TORSION ENDO-TRIVIAL MODULES

JON F. CARLSON^{∗} AND JACQUES THÉVENAZ

Abstract. We prove that the group T (G) of endo-trivial modules for a noncyclic finite p-group G is detected on restriction to the family of subgroups which are either elementary abelian of rank 2 or (almost) extraspecial. This result is closely related to the problem of finding the torsion subgroup of T (G). We give the complete structure of T (G) when G is dihedral, semi-dihedral, or quaternion.

1. Introduction

Endo-trivial modules play a crucial role in the representation theory of finite groups. They appear in particular in connection with self-equivalences of the stable category of a block (see [CaRo]). The group of endotrivial modules is a fundamental part of the system of endopermutation modules that are important in local representation theory (e.g. see [Pu2]). Indeed, many questions about endo-permutation modules can be reduced to the special case of endo-trivial modules.

Recall that if G is a finite p-group and if k is a field of characteristic p, then a kG-module M is an endo-trivial module if End_{k}(M) is isomorphic to the direct sum of a trivial module k and a free module. One is particularly interested in indecomposable endo-trivial modules. The group T (G) of endo-trivial kG-modules can be viewed either as a group of suitable equivalence classes of endo-trivial kG-modules (with multiplication induced by tensor product) or as the group of indecomposable endo-trivial kG-modules (because there is one such indecomposable module in each equivalence class, up to isomorphism).

It is known that T (G) is a finitely generated abelian group (Puig [Pu]) and its torsion-free rank has been determined recently by Alperin [Al1]. More than twenty years ago, it was proved by Dade [Da2] that T (G) ≅ ℤ if G is a noncyclic abelian p-group and T (G) ≅ ℤ/2ℤ if G is cyclic of order ≥ 3. Moreover T (G) ≅ ℤ/4ℤ ⊕ ℤ/2ℤ if G is a quaternion 2-group (Dade [Da1]). However, no progress was made on the tantalizing question of describing the torsion subgroup of T (G) for an arbitrary p-group G.

By a fundamental result of Puig [Pu], the restriction map to all elementary abelian subgroups

Res : T (G) ⟶ ∏_{E} T (E)

has a finite kernel. If G is not cyclic and p is odd, it follows that this kernel is precisely the torsion subgroup of T (G). We conjecture that this restriction map is actually injective for odd p, or in other words that the torsion subgroup of T (G) is

Date: January 14, 2000.

* Partly supported by a grant from NSF.

2 JON CARLSON AND JACQUES THÉVENAZ

trivial if G is not cyclic. When p = 2 (and G is not cyclic), we certainly have to enlarge the family of subgroups and include the quaternion group Q_{8} of order 8. We conjecture that the restriction map to this larger family of subgroups is injective.

In view of this, the problem of finding torsion elements in T (G) is now replaced by the question of finding the right class X of groups such that the restriction map to subgroups in X is injective, in which case we say that T (G) is detected on restriction to X. Our first result (Section 3) asserts that T (G) is detected on restriction to the family of groups consisting of elementary abelian groups of rank 2, extraspecial groups, and almost extraspecial groups (see Section 2 for the definitions). Moreover, for odd p, we prove in Section 4 that one can also get rid of all extraspecial groups of exponent p^{2} and all almost extraspecial groups, using a cohomological argument which resembles Chouinard’s theorem. Thus the only remaining problem for proving the conjecture above for p odd lies with the family of extraspecial groups of exponent p. When p = 2, we show that one can get rid of the dihedral group D_{8} of order 8, but the problem remains open for all the other (almost) extraspecial 2-groups (except Q_{8} which has to remain in the detecting family).

When p is odd, the conjecture above asserts that there is no torsion element in T (G), except if G is cyclic. However, when p = 2, there are examples of torsion elements for all quaternion 2-groups and all semi-dihedral 2-groups. In Sections 5, 6, and 7, we actually give explicitly the structure of T (G) for these groups, and also for the dihedral 2-groups in which case there is no torsion.

When G is extraspecial or almost extraspecial, one would like to prove the injectivity of the restriction map to all proper subgroups, in order to settle the conjecture above. Although we are unable to do this in general, we show in Section 8 that there is a very strong condition on an endo-trivial kG-module M in order that the class of M lies in the kernel of the restriction map to all proper subgroups. We also explain in Section 9 why the cohomological argument of Section 4 does not work for extraspecial p-groups of exponent p.

Finally, we show in Section 10 that our result on T (G) implies an injectivity result for the whole Dade group D(G) of endo-permutation modules. We also compute the Dade group for dihedral, semi-dihedral, and quaternion 2-groups.

The first author would like to thank the Alexander von Humboldt Stiftung and the Universität Stuttgart for support during the time when part of the results of this paper were written. He also wants to thank the Université de Lausanne for supporting his visit to Lausanne.

2. Preliminaries and Statement of the Main Result

Let k be a field of characteristic p and let G be a finite p-group. Since kG is a local algebra, every projective kG-module is free. We first recall some basic facts concerning endo-trivial modules. Details can be found in Section 7 of [Da2].

A kG-module M is called endo-trivial if there is an isomorphism of kG-modules

End_{k}(M) ≅ k ⊕ F ,

where k denotes the trivial module and F is a free kG-module. The trivial summand can actually always be chosen to be k· Id_{M}. If M is endo-trivial, then so is M ⊕P for every free kG-module P. Also, if M is endo-trivial and decomposes as M = M_{1}⊕M_{2}, then one of the modules is free and the other one is endo-trivial. This follows from the decomposition

k ⊕ F ≅ End_{k}(M) ≅ End_{k}(M_{1}) ⊕ Hom_{k}(M_{1}, M_{2}) ⊕ Hom_{k}(M_{2}, M_{1}) ⊕ End_{k}(M_{2})

together with the Krull-Schmidt theorem and Higman’s criterion. Consequently, every endo-trivial module M is a direct sum M = M_{0} ⊕ P, where P is free and M_{0} is an indecomposable endo-trivial module, called the cap of M.

Two endo-trivial modules M and N are said to be equivalent if one of them is a direct summand of the other, so that

either M ≅ N ⊕ (free) , or N ≅ M ⊕ (free) .

We shall write [M] for the equivalence class of an endo-trivial module M. In the equivalence class [M] of M, there is, up to isomorphism, a unique indecomposable module M_{0} and every module in the class is isomorphic to M_{0} ⊕ (free). We let T (G) be the set of equivalence classes of endo-trivial kG-modules. We write T (G) = T_{k}(G) when we need to emphasize the base field k (but we note that all known results on T_{k}(G) are independent of k, except for one specific result about the quaternion group Q_{8}).

