Endo‐trivial modules for finite
groups
with dihedral
Sylow
2‐subgruops
Department
of Mathematics andInformatics,
Graduate School ofScience,
ChibaUniversity
e‐mail[email protected]‐u.ac.jp
千葉大学大学院理学研究科
Shigeo
Koshitani越谷重夫
1. ENDO‐TRIVIAL MODULES
This is
joint
work with Caroline Lassueur(see [8], [7]).
We assume thatk is afield with characteristic
p>0
, and G is afinite group withp||G|.
A
finitely generated right
kG‐module M is called endo‐trivial if\mathrm{E}\mathrm{n}\mathrm{d}_{k}(M)\cong k_{G}\oplus(
projective)
as
right
kG‐modules wherek_{G}
is the trivial kG‐module. Then in thestable module
category
stmod(kG)
offinitely generated
kG‐modules,
the setT(G)
:={ [M]\in
stmod(kG)|M
isendo‐trivial}
has an abelian group structure
by making
use of the tensorproduct
over k.Namely,
we define an addition + inT(G)
by
[M]+[N]:=[M\otimes_{k}N]
(note
thatM\otimes_{k}N
isagain
an endo‐trivial kG‐module ifso are M andN, and recall also that
\mathrm{E}\mathrm{n}\mathrm{d}_{k}(M)\cong M^{*}\otimes_{k}M
as kG‐modules whereM^{*}
:=\mathrm{H}\mathrm{o}\mathrm{m}_{k}(M, k)
. Remind also that forright
kG‐modules M andN,
M\otimes_{k}N
is considered as left kG‐moduleby
thediagonal
action,
namely
(m\otimes n)g
:=mg\otimes ng forg\in G,
m\in M and n\in N, andthat M^{*} is
again
a left kG‐modlue via( $\phi$ g)(m)
:= $\phi$(mg^{-1})
forg\in G,
m\in M and
$\phi$\in M^{*}
Then it is easy to know that the zero element inT(G)
is[k_{G}]
and the inverse element-[M]
of[M]
inT(G)
is[M^{*}].
The endo‐trivial modules go back to Dade in 1978[6].
Since then endo‐trivial modules showup in manyplaces
inthe modularrepresenta‐
tiontheory
of finitegroupsandthey
do haveplayed
veryimportant
rolein this area.
Actually,
Dade classifies all the endo‐trivial kG‐modulesfor the case where G are finite abelian p‐groups
(see
[6]).
Since thenthe determination of the structure of
T(G)
has been done for the casewhere G are finite p‐groups
by
Carlson and Thévenaz(see [13], [2]).
Then,
wemight
be,interested
inT(G)
forartibrary
finite groups G.This
problem
is still open as far as the author knows. It is known that‐T(G)
isfinitely generated
by
a result of[3] (depending
on initiatedwork of
Puig).
From now on. letTT(G)
andTF(G)
be the torsionpart
and the torsion freepart
ofT(G)
respectively,
and let P be aSylow
p\leftrightarrow‐subgroup
of G.First,
if P iscyclic,
thenT(G)
is considered in[9].
This is of courseexactly
the case where the groupalgebra
kG is offinite‐representation
type
as is well‐known since many years ago.Further,
ifp=2
and Pis
generalized
quaternion
orsemidihedral,
thenT(G)
is treatedby
[4].
Recall that these cases cover almost of all the cases where the group
algebras
kGareoffinite‐representation
type
andoftame‐representation
type.
Then,
whatsmissing?
Yes,
the case where P is dihedral(possibly
the Klein‐four group of order4)
ismissing.
This isactually
our motivaion for the work
(see [7]
and[8]).
Our main result is the
folllowing:
Theorem 1.1.
Suppose
that G is afinite
group with a dihedralSylow
2‐subgroup
Pof
order at least8,
and assume thatT(G)
is the abeliangroup
of
endo‐trivialkG‐modules over analgebraically
closedfield
kof
characteristic 2. Set
\overline{G}
:=G/O_{2'}(G)
. Then we have thefollowing:
(i)
If
\overline{G}\not\cong \mathfrak{A}_{6}
(here
\mathfrak{A}_{6}
is thealternating
group on 6letters),
thenTT(G)=X(G)
, whereX(G)
:=\{[M]\in T(G)|\dim_{k}(M)=
1\}.
(ii)
If
\overline{G}\cong \mathfrak{A}_{6}
, then either(a)
TT(G)=X(G)
, or(b) TT(G)/X(G)
is anelementary
abelian3‐group
and eachindecomposable
endo‐trivial kG ‐module M with[M]
\inTT(G)\backslash X(G)
is a 9‐dimensionalsimple
module.Proof.
See[8,
Theorem1.2].
\squareRemark 1.2. For the case where the
Sylow 2‐subgroup
P of G is theActually
we do have apretty
muchgeneral
result tocompute
T(G)
.It is stated as follows:
Theorem 1.3.
