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On Strongly Associative Group AlgebrasJonas Gonçalves Lopes aa University of Rio Grande do Norte , Natal, BrazilPublished online: 07 Apr 2008.

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Communications in Algebra®, 36: 478–492, 2008Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870701716157

ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS

Jonas Gonçalves LopesUniversity of Rio Grande do Norte, Natal, Brazil

Given a partial action � of a group G on the group algebra FH , where H is a finitegroup and F is a field whose characteristic p divides the order of H , we investigate theassociativity question of the partial crossed product FH ∗� G. If FH ∗� G is associativefor any G and any �� then FH is called “strongly associative.” Using a result ofDokuchaev and Exel (2005) we characterize the strongly associative modular groupalgebras FH for several classes of groups H .

Key Words: Group algebras; Partial actions.

2000 Mathematics Subject Classification: Primary 16S34; Secondary 20C05.

1. INTRODUCTION

Partial actions formally appeared in theory of operator algebras and becamean important tool of their study (see, for example, Abadie, 2003; Exel, 1994, 1997,1998; Exel et al., 2002). Algebraic results on partial actions and related topics havebeen obtained in Dokuchaev and Exel (2005); Dokuchaev et al. (2000, 2007a,b),Dokuchaev and Polcino Milies (2004), Dokuchaev and Zhukavets (2004), Ferrero(2006), and Kellendonk and Lawson (2004). Following Dokuchaev and Exel (2005)we give the next definition.

Definition 1.1. Let G be a group with identity element 1 and � an associative non-unital (i.e., non-necessarily unital) algebra. A partial action � of G on � is a pair� = ���g�g∈G� ��g�g∈G�, formed by a collection of (two-sided) ideals �g of �, and acollection of isomorphisms of algebras

�g � �g−1 −→ �g

which satisfy the following conditions for every g� h ∈ G:

(i) �1 = � and �1 is the identity automorphism of �;(ii) ��gh�−1 ⊇ �−1

h ��h ∩ �g−1�;(iii) �g � �h�x� = �gh�x� for each x ∈ �−1

h ��h ∩ �g−1�.

Received July 18, 2006; Revised January 26, 2007. Communicated by I. P. Shestakov.Address correspondence to Jonas Gonçalves Lopes, Rua Ind. João Motta, 1637 Apto. 401-

Capim Macio, Natal-RN, CEP 59082-410, Brazil; Fax: +84-32119219; E-mail: jonas@ccet.ufrn.br

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ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS 479

The next concept generalizes that of a skew group rings by a (global) action(see Dokuchaev and Exel, 2005).

Definition 1.2. Given a partial action � of a group G on a F -algebra �, the partialskew group ring of � and G by �, written � ∗� G, is the set of all finite formal sums{∑

g∈G ag�g � ag ∈ �g

}, where �g are symbols. Addition is defined in the obvious

way, and multiplication is determined by

�ag�g� · �bh�h� = �g��g−1�ag�bh��gh

The first question which naturally arises is whether or not � ∗� G is associative.Using the existence of approximate units, Exel proved that � ∗� G is associative, inC∗-algebraic context (see Exel, 1994). Recently Dokuchaev and Exel showed that� ∗� G is always associative if � is a semiprime algebra (see Proposition 2.6 andTheorem 3.1 of Dokuchaev and Exel, 2005).

Definition 1.3. We say that an algebra � is strongly associative if for any groupG and an arbitrary partial action � of G on � the partial skew group ring � ∗� G isassociative.

Thus by the above mentioned result, a semiprime algebra is strongly associative.In Dokuchaev and Exel (2005), the authors also gave an example of a partial

action �� with G cyclic of order 2� on a 4-dimensional algebra � over a field Fsuch that � ∗� G is not associative. It is pointed out that if char F = 2, then � isisomorphic the group algebra of the Klein 4-group over F They also suggestedto consider the problem of characterization of strongly associative group algebras.Given a finite group H and a field F whose characteristic does not divide the orderof H� the group algebra FH is semisimple and thus is strongly associative.

The purpose of this article is to study the question of strong associativity formodular group algebras, and in what follows H will stand for a finite group andF for a field whose characteristic divides the order of H

In Section 2 we consider our problem for modular group algebras FH forthe following three cases: H is a nilpotent group, H is a Frobenius group whosecomplement is a p-group, and H is a metabelian group. In Section 3, we applyour results for H = Sn� the symmetric group on n letters, and H = Dn� the dihedralgroup of order 2n.

