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City, University of London Institutional Repository

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Kessar, R., Koshitani, S. and Linckelmann, M. (2015). On symmetric quotients

of symmetric algebras. Journal of Algebra, 442, pp. 423-437. doi:

10.1016/j.jalgebra.2014.05.035

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version.

Permanent repository link: http://openaccess.city.ac.uk/6903/

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http://dx.doi.org/10.1016/j.jalgebra.2014.05.035

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RADHA KESSAR, SHIGEO KOSHITANI, AND MARKUS LINCKELMANN

Abstract. We investigate symmetric quotient algebras of symmetric algebras, with an emphasis on finite group algebras over a complete discrete valuation ringO. Using elementary methods,

we show that if an ordinary irreducible characterχof a finite groupGgives rise to a symmetric quotient overO which is not a matrix algebra, then the decomposition numbers of the row labelled byχare all divisible by the characteristicpof the residue field ofO.

1. Introduction

Letpbe a prime andOa complete discrete valuation ring having a residue fieldkof characteristic

pand a quotient field K of characteristic zero. Unless stated otherwise, we assume thatK andk

are splitting fields for all finite groups under consideration. Let Gbe a finite group. Any subset

M of the set IrrK(G) of irreducible K-valued characters of G gives rise to an O-free quotient

algebra, namely the image of a structural homomorphismOG→EndO(V), where V is anO-free OG-module having characterPχMχ. This image is isomorphic toOG(PχMe(χ)), wheree(χ) denotes the primitive idempotent in Z(KG) corresponding to χ. Any O-free quotient algebra of

OG arises in this way; in particular, OG has only finitely many O-free algebra quotients. Any quotient of OG admits a decomposition induced by the block decomposition, and hence finding symmetric quotients of OGis equivalent to finding symmetric quotients of the block algebras of

OG. We denote by IBrk(G) the set of irreducible Brauer characters ofG, and bydG:ZIrrK(G)→

ZIBrk(G) the decomposition map, sending a generalised character of G to its restriction to the

set Gp′ of p′-elements in G. For B a block algebra of OG, we denote by Irr

K(B) and IBrk(B)

the sets of irreducible K-characters and Brauer characters, respectively, associated with B. We denote by dB :ZIrrK(B)→ ZIBrk(B) the decomposition map obtained from restrictingdG. We

denote byDG = (dχϕ) the decomposition matrix of OG, with rows indexed by χ ∈ IrrK(G) and

columns indexed by IBrk(G); that is, the dχϕ are the nonnegative integers satisfying dG(χ) =

P

ϕ∈IBrk(G) d

χ

ϕ·ϕ, for anyχ∈IrrK(G). Equivalently,dχϕ=χ(i), whereiis a primitive idempotent

inOGsuch thatOGiis a projective cover of a simplekG-module with Brauer character ϕ. Ifχ,

ϕbelong to different blocks, then dχ

ϕ = 0. For B a block algebra of OG, we denote by DB the

submatrix of DG labelled byχ ∈ IrrK(B) andϕ∈ IBrk(B). We say that χ ∈ IrrK(G) lifts the

irreducible Brauer characterϕ∈IBrk(G) ifdG(χ) =ϕ, or equivalently, ifdχϕ= 1 andd χ

ϕ′ = 0 for

allϕ′IBr

k(G) different fromϕ. In that case,OGe(χ) is a matrix algebra overO, hence trivially

symmetric (see 4.1 below).

Theorem 1.1. Let G be a finite group and χ ∈ IrrK(G). Suppose that OGe(χ) is symmetric.

Then eitherχ lifts an irreducible Brauer character, ordχ

ϕ is divisible by pfor allϕ∈IBrk(G).

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Corollary 1.2. Let Gbe a finite group andχ∈IrrK(G)such that dχϕ is prime topfor someϕ∈

IBrk(G). Then the O-algebraOGe(χ)is symmetric if and only ifχ lifts ϕ.

Corollary 1.3. Let Gbe a finite group, B a block of OG, and χ∈IrrK(B). Suppose that χ has

height zero. Then the O-algebraOGe(χ) is symmetric if and only if χ lifts an irreducible Brauer character.

Corollary 1.4. Let G be a finite group and B a block of OG. Suppose that each row of the decomposition matrix DB has at least one entry prime to p. Then for any χ ∈ IrrK(B), the

algebraOGe(χ)is symmetric if and only ifχ lifts an irreducible Brauer character inIBrk(B).

The hypotheses of 1.4 apply to many blocks of quasi-simple finite groups. By a result of Dade [2] they also apply to blocks with cyclic defect groups.

Corollary 1.5. Let Gbe a finite group andB a block with cyclic defect groups. Letχ∈IrrK(B).

ThenOGe(χ)is symmetric if and only if χcorresponds to a nonexceptional vertex at the end of a branch of the Brauer tree ofB.

