On the Brauer
indecomposability
of Scott modules東京理科大学理学研究科 石岡 大樹
Hiroki Ishioka DepartmentofMathematics,
Tokyo Universityof Science
1. INTRODUCTION
Let k be an
algebraically
closed field ofprime
characteristic p. Let G be a finite group. For a finite dimensional kG‐module M and a‐subgroup Q
of G, we denoteby
M(Q)
the Brauerquotient
of M withrespect
toQ
. The Brauerquotient
M(Q)
isnaturally
akN_{G}(Q)
‐module. A kG‐module M is said to be Brauerindecomposable
ifM(Q)
isindecomposable
or zero as akQC_{G}(Q)
‐modulefor anyp‐‐subgroup Q
ofG([4]).
Brauer
indecomposability
ofp‐‐permutation
modules isimportant
forconstructing
stableequivalences
ofMoritatype
between blocks offinite groups(see
[1]).
In
[4],
arelationship
between Brauerindecomposability
of p‐permutation
modules andsaturated fusion
systems
wasgiven.
For ap‐‐subgroup
P of G, we denoteby
\mathcal{F}_{P}(G)
the fusionsystem
of G overP. One ofthe main result in[4]
isthefollowing.
Theorem 1
([4,
Theorem1.1| ).
Let P be a p‐subgroup of
G and M anindecomposable
p
‐permutation
kG‐module with vertexP.If
M is Brauerindecomposable,
then\mathcal{F}_{P}(G)
isa saturated
fusion
system.
In the
special
case that P is abelian and M is the Scott kG‐moduleS(G, P)
, the converse of theabove theorem holds.Theorem 2
([4,
Theorem1.2]).
Let P be an abelian p‐subgroup
of
G.If
\mathcal{F}_{P}(G)
issaturated,
thenS(G, P)
is Brauerindecomposable.
In
general,
the above theorem does not hold for non‐aUelian P.However,
there aresome cases inwhich theScott kG‐module
S(G, P)
is Brauerindecomposable,
even if P isnot
necessarily
abelian.We
study
the condition thatS(G, P)
to be Brauerindecomposable
where P is notnecessarily
abelian. Thefollowing
resultgives
anequivalent
condition for Scott kG‐module withvertexPto be Brauer
indecomposable.
Theorem 3. LetG be a
finite
group and Pap‐subgroup of
G.Suppose
thatM=S(
G)P)
and that\mathcal{F}_{P}(G)
issaturated. Then thefollowing
areequivalent.
(i)
M is Brauerindecomposable.
(ii)
Foreachfully
normalizedsubgroup Q of
P, the module{\rm Res}_{QC_{G}(Q)}^{N_{G}(Q)}S(N_{G}(Q), N_{P}(Q))
isindecomposable.
If
these conditionsaresatisfied,
thenM(Q)\cong S(N_{G}(Q), N_{P}(Q))
for
eachfully
normalizedsubgroup Q\leq P.
数理解析研究所講究録
A similar result is obtained
independently
in[3]
by
R.Kessar,
S. Koshitani and M. Linckelmann. In their theorem([3,
Theorem1.1]),
they
obtain a better condition than ourssincethey
assume that\mathcal{F}_{P}(G)=\mathcal{F}_{P}(N_{G}(P))
which we donot assume.The
following
theorem shows that{\rm Res}_{QC_{G}(Q)}^{N_{G}(Q)}S(N_{G}(Q), N_{P}(Q))
isindecomposable
ifQ
satisfiessome conditions.Theorem 4. Let G be a
finite
group, P a p‐subgroup of
G andQ
afully
normalizedsubgroup of
P.Suppose
that\mathcal{F}_{P}(G)
is saturated.Moreover,
we assume that there is asubgroup
H_{Q} of N_{G}(Q) satisfying following
two conditions:(i)
N_{P}(Q)\in Syl_{p}(H_{Q})
(ii) |N_{G}(Q)
:H_{Q}|=p^{a}
(a\geq 0)
Then
{\rm Res}_{QC_{G}(Q)}^{N_{G}(Q)}S(N_{G}(Q), N_{P}(Q))
isindecomposable.
The
following
is aconsequence of above twotheorems.Corollary
5. Let G be afinite
group and P ap‐subgroup of
G.Suppose
that\mathcal{F}_{P}(G)
issaturated.
If for
everyfully
normalizedsubgroup Q of
P there isasubgroup
H_{Q}
of
N_{G}(Q)
satisfies
the conditionsof
Theorem4,
thenS(G, P)
is Brauerindecomposable.
