A note on Olsson’s Conjecture

A note on Olsson’s Conjecture

Lászlo Héthelyi, Burkhard Külshammer and Benjamin Sambale March 31, 2013

Dedicated to Geoffrey Robinson on the occasion of his 60th birthday

Abstract

For a p-blockB of a finite group Gwith defect groupD Olsson conjectured thatk0(B)≤ |D:D⁰|, where k0(B)is the number of characters inB of height0andD⁰ denotes the commutator subgroup ofD. Brauer deduced Olsson’s Conjecture in the case whereDis a dihedral2-group using the fact that certain algebraically conjugate subsections are also conjugate inG. We generalize Brauer’s argument for arbitrary primespand arbitrary defect groups. This extends two results by Robinson. Forp >3we show that Olsson’s Conjecture is satisfied for defect groups ofp-rank2and for minimal non-abelian defect groups.

1 Introduction

In order to state Olsson’s Conjecture we need some notations. LetRbe a complete discrete valuation ring with quotient fieldKof characteristic0. Moreover, let(π)be the maximal ideal ofRandF:=R/(π). We assume that Fis algebraically closed of characteristicp >0. We fix a finite groupG, and assume thatK contains all|G|-th roots of unity. LetBbe ap-block ofRG(or simply ofG) with defect groupD. We denote the set of irreducible ordinary characters byIrr(B)and its cardinality byk(B). These characters split inki(B)characters of height i ∈N0. Here the height of a character χ in B is the largest integer h(χ)≥ 0 such that p^h(χ)|G: D|p

χ(1), where|G:D|pdenotes the highestp-power dividing|G:D|. We setIrr0(B) :={χ∈Irr(B) :h(χ) = 0}. Finally, letIBr(B)be the set of irreducible Brauer characters andl(B) :=|IBr(B)|.

In the situation above, Olsson conjectured in 1975 that we always have k0(B) ≤ |D : D⁰|, where D⁰ denotes the derived subgroup of D (see [42]). This conjecture has been verified in some cases, but remains open in general. For example it was shown in [30] that Olsson’s Conjecture forBwould follow from the Alperin-McKay Conjecture for B (see also [56, 21]). Recall that the Alperin-McKay Conjecture predicts that k₀(B) = k₀(b), where b is the Brauer correspondent ofB inRN_G(D). In particular Olsson’s Conjecture holds forp-solvable, symmetric or alternating groups by [41, 44, 36]. IfDis abelian, Olsson’s Conjecture follows from Brauer’sk(B)- Conjecture k(B)≤ |D|. Moreover, Olsson’s Conjecture is satisfied ifD is metacyclic (see [55, 61]) or if p= 2 and D is minimal non-abelian (see [52]). Hendren verified Olsson’s Conjecture for some, but not all p-blocks with a non-abelian defect group of order p³ (see [24, 23]).

This paper is organized as follows. In the second section we introduce two results by Robinson and extend them in some sense using ideas of [53, 54]. In the third and fourth sections we generalize an argument of Brauer regarding a Galois action on subsections. In Section 5 we show that Olsson’s Conjecture is fulfilled for controlled blocks with certain defect groups. In the last section we use the classification of finite simple groups to prove Olsson’s Conjecture for defect groups ofp-rank 2and for minimal non-abelian defect groups ifp >3 (in both cases). In particular, our results here settle most of the cases of Olsson’s Conjecture left open in Hendren’s papers [24, 23].

2 Subsection

The notion ofB-subsections provides one tool in order to attack Olsson’s Conjecture. Here aB-subsection is a pair(u, bu), whereu∈D and bu is a Brauer correspondent of B in RCG(u). Robinson showed the following proposition (see [47]).

Proposition 2.1 (Robinson). If bu has defectd, thenk0(B)≤p^dp l(bu).

We mention another result by Robinson which will be improved later (see Theorem 3.4 in [46]). Recall that a B-subsection(u, bu)is calledmajor ifbu andB have the same defect.

Proposition 2.2 (Robinson). If (u, bu)is a major B-subsection such thatl(bu) = 1, then

∞

X

i=0

p²ⁱk_i(B)≤ |D|.

In order to make these propositions clearer, we introduce the fusion systemFofB. For this we use the notation of [43, 34], and we assume that the reader is familiar with these articles. Let b_D be a Brauer correspondent of B inRDCG(D). Then for every subgroupQ≤D there is a unique blockbQofRQCG(Q)such that(Q, bQ)≤ (D, bD). We denote the inertial group ofbQ inNG(Q)byNG(Q, bQ). ThenAut_F(Q)∼= NG(Q, bQ)/CG(Q).

The fusion of subsections is given by the following proposition (see [51]).

Proposition 2.3. LetRbe a set of representatives for theF-conjugacy classes of elements of D such thathui is fully F-normalized foru∈ R(Ralways exists). Then

(u, bu) :u∈ R

is a set of representatives for theG-conjugacy classes ofB-subsections, wherebu:=b_huihas defect groupCD(u).

Brauer proved Olsson’s Conjecture for2-blocks with dihedral defect groups using a Galois action on the gen- eralized decomposition numbers (see [10]). We provide the necessary definitions for that purpose. Let p^k be the order of u, and let ζ := ζ_pk be a primitive p^k-th root of unity. Then the generalized decomposition numbers d^u_χϕ for χ ∈ Irr(B) and ϕ ∈ IBr(b_u) lie in the ring of integers Z[ζ]. Hence, there exist integers a^ϕ_i := (a^ϕ_i(χ))_χ∈Irr(B)∈Z^k(B)such that

d^u_χϕ =

ϕ(p^k)−1

X

i=0

a^ϕ_i(χ)ζⁱ (2.1)

(see Satz I.10.2 in [39]). Hereϕ(p^k)denotes Euler’s totient function.

Let G be the Galois group of the cyclotomic field Q(ζ)over Q. Then G ∼= Aut(hui)∼= (Z/p^kZ)^× and we will often identify these groups. We will also interpret the elements ofG as integers in{1, . . . , p^k}by a slight abuse of notation. Then(u^γ, bu)forγ∈ G is also a (algebraically conjugate) subsection and

γ(d^u_χϕ) =d^u_χϕ^γ =

ϕ(p^k)−1

X

i=0

a^ϕ_i(χ)ζ^iγ.