By Erdmann’s results in [3], the hypotheses of 1.4 apply to all nonnilpotent tame blocks.

Corollary 1.6. Suppose that p= 2. LetG be a finite group and B a nonnilpotent block of OG

having a defect group P which is either generalised dihedral, quaternion, or semidihedral. Then for anyχ∈IrrK(B), the algebraOGe(χ)is symmetric if and only ifχlifts an irreducible Brauer

character inIBrk(B).

Proposition 1.7. Let P be a finite p-group having a normal cyclic subgroup of index p. Then

OP e(χ)is symmetric for anyχ∈IrrK(P).

This will be proved in§3. Since a nilpotent block is isomorphic to a matrix algebra over one of its defect group algebras, this proposition, specialised top= 2, has the following consequence (which includes nilpotent tame blocks).

Corollary 1.8. Suppose thatp= 2. LetGbe a finite group andBa nilpotent block of OGhaving a defect groupP which is either generalised dihedral, quaternion, semidihedral, or quasidihedral. Then for anyχ∈IrrK(B), the algebra OGe(χ)is symmetric.

The symmetric algebras arising in 1.4 and subsequent corollaries are matrix algebras (see 4.1 below). By contrast, the symmetric algebras obtained from nonlinear characters in 1.7 are not isomorphic to matrix algebras. Further examples of charactersχwith symmetric quotientOGe(χ) which are not isomorphic to matrix algebras can be obtained from characters of central type. An irreducible characterχof a finite groupGisof central typeifχ(1)2=|G:Z(G)|.

Proposition 1.9. Let G be a finite group and χ ∈ IrrK(G) a character of central type. Then

OGe(χ)is symmetric.

This is shown as a special case of a slightly more general situation in 2.4. As a consequence of 1.7 and 1.9, ifP is a finitep-group of order at mostp3, thenOP e(χ) is symmetric for allχ IrrK(P). In §6 we give an example showing that this is not the case in general for irreducible

characters of finitep-groups of orderp4.

If M is the set of all nontrivial characters of G, then the corresponding quotient algebra is

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characters ofGappear, andO(PxGx) is the annihilator inOGofI(OG). The hypothesis ofK,

kbeing large enough is not necessary for the following result.

Proposition 1.10. Let Gbe a finite group. The following are equivalent. (i) TheO-algebraOG/O(PxGx)is symmetric.

(ii) The groupGisp-nilpotent and has a cyclic Sylow-p-subgroup.

The arguments used in the proof of 1.10 can be adapted to yield the following block theoretic version; we needkto be large enough for the blockB in the next result.

Proposition 1.11. LetGbe a finite group,Ba block algebra ofOG, and letχ∈IrrK(B). Suppose

thatχ lifts an irreducible Brauer character. SetI=B∩(K⊗OB)e(χ), where we identifyB with

its image1⊗B in K⊗OB. The following are equivalent.

(i) The algebraB/I is symmetric.

(ii) The blockB is nilpotent with cyclic defect groups.

2. Notation and basic facts

If A is an O-algebra which is free of finite rank as an O-module, we denote by IrrK(A) the

set of characters of the simpleK⊗OA-modules. Taking characters ofK⊗OA-modules yields an

isomorphism between the Grohendieck group RK(A) of finitely generated K⊗O A-modules and

the free abelian group with basis IrrK(A), inducing a bijection between the isomorphism classes of

simpleK⊗OA-modules and IrrK(A). If in addition theK-algebraK⊗OAis semisimple, hence a

direct product of simpleK-algebras corresponding to the isomorphism classes of simpleK⊗OA

-modules, we denote bye(χ) the primitive idempotent ofZ(K⊗OA) which acts as identity on the

simpleK⊗OA-modules with characterχand which annihilates all other simpleK⊗OA-modules.

In this case we have K⊗OA ∼=QχIrrK(A) (K⊗O A)e(χ), and each factor (K⊗O A)e(χ) is a

simpleK-algebra. AnO-algebraAissymmetricifAis finitely generated free as anO-module and ifAis isomorphic to itsO-dualA∨= Hom

O(A,O) as anA-A-bimodule. LetAbe an O-algebra.

A submoduleV of anA-moduleU is calledO-pureifV is a direct summand ofU as anO-module. We use without further comments the following well-known facts. See e. g. [4, 17.2], and also [7, Theorem 1] for an application in the context of blocks with cyclic defect groups.

Lemma 2.1. Let A be an O-algebra and letU be an A-module which is finitely generated free as an O-module. Let V be a submodule of U. Then U ∩(K⊗O V) is the unique minimal O-pure

submodule ofU containing V, where we identify U with its image 1⊗U in K⊗O U. Moreover,

the following are equivalent.

(i) TheA-moduleV isO-pure inU. (ii) TheA-moduleU/V isO-free. (iii) We have J(O)V =J(O)U∩V.

(iv) The image ofV ink⊗OU is isomorphic to k⊗OV.

