Broue's Conjecture Holds for Principal 3-Blocks with Elementary Abelian Defect Group of Order 9

30 Pages·2002·0.22 MB·English

JournalofAlgebra248,575–604(2002)
doi:10.1006/jabr.2001.9048,availableonlineathttp://www.idealibrary.comon
Brou´e’s Conjecture Holds for Principal 3-Blocks with
Elementary Abelian Defect Group of Order 9
Shigeo Koshitani1
DepartmentofMathematics,FacultyofScience,ChibaUniversity,
Yayoi-cho,Inage-ku,Chiba263-8522,Japan
E-mail:[email protected]
and
Naoko Kunugi
c/oS.Koshitani,DepartmentofMathematics,FacultyofScience,
ChibaUniversity,Yayoi-cho,Inage-ku,Chiba263-8522,Japan
E-mail:[email protected]
CommunicatedbyMichelBrou´e
ReceivedMay2,2000
In representation theory of finite groups, there is a well-known and important
conjectureduetoM.Brou´e.Heconjecturesthat,foranyprimep,ifafinitegroup
GhasanabelianSylowp-subgroupP,thenthederivedcategoriesoftheprincipal
p-blocks of G and of the normalizer N (cid:1)P(cid:2) of P in G are equivalent. We prove
G
inthispaperthatBrou´e’sconjectureholdsfortheprincipal3-blockofanarbitrary
finitegroupGwithanelementaryabelianSylow3-subgroupP oforder9,byusing
initiated works for the case where G is simple, which are due to Puig, Okuyama,
Waki, Miyachi, and the authors. The result depends on the classification of finite
simplegroups. 2002ElsevierScience(USA)
0. INTRODUCTION AND NOTATION
Inrepresentationtheoryoffinitegroups,oneofthemostimportantprob-
lems now is to solve a conjecture given by M. Brou´e [2, 4]. He conjectures
1PartiallysupportedbytheJSPS(JapanSocietyforpromotionofScience)andtheMinistry
ofEducation,Science,SportsandCulture,Japan.
575
0021-8693/02$35.00
2002ElsevierScience(USA)
Allrightsreserved.
576 koshitani and kunugi
the following:
(0.1) Conjecture (Brou´e [2, 6.2. Question], [4, 4.9. Conjecture]). Let G
be a finite group with a Sylow p-subgroup P, where p is a prime, and
let H =N (cid:1)P(cid:2), the normalizer of P in G. Then, is it true that the derived
G
categories of the principal p-blocks of G and H are equivalent provided P
is abelian?
There are only several special cases where Brou´e’s conjecture (0.1) has
been checked, see [4, p. 15] and [14, p. 136] (see also [5, 6, 15–17, 19, 23,
26, 27]).
The purpose of this paper is to present that Brou´e’s conjecture (0.1) is
true for an arbitrary finite group G with an elementary abelian Sylow 3-
subgroup P of order 9. As a matter of fact, it recently has been checked
that Brou´e’s conjecture (0.1) holds if the above G is, moreover, simple
by the works of Puig [26],Okuyama-Waki [25], Okuyama [23],Kunugi [19],
Koshitani–Kunugi[15],andKoshitani–Miyachi[16,17].Then,itlooksquite
naturaltoaskwhetherBrou´e’sconjectureholdsforanarbitraryfinitegroup
G with the same Sylow 3-subgroup P, by making use of the classification
of finite simple groups. On the other hand, there is a work by Marcus [21]
in which he proves that Brou´e’s conjecture is liftable from a finite group G
to a finite group G(cid:1)if G is a normal subgroup of G(cid:1)such that G(cid:1)/G is a p(cid:4)-
group under some hypotheses. However, in general it is not trivial to know
that the hypotheses in [21] are satisfied. Nevertheless, in the case where
P is elementary abelian of order 9, we can show that Brou´e’s conjecture
holds by preparing tools (lemmas) which are a sort of generalization or
application of theorems of Dade [10] and Marcus [21].
The strategy of this paper is the following. Assume G(cid:5)G(cid:1), G(cid:1)/G is a
p(cid:4)-group, and P is an abelian Sylow p-subgroup of G. By a theorem of
Alperin and Dade [1, 9], Brou´e’s conjecture (0.1) is liftable from G to a
normalsubgroupG·C (cid:1)P(cid:2)ofG(cid:1).Itiseasytoknowthatthequotientgroup
G(cid:1)
G(cid:1)/(cid:7)G·C (cid:1)P(cid:2)(cid:8) is a subquotient group of N (cid:1)P(cid:2)/C (cid:1)P(cid:2) (cid:8)→ Aut(cid:1)P(cid:2). So, if
G(cid:1) G(cid:1) G(cid:1)
P is small, then we might have a chance to lift Brou´e’s conjecture from
G·C (cid:1)P(cid:2) to G(cid:1). Actually, this is the case, though it is not yet easy.
G(cid:1)
Our main theorem is the following:
(0.2) Theorem. Let (cid:1) be a complete discrete valuation ring of rank 1,
and let k be its residue field (cid:1)/J(cid:1)(cid:1)(cid:2) such that k is an algebraically closed
field of characteristic 3. Let G be an arbitrary finite group with an elementary
abelianSylow3-subgroupP oforder9,andletH =N (cid:1)P(cid:2),thenormalizerof
G
P in G. Then, the principal block algebras B (cid:1)kG(cid:2) and B (cid:1)kH(cid:2) of kG and
0 0
kH, respectively, are splendidly equivalent, and therefore so are B (cid:1)(cid:1)G(cid:2) and
0
B (cid:1)(cid:1)H(cid:2).