2. THE PROBLEM OF STRONG ASSOCIATIVITY OF FH

We start by recalling the concept of the algebra of multipliers. Let F be a fieldand � an F -algebra.

Definition 2.1. A multiplier of � is a pair �L� R�, where L and R are lineartransformations of � which satisfy the following conditions for every a� c ∈ �:

i) L�ac� = L�a�c;ii) R�ac� = aR�c�;iii) aL�c� = R�a�c.

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480 LOPES

We say that L is the left action and R the right action of the multiplier �L� R�.We will denote by W��� the set of all multipliers of �. It is immediately verifiedthat W��� is an associative algebra with unity element �L1� R1�, where L1 and R1 areidentity maps, under the following operations:

�L� R�+ �L′� R′� = �L+ L′� R+ R′��

�L�R� = �L� R�� � ∈ F��

�L�R��L′� R′� = �L � L′� R′ � R�

With this operations W��� is called the algebra of multipliers of �.Now suppose that I is an ideal of �. Note that each x ∈ � determine a

multiplier �x = �Lx� Rx� of I defined by

Lx�b� = xb� Rx�b� = bx� ∀b ∈ I

Define the map � � � → W��� by putting ��x� = �Lx� Rx�� x ∈ �. This isa homomorphism of algebras since it is F -linear and, moreover, Lxy = Lx � Ly,Rxy = Ry � Rx, which gives ��xy� = �Lx � Ly� Ry � Rx� = ��x���y�. The followingstatements hold (see the Proposition 2.3 of Dokuchaev and Exel, 2005):

• ���� is an ideal of W���;• � � � → W��� is an isomorphism if and only if � is a unital algebra.

Dokuchaev and Exel (2005, Theorem 3.1) proved that if � = ���g�g∈G� ��g�g∈G�is an arbitrary partial action of a group G on �, then � is strongly associative ifand only if the equality

��g � Rc � �g−1� � La = La � ��g � Rc � �g−1� (1)

is valid on �g for every g ∈ G and all a� c ∈ �.

Remark. We can consider La as the left action of the multiplier �a = �La� Ra� of�g and �g � Rc � �g−1 as the right action of the multiplier

�g�Lc� Rc� = ��g � Lc � �g−1� �g � Rc � �g−1�

of �g (see the proof of the Theorem 3.1 of Dokuchaev and Exel, 2005).We will check condition (1) whenever we shall need to prove that a group

algebra is strongly associative. Since (1) is obvious if �g = 0, we shall tacitly supposethat �g = 0 when dealing with (1).

The following result will be frequently used.

Lemma 2.1. Let H be a finite group and F a field. Let G be an arbitrary groupand � = (

��g�g∈G� ��g�g∈G)an arbitrary partial action of G on the group algebra

FH . Suppose that g ∈ G is such that �g = FH or �g = �H� (the augmentation idealof FH). Then the equality (1) is verified in �g, with arbitrary a� c ∈ FH .

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ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS 481

Proof. We want to prove that

��g � Rc � �g−1� � La = La �(�g � Rc � �g−1

)is valid on �g, ∀a� c ∈ FH .

If �g = FH , then �g is an automorphism of FH . Since a ∈ �g we obtain forb ∈ �g� c ∈ FH , that

�g(�g−1�ab�c

) = �g(�g−1�a��g−1�b�c

)= �g

(�g−1�a�

)�g(�g−1�b�c

)= a�g

(�g−1�b�c

)�

and the equality (1) is verified in this case.Now, if �g = �H�, let b ∈ �H� and a� c ∈ FH . Since FH = �H�⊕ F (direct

sum of F -spaces), write a = a′ + , where a′ ∈ �H�� ∈ F .Then

�g(�g−1�ab�c

) = �g(�g−1�a′b�c

)+ �g(�g−1�b�c

)= �g

(�g−1�a′��g−1�b�c

)+ �g(�g−1�b�c

)= a′�g

(�g−1�b�c

)+ �g(�g−1�b�c

)= �a′ + ��g

(�g−1�b�c

)= a�g

(�g−1�b�c

)�

so that the equality (1) is satisfied on �g = �H�. �

Next we give two examples of strongly associative group algebras.

