NADIA MAZZA AND JACQUES TH ´EVENAZ

1. _{Introduction}

The purpose of this note is to determine all endotrivial modules in prime characteristicp, for a finite group having a cyclic Sylow p-subgroup. In other words, we describe completely the group of endotrivial modules in that case.

The endotrivial modules are a family of finitely generated modules for a group algebra of a finite group which appear naturally in modular representation theory. Several contributions towards a general classification have already been obtained (cf. [2], [3], [4], [5], [6], [7] and [8]).

We writeT(G) for the group of endotrival modules for the finite groupG. Assume thatGhas a cyclic Sylow p-subgroup P and let H be the normalizer in G of the unique subgroup of P of orderp. We prove here that induction and restriction induce inverse isomorphismsT(G)∼=T(H) and thatT(H) is generated by one-dimensional representations ofH together with the class of the first syzygy of the trivial module, subject to a single explicit relation.

2. _{Preliminaries}

Throughout this note, we letkbe an algebraically closed field of prime characteristicp. IfGis a finite group, allkG-modules are finitely generated (left) modules and the symbol ⊗stands for the tensor product⊗k of modules over the ground field, with diagonal group action ofG.

Definition 2.1. Let G be a finite group. A kG-module M is endotrivial if its endomorphism algebra EndkM is isomorphic (as kG-module) to the direct sum of the one-dimensional trivial

module and a projective module. Alternatively,M is endotrivial if and only ifM∗⊗M ∼=k⊕(proj), whereM∗ denotes thek-dual ofM.

Here are some of the basic properties of endotrivial modules (cf. [7], [4], or [2]).

Lemma 2.2. Let Gbe a finite group.

(1) If M is an endotrivial kG-module, then M splits as the direct sum M0⊕(proj) for an indecomposable endotrivialkG-moduleM0, unique up to isomorphism. Here (proj) denotes some projective module.

(2) The relation

M ∼N ⇐⇒ M0∼=N0

on the class of endotrivialkG-modules is an equivalence relation. We let T(G) be the set of equivalence classes. Every equivalence class contains a unique indecomposable module up to isomorphism.

(3) The tensor product ⊗ induces an abelian group structure on the set T(G):

[M] + [N] = [M ⊗N] .

Date: May 21, 2007.

The zero element ofT(G)is the class[k]of the trivial module, consisting of all modules of the formk⊕(proj).

The group T(G) is called the group of endotrivial kG-modules. It is known to be a finitely generated abelian group. In particular, the torsion subgroup Tt(G) ofT(G) is finite. The rank ofT(G) can be described explicitly (see [2]). For groups with a cyclic Sylow p-subgroup,T(G) =

Tt_{(}_{G}_{) is a finite group, as we shall see in the main theorem. This also follows immediately from}

the fact that there are finitely many indecomposablekG-modules up to isomorphism. An important family of endotrivial modules is provided by the syzygies Ωn

G(k) of the trivial

module. Recall that Ωn_{G}(k) is the kernel of the (n−1)st differential in a minimal kG-projective
resolution of the trivial modulek.

Notation 2.3. We write ΩG for the class of Ω1G(k) inT(G).

Recall that we havenΩG = [ΩnG(k)] in T(G), for alln∈Z.

For p-groups, it turns out in many cases that the syzygies of the trivial module are the only indecomposable endotrivial modules (up to isomorphism). It is for instance the case in the following result of Dade (cf. [7]).

Theorem 2.4. Let P be an abelianp-group. Then

T(P) =hΩPi ∼=

0 if |P| ≤2,

Z/2Z if P is cyclic and |P| ≥3, Z otherwise.

Notation 2.5. If Gis a finite group, we denote by X(G) the abelian group of all isomorphism classes of one-dimensionalkG-modules (for the group law induced by⊗). This group can also be identified with the group Hom(G, k×) ofk×-valued linear characters of G, where k× =k− {0}. The groupX(G) is ap0-group, isomorphic to thep0-part of the abelianizationG/[G, G].

It is obvious that any one-dimensional module is endotrivial and we letX0(G) be the image of the embedding

X(G)−→T(G), µ7→[µ]

mapping a one-dimensional module to its class inT(G). Note thatX0(G) is isomorphic toX(G), but it is written additively : [λ] + [µ] = [λ·µ].