0
broue´’s conjecture for principal 3-blocks 577
(0.3) Remark. The main result (0.2) depends on the classification of
finite simple groups.
Throughout this paper we use the following notation and terminology.
Let k be an algebraically closed field of characteristic p>0. In this paper
a module always means a finitely generated right module, unless stated
otherwise. Let G and H be finite groups. Even in the case where M is a
(cid:1)kG(cid:14)kH(cid:2)-bimodule, we say that M is a (right) k(cid:7)G×H(cid:8)-module via an
action m·(cid:1)g(cid:14)h(cid:2) =g −1mh for m ∈ M(cid:14)g ∈ G, and h ∈ H. For the same
M, let M∨ =Hom (cid:1)M(cid:14)k(cid:2) be the k-dual of M, so that M∨ is a (cid:1)kH(cid:14)kG(cid:2)-
k
bimodule. We say that G and H have the same p-local structure in the
sense of Brou´e (see [4, 6.5. Theorem], [27, p. 342], and [21, p. 374]). We
write δ(cid:1)G(cid:2) for (cid:14)(cid:1)g(cid:14)g(cid:2) ∈G ×G (cid:15) g ∈ G(cid:16). Let G(cid:1)(cid:17)G, and let N be a right
kG-module. Then, we write Ng˜ or Ng˜ for the g˜-conjugate of N where
g˜ ∈ G(cid:1), namely, Ng˜ = N as k-vector spaces and the action of G on Ng˜ is
defined by n∗g = n·g˜−1gg˜ for n ∈ N and g ∈ G. For an abelian Sylow
p-subgroup P of G, let E(cid:1)G(cid:2) =N (cid:1)P(cid:2)/C (cid:1)P(cid:2), so that this is the inertial
G G
quotient of the principal p-block of G (note that E(cid:1)G(cid:2) is different from
thelargestsemi-simplenormalsubgroupof G).Letnbeapositiveinteger.
Then, C , (cid:21) , and A , respectively, denote the cyclic group of order n,
n n n
the symmetric group on n letters, and the alternating group on n letters.
We write n = pm for a nonnegative integer m if pm(cid:15)n and pm+1 (cid:21) (cid:15)n. We
p
denote by D , Q , and SD , respectively, the dihedral, the quaternion, and
n n n
the semi-dihedral groups of order n. For two groups G and H we denote
by G(cid:1)H a semi-direct product of G by H, namely G(cid:5)G(cid:1)H. We denote
by k the trivial kG-module of k-dimension 1. We write B (cid:1)kG(cid:2) for the
G 0
principal block algebra of the group algebra kG. For a ring R, we denote
by 1 , R×, Z(cid:1)R(cid:2), and J(cid:1)R(cid:2), respectively, the unit element, the set of all
R
units, the center, and the Jacobson radical of R. We say that two block
algebras A and B of kG and kH, respectively, are Puig (splendidly Morita)
equivalent,followingBrou´e[4,p.22](seealso[21,1.6.Theorem]),andthat
A and B are splendidly equivalent, following Rickard [27, p. 337].
Let P be a Sylow p-subgroup of G, and let H be a subgroup of
G with H ⊇ N (cid:1)P(cid:2). Then, we can define the Green correspondences
G
f↓G×G: G×G → G×H and g↑H×G: H ×H → H ×G with respect
G×H H×H
to a p-subgroup δ(cid:1)P(cid:2) of G×G. Then, let M(cid:1)1(cid:2) and N(cid:1)1(cid:2) be M(cid:1)1(cid:2) =
M(cid:1)1(cid:2) = f↓G×G(cid:1)B (cid:1)kG(cid:2)(cid:2) and N(cid:1)1(cid:2) =N (cid:1)1(cid:2) = g↑H×G(cid:1)B (cid:1)kH(cid:2)(cid:2).
(cid:1)G(cid:14)H(cid:2) G×H 0 (cid:1)H(cid:14)G(cid:2) H×H 0
Note that M(cid:1)1(cid:2) and N(cid:1)1(cid:2) are p-permutation (trivial source) k(cid:7)G×H(cid:8)-
and k(cid:7)H ×G(cid:8)-modules with vertex δ(cid:1)P(cid:2), respectively. We use the nota-
tions f↓G×H and g↑G×G in similar ways. For any subgroup H of G we
H×H H×G
denote by Scott (cid:1)H(cid:2) the Alperin–Scott module of H in G (see [30, Sect. 3,
G
p. 1549] and [22, Chap. 4, Theorem 8.4 and p. 297]), and we denote by
P (cid:1)M(cid:2) the (relatively) H-projective cover of a right kG-module M (see
H
578 koshitani and kunugi
[30, Sect. 1, p. 1538]). For kG-modules X and Y we write (cid:7)X(cid:14)Y(cid:8)G for
dim (cid:7)Hom (cid:1)X(cid:14)Y(cid:2)(cid:8). LetA be a finite-dimensional k-algebra and let M
k kG
be an A-module. Then, we denote by P(cid:1)M(cid:2) the projective cover of M,
by Soc(cid:1)M(cid:2) the socle of M, by(cid:31) (cid:1)M(cid:2) the kernel of the projective cover
P(cid:1)M(cid:2)→→M of M (namely (cid:31) is the Heller functor (operator)), by IBr(cid:1)A(cid:2)
the set of all non-isomorphic simple A-modules, by mod-A the category of
all finitely generated right A-modules, and by mod-A the stable module
category of finitely generated right A-modules (see [14, p. 193, Defini-
tion 9.3.1]). We denote by c (cid:1)S(cid:14)T(cid:2) or just by c(cid:1)S(cid:14)T(cid:2) the Cartan invariant
A
with respect to simple A-modules S and T. We write Y(cid:15)X when Y is a
direct summand of X. For other notation and terminology, see the book
of Nagao and Tsushima [22].