Example 2.1. Let H = �y � y2 = 1� be a cyclic group of order 2 and F a fieldof characteristic 2. Then the group algebra FH is strongly associative. Indeed,the unique nontrivial ideal of FH is J�FH� = FH�y − 1� (the Jacobson radicalof FH) (see Lemma 17.13(ii) of Karpilovsky, 1987). Let G be any group and�= ���g�g∈G� ��g�g∈G� an arbitrary partial action of G on FH . Then, for every g ∈G,we have �g = FH or �g = J�FH�. Since H is a 2-group, it follows that J�FH� = �H�. By Lemma 2.1 we obtain that the condition (1) is satisfied. Therefore FH isstrongly associative.

Example 2.2. Let H = �y � y3 = 1� be the cyclic group of order 3 and F a field ofcharacteristic 3. Then the group algebra FH is strongly associative. Indeed, the uniquenontrivial ideals of FH are J�FH� = FH�y − 1� and J�FH�2 = FH�y − 1�2. Let G beany group and � = ���g�g∈G� ��g�g∈G� an arbitrary partial action of G on FH . Inthis case, ∀g ∈ G, we have �g = FH or �g = J�FH�, or �g = J�FH�2. Since H is a3-group it follows that J�FH� = �H�. Thus by Lemma 2.1, the equality (1) is validon �g = FH and on �g = J�FH�. Now, let �g = J�FH�2 Since dimFJ�FH�2 = 1, weeasily see that �g � Rc � �g−1 and La commute on �g, ∀a� c ∈ FH . Therefore FH isstrongly associative.

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482 LOPES

In general, if H is a cyclic group of prime order p, obtain the following result.

Proposition 2.1. Let H = �y � yp = 1� be a cyclic group of order p and F a field ofcharacteristic p > 0. Then the group algebra FH is strongly associative if and only if H = 2 or H = 3.

Proof. If H = 2 or H = 3, then by Examples 2.1 and 2.2, respectively, FH isstrongly associative.

Conversely, assume that FH is strongly associative. Suppose that H ≥ 5.Since H is cyclic it follows that J�FH�p−2 has dimension 2 over F and we have thatJ�FH� � J�FH�p−2.

Then

B = {1� �y − 1�� � �y − 1�p−2� �y − 1�p−1

}is an F -basis of FH , then J�FH�p−2 = FH�y − 1�p−2 is the subspace generated by�y − 1�p−2 and �y − 1�p−1.

Let G = �g � g2 = 1� be the cyclic group of order 2, I = J�FH�p−2 and considerthe partial action � of G on FH given by �g−1 = �g = I ,

�g � �g−1 → �g

�y − 1�p−2 → �y − 1�p−1

�y − 1�p−1 → �y − 1�p−2

(By definition �1 = FH and �1 is the identity automorphism of FH).Taking x = �y − 1��1 + �y − 1�p−2�g (note that �y − 1� ∈ J�FH� ⊂ FH = �1

and �y − 1� � J�FH�p−2 = �g), we obtain

xx = (�y − 1��1 + �y − 1�p−2�g

)(�y − 1��1 + �y − 1�p−2�g

)= �y − 1�2�1 + �y − 1�p−1�g + 0+ 0

On the one hand, we have

�xx�x = [�y − 1�2�1 + �y − 1�p−1�g

](�y − 1��1 + �y − 1�p−2�g

)= �y − 1�3�1 + 0+ �y − 1�p−2�g + 0

= �y − 1�3�1 + �y − 1�p−2�g�

while

x�xx� = (�y − 1��1 + �y − 1�p−2�g

)[�y − 1�2�1 + �y − 1�p−1�g

]= �y − 1�3�1 + 0+ 0+ 0

= �y − 1�3�1

Thus, the partial crossed product FH ∗� G is not associative, a contradiction.Therefore H = 2 or H = 3. �

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ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS 483

The next result gives a condition for the group algebra not to be stronglyassociative.

Lemma 2.2. Let H be a finite group and M = H a normal subgroup of H . Let F bea field of characteristic p. If p divides M , then the group algebra FH is not stronglyassociative.

Proof. Denote by T = �1 = h1� h2� � hk� a left transversal for M in H , where theidentity element 1 of H is choosed as the representative of the class M . Let h =1+ h2 + · · · + hk, and consider M = ∑

x∈M x, which is a nonzero central element ofFH . It is easy to see that

hM = Mh = H� h2M = kH� M2 = 0 �since M2 = M M� and p∣∣ M �

Let I = �M� be the ideal of FH generated by M . It has the following F -basis

�H� M� h2M� � hk−1M�

Take G = �g � g2 = 1�� the cyclic group of order 2, and consider the partialaction � of G on FH given by �g−1 = �g = I ,

�g � �g−1 → �g

if k = 2

{H → M

M → H

and if k ≥ 3

H → M

M → H

· · ·Fixes other elements

of the basis.