If M and N are endo-trivial kG-modules, then so are

M ⊗ N = M ⊗_{k} N and M^{∗} = Hom_{k}(M, k)

with their usual kG-module structure. Moreover, the tensor product induces a structure of abelian group on T (G), defined by

[M] + [N] = [M ⊗ N] ,

using an additive notation which is consistent with the choice made in [BoTh]. The zero element is [k] and we have −[M] = [M^{∗}]. We call T (G) the group of endotrivial modules for the p-group G. By a theorem of Puig [Pu] (see below), T (G) is a finitely generated abelian group.

There is another approach to this group as a subgroup of the Dade group D(G), which is the group of equivalence classes of endo-permutation kG-modules. We refer to [Da2], [Th1] or [BoTh] for the definition of the Dade group of G. If H is a subgroup of G and K is a normal subgroup of H, there is a restriction map

Res_{H}^{G} : D(G) ⟶ D(H)

and a deflation map

Def_{H/K}^{H} : D(H) ⟶ D(H/K) .

We define

T′(G) = ⋂_{K} Ker(Def_{N_{G}(K)/K}^{N_{G}(K)} Res_{N_{G}(K)}^{G})

where K runs over all non-trivial subgroups of G. Here N_{G}(K) denotes the normalizer of K in G.

4 JON CARLSON AND JACQUES THÉVENAZ

Lemma 2.1. (Puig [Pu]) Let G be a p-group. The map sending an endo-trivial kG-module M to its class in the Dade group D(G) induces an isomorphism

T (G) ≅ T′(G) .

It should be mentioned that the class of an endo-trivial module M in D(G) usually contains more modules than those belonging to the class of M in T (G), that is, it contains endo-permutation modules which are not endo-trivial. However, in any class belonging to T′(G), the unique indecomposable module in the class is endotrivial.

The main examples of endo-trivial modules are the Heller translates Ω_{G}^{n}(k) of the trivial module k. Recall that Ω_{G}^{n}(k) = Ker(δ_{n−1}), where

… ⟶ D ⟶^{∂_{2}} D ⟶^{∂_{1}} D ⟶^{∂_{0}} k ⟶ 0

is a minimal projective resolution of the trivial module k as a kG-module. In the group T (G), we have

[Ω_{G}^{n}(k)] = n [Ω_{G}^{1}(k)] = n Ω_{G} ,

where Ω_{G} denotes the class of Ω_{G}^{1}(k). The structure of T (G) is known in the abelian case by a theorem of Dade (see also [Ca1]).

Theorem 2.2. (Dade [Da2]) Let G be an abelian p-group.

(a) If G is trivial or cyclic of order 2, then T (G) = 0.

(b) If G is cyclic of order ≥ 3, then T (G) ≅ ℤ/2ℤ, generated by Ω_{G}.

(c) If G is abelian and non-cyclic, then T (G) ≅ ℤ, generated by Ω_{G}.

The next result shows that the torsion-free part of T (G) is detected on restriction to elementary abelian subgroups.

Theorem 2.3. (Puig [Pu]) Let G be a p-group. Then the restriction map

Res : T (G) ⟶ ∏_{E} T (E)

to all elementary abelian subgroups E of G has a finite kernel.

Corollary 2.4. T (G) is finitely generated.

The torsion-free rank of T (G) has been determined recently by Alperin [Al1], using Puig’s theorem and a construction of endo-trivial modules via relative syzygies. Another proof of Alperin’s result appears in [BoTh], using tensor induction.

In this paper, we shall focus on the question of finding more information on the kernel of the restriction map to elementary abelian subgroups. In view of Puig’s theorem 2.3, this kernel consists of torsion elements and this motivates our title, but we shall actually never use Puig’s theorem in this paper. However, for completeness, we emphasize the following result.

Corollary 2.5. If p is odd and G is a non-cyclic p-group, then the torsion subgroup of T (G) coincides with the kernel of the restriction map to all elementary abelian subgroups of G, and also with the kernel of the restriction map to all elementary abelian subgroups of G of rank 2.

Proof. We first prove that both kernels in the statement coincide. We write E_{n} for an elementary abelian subgroup of rank n. If [M] lies in the kernel of the restriction map to all subgroups E_{2} of rank 2, then it also lies in the kernel of Res_{E_{n}}^{G} for n ≥ 3, because Res_{E_{2}}^{E_{n}} on T (E_{n}) is an isomorphism by Dade’s theorem (since the generator Ω_{E_{n}} restricts to the generator Ω_{E_{2}}). Moreover, Res_{E_{1}}^{G}[M] = 0 for any cyclic subgroup E_{1} of order p, because E_{1} is contained in some subgroup E_{2} of rank 2 (since p is odd and G is non-cyclic).

Now if [M] lies in the above kernel, then [M] is a torsion element by Puig’s Theorem 2.3. For the reverse inclusion, we note that it suffices to consider the restriction to maximal elementary abelian subgroups E of G. Since G is non-cyclic and p is odd, any such subgroup E has rank ≥ 2, so that T (E) ≅ ℤ by Dade’s Theorem 2.2. Therefore the restriction to E of any torsion element of T (G) must be zero.

Puig’s theorem 2.3 obviously raises the question of finding the right class of groups such that T (G) is detected on restriction to this class. If G = C_{p^{n}} is cyclic of order p^{n}, then T (G) is detected on restriction to C_{p} if p is odd and C_{4} if p = 2 (by Dade’s theorem 2.2). Otherwise it turns out that cyclic groups are not necessary in the detecting class of groups, so we assume from now on that G is non-cyclic. We conjecture the following.

Conjecture 2.6. Let G be a non-cyclic p-group. Then the restriction map

Res : T (G) ⟶ ∏_{E} T (E)

is injective, where E runs over all subgroups of G which are either elementary abelian of rank 2 or, in addition if p = 2, isomorphic to the quaternion group Q_{8} of order 8.

If G is non-cyclic and p is odd, another form of the conjecture says that the torsion subgroup of T (G) is trivial. This follows immediately from Corollary 2.5. When p = 2, the situation is more complicated since there are known torsion elements in T (G) when G is a quaternion group or a semi-dihedral group (see Sections 6 and 7).