Suppose
that k is analgebraically
closedfield of
charac‐teristic
p>0
, and that G is afinite
groupofp‐rank
at least2,
namely
G contains a
subgroup
C_{p}\times C_{p}
whereC_{p}
is thecyclic
groupof
orderp. Further assume thatG has no
strongly
p‐embeddedsubgroups. Now_{f}
we suppose thatH\triangleleft G with
p\displaystyle \int|H|
. Set\overline{G}:=G/H
, and assume more
over that
\mathrm{H}^{2}(\overline{G}, k^{\times})=1
.Then,
we havek(G)\cong X(G)+k(\overline{G})
where
K(G)
is the kernelof
the restriction map{\rm Res}_{P}^{G}:T(G)\rightarrow T(P)
given
by
[1\mathrm{J}4i]\mapsto[M\downarrow_{P}^{G}],
and
further
k(G) :=\{[M]\in T(G)|M=f^{-1}(L)
for
a 1‐dimensionalkN_{G}(P)
‐module L}
where
f
is the Greencorrespondence
withrespect
to(G, P, N_{G}(P))
.Proof.
See[8,
Theorem1.1].
\squareRemark 1.4. The
point
in Theorem 1.3 is thefollowing.
Weactually
would like tocompute
TT(G)
, and it is mostlikely
thesame asK(G)
.Precisely speaking,
K(G)=TT(G)
unless P iscyclic
orgeneralized
quaternion.
Soat least forourpurpose this is thecase. Soourfinal aimis recuded from the
computation
ofTT(G)
to that ofK(G)
.Now,
letus lookat the
right
hand side.Then,
first ofall,
X(G)
iscomputable
(it
is
nothing
but the p‐part
ofG/[G, G]
where[G, G]
is the commutatorsubgroup
of G. Then what about the second termK(\overline{G})
. Since wecan assume that
H\neq 1
, we are able to use inductiveargument.
So,
itworks!
2. OKUYAM\mathrm{A}^{}\mathrm{S} THEOREM
The author would like to introduce a theorem of
Okuyama
in 1981with a
proof,
which showed uponly
inJapanese
[12,
Theorem1],
be‐cause the author
hopes/believes
thatOkuyamas
theoremmay/should
be useful even for
understanding
endo‐trivial modules and also endo‐permutation
modules and so on. Who knows?Okuyamas
theoremdimentional k
‐algebra
A andI\subseteq A
we denoteby
\mathrm{A}\mathrm{n}\mathrm{n}_{A}(I)
the(right)
annihilator of I in A,
namelyì
\mathrm{A}\mathrm{n}\mathrm{n}_{A}(I) :=\{a\in A|Ia=\{0\}\}
, and wedenote
by
Z(A)
the center of A.Theorem 2.1
(Okuyama (1981)
[12]).
Let B be a blockalgebra
of
kGof
afinite
group G over analgebraically
closedfield
kof
character‐istic
p>0
(actually,
this theorem holds evenfor
anarbitrary finite
dimensional
symmetric
k‐algebra
B).
Further,
let\ell(B)
be the numberof
non‐isomorphic simple
right
B‐modules,
and letS_{1},
\cdotsì
S_{\ell(B)}
be theall
non‐isomorphic simple
right
B ‐modules.Then,
it holds\displaystyle \dim_{k}[\mathrm{A}\mathrm{n}\mathrm{n}_{B}(J(B)^{2})\cap Z(B)]=\ell(B)+\sum_{i=1}^{\ell(B)}\dim_{k}[\mathrm{E}\mathrm{x}\mathrm{t}_{B}^{1}(S_{i}, Si)].
Proof.
(Okuyama
in[12,
Theorem1]).
Set \ell:=\ell(B)
, and let e_{1}, \cdots, e_{\ell}be the set of all
primitive
idempotents
of B such thatP_{i}
:=P(S_{i})
:=e_{i}B
is theprojective
cover ofS_{i}
fori=1,
\cdots,\ell. Set e :=e_{1}+\cdots+e_{\ell},and A :=eBe, and hence A is the basic
ring
(algebra)
of B, and Ais a finite dimensional
symmetric
k‐algebra
(recall
that A and B areMorita
equivalent).
Further,
set J:=J(B)
the Jacobson radical of B.Step
1.\dim_{k}[\mathrm{E}\mathrm{x}\mathrm{t}_{B}^{1}(S_{i}, S_{i})]=\dim_{k}(e_{i}Je_{i}/e_{i}J^{2}e_{i})
.Step
2. IfC is a finite dimensional k‐algebra
such thatC/J(C)\cong k,
then
\mathrm{A}\mathrm{n}\mathrm{n}_{C}(J(C)^{2})\subseteq Z(C)
.Proof of
Step
2. Followsby
[10,
Lines 7‐8 of theproof
of Lemma2],
Step
3. For any two‐sided ideal I of B, we have that\dim_{k}[\mathrm{A}\mathrm{n}\mathrm{n}_{B}(I)\cap Z(B)]=\dim_{k}[\mathrm{A}\mathrm{n}\mathrm{n}_{A}(eJe)\cap Z(A)].
Proof of
Step
3. Recall that B and A are Moritaequivalent.
Step
4. If0\neq f=f^{2}\in C
for a finite‐dimensionalsymmetric
k‐algebra
C, thenfCf
isagain symmetric.
Proof of
Step
4. Take a similar way to prove that a Moritaequiva‐
lence preserves
being
symmetric,
though
fCf
isnotnecessarily
Moritaequivalent
to C, of course.Step
5. SetT:=\displaystyle \sum_{i}e_{i}J^{2}e_{i}+\sum_{i\neq j}e_{i}Je_{j}.
Then T is a two‐sided ideal of A.
Acknowledgements.
The author would like to thank Professor Hiroki Shimakura forgiving
him anopportunity
togive
a talk in themeeting
held in the RIMS of theUniversity
ofKyoto,
January
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Edited by S.Endo