1. ELEMENTARY GROUP THEORY
In this section we give elementary facts on group theory. The proofs are
easy. But, they are helpful for obtaining our main result.
(1.1) Notation. We fix a finite group G and assume that G has an
abelian Sylow p-subgroup P.
(1.2) Lemma. (i) The group E(cid:1)G(cid:2) is a p(cid:4)-subgroup of Aut(cid:1)P(cid:2). In par-
ticular, if P =C ×C , then E(cid:1)G(cid:2)(cid:8)→SD .
3 3 16
In the following assume that L is a subgroup of a finite group G(cid:1) such
that p (cid:21) (cid:15)(cid:15)G(cid:1)(cid:25) L(cid:15), P ∈ Syl (cid:1)L(cid:2), G(cid:5)G(cid:1), and that G(cid:1)/G is a p(cid:4)-group (so that
p
P ∈Syl (cid:1)G(cid:1)(cid:2)), and let G =G·C (cid:1)P(cid:2).
p 0 G(cid:1)
(ii) A correspondence ϕ (cid:25) E(cid:1)L(cid:2) →E (cid:1)G(cid:1)(cid:2) given by $ · C (cid:1)P(cid:2) (cid:26)→
L L
$·C (cid:1)P(cid:2) for $ ∈ N (cid:1)P(cid:2) is a group-monomorphism, and hence we can con-
G(cid:1) L
sider E(cid:1)L(cid:2) ≤E (cid:1)G(cid:1)(cid:2) via ϕ . Moreover, ϕ (cid:1)E(cid:1)G(cid:2)(cid:2)(cid:5)E(cid:1)G(cid:1)(cid:2), so that we may
L G
consider E(cid:1)G(cid:2)(cid:5)E(cid:1)G(cid:1)(cid:2) via ϕ .
G
(iii) G (cid:5)G(cid:1)and E(cid:1)G(cid:2)∼=E(cid:1)G (cid:2).
0 0
(iv) G(cid:1)/G∼=N (cid:1)P(cid:2)/N (cid:1)P(cid:2).
G(cid:1) G
(v) If, moreover, C (cid:1)P(cid:2)⊆G, then we have that C (cid:1)P(cid:2)=C (cid:1)P(cid:2) and
G(cid:1) G(cid:1) G
G(cid:1)/G∼=N (cid:1)P(cid:2)/N (cid:1)P(cid:2)∼=E(cid:1)G(cid:1)(cid:2)/E(cid:1)G(cid:2).
G(cid:1) G
(vi) If C (cid:1)P(cid:2)⊆G and (cid:15)E(cid:1)G(cid:1)(cid:2)(cid:15)=(cid:15)E(cid:1)G(cid:2)(cid:15), then we have G(cid:1)=G.
G(cid:1)
(vii) E(cid:1)G(cid:2) =E (cid:1)G (cid:2) and E(cid:1)G(cid:1)(cid:2)/E(cid:1)G (cid:2) ∼= G(cid:1)/G . In particular, if
0 0 0
E(cid:1)G(cid:1)(cid:2)∼=E(cid:1)G (cid:2), then G(cid:1)=G .
0 0
(viii) If G(cid:1)/G ∼= E(cid:1)G(cid:1)(cid:2)/E(cid:1)G(cid:2) ∼= C for a prime q with q (cid:21)= p, then
q
C (cid:1)P(cid:2)⊆G, namely, G =G.
G(cid:1) 0
broue´’s conjecture for principal 3-blocks 579
(ix) If G(cid:1)/G ∼= C for a prime q with q (cid:21)= p and E(cid:1)G(cid:1)(cid:2) ∼= E(cid:1)G(cid:2), then
q
G(cid:1)=G .
0
Proof. (i) Clear since Aut(cid:1)C ×C (cid:2)∼=GL (cid:1)3(cid:2).
3 3 2
(ii) Clearly, ϕ is well defined and is a group-homomorphism. If
L
ϕ (cid:1)$·C (cid:1)P(cid:2)(cid:2) = 1 for $ ∈ N (cid:1)P(cid:2), then $ ∈ C (cid:1)P(cid:2), so that $ ∈ C (cid:1)P(cid:2).
L L L G(cid:1) L
Hence, ϕ is a monomorphism.
L
Take any x ∈ E(cid:1)G(cid:2) and x˜ ∈ E(cid:1)G(cid:1)(cid:2). Write x = g ·C (cid:1)P(cid:2) and x˜ =
G
g˜ ·C (cid:1)P(cid:2) for some g ∈ N (cid:1)P(cid:2) and g˜ ∈ N (cid:1)P(cid:2). Then, x˜−1 ·ϕ (cid:1)x(cid:2)·x˜ =
G(cid:1) G G(cid:1) G
g˜−1gg˜ ·C (cid:1)P(cid:2) ∈ (cid:7)N (cid:1)P(cid:2)·C (cid:1)P(cid:2)(cid:8)/C (cid:1)P(cid:2) =Im (cid:1)ϕ (cid:2) since G(cid:5)G(cid:1). Hence,
G(cid:1) G G(cid:1) G(cid:1) G
ϕ (cid:1)E(cid:1)G(cid:2)(cid:2)(cid:5)E(cid:1)G(cid:1)(cid:2).