In order to state our main result, we need some terminology. Recall that a p-group G is called extraspecial if Φ(G) has order p and is equal to Φ(G) = [G, G] = Z(G), where Φ(G) is the Frattini subgroup, [G, G] the commutator subgroup, and Z(G) the center of G. In that case G has order p^{2n+1} for some n ≥ 1 and there are two isomorphism classes of extraspecial groups of that order. If p is odd, one group has exponent p and the other one has exponent p^{2}. If p = 2, one group is the central product of n copies of the dihedral group D_{8} of order 8 and the other is the central product of n−1 copies of D_{8} and one copy of Q_{8}. If p is odd, the two extraspecial groups of order p^{2n+1} are presented as follows. In the first case G = ⟨ x_{1}, …, x_{2n}, z ⟩ where

x_{i}^{p} = 1 for 1 ≤ i ≤ 2n,  z^{p} = 1,  [x_{2i−1}, x_{2i}] = z for 1 ≤ i ≤ n ,(R2.1)

and otherwise x_{i} and x_{j} commute. The center of the group is generated by z. The group of exponent p^{2} is given as G = ⟨ x_{1}, …, x_{2n} ⟩ where

x_{i}^{p} = 1 for 1 ≤ i ≤ 2n − 1,  x_{2n}^{p^{2}} = 1,  [x_{2i−1}, x_{2i}] = x_{2n}^{p} for 1 ≤ i ≤ n , (R2.2)

and otherwise x_{i} and x_{j} commute. The center of the group is generated by z = x_{2n}^{p}.

We shall say that G is almost extraspecial if Φ(G) has order p, Φ(G) = [G, G] ≅ C_{p}, and Z(G) is cyclic of order p^{2}. In that case G has order p^{2n+2} for some n ≥ 1 and G is unique up to isomorphism for any such n. The subgroup H generated by all elements of G of order p has order p^{2n+1} and is extraspecial of exponent p if p is odd, respectively a central product of copies of D_{8} if p = 2. The group G is the central product of H and a cyclic group of order p^{2}. If p is odd, it is presented as G = ⟨x_{1}, …, x_{2n+1}⟩, where

x_{i}^{p} = 1 for 1 ≤ i ≤ 2n,  x_{2n+1}^{p^{2}} = 1,  [x_{2i−1}, x_{2i}] = x_{2n+1}^{p}, for 1 ≤ i ≤ n , (R2.3)

and otherwise x_{i} and x_{j} commute. The center of the group is generated by x_{2n+1}.

Our main result provides a step towards Conjecture 2.6 and it takes the following form.

Theorem 2.7. Let G be a non-cyclic p-group. Let Y be the class of p-groups consisting of all elementary abelian groups of rank 2 and all extraspecial p-groups of exponent p if p is odd, and consisting of all elementary abelian groups of rank 2, all almost extraspecial 2-groups and all extraspecial 2-groups except D_{8} if p = 2. Then the restriction map

Res : T (G) ⟶ ∏_{E} T (E)

is injective, where E runs over all subgroups of G belonging to Y.

By an easy induction argument, it is clear that this theorem is equivalent to the following.

Theorem 2.8. Let G be a non-cyclic p-group and let Y be the class of p-groups defined above. If G does not belong to Y, then the restriction map

Res : T (G) ⟶ ∏_{H} T (H)

is injective, where H runs over all proper subgroups of G.

In Section 3, we shall prove this theorem in the case where G is neither extraspecial nor almost extraspecial. When p is odd, the proof for extraspecial p-groups of exponent p^{2} and almost extraspecial p-groups will be given in Section 4. The case of D_{8} is treated in Section 5.

In order to settle Conjecture 2.6, one would need to prove that Theorem 2.8 still holds when G is an extraspecial group of exponent p (for p odd) or when G is an (almost) extraspecial 2-group different from Q_{8}. An additional restrictive condition is given in Section 8.

We end this section with a few useful observations concerning endo-trivial modules.

Lemma 2.9. Let G be a p-group and let M be a kG-module. If the restriction of M to every elementary abelian subgroup of G is endo-trivial, then M is endo-trivial.

Proof. Let P = End_{k}(L)/k· Id_{L}. By assumption, on restriction to every elementary abelian subgroup, End_{k}(L) splits as the direct sum of k· Id_{L} and a free module. Therefore P is free on restriction to elementary abelian subgroups and so P is free by Chouinard’s theorem (see [Ch] or Theorem 5.2.4 in Vol. II of [Be]). It follows that End_{k}(L) ≅ k ⊕ P as required.

Finally, we prove an elementary result about the dimension of endo-trivial modules.

Lemma 2.10. Let G be a p-group of order p^{n} with n ≥ 2 and let M be an endotrivial kG-module.

(a) If p is odd, then dim(M) ≡ ±1 (mod p^{n}).

(b) If p = 2, then dim(M) ≡ ±1 (mod 2^{n−1}).

Proof. We have

dim(M)^{2} = dim(End_{k}(M)) = dim(k ⊕ (free)) ≡ 1 (mod p^{n}) .

Thus dim(M) is a square root of 1 modulo p^{n} and the result follows.

It is clear that the dimension modulo p^{n} is an invariant of the class of M in T (G). Therefore we obtain a dimension homomorphism from T (G) to a cyclic group of order 2 if p is odd, respectively a Klein four group if p = 2.

Note that if p = 2 and dim(M) ≡ 2^{n−1} ± 1 (mod 2^{n}), then the class of M cannot be the restriction of a class of endo-trivial modules for an overgroup of G.

3. Reduction to (Almost) Extraspecial Groups

The purpose of this section is to prove Theorem 2.8 if G is neither extraspecial nor almost extraspecial. For our main problem, this provides a reduction to the (almost) extraspecial case. The explicit statement also includes an elimination of all elementary abelian p-groups of rank ≥ 3. We first need a (well-known) lemma which explains the role played by the (almost) extraspecial groups.

Lemma 3.1. Let G be a non-cyclic p-group such that G/N is elementary abelian for every non-trivial normal subgroup N of G. Then G is either elementary abelian, or extraspecial, or almost extraspecial.

Proof. The result is easy if G is abelian, so we assume [G, G] ≠ 1, hence Φ(G) ≠ 1. Suppose that |Φ(G)| = p, so that [G, G] = Φ(G). Then either Φ(G) = Z(G) and G is extraspecial, or Z(G) strictly contains Φ(G) and G must be almost extraspecial. Indeed Z(G)/Φ(G) is elementary abelian and any non-trivial subgroup of Z(G) contains Φ(G) by our assumption, so that Z(G) must be cyclic of order p^{2}. Suppose now that Φ(G) has order at least p^{2}. Since the intersection of Φ(G) with the center of G is non-trivial (because G is a p-group), we can choose a central subgroup Z of order p contained in Φ(G). As Z is a proper subgroup of Φ(G) (because |Φ(G)| ≥ p^{2}), G/Z is not elementary abelian, contrary to our assumption.