(By definition �1 = FH and �1 = IdFH ).Finally, taking x = h�1 + M�g, we obtain

xx = �h�1 + M�g��h�1 + M�g�

= h2�1 + H�g + kM�g + 0

On the one hand, we have

�xx�x = h3�1 +[kH + �1+ k2�M

]�g�

while

x�xx� = h3�1 +[2kH + k2M

]�g

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484 LOPES

Note that �xx�x − x�xx� = �M − kH��g = 0. Therefore FH is not stronglyassociative. �

The next result follows from Proposition 2.1 and Lemma 2.2.

Theorem 2.1. Let H be a nilpotent group and F a field whose characteristic p dividesthe order of H . Then the group algebra FH is strongly associative if and only if H = 2or H = 3.

Proof. Suppose that FH is strongly associative.

Claim. H = p.

Indeed, since H is finite nilpotent and p divides H , we have that the Sylowp-subgroup P of H is nontrivial and normal. If P � H it follows by Lemma 2.2that FH is not strongly associative, a contradiction. Thus H = P, and there exists anormal subgroup M of H such that H/M = p (see Theorem 7.2 of Huppert, 1967).If M = �1�, by Lemma 2.2 we again obtain that FH is not strongly associative,a contradiction. Therefore H = p as claimed.

It follows now by Proposition 2.1 that H = 2 or H = 3.The converse we already know by Examples 2.1 and 2.2. �

We prove next two auxiliary results.

Lemma 2.3. Let � = ��, where � and � are F -algebras. If the F -algebra � isstrongly associative, then � and � are strongly associative.

Proof. Suppose that � is not strongly associative. Then there exists a group Gand a partial action � = (

��g�g∈G� ��g�g∈G� of G on � such that � ∗� G is notassociative. By (1) this implies that there exist a ideal �go

= � with go = 1 such that(�go

� Rc � �g−1o

) � La = La �(�go

� Rc � �g−1o

)on �go

for some a ∈ � and for some c ∈ �.But � = ���g�g∈G� ��g�g∈G� given by

�g = �g� ∀g ∈ G� g = 1� �1 = �� �g = �g� ∀g ∈ G� g = 1�

�1 = Id�, is a partial action of G on �, since for each g ∈ G��g is an ideal of �.Consequently, the equality (1) is not valid on �go

= �gofor the elements a and c in

�, a contradiction. Therefore � is strongly associative and the argument for � isthe same. �

Lemma 2.4. Let � be an F -algebra, F a field, G any group and�= ���g�g∈G� ��g�g∈G� an arbitrary partial action of G on �. If the ideal �g� g ∈ G isa direct summand, then the equality (1) is satisfied on �g, for every a� c ∈ �.

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ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS 485

Proof. Suppose that � is an ideal of � such that

� = �go⊕ �

We want to prove that

��go � Rc � �g−1o� � La = La � ��go � Rc � �g−1

o�

is valid on �go, ∀a� c ∈ �.

For let b ∈ �go� a� c ∈ �. Write a = a1 + a2, where a1 ∈ �go

� a2 ∈ � .Then

�go(�g−1

o�ab�c

) = �go(�g−1

o�a1 + a2�b�c

)= �go

(�g−1

o�a1b�c

)= �go

(�g−1

o�a1��g−1

o�b�c

)= a1�go

(�g−1

o�b�c

)= �a1 + a2��go

(�g−1

o�b�c

)= a�go

(�g−1

o�b�c

)

Therefore the equality (1) is satisfied in every direct summand �g of �, for everya� c ∈ �. �

Now we recall some definitions.A group H is called p-solvable (p is a prime) if H has a subnormal series

Ho = H ⊇ H1 ⊇ H2 ⊇ · · · ⊇ Hr = �1�

such that Hi−1/Hi is a p-group or a p′-group for each i.A transitive permutation group H in which only the identity fixes more than

one letter, but the subgroup fixing a letter is nontrivial, is called a Frobenius group(see Karpilovsky, 1987, p. 183).

Let � be a finite-dimensional algebra over a field F . A central idempotente∈� is called centrally primitive if e = 0 and e can not to be written as a sum oftwo nonzero ortogonal central idempotents in �.