(v) We haveV =U∩(K⊗OV), where we identify U to its image 1⊗U in K⊗OU.

Thus ifIis an ideal in anO-algebraAwhich is finitely generated free as anO-module, then the quotient algebraA/I isO-free if and only ifIisO-pure inA. AnyO-pure idealI ofAis equal to

A∩MI for a unique idealMI inK⊗OA, where we have identifiedAto its canonical image 1K⊗A

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every ideal ofK⊗OAis a product of a subset of those simple algebras. In particular, in that case

the set ofO-pure ideals in Ais finite and corresponds bijectively to the set of subsets of a set of representatives of the isomorphism classes of simpleK⊗OA-modules.

For symmetric algebras over a field, the following result is due to Nakayama [5]; the generalisation toO-algebras is straightforward (we include a proof for the convenience of the reader). Note that the left and right annihilators of an idealI in a symmetric algebraAare always equal, denoted by ann(I).

Proposition 2.2(cf. [5, Theorem 13]). Let A be a symmetricO-algebra, and let I be an O-pure ideal in A. The quotient algebra A/I is a symmetric O-algebra if and only if there is an element

z∈Z(A)such that ann(I) =Az.

Proof. Set ¯A = A/I, and denote by π : A → A¯ the canonical surjection. Let s : A → O be a symmetrising form for A; that is, s is symmetric and the map sending a ∈ A to the form a·s

defined by (a·s)(b) = s(ab) for all b∈A is an isomorphismA∼=A∗= Hom

O(A,O). (Sinces is

symmetric, this map is automatically a homomorphism ofA-A-bimodules.) Suppose that ann(I) =

Azfor somez∈Z(A). Thent=z·sannihilatesI, hence induces a form ¯t∈A¯∗defined by ¯ta) =

s(za), wherea∈A and ¯a=a+I∈A¯. Sincesis symmetric andz∈Z(A), the formstand ¯t are again symmetric. It suffices to show that the map sending ¯a∈A¯ to ¯a·¯t ∈A¯∗ is surjective. Let

¯

u∈A¯∗. Thenu= ¯uπA, henceu=a·sfor a uniqueley determined elementaA. SinceI

ker(u), we haves(aI) ={0}, hence a∈ann(I). Thusa=cz for somec∈A. It follows thatu=

c·(z·s) =c·t, and hence ¯u= ¯c·¯t, which shows that ¯Ais symmetric. Suppose conversely that ¯Ais symmetric. Let ¯t: ¯A→ Obe a symmetrising form, and sett= ¯t◦π. Thentis symmetric (because ¯

t is) and hence t = z·s for some z ∈ Z(A). Since I ⊆ker(t) we have s(zI) = {0}, hence z ∈

ann(I). Let b∈ann(I). Thenu=b·shasI in its kernel, hence induces a form ¯uon ¯Asatisfying ¯

u(¯a) =s(ab). Since ¯A is symmetric, there is ¯c∈A¯such that ¯u= ¯c·¯t, hence such thatu=c·t= (cz)·s. This (cz)·s=b·s, and henceb=cz∈Az, whence the equality ann(I) =Az.

LetAbe a symmetric O-algebra. Ifz ∈Z(A) such that Az is the annihilator of an idealI in

A, thenAzisO-pure. Thus finding symmetric quotients ofAis equivalent to finding elementsz∈

Z(A) with the property thatAz isO-pure inA.

Corollary 2.3. Let A be a symmetric O-algebra andz∈ Z(A). If Az isO-pure, then the anni-hilatorI= ann(z) = ann(Az) is an O-pure ideal satisfying ann(I) =Az, and the O-algebraA/I

is symmetric. Moreover, any symmetricO-algebra quotient ofAarises in this way.

Proof. This follows from 2.2 and the preceding remarks.

Proposition 2.4. Let G be a finite group and N a normal subgroup of G. Suppose that K is a splitting field for N. Let η ∈ IrrK(N), and suppose thatη is G-stable. If ON e(η) is symmetric,

thenOGe(η)is symmetric.

Proof. Suppose thatON e(η) is symmetric. By 2.2 there exists an elementz∈Z(ON) such that

ON zis the annihilator of the kernel of the mapON → ON e(η). Thusz∈Z(ON)e(η) =Oe(η), where we use thatKis large enough forN. In particular, there isλ∈ Osuch thatz=λe(η), and hencez∈Z(OG). SinceOGis free as a rightON-module, it follows thatOGzis O-pure inOG. The annihilator of the kernel of the mapOG→ OGe(η) is therefore equal to OGz. The result

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Proof of Proposition 1.9. Sinceχis of central type, there is a unique (henceG-stable) linear char-acterζ:Z(G)→ O×such thate(χ) =e(ζ). We haveOZ(G)e(ζ)=O, which is trivially symmetric,

and henceOGe(ζ) =OGe(χ) is symmetric by 2.4.