We use the opportunity to present a slight generalization of Lemma 1 in [54]. Here we call two matrices A, B∈Z^l×l equivalent if there exists a matrixS∈GL(l,Z)withA=S^TBS, whereS^T denotes the transpose ofS. This is just Brauer’s notion of basic sets.

Theorem 2.4. Let B be a p-block of G, and let (u, b_u) be a B-subsection. Let C_u = (c_ij) be the Cartan matrix of b_u up to equivalence. Then for every positive definite, integral quadratic form q(x₁, . . . , x_l(bu)) = P

1≤i≤j≤l(b_u)q_ijx_ix_j we have

k0(B)≤ X

1≤i≤j≤l(bu)

qijcij.

In particular

k0(B)≤

l(b_u)

X

i=1

cii−

l(b_u)−1

X

i=1

ci,i+1.

If (u, bu)is major, we can replace k0(B)by k(B)in these formulas.

Proof. ¹ First of all assume, that Cu is the Cartan matrix of bu (not only up to equivalence!). Let ϕ1, . . . , ϕl

(l:=l(bu)) be the irreducible Brauer characters ofbu. Then we have rowsdχ:= (d^u_χϕ1, . . . , d^u_χϕl)forχ∈Irr(B).

LetQ= (qeij)^l_i,j=1 with

qe_ij :=

(q_ij ifi=j, qij/2 ifi6=j . Then we have

X

1≤i≤j≤l

qijcij = X

1≤i,j≤l

eqijcij = X

1≤i,j≤l

X

χ∈Irr(B)

eqijd^u_χid^u_χj

= X

χ∈Irr(B)

dχQdχ

T≥ X

χ∈Irr0(B)

dχQdχ T,

sinceQis positive definite. Thus, it suffices to show X

χ∈Irr₀(B)

d_χQd_χ^T≥k₀(B).

For this, letpⁿ be the order ofu, and letf :=pⁿ⁻¹(p−1)−1. We fix a character χ∈Irr₀(B)and setd:=d_χ. Then there are integral rowsam∈Z^l(m= 0, . . . , f)such thatd=Pf

m=0amζ^m. By Corollary 2 in [11] at least one of the rowsamdoes not vanish.

It is known that for everyγ∈ G there is a characterχ⁰ ∈Irr(B)such thatγ(d) =dχ⁰. Thus, it suffices to show X

γ∈G

γ(d)Qγ(d)^T=X

γ∈G

γ(dQd^T)≥ |G|=f+ 1.

We have

X

γ∈G

γ(dQd^T) =X

γ∈G

γ

f

X

i=0

a_iQa^T_i +

f

X

j=1 f−j

X

m=0

a_mQa^T_m+j(ζ^j+ζ^j)

!

= (f+ 1)

f

X

i=0

aiQa^T_i + 2

f

X

j=1 f−j

X

m=0

amQa^T_m+jX

γ∈G

γ(ζ^j).

Thep^m-th cyclotomic polynomialΦp^m has the form

Φp^m =X^{pm−1(p−1)}+X^{pm−1(p−2)}+. . .+X^pm−1+ 1.

This gives

X

γ∈G

γ(ζ^j) =

(−pⁿ⁻¹ ifpⁿ⁻¹|j

0 otherwise

forj∈ {1, . . . , f}. It follows that X

γ∈G

γ(dQd^T) = (f + 1)

f

X

i=0

aiQa^T_i −2pⁿ⁻¹

p−2

X

j=1

f−jpⁿ⁻¹

X

m=0

amQa^T_m+pn−1j

=pⁿ⁻¹ (p−1)

f

X

i=0

aiQa^T_i −2

p−2

X

j=1

f−jpⁿ⁻¹

X

m=0

amQa^T_m+pn−1j

!

. (2.2)

1Proof corrected on March 31, 2013

Forp= 2the claim follows immediately, since thenf+ 1 = 2ⁿ⁻¹. Thus, suppose p >2. Then we have 0,1, . . . , f−jpⁿ⁻¹ ∪˙

(p−1−j)pⁿ⁻¹,(p−1−j)pⁿ⁻¹+ 1, . . . , f ={0,1, . . . , f}

for allj∈ {1, . . . , p−2}. This shows that every row amoccurs exactlyp−2times in the second sum of (2.2).

Hence,

X

γ∈G

γ(dQd^T) =pⁿ⁻¹

f

X

i=0

aiQa^T_i +

p−2

X

j=1

f−jpⁿ⁻¹

X

m=0

(am−a_m+jpⁿ⁻¹)Q(am−a_m+jpⁿ⁻¹)^T

! .

Now assume thatam does not vanish for somem∈ {0, . . . , f}. Then we haveamQa^T_m≥1, sinceQis positive definite. Again,am occurs exactly p−2 times in the second sum. Let am−am⁰ (resp.am⁰−am) be such an occurrence. Then we have

am⁰Qa^T_m0+ (am−am⁰)Q(am−am⁰)^T≥1.

Now the first inequality of the theorem follows easily.

The result does not depend on the basic set for Cu, since changing the basic set is essentially the same as taking another quadratic form q(see [32]). For the second claim we take the quadratic form corresponding to the Dynkin diagram of type Al for q. If (u, bu) is major, then all rows dχ for χ ∈ Irr(B) do not vanish (see Theorem V.9.5 in [18]). Hence, we can replacek₀(B)byk(B).

We present an application.

Proposition 2.5. Let (u, b_u)be a B-subsection such thatb_u has defect groupQ. Then the following hold:

(i) IfQ/huiis cyclic, we have

k0(B)≤

|Q/hui| −1

l(b_u) +l(bu)

|hui| ≤ |Q|.

(ii) If|Q/hui| ≤9, we havek0(B)≤ |Q|.

(iii) Suppose p= 2. If Q/huiis metacyclic or minimal non-abelian or isomorphic to C4oC2, we havek0(B)≤

|Q|.

Proof.

(i) It is well-known thatb_u dominates a block b_u ofC_G(u)/hui with cyclic defect groupQ/hui andl(b_u) = l(b_u). By [14, 48] the Cartan matrix b_u has the form |hui|(m+δ_ij)_{1≤i,j≤l(bu)} up to equivalence, where m:= (|Q/hui| −1)/l(b_u)is the multiplicity ofb_u. Now the claim follows from Theorem 2.4.

