋊d

C4)⋊e

C2

)

,

where f is central and [c, d] = f and all the other relations are the same as in H .Then, again, as in Lemma 4.6, by Theorem 2.14 the degrees of all the charactersof Gβ2 and Hβ2 are powers of 2. And also, both groups Gβ2 and Hβ2 admit anormal abelian subgroup of index 4. For Gβ2 we can choose 〈a, b, c, d2, f〉 and forHβ2 we can choose 〈a, b, c2, d2, e, f〉. Hence, it follows from Theorem 2.15 that the


maximal degree of an irreducible character of Gβ2 and Hβ2 is smaller or equal to 4.A simple calculation shows that Gβ2 admits 96 conjugacy classes while Hβ2 admits84 conjugacy classes. It follows that

CGβ2∼= 16C⊕ 60C2×2 ⊕ 20C4×4

andCHβ2

∼= 16C⊕ 44C2×2 ⊕ 24C4×4.

The result follows by Lemma 4.6 and by noticing that

CGβ2∼= CG⊕ Cβ

2

G and CHβ2∼= CH ⊕ Cβ

2

H.

�

Corollary 4.8. With the above notation G 6∼C H.

Proof. Assume that ψ : M(G) →M(H) is an isomorphism. As the squares inM(G)and M(H) form a characteristic subgroup, ψ([β2]) = [β2]. Hence by Lemma 4.7we can not have an isomorphism CγG ∼= Cψ(γ)H for all [γ] ∈M(G). �

In the rest of this section we will prove that the groups G and H satisfy theC-bijectivity condition.

Lemma 4.9. Let [γ] ∈ {[β], [β3], [αβ], [αβ3]} in M(G) or alternatively M(H).Then

CγG ∼= CγH ∼= 6C4×4 ⊕ 3C8×8.

Proof. We will use Proposition 2.12. By the construction of Gω, we get that

Gβ ∼=(

a

C3 ×b

C3

)

⋊

(

(f

C4 ×c

C4)⋊d

C4)⋊e

C2

)

,

where f is central and [c, d] = f and all the other relations are the same as in G.Similarly,

Hβ∼=

(

a

C3 ×b

C3

)

⋊

(

(f

C4 ×c

C4)⋊d

C4)⋊e

C2

)

,

where f is central and [c, d] = f and all the other relations are the same as in H .To prove the lemma we can argue as in the proofs of Lemma 4.6 and Lemma 4.7using that the degree of any β-representation is divisible by 4, that 〈a, b, c2, d2, f〉is an abelian normal subgroup in both G and H and that CβG ∼= Cβ

3

G and

CβH ∼= Cβ3

H . Analogues arguments also handle the cohomology classes [αβ] and[αβ3]. �

We are left with the cohomology classes [α] and [αβ2] in both M(G) and M(H).Using similar techniques to the above we can show the following.


CαG ∼= Cαβ2

G ∼= 16C2×2 ⊕ 14C4×4,

CαH ∼= 16C2×2 ⊕ 14C4×4

andCαβ

2

H ∼= 32C2×2 ⊕ 10C4×4.

Corollary 4.11. G and H satisfy the C-bijectivity condition.

Proof. We can take ϕ :M(G) →M(H) to be determined by


(1) ϕ([γ]) = [γ] for [γ] ∈ {1, [α], [β], [β3], [αβ], [αβ3]}.(2) ϕ([β2]) = [αβ2].(3) ϕ([αβ2]) = [β2].

Then, the result follows from Lemmas 4.6-4.10. �

Notice now that Theorem 4.(2) follows from Corollary 4.5, Corollary 4.8 andCorollary 4.11.

5. Structure theorem for twisted group algebras of central

extensions

The aim of this section is to generalize a result [Kar85, Theorem 3.2.9] whichdescribes the decomposition of a group algebra of a group S into a direct sum oftwisted group algebras of a quotient of S by a central subgroup. Note that it isknown that FαG only maps naturally to F βG/N , for N some normal subgroupof G and [β] ∈ H2(G/N,F ∗), if [α] is in the image of the inflation map fromH2(G/N,F ∗) to H2(G,F ∗) [Kar85, Lemma 3.2.12].

We first introduce some notation. Let

(5) 1 → A→ S → G→ 1

be a central extension. Let α′ ∈ Z2(G,F ∗) and let α ∈ Z2(S, F ∗) such that αis inflated from α′. Now, fix a 2-cocycle σ ∈ Z2(G,A) such that [σ] ∈ H2(G,A)corresponds to the central extension in (5). Note that in this way we also fix asection G → S. Define for any field K containing F and χ ∈ Hom(A,K∗) thefollowing function

αχ : G×G→ K∗, αχ(g, h) = α′(g, h)χ(σ(g, h)).

Lemma 5.1. With the above notation αχ ∈ Z2(G,K∗).

Proof. It is straightforward to verify the 2-cocyle condition for αχ using that α′

and σ are 2-cocycles and χ a group homomorphism from A to K∗. �

Denote by F (χ) the smallest field extension of F which contains all the valuesof χ. Note that then αχ has only values in F (χ)∗, so αχ ∈ Z2(G,F (χ)∗). We viewG as a subset of S via the section given by σ.

Proposition 5.2. Let {us}s∈S be a basis of FαS corresponding to α and {vg}g∈Ga basis of F (χ)αχG corresponding to αχ. Define

ψχ : FαS → F (χ)αχG, kuag 7→ kχ(a)vg

for any k ∈ F , a ∈ A and g ∈ G and by extending ψχ linearly to all of FαS. Thenψχ is a surjective ring homomorphism.

Proof. We set ψ = ψχ. Since F is central in FαS and F (χ) is central in F (χ)αχGit is sufficient to prove that

ψ(ua1g1ua2g2) = ψ(ua1g1)ψ(ua2g2)

for any g1, g2 ∈ G and a1, a2 ∈ A. Notice that

ua1g1ua2g2 = α(a1g1, a2g2)ua1g1a2g2 = α(a1g1, a2g2)ua1a2σ(g1,g2)g1g2 .

Therefore,

ψ(ua1g1ua2g2) = α(a1g1, a2g2)χ(a1a2σ(g1, g2))vg1g2 .


Using the fact that α is inflated from α′ and the definition of αχ we get that

ψ(ua1g1ua2g2) = α′(g1, g2)χ(a1a2σ(g1, g2))vg1g2 = χ(a1a2)αχ(g1, g2)vg1g2 .

On the other hand,

ψ(ua1g1)ψ(ua2g2) = χ(a1)vg1χ(a2)vg2 = χ(a1a2)vg1vg2 = χ(a1a2)αχ(g1, g2)vg1g2 .