The argument in the proof of 2.4 to describe a central elementz which generates a pure ideal admits the following generalisation.

Proposition 2.5. Let A be a symmetric O-algebra such that K⊗O A is semisimple. Let χ ∈

IrrK(A) be the character of an absolutely simple K⊗O A-module, and denote by e(χ) the

corre-sponding primitive idempotent inZ(K⊗OA). Letλ∈ O be an element having minimal valuation

such that λe(χ) ∈ A, where we identify A with its image 1⊗A in K⊗O A. The following are

equivalent.

(i) TheO-algebraAe(χ)is symmetric. (ii) TheA-moduleAλe(χ) isO-pure inA. (iii) We have Aλe(χ) =A∩(K⊗OA)e(χ).

Proof. The equivalence of (ii) and (iii) is a general fact (cf. 2.1). LetIbe the kernel of the algebra homomorphismA→Ae(χ) sendinga∈Atoae(χ). Multiplication bye(χ) inK⊗OAyields the

projection ofK⊗OA=Qψ∈IrrK(A)(K⊗OA)e(ψ) onto the factor (K⊗OA)e(χ); this is a matrix

algebra asχis the character of an absolutely simpleK⊗OA-module. Thus I is theO-pure ideal

corresponding to the complement of{χ} in IrrK(A); that is,I=A∩Qψ(K⊗OA)e(ψ), whereψ

runs over the set IrrK(A)− {χ}. It follows that the annihilator ofIis equal toA∩(K⊗OA)e(χ),

and henceZ(A)∩(K⊗OA)e(χ)⊆ Oe(χ). ThusA/I is symmetric if and only ifAz isO-pure for

some z∈ A∩ Oe(χ), hence if and only ifAλe(χ) is O-pure for someλ∈ O. In that case, λhas necessarily the smallest possible valuation such thatλe(χ)∈A. The result follows.

Remark 2.6. LetA be anO-algebra which is finitely generated free as anO-module, and letI

be anO-pure ideal inA. Then the image of the canonical mapA→EndO(A/I) sendinga∈Ato

left multiplication bya+IinA/I has kernelI, hence image isomorphic toA/I. Thus anyO-free algebra quotient ofAis isomorphic to the image of the structural homomorphismA→EndO(V)

sending a ∈ A to left multiplication by a on V, for some A-module V which is free of finite rank as anO-module. SinceV is O-free, this image is isomorphic to the canonical image of Ain EndK(K⊗OV). Thus, ifK⊗OV is a semisimpleK⊗OA-module, then this image depends only

on the isomorphism classes of simpleK⊗OA-modules occurring in a decomposition ofK⊗OV,

but not on the multiplicity of the simple factors ofK⊗OV.

Remark 2.7. It follows from the formal properties of Morita equivalences that a Morita equiva-lence between two algebras induces a bijection between quotients of the two algebras, in such a way that quotients corresponding to each other are again Morita equivalent. In particular, symmetric quotients are preserved under Morita equivalences. We describe this briefly for the convenience of the reader. Let A,B be Morita equivalent O-algebras; that is, there is anA-B-bimoduleM and a B-A-bimodule N such that M, N are finitely generated projective as left and right modules, and such that we have isomorphisms of bimodulesM ⊗BN ∼=AandN⊗AM ∼=B. The functor

M⊗B−induces an equivalence between the categories ofB-B-bimodules and ofA-B-bimodules,

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and subbimodules inM sending an ideal inIinAto the subbimoduleIM ∼=I⊗AM. Combining

these bijections yields a bijection between ideals inAand ideals inB, with the property that the idealI inA corresponds to the idealJ in B if and only if IM =M J, which in turn holds if and only ifJN =N I. IfI,J correspond to each other through this bijection, thenM/IM andN/JN

induce a Morita equivalence betweenA/I andB/J. Indeed,M/IMis finitely generated projective as a leftA/I-module, sinceM is finitely generated as a leftA-module. UsingIM =M J, this argu-ment shows thatM/IMis finitely generated projective as a rightB/J-module. Similarly,N/JN is finitely generated projective as a left and right module. Moreover, we haveM/IM⊗B/JN/JN∼=

M/M J ⊗B N/JN ∼=M ⊗BN/JN ∼=M ⊗BN/M⊗BJN ∼=M ⊗BN/M ⊗BN I ∼=A/I. The

same argument with reversed roles yields N/N I⊗A/I M/IM ∼=B/J. Since Morita equivalences

preserve the property of being symmetric, this shows thatA/I is symmetric if and only ifB/J is symmetric.

3. Proof of Proposition 1.7

LetGbe a finitep-group having a cyclic normal subgroupH of indexp, and letχ∈IrrK(G).