There exists a direct decomposition

� = B1 ⊕ · · · ⊕ Bn

of � into indecomposable two-sided ideals Bi = 0 with BiBj = 0 for i = j.Write 1 = e1 + · · · + en with ei ∈ Bi. Then the e�is are mutually orthogonal

centrally primitive idempotents and Bi = �ei = ei� (see Proposition 10.8 ofKarpilovsky, 1987, p. 57).

The ideal Bi is called a block of � and ei is called a block idempotent of �.Let H be a finite group and F an arbitrary field of characteristic p > 0. The

principal block of FH is the block that contains the trivial FH-module 1H or 1FH(the module for the 1-representation, i.e., the trivial representation of H of degree 1).

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486 LOPES

In what follows we write Op′�H� for the maximal normal p′-subgroup of H ,while Op�H� stands for the maximal normal p-subgroup of H . The inverse imageof Op�H/Op′�H�� in H is denoted by Op′�p�H�. A group H is called p-constrained ifthe inverse image of CH/Op′ �H��Op′�p�H�/Op′�H�� in H is contained in Op′�p�H�. It iswell known (see Suzuki, 1986, p. 170) that any p-solvable group is p-constrained.

We are now ready to prove the following result.

Theorem 2.2. Let H be a Frobenius group with complement P, where P is a Sylowp-subgroup of H , and let F be a field of characteristic p > 0. Then the group algebraFH is strongly associative if and only if P = 2 or P = 3.

Proof. Since H is a Frobenius group with complement P, there exists a normalsubgroup M of H such that

H = MP� M ∩ P = �1�

Now as H = M P , we have that M is a p′-group. It obviously follows thatM =Op′�H� and that H is p-constrained.

Let e = M −1 ∑x∈M x. It is well known (see Theorem 1.2.9(a) of Huppert and

Blackburn, 1982, p. 182) that

FH = FHe⊕ FH�1− e�� (2)

where FHe � F�H/M� � FP and FH�1− e� = �H�M�.Suppose that FH is strongly associative. We have by Lemma 2.3 that FHe is

strongly associative. Since FHe � FP it follows that FP is strongly associative. ByTheorem 2.1 we conclude that P = 2 or P = 3.

Conversely, if P = 2 or P = 3, it follows by (2) and by a result of thetheory of blocks (see Karpilovsky, 1987, p. 59, Proposition 10.10 or Huppert andBlackburn, 1982, Theorem 12.1) that the ideals FHe and FH�1− e� of FH aredirect sums of blocks. Since H is p-constrained, M = Op′�H� and e = M −1 ∑

x∈M xit follows that FHe is the principal block of FH (see Proposition 1.20 of Karpilovsky,1987, p. 115). Hence

FH = FHe⊕ FHe1 ⊕ · · · ⊕ FHen

is the direct decomposition of FH into a sum of blocks, where e� e1� e2� � en aremutually ortogonal centrally primitive idempotents.

Now by a result of Wallace (see Corollary 7.8 of Karpilovsky, 1987, p. 189) itfollows that

J�FH� = �P�e = J�FP�e�

Since FHe � FP we have J�FHe� = J�FP�e, and J�FH� = J�FHe�. Thus, J�FHei� = 0,∀i = 1� � n. This implies that each block FHei, i = 1� � n, is a simple F -algebra.

We want to prove that FH is strongly associative. For let G be any group and� = ���g�g∈G� ��g�g∈G� an arbitrary partial action of G on FH .

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ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS 487

If P = 2, the possibilities for �g are either FH� FHe, J�FH� = J�FH�e,FHei1 ⊕ · · · ⊕ FHeit , where 1 ≤ i1 < · · · < it ≤ n� or FHe⊕ FHei1 ⊕ · · · ⊕ FHeit orJ�FH�e⊕ FHei1 ⊕ · · · ⊕ FHeit .

By Lemma 2.1 we have that the equality (1) is satisfied in �g = FH . Nowin �g = FHe, FHei1 ⊕ · · · ⊕ FHeit and FHe⊕ FHei1 ⊕ · · · ⊕ FHeit , we obtain byLemma 2.4 that the equality (1) is verified. Since dimFJ�FH�e = 1 it follows that thisequality is valid on �g = J�FH�e.