We proved that ψ is indeed a homomorphism. The surjectivity of ψ is clear sincevg is in the image of ψ for any g ∈ G and from the definition of F (χ). �

We now proceed to generalize [Kar85, Theorem 3.2.9] for cohomology classes ofS which are inflated from cohomology classes of G. Assume that the characteristicof F does not divide the order of A, let K be a field containing F and a primitiveexp(A)-th root of unity. Denote by M the set of simple FA-modules up to isomor-phism. Note that as A is abelian we can identify Hom(A,K∗) with the irreducibleK-characters of A. Let Irr(A,F ) be a subset of K-characters of A such that itcontains for each M ∈ M one character which corresponds to a composition factorof K ⊗F M . In particular, |M| = |Irr(A,F )|.Theorem 5.3. Let 1 → A → S → G → 1 be a central extension of finite groups,F a field of characteristic not dividing the order of A and [α] ∈ H2(S, F ∗) a coho-mology class inflated from an element in H2(G,F ∗). Then

FαS ∼=⊕

χ∈Irr(A,F )

F (χ)αχG.

Proof. Define

ψ : FαS →⊕

χ∈Irr(A,F )

F (χ)αχG, x 7→ (ψχ(x))χ∈Irr(A,F )

where ψχ is defined as in Proposition 5.2 for every χ ∈ Irr(A,F ) and (ψχ(x))χ∈Irr(A,F )

denotes a tuple with entries indexed by the elements of Irr(A,F ). By Proposi-tion 5.2 the map ψ is a surjective ring homomorphism on every direct summandand as the dimensions of FαS and

⊕

χ∈Irr(A,F ) F (χ)αχG are equal, we are done

once we show that ψ is injective.We view G embedded in S, as a transversal of the cosets of A. Then a general

element x ∈ FαS can be written as x =∑

g∈G agg where ag ∈ FA. Fix some

χ ∈ Irr(A,F ). Then, by the definition of ψχ, the element x lies in the kernel of ψχif and only if χ(ag) = 0 for all g ∈ G. Hence x lies in the in the kernel of ψ if andonly if χ(ag) = 0 for all g ∈ G and all χ ∈ Irr(A,F ). By our assumption on F thealgebra FA is semisimple, so FA is isomorphic to the direct sum of the elements inM, the simple FA-modules. Thus by the definition of Irr(A,F ) we have, for a fixedg ∈ G, that χ(ag) = 0 for all χ ∈ Irr(A,F ) if and only if ag annihilates all simpleFA-modules, i.e. ag = 0. Hence χ(ag) = 0 for all all g ∈ G and χ ∈ Irr(A,F ) ifand only if ag = 0 for all g ∈ G, i.e. x = 0. Consequently ker(ψ) = 0 and ψ isinjective. �

6. Twisted group algebras of p-groups

In this section we study the twisted group algebras of p-groups starting withelementary abelian groups of rank 2. We then proceed to prove Theorems 1 and 2.

Throughout this section p denotes a prime and F a field of characteristic differentfrom p. Moreover ζ denotes a primitive p-th root of unity in F or an extension of F .


Let [F (ζ) : F ] = p−1d for some divisor d of p−1 and let ζ1 . . . , ζd be representatives of

the orbits of primitive p-th roots of unity in F (ζ) under the action of Aut(F (ζ)/F ).Note that F (ζi) = F (ζj) for any 1 ≤ i, j ≤ d. We will use Theorem 2.1 throughoutthis section without mentioning it explicitly.

6.1. Twisted group algebras of elementary abelian groups of rank 2. Westart by investigating the twisted group algebras of elementary abelian groups ofrank 2. This example will turn out to be crucial in the sequel.

Let G = 〈a〉 × 〈b〉 ∼= Cp × Cp. The Schur multiplier of G is isomorphic toCp, a generator being explained by the relation [ua, ub] = ζ. Therefore, if ζ ∈F , then Hom(M(G), F ∗) ∼= Cp and Hom(M(G), F ∗) = 1 otherwise. Moreover,Ext(G/G′, F ∗) = Ext(G,F ∗), so by Section 2.1 a generic cohomology class [α] ∈H2(G,F ∗) is determined by parameters λ, µ ∈ F ∗ and an integer 0 ≤ ℓ ≤ p − 1with corresponding relations in the twisted group algebra given by

upa = λ, upb = µ, [ua, ub] = ζℓ

where ℓ = 0 if ζ /∈ F .Assuming ℓ 6= 0, in particular ζ ∈ F , the relations in FαG are exactly the

defining relations of the symbol algebra (λ, µ;F ; ζℓ)p. Hence, FαG is a centralsimple F -algebra which is by Lemma 2.8 isomorphic to a matrix algebra over F ifand only if µ ∈ Nr(F ( p

√λ)/F ).

Next assume ℓ = 0. We will distinguish several cases in the description of FαG.

(1) Assume λ, µ ∈ (F ∗)p, i.e. the polynomials Xp − λ and Xp − µ have rootsin F [X ]. Then [α] is the neutral element, as it corresponds to the trivialhomomorphism in Hom(G/G′, F ∗/(F ∗)p). So FαG ∼= FG. Now FG hasa trivial module F and non-trivial modules all of which are isomorphic toF (ζ) as a field. There are p + 1 possible kernels for a non-trivial moduleas this is the number of non-trivial subgroups of G. Moreover for a fixedkernel, say 〈b〉, there are d non-equivalent representations of FG on F (ζ)given by sending a to ζi for 1 ≤ i ≤ d. Hence

FαG ∼= FG ∼= F ⊕ (p+ 1)dF (ζ).

(2) Assume λ /∈ (F ∗)p and µ ∈ (F ∗)p. Then [F ( p√λ) : F ] = p, as Xp − λ is

irreducible in F [X ] if it has no roots [R67, Theorem 427]. F ( p√λ, ζ) is a

field of dimension p(p−1)d over F . The following are pairwise non-equivalent

projective irreducible α-representations of G:

η : G→ F (p√λ), x 7→ p

√λ, y 7→ p

√µ

ηi : G→ F (p√λ, ζ), x 7→ p

√λ, y 7→ ζi p

√µ, for 1 ≤ i ≤ d.

Note that [F ( p√λ) : F ] + d[F ( p

√λ, ζ) : F ] = p+ d

(

p(p−1)d

)

= p2 = |G|. So

FαG ∼= F ( p√λ)⊕ dF ( p

√λ, ζ).

(3) Assume λ, µ /∈ (F ∗)p and that F ( p√λ) 6= F ( p

√µ). Note, that if F ( p

√λ) =

F ( p√µ) then after a base change in FαG we can assume that we are in the

previous case. The following is a projective irreducible α-representation ofG:

η : G→ F (p√λ, p

√µ), x 7→ p

√λ, y 7→ p

√µ.

Note that the dimension of F (p√λ, p

√µ) over F is p2. So FαG ∼= F (

p√λ, p

√µ).


6.2. Non-abelian groups of order p3. In this section we prove Theorem 1.Throughout this section p is an odd prime. Abusing notation, we denote the twonon-abelian groups of order p3 by

G = 〈b〉⋊ 〈a〉, ap = bp2

= 1, bp =: c, ba = bc, c ∈ Z(G)

and

H = (〈c〉 × 〈b〉)⋊ 〈a〉, ap = bp = cp = 1, ba = bc, c ∈ Z(H).