Theorem 3.2. Let G be a non-cyclic p-group. Assume that G is neither elementary abelian of rank 2, nor extraspecial, nor almost extraspecial. Then the restriction map

Res : T (G) ⟶ ∏_{H} T (H) ,

where H runs over all proper subgroups of G, is injective.

Proof. If G is elementary abelian of rank ≥ 3, then T (G) ≅ ℤ by Dade’s Theorem 2.2 and the restriction map

Res_{H}^{G} : T (G) ⟶ T (H)

is an isomorphism for any maximal subgroup H of G (because the generator Ω_{G} restricts to the generator Ω_{H}). Thus we can assume that G is not elementary abelian. By Lemma 3.1 and our assumption, there exists a normal subgroup Z such that G/Z is not elementary abelian. This will allow us to use Chouinard’s Theorem for the group G/Z. (Since the intersection of Z with the center of G is non-trivial, we could assume that Z is central of order p, but this does not play any role.)

Now let M be an endo-trivial kG-module whose class in T (G) is in the kernel of the restriction to proper subgroups. This means that

M↓_{H}^{G} ≅ k ⊕ (free)

for every proper subgroup H of G. We have to prove that M ≅ k ⊕ (free). Consider the subspace M^{Z} of fixed points under the action of Z and its subspace M_{1}^{Z} consisting of all relative traces

t_{1}^{Z}(m) = ∑_{z∈Z} z · m , m ∈ M .

It is clear that M^{Z} and M_{1}^{Z} are k[G/Z]-modules.

Let H be a maximal subgroup of G containing Z. Then M↓_{H}^{G} = k ⊕ F where F is free and so

M^{Z} = k ⊕ F^{Z} and M_{1}^{Z} = F_{1}^{Z} = F^{Z} ,

the last equality being a consequence of the fact that F is also free on restriction to Z. Since F is a free kH-module, F^{Z} is a free k[H/Z]-module (with a basis obtained by applying t_{1}^{Z} to a basis of F). Therefore M_{1}^{Z} is a free k[H/Z]-module, for every maximal subgroup H of G containing Z. Using the fact that G/Z is not elementary abelian, we see by Chouinard’s theorem (see [Ch] or Theorem 5.2.4 in Vol. II of [Be]), that M_{1}^{Z} is a free k[G/Z]-module. This, with the following lemma, completes the proof of the theorem.

Lemma 3.3. Let Z be a normal subgroup of a p-group G and let M be a kG-module such that M_{1}^{Z} is free as k[G/Z]-module and M↓_{Z}^{G} ≅ k ⊕(free). Then M ≅ k ⊕ (free) as kG-module.

Proof. Let t_{1}^{Z}(m_{1}), …, t_{1}^{Z}(m_{n}) be a basis of M_{1}^{Z} as a k[G/Z]-module. Then

t_{Z}^{G}(t_{1}^{Z}(m_{1})) = t_{1}^{G}(m_{1}), …, t_{Z}^{G}(t_{1}^{Z}(m_{n})) = t_{1}^{G}(m_{n})

form a basis of (M_{1}^{Z})^{G/Z}. Since they are linearly independent, this forces m_{1}, …, m_{n} to generate a free kG-submodule L of M. Now we have that

L^{Z} = L_{1}^{Z} = ⟨ t_{1}^{Z}(m_{1}), …, t_{1}^{Z}(m_{n}) ⟩ = M_{1}^{Z} = F^{Z} ,

where F is a free kZ-module such that M↓_{Z}^{G} = k⊕F. Therefore dim(L) = dim(F) = n|G| since both L and F are free kZ-modules. Thus dim(L) = dim(M) − 1 and since L is free, hence injective, as a kG-module, we get that M ≅ L ⊕ k, as was to be shown.

4. (Almost) Extraspecial Groups of Exponent p^{2} (p odd)

In this section we wish to show that, if p is odd, the group of endo-trivial modules for the group algebra of an extraspecial or almost extraspecial group G of exponent p^{2} is detected on restriction to the proper subgroups of G. This will complete the proof of Theorem 2.8 for odd p. The technique is to look at a quotient algebra of the group algebra and prove a theorem similar to that of Chouinard [Ch] for the quotient algebra. Specifically we show that the projectivity of modules for the quotient algebra is detected on a certain subalgebra of the quotient. The result may have some independent interest. Unfortunately, the same theorem is not true for the corresponding quotients algebras of group algebras of extraspecial groups of exponent p. More discussion of this is given in Section 9.

We assume throughout the section that p is an odd prime. Let G be either an extraspecial group of exponent p^{2} or an almost extraspecial group. We have that G = ⟨ x_{1}, …, x_{2n} ⟩ in the first case and G = ⟨ x_{1}, …, x_{2n+1} ⟩ in the second, with relations specified as in Section 2 (R2.1, R2.2 and R2.3). For convenience we let z = x_{2n}^{p} in the first case and z = x_{2n+1}^{p} in the second, so that z is a generator for the unique central subgroup of order p in both cases.

In both cases let H denote the subgroup of G generated by the elements of order p. So if G is extraspecial then H = ⟨x_{1}, …, x_{2n−1}⟩ is the direct product of an extraspecial group of order p^{2n−1} with a cyclic group of order p. In the almost extraspecial case H = ⟨x_{1}, …, x_{2n}⟩ is an extraspecial group of order p^{2n+1}. In both cases H has exponent p and if x is any element not in H then x^{p} ≠ 1.

Now let A = kG, B = kG/kG(z − 1)^{p−1} and let C be the algebra

C = k(G/⟨z⟩) ≅ A/A(z − 1) ≅ B/B(z − 1) .

Note here that we are identifying z with its image in B when appropriate. All three of the algebras A, B and C are local k-algebras and projective modules are free modules for each. Now let (P_{∗}, ϵ) be an A-projective resolution of k, (Q_{∗}, ϵ̂) be a B-projective resolution of k and (R_{∗}, δ) a C-projective resolution of k. We may regard (Q_{∗}, ϵ̂) as a complex of A-modules and hence we have a chain map ψ

… ⟶ P_{2}  ⟶^{∂_{2}} P_{1}  ⟶^{∂_{1}} P_{0}  ⟶^{ϵ} k ⟶ 0
↓ψ_{2}  ↓ψ_{1}  ↓ψ_{0}  ∥
… ⟶ Q_{2}  ⟶^{∂_{2}} Q_{1}  ⟶^{∂_{1}} Q_{0}  ⟶^{ϵ̂} k ⟶ 0

10 JON CARLSON AND JACQUES THÉVENAZ

The double complex ((R ⊗ P)_{∗}, δ ⊗ ϵ) is also an A-projective resolution of k. The tensor product here is taken over the base field k. Similarly, we claim that the double complex ((R ⊗ Q)_{∗}, δ ⊗ ϵ̂) is also a B-projective resolution of k. First of all, B is not a Hopf algebra and hence there is no natural way to define an action of B on the tensor product of two arbitrary B-modules. On the other hand, the free C-module C is naturally a B-module (and an A-module) on which z acts trivially. Hence if we take the tensor product C ⊗_{k} B and define the structure as A-modules, then the result is a B-module because (z −1)^{p−1}(C ⊗B) = C ⊗((z −1)^{p−1}B) = 0. Also there is the issue of projectivity. This we state as a lemma.