(ii) See Theorem 1 in [54].

(iii) IfQ/hui is metacyclic, the claim follows as in Theorem 2 of [53]. If Q/hui is minimal non-abelian, the claim can easily deduced from the results in [52, 16]. Finally, forD/hui ∼=C₄oC₂the result follows from [29].

Sinceu∈Z(Q)in Proposition 2.5(i), the condition implies thatQis abelian of rank at most2. It is known that the numberl(bu)in Proposition 2.5(i) equals the inertial index ofbu(see [14]).

3 The case p = 2

Let p = 2, and let (u, bu) be a B-subsection for a block B of G. Then by Proposition 2.3 we may assume that hui is fullyF-normalized, where F is the fusion system of B. By Proposition 2.5 in [34] hui is also fully F-centralized and

Aut_F(hui) = AutD(hui) = ND(hui) CG(u)/CG(u)∼= ND(hui)/CD(u).

Hence, Theorem 2.4(ii) in [33] implies thatCD(u)is a defect group of bu.

Theorem 3.1. LetB be a2-block of a finite groupGwith defect group D and fusion systemF, and let(u, bu) be a B-subsection such that hui is fully F-normalized and bu has Cartan matrix Cu = (cij). Let IBr(bu) = {ϕ1, . . . , ϕl(b_u)} such that ϕ1, . . . , ϕm are stable under ND(hui) and ϕm+1, . . . , ϕl(b_u) are not. Then m ≥ 1.

Suppose further thatuis conjugate to u⁻⁵ⁿ for somen∈Z inD. Then k0(B)≤ |ND(hui)/CD(u)|

ϕ |hui|

X

1≤i≤j≤m

qijcij (3.1)

for every positive definite, integral quadratic formq(x₁, . . . , x_m) =P

1≤i≤j≤mq_ijx_ix_j. In particular ifl(b_u) = 1, we get

k0(B)≤|ND(hui)|

ϕ |hui| . (3.2)

If l(bu) = 2, we may replace Cu by an equivalent matrix such that |CD(u)|c11/detCu is even and as small as possible. In this case (with the hypothesis above) we have

k₀(B)≤|ND(hui)/CD(u)|c11

ϕ |hui| ≤ |ND(hui)|

ϕ |hui| . (3.3)

Proof. Letχ∈Irr0(B)and|hui|= 2^k for some k≥0. We writed^u_χ := (d^u_χϕ1, . . . , d^u_χϕl), wherel:=l(bu). Then we have|CD(u)|m^(u,bχχ ^u)=d^u_χ|CD(u)|C_u⁻¹d^u_χ for the contributionm^(u,bχχ ^u)(see Eq. (5.2) in [9]). By Corollary 2 in [11] it follows that

|C_D(u)|m^(u,b_χχ ^u)=|C_D(u)| χ^(u,bu), χ

G 6≡0 (mod (π)).

Sinceζ≡1 (mod (π)), we see that

d^u_χϕi ≡γ(d^u_χϕi)≡

ϕ(2^k)−1

X

j=0

aⁱ_j(χ) (mod (π))

forγ∈ G. In particulard^u_χϕ

i ≡d^u_χϕ

i (mod (π)). We write|CD(u)|C_u⁻¹= (ec_ij). Then it follows that 06≡ |CD(u)|m^(u,b_χχ^u)≡ X

1≤i,j≤l

ec_ijd^u_χϕ

id^u_χϕ

j ≡ X

1≤i≤l

ec_ii(d^u_χϕ

i)²

≡ X

1≤i≤l

ecii ϕ(2^k)−1

X

j=0

aⁱ_j(χ)²≡ X

1≤i≤l

ecii ϕ(2^k)−1

X

j=0

aⁱ_j(χ) (mod (π))

Now every g ∈ND(hui) induces a permutation onIBr(bu). LetPg be the corresponding permutation matrix.

Thengalso acts on the rowsd^u_i := (d^u_χϕi:χ∈Irr(B))fori= 1, . . . , l, and it follows thatCuPg=PgCu. Hence, we also have C_u⁻¹Pg =PgC_u⁻¹ for all g ∈ND(hui). If {ϕm₁, . . . , ϕm₂} (m < m1 < m2 ≤l) is an orbit under ND(hui), it follows that d^u_χϕm

1 ≡ . . . ≡d^u_χϕm

2 (mod (π))and ecm₁m₁ =. . . =ecm₂m₂. Since the length of this orbit is even, we get

X

1≤i≤m

ec_ii

ϕ(2^k)−1

X

j=0

aⁱ_j(χ)6≡0 (mod 2).

In particularm≥1. In case |hui| ≤2this simplifies to X

1≤i≤m

eciiaⁱ₀(χ)6≡0 (mod 2).

We show that this holds in general. Thus, letk≥2 andi∈ {1, . . . , m}. Since(u, bu)is conjugate to(u⁻⁵ⁿ, bu) andϕi is stable, we have

ϕ(2^k)−1

X

j=0

aⁱ_j(χ)ζ^j =d^u_χϕi=d^u−5

n

χϕi =

2^k−1−1

X

j=0

aⁱ_j(χ)ζ^−5nj.

Moreover, for every j ∈ {0, . . . , ϕ(2^k)−1} there is some j₁ ∈ {0, . . . , ϕ(2^k)−1} such that ζ^−5nj =±ζ^j1. In order to compare coefficients observe that

ζ^j=ζ^−5nj =⇒j≡ −5ⁿj (mod 2^k) =⇒1≡ −5ⁿ (mod 2^k/gcd(2^k, j)) =⇒j= 0.

Hence, the set{±ζ^j : j = 1, . . . , ϕ(2^k)−1} splits under the action of h−5ⁿ+ 2^kZiinto orbits of even length.

This showsPϕ(2^k)−1

j=0 aⁱ_j(χ)≡aⁱ₀(χ) (mod 2). Hence, X

1≤i≤m

eciiaⁱ₀(χ)6≡0 (mod 2) (3.4)

for everyχ∈Irr0(B). In particular, there is ani∈ {1, . . . , m} such thataⁱ₀(χ)6= 0. This gives k₀(B)≤ X

1≤i≤j≤m

q_ij(aⁱ₀, a^j₀)

(see proof of Theorem 2.4).