In both groups the center is cyclic of order p generated by c and the quotient bythe derived subgroup is elementary abelian of order p2 generated by the images ofa and b. We denote by a and b the images of a and b in G/G′ = G/Z(G) andH/H ′ = H/Z(H). We will study the twisted group algebras of G and H over Fstarting with cohomolgy classes in the images of Ext(G/G′, F ∗) and Ext(H/H ′, F ∗)under the inflation map respectively.

By [Kar85, Theorem 4.7.3] we have M(G) = 1 and M(H) ∼= Cp × Cp wheregenerators of M(H) are given by

[βa] : [uc, ua] = ζ, [uc, ub] = 1, [ub, ua] = uz,

[βb] : [uc, ua] = 1, [uc, ub] = ζ, [ub, ua] = uz.

First let [α] ∈ H2(G,F ∗) be such that α is symmetric and given by upa = λ,upb = µuc. Then [α] is inflated from [γ] ∈ H2(G/〈c〉, F ∗) which is given by upa = λ,upb= µ, [ua, ub] = 1. By Theorem 5.3 we have

FαG ∼=⊕

χ∈Irr(〈c〉,F )

F (χ)αχ(G/〈c〉).

Now χ ∈ Irr(〈c〉, F ) is determined by χ(c) = ζℓ for some 0 ≤ ℓ ≤ p − 1 and|Irr(〈c〉, F )| = 1 + d. For ℓ = 0 we have F (χ)αχ(G/〈c〉) = F γ(G/〈c〉) which is adirect sum of fields described in Section 6.1. If ℓ 6= 0, then [αχ] can be describedby relations upa = λ, up

b= µζℓ, [ua, ub] = ζℓ. The corresponding direct summand

F (χ)αχG is then the central simple F (ζ)-algebra (λ, µζℓ;F (ζ); ζℓ)p as described inSection 2.2. Note that M(G) is trivial and hence by Section 2.1 any cohomologyclass inH2(G,F ∗) contains a symmetric cocycle. Thus we have described all twistedgroup algebras of G over F .

Next we describe all twisted group algebras of the Heisenberg group H . By thedescription of M(H) above and Section 2.1 we can write any [α] ∈ H2(H,F ∗) as

[α′βiaβjb ] where 0 ≤ i, j ≤ p−1 and α′ is a symmetric cocycle explained by relations

upa = λ, upb = µ. There are essentially two cases which we will handle separately.Case 1: i = j = 0, i.e. [α] contains a symmetric cocycle.This case is similar to the above description of the twisted group algebra of

G. More precisely, [α] is inflated from [γ] ∈ H2(H/〈c〉, F ∗) with [γ] explained byrelations upa = λ, up

b= µ, [ua, ub] = 1. Note that this is exactly the [γ] from the

description of twisted group algebras for G. Again by Theorem 5.3 we have

FαH ∼=⊕

χ∈Irr(〈c〉,F )

F (χ)αχ(H/〈c〉).

For χ(c) = 1 we get the summand F γ(H/〈c〉) which is isomorphic to F γ(G/〈c〉).For χ(c) = ζℓ we get the central simple F (ζ)-algebra (λ, µ;F (ζ); ζℓ)p. Notice thatthe parameters of this algebra are not the same as in the description of twistedgroup algebras of G.


Case 2: (i, j) 6= (0, 0).Assume first i 6= 0. Let k be an integer such that ki + j ≡ 0 mod p and define

in FαH the element v = uakub. Then a straightforward computation shows that[uc, v] = 1 and vp = δ ∈ F ∗ for δ = λkµ. Next observe that [v, ua] = uc. So FαHis generated by the elements uc, v, ua with relations given by

upc = 1, vp = δ, upa = λ, [v, ua] = uc, [v, uc] = 1, [uc, ua] = ζi.

Now, notice that e = 1p (1 + uc + . . .+ up−1

c ) is an idempotent, and the set

e, eua , . . . , eup−1

a

is a set of orthogonal idempotents such that their sum is 1. Then, by Lemma 2.13we have FαH ∼= Mp(e(F

αH)e). Noticing that euce = e and euae = 0 we get thate(FαH)e ∼= 〈ve|(ve)p = δ〉. Notice that 〈ve〉 is not only a subalgebra by dimensionconsiderations. Hence

FαH ∼={

Mp(F (p√δ)), if p

√δ /∈ F

pMp(F ), otherwise.

In case i = 0 we can make an analogues calculation interchanging a and b (andchoosing k = 0) so that we then obtain

FαH ∼={

Mp(F (p√λ)), if p

√λ /∈ F

pMp(F ), otherwise

We note that in this case all components are matrix algebras over fields.

Proof of Theorem 1: We show that G ∼F H if and only if ζ /∈ F and ζ ∈Nr(F ( p

√λ, ζ)/F (ζ)) for all λ ∈ F ∗. Note that ζ ∈ F if and only if there is an

[α] ∈ H2(H,F ∗) such that FαH has only non-commutative components by Propo-sition 2.10. On the other hand F βG has a commutative component for every[β] ∈ H2(G,F ∗). Thus G ∼F H implies ζ /∈ F .

So assume ζ /∈ F from now on. Moreover, assume first that ζ ∈ Nr(F ( p√λ, ζ)/F (ζ))

for all λ ∈ F ∗. Let ψ : H2(H,F ∗) → H2(G,F ∗) be defined by sending an element[α] defined by the relations upa = λ, upb = µ to the element defined by upa = λ,upb = µuc. Then by the above

FαH ∼= F γ(Cp × Cp)⊕ d(λ, µ)p

andFψ(α)G ∼= F γ(Cp × Cp)⊕ d(λ, µζ)p

for a certain [γ] ∈ H2(Cp × Cp, F∗) and where the symbol algebras are defined

over F (ζ). Now by Lemma 2.8 we have (λ, µ)p ∼= (λ, µζ)p if and only if ζ ∈Nr(F ( p

√λ, ζ)/F (ζ)). So indeed G ∼F H via ψ.

Finally assume there is λ ∈ F such that ζ /∈ Nr(F ( p√λ, ζ)/F (ζ)). Then also

p√λ /∈ F (ζ). Define [β] ∈ H2(G,F ∗) via the relations upa = λ, upb = 1. So by the

aboveF βG ∼= F (

p√λ)⊕ dF (

p√λ, ζ)⊕ d(λ, ζ)p.