G
(iii) By the Frattini argument, G(cid:1) = G·N (cid:1)P(cid:2). Then, G is a sub-
G(cid:1) 0
group of G(cid:1) since G(cid:5)G(cid:1). Take any g˜ ∈ G(cid:1) and g ∈ G . We can write
0 0
g˜ = gn and g = g(cid:4)c for some g(cid:14)g(cid:4) ∈ G(cid:14)n ∈ N (cid:1)P(cid:2), and c ∈ C (cid:1)P(cid:2). Now
0 G(cid:1) G(cid:1)
g˜g g˜−1 = g˜(cid:1)g(cid:4)c(cid:2)g˜−1 = g˜g(cid:4)g˜−1 ·g˜cg˜−1. Moreover, g˜cg˜−1 = gncn−1g−1 ∈
0
G·C (cid:1)P(cid:2)·G =G·C (cid:1)P(cid:2)=G since C (cid:1)P(cid:2)(cid:5)N (cid:1)P(cid:2) and G(cid:5)G(cid:1). Hence,
G(cid:1) G(cid:1) 0 G(cid:1) G(cid:1)
g˜g g˜−1 ∈G·G·C (cid:1)P(cid:2)=G , which shows G (cid:5)G(cid:1).
0 G(cid:1) 0 0
Now, let ψ: E(cid:1)G(cid:2) →E (cid:1)G (cid:2) as in (ii). Then, by (ii), ψ(cid:1)E(cid:1)G(cid:2)(cid:2)(cid:5)E(cid:1)G (cid:2).
0 0
Takeanyx ∈E(cid:1)G (cid:2).Then,wecanwritex =g ·C (cid:1)P(cid:2)forg ∈N (cid:1)P(cid:2).
0 0 0 0 G 0 G
0 0
We can write also that g = gc for g ∈ G and c ∈ C (cid:1)P(cid:2), so that c =
0 G(cid:1)
g−1g ∈G . Hence, c ∈C (cid:1)P(cid:2). Thus, x =gc·C (cid:1)P(cid:2)=g ·C (cid:1)P(cid:2). Fur-
0 0 G 0 G G
0 0 0
thermore, g = g c−1 ∈ N (cid:1)P(cid:2)·C (cid:1)P(cid:2) ⊆N (cid:1)P(cid:2), so that g ∈ N (cid:1)P(cid:2)∩
0 G G G G
0 0 0 0
G=N (cid:1)P(cid:2). Therefore, x =g·C (cid:1)P(cid:2)=ψ(cid:1)g ·C (cid:1)P(cid:2)(cid:2). Hence, ψ is onto.
G 0 G G
0
Then, by (ii), ψ is an isomorphism.
(iv) By the Frattini argument, G(cid:1)/G = G · N (cid:1)P(cid:2)/G ∼=
G(cid:1)
N (cid:1)P(cid:2)/(cid:7)G∩N (cid:1)P(cid:2)(cid:8)=N (cid:1)P(cid:2)/N (cid:1)P(cid:2).
G(cid:1) G(cid:1) G(cid:1) G
(v) By the assumption, C (cid:1)P(cid:2)=C (cid:1)P(cid:2). Hence, by (iv),
G(cid:1) G
G(cid:1)/G∼=N (cid:1)P(cid:2)/N (cid:1)P(cid:2)∼=(cid:7)N (cid:1)P(cid:2)/C (cid:1)P(cid:2)(cid:8)/(cid:7)N (cid:1)P(cid:2)/C (cid:1)P(cid:2)(cid:8)=E(cid:1)G(cid:1)(cid:2)/E(cid:1)G(cid:2)(
G(cid:1) G G(cid:1) G(cid:1) G G(cid:1)
(vi) Easy by (v).
(vii) This follows from (iii), (v), and (vi).
(viii) Assume C (cid:1)P(cid:2)(cid:21)⊆G. Then, G ≤ G·C (cid:1)P(cid:2) =G , so that G =
G(cid:1) G(cid:1) 0 0
G(cid:1) since (cid:15)G(cid:1) (cid:25) G(cid:15) is a prime. Hence, E(cid:1)G(cid:1)(cid:2) =E (cid:1)G (cid:2) =E (cid:1)G(cid:2) by (iii), a
0
contradiction.
(ix) If G = G, then C (cid:1)P(cid:2) ⊆G, so that G(cid:1)/G ∼= E(cid:1)G(cid:1)(cid:2)/E(cid:1)G(cid:2) from
0 G(cid:1)
(v), a contradiction. Hence G(cid:1)G , which implies G =G(cid:1)since (cid:15)G(cid:1)(cid:25)G (cid:15) is
0 0 0
a prime.
(1.3) Lemma (Gaschu¨tz). AssumethatGhasaSylowp-subgroupP such
that P ∼= C ×C and that C (cid:8)→ Z(cid:1)G(cid:2). Then, there is a normal subgroup
p p p
L of G such that G=C ×L.
p
580 koshitani and kunugi
Proof. By Sylow’s theorem we can assume P =P ×P such that P ∼=
1 2 1
P ∼=C and P ⊆Z(cid:1)G(cid:2). Hence P (cid:5)G and P is abelian. Namely, P splits
2 p 1 1 1
on P∩P .
1
Take any prime q (cid:21)= p, and any Q ∈ Syl (cid:1)G(cid:2). Then, Q∩P = 1, which
q 1
meansthatQsplitsonQ∩P .Thus,itfollowsfromatheoremofGaschu¨tz
1
[29, Chap. 2, Theorem 8.6] that G splits on P . That is, G = P (cid:1)L for a
1 1
subgroup L of G. Since P ⊆Z(cid:1)G(cid:2), P (cid:1)L=P ×L.