Now letk again be arbitrary. Observe thataⁱ₀ =ϕ(2^k)⁻¹P

γ∈Gγ(d^u_i)for i∈ {1, . . . , m}. By the orthogonality relations for generalized decomposition numbers we have(d^u_i^γ, d^u_j^δ) =cij forγ, δ∈ G ifu^γ andu^δ are conjugate underN_D(hui)(see Theorem 5.4.11 in [37] for example). Otherwise we have (d^u_i^γ, d^u_j^δ) = 0. This implies

(aⁱ₀, a^j₀) = 1 ϕ(2^k)²

X

γ,δ∈G

(d^u_i^γ, d^u_j^δ) =|ND(hui)/CD(u)|

ϕ(2^k) cij,

and (3.1) follows. In casel= 1we have C= (|CD(u)|), and (3.2) is also clear.

Now assumel= 2. Here we can use (3.4) in a stronger sense. We havem= 2. Since|C_D(u)|occurs as elementary divisor ofCu exactly once, we see that the rank of ^|C_det^D(u)|_C

u Cu (mod 2)is1. Hence, ^|C_det^D(u)|_C

u Cu (mod 2)has the form

1 0 0 0

(mod 2),

0 0 0 1

(mod 2), or

1 1 1 1

(mod 2).

Now it is easy to see that we may replaceCu by an equivalent matrix (still denoted by Cu = (cij)) such that

|CD(u)|c11/detCu is even and as small as possible. Then we also have to replace the rowsd^u₁ andd^u₂ by linear combinations of each other. This gives rowsdb^u_i and baⁱ_j fori= 1,2 andj = 0, . . . , ϕ(2^k)−1. Observe that the contributions do not depend on the representative of the equivalence class ofCu. Moreover, ec11 is odd andec22

is even. Hence, (3.4) takes the form

ba¹₀(χ)6≡0 (mod 2)

for allχ∈Irr0(B). Since bothϕ1 andϕ2are stable underND(hui), we haveγ(db^u₁) =db^u₁ for allγ∈AutF(hui).

Hence,

k₀(B)≤(ba¹₀,ba¹₀) = |ND(hui)/CD(u)|c11

ϕ(2^k)

as above. It remains to show thatc11≤ |CD(u)|. The reduction theory of quadratic forms gives an equivalent matrixC_u⁰ = (c⁰_ij) such that0 ≤2c⁰₁₂ ≤min{c⁰₁₁, c⁰₂₂} (see [12] for example). In case c⁰₁₂ = 0 we may assume

c11≤c⁰₁₁=|CD(u)|, since|CD(u)|is the largest elementary divisor ofC_u⁰. Hence, letc⁰₁₂>0. Since the entries ofCu and thus also ofC_u⁰ are divisible byα:= detCu/|CD(u)|, we even havec⁰₁₂≥α. It follows that

3α²≤3(c⁰₁₂)²≤c⁰₁₁c⁰₂₂−(c⁰₁₂)²= detC_u⁰ ≤|CD(u)|² 2 andα≤ |CD(u)|/4. It was shown in the proof of Theorem 1 of [53] that

max{c⁰₁₁, c⁰₂₂} ≤c⁰₁₁+c⁰₂₂−c⁰₁₂≤c⁰₁₁+c⁰₂₂−α≤α|CD(u)|/α+ 3

2 = |CD(u)|+ 3α

2 ≤ |CD(u)|.

Ifα⁻¹c⁰₁₁ orα⁻¹c⁰₂₂ is even, the result follows from the minimality ofc11. Otherwise we replace C_u⁰ by 1 −1

0 1

C_u⁰

1 0

−1 1

=

c⁰₁₁+c⁰₂₂−2c⁰₁₂ c⁰₁₂−c⁰₂₂ c⁰₁₂−c⁰₂₂ c⁰₂₂

.

Thenc₁₁≤c⁰₁₁+c⁰₂₂−2c⁰₁₂≤ |CD(u)|. This finishes the proof.

In the situation of Theorem 3.1 we haveu∈Z(C_G(u)). Hence, all Cartan invariantsc_ij are divisible by |hui|.

This shows that the right hand side of (3.1) is always an integer. It is also known that k₀(B)is divisible by4 unless|D| ≤2.

Observe that the subsection (u, bu) in Theorem 3.1 cannot be major unless |hui| ≤ 2, since then uwould be contained inZ(D).

If m=l(bu)in Theorem 3.1, it suffices to know the Cartan matrix Cu only up to equivalence. For, replacing Cu by an equivalent matrix is essentially the same as taking another quadratic formq. However, form < l(bu) we really have to use the “exact” Cartan matrixCu which is unknown in most cases. For p >2 there are not always stable characters inIBr(bu)(see Proposition (2E)(ii) and the example following it in [28]).

We give an example. LetDbe a (non-abelian) 2-group of maximal class. Then there is an elementx∈D such that |D:hxi|= 2 andxis conjugate tox⁻⁵ⁿ for somen∈ {0,|hxi|/8}under D. SincehxiED, the subgroup hxi is fullyF-normalized, andb_x has cyclic defect groupC_D(x) = hxi. Thus, Dade’s Theorem on blocks with cyclic defect groups implies l(b_x) = 1. Hence, Theorem 3.1 shows Olsson’s Conjecture k₀(B)≤ 4 =|D : D⁰|.

This was already proved in [10, 42].

On the other hand, we cannot improve Theorem 3.1 or Theorem 2.4 if u is not conjugate to u⁻⁵ⁿ in D.

Indeed, if D a modular 2-group and x ∈ D such that |D : hxi| = 2, then B is nilpotent (see [17]) and k0(B) =|D:D⁰|=|D|/2 =|CD(x)|.

We give a more general example.

Proposition 3.2. Let D be a2-group andx∈D such that|D:hxi| ≤4, and suppose that one of the following holds:

(i) xis conjugate tox⁻⁵ⁿ in D for somen∈Z, (ii) hxiED.

Then Olsson’s Conjecture holds for all blocks with defect groupD.

Proof. Let B be a block with defect group D and fusion system F. By [55] we may assume that D is non- metacyclic.

(i) By hypothesis,xis conjugate tox⁻⁵ⁿinF. This condition is preserved if we replacexby anF-conjugate.