Throughout
thisarticle,
wedenoteby
L\displaystyle \bigcap_{G}H
the set\{^{g}L\cap H|g\in G}
forsubgroups
L and Kof G.
2. PRELIMINARIES
2.1. Scott modules.
First,
We recall the definition of Scott modules and some of itsproperties:
Definition 6. Fora
subgroup
H of G, the Scott kG‐moduleS(G, H)
withrespect
to His the
unique
indecomposable
summand of\mathrm{I}\mathrm{n}\mathrm{d}_{H}^{G}k_{H}
that contains the trivial kG‐module. If P is aSylow ‐subgroup
of H,thenS(G, H)
isisomorphic
toS(G, P)
.By definition,
the Scott kG‐module
S(G, P)
is ap‐permutation
kG‐module.By
Greensindecomposability
criterion,
thefollowing
result holds.Lemma 7. Let H be a
subgroup of
G such that|G
:H|
=p^{a}
(for
somea\geq 0
).
Then\mathrm{I}\mathrm{n}\mathrm{d}_{H}^{G}k_{H}
isindecomposable.
Inparticular,
we have thatS(G, H)\cong \mathrm{I}\mathrm{n}\mathrm{d}_{H}^{G}.
Hence,
forp‐‐subgroup
P of G, if there is asubgroup
H of G such that P is aSylow
‐subgroup
of H and|G
:H|=p^{a}
, then wehave thatS
(
G)P)
\cong \mathrm{I}\mathrm{n}\mathrm{d}_{H}^{G}k_{H}.
The
following
theoremgives
usinformation ofrestrictions ofScottmodules.Theorem 8
([2,
Theorem1.7]).
Let H be asubgroup of
G and P ap‐subgroup of
G.If
Q
is a maximal elementof
P\displaystyle \bigcap_{G}H
, thenS(
H)Q
)
is a direct summandof
{\rm Res}_{H}^{G}S(G, P)
.2.2. Brauer
quotients.
Let M be a kG‐module and H asubgroup
of G. LetM^{H}
be thesetof H‐fixed elements in M. Forsubgroups
Lof H)wedenoteby
\mathrm{T}\mathrm{r}_{H}^{G}
thetrace map\mathrm{T}\mathrm{r}_{L}^{H}
:M^{L}\rightarrow M^{H}
. Brauerquotients
aredefined as follows.Definition 9. Let M be akG‐module. For a
‐subgroup Q
of G, the Brauerquotient
of M withrespect
toQ
is the k‐vector spaceM(Q):=M^{Q}/(\displaystyle \sum_{R<Q}\mathrm{T}\mathrm{r}_{R}^{Q}(M^{R}))
. This k‐vector space has anatural structureofkN_{G}(Q)
‐module.Proposition
10. Let P be a p‐subgroup
of
G and M =S(G, P)
. ThenM(P)
\congS(N_{G}(P), P)
.Proposition
11. Let M be anindecomposable
p‐permutation
kG‐module with vertexP.Let
Q
be ap‐subgroup of
G. ThenQ\leq c^{P} if
andonly
if
M(Q)\neq 0.
2.3. Fusion
systems.
Fora 1\succ‐subgroup
PofG, the fusionsystem
\mathcal{F}_{P}(G)
ofG overP isthe
category
whoseobjects
arethesubgroups
of P, and whosemorphisms
arethe grouphomomorphisms
inducedby conjugation
in G.Definition 12. Let P be ap
‐subgroup
of G(i)
Asubgroup
Q
of P is saidto befully
normalized in\mathcal{F}_{P}(G)
if|N_{P}(^{x}Q
) |
\leq|N_{P}(Q)|
for all x\in G such that
XQ\leq P.
(ii)
Asubgroup Q
of P is said to befully
automized in\mathcal{F}_{P}(G)
if p$\dagger$ |N_{G}(Q)
:N_{P}(Q)C_{G}(Q)|.
(iii)
Asubgroup Q
of Pissaidtobereceptive
in\mathcal{F}_{P}(G)
ifithas thefollowing
property:
for each
R\leq P
and$\varphi$\in \mathrm{I}\mathrm{s}\mathrm{o}_{\mathcal{F}_{P}(G)}(R, Q)
, ifwe setN_{ $\varphi$} :=\{g\in N_{P}(Q) |\exists h\in N_{P}(R), c_{g}\circ $\varphi$= $\varphi$\circ c_{h}\},
then there is\overline{ $\varphi$}\in \mathrm{H}\mathrm{o}\mathrm{m}_{\mathcal{F}_{P}(G)}(N_{ $\varphi$}, P)
such that\overline{ $\varphi$}|_{R}= $\varphi$.