Finally, let �g = J�FH�e⊕ FHei1 ⊕ · · · ⊕ FHeit . We may simplify the notationby supposing that i1 = 1� , it = t < n. Since FHe � FP we have that

J�FH�e = J�FHe� � J�FP�

Hence, since FP = J�FP�⊕ F (as F -spaces), FHe = J�FH�e⊕ Fe (as F -spaces). Letb ∈ �g, a� c ∈ FH . Write

a = a′ + a1 + · · · + at + at+1 + · · · + an� where

a′ ∈ FHe� aj ∈ FHej� j = 1� � n

Let a = a1 + · · · + at, then

�g(�g−1�ab�c

) = �g(�g−1�a′b�c

)+ �g(�g−1�ab�c

)

Now write a′ = a′′ + e, where a′′ ∈ J�FH�e� e ∈ Fe. We obtain

�g(�g−1�ab�c

) = �g(�g−1�a′′b�c

)+ �g(�g−1�eb�c

)+ �g(�g−1�ab�c

)

Note that a′′b = 0 since �J�FH�e�2 = 0, and that J�FH�e ∩ �FHe1 ⊕ · · · ⊕ FHet� = 0.Hence, as a ∈ �g, we have

�g(�g−1�ab�c

) = 0+ �g(�g−1�eb�c

)+ a�g(�g−1�b�c

)

Claim. �g(�g−1�eb�c

) = e�g(�g−1�b�c

)

Indeed, write b = b′ + b′′, where b′ ∈ J�FH�e� b′′ ∈ FHe1 ⊕ · · · ⊕ FHek.

Case 1. b′′ = 0. Then b = b′, so that eb = eb′ = b′, since e is the identity ofJ�FH�e, and we have

�g(�g−1�eb�c

) = �g(�g−1�eb′�c

) = �g(�g−1�b′�c

)

On the other hand, notice that �g−1 = J�FH�e⊕ �g−1�FHe1 ⊕ · · · ⊕ FHet�, and that�g−1�FHe1 ⊕ · · · ⊕ FHet�, is a ideal of the form FHej1 ⊕ · · · ⊕ FHejt . Since e is theidentity of J�FH�e, we obtain

e�g(�g−1�b�c

) = e�g(�g−1�b′�c

) = �g(�g−1�b′�c

)

Hence

�g(�g−1�eb�c

) = e�g(�g−1�b�c

)

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Case 2. b′ = 0. Then b = b′′ and eb = eb′′ = 0, so that

�g(�g−1�eb�c

) = �g(�g−1�eb′′�c

) = 0

On the other hand, since

�g � J�FH�e⊕ �g−1�FHe1 ⊕ · · · ⊕ FHet� −→ J�FH�e⊕ FHe1 ⊕ · · · ⊕ FHet

is an isomorpism, we also have e�g(�g−1�b�c

) = 0. Thus, our claim is valid in thiscase.

Finally, if b = b′ + b′′� b′� b′′ = 0, by Cases 1 and 2 we have that

�g(�g−1�eb�c

) = �g(�g−1�eb′�c

)+ �g(�g−1�eb′′�c

)= e�g

(�g−1�b′�c

)+ e�g(�g−1�b′′�c

)= e�g

(�g−1�b′ + b′′�c

)= e�g

(�g−1�b�c

)�

and the claim follows.Note that the proof of this claim does not depend on the nilpotency index of

the radical.Therefore,

�g(�g−1�ab�c

) = e�g(�g−1�b�c

)+ a�g(�g−1�b�c

)= �a′′ + e��g

(�g−1�b�c

)+ a�g(�g−1�b�c

)= �a′ + a��g

(�g−1�b�c

)= a�g

(�g−1�b�c

)�

so that equality (1) is satisfied on �g = J�FH�e⊕ �g−1 ⊕ FHei1 ⊕ · · · ⊕ FHeit . Thus,the condition (1) is verified, and FH is strongly associative.

Assume now that P = 3. Here, the possibilities for �g are either FH� FHe�J�FH�e� �J�FH�e�2� FHei1 ⊕ · · · ⊕ FHeik � 1 ≤ i1 < · · · < ik ≤ n� or FHe⊕ FHei1 ⊕· · · ⊕ FHeik � J�FH�e⊕ FHei1 ⊕ · · · ⊕ FHeik or �J�FH�e�2 ⊕ FHei1 ⊕ · · · ⊕ FHeik

Note that dimF�J�FH�e�2 = 1 and �J�FH�e�3 = 0.It follows by the same arguments as in the case P = 2 that condition (1) is

verified. Therefore, FH is strongly associative. This completes the proof. �

Recall that a group H is called metabelian if H contains a normal subgroup Msuch that M and H/M both are abelian.

Now a group H is called p-nilpotent if H has a normal p′-subgroup N suchthat H/N is a p-group.

Next we give some more definitions (more details can be found, for example,in Karpilovsky, 1987).