By Lemma 2.8 the symbol algebra (λ, ζ)p is not a matrix algebra over a field as

ζ /∈ Nr(F ( p√λ, ζ)/F (ζ)). We claim that there is no [α] ∈ H2(H,F ∗) such that

FαH ∼= F βG and thus G 6∼F H . Indeed if FαH ∼= F βG, then FαH has exactlyd+1 summands which are fields and one of these fields is F ( p

√λ). Using the direct

summand F γ(H/〈c〉) described above, by Section 6.1 this means that we can assume


that [α] is given by relations upa = λ and upb = 1. Note here that we can interchangea and b by an isomorphism of H . Then the non-commutative summands in theWedderburn decomposition of FαH are isomorphic to (λ, 1)p which is a matrixring over a field by Lemma 2.8, so FαH 6∼= F βG. It now follows directly fromLemma 2.9 that G 6∼Q H . Hence Theorem 1 holds. �

6.3. Answer to Question 1.2 and groups of order p4. We next study thosegroups of order p4 which have isomorphic group algebras over Q, and hence overall fields of characteristic different from p. This includes the proof of Theorem 2which is a direct combination of Propositions 6.2, 6.3 and 6.5. We consider all suchgroups with the exception of the direct product of a cyclic group of order p and anon-abelian group of order p3, as this case follows from Section 6.2. There are somedifferences between the cases p = 3 and p > 3 and we will treat them separately,although the arguments are similar in both cases. We remark that the complextwisted group algebras for all groups of order p4 were described in [HH06].

First assume p = 3 and consider the following groups of order 81:

G8 = 〈a, b, c | a9 = b3 = c3 = 1, [a, b] = c, [c, a] = 1, [c, b] = a3〉,

G9 = 〈a, b, c | a9 = b3 = c3 = 1, [a, b] = c, [c, a] = 1, [c, b] = a−3〉,G10 = 〈a, b, c | a9 = c3 = 1, b3 = a−3, [a, b] = c, [c, a] = 1, [c, b] = a−3〉.

The numbering of the groups is coming from the Small Group Library. For alli ∈ {8, 9, 10} we have Z(Gi) = 〈a3〉 and Gi/Z(Gi) ∼= H where H denotes theHeisenberg group of order 27. Furthermore, G′

i = 〈a3, c〉 and Gi/G′i = 〈a〉 × 〈b〉 ∼=

Cp × Cp. Moreover, by [HH06, Theorem 3.1] we have M(G8) ∼=M(G10) ∼= Cp andM(G9) ∼= Cp × Cp.

We first handle cohomology classes containing symmetric cocycles. Let [αi] ∈Ext(Gi/G

′i, F

∗). So by the properties of Gi mentioned above [αi] is determined bythe following relations in the twisted group algebra for i ∈ {8, 9}

u3a = λua3 and u3b = µ for λ, µ ∈ F ∗

and for i = 10 by

u3a = λua3 and u3b = µua−3 for λ, µ ∈ F ∗.


Fα8G8∼= Fα9G9

∼= Fα10G10.

Proof. We will apply Theorem 5.3 with the central subgroup 〈a3〉. So we have

FαiGi ∼=⊕

χ∈Irr(〈a3〉,F )

F (χ)(αi)χH,

where

H = (〈c〉 × 〈a〉) ⋊ b, a3 = b3 = c3 = 1, ab = ac, c ∈ Z(H)

is the Heisenberg group. Recall from Section 6.2 that an element in H2(H,F ∗) isexplained by relations u3a = λ′, u3

b= µ′, [uc, ua] = ζk and [uc, ub] = ζℓ for some

λ′, µ′ ∈ F and where k = ℓ = 0 if ζ /∈ F .First consider χ to be the principal character. Then [(αi)χ] is, independently of

i, explained by u3a = λ, u3b= µ and uc is central. So the isomorphism type of this

direct summand of FαiGi does not depend on i.


Next let χ(a3) = ζℓ where ζℓ 6= 1. So in particular, F (χ) = F (ζ). Then [(αi)χ]is explained by u3a = λζℓ, [uc, ua] = 1 and

u3b =

{

µ, if i ∈ {8, 9}µζ−ℓ, if i = 10

, [uc, ub] =

{

ζ−ℓ, if i ∈ {9, 10}ζℓ, if i = 8

.

By the last case considered in Section 6.2 we then have, again independently of i,that

F (χ)(αi)χH ∼={

Mp(F (ζ,p√

λζℓ)), if p√

λζℓ /∈ FpMp(F (ζ)), otherwise

.

So also in this case the isomorphism type of the direct summand corresponding toχ is independent of i and the lemma follows. �

This allows us to determine over which fields these groups are in relation.

Proposition 6.2. Let F be a field of characteristic different from 3 and assumethat F contains no primitive 3-rd too of unity. Then G8 ∼F G9 ∼F G10.

Proof. By the assumption on F we have Hom(M(Gi), F∗) = 1, so any cohomology

class [αi] ∈ H2(Gi, F∗) contains a symmetric cocycle. Such a cohomology class can

be explained by the relations

u3a = λua3 and u3b = µ for λ, µ ∈ F ∗

for i ∈ {8, 9} and for i = 10 by

u3a = λua3 and u3b = µua−3 for λ, µ ∈ F ∗.

Denote this cohomology class by [αλ,µi ]. Hence Lemma 6.1 implies that for i, j ∈{8, 9, 10} the map

ϕi,j : H2(Gi, F

∗) → H2(Gj , F∗), [αλ,µi ] 7→ [αλ,µj ]

is an isomorphism of groups such that Fαλ,µi Gi ∼= Fα

λ,µj Gj . We conclude that

G8 ∼F G9 ∼F G10. �

Proposition 6.3. Let F be a field of characteristic different from 3 containing aprimitive 3-rd root of unity. Then G8 6∼F G9 and G8 ∼F G10.

Proof. As M(G8) ∼= C3 and M(G9) ∼= C3 × C3, we obtain Hom(M(G8), F∗) ∼= C3

while Hom(M(G9), F∗) ∼= C3 × C3. Hence G8 6∼F G9 by Proposition 2.11.

We need to show that G8 ∼F G10. M(G8) and M(G10) are both explained bythe following cohomology classes:

[β8] : [ua, ub] = uc, [uc, ua] = ζ, [uc, ub] = ua3 ,

[β10] : [ua, ub] = uc, [uc, ua] = ζ, [uc, ub] = ua−3 .

To show that [β8] is indeed a cohomology class notice that, G8 = (〈a〉 × 〈c〉)⋊ 〈b〉,and by our knowledge on the cohomology of an abelian group there is a cohomologyclass [β8] defined on 〈a〉 × 〈c〉 by [uc, ua] = ζ. Hence, in order to show that [β8]

is indeed a cohomology class it is sufficient, by Lemma 2.2, to show that [β8] isinvariant to the action of b on 〈a〉 × 〈c〉. This is now a straightforward calculationto check that

[ub(c), ub(a)] = [ua3c, uac] = ζ

and hence [β8] is indeed a cohomology class of G8. By similar arguments we canshow that [β10] is a cohomology class of G10.