In order to prove Proposition 1.7 we may assume thatGis nonabelian, henceH has order at least

p2. Suppose first that pis odd. Then the automorphism ofH induced by an element tGH acts trivially on the subgroup Hp of index p in H, hence Z(G) = Hp has index p2 in G. Any nonlinear character ofGhas degreep, hence is a character of central type. It follows from 1.9 that

OGe(χ) is symmetric.

Suppose now thatp= 2. The previous arguments remain valid so long as the action ofGon the cyclic normal 2-subgroupHof index 2 is trivial on the subgroupH2of index 2 inH. This includes the case of semidihedral 2-groups (wheret is an involution which acts on the cyclic subgroup H

of order 2n by sending a generator s of H to s1+2n−1

). If n = 2, then |G| = 8, hence χ is a character of central type, and so the symmetry of OGe(χ) follows from 1.9. Suppose that n≥3 and thatGdoes not act trivially on the subgroup of index 2 inH. ThenGis generalised dihedral or generalised quaternion (corresponding in both cases to the action of an element t ∈ G−H

of order either 2 or 4 on H sending s to s−1) or quasidihedral (corresponding to the action of t sending sto s−1+2n−1

). We will need the following elementary facts (a proof is included for the convenience of the reader).

Lemma 3.1. Let n≥3 and letζ be a primitive 2n-th root of unity in O. Leta,b Zsuch that

b−a is even. The following hold.

(i) The numbersζa±ζb are divisible by(1ζ)2 in O. (ii) The numbers (1ζ±ζζ)12 are invertible inO.

Proof. The integer 2 is divisible by (1−ζ)3 in O, as n≥ 3. In particular, (12ζ)2 ∈ J(O), and hence, in order to prove (i), it suffices to show thatζa+ζb is divisible by (1−ζ)2. Writeb−a= 2cfor some integerc. Thenζa+ζb=ζa(1 +ζb−a) =ζa(1 +ζ2c). Thus we may assumea= 0 and

b= 2c. We have 1 +ζ2c = 1ζ2c+ 2ζ2c= (1ζc)(1 +ζc) + 2ζ2c= (1ζc)(1ζc+ 2ζc) + 2ζ2c,

wich is divisible by (1−ζ)2 because 1ζ divides 1ζc and (1ζ)2 divides 2. This shows (i). For (ii), observe that (11−ζζ2)2 =

1+ζ

1−ζ = 1 +

1−ζ is invertible inO. Multiplying this by ζ

−1 shows

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We complete the proof of 1.7. Let G be a generalised dihedral or quaternion group of order 2n+1. In order to show thatOGe(χ) is symmetric, we may assume that χ(1)>1, henceχ(1) = 2. LetH be the cyclic subgroup of order 2n ofG. Since χ(1) = 2 we haveχ= IndG

H(η) for some

nontrivial linear characterη ∈IrrK(H). Arguing by induction, we may assume that η is faithful.

Letζbe a root of unity of order 2n and letsbe a generator ofH such thatη(s) =ζ. Let ¯η be the

character ofH sendingstoζ−1. Then ResG

H(χ) =η+ ¯η, andχvanishes outsideH. In particular,

e(χ) =e(η) +e(¯η) = 1 2n

2n1

X

a=0

(ζa+ζ−a)sa .

By 3.1 the coefficientsζa+ζ−a in this sum are all divisible by (1ζ)2. Set λ= 2n

(1−ζ)2, and set

z=λe(χ). By the above, we have z∈ OG. We need to show thatOGz isO-pure in OG. Since

e(χ)∈ OH, and hence z ∈ OH, it suffices to show thatOHz is O-pure in OH. For anya such that 0≤a≤2n−1 we have

saz=λ(sae(η) +sae(¯η)) =λ(ζae(η) +ζ−ae(¯η)).

We claim that the set{z,2ne(η)}is anO-basis ofOHz. Note first that

sz−ζ−1z=ζ−1(ζ21)λe(η) =µ2ne(η)

for someµ∈ Obecause ζ21 is divisible, inO, by (1ζ)2. Thus bothz and 2ne(η) belong to

OHz. We need to show that any of the elementssazwithaas before is an O-linear combination

ofzand 2ne(η). We have

saz−ζ−az=λζ−a(ζ12a)e(η) =ν2ne(η)

for some ν ∈ O, whence the claim. It remains to show that OHz is O-pure in OH. Let u =

P2n1

a=0 µasa be an element inOHz such that all coefficientsµa∈ O are divisible by 1−ζ. Write

u=βz+γ2ne(η) with β, γ∈ O. We need to show thatβ, γ are divisible by 1ζ. Comparing

coefficients for the two expressions ofuabove yields

µa=β

ζa+ζ−a

(1−ζ)2 +γζ

a

for 0 ≤ a ≤2n1. If a = 2n−2, then ζa+ζ−a = 0, hence µ

a = γζa, which implies that γ is

divisible by 1−ζ, as µa is divisible by 1−ζ. We consider the above equation fora = 1, which

reads

µ1=β

ζ+ζ−1 (1−ζ)2 +γζ . By 3.1 we have (1ζ+ζζ)12 ∈ O

×. Sinceµ

1 andγare divisible by 1−ζ, this implies thatβ is divisible by 1−ζ. ThusOHzisO-pure inOH, and henceOGe(χ) is symmetric.