Let � be a conjugacy class of H and let h ∈ �. A Sylow p-subgroup ofCH�h� iscalled a defect group of� (with respect to p). Thus all defect groups of� are conjugateand so have a common order, say pd. The integer d is called the defect of �.

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ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS 489

Let e be a block idempotent of FH and let B = B�e� (the block containing e).Since e ∈ Z�FH� (center of FH), it follows that

Supp e = �1 ∪�2 ∪ · · · ∪�r

for some �i ∈ Cl�H�. The largest of the defect groups of �i� 1 ≤ i ≤ r, is called adefect group of e (or of B). It is easy to see that all defect groups of e are conjugateand so have a common order, saypd. The integer d is called the defect of e (or of B).

Finally, we are ready to prove the last result of this section.

Theorem 2.3. Let H be a metabelian group and let F be an algebraically closed fieldwhose characteristic p divides the order of H . Then the group algebra FH is stronglyassociative if and only if H is one of the following two types:

(i) H is cyclic of order 2 or 3;(ii) H is a Frobenius group with complement P and kernel H ′, where P is a Sylow

p-subgroup of H� P = 2 or P = 3.

Proof. Suppose that FH is strongly associative. If H is abelian, then byTheorem 2.1 H = 2 or H = 3; thus, H is of the type (i). Assume now that H isnot abelian. Since H is metabelian, H contains a normal abelian subgroup M suchthat H/M is also abelian. We have that M = �1� and H = M H/M . Note that pdoes not divide M , otherwise, by Lemma 2.2, FH is not strongly associative, acontradiction. Since p divides H , it follows that p divides H/M .

Let e1 = M −1 ∑x∈M x. Then

FH = FHe1 ⊕ FH�1− e1�� (3)

where FHe1 � F�H/M� and FH�1− e1� = �H�M�.Since FH is strongly associative, we have by Lemma 2.3 that FHe1 is strongly

associative. Now FHe1 � F�H/M� implies that F�H/M� is strongly associative. SinceH/M is abelian and p divides H/M we have by Theorem 2.1 that H/M = 2 or H/M = 3. Then M = Op′�H� and H = MP�M ∩ P = �1�, where P is a Sylow p-subgroup of H with p = P = 2 or p = P = 3.

Note that p divides P , hence P is not normal on H , and consequently H isnot nilpotent.

It follows from (3) that the ideals FHe1 and FH�1− e1� = �H�M� are directsums of blocks. But since H is metabelian, this implies that H is solvable, thus it isp-solvable, and consequently H is p-constrained, so that FHe1 is the principal blockof FH . Hence,

FH = FHe1 ⊕ FHe2 ⊕ · · · ⊕ FHek

is the direct decomposition of FH into sum of blocks, where e1� � ek are mutuallyortogonal centrally primitive idempotents.

By definition we have that the blocks of FH have defect d = 1 or d = 0. Nowit is well known that the Sylow p-subgroups of H are defect groups of the principalblock FHe1. Therefore, FHe1 is a block of defect 1.

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Claim. FH�1− e1� = �H�Op′�H�� is a semisimple F -algebra.

Suppose that there exists j� 2 ≤ j ≤ k, such that FHej is a block of defect 1.Note that H is p-nilpotent. It follows that every block of FH contains only oneirreducible FH-module (see Theorem 14.9 of Huppert and Blackburn, 1982, p. 198).In particular the principal block FHe1 contains only one irreducible FH-module,which necessarily is the trivial FH-module 1H .

If L is a field, it is well known that there exists a 1-1 correspondencebetween the different unidimensional L-representations of H and the differentunidimensional L-representations of H/H ′.

Note that since H/M is abelian it follows that �1� = H ′ ⊆ M . We have thatp does not divide H ′ , otherwise, by Lemma 2.2, FH is not strongly associative, acontradiction. Hence H ′ ∩ P = �1�. If H ′ � M , since H ′P = H is a normal subgroupof H and p divides H ′P = H ′ P , we again have by Lemma 2.2 that FH is notstrongly associative, a contradiction. Therefore, M = Op′�H� = H ′.

Since H/H ′ � P is a (abelian) p-group it follows that the unique irreduciblemodular representation of H/H ′ is the 1-representation. Thus,

there exists a unique unidimensional F -representation of H�

which is the 1-representation.(4)

Now let Wj be the unique irreducible FH-module contained in the block FHej .Let zj = dimF Wj . Since H is p-solvable and M = Op′�H� is abelian it follows (seeCorollary 4.13 of Karpilovsky, 1987, p. 154) that zj divides �H � M� = p. Thus, zj =1 or zj = p.