Let a general element [α8] ∈ H2(G8, F∗) be explained by the product of [β8]

r

and a cohomology class containing a symmetric cocycle such that u3a = λua3 andu3b = µ. So overall we can assume that [α8] is explained by

[ua, ub] = uc, [uc, ua] = ζ, [uc, ub] = ua3 , u3a = λua3 , u

3b = µ.

Let ψ : H2(G8, F∗) → H2(G10, F

∗) be a map defined by ψ([α8]) = [α10] such that[α10] is explained by the product of [β10]

r and a cohomology class containing asymmetric cocycle such that u3a = ζrλua3 and u3b = µ−1ua−3 . So [α10] is explainedby

[ua, ub] = uc, [uc, ua] = ζ, [uc, ub] = ua−3 , u3a = ζrλua3 , u3b = µ−1ua−3 .

We claim that ψ is an isomorphism realizing the relation G8 ∼F G10. It is easyto check that ψ is an isomorphism of groups, so it remains to check that Fα8G8

∼=Fα10G10. If r ≡ 0 mod 3, then we know Fα8G8

∼= Fα10G10 from Lemma 6.1. Notehere that in Lemma 6.1 we had [α10] explained by u3b = µua−3 , but this differencedoes not influence the calculations. So assume r 6≡ 0 mod 3.

Set A = Fα8G8. The element ua3 is central in A and satisfies u3a3 = 1. Hence thecentral subalgebra F 〈ua3〉 is isomorphic to F ⊕F ⊕F and contains three primitivecentral idempotents f0, f1 and f2, such that

A = Af0 ⊕Af1 ⊕Af2

and fiua3 = fiζi for 0 ≤ i ≤ 2. Set Ai = Afi. Then A0 is a twisted group algebra

of the Heisenberg group G8/〈a3〉 which is by Section 6.2 isomorphic toM3(F ( 3√µ)),

if 3√µ /∈ F , and to 3M3(F ) otherwise.

For this paragraph let i ∈ {1, 2}. Recall that for an element x in an F -algebraX and a unit u ∈ X we write xu = u−1xu. Let ei =

13 (1 + uc + uc2)fi. As u3c = 1

this is an idempotent in Ai. Note that uafi is a unit in Ai. As [uc, ua] = ζr the

elements ei, euafii , e

u2

afii are pairwise orthogonal idempotents which sum up to fi.

Hence Ai ∼=M3(eiAiei) by Lemma 2.13. Note that eiuaei = 0 and eiucei = ei. Letvi = uau

−irb fi. We claim that (eifi)

vi = eifi. Indeed

(eifi)vi =

1

3(1 + uc + uc2)

uau−irb fi =

1

3(1 + ζruc + ζ2ruc2)

u−irb fi

=1

3(1 + ζru−ira3 uc + ζ2ru−2ir

a3 uc2)fi =1

3(1 + ζrζ−i

2ruc + ζ2rζ−2i2ruc2)fi = eifi.

Hence eiviei = eivi and (eivi)3 = eifiζ

iλµ−ir . So

eiAiei =

{

F ( 3

√

ζiλµ−ir), if 3

√

ζiλµ−ir /∈ F,3F, otherwise

.

In a similar way we can consider Fα10G10. Here we can also define the fi, Aiand ei as for G8. So we have A0

∼= M3(F (3

√

µ−1)), if 3

√

µ−1 /∈ F , and 3M3(F )otherwise. We define vi = uau

irb fi, so that again (eifi)

vi = eifi. Moreover,

(eiviei)3 = eiλζ

rua3µ−iruira−3fiei = eiζ

r+i−i2rλµ−irfiei = eiζiλµ−irfiei.

So in both cases the three direct summands of the algebras are isomorphic. �

We will now consider the case p > 3. The arguments here are similar and we willnot give as many details as before, but now there are four groups involved. Again


we use the numbering of the Small Groups Library for our groups and set

G7 = 〈a, b, c, d | ap = bp = cp = dp = 1, [a, b] = c, [c, a] = 1, [c, b] = d, d ∈ Z(G7)〉,G8 = 〈a, b, c | ap2 = bp = cp = 1, [a, b] = c, [c, a] = ap, [c, b] = 1〉

and

G9 = 〈a, b, c | ap2 = bp = cp = 1, [a, b] = c, [c, a] = 1, [c, b] = ap〉G10 = 〈a, b, c | ap2 = bp = cp = 1, [a, b] = c, [c, a] = 1, [c, b] = apm〉

where m is a quadratic nonresidue modulo p.Then Z(G7) = 〈d〉 and Z(Gi) = 〈ap〉 for i ∈ {8, 9, 10}. Moreover Gi/Z(Gi) ∼= H

where H denotes the Heisenberg group for all i ∈ {7, 8, 9, 10}. Furthermore G′i =

〈Z(Gi), c〉 and Gi/G′i = 〈a〉 × 〈b〉 ∼= Cp × Cp, for all i. Finally M(G7) ∼= Cp × Cp

while M(Gi) ∼= Cp for i ∈ {8, 9, 10} by [HH06, Theorem 3.1].Again we will first consider cohomology classes containing symmetric cocycles.

By the above we can assume that [αi] ∈ Ext(Gi, F∗) is determined by relations

upa =

λ, if i = 7µ−1uap , if i = 8λuap , if i ∈ {9, 10}

and upb =

{

µ, if i ∈ {7, 9, 10}λ, if i = 8

for λ, µ ∈ F ∗.


Fα7G7∼= Fα8G8

and

Fα9G9∼= Fα10G10.

Proof. As in the situation for p = 3 we will apply Theorem 5.3 with the centralsubgroup Z(Gi). Set Z(Gi) = 〈z〉. So we have

FαiGi ∼=⊕

χ∈Irr(〈z〉,F )

F (χ)(αi)χH,

where

H = (〈c〉 × 〈a〉) ⋊ b, a3 = b3 = c3 = 1, ab = ac, c ∈ Z(H)

is the Heisenberg group.Let χ be the principal character. Then [(αi)χ] is explained by

upa =

{

λ, if i ∈ {7, 9, 10}µ−1, if i = 8

and upb=

{

µ, if i ∈ {7, 9, 10}λ, if i = 8

and uc is central in all cases. From our previous considerations for the twistedgroup algebras of the Heisenberg group and the elementary abelian group of orderp2 we see that the isomorphism type of this direct summand does not depend on i.This follows from Lemma 2.8, as this gives (λ, µ)p ∼= (µ−1, λ)p.

Next let χ(z) = ζℓ where ζℓ 6= 1. Then [(αi)χ] is explained by

upa =

λ, if i = 7µ−1ζℓ, if i = 8λζℓ, if i ∈ {9, 10}

, upb=

{

µ, if i ∈ {7, 9, 10}λ, if i = 8


and

[uc, ua] =

{

1, if i ∈ {7, 9, 10}ζℓ, if i = 8

, [uc, ub] =

ζℓ, if i ∈ {7, 9}1, if i = 8ζℓm, if i = 10

.