Saturatedfusion
systems
are definedasfollows.Definition 13. Let P be a
p‐‐subgroup
of G. The fusionsystem
\mathcal{F}_{P}(G)
is saturated ifthe
following
twoconditions are satisfied:(i)
P isfully
normalized in\mathcal{F}_{P}(G)
.(ii)
For eachsubgroup Q
ofP, ifQ
isfully
normalized in\mathcal{F}_{P}(G)
, thenQ
isreceptive
in
\mathcal{F}_{P}(G)
.For
example,
if P is aSylow
p‐subgroup
of G, then\mathcal{F}_{P}(G)
issaturated. 3. SKETCH OF PROOFIn this
section,
let P be ap‐‐subgroup
ofG and M the ScottmoduleS(G, P)
.Lemma 14.
If Q\leq P
isfully
normalized in\mathcal{F}_{P}(G)
, thenN_{P}(Q)
is a maximal elementof
P\displaystyle \bigcap_{G}N_{G}(Q)
.By
abovelemma,
we can show thatS(N_{G}(Q), N_{P}(Q))
is a direct summand ofM(Q)
for each
fully
normalizedsubgroup Q
of P.Therefore,
we have that(i)
implies
(ii)
in Theorem 3.Assume that Theorem 3
(ii)
holds. Weprovethat{\rm Res}_{QC_{G}(Q)}^{N_{G}(Q)}(M(Q))
isindecomposable
for eachQ\leq P by
inductionon|P
:Q|
. Without loss ofgenerality,
we can assume thatQ
isfully
normalized. IfM(Q)
isdecomposable,
thenby
thefollowing
lemma,
we canshow that there is a
subgroup
R such thatQ
<R\leq
P and{\rm Res}_{RC_{G}(R)}^{N_{G}(R)}
isdecomposable,
thiscontradicts the induction
hypothesis.
Lemma 15.
Suppose
that asubgroup Q of
P isfully
automized andreceptive.
Thenfor
any
g\in G
such thatQ\leq 9P
, we have thatN_{gP}(Q)
\leq_{N_{G}(Q)}N_{P}(Q)
.Hence,
M(Q)
isindecomposable,
andisomorphic
toS(N_{G}(Q), N_{P}(Q))
.Consequently,
Theorem 3(ii)
implies
3(i).
Theorem 4 is
proved by using properties
of Scottmodules and thefollowing
lemma. Lemma 16.If
Q
isfully
automizedsub_{9}roup of
P, and there is asub_{9}roupH_{Q}\leq N_{G}(Q)
containing
N_{P}(Q)
sttch that|N_{G}(Q):H_{Q}|=p^{a}
, thenC_{G}(Q)H_{Q}=N_{G}(Q)
. 4. EXAMPLESuppose
thatp=2
. Let Gbeagroup definedby
G :=\langle a, x, y|a^{4}=x^{2}=e, a^{2}=y^{2},
xax=a^{-1}, ay=ya, xy=yx\rangle,
and let P bea
subgroup
\{a, xy\rangle
of G. Thenwe caneasily
verify
that\mathcal{F}_{P}(G)
issaturated. For eachfully
normalizedsubgroup
Q
ofP, ifwe chooseH_{Q}
as P, thenH_{Q}
satisfies two conditions in Theorem 4.Therefore,
S(G, P)
isBrauerindecomposable by Corollary
5.In
particular,
if G is a p‐‐group and\mathcal{F}_{P}(G)
is saturated for a p‐subgroup
P ofG,
then G and P
satisfy
thehypothesis
of theCorollary
5)
and henceS(G, P)
is Brauerindecomposable.
REFERENCES
[1]
M.Broué,On Scott Modules and\mathrm{p}‐permutationmodules: anapproach throughthe Brauermorphism,Proc. Amer. Math. Soc. 93
(1985))
401‐408.[2]
H. Kawai. Onindecomposablemodules and blocks.Osaka J. Math.,23(1):201-205
, 1986.[3]
R.Kessar,S.Koshitani,andM.Linckelmann,On the BrauerindecomposabilityofScottmodules,Q.J. Math. 66
(2015),
895‐903.[4]
R. Kessar,N.Kunugi,N.Mitsuhashi, On Saturated fusionsystemsand Brauerindecomposabilityof Scott modules,J.Algebra340(2011),
90‐103.DEPARTMENT OFMATHEMATICS
TOKYO UNIVERSITYOFSCIENCE
1‐3KAGURAZAKA, SHINJUKU‐KU, TOKYO 162‐8601
JAPAN
E‐mail address: [email protected]