Lemma 4.1. As a B-module, C ⊗ B ≅ B^{dim(C)}, the direct sum of dim(C) copies of B. In particular, the tensor product of a projective C-module and a projective B-module is a projective B-module.

Proof. Note that (z − 1)(C ⊗ B) = C ⊗ ((z − 1)B) is in the radical of C ⊗ B. But then

(C ⊗ B)/(z − 1)(C ⊗ B) ≅ C ⊗ (B/(z − 1)B) ≅ C ⊗ C ≅ C^{dim(C)} .

So C ⊗ B is generated by dim(C) elements and its projective cover as a B-module is B^{dim(C)}. That is, we have a surjective homomorphism B^{dim(C)} ⟶ C ⊗ B. However, because the dimensions of these two modules are the same, the homomorphism is an isomorphism.

Now we return to the double complexes. The chain map

1 ⊗ ψ : (R ⊗ P)_{∗} ⟶ (R ⊗ Q)_{∗}

is a chain map of A-modules and lifts the identity on k. Applying the functors Hom_{A}( , k) and Hom_{B}( , k), we get natural spectral sequences whose E_{0} terms are

E_{0}^{r,s} = Hom_{A}(R_{r} ⊗ P_{s}, k) and Ê_{0}^{r,s} = Hom_{B}(R_{r} ⊗ Q_{s}, k)

and which converge to Ext_{A}(k, k) and Ext_{B}(k, k), respectively. Moreover the map 1⊗ψ : (R⊗P)_{∗} ⟶ (R⊗Q)_{∗} induces a map ψ̃ : Ê_{∗}^{∗,∗} ⟶ E_{∗}^{∗,∗} of spectral sequences.

Our next task is to analyze the spectral sequences and the map ψ̃. The first spectral sequence is well known. That is, we have the E_{0} term of the classical Lyndon-Hochschild-Serre (= LHS) spectral sequence of the group extension of G/⟨z⟩ by ⟨z⟩. Its E_{2} term has the familiar form

E_{2}^{r,s} = H^{r}(G/⟨z⟩, H^{s}(⟨z⟩, k)) ≅ Ext_{k(G/⟨z⟩)}^{r}(k, Ext_{⟨z⟩}^{s}(k, k)).

The other spectral sequence needs some work.

Let Z = z − 1 and let D be the subalgebra of B generated by 1 and Z. Then D ≅ k[Z]/(Z^{p−1}) is a truncated polynomial ring. As in the case of group algebras, we have an isomorphism

Hom_{B}(R_{r} ⊗ Q_{s}, k) ≅ Hom_{C}(R_{r}, Hom_{D}(Q_{s}, k))

given by f ↦ f′ where f′(u)(v) = f(u ⊗ v) for all u ∈ R_{r} and v ∈ Q_{s}. We leave it to the reader to check that the map is well defined and is an isomorphism. We also leave the exercise of showing that the isomorphism is natural. The naturality implies that we really have an isomorphism of double complexes

Hom_{B}(R_{∗} ⊗ Q_{∗}, k) ≅ Hom_{C}(R_{∗}, Hom_{D}(Q_{∗}, k)) ,

and a map of the E_{0} terms of the spectral sequences

Hom_{C}(R_{∗}, Hom_{D}(Q_{∗}, k)) ⟶^{1⊗ψ^{∗}} Hom_{C}(R_{∗}, Hom_{⟨z⟩}(P_{∗}, k)) .

This point is crucial to our arguments.

To get the E_{1} page of the spectral sequence we take the homology with respect to the boundary map on Q_{∗}. Notice that Hom_{C}(R_{r}, ) is an exact functor since R_{r} is projective. Also Q_{∗} is a D-projective resolution of the trivial module k, because B is a free D-module (with basis consisting of cosets representatives of G/⟨z⟩). Therefore the homology with respect to the differential induced by the boundary on Q_{∗} is

Ê_{1}^{r,s} = Hom_{C}(R_{r}, Ext_{D}^{s}(k, k)) .

Then the E_{2} page is obtained by taking the homology with respect to the differential induced by the boundary map on R_{∗}. As R_{∗} is a C-projective resolution of k, we get

Ê_{2}^{r,s} = Ext_{C}^{r}(k, Ext_{D}^{s}(k, k)) .

Thus we have the following.

Proposition 4.2. There is a spectral sequence whose E_{2} page is

Ê_{2}^{r,s} = Ext_{C}^{r}(k, Ext_{D}^{s}(k, k)) ⇒ Ext_{B}^{r+s}(k, k) ,

and a homomorphism to the LHS spectral sequence of the group extension

Ê_{2}^{r,s} ⟶^{μ} E_{2}^{r,s} = Ext_{C}^{r}(k, Ext_{k⟨z⟩}^{s}(k, k))

induced by the map on homology

Ext_{D}^{∗}(k, k) ⟶ Ext_{k⟨z⟩}^{∗}(k, k)

which in turn arises from the ring homomorphism k⟨z⟩ ⟶ D.

Of course, the map on spectral sequences also comes from the homomorphism on double complexes induced by ψ.

The minimal projective resolution of k as a D-module has the form

… ⟶ D ⟶^{∂_{2}} D ⟶^{∂_{1}} D ⟶^{ϵ} k ⟶ 0

where ∂_{i} is multiplication by Z if i is odd and multiplication by Z^{p−2} if i is even. So Ext_{D}^{∗}(k, k) has dimension 1 in every degree. In degree one, the nonzero class θ can be taken as represented by the extension (exact sequence)

0 ⟶ k ⟶ D/Z^{2}D ⟶ k ⟶ 0 .

The thing to notice is that it can also be viewed as a sequence of k⟨z⟩-modules and the sequence represents the nonzero class in Ext_{k⟨z⟩}^{1}(k, k). So now let θ̃ be the element of Ê_{2}^{0,1} ≅ Hom_{C}(k, Ext_{D}^{1}(k, k)) that sends 1 to θ. Then we have that μ(θ̃) ∈ E_{2}^{0,1} is not zero.