From our considerations for the Heisenberg group we see that F (χ)(α7)χH ∼=F (χ)(α8)χH and F (χ)(α9)χH ∼= F (χ)(α10)χH which implies the lemma. �

We obtain a result similar to the one for the case p = 3.

Proposition 6.5. Let p be a prime bigger than 3 and let F be a field of character-istic different from p. Then

(1) G7 ∼F G8 if and only if F does not contain a primitive p-th root of unity.(2) G9 ∼F G10.

Proof. First assume F contains no primitive p-th root of unity. Then Hom(M(Gi), F∗)

is the trivial group for all i and it follows from Lemma 6.4 that G7 ∼F G8 andG9 ∼F G10. Now assume F does contain a primitive p-th root of unity ζ. AsM(G7) ∼= Cp × Cp while M(G8) ∼= Cp we conclude from Proposition 2.11 thatG7 6∼F G8. So it remains to show that G9 ∼F G10.

We have M(G9) ∼= M(G10) ∼= Cp and the Schur multipliers are generated by acohomology class explained by

[β9] : [ua, ub] = uc, [uc, ua] = ζ, [uc, ub] = uap ,

[β10] : [ua, ub] = uc, [uc, ua] = ζ, [uc, ub] = uapm

for G9 and G10 respectively. It can be checked as in the proof of Proposition 6.3that [β9] and [β10] indeed define cohomology classes of the corresponding group.

Let a general cohomology class [α9] in H2(G,F ∗) be explained by the product

of [β9]r with the cohomology class containing a symmetric cocycle defined by upa =

λuap and upb = µ. Let ψ : H2(G9, F∗) → H2(G10, F

∗) be a map and ψ([α9]) = [α10]where [α10] is explained by the product of [β10]

r and a cohomology class containinga symmetric cocycle defined by the relations upa = λuap and upb = µm. We claimthat ψ is a group isomorphism realizing the relation G9 ∼F G10. If r ≡ 0 mod p,then Fα9G9

∼= Fα10G10 by Lemma 6.4.So let r 6≡ 0 mod p and set A = Fα9G9. As uap is central in A satisfying upap = 1

the subalgebra F 〈uap〉 is isomorphic to the direct sum of p copies of F such thateach direct summand contains an idempotent fi which is central in A and such thatuapfi = ζifi for 0 ≤ i ≤ p − 1. Hence A = ⊕p−1

i=0Afi. Set Ai = Afi. Then A0 isa twisted group algebra of the Heisenberg group G9/〈ap〉 which is by Section 6.2isomorphic with Mp(F ( p

√µ)), if p

√µ /∈ F , and pMp(F ) otherwise.

For this paragraph let 1 ≤ i ≤ p− 1. Set ei =1p (1 + uc + . . .+ up−1

c )fi which is

an idempotent in Ai. Moreover ei, euafii , . . . , e

up−1

a fii is a set of pairwise orthogonal

idempotents which sum up to fi. Hence by Lemma 2.13 we know Ai ∼=Mp(eiAiei).

Let j be an integer such that ij ≡ 1 mod p and set vi = uau−rjb fi. Then vi is


centralizing eifi as

(eifi)vi =

1

p(1 + uc + . . .+ up−1

c )uau−rj

b fi =1

p(1 + ζruc + . . .+ ζ(p−1)rup−1

c )u−rj

b fi

=1

p(1 + ζru−rjap uc + . . .+ ζ(p−1)ru

−(p−1)rjap up−1

c )fi

=1

p(1 + ζr−rijuc + . . .+ ζ(p−1)r−(p−1)rijup−1

c )fi = eifi.

Moreover vpi = ζiλµ−rj , so Ai ∼= Mp(F (p√

ζiλµ−rj)), if p√

ζiλµ−rj /∈ F , andpMp(F ) otherwise.

To determine the isomorphism type of Fα10G10 we can define the fi, Ai and eiin the same way. Then A0

∼= Mp(F ( p√µm)), if p

√µm /∈ F , and pMp(F ) otherwise.

Note that F ( p√µ) = F ( p

√µm). To define the vi choose j

′ such that j′im ≡ 1 mod p

and set viuau−rj′

b . Then a similar computation as before shows that vi centralizes

eifi and we have vpi = ζiλµ−mrj′fi. Note that µ−mrj′ ≡ µ−rj mod (F ∗)p. Hence

F ( p√

ζiλµ−rj) = F ( p√

ζiλµ−mrj′ ) and we conclude that indeed Fα8G9∼= Fα10G10.

�

7. Dade’s examples and proof of Theorem 3

In 1971 Dade answered a question of Brauer by describing a series of finite groupsG and H such that FG ∼= FH for all fields F [Dad71]. In [MS20, Section 5] westudied the (TGRIP) for the examples of Dade and could show for a subclass ofthese examples that there exists a finite field F such that G 6∼F H . Our goal herewill be to show that in fact G 6∼Q H for any of Dade’s examples, i.e. to proveTheorem 3.

Let us first describe the groups of Dade. Let p and q be primes such that q ≡ 1mod p2 and let w be an integer such that w 6≡ 1 mod q2, but wp ≡ 1 mod q2. LetQ1 and Q2 be the following two non-abelian groups of order q3.

Q1 = (〈τ1〉 × 〈σ1〉)⋊ 〈ρ1〉,Q2 = 〈σ2〉⋊ 〈ρ2〉,

τq1 = σq1 = ρq1 = σq2

2 = ρq2 = 1, σq2 =: τ2,

τρ11 = τ1, σρ11 = τ1σ1, σ

ρ22 = τ2σ2.

So Q1 and Q2 are just the two non-abelian groups of order q3 such that Q1 hasexponent q. These are exactly the groups for which we studied the (TGRIP) inSection 6.2 and the difference between the groups we encountered in Theorem 1 isgoing to be crucial here.

Let 〈π1〉 ∼= Cp2 , 〈π2〉 ∼= Cp and for i, j ∈ {1, 2} let

ρπj

i = ρi, σπj

i = σwj , τπj

i = τwi .

Define two groups by

G = (Q1 ⋊ 〈π1〉)× (Q2 ⋊ 〈π2〉),H = (Q1 ⋊ 〈π2〉)× (Q2 ⋊ 〈π1〉).

These are the groups constructed by Dade as a counterexample to Brauer’s questionand we will fix them throughout this section.


