128

数理解析研究所講究録
第2003巻 2016年 128-132

Endo‐trivial modules for finitegroups with dihedral Sylow

2‐subgruops

Department of Mathematics and Informatics,
Graduate School of Science, Chiba Universitye‐mailkoshitan@math.s.chiba‐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 a field with characteristic p>0 , and G is a finite group with
A finitely generated right kG‐module M is called endo‐trivial if

p||G|.

\mathrm{E}\mathrm{n}\mathrm{d}_{k}(M)\cong k_{G}\oplus( projective)
right kG ‐modules where k_{G} is the trivial kG‐module. Then in thestable module category stmod (kG) of finitely generated kG‐modules,
asthe setT(G){ [M]\in stmod (kG)|M is endo‐trivial}:=

has an abelian group structure by making use of the tensor productoverk.

Namely, we define an addition + in T(G) by[M]+[N]:=[M\otimes_{k}N](note that M\otimes_{k}N is again an endo‐trivial kG‐module if so 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 for right kG‐modules M and,.

N, M\otimes_{k}N is considered as left kG ‐module by the diagonal action,
namely (m\otimes n)g :=mg\otimes ng for g\in G, m\in M and n\in N andthat M^{*} is again a left kG‐modlue via ( $\phi$ g)(m) := $\phi$(mg^{-1}) for g\in G,m\in M and $\phi$\in M^{*} Then it is easy to know that the zero element in,T(G) is [k_{G}] and the inverse element -[M] of [M] in T(G) is [M^{*}].
The endo‐trivial modules go back to Dade in 1978 [6]. Since thenendo‐trivial modules show up in many places in the modular representa‐tion theory of finite groups and they do have played very important role129

in this area.

Actually, Dade classifies all the endo‐trivial kG ‐modules

for 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, we might be,interested in T(G) for artibrary finite groups G.
This problem is still open as far as the author knows. It is known that‐T(G) is finitely generated by a result of [3] (depending on initiatedwork of Puig).

From now on. let TT(G) and TF(G) be the torsion

part and the torsion free part of T(G) respectively, and let P be aof G.
Sylow p\leftrighta‐subgroupr ow
First, if P is cyclic, then T(G) is considered in [9]. This is of course
exactly the case where the group algebra kG is of finite‐representation
type as is well‐known since many years ago. Further, if p=2 and P
is generalized quaternion or semidihedral, then T(G) is treated by [4].
Recall that these cases cover almost of all the cases where the groupalgebras kG are of finite‐representation type and of tame‐representation
type. Then, whats missing? Yes, the case where P is dihedral(possibly the Klein‐four group of order 4) is missing. This is actually
ourmotivaion for the work (see[7] and [8]).

Our main result is the folllowing:
Theorem 1.1.

Suppose that G is a finite group with a dihedral Sylow
2‐subgroup P of order at least 8, and assume that T(G) is the abeliangroup of endo‐trivial kG ‐modules over an algebraically closed field k ofcharacteristic 2. Set \overline{G} :=G/O_{2'}(G) Then we have the following:.

(i) If \overline{G}\not\cong \mathfrak{A}_{6} (here \mathfrak{A}_{6} is the alternating group on 6 letters), thenTT(G)=X(G) where X(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 an elementary abelian 3‐group and eachindecomposable endo‐trivial kG ‐module M with [M] \in
TT(G)\backslash X(G) is a 9‐dimensional simple module.
,,,Proof. See [8, Theorem 1.2].

\squareRemark 1.2. For the case where the Sylow 2‐subgroup P of G is theKlein‐four group C_{2}\times C_{2},see[8, Theorem 1.5] and also [7].

130Actually we do have a pretty much general result to compute T(G).

It is stated as follows:

Suppose that k is an algebraically closed field of charac‐
teristic p>0 and that G is a finite group ofp‐rank at least 2, namely
G contains a subgroup C_{p}\times C_{p} where C_{p} is the cyclic group of orderp Further assume that G has no strongly p ‐embedded subgroups. Now_{f}Theorem 1.3.
,.

wesuppose that H\triangleleft G withoverthat \mathrm{H}^{2}(\overline{G}, k^{\times})=1.

p\displaystyle \int|H| Set \overline{G}:=G/H.