Let finallyGbe quasidihedral; that is, conjugation by the involutiontsendsstos−1+2n−1

. The calculations are similar to the previous case; we sketch the modifications. Let ¯ηbe the character of

H sendingstoζ−1+2n−1

=−ζ−1. Thene(χ) =e(η) +eη) =P2n1

a=0 (ζa+ (−1)aζ−a). As before, setting z = (12nζ1)2e(χ), it suffices to show that OHz is O-pure in OH. One verifies as before, that{z,2ne(η)} is anO-basis of OHz. Let u=P2n1

a=0 µasa be an element inOHzsuch that all

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show thatβ, γ are divisible by 1−ζ. Comparing coefficients yieldsµa =βζ

a+(1)aζ−a

(1−ζ)2 +γζa. If

a= 2n−2, then a is even andζa+ζ−a = 0, implying µ

a = γζa, which in turn implies that γ is

divisible by 1−ζ. Comparing coefficients for a= 1 yields thatβ is divisible by 1−ζ. ThusOHz

isO-pure in OH, and henceOGe(χ) is symmetric. This completes the proof of 1.7.

4. On symmetric subalgebras of matrix algebras

Let Gbe a finite group. By the above, the O-free O-algebra quotients of OG correspond to

O-pure ideals, hence to subsets of IrrK(G), and any O-free quotient of OG is the image of a

structural mapα:OG→EndO(V) for some finitely generatedO-freeOG-moduleV. By 2.6, the

kernel of this map depends only on the set of irreducible characters ofGarising as constituents of the character ofV. If the characterχofV is irreducible, then the image ofαis isomorphic to the algebraOGe(χ), where e(χ) = χ|G(1)| Px∈Gχ(x−1)x. Moreover, in that case, Im(α)∼=OGe(χ) has

the sameO-rankχ(1)2as End

O(V). If χ∈IrrK(G) has degree one, thenOGe(χ)∼=Ois trivially

symmetric. Ifχ has defect zero, thenOGe(χ) is a matrix algebra overO, hence also symmetric. Both examples are special cases of the following well-known situation:

Proposition 4.1. LetGbe a finite group,Ka splitting field forG, andχ∈IrrK(G). The algebra

OGe(χ) is isomorphic to a matrix algebra over O if and only if χ lifts an irreducible Brauer character.

Proof. We include a proof for the convenience of the reader. Let X be an O-free OG-module with character χ. The characterχ lifts an irreducible Brauer character if and only if k⊗OX is

a simplekG-module. The kG-modulek⊗OX is simple if and only if the structural mapkG →

Endk(k⊗OX) is surjective. By Nakayama’s Lemma, this is the case if and only if the structural

mapOG→EndO(X) is surjective. The result follows.

IfOGe(χ) is symmetric but not a matrix algebra, then the following observation narrows down the possible symmetrising forms.

Proposition 4.2. Let V be a freeO-module of finite rankn, and letAbe a symmetric subalgebra of EndO(V)of rank n2. Then Z(A)∼=O, and there is an integerr≥0 such that the restriction

toA of the map π−rtr

V : EndO(V)→ K sends A toO and induces a symmetrising form on A.

Moreover,r is the smallest nonnegative integer satisfyingπrEnd

O(V)⊆A.

Proof. Since theO-rank ofAisn2, we haveK

OA∼= EndK(K⊗OV), whence Z(A)∼=O. Any

symmetrising form on EndK(K⊗OV) is a nonzero linear multiple of the trace map trK⊗OV, and

hence a symmetrising form onAis of the formπ−rtr

V for some integerr such thatπ−rtrV(A)⊆

O. This forcesr≥0. Letsbe the smallest nonnegative integer satisfyingπsEnd

O(V)⊆A. Lete

be a primitive idempotent in EndO(V). Then trV(e) = 1. We haveπse∈A, henceπ−rtrV(πse) =

πs−r∈ O. This implies thats≥r. Sinceπs−1EndO(V) is not contained inA, there is an element

c∈Asuch thatπ−1c6∈AandcπsEnd

O(V). Thus Ocis a pureO-submodule ofA. Thus there

is anO-basis ofAcontainingc. By considering the dualO-basis with respect to the symmetrising form π−rtr

V, it follows that π−rtrV(cA) = O. Since c ∈ πsEndO(V), we have trV(cA)⊆ πsO,

henceπ−rtr

V(cA)⊆πs−rO. This yields the inequalitys≤r, whence the equalitys=r.