12 JON CARLSON AND JACQUES THÉVENAZ

Now recall that C = k(G/⟨z⟩) is the group algebra of an elementary abelian p-group. So

Ext_{C}^{∗}(k, k) ≅ k[ζ_{1}, …, ζ_{m}] ⊗ Λ(η_{1}, …, η_{m})

is the tensor product of a polynomial ring in elements ζ_{i} of degree 2 with an exterior algebra, Λ, generated by elements in degree 1. Here m is 2n or 2n + 1 depending on whether G is extraspecial or almost extraspecial. Moreover we can choose ζ_{m} and η_{m} to be inflated from the generators of the cohomology H^{∗}(G/H, k) of the cyclic group G/H of order p. If k ≅ ℤ/pℤ is the prime field, then in the LHS spectral sequence the differential on the E_{2} page maps E_{2}^{0,1} to E_{2}^{2,0} ≅ Ext_{C}^{2}(k, k) ≅ H^{2}(G/⟨z⟩, k) and sends the nonzero class in E_{2}^{0,1} to the class of the group extension

0 ⟶ ℤ/pℤ ⟶ G ⟶ G/⟨z⟩ ⟶ 1 .

If k is an extension of the prime field, then the class in E_{2}^{0,1} is mapped to a nonzero multiple of the extension class in H^{2}(G/⟨z⟩, k) ≅ k ⊗ H^{2}(G/⟨z⟩, ℤ/pℤ). In the case in question we know exactly what the extension class is. Given the correct choice of the variables it can be taken to be

ξ = ∑_{i=1}^{n} η_{2i−1}η_{2i} + ζ_{m} ∈ Ext_{C}^{2}(k, k) ,

using our choice of ζ_{m} made above (see Section 10 of [BeCa] for details).

Thus d_{2}(μ(θ̃)) = inf_{1}(ξ) where inf_{1} is the inclusion map of Ext_{C}^{∗}(k, k) as bottom row of the E_{2} page of the spectral sequence. It follows that μ(d′_{2}(θ̃)) = inf_{1}(ξ) where d′_{2} is the differential on Ê_{2}^{∗,∗}. Moreover μ is an isomorphism on the edge Ê_{2}^{∗,0}. Therefore, d′_{2}(θ̃)) = inf_{2}(ξ) where inf_{2} is the inclusion map of Ext_{C}^{∗}(k, k) as bottom row of Ê_{2}. In particular, if inf : Ext_{C}^{∗}(k, k) ⟶ Ext_{B}^{∗}(k, k) is the inflation map, then inf(ξ) = 0 and, more importantly,

inf(ζ_{m}^{p}) = inf(ξ^{p}) = 0,

because (η_{2i−1}η_{2i})^{p} = 0 in the exterior algebra Λ(η_{1}, …, η_{m}). We use this to prove the following. For notation, let U ⊆ B be the subalgbra U = kH/kH(z − 1)^{p−1}.

Lemma 4.3. Suppose that M is a B-module with the property that its restriction M↓_{U}^{B} is a free U-module. Then there exists an exact sequence of B-modules

E : 0 ⟶ M ⟶ W_{2p−1} ⟶ … ⟶ W_{0} ⟶ M ⟶ 0

such that every B-module W_{i} is free and the class of the sequence E in Ext_{B}^{2p}(M, M) is zero.

Proof. What we need to notice is that ζ_{m} ∈ Ext_{C}^{2}(k, k) is represented by the sequence

0 ⟶ k ⟶ k↑\overline{G} ; \overline{H} ⟶^{(x_{m}−1)} k↑\overline{G} ; \overline{H} ⟶ k ⟶ 0

where \overline{G} = G/⟨z⟩, \overline{H} = H/⟨z⟩ and the middle map is multiplication by x_{m} − 1. Its inflation to an A-sequence is the same except for the cosmetic change that we replace the induced module k↑\overline{G} ; \overline{H} by its inflation, the induced module k↑_{H}^{G}. This is a sequence of B-modules because z acts trivially on all of the modules. Moreover if we splice the sequence together p times we get a sequence

E′ : 0 ⟶ k ⟶ k↑_{H}^{G} ⟶ … ⟶ k↑_{H}^{G} ⟶ k ⟶ 0

that represents the inflation of ζ_{m}^{p} and hence the zero class in Ext_{B}^{2p}(k, k).

The sequence E in the lemma can be taken to be M ⊗ E′, whose class is zero in Ext_{B}^{2p}(M, M). Clearly the ends of the sequence are isomorphic to M. It needs only be checked that M ⊗ k↑_{H}^{G} is a free B-module. We should remember that the B-action on the tensor product is inherited from the action of A = kG. The tensor product is a B-module only because (z − 1)^{p−1} annihilates it. As an A-module, M ⊗ k↑_{H}^{G} ≅ M↓_{H}^{G}↑_{H}^{G} by Frobenius reciprocity. But M↓_{H}^{G} is a free U-module, that is, a direct sum of copies of U = kH/(z − 1)^{p−1}kH. So we need only verify that U↑_{H}^{G} ≅ B. But it is clear that kG ⊗_{kH} kH/kH(z − 1)^{p−1} ≅ kG/kG(z − 1)^{p−1}.

We are now prepared to prove the main theorem of this section.

Theorem 4.4. Let B = kG/kG(z − 1)^{p−1} and U = kH/kH(z − 1)^{p−1} ⊆ B. If M is a B-module and if M↓_{U}^{B} is a free U-module, then M is free as a B-module.

Proof. The proof follows precisely the lines of the proof of the main theorem of [Ca2]. For that reason we will present only a sketch of it here. The idea is the following. Let W_{∗} denote the complex

W_{∗} : 0 ⟶ W_{2p−1} ⟶ … ⟶ W_{0} ⟶ 0

obtained by truncating the ends of the exact sequence of the last lemma. Hence the homology of W_{∗} is given by H_{s}(W_{∗}) ≅ M if s = 0 or s = 2p − 1. Otherwise H_{s}(W_{∗}) = 0. Let (P_{∗}, ϵ) be a minimal projective resolution of the complex W_{∗}. If N is the cokernel of ∂_{2p+1}, then the sequence

… ⟶ P_{2p+2} ⟶^{∂_{2p+2}} P_{2p+1} ⟶^{∂_{2p+1}} P_{2p} ⟶ N ⟶ 0

is exact.