1 1 1
(1.4) Lemma. AssumethatG(cid:5)G(cid:1),G(cid:1)/Gisap(cid:4)-group,Gisanon-abelian
simple group, and O (cid:1)G(cid:1)(cid:2)=1. Then, G(cid:5)G(cid:1)(cid:8)→Aut(cid:1)G(cid:2)=G(Out(cid:1)G(cid:2).
p(cid:4)
Proof. Since G is non-abelian simple, C (cid:1)G(cid:2)∩G =Z(cid:1)G(cid:2)=1. Hence,
G(cid:1)
there is a direct product C (cid:1)G(cid:2) ×G (cid:5) G(cid:1). Since G(cid:1)/G is a p(cid:4)-group,
G(cid:1)
p(cid:21)(cid:15)(cid:15)C (cid:1)G(cid:2)(cid:15). Hence, C (cid:1)G(cid:2) =1 since O (cid:1)G(cid:1)(cid:2) =1. Therefore, G(cid:1) ∼=
G(cid:1) G(cid:1) p(cid:4)
N (cid:1)G(cid:2)/C (cid:1)G(cid:2)(cid:8)→Aut(cid:1)G(cid:2).
G(cid:1) G(cid:1)
(1.5) Lemma. LetG,H(cid:1),andH besubgroupsofG(cid:1)suchthatG∩H(cid:1) ⊇H,
and let M be a right kH-module, where
G(cid:1) ≥ H(cid:1)
(cid:15) (cid:15)
G ≥ H(
Let h˜ ∈H(cid:1). Then, a map
M⊗ kG·h˜ −→(cid:1)M⊗ kH·h˜(cid:2)⊗ k(cid:7)h˜−1Gh˜(cid:8)
kH kH k(cid:7)h˜−1Hh˜(cid:8)
given by
m⊗ gh˜ (cid:26)→(cid:1)m⊗ h˜(cid:2)⊗ h˜−1gh˜ for m∈M(cid:14)g∈G(cid:14)
kH kH k(cid:7)h˜−1Hh˜(cid:8)
is an isomorphism of right k(cid:7)h˜−1Gh˜(cid:8)-modules.
(1.6) Lemma. Let P ∈Syl (cid:1)G(cid:2).
p
(i) B (cid:1)kG(cid:2) =P (cid:1)k (cid:2)=Scott (cid:1)δ(cid:1)P(cid:2)(cid:2).
0 G×G δ(cid:1)P(cid:2) G×G G×G
(ii) Let H be a subgroup of G with H ⊇ N (cid:1)P(cid:2). Then, it holds that
G
M(cid:1)1(cid:2) =P (cid:1)k (cid:2) =Scott (cid:1)δ(cid:1)P(cid:2)(cid:2), and that there is a primitive idem-
δ(cid:1)P(cid:2) G×H G×H
potent e of (cid:1)kG(cid:2)H such that M(cid:1)1(cid:2)=kGe , where (cid:1)kG(cid:2)H =C (cid:1)H(cid:2).
1 1 kG
Proof. (i) Let Y be a subgroup of a finite group X. LetX =∪n Yx
i=1 i
be a coset decomposition of Y in X, where n = (cid:15)X (cid:25) Y(cid:15), x ∈ X,
i
and x = 1. A map ϕ: k ↑X = ⊕n (cid:1)k ⊗ x(cid:2) →k given by
(cid:2) 1 (cid:2) Y i=1 Y kY i X
n α ⊗ x (cid:26)→ n α(cid:14)α ∈k, is aY -split epimorphism of kX-modules
i=1 i kY i i=1 i i
sinceamapψ:k →k ↑X givenbyα(cid:26)→α⊗ 1isakY -homomorphism
Y Y kY
which satisfies ϕ ◦ ψ = id . Now, take X = G × G and Y = δ(cid:1)P(cid:2).
k
Then, k ↑G×G →→k is a δ(cid:1)P(cid:2)-split k(cid:7)G×G(cid:8)-epimorphism. Let
δ(cid:1)P(cid:2) G×G
broue´’s conjecture for principal 3-blocks 581
A=B (cid:1)kG(cid:2).Then,Aisatrivialsourcek(cid:7)G×G(cid:8)-modulewithvertexδ(cid:1)P(cid:2)
0
by a theorem of Green [22, Chap. 5, Theorem 10.8]. Hence, A(cid:15)k ↑G×G.
δ(cid:1)P(cid:2)
Since (cid:7)k (cid:14)k ↑G×G(cid:8)G×G = (cid:7)k ↑G×G(cid:14)k (cid:8)G×G = 1 and since
G×G δ(cid:1)P(cid:2) δ(cid:1)P(cid:2) G×G
there are a monomorphism k (cid:2)A and an epimorphism θ: A→→k
G×G G×G
of k(cid:7)G×G(cid:8)-modules, θ is a δ(cid:1)P(cid:2)-split k(cid:7)G×G(cid:8)-epimorphism. Hence,
P (cid:1)k (cid:2)(cid:15)A by a standard argument. Since A is indecomposable as a
δ(cid:1)P(cid:2) G×G
k(cid:7)G×G(cid:8)-module, we have A=P (cid:1)k (cid:2). The second equality is given
δ(cid:1)P(cid:2) G×G
by [30, Proposition 3.1].
(ii) By the definition, M(cid:1)1(cid:2) is a unique indecomposable direct sum-
mand of B (cid:1)kG(cid:2) ↓G×G as a right k(cid:7)G×H(cid:8)-module with a vertex δ(cid:1)P(cid:2).