Hence, we may assume that hxi is fully F-normalized. Then x is conjugate tox⁻⁵ⁿ in D. In particular

|CD(x)/hxi| ≤ 2. Hence, bx dominates a block of CG(x)/hxi with cyclic defect group CD(x)/hxi. This showsl(bx) = 1. Now we can apply Theorem 3.1 which givesk0(B)≤8. In case|D:D⁰|= 4a theorem of Taussky (see for example Proposition 1.6 in [8]) implies thatD has maximal class which was excluded.

(ii) We consider the order ofCD(x).

Case 1:CD(x) =hxi.

SinceD is non-metacyclic,D/hxiis non-cyclic. Hence, we are in case (i).

Case 2:x∈Z(D).

IfDis abelian, the result follows from Theorem 2 in [53]. Thus, we may assume thatDis non-abelian. Then every conjugacy class of D has length at most2. By a result of Knoche (see for example Aufgabe III.24b in [25]) this is equivalent to |D⁰| = 2. Let y∈D\Z(D). ThenCD(y)is non-cyclic. After replacing y by xyif necessary, we have|hxi|=|hyi|. By Proposition 2.5 it suffices to show thathyiis fullyF-normalized.

By Alperin’s Fusion Theorem (see [34]) every F-isomorphism on hyi is a composition of automorphisms ofF-essential subgroups containingyor ofDitself. Assume thatE < DisF-essential such thathyi ≤E.

Since E is metacyclic and Aut(E) is not a 2-group, Lemma 1 in [35] implies E ∼=Q₈ or E ∼=C₂×C₂; in particular |D| ≤16. Moreover, Proposition 1.8 and Proposition 10.17 in [8] imply thatD has maximal class, because every F-essential subgroup is self-centralizing. This contradiction shows that there are no F-essential subgroups containingy. Then of coursehyiis fullyF-normalized.

Case 3:|CD(x)/hxi|= 2.

Lety ∈CD(x)\ hxibe of order2. Ifz∈D\CD(x), we may assume thathx, ziis a modular 2-group by (i). In particular we have |hzi|= 2after replacing z byzx^m for some m∈Zif necessary. Let |hxi|= 2^r for somer∈N. SincehxiED, we havezyz⁻¹∈ {y, yx^2r−1}. In casezyz⁻¹=yx^2r−1 it is easy to see that

|D : hxyi|= 4and xy ∈Z(D). Then we are done by Case 2. Thus, we may assume that zyz⁻¹ =y and y∈Z(D). ThenD is given as follows:

D=hx, zi × hyi ∼=M₂r+1×C2,

whereM₂r+1denotes the modular2-group of order2^r+1andC2denotes a cyclic group of order2. Now we have |D⁰| = 2and the claim follows from Proposition 2.5 applied to the subsection(x, bx). Here observe that hxiis fullyF-normalized, sincehxiED.

We like to point out that every subgroup ofD is fullyF-normalized wheneverFis controlled byAut_F(D). The groups in Proposition 3.2 were given explicitly by generators and relations in [40].

By the propositions in [54] it is easy to see that Olsson’s Conjecture holds for2-blocks with defect at most4.

For defect groupsD of order32one can show with GAP [19] that there is always an element x∈D such that

|CD(x)|=|D :D⁰|. If in addition D is abelian, Olsson’s Conjecture follows from Corollary 2 in [54] for every block with defect group D. If D is non-abelian, then |CD(x)/hxi| ≤8. Thus, by Proposition 2.5(ii) Olsson’s Conjecture also holds for controlled2-blocks of defect5.

4 The case p > 2

Now we turn to the case where B is ap-block ofGfor an odd primep. We fix some notations for this section.

As before(u, bu)is aB-subsection such that|hui|=p^k. Moreover,ζ∈Cis a primitivep^k-th root of unity. Since the situation is more complicated for odd primes, we assume further thatl(bu) = 1. We writeIBr(bu) ={ϕu}.

Then the generalized decomposition numbers d^u_χϕu for χ∈Irr(B)form a column d(u). Let dbe the defect of bu. Sinceu∈Z(CG(u)),uis contained in every defect group ofbu. In particular,k≤d. As in the casep= 2we can write

d(u) =

ϕ(p^k)−1

X

i=0

a^u_iζⁱ

witha^u_i ∈Z^k(B)(change of notation!). We define the following matrix A:= a^u_i(χ) :i= 0, . . . , ϕ(p^k)−1, χ∈Irr(B)

∈Z^ϕ(p

k)×k(B).

The next lemma uses the same idea as in casep= 2.

Lemma 4.1. Let (u, b_u)be a B-subsection with |hui|=p^k andl(b_u) = 1.

(i) For χ∈Irr0(B)we have

ϕ(p^k)−1

X

i=0

a^u_i(χ)6≡0 (mod p).

(ii) If(u, b_u)is major and χ∈Irr(B), thenp^h(χ)|a^u_i(χ)fori= 0, . . . , ϕ(p^k)−1 and

ϕ(p^k)−1

X

i=0

a^u_i(χ)6≡0 (mod p^h(χ)+1).

Proof.

(i) Since l(bu) = 1, we have p^dm^(u,bχχ ^u) = d^u_χϕud^u_χϕu for the contribution m^(u,bχχ^u) (see Eq. (5.2) in [9]). By Corollary 2 in [11] it follows that

p^dm^(u,b_χχ ^u)=p^d χ^(u,bu), χ

G 6≡0 (mod (π)) andd^u_χϕ

u 6≡0 (mod (π)). Sinceζ≡1 (mod (π)), the claim follows from (2.1).

(ii) Letψ∈Irr₀(B). Then (5G) in [9] implies

h(χ) =ν |D|m^(u,b_χψ ^u)

=ν(d^u_χϕu) +ν(d^u_ψϕ

u),

where ν is thep-adic valuation. Thus,h(χ) =ν(d^u_χϕu)follows from (i). Now the claim is easy to see.

The proof of the main theorem of this section is an application of the next proposition.

Proposition 4.2. For every positive definite, integral quadratic formq(x1, . . . , x_ϕ(pk)) =P

1≤i≤j≤ϕ(p^k)qijxixj

we have

k₀(B)≤ X

1≤i≤j≤ϕ(p^k)

q_ij(a^u_i−1, a^u_j−1). (4.1)

If (in addition) (u, b_u)is major, we can replace k₀(B)by P∞

i=0p²ⁱk_i(B) in (4.1).