Corollary 4.3. Let V be a free O-module of finite rank n, and let A be a proper symmetric subalgebra of EndO(V)of rank n2. Then for any idempotent i∈ A, the integer trV(i)is divisible

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Proof. IfAis a proper symmetric subalgebra of EndO(V), then the smallest nonnegative integerr

satisfyingπrEnd

O(V)⊆Ais positive. It follows from 4.2 thatA has a symmetrising form which

is the restriction toA of π−rtr

V for some positive integer r. In particular, if i is an idempotent

inA, thenπ−rtr

V(i) is an element in O. Since trV(i) is an integer and r >0, this implies thatp

divides trV(i), Applied toi= 1A= IdV yields that pdividesn= trV(IdV).

Corollary 4.4. LetV be a freeO-module of finite rankn. Suppose that(n, p) = 1. ThenEndO(V)

has no proper symmetric subalgebra ofO-rankn2.

Proof. This is clear by 4.3.

Corollary 4.5. Let V be a freeO-module of finite rankp. Every proper symmetric subalgebra of rank p2 of End

O(V)is local.

Proof. If i is an idempotent in a proper symmetric subalgebra A of EndO(V), then p divides

trV(i)≤trV(IdV) =p. Thusi= IdV. The result follows.

Proof of Theorem 1.1. LetV be anO-freeOG-module with an irreducible characterχ∈IrrK(G).

Suppose thatχdoes not lift an irreducible Brauer character. Then the image E of the structural mapOG→EndO(V) is a proper subalgebra of EndO(V). Note thatE∼=OGe(χ), by the preceding

remarks. Thus, ifOGe(χ) is symmetric, then it follows from 4.3 that for any idempotenti∈ OG, the imagej ofiin E has the property that trV(j) is divisible by p. By the choice of V, we have

χ(i) = trV(j). Thusχ(i) is divisible byp, for any primitive idempotenti, and hence so isdχϕ for

anyϕ∈IBrk(G).

5. Proofs of Propositions 1.10 and 1.11

Ifχ:G→ O× is a linear character, then the correspondingO-pure idealI

χ =KGe(χ)∩ OG

hasO-rank 1 and is equal toO(PxGχ(x−1)x). There is a uniqueO-algebra automorphism of

OG sending x ∈ G to χ(x)x. This automorphism sends PxGχ(x−1)x to P

x∈Gx. Thus the

linear characters of OG are permuted transitively by the group of O-algebra automorphisms of

OG, and therefore, in order to address the question whether OG/Iχ is symmetric, it suffices to

consider the case whereχ= 1 is the trivial character, in which case the corresponding pure ideal is

I1=OG(PxGx) =O(PxGx). The annihilator of I1 in OGis the augmentation idealI(OG). Combining these observations with Proposition 2.2 and some block theory yields a proof of 1.10, which we restate in a slightly more precise way.

Proposition 5.1. Let Gbe a finite group. The following are equivalent. (i) TheO-algebraOG/O(PxGx)is symmetric.

(ii) There exists an elementz∈Z(OG) such thatI(OG) =OGz. (iii) The groupGisp-nilpotent and has a cyclic Sylow-p-subgroup.

Proof. The equivalence of (i) and (ii) is clear by 2.2. Suppose that (ii) holds. Letbbe the principal block idempotent ofOG. ThenI(OG)b=OGzbis a proper ideal inOGb; in particular,zbis not invertible inZ(OGb). Since Z(OGb) is local, it follows thatzb is in the radical ofZ(OGb), and hence the idealOGzbis contained inJ(OGb). SinceOGb/I(OG)b∼=OG/I(OG)∼=O, this implies thatOGbis a local algebra, and thatkG¯z¯b=J(kG¯b), where ¯z, ¯bare the canonical images ofz,bin

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from a result of Nakayama [6] thatkG¯b is uniserial, and hence P is cyclic. Thus (ii) implies (iii). Conversely, if (iii) holds, then the principal block algebraOGbofOGis isomorphic toOP, where

P is a Sylow-p-subgroup of G, and ify is a generator ofP, then I(OP) = OP(y−1). Note that

I(OG) contains all nonprincipal block algebras of OG. Set z = (y−1)b+Pb′ b′, where in the

sumb′ runs over all nonprincipal block idempotents. By the above, this is an element in Z(OG)

satisfyingI(OG) =OGz, completing the proof.

In a similar way, we obtain a proof of 1.11.

Proof of Proposition 1.11. Suppose that B/I is symmetric. By 2.2 the annihilator J of I is of the form J = Bz for some z ∈Z(B). Let X be an O-free B-module with character χ. By the assumptions, ¯X=k⊗OX is a simple module over ¯B=k⊗OB. By 4.1, and using thatkis large

enough, the structural map B →EndO(X) is surjective, and the kernel of this map is J. Since

J =Bz is a proper ideal inB andZ(B) is local, it follows thatz∈J(B). Thus the image ¯J ofJ

in ¯B is equal toJ( ¯B), and ¯B has a single isomorphism class of simple modules. SinceJ( ¯B) = ¯Bz¯, where ¯z is the image of z in Z( ¯B), it follows as before from [6] that ¯B is uniserial. Thus B has cyclic defect groups. A block with cyclic defect and a single isomorphism class of simple modules is nilpotent, which shows that (i) implies (ii). Conversely, if (ii) holds, thenBis Morita equivalent to OP, where P is a defect group of B (and P is cyclic by the assumptions). Since χ lifts an irreducible Brauer character it follows that under some Morita equivalence,χ corresponds to the trivial character ofOP, and hence (i) holds by 1.10 applied to P.