If H is a subgroup of G, we write H ≤ G. For a kG-module M, we denote by M ↓G H the

restriction ofM tokH, and for akH-moduleN, we writeN↑G

H for the induction ofN tokG. It

is clear that restriction induces a group homomorphism

ResG_{H} :T(G)−→T(H)

and we note that ResG_{H}(ΩG) = ΩH, because Ω1G(k)↓
G
H ∼= Ω

1

H(k)⊕(proj).

IfH, K≤G, we write [G/H] and [K\G/H] for sets of representatives of the leftH−cosets and
of the (K−H)−double cosets inG, respectively. Ifx∈GandH ≤G, we writex_{y}_{and}x_{H} _{for the}

elementxyx−1 and the subgroupxHx−1ofG, respectively.

Lemma 2.6. Let Gbe a finite group and let P be a Sylow p-subgroup of G. Assume that for all

x∈Gthe subgroupx_{P}_{∩}_{P} _{is non trivial. Then, the kernel of the restriction map}_{Res}G

P :T(G)→ T(P)is equal to X0(G).

Proof. AssumeM is an indecomposablekG-module such that M↓G

P =k⊕(proj). By the Mackey

decomposition formula and relative projectivity, we have thatM↓G

P is also a direct summand of

k↑G P↓

G P ∼= k|

NG(P):P|_{⊕}M

x∈A k↑P

whereA={x∈[P\G/P]|x6∈NG(P)}and whereV has no projective summand. Indeed, since

thep-groupxP∩P is nontrivial, each indecomposablekP-modulek↑P

x_{P}_{∩}_{P} appearing above is not

projective and has vertexxP∩P strictly contained inP, for allx6∈NG(P). Now on the one hand M↓G

P =k⊕(proj), and on the other handM↓GP is a direct summand ofk|NG(P):P|⊕V. By the

Krull-Schmidt theorem, this forcesM↓G

P =k. ThusM belongs toX(G).

Note that Lemma 2.6 applies in particular whenOp(G) is nontrivial.

For our next lemma, we recall that a subgroupH of Gisstrongly p-embedded ifpdivides the
order ofH and ifx_{H}_{∩}_{H} _{is a group of order prime to}_{p}_{, for all}_{x}_{∈}_{G}_{−}_{H}_{. Note that any strongly}
p-embedded subgroup of Gcontains the normalizer inGof a Sylowp-subgroup ofG.

Lemma 2.7. LetH be a subgroup of Gcontaining the normalizerN of a Sylow p-subgroup ofG. (1) The restriction map ResGH: T(G)−→ T(H)is injective. More precisely, ifM is an

inde-composable endotrivialkG-module and ifM↓G

H =L⊕(proj) whereLis an indecomposable kH-module, thenM is the Green correspondent ofL.

(2) Assume that H is strongly p-embedded in G. Then ResGH :T(G)−→T(H) is an

isomor-phism. Moreover, the inverse map is induced by induction. On indecomposable endotrivial modules, the inverse map is induced by the Green correspondence.

Proof. The first statement is proven in [2, Proposition 2.6]. Let us give the argument again. SinceH

containsN, the Green correspondence yields a one-to-one correspondence between indecomposable
modules forG andH having a common vertex. The Green correspondent of an indecomposable
endotrivialkH-moduleLis an indecomposablekG-moduleM with vertex a Sylow subgroup. This
moduleM might or might not be endotrivial. If it is, then ResG_{H}([M]) = [L] and the uniqueness
ofM shows the injectivity of the map ResGH.

The second statement is proven in [2, Proposition 2.8]. By (1), the restriction map ResG_{H} is
injective. Now letLbe an indecomposable endotrivialkH-module. The Mackey formula yields

L↑G H↓

G H ∼=

M

x∈[H\G/H]

x_{L}_{↓}xH

x_{H}_{∩}_{H}↑Hx_{H}_{∩}_{H} ∼=L⊕(proj).