Since FHej has defect d = 1, it follows (see Theorem 2.7 of Karpilovsky,1987, p. 79) that 1 is the smallest integer such that p1−1 = 1 divides zj . Thisimplies that zj = 1, otherwise zero is the smallest integer such that p1−0 = p divideszj , a contradiction. But zj = 1 contradicts (4) and it follows that for every i =2� � k, FHei has defect zero. Then FHe2� � FHek are simple F -algebras (seeTheorem 6.4(i) of Karpilovsky, 1987, p. 179), so that J�FHei� = 0, ∀i = 2� � k

Hence

J� �H�Op′�H��� = J�FHe2�⊕ · · · ⊕ J�FHek� = 0�

proving the claim.Thus

J�FH� = J�FHe1�⊕ J�FH�1− e1��

= J�FHe1� � J�FP�

This implies that dimFJ�FH� = P − 1. Since H ′ = Op′�H� = �1�, we have H = P,and by Theorem of Wallace (see Theorem 8.11 of Karpilovsky, 1987, p. 198) itfollows that H is a Frobenius group with complement P (and kernel H ′, by theuniqueness of the Frobenius kernel). This completes the proof of the “only if” part.

Conversely, if H = 2 or H = 3 we have by Examples 2.1 and 2.2,respectively, that FH is strongly associative.

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ON STRONGLY ASSOCIATIVE GROUP ALGEBRAS 491

Now if H is a Frobenius group with complement P, where P is a Sylow p-subgroup of H , P = 2 or P = 3, it follows by Theorem 2.2 that FH is stronglyassociative. This completes the proof. �

3. THE MODULAR GROUP ALGEBRAS OF Sn AND Dn

Now we are ready to apply our results for group algebras FSn and FDn� whereSn is the symmetric group, n ≥ 2, and

Dn = �a� b � an = 1 = b2� ba = a−1b��

the dihedral group of order 2nTake first H = Sn and suppose that the characteristic p of the field F divides

Sn = n!.(i) If n = 2 and p = 2, then Sn = 2 and by Example 2.1 we have that FS2 is

strongly associative.(ii) If n = 3 and p = 2, we know that

S3 = �id� �123�� �132�� �12�� �13�� �23��

is a Frobenius group with complement P = ��12��, where P is a Sylow 2-subgroup of S3. Since P = 2, it follows by Theorem 2.2 that FS3 is stronglyassociative.

(iii) For n = 3 and p = 3, take

M = �id� �123�� �132�� � S3

Note that p = 3∣∣ M = 3, so that by Lemma 2.2 FS3 is not strongly associative.

(iv) If n = 4, we know that the only normal proper subgroups of S4 are A4 (thealternating group of degree 4) and the Klein 4-group

K = �id� �12��34�� �13��24�� �14��23��

Hence, if p = 2 we have 2∣∣ K = 4 (2

∣∣ A4 = 12 also), while if p = 3, 3∣∣ A4 = 12.

In each case it follows by Lemma 2.2 that FS4 is not strongly associative.(v) Finally, for n ≥ 5, we know that An, the alternating group of degree n, is the

unique normal proper subgroup of Sn. Note that p∣∣n! implies p

∣∣ An = n!2 , and

therefore by Lemma 2.2, FSn� n ≥ 5, is not strongly associative.

Remark. Item (v) gives us examples of nonsolvable groups whose group algebrasare not strongly associative.

Suppose now that H = Dn and p = char F divides 2n.

(i) If p = 2 and n is odd, we know that Dn is a Frobenius group withcomplement P = �b � b2 = 1�, where P is a Sylow 2-subgroup of Dn. Since P = 2,it follows by Theorem 2.2 that FDn is strongly associative.

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Now considering M = �a � an = 1� �Dn, notice that for any other prime p> 2,obviously p

∣∣2n implies p∣∣ M . Therefore, by Lemma 2.2, FDn is not strongly

associative.

(ii) If n is even, then evidently p∣∣2n implies p

∣∣ M . Therefore, by Lemma 2.2,FDn is not strongly associative.

ACKNOWLEDGMENTS

The present article is based on a part of my Ph.D. thesis (Lopes, 2005) whichwas directed by Prof. M. Dokuchaev at the IME-USP (São Paulo, Brazil). I wishto thank him for his attention and encouragement. This article is supported byPICDT/CAPES of Brazil.

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