,and assume more

Then, we havek(G)\cong X(G)+k(\overline{G})where K(G) is the kernel of the restriction map{\rm Res}_{P}^{G}:T(G)\rightarrow T(P)given by[1\mathrm{J}4i]\mapsto[M\downarrow_{P}^{G}],and furtherk(G) :=\{[M]\in T(G)|M=f^{-1}(L)for a 1‐dimensional kN_{G}(P) ‐module L }where f is the Green correspondence with respect to (G, P, N_{G}(P))

Proof. See [8, Theorem 1.1].

.

\squareRemark 1.4. The point in Theorem 1.3 is the following. We actuallywould like to compute TT(G) , and it is most likely the same as K(G).

Precisely speaking, K(G)=TT(G) unless P is cyclic or generalized
quaternion. So at least for our purpose this is the case. So our final aim
is recuded from the computation of TT(G) to that of K(G) Now, let
us look at the right hand side. Then, first of all,X(G) is computable (itis nothing but the p ‐part of G/[G, G] where [G, G] is the commutator
subgroup of G Then what about the second term K(\overline{G}) Since wecan assume that H\neq 1
we are able to use inductive argument. So, it.

.

.

,works!
2. OKUYAM \mathrm{A}^{} \mathrm{S} THEOREM

The author would like to introduce a theorem of Okuyama in 1981

[12, Theorem 1], be‐cause the author hopes/believes that Okuyamas theorem
may/shouldwith a proof, which showed up only in Japanese

be useful even for understanding endo‐trivial modules and also endo‐

permutation modules and so on. Who knows?
here makes Brandts resultOkuyamas theorem[1, Theorem \mathrm{B} ] more precise. For a finite

131

dimentional k ‐algebra A and I\subseteq A we denote by \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.

(Okuyama (1981) [12]). Let B be a block algebra of kG

Theorem 2.1

of a finite group G over an algebraically closed field k of character‐istic p>0 (actually, this theorem holds even for an arbitrary finite
dimensional symmetric k ‐algebra B ). Further, let \ell(B) be the numberof non‐isomorphic simple right B ‐modules, and let S_{1},S_{\ell(B)} be the
all non‐isomorphic simple right B ‐modules. Then, it holds\cdotsì\displaystyle \dim_{k}[\mathrm{A}\mathrm{n}\mathrm{n}_{B}(J B)^{2})\cap Z(B)]=\el (B)+\sum_{i=1}^{\el (B)}\dim_{k}[\mathrm{E}\mathrm{x}\mathrm{t}_{B}^{1}(S_{i}, Si)].
Proof. (Okuyama in [12, Theorem 1]). Set \ell :=\ell(B),and let e_{1}, \cdots, e_{\ell}be the set of all primitive idempotents of B such that P_{i} :=P(S_{i}):=e_{i}B is the projective cover of S_{i} for i=1, \cdots, \ell Set e :=e_{1}+\cdots+e_{\ell},.

and A :=eBe , and hence A is the basic ring (algebra) of B , and A
is a finite dimensional symmetric k‐algebra

Morita equivalent).

(recall that A and B are

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. If C is a finite dimensional k ‐algebra such that C/J(C)\cong k,then \mathrm{A}\mathrm{n}\mathrm{n}_{C}(J(C)^{2})\subseteq Z(C).

Proof of Step 2. Follows by [10, Lines 7‐8 of the proof of Lemma 2],

Step

3.

For

any

two‐sided

idealIofB,

wehavethat\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 Morita equivalent.

Step 4. If 0\neq f=f^{2}\in C for a finite‐dimensional symmetric k‐
algebra C then fCf is again symmetric.
Proof of Step 4. Take a similar way to prove that a Morita equiva‐
lence preserves being symmetric, though fCf is not necessarily Morita
equivalent to C of course.
,,

Step 5. Set

T:=\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.

Step 6. \mathrm{A}\mathrm{n}\mathrm{n}_{A}(T)=\mathrm{A}\mathrm{n}\mathrm{n}_{A}[(e\cdot J(A)^{2}\cdot e)\cap Z(A)].

\square132

Acknowledgements. The author would like to thank Professor
Hiroki Shimakura for giving him an opportunity to give a talk in themeeting held in the RIMS of the University of Kyoto, January 2016.
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\grave{}\backslash

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