Proof. By Lemma 4.1(i) every columna^u(χ)ofA corresponding to a characterχ of height0 does not vanish.

Hence, we have k0(B)≤ X

χ∈Irr(B)

q(a^u(χ)) = X

χ∈Irr(B)

X

1≤i≤j≤ϕ(p^k)

qija^u_i−1(χ)a^u_j−1(χ) = X

1≤i≤j≤ϕ(p^k)

qij(a^u_i−1, a^u_j−1).

If (u, bu)is major and χ ∈Irr(B), then p^−h(χ)a^u(χ) is a non-vanishing integral column by Lemma 4.1(ii). In this case we have

∞

X

i=0

p²ⁱki(B)≤ X

χ∈Irr(B)

p^2h(χ)q(p^−h(χ)a^u(χ)) = X

1≤i≤j≤ϕ(p^k)

qij(a^u_i−1, a^u_j−1).

The second claim follows.

Notice that we have used only a weak version of Lemma 4.1 in the proof above.

In order to find a suitable quadratic form it is often very useful to replaceA byU A for some integral matrix U ∈GL(ϕ(p^k),Q)(observe that the argument in the proof of Proposition 4.2 remains correct).

However, we need a more explicit expression of the scalar products (a^u_i, a^u_j). For this reason we introduce an auxiliary lemma about inverses of Vandermonde matrices. LetG={σ1, . . . , σ_ϕ(pk)}. For an integer i∈Zthere isi⁰ ∈ {1, . . . , p^k−1}such that −i≡i⁰ (mod p^k−1). We will use this notation for the rest of the paper.

Lemma 4.3. The inverse of the Vandermonde matrixV := σi(ζ)^j−1ϕ(pk)

i,j=1 is given by V⁻¹=p^−k σj(ti−1)ϕ(p^k)

i,j=1, whereti=ζ⁻ⁱ−ζⁱ⁰.

Proof. Fori, j∈ {0, . . . , ϕ(p^k)−1}we have

ϕ(p^k)

X

l=1

σ_l(t_i)σ_l(ζ)^j =

ϕ(p^k)

X

l=1

σ_l(ζ^j−i−ζ^j+i0).

Assume first that i = j. Then ζ^j−i = 1 and j+i⁰ = i+i⁰ is divisible by p^k−1 but not by p^k. Hence, ζ^j+i0 is a primitive p-th root of unity. Since the second coefficient of the p-th cyclotomic polynomial Φp(X) = X^p−1+X^p−2+. . .+X+ 1is1, we getP^ϕ(pk)

l=1 σl(ζ^j+i0) =−p^k−1. This shows that

ϕ(p^k)

X

l=1

σl(1−ζⁱ⁺ⁱ⁰) =ϕ(p^k) +p^k−1=p^k.

Now leti6=j. Thenj−i6≡0 (modp^k)andj+i⁰6≡0 (modp^k). Moreover,j−i≡j+i⁰ (mod p^k−1), sincei+i⁰≡ 0 (modp^k−1). Assume first thatj−i6≡0 (modp^k−1). Thenζ^j−iis a primitivep^s-th root of unity for somes≥2.

Since the second coefficient of thep^s-th cyclotomic polynomialΦp^s(X) =X^{(p−1)ps−1}+X^{(p−2)ps−1}+. . .+X^ps−1+1 (see Lemma I.10.1 in [39]) is 0, we have Pϕ(p^k)

l=1 σ_l(ζ^j−i) = 0. The same holds forj+i⁰. Finally letj−i ≡0 (modp^k−1). Then we have (as in the first part of the proof)

ϕ(p^k)

X

l=1

σ_l(ζ^j−i−ζ^j+i0) =−p^k−1+p^k−1= 0.

This proves the claim.

Now letA:= AutF(hui)≤ G. The next proposition shows that the scalar products(a^u_i, a^u_j)only depend onp, k−dandA.

Proposition 4.4. We have

p^k−d(a^u_i, a^u_j) =|{τ∈ A:p^k |i−jτ}| − |{τ∈ A:p^k |i+j⁰τ}|+

|{τ∈ A:p^k |i⁰−j⁰τ}| − |{τ ∈ A:p^k|i⁰+jτ}|. (4.2) Proof. LetW := d^σχϕ^i(u)u :i= 1, . . . , ϕ(p^k), χ∈Irr(B)

be a part of the generalized decomposition matrix. IfV is the Vandermonde matrix in Lemma 4.3, we haveV A=W andA=V⁻¹W. This shows

(a^u_i−1, a^u_j−1)ϕ(p^k)

i,j=1 =AA^T=V⁻¹W W^TV^−T=V⁻¹W W^TV^−T. Now letS:= (sij)^ϕ(p_i,j=1^k), where

sij :=

(1 ifσiσ_j⁻¹∈ A, 0 otherwise.

Then the orthogonality relations (see proof of Theorem 3.1) implyW W^T=p^dS. It follows that p^2k−d(a^u_i, a^u_j) =

ϕ(p^k)

X

l=1

σ_l(t_i)

ϕ(p^k)

X

m=1

s_lmσ_m(t_j) =

ϕ(p^k)

X

l=1

X

τ∈A

σ_l(t_iτ(t_j))

=X

τ∈A ϕ(p^k)

X

l=1

σ_l((ζ⁻ⁱ−ζⁱ⁰)τ(ζ^j−ζ^−j0))

=X

τ∈A ϕ(p^k)

X

l=1

σl(ζ^jτ−i+ζ^i0−j0τ−ζ^−i−j0τ−ζ^i0+jτ). (4.3)

As in the proof of Lemma 4.3 we have

ϕ(p^k)

X

l=1

σl(ζ^jτ−i) =







ϕ(p^k) ifp^k|jτ−i, 0 ifp^k−1-jτ−i,

−p^k−1 otherwise.

This can be combined to X

τ∈A ϕ(p^k)

X

l=1

σl(ζ^jτ−i) =p^k|{τ∈ A:p^k|jτ−i}| −p^k−1|{τ∈ A:p^k−1|jτ−i}|.

We get similar expressions for the other numbers i⁰ −j⁰τ, −i−j⁰τ and i⁰+jτ. Since i+i⁰ ≡ j+j⁰ ≡ 0 (modp^k−1), we havejτ−i≡i⁰−j⁰τ≡ −i−j⁰τ≡i⁰+jτ (modp^k−1). Thus, the terms of the formp^k−1|{. . .}|

in (4.3) cancel out each other. This proves the proposition.