6. Examples

Example 6.1. LetP be a finite abelianp-group andEan abelianp′-subgroup of Aut(P). Denote

by G = P ⋊E the corresponding semidirect product, and let χ ∈ IrrK(G). Then OGe(χ) is

symmetric if and only if χ(1) = 1. Indeed, since P is abelian, it follows that χ(1) divides |E|, which is prime top, and hencedχ

ϕis prime topfor someϕ∈IBrk(G). The statement follows from

1.1.

The next example shows that there are finitep-groups having at least one irreducible character which does not have the symmetric quotient property.

Example 6.2. Let pbe an odd prime. G=Q≀R =H⋊R, where Q, R are cyclic of order p, and whereH is a direct product of pcopies of Qwhich are transitively permuted byR. Letsbe a generator ofQand ζ be a primitivep-th root of unity. For 1≤i≤pletψi :H → O× be the

linear character sending (sa1, sa2, .., sap)H to ζai; that is, the kernel of ψ

i contains all but the

i-th copy ofQin H, and theψi are permuted transitively by the action ofG. Setχ= IndGH(ψ1). Thenχ∈IrrK(G), and theO-algebra OGe(χ) is not symmetric.

In order to show this, observe first that ResGH(χ) =

Pp

i=1ψi because the ψi form aG-orbit in

IrrK(H). We have

e(χ) =

p

X

i=1

e(ψi) =

1

|H|

X

h∈H

(

p

X

i=1

ψi(h−1))h .

The coefficientsPpi=1ψi(h−1) are divisible by 1−ζbecause they are sums ofp(arbitrary) powers

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|H|

1−ζe(χ). Since z ∈ OH, this is the case if and only if OHz is pure in OH. We will show that

OHzis not pure inOH. Ifu= (sa1, sa2, .., sap)H, then

uz= |H| 1−ζ

p

X

i=1

ue(ψi) = |H|

1−ζ

p

X

i=1

ζaie(ψ

i).

Applied to the identity element andv= (1,1, ..1, s,1, ..1), withsin thei-th component, and taking the difference yieldsz−vz=|H|e(ψi)∈ OGz. For anyu∈H, the above formula yieldsuz−z∈

⊕pi=1O|H|e(ψ). Thus the set

{z,|H|e(ψi) (2≤i≤p)}

is anO-basis of OHz. Since pis odd, this basis has at least three elements. Suppose thatw =

P

h∈Hµhhis an element inOHz. Writew=αz+

Pp

i=2βi|H|e(ψi) withα,βi∈ O. Thus

µh=α

Pp

i=1 ψ(h−1) 1−ζ +

p

X

i=2

βiψi(h−1).

Note that ifPpi=2βi is divisible by 1−ζ then any sum of the form Ppi=2 βiψi(h−1) is divisible

by 1−ζ because any character value ψi(h−1) is a power of ζ. This shows that if 1−ζ divides

bothαand the sumPpi=2βi, then 1−ζ dividesµh for allh∈H. But sincep >2 we may choose

invertible elementsβisatisfyingPpi=2βi= 0. This shows that even if allµhare divisible by 1−ζ,

this does not imply thatαand allβi are divisible by 1−ζ, henceOHzis notO-pure inOH.

References

[1] M. Brou´e, L. Puig,A Frobenius theorem for blocks.Invent. Math.56(1980), 117–128.

[2] E. C. Dade,Blocks with cyclic defect groups, Ann. Math.84(1966), 20–48.

[3] K. Erdmann,Blocks of tame representation type and related algebras, Lecture Notes Math.1428, Springer

Verlag, Berlin Heidelberg (1990).

[4] W. Feit,The Representation Theory of Finite Groups.North Holland, Amsterdam (1982). [5] T. Nakayama,On Frobeniusean Algebras. I. Annals Math.40(1939), 611–633.

[6] T. Nakayama,Note on uni-serial and generalized uni-serial rings. Proc. Imp. Acad.Tokyo XVI (1940) 285–289.

[7] J. Thompson,Vertices and sources. J. Algebra6(1967), 1–6.

City University London, Department of Mathematical Science, Northampton Square, London, EC1V 0HB, United Kingdom

Department of Mathematics, Graduate School of Science, Chiba University, 1-33 Yayoi-cho, Inage-ku, Chiba, 263-8522, Japan

References

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