0 G×H
On the other hand, by the definition of Alperin–Scott modules, we know
that Scott (cid:1)δ(cid:1)P(cid:2)(cid:2) (cid:15) (cid:7)Scott (cid:1)δ(cid:1)P(cid:2)(cid:2)(cid:8) ↓G×G ( Therefore, by (i), we get
G×H G×G G×H
M(cid:1)1(cid:2)=Scott (cid:1)δ(cid:1)P(cid:2)(cid:2) by looking at the vertices. The rest is easy.
G×H
2. FINITE GROUPS WITH AN ELEMENTARY
ABELIAN SYLOW 3- SUBGROUP OF ORDER 9
InthissectionwegivealistoffinitegroupsGsuchthatGhasanelemen-
tary abelian Sylow 3-subgroup of order 9, O (cid:1)G(cid:2)=1 and O3(cid:4)(cid:1)G(cid:2)=G, by
3(cid:4)
making use of the classification of finite simple groups. Namely, we need
to use the classification of finite simple groups to prove (2.1), (2.2), and
(2.4). The following two propositions, (2.1) and (2.2), were provided by
S. Yoshiara.
(2.1) Proposition. Let G be a finite group with an elementary abelian
Sylow 3-subgroup of order 9 such that O (cid:1)G(cid:2)=1 and O3(cid:4)(cid:1)G(cid:2)=G. Then, G
3(cid:4)
is one of (i) or (ii).
(i) G = X ×Y for finite simple groups X and Y such that both of
them have cyclic Sylow 3-subgroups of order 3.
(ii) G is a non-abelian finite simple group with an elementary abelian
Sylow 3-subgroup of order 9.
Proof. See [17, (1.1) Proposition].
(2.2) Proposition. If G is a non-abelian finite simple group with an ele-
mentary abelian Sylow 3-subgroup of order 9, then G is one of the following
nine types:
(i) A (cid:14)A (cid:14)A (cid:14)M (cid:14)M (cid:14)M (cid:14)HS;
6 7 8 11 22 23
(ii) PSL (cid:1)q(cid:2) for a power q of a prime with q≡4 or 7 (mod 9);
3
(iii) PSU (cid:1)q2(cid:2) for a power q of a prime with 2<q≡2 or 5 (mod 9);
3
(iv) PSp (cid:1)q(cid:2) for a power q of a prime with q≡4 or 7 (mod 9);
4
582 koshitani and kunugi
(v) PSp (cid:1)q(cid:2) for a power q of a prime with 2<q≡2 or 5 (mod 9);
4
(vi) PSL (cid:1)q(cid:2) for a power q of a prime with 2<q≡2 or 5 (mod 9);
4
(vii) PSU (cid:1)q2(cid:2) for a power q of a prime with q≡4 or 7 (mod 9);
4
(viii) PSL (cid:1)q(cid:2) for a power q of a prime with q≡2 or 5 (mod 9);
5
(ix) PSU (cid:1)q2(cid:2) for a power q of a prime with q≡4 or 7 (mod 9).
5
Proof. See, e.g., [17, (1.2) Proposition].
(2.3) Remark. Note that in (iii) of (2.2) PSU (cid:1)22(cid:2)∼=(cid:1)C ×C (cid:2)(cid:1)Q , in
3 3 3 8
(v)of(2.2)PSp (cid:1)2(cid:2)=Sp (cid:1)2(cid:2)∼=(cid:21) ,andin(vi)of(2.2)PSL (cid:1)2(cid:2)=SL (cid:1)2(cid:2)∼=
4 4 6 4 4
A from [12, II 10.14 Satz], [12, II 9.21 Satz, II 9.22 Hauptsatz], and [12,
8
II 6.14 Satz (5)], respectively.
(2.4) Proposition. Assume that G is a finite group with elementary
abelianSylow3-subgroupP oforder9suchthatO (cid:1)G(cid:2)=1 andO3(cid:4)(cid:1)G(cid:2)=G.
3(cid:4)
Then we get the following:
(i) If G=C ×C , then E(cid:1)G(cid:2)=1.
3 3
(ii) If G=C ×X for a non-abelian simple group X with (cid:15)X(cid:15) =3,
3 3
then E(cid:1)G(cid:2)=C .
2
(iii) If G = X ×Y for non-abelian simple groups X and Y with
(cid:15)X(cid:15) =(cid:15)Y(cid:15) =3, then E(cid:1)G(cid:2)=C ×C .
3 3 2 2
(iv) If G=A or A , then E(cid:1)G(cid:2)=C .
6 7 4
(v) If G=A , then E(cid:1)G(cid:2)=D .
8 8
(vi) If G=M , M , orHS , then E(cid:1)G(cid:2)=SD .
11 23 16
(vii) If G=M (cid:1)∼=PSL (cid:1)4(cid:2)) orM , then E(cid:1)G(cid:2)=Q .
21 3 22 8
(viii) If G=PSL (cid:1)q(cid:2) such that q≡4 or 7 (mod 9), then E(cid:1)G(cid:2)=Q .
3 8
(ix) If G = PSU (cid:1)q2(cid:2) such that 2 < q ≡ 2 or 5 (mod 9), then
3
E(cid:1)G(cid:2)=Q .
8
(x) If G=PSp (cid:1)q(cid:2) such that q≡4 or 7 (mod 9), then E(cid:1)G(cid:2)=D .
4 8
(xi) If G = PSp (cid:1)q(cid:2) such that 2 < q ≡ 2 or 5 (mod 9), then
4
E(cid:1)G(cid:2)=D .
8
(xii) If G=PSL (cid:1)q(cid:2) such that q≡2 or 5 (mod 9), then E(cid:1)G(cid:2)=D .