Since the groupAut(hui)is cyclic, Ais uniquely determined by its order. We introduce a notation.

Definition 4.5. LetAbe as in Proposition 4.4. Then we defineΓ(d, k,|A|)as the minimum of the expressions X

1≤i≤j≤ϕ(p^k)

qij(a^u_i−1, a^u_j−1),

whereqranges over all positive definite, integral quadratic forms. By Proposition 4.2 we havek0(B)≤Γ(d, k,|A|), andP∞

i=0p²ⁱki(B)≤Γ(d, k,|A|)if (u, bu)is major.

We will calculateΓ(d, k,|A|)by induction onk. First we collect some easy facts.

Lemma 4.6. Let H ≤ (Z/p^kZ)^× where we regard H as a subset of {1, . . . , p^k}. Then |{σ ∈ H : σ ≡ 1 (modp^j)}|= gcd(|H|, p^k−j) for1≤j≤k.

Proof. The canonical epimorphism(Z/p^kZ)^× →(Z/p^jZ)^× has kernelK of order p^k−j. Hence,|{σ∈ H:σ≡1 (modp^j)}|=|H ∩ K|= gcd(|H|, p^k−j), since the p-subgroups of the cyclic group(Z/p^kZ)^× are totally ordered by inclusion.

Lemma 4.7. Let |A|p be the order of a Sylow p-subgroup of A. Then we have (a^u₀, a^u₀) = |A|+|A|p

p^d−k

and

p^k−d

gcd(|A|p, j)(a^u_i, a^u_j)∈ {0,±1,±2}

for i+j > 0. If a^u_i 6= 0 for some i ≥1, then(a^u_i, a^u_i) = 2p^d−kgcd(|A|p, i). Moreover, (a^u_i, a^u_j) = 0 whenever gcd(i, p^k−1)6= gcd(j, p^k−1).

Proof. Fori =j = 0 we have i+j⁰τ =p^k−1τ 6≡0 (modp^k)and i⁰+jτ =p^k−1 6≡0 (modp^k) for allτ ∈ A.

Moreover, by Lemma 4.6 there are precisely|A|pelementsτ∈ Asuch thati⁰−j⁰τ =p^k−1(1−τ)≡0 (modp^k).

The first claim follows from Proposition 4.4.

Now leti+j >0 andτ ∈ Asuch thati≡jτ (modp^k). Then we havej6= 0. Assume that alsoτ1∈ Asatisfies i≡jτ1 (modp^k). Thenj(τ−τ1)≡0 (modp^k)andτ⁻¹τ1≡1 (modp^k/gcd(p^k, j)). Thus, Lemma 4.6 implies

|{τ∈ A:p^k |i−jτ}| ∈ {0,gcd(|A|p, j)}.

The same argument also works for the other summands in (4.2), sincegcd(|A|p, j) = gcd(|A|p, j⁰). This gives p^k−d(a^u_i, a^u_j)∈ {0,±gcd(|A|p, j),±2 gcd(|A|p, j)}

wheneveri+j >0.

Supposei≥1 andi≡iτ (modp^k) for someτ ∈ A. Thenτ ≡1 (modp)and thus i≡iτ−(i+i⁰)(τ−1)≡

−i⁰τ+i+i⁰ (modp^k). Hencei⁰ ≡i⁰τ (modp^k). This shows|{τ ∈ A:p^k |i−iτ}|=|{τ ∈ A:p^k |i⁰−i⁰τ}|.

Moreover, we have|{τ ∈ A: p^k |i+i⁰τ}|=|{τ ∈ A :p^k |iτ⁻¹+i⁰}|=|{τ ∈ A :p^k | i⁰+iτ}|. This shows a^u_i = 0or(a^u_i, a^u_i) = 2p^dgcd(|A|p, i)/p^k.

Finally suppose thatgcd(i, p^k−1)6= gcd(j, p^k−1). Then i6≡jτ (modp^k−1) and thusp^k -i−jτ for all τ ∈ A.

The same holds for the other terms in (4.2), sincei+i⁰≡j+j⁰ ≡0 (modp^k−1). The last claim follows.

Proposition 4.8. We have

Γ(d,1,|A|) = |A|+ (p−1)/|A|

p^d−1.

Proof. Since|A| |p−1, we have|A|p= 1. Hence,(a^u₀, a^u₀) = (|A|+ 1)p^d−1 and (a^u_i, a^u_j)∈ {0,±p^d−1,±2p^d−1} fori+j >0 by Lemma 4.7. First we determine the indicesisuch thata^u_i = 0. For this we use Proposition 4.4.

Observe that we always havei⁰ = 1. In particular for alli, j we havep|i⁰−j⁰τforτ= 1. It follows thata^u_i = 0 if and only if−i≡τ (mod p)for someτ ∈ A. We write this condition in the form−i∈ A. This gives exactly

|A| −1 vanishing rows and columns. Thus, all the scalar products (a^u_i, a^u_j) with −i ∈ A or −j ∈ A vanish.

Hence, assume that−i /∈ Aand −j /∈ A. Then(a^u_i, a^u_j)∈ {p^d−1,2p^d−1}fori+j >0. In case(a^u_i, a^u_j) = 2p^d−1 we have a^u_i =a^u_j. This happens exactly whenj 6= 0and ij⁻¹∈ A. Since −i /∈ A, the cosetiA in G does not contain−1. Hence, there are precisely|A|choices forj such thatij⁻¹∈ A.

Hence, we have shown that the rowsa^u_i fori= 1, . . . , p−2split in|A| −1zero rows and(p−1)/|A| −1 groups consisting of|A|equal rows each. If we replace the matrixAbyU Afor a suitable matrixU ∈GL(p−1,Z), we get a new matrix with exactly(p−1)/|A|non-vanishing rows (this is essentially the same as taking another (positive definite) quadratic form in (4.1), see [32]). After leaving out the zero rows we get a(p−1)/|A| ×(p−1)/|A|

matrix

AA^T=p^d−1













|A|+ 1 1 . . . 1 1 2 . .. ... ... . .. . .. 1 1 . . . 1 2











 .