As in the proof of the main theorem of [Ca2] we need only note two things about the module N. First of all, N is free as a B-module, because every W_{i} is a projective B-module and every projective B-module is free. Secondly, because the sequence of Lemma 4.3 represents the zero element in Ext_{B}^{2p}(M, M), we have that

N ≅ M ⊕ Ω^{2p−1}(M) ⊕ Q

for some free module Q. Now by the Krull-Schmidt Theorem we must have that M is a free B-module.

Corollary 4.5. Suppose that M is a kG-module with the property that M↓_{H}^{G} ≅ k⊕Q for some free kH-module Q. The M ≅ k ⊕ P where P is a free kG-module.

Proof. Let N = (z − 1)M and notice first that N is a B-module, where B and U are as in the theorem. We claim that the restriction N↓_{U}^{B} of N to U is a free U-module. The reason is that N↓_{U}^{B} = (z − 1)Q and N↓_{U}^{B} ≅ Q/(z − 1)^{p−1}Q is a free U-module. So by Theorem 4.4, N is free as a B-module. Consequently, we have also that (z −1)^{p−1}M = (z −1)^{p−2}N is free as a k[G/⟨z⟩]-module. Thus the assumptions of Lemma 3.3 are satisfied (because we also have M↓_{Z}^{G} ≅ k ⊕ (free)) and the result follows from this lemma.

Corollary 4.5 completes the proof of Theorem 2.8 for p odd.

5. Dihedral 2-Groups

The first purpose of this section is to prove Theorem 2.8 for the dihedral group D_{8} of order 8. In view of the results of the previous two sections, this will complete the proof of Theorem 2.8. Then we will describe explicitly the group of endo-trivial modules for dihedral 2-groups. The result has been known for some years, but the proof that we give here is somewhat different. The description of the whole group of endo-permutation modules in these cases is postponed until Section 10.

Let D_{8} = ⟨ x, y | x^{2} = y^{2} = (xy)^{4} = 1 ⟩ and let a = x + 1, b = y + 1 in the group algebra kD_{8}, where k is a field of characteristic 2. Then a^{2} = b^{2} = 0 and kD_{8} has a basis { 1, a, b, ab, ba, aba, bab, abab }. Note that abab = baba = ∑_{g∈D_{8}} g, so that if M is a kD_{8}-module and abab·m ≠ 0 for some m ∈ M, then m generates a free submodule of M.

Let H = ⟨ x, (xy)^{2} ⟩ and H′ = ⟨ y, (xy)^{2} ⟩ be the two Klein four-groups in D_{8}. The kD_{8}-module with basis {v_{1}, v_{2}} such that b·v_{1} = v_{2} and a·v_{1} = a·v_{2} = b·v_{2} = 0 is isomorphic to k↑_{H}^{D_{8}} because H acts trivially and ⟨ y ⟩ acts freely. Similarly we get k↑_{H′}^{D_{8}} if we exchange the role of a and b.

The following lemma provides an explicit form of the methods of [Ca2] for the dihedral group.

Lemma 5.1. The kD_{8}-module V = kD_{8} ⊕ k ⊕ k has a filtration

0 = V_{0} ⊂ V_{1} ⊂ V_{2} ⊂ V_{3} ⊂ V_{4} ⊂ V_{5} = V

such that each quotient V_{i}/V_{i−1} is isomorphic to either k↑_{H}^{D_{8}} or k↑_{H′}^{D_{8}}.

Proof. Let u be a generator of the free submodule kD_{8} and let v and w be generators of the two trivial summands. We define

\begin{matrix}V_{1} = ⟨ bab·u+v ⟩ , & V_{2} = V_{1} + ⟨ ab·u+w ⟩ , \\ V_{3} = V_{2} + ⟨ b·u ⟩ , & V_{4} = V_{3} + ⟨ ba·u ⟩ . \end{matrix}

We have a·(bab·u + v) = abab·u, but b·(bab·u + v) = 0 and therefore V_{1} ≅ k↑_{H′}^{D_{8}}. Similarly, one can check that the other quotients of the filtration have dimension 2 and are killed by one of a or b but not by the other. The result follows easily from this.

Lemma 5.2. Let M be a kD_{8}-module such that M↓_{H}^{D_{8}} and M↓_{H′}^{D_{8}} both have a nonzero free submodule. Then dim(M) ≥ 6.

Proof. If abab·M ≠ 0, then M has a free submodule and so dim(M) ≥ 8. So we assume now that M is killed by abab = baba. A direct computation shows that

∑_{h∈H} h = baba + aba and  ∑_{h∈H′} h = abab + bab .

Therefore, since (∑_{h∈H} h)M ≠ 0 and (∑_{h∈H′} h)M ≠ 0 by our assumption, we obtain aba·M ≠ 0 and bab·M ≠ 0. Let u, v ∈ M such that aba·u ≠ 0 and bab·v ≠ 0. We claim that u, a·u, ba·u, aba·u, b·v, ab·v are linearly independent, so that dim(M) ≥ 6.

Now consider a linear combination with coefficients in k

λ_{1}u + λ_{a}a·u + λ_{ba}ba·u + λ_{aba}aba·u + λ_{b}b·v + λ_{ab}ab·v = 0 .

Multiplying by aba, one gets λ_{1}aba·u = 0, hence λ_{1} = 0. Similarly, multiplying by ab or ba, one obtains λ_{a} = 0 and λ_{b} = 0, and multiplying next by a or b, one gets λ_{ab} = λ_{ba} = 0. We are left with λ_{aba}aba·u = 0 and so λ_{aba} = 0.

Theorem 5.3. Let M be a kD_{8}-module such that M↓_{H}^{D_{8}} ≅ k ⊕ (free) and M↓_{H′}^{D_{8}} ≅ k ⊕ (free). Then M ≅ k ⊕ (free).

Proof. We can assume that M is indecomposable and we have to prove that M ≅ k. Let V be as in Lemma 5.1 and consider M ⊗ V ≅ M ⊕ M ⊕ (free). By Lemma 5.1, M ⊗ V has a 5-term filtration with quotients isomorphic to

M ⊗ k↑_{Q}^{D_{8}} ≅ M↓_{Q}^{D_{8}}↑_{Q}^{D_{8}} ≅ (k ⊕ (free))↑_{Q}^{D_{8}} ≅ k↑_{Q}^{D_{8}} ⊕ (free) ,

where Q is either H or H′. Since every free module is injective, all the free modules of these quotients split off and we are left with a 10-dimensional module having a 5-term filtration with quotients isomorphic to either k↑_{H}^{D_{8}} or k↑_{H′}^{D_{8}}. Since M is a nonfree indecomposable module, M ⊕M is still a direct summand of this 10-dimensional module and so dim(M) ≤ 5. This implies that M ≅ k, otherwise both M↓_{H}^{D_{8}} and M↓_{H′}^{D_{8}} have a non-zero free submodule and we can apply Lemma 5.2 to conclude that dim(M) ≥ 6. The contradiction proves the theorem.