4 8
(xiii) If G=PSL (cid:1)q(cid:2) such that q≡2 or 5 (mod 9), then E(cid:1)G(cid:2)=D .
5 8
(xiv) If G=PSU (cid:1)q(cid:2) or PSU (cid:1)q(cid:2) such that q≡4 or 7 (mod 9), then
4 5
E(cid:1)G(cid:2)=D .
8
broue´’s conjecture for principal 3-blocks 583
3. LEMMAS
Inthissectionwegiveseverallemmaswhichareusefulforobtainingour
main result. Throughout this section we assume that k is an algebraically
closed field of characteristic p >0, and that G, G(cid:1), G(cid:4), G(cid:1)(cid:4), X, X(cid:1), H, H(cid:4),
H(cid:1), and H(cid:1)(cid:4) are all finite groups.
(3.1) Lemma (Alperin–Dade). Suppose that G(cid:5)G(cid:1), G(cid:1)/G is a p(cid:4)-group,
P ∈ Syl (cid:1)G(cid:2), and G(cid:1)= G·C (cid:1)P(cid:2). Let e˜ and e be the block idempotents of
p G(cid:1)
B (cid:1)kG(cid:1)(cid:2) and B (cid:1)kG(cid:2), respectively. Then, B (cid:1)kG(cid:1)(cid:2) and B (cid:1)kG(cid:2) are category-
0 0 0 0
isomorphic, namely, a map B (cid:1)kG(cid:2) →B (cid:1)kG(cid:1)(cid:2) defined by a (cid:26)→ ae˜ for a ∈
0 0
B (cid:1)kG(cid:2) is a k-algebra-isomorphism, so that ee˜ = e˜e = e˜. Moreover, a right
0
k(cid:7)G(cid:1)×G(cid:8)-module B (cid:1)kG(cid:1)(cid:2)=e˜kG(cid:1)=e˜kG(cid:1)e gives a Puig equivalence between
0
B (cid:1)kG(cid:1)(cid:2) and B (cid:1)kG(cid:2).
0 0
Proof. This is proved in [1, Theorems 1 and 2] and [9, Theorem]. Note
that (cid:1)kG(cid:1)e˜(cid:2)↓G(cid:1)×G(cid:1)∼=kGe as right k(cid:7)G×G(cid:8)-modules as in [11, Sect. 4] (see
G×G
also [3, Theorems 0.1 and 0.2] and [21, (5.4)]).
(3.2) Lemma. Suppose that G(cid:5)G(cid:1), G(cid:1)/G is a p(cid:4)-group, and P ∈Syl (cid:1)G(cid:2)
p
such that G(cid:1)=G·C (cid:1)P(cid:2). Let H(cid:1) be a subgroup of G(cid:1)with H(cid:1) ⊇N (cid:1)P(cid:2), and
G(cid:1) G(cid:1)
let H =G∩H(cid:1). Assume that H(cid:1) =H·C (cid:1)P(cid:2). Assume, moreover, that a Puig
H(cid:1)
equivalence between B (cid:1)kG(cid:2) and B (cid:1)kH(cid:2) is given by the right k(cid:7)G×H(cid:8)-
0 0
module M(cid:1)1(cid:2) . Then, the right k(cid:7)G(cid:1)×H(cid:1)(cid:8)-module M(cid:1)1(cid:2) gives a Puig
(cid:1)G(cid:14)H(cid:2) (cid:1)G(cid:1)(cid:14)H(cid:1)(cid:2)
equivalence between B (cid:1)kG(cid:1)(cid:2) and B (cid:1)kH(cid:1)(cid:2).
0 0
Proof. Clearly, H(cid:5)H(cid:1) and G(cid:1)=G·H(cid:1), and
G(cid:1) = G·C (cid:1)P(cid:2) =G ·H(cid:1) --(cid:2) H(cid:1) = H·C (cid:1)P(cid:2) ⊇ N (cid:1)P(cid:2)
G(cid:1) H(cid:1) G(cid:1)
(cid:3) (cid:3)
G −→ H = G∩H(cid:1) ⊇ N (cid:1)P(cid:2)(
G
Let A(cid:1)= B (cid:1)kG(cid:1)(cid:2), A = B (cid:1)kG(cid:2), B(cid:1)= B (cid:1)kH(cid:1)(cid:2), and B = B (cid:1)kH(cid:2), and let e˜
0 0 0 0
and f˜ be the block idempotents of A(cid:1)and B(cid:1), respectively. By (3.1), A(cid:1)
A(cid:1) A
realizes a Puig equivalence between A(cid:1)and A, and so does B(cid:1) between
B B(cid:1)
B and B(cid:1). LetM (cid:1)1(cid:2) =M (cid:1)1(cid:2) . By the hypothesis, M(cid:1)1(cid:2) realizes a Puig
(cid:1)G(cid:14)H(cid:2)
equivalence between A and B. Let M(cid:1) =A(cid:1)⊗ M(cid:1)1(cid:2)⊗ B(cid:1). Then, M(cid:1) real-
A B
izes a Puig equivalence between A(cid:1)and B(cid:1). Since M(cid:1)↓G(cid:1)×H(cid:1) =M(cid:1)1(cid:2) by (3.1),
G×H (cid:3) (cid:4)
M(cid:1) is an indecomposable (cid:1)A(cid:1)(cid:14)B(cid:1)(cid:2)-bimodule. Since M(cid:1) = e˜ M(cid:1)1(cid:2)↑G(cid:1)×H(cid:1) f˜,
(1.6)(ii) implies that (cid:7)M(cid:1)1(cid:2)(cid:14)k (cid:8)G×H = 1, so that (cid:7)M(cid:1)(cid:14)k (cid:8)G(cid:1)×H(cid:1) (cid:21)= 0.