Now we can apply the quadratic formqcorresponding to the Dynkin diagramA_(p−1)/|A|in Eq. (4.1). This gives Γ(d,1,|A|)≤ |A|+ (p−1)/|A|

p^d−1.

On the other handp^1−dAA^T is the square of the matrix











1 · · · 1

1 1

... . ..

1 1











which has exactly|A|+ (p−1)/|A|columns. This shows thatΓ(d,1,|A|)cannot be smaller.

The next proposition gives an induction step.

Proposition 4.9. If |A|_p6= 1, then

Γ(d, k,|A|) = Γ(d, k−1,|A|/p).

Proof. Since |A|p 6= 1, we have k ≥ 2. Let i ∈ {1, . . . , ϕ(p^k)−1} such that gcd(i, p) = 1. We will see that (a^u_i, a^u_i) = 0 and thusa^u_i = 0. By Lemma 4.7 and Eq. (4.2) it suffices to show that there is someτ ∈ Asuch thatp^k|i⁰+iτ. We can write this in the form−i⁻¹i⁰∈ A, sinceirepresents an element of(Z/p^kZ)^×. Now let

−i⁰ =i+αp^k−1 for someα∈Z. Then −i⁻¹i⁰= 1 +i⁻¹αp^k−1 is an element of order pinG. Since G has only one subgroup of orderp, it follows that−i⁻¹i⁰ ∈ A.

Hence, in order to apply Proposition 4.2 it remains to consider the indices which are divisible by p. Let A be the image of the canonical map(Z/p^kZ)^×→(Z/p^k−1Z)^× underA. Then|A|=|A|/p(cf. Lemma 4.6). Ifiand j are divisible byp, we have

|{τ∈ A:p^k|i+jτ}|=p· |{τ∈ A:p^k−1|(i/p) + (j/p)τ}|.

A similar equality holds for the other summands in (4.2). Here observe that(i/p)⁰ =i⁰/p, where the dash on the left refers to the casep^k−1. Thus, the remaining matrix is just the matrix in casep^k−1. HenceΓ(d, k,|A|) = Γ(d, k−1,|A|) = Γ(d, k−1,|A|/p).

Now we are in a position to prove the main theorem of this section.

Theorem 4.10. LetB be ap-block of a finite groupGwherepis an odd prime, and let(u, b_u)be aB-subsection such thatl(b_u) = 1andb_u has defectd. Moreover, letF be the fusion system ofB and|Aut_F(hui)|=p^sr, where p-r ands≥0. Then we have

k0(B)≤|hui|+p^s(r²−1)

|hui| ·r p^d. (4.4)

If (in addition) (u, bu)is major, we can replace k0(B)by P∞

i=0p²ⁱki(B) in (4.4).

Proof. As before let|hui|=p^k. We will prove by induction onkthat Γ(d, k, p^sr) = p^k+p^s(r²−1)

p^kr p^d.

By Proposition 4.8 we may assume k ≥ 2. By Proposition 4.9 we can also assume that s = 0. As before we consider the matrixA. Like in the proof of Proposition 4.9 it is easy to see that the indices divisible bypform a block of the matrix AA^T which contributes Γ(d, k−1, r)/p to Γ(d, k, r). It remains to deal with the matrix Ae:= a^u_i : gcd(i, p) = 1

. By Lemma 4.7 the entries of p^k−dAeAe^Tlie in {0,±1,±2}. Moreover, if gcd(i, p) = 1 we have(a^u_i, a^u_i) = 2p^d−k (see proof of Proposition 4.9).

With the notation of the proof of Proposition 4.4 we haveV A=W. In particularrkAA^T= rkA= rkW =|G: A|. If we setA1 := a^u_i : gcd(i, p)>1

, it also follows thatrkA1A^T₁ = rkA1=ϕ(p^k−1)/r. Since the rows of Ae are orthogonal to the rows ofA1(see Lemma 4.7), we see thatrkAe= (ϕ(p^k)−ϕ(p^k−1))/r=p^k−2(p−1)²/r.

Now we will findp^k−2(p−1)²/rlinearly independent rows ofA. For this observe thate Aacts onΩ :={i: 1≤i≤ p^k−1, gcd(i, p) = 1}by^τi:=τ·i (modp^k−1)forτ∈ A. Since p-r, every orbit has lengthr(see Lemma 4.6).

We choose a set of representatives ∆ for these orbits. Then |∆| = p^k−2(p−1)/r. Finally for i ∈ ∆ we set

∆i :={i+jp^k−1:j= 0, . . . , p−2}. We claim that the rowsa^u_i withi∈S

j∈∆∆j are linearly independent. We do this in two steps.

Step 1:(a^u_i, a^u_j) = 0fori, j∈∆,i6=j.

We will show that all summands in (4.2) vanish. First assume thati≡jτ (mod p^k)for someτ ∈ A. Then of course we also have i≡jτ (mod p^k−1)which contradicts the choice of∆. Exactly the same argument works for the other summands. For the next step we fix somei∈∆.

Step 2:a^u_j forj∈∆_i are linearly independent.

It suffices to show that the matrixA⁰ :=p^k−d(a^u_l, a^u_m)_l,m∈∆i is invertible. We already know that the diagonal entries ofA⁰equal2. Now writem=l+jp^k−1for somej6= 0. We consider the summands in (4.2). Assume that there is someτ∈ Asuch thatl≡mτ≡(l+jp^k−1)τ (mod p^k). Then we haveτ ≡1 (modp^k−1)which implies τ= 1. However, this contradictsj 6= 0. On the other hand we havel⁰≡m⁰τ≡l⁰τ (modp^k)forτ= 1∈ A. Now assume−l ≡m⁰τ (mod p^k). Then the argument above impliesτ = 1 andl+l⁰ ≡0 (modp^k) which is false.

Similarly the last summand in (4.2) equals0. Thus, we have shown thatA⁰= (1 +δlm)_l,m∈∆i is invertible.

This implies that the rank ofAeisp^k−2(p−1)²/r. Hence, there exists an integral matrixU ∈GL(p^k−2(p−1)²,Q) such that the only non-zero rows ofUAearea^u_i fori∈S

j∈∆∆j. Then we can leave out the zero rows and obtain a matrix (still denoted byA) of dimensione p^k−2(p−1)²/r. Moreover, the two steps above show thatp^k−dAeAe^T