Notice that G = G1 ×G2 and H = H1 ×H2 for

G1 = Q1 ⋊ 〈π1〉, G2 = Q2 ⋊ 〈π2〉, H1 = Q1 ⋊ 〈π2〉, H2 = Q2 ⋊ 〈π1〉.The Schur multipliers of G and H are described in [MS20, Proposition 5.3], but

it will turn out that they are in fact not needed to obtain G 6∼Q H . Note that,using the bar notation for taking elements modulo the commutator subgroup, fori ∈ {1, 2} we have

Gi/G′i = 〈ρi〉 × 〈πi〉

and

Hi/H′i = 〈ρi〉 × 〈πj〉

where j ∈ {1, 2} such that i 6= j. Let λ ∈ F ∗. Abusing notation we define acohomology class [αλ] in H

2(G1, F∗) and H2(H1, F

∗) and a cohomology class [βλ]in H2(G2, F

∗) and H2(H2, F∗). These cohomology classes are in the image of

Ext(Gi/G′i, F

∗) and Ext(Hi/H′i, F

∗) under the inflation map, respectively, and aredescribed by the relations

[αλ] : uqρ1 = λ and [βλ] : u

qρ2 = λ.

We will also write simply [αλ] and [βλ] when we restrict these cohomology classesto subgroups of G and H which contain ρ1 or ρ2 respectively.

Notation: For a positive integer n let ζn denote a primitive complex n-th rootof unity. Let F be a field of characteristic 0 such that F ∩ Q(ζq2) = Q. We

denote ξ =∑p−1

i=0 ζwiq . Moreover, we denote δp ∈ Gal(F (ζq)/F (ξ)), explained by

δp(ζq) = ζwq , and δpq ∈ Gal(F (ζq2 )/F (ξ)), explained by δpq(ζpq) = ζ(q+1)wq2 .

We first determine the isomorphism types of the twisted group algebras of thosegroups which form Dades’s examples.

Lemma 7.1. Let λ ∈ Q. Assume λ 6∈ (Q∗)q and set z = q√λ. With the cohomology

classes defined above the following isomorphisms hold

QH1∼= QG2

∼= Q ⊕Q(ζp)⊕Q(ζq)⊕Q(ζp, ζq)⊕Mp(Q(ξ))⊕q − 1

pMp(Q(ζq))⊕Mpq(Q(ξ)),

QαλH1

∼= Q(z)⊕Q(z, ζp)⊕Mp(Q(z, ξ))⊕Mpq(Q(ξ)),

QβλG2

∼= Q(z)⊕Q(z, ζp)⊕Mp(Q(z, ξ))⊕ (λ,Q(ζq2)/Q(ξ), δpq),

QG1∼= QH2

∼= QH1 ⊕Q(ζp2)⊕Q(ζq, ζp2)⊕ (ζp,Q(ζq, ζp)/Q(ξ, ζp), δp)⊕q − 1

pMp(Q(ζq, ζp))

⊕Mq((ζp,Q(ζq, ζp)/Q(ξ, ζp), δp)),

QαλG1

∼= QαλH1 ⊕Q(z, ζp2)⊕ (ζp,Q(ζq, z, ζp)/Q(ξ, z, ζp), δp)⊕Mq((ζp,Q(ζq, ζp)/Q(ξ, ζp), δp)),

QβλH2

∼= QβλG2 ⊕Q(z, ζp2)⊕ (ζp,Q(ζq, z, ζp)/Q(ξ, z, ζp), δp)⊕ (ζpλ,Q(ζq2 , ζp)/Q(ξ, ζp), δpq).

Proof. We will not prove the isomorphisms of the non-twisted group algebras, asthese can be derived by standard methods, e.g. this can be be achieved using thetheory of Shoda pairs as presented in [JdR16, Chapter 3].

Note that by assumption z /∈ Q. Set e = 1q (1+ τ1+ . . .+ τ

q−11 ) which is a central

idempotent in QαλH1. So we have

(6) QαλH1∼= eQαλH1e ⊕ (1− e)QαλH1(1− e).

Now

eQαλH1e ∼= Qαλ(〈σ1〉⋊ 〈π2〉 × 〈ρ1〉) ∼= Q(〈σ1〉⋊ 〈π2〉)⊗Qαλ〈ρ1〉.


While Qαλ〈ρ1〉 ∼= Q(z), the isomorphism type of the untwisted group algebra isgive by

(7) Q(〈σ1〉⋊ 〈π2〉) ∼= Q⊕Q(ζp)⊕Mp(Q(ξ)).

Hence, we obtain

eQαλH1e ∼= Q(z)⊕Q(z, ζp)⊕Mp(Q(z, ξ)).

So by (6) it remains to consider the algebra (1− e)QαλH1(1− e) and show that

it is isomorphic to Mpq(Q(ξ)). Let f = 1q (1 + σ1 + . . . + σq−1

1 ), so (1 − e)f(1 − e)

is an idempotent. The idempotents

(1 − e)f(1− e), (1 − e)fρ1(1− e), . . . , (1− e)fρq−1

(1 − e)

are then pairwise orthogonal and sum up to (1−e). Hence by Lemma 2.13 we have

(1− e)QαλH1(1 − e) ∼=Mq((1− e)fQαλH1f(1− e)).

As fρ1f = 0, we have

(1− e)fQαλH1f(1− e) ∼= (1− e)Q(〈τ1〉⋊ 〈π2〉)(1− e).

By (7) this algebra is isomorphic to Mp(Q(ξ)), where we use that 〈σ1〉⋊ 〈π2〉 ∼=〈τ1〉⋊ 〈π2〉 . Overall we obtain the claimed isomorphism type of QαλH1.

Now set e = 1q (1 + τ2 + . . .+ τq−1

2 ). Similarly as before we have

QβλG2∼= eQβλG2e⊕(1−e)QβλG2(1−e) ∼= Q(z)⊕Q(z, ζp)⊕Mp(z, ξ)⊕(1−e)QβλG2(1−e)

and we need to prove that

(1− e)QβλG2(1− e) ∼= (λ,Q(ζq2 )/Q(ξ), δpq).

Let a ∈ (λ,Q(ζq2 )/Q(ξ), δpq) such that an = λ and µa = aδpq(µ) for all µ ∈ Q(ζq2).

The algebra (1−e)QβλG2(1−e) has a faithful representation in (λ,Q(ζq2 )/Q(ξ), δpq)by sending σ2 to ζq2 and ρ2π2 to a. As both these algebras have the same dimension,we conclude that they are in fact isomorphic.

The isomorphism types of QαλG1 and QβλH2 follow by similar arguments andusing

Q(〈σ1〉⋊ 〈π1〉) ∼= Q(〈σ1〉⋊ 〈π2〉)⊕Q(ζp2)⊕ (ζp,Q(ζq, ζp)/Q(ξ, ζp), δp)

where we used (7) before. �

We will also use another field than Q to simplify our argument later.

Corollary 7.2. Let F = Q(ζq, ζp), λ ∈ Q and assume λ 6∈ (F ∗)q. Set z = q√λ.