G×H G(cid:1)×H(cid:1)
Since (cid:7)k ↑G(cid:1)×H(cid:1)(cid:14)k (cid:8)G(cid:1)×H(cid:1) = (cid:7)Scott (cid:1)δ(cid:1)P(cid:2)(cid:2)(cid:14)k (cid:8)G(cid:1)×H(cid:1) = 1, it
δ(cid:1)P(cid:2) G(cid:1)×H(cid:1) G(cid:1)×H(cid:1) G(cid:1)×H(cid:1)
follows from a theorem of Krull–Schmidt that M(cid:1) ∼= Scott (cid:1)δ(cid:1)P(cid:2)(cid:2).
G(cid:1)×H(cid:1)
584 koshitani and kunugi
(Note that we get this also by [28, Lemma 2].) Hence we get the assertion
from (1.6)(ii).
(3.3) Lemma (Alperin–Dade–Marcus). Assume as in (3.1). Moreover,
suppose that H (cid:5) H(cid:1), P ∈ Syl (cid:1)G(cid:2) ∩Syl (cid:1)H(cid:2), H(cid:1) = H · C (cid:1)P(cid:2), and
p p H(cid:1)
G(cid:1)/G ∼= H(cid:1)/H. IfB (cid:1)kG(cid:2) and B (cid:1)kH(cid:2) are Morita (resp. Puig, derived
0 0
and splendidly) equivalent, then B (cid:1)kG(cid:1)(cid:2) and B (cid:1)kH(cid:1)(cid:2) are also Morita (resp.
0 0
Puig, derived and splendidly) equivalent.
Proof. Easy by (3.1) since a Puig equivalence is a Morita, derived, and
splendid equivalence (see [21, (5.4)]).
(3.4) Lemma (Dade). Let G(cid:5)G(cid:1) such that G(cid:1)/G is a p(cid:4)-group, and let
P ∈Syl (cid:1)G(cid:2). Suppose, furthermore, that G(cid:1)has a subgroup H(cid:1) such that H(cid:1) ⊇
p
N (cid:1)P(cid:2), and let H =G∩H(cid:1), so that G(cid:1)=GH(cid:1) and G(cid:1)/G∼=H(cid:1)/H, and
G(cid:1)
G(cid:1)=GH(cid:1) → H(cid:1) ⊇N (cid:1)P(cid:2)
G(cid:1)
(cid:3) (cid:3)
G → H =G∩H(cid:1) ⊇N (cid:1)P(cid:2)(
G
Let f↓G×H(cid:25) G×H → H×H be the Green correspondence with respect to
H×H
δ(cid:1)P(cid:2), and let A=B (cid:1)kG(cid:2), B=B (cid:1)kH(cid:2), and M(cid:1)1(cid:2)=M (cid:1)1(cid:2) .
0 0 (cid:1)G(cid:14)H(cid:2)
(i) Itholdsthatf↓G×H(cid:1)M(cid:1)1(cid:2)(cid:2)=Basrightk(cid:7)H×H(cid:8)-modules.More-
H×H
over, B is δ(cid:1)H(cid:1)(cid:2)-invariant, namely B·(cid:1)h˜(cid:14)h˜(cid:2)=B as right k(cid:7)H×H(cid:8)-modules
for any h˜ ∈H(cid:1), and B is extendible to a right k(cid:7)(cid:1)H×H(cid:2)·δ(cid:1)H(cid:1)(cid:2)(cid:8)-module.
(ii) It holds that M(cid:1)1(cid:2) is δ(cid:1)H(cid:1)(cid:2)-invariant as a right k(cid:7)G×H(cid:8)-module,
and that M(cid:1)1(cid:2) is extendible to a right k(cid:7)(cid:1)G×H(cid:2)·δ(cid:1)H(cid:1)(cid:2)(cid:8)-module.
Proof. (i) The first assertion is easy by the definition of Green cor-
respondences and the theorem of Krull–Schmidt. We easily know that B
is δ(cid:1)H(cid:1)(cid:2)-invariant since B is the principal block. Now, let 1 be the block
B
idempotent of B in kH, and let h∈H and h˜ ∈H(cid:1). Then,
h1 ·(cid:1)h˜(cid:14)h˜(cid:2)=h˜−1h1 h˜ =h˜−1hh˜ ·h˜−11 h˜ =h˜−1hh˜ ·1 ∈kH1 (cid:14)
B B B B B
sinceh˜−11 h˜ =1 andH(cid:5)H(cid:1).ThismeansthatB extendstoak(cid:7)(cid:1)H×H(cid:2)·
B B
δ(cid:1)H(cid:1)(cid:2)(cid:8)-module.
(ii) The first assertion is obtained as in (i). Then, M(cid:1)1(cid:2) extends
to a k(cid:7)(cid:1)G×H(cid:2)·δ(cid:1)H(cid:1)(cid:2)(cid:8)-module by use of (i) and a result of Dade [10,
(6.5)Corollary, p. 94].
(3.5) Lemma. Let G(cid:1), G, H(cid:1), H, P, and M(cid:1)1(cid:2) be the same as in (3.4).
Assume, moreover, that C (cid:1)P(cid:2) ⊆G, and that the right k(cid:7)G×H(cid:8)-module
G(cid:1)
M(cid:1)1(cid:2) gives a Puig equivalence between B (cid:1)kG(cid:2) and B (cid:1)kH(cid:2). Then, B (cid:1)kG(cid:1)(cid:2)
0 0 0
and B (cid:1)kH(cid:1)(cid:2) are Puig equivalent.
0