Indeed, for anyx∈G−H, the subgroupxH∩H ofH has order prime top, which implies that the

kH-moduleL↓xH

x_{H}_{∩}_{H}↑Hx_{H}_{∩}_{H} is projective. It follows thatL↑G_{H} is an endotrivialkG-module. Thus

thekG-Green correspondent of any indecomposable endotrivialkH-module is an indecomposable

endotrivialkG-module. _{}

3. _{Cyclic Sylow} p_{-subgroups}

In this section, we determine completelyT(G) for a finite groupG having a nontrivial cyclic Sylowp-subgroupP. We letZ be the unique subgroup ofP of orderp, and we letH =NG(Z) be

its normalizer inG. Our first purpose is to work out the structure ofT(H).

The quotientH/CG(Z) embeds as a subgroup of Aut(Z) and is therefore a cyclic groupCe of

ordere, for some integeredividingp−1. Givenc∈H, then for allu∈Z we have

c_{u}_{=}_{u}ν(c)_{,} _{for some}_{ν}_{(}_{c}_{)}_{∈}_{(}

Z/pZ)×,

and we viewν(c) as an element ofk×via the embedding_{Z}/p_{Z}→k. This defines a linear character

ν:H →k× with kernelCG(Z). For simplicity of notation, we identifyν with a one- dimensional kH-module, hence an element ofX(H).

The main ingredient is the following well-known result.

Proposition 3.1. LetH =NG(Z). The projective coverFkof the trivialkH-modulekis uniserial,

Proof. The result is part of the structure of blocks with cyclic defect groups and can be found for instance in [1, Proposition 6.5.4], except for the explicit description of Rad(Fk)/Rad2(Fk) as the

above one-dimensional moduleν.

We sketch here a direct proof. We first show that the group K = CG(Z) has a normal p

-complement. The group NK(P)/CK(P) is ap0- group of automorphisms of P, hence completely

determined by its restriction toZ. ButK=CG(Z) acts trivially onZand thereforeNK(P)/CK(P)

is trivial. ThusNK(P) =CK(P) and Burnside’s transfer theorem applies, proving that K has a

normalp-complementM.

SinceM =Op0(K) andKis normal inH,M is a normal subgroup ofH. We haveK/M∼=Pand

H/M∼=PoCe, becauseH/K=∼CeandP has a complementCeinH/M by the Schur-Zassenhaus

theorem.

The projective cover of kas a k[H/M]-module is also a projective cover ofk as a kH-module becauseMis a normalp0-subgroup. Thus it suffices to prove the result for the groupH/M∼=PoC,

where we writeC=Ce.

Now P acts on P by left multiplication and C acts on P by conjugation. These actions are compatible in the sense that they define an action ofPoConP. ThereforeF =kP has a structure

ofk[PoC]-module (actually a permutation modulek↑PCoC). The unique series of submodules ofF

as akP-module is invariant underC and thereforeF is uniserial of length|P|and dimension|P|.
Explicitly, Radj(F)/Radj+1(F) = (x−1)j_{F/}_{(}_{x}_{−}_{1)}j+1_{F}_{, where} _{x}_{is a generator of}_{P}_{.}

By expanding (x−1) + 1m

, one can see thatxm_{−}_{1}_{≡}_{m}_{(}_{x}_{−}_{1) (mod (}_{x}_{−}_{1)}2_{) and hence}
(xm_{−}_{1)}j _{≡}_{m}j_{(}_{x}_{−}_{1)}j _{(mod (}_{x}_{−}_{1)}j+1_{). Now the conjugation action of}_{c}_{∈}_{C} _{on}_{P} _{is given by}

c_{x}_{=}_{x}ν(c) _{by definition of}_{ν}_{. The congruence above with} _{m}_{=}_{ν}_{(}_{c}_{) shows that the action of} _{c} _{on}

(x−1)jF/(x−1)j+1F is multiplication byν(c)j. Therefore (x−1)jF/(x−1)j+1F ∼=νj. _{}
Theorem 3.2. Let G be a finite group with a cyclic Sylow p-subgroup P. Let Z be the unique
subgroup ofP of orderp, and letH =NG(Z).

(1) There is an exact sequence

0 //X0(H) //T(H) Res

H P //

T(P) //0.

(2) If|P|= 2, thenT(P) ={0} andT(H) =X0(H). (3) If|P|>2, thenT(P) =hΩPi ∼=Z/2Zand

T(H) =hX0(H), ΩHi ∼= X0(H)⊕ hΩHi

([ν]−2ΩH),

where[ν]∈X0(H)is the class of ν∈X(H).