With the cohomology classes described above the following isomorphisms hold

FH1∼= FG2

∼= pqF ⊕ q(q − 1)

pMp(F )⊕

q − 1

pMpq(F ),

FαλH1∼= pF (z)⊕ q − 1

pMp(F (z))⊕

q − 1

pMpq(F ),

F βλG2∼= pF (z)⊕ q − 1

pMp(F (z))⊕ (q − 1)p(λ, ζq;F ; ζq)q,

FG1∼= FH2

∼= pqF ⊕ q(p− 1)F (ζp2)⊕ q(q − 1)Mp(F )⊕ (q − 1)Mpq(F ),

FαλG1∼= pF (z)⊕ (p− 1)F (z, ζp2)⊕ (q − 1)Mp(F (z))⊕ (q − 1)Mpq(F ),

F βλH2∼= pF (z)⊕ (p− 1)F (z, ζp2)⊕ (q − 1)Mp(F (z))⊕ p2(q − 1)(λ, ζq;F ; ζq)q.


Proof. The claimed isomorphism types follows from Lemma 7.1 by tensoring withF . We will just comment on how the cyclic algebras appearing in Lemma 7.1behave under this change of coefficients. As [F : Q(ξ, ζp)] = p, by Proposition 2.6we know that F ⊗ (ζp,Q(ζq, ζp)/Q(ξ, ζp), δp) is Brauer equivalent to (ζp, F/F, id)which is isomorphic to F . So

F ⊗ (ζp,Q(ζq, ζp)/Q(ξ, ζp), δp) ∼=(q − 1)(p− 1)

pMp(F ).

Moreover, also by Proposition 2.6, F ⊗ (λ,Q(ζq2)/Q(ξ), δpq) is Brauer equivalentto (λ,Q(ζq2 )/F, δq) where δq is a generating element of Gal(Q(ζq2 )/F )). The latteralgebra is just the symbol algebra (λ, ζq ;F ; ζq)q. Finally in the same manner

F ⊗ (λ,Q(ζq2 )/Q(ξ), δpq) ≡ (ζpλ, ζq;F ; ζq)q.

Now NrQ(ζq2 )/F(ζp) = ζqp , so ζp ∈ Nr(Q(ζq2)/F ). We conclude from Lemma 2.8

that

(ζpλ, ζq;F ; ζq)q ∼= (λ, ζq ;F ; ζq)q.

�

In order to prove Theorem 3 we will use the following lemma.

Lemma 7.3. There exists a λ ∈ F ∗ such that (λ, ζq ;F ; ζq)q is not Brauer equivalentto a field.

Proof. By Lemma 2.8 we have (λ, ζq;F ; ζq)q ≡ 1 if and only if ζq ∈ Nr(F (z)/F ).By Lemma 2.9 however, there exists λ ∈ Q∗ such that ζq /∈ Nr(F (z)/F ). Thus, theproof is complete. �

Proof of Theorem 3: Let λ ∈ F such that (λ, ζq;F ; ζq)q is not Brauer equivalentto a field, i.e. (λ, ζq ;F ; ζq)q is not isomorphic to a matrix ring over a field. Such aλ exists by Lemma 7.3. We claim that there exists no [γ] ∈ H2(G,Q∗) such thatQγG ∼= QαλH . If such a [γ] exists, then also F γG ∼= FαλH , so it will suffice to provethat F γG 6∼= FαλH for any [γ]. So assume for a contradiction that F γG ∼= FαλHfor some [γ] ∈ H2(G,F ∗).

By Corollary 7.2 we have that FαλH ∼= FαλH1 ⊗ FH2 is isomorphic to(

pF (z)⊕ q − 1

pMp(F (z))⊕

q − 1

pMpq(F )

)

⊗(

pqF ⊕ q(p− 1)F (ζp2)⊕ q(q − 1)Mp(F )⊕ (q − 1)Mpq(F ))

.

We first note that the Wedderburn decomposition of FαλH consists solemnlyof components isomorphic to matrix rings over fields. Moreover it contains thecommutative component F (z). It follows from Proposition 2.10 that [γ] is in theimage of the inflation map of Ext(G/G′, F ∗) in H2(G,F ∗). Recall that

G/G′ = 〈ρ1〉 × 〈ρ2〉 × 〈π1〉 × 〈π2〉.So Ext(G/G′, F ∗) has exponent pq. Assume that the order of [γ] is divisible by p.Then there exists a µ ∈ F ∗ such that µ /∈ (F ∗)p and we could assume that in F γGone has upρ1 = µ or upρ2 = µ. Then F γG would contain a Wedderburn componentwhich is a field containing p

√µ which is not the case. Hence [γ] must have order q.

So we can assume that [γ] = [αλ1βλ2

] for some λ1, λ2 ∈ F . By Corollary 7.2 wethen know that F γG contains a component isomorphic to F ( q

√λ1,

q√λ2). Looking

at the components of FαλH we see that F ( q√λ1,

q√λ2) ∼= F (z) which implies that


we can assume {λ1, λ2} ⊆ {1, λ}. Note that FαλH contains exactly p2q componentsof type F (z). If λ1 = λ2 = 1, then by Corollary 7.2 the algebra F γG contains nosuch component, while when λ1 = λ2 = λ it contains exactly p2 such components.So we conclude {λ1, λ2} = {1, λ}.

Assume that λ2 = λ. Then by Corollary 7.2 the algebra F γG contains a compo-nent (λ, ζq ;F ; ζq)q. By our assumption on λ this component is not isomorphic toa matrix ring over a field, contradicting the fact that all components of FαλH areisomorphic to matrix rings over fields.

Hence we can assume λ2 = 1 and λ1 = λ. By Corollary 7.2 we then know thatF γG = FαλG ∼= FαλG1 ⊗ FG2 contains a component

F (z, ζp2)⊗Mpq(F ) ∼=Mpq(F (z, ζp2)).

But analyzing the Wedderburn decomposition of FαλH we obtain that it does notcontain such a component, namely the only components which are isomorphic to(pq × pq)-matrix rings over a field are Mpq(F ) and Mpq(F (z)). Hence FαλH 6∼=FαλG and overall FαλH is not isomorphic to F γG for any [γ] ∈ H2(G,F ∗). �

Remark 7.4. To our knowledge there are two more examples in the literatureof pairs of groups which have isomorphic group algebras over any field. Theseare on one hand the famous counterexamples to the integral isomorphism problemby Hertweck [Her01] and on the other hand some much simpler and less knownexamples of Roggenkamp [RT92, Chapter VIII]. It remains to be seen whetherthese examples might satisfy the relation of the (TGRIP) over any field.

Thanks: We thank Adam Chapman, Yuval Ginosar and Danny Neftin for manyfruitful conversations. We also express our gratitude to the Technion and the VrijeUniversiteit Brussel, and especially Eric Jespers, which helped arrange visits of theauthors to each others institutions.

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Departament of Mathematics, Free University of Brussels, 1050 Brussels, Belgium

Email address: [email protected]

Department of Mathematics, ORT Braude College, 2161002 Karmiel, Israel

Email address: [email protected]

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TWISTED GROUP RING ISOMORPHISM PROBLEM AND ...Dade [Dad71] constructed a family of pairs of groups Gand H of order p3q6 which provided a counterexample for the problem of Brauer. Here