(4) The exact sequence above splits if and only if ν is a square in X(H). In particular, the sequence splits if the integere=|NG(Z)/CG(Z)|is odd.

Proof. SinceP is cyclic, we haveT(P) =hΩPi, by Theorem 2.4. It is trivial if |P|= 2 and has

order 2 otherwise. Moreover, ResH_{P} is surjective because ResH_{P}(ΩH) = ΩP. SinceZ is a non-trivial

normal subgroup ofH, Ker(ResH_{P}) =X0(H) by Lemma 2.6. This proves (1) and (2).

Assume|P|>2. Since ΩH is a preimage of the generator ΩP ofT(P), we need to identify 2ΩH

as an element ofX0(H). We claim that 2ΩH = [ν], from which (3) follows. We need to compute

Ω2

H(k) (since its class in T(H) is 2ΩH). From Proposition 3.1, we see that Ω1H(k) = Rad(Fk)

(with compositions factors ν, ν2, . . . , νp−2, k). Its projective cover is ν ⊗Fk (with compositions

factorsν, ν2, . . . , k, ν) and therefore Ω2_{H}(k) is the bottom composition factor ofν⊗Fk, namelyν,

as claimed.

We are left with the proof of (4). If |P| = 2, then the sequence splits trivially and on the other hand ν is a square because it is trivial (e= 1). Assume now that |P|>2. Since ΩH is a

such that [µ]−ΩH has order 2, or in other words 2[µ] = [ν]. Going back to X(H) and its

multiplicative notation, we see that the condition is that ν must be a square in X(H). For the special case, note thatν is an element of ordere, because it is a generator of the character group ofNG(Z)/CG(Z)∼=Ce. Ifeis odd, thenν is (necessarily) a square.

Remark 3.3. ΩH has order 2einT(H), becauseν has ordereand 2ΩH= [ν].

Remark 3.4. Any element ofT(H) has the form [λ] +jΩH for someλ∈X(H). This is the class

of the endotrivial module Ωj_{H}(λ). Thus we can also describe T(H) as the group whose elements
are the equivalence classes of the syzygies of the one-dimensional modules.

In order to pass from the subgroupH =NG(Z) to the whole groupG, we need the following

observation.

Lemma 3.5. The subgroupH=NG(Z)is stronglyp-embedded inG. Hence,ResGH :T(G)→T(H)

is an isomorphism (with inverse map induced by induction or Green correspondence).

Proof. Let x ∈ G such that pdivides the order of x_{H} _{∩}_{H}_{, that is, there exists a nontrivial} _{p}_{}

-subgroupQofGsuch that Q≤x_{P}_{∩}_{P}_{. Since}_{Z} _{is the unique subgroup of}_{P} _{of order}_{p}_{, we have}

thatZ ≤x_{H}_{∩}_{H}_{. This forces the equality}x_{Z}_{=}_{Z}_{, and hence}_{x}_{∈}_{H}_{, as was to be shown.}

The second claim is a direct application of Lemma 2.7. _{}

Our final result follows immediately from the previous lemma.

Theorem 3.6. Let G be a finite group with a cyclic Sylow p- subgroup P. Let Z be the unique subgroup ofP of orderp, and letH =NG(Z). Then

T(G) ={[IndGHM] | [M]∈T(H)} ∼=T(H).

Alternatively, ifΓ(M)denotes the Green correspondent of an indecomposablekH-moduleM, then

T(G) ={[Γ(M)] |M is an indecomposable endotrivialkH-module} ∼=T(H) .

We see in particular that there exist often indecomposable endotrivialkG-modules which are neither one-dimensional nor syzygies of a one-dimensional module. Indeed the Green correspondent of a one-dimensionalkH-module is in general not one-dimensional.

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Nadia Mazza, Department of Mathematical Sciences, Meston Building, Aberdeen University, Ab-erdeen AB24 3UE, Scotland.

Jacques Th´evenaz, Institut de G´eom´etrie, Alg`ebre et Topologie, EPFL, Bˆatiment BCH, CH-1015 Lausanne, Switzerland.