Now we can deduce the precise structure of the group T (D_{8}) and show in particular that is has no torsion. We do this more generally for the dihedral group D_{2^{n}} generated by two involutions x and y such that (xy)^{2^{n−1}} = 1 (where n ≥ 3). The results are reasonably well known but not reasonably recorded. Following Alperin, we first construct a specific endo-trivial module. Let L be the relative syzygy of the trivial module with respect to the subgroup ⟨ y ⟩, that is,

0 ⟶ L ⟶ k[D_{2^{n}}/⟨ y ⟩] ⟶ k ⟶ 0 .

By a general result of Alperin, L is an endo-permutation module (see [Al1] or [Bo]), and in fact in this case it is an endo-trivial module. This can be seen directly by applying Lemma 2.9 because we are going to see that the restriction of L to each Klein four-group is endo-trivial. More precisely we have

L↓_{H}^{D_{2^{n}}} ≅ Ω_{H}^{1}(k) ⊕ (free) and L↓_{H′}^{D_{2^{n}}} ≅ Ω_{H′}^{−1}(k) ⊕ (free) ,

where H and H′ are representatives of the two conjugacy classes of Klein fourgroups in D_{2^{n}}. As before, we choose H = ⟨ x, z ⟩ and H′ = ⟨ y, z ⟩, where z = (xy)^{2^{n−2}} is the central involution of D_{2^{n}}. The first isomorphism above is clear because k[D_{2^{n}}/⟨ y ⟩] ↓_{H}^{D_{2^{n}}} is free. For the second isomorphism, we first note that

k[D_{2^{n}}/⟨ y ⟩] ↓_{H′}^{D_{2^{n}}} ≅ k[H′/⟨ y ⟩] ⊕ k[H′/⟨ zy ⟩] ⊕ (free) .

16 JON CARLSON AND JACQUES THÉVENAZ

On the other hand, there is an exact sequence

0 ⟶ k ⟶ kH′ ⟶ k[H′/⟨ y ⟩] ⊕ k[H′/⟨ zy ⟩] ⟶ k ⟶ 0

which shows that the kernel of the right hand side surjection is isomorphic to Ω_{H′}^{−1}(k). By adding a suitable free module to the two middle terms of this sequence, the right hand side surjection becomes the restriction to H′ of the map k[D_{2^{n}}/⟨ y ⟩] → k, and so L↓_{H′}^{D_{2^{n}}} ≅ Ω_{H′}^{−1}(k) ⊕ (free).

Theorem 5.4. If n ≥ 3, the group T (D_{2^{n}}) is a free abelian group of rank 2 generated by Ω_{D_{2^{n}}} and the class of the above module L.

Proof. First recall that T (H) ≅ ℤ, generated by Ω_{H}, and similarly for H′ (see Theorem 2.2). Consider the restriction map

T (D_{2^{n}}) ⟶ T (H) ⊕ T (H′) ≅ ℤ ⊕ ℤ .

This is injective by our main theorem 2.8, because D_{2^{n}} has only H and H′ as elementary subgroups of rank 2 (up to conjugacy) and the only extraspecial subgroups are isomorphic to D_{8}, which has been removed from our detecting family. Moreover, the image of this restriction map is contained in the “even” part of T (H) ⊕ T (H′) generated by (Ω_{H}, Ω_{H′}) and (Ω_{H}, −Ω_{H′}). The reason for this is that we have dim(Ω_{H}^{n}(k)) ≡ (−1)^{n} (mod 4), while the dimension of an endo-trivial kD_{2^{n}}-module M is ±1 (mod 4) (by Lemma 2.10) and the signs on both restrictions to H and H′ must be equal.

Now Ω_{D_{2^{n}}} restricts to (Ω_{H}, Ω_{H′}) and [L] restricts to (Ω_{H}, −Ω_{H′}) by the computation above. Therefore the restriction map is surjective onto the even part of T (H) ⊕ T (H′).

The same result can be obtained using another construction of an endo-trivial kD_{2^{n}}-module instead of L. By the method of Theorem 5.2 of [BoTh], one can prove that the tensor-induced module Ten_{H}^{D_{2^{n}}}(Ω_{H}^{1}(k)) is an endo-permutation which belongs to the class of an indecomposable endo-trivial module M, and the restrictions of [M] to H and H′ are 2Ω_{H} and 0 respectively. Thus Ω_{D_{2^{n}}} and [M] also form a basis of T (D_{2^{n}}). Actually dim(L) = 2^{n−1}−1, dim(M) = 2^{n−1}+1, and M ≅ Ω_{D_{2^{n}}}^{1}(L) because both modules have the same restrictions to H and H′.

6. Quaternion 2-Groups

We describe in this section the group of endo-trivial modules for the generalized quaternion 2-group Q_{2^{n}} of order 2^{n} (but we postpone to Section 10 the description of the whole Dade group). The result is actually due to Dade [Da1] but is not explicitly stated in terms of endo-trivial modules, so that some work is needed in order to restate it as a theorem about T (Q_{2^{n}}). Instead of doing this, we give a completely different proof using methods similar to those of the previous section.

We start with the quaternion group

Q_{8} = ⟨ x, y | x^{2} = y^{2} = (xy)^{2}, x^{4} = 1 ⟩

of order 8. Let k be a field of characteristic 2. If we let X = x + 1 and Y = y + 1 then it is easy to see that the group algebra kQ_{8} = k⟨X, Y ⟩/I where I is the ideal generated by the relations X^{2} = Y^{2}, X^{4} = 0 and Y X = X^{2}+XY +X^{3}+X^{2}Y +X^{3}Y. Then the first two terms of a projective resolution of k have the form

… ⟶ kQ_{8}^{2} ⟶^{∂_{1}} kQ_{8} ⟶^{ϵ} k ⟶ 0.

The kernel of ϵ is generated by X and Y, and ∂_{1}(u, v) = uX + vY for u, v ∈ kQ_{8}. From this we can get that Ω_{Q_{8}}^{2}(k) = Ker(∂_{1}) is the submodule of kQ_{8}^{2} generated by the elements

a = (X, Y) and b = (Y, X + Y + Y X) .

It can be further checked that the two elements satisfy the relation

(Y + X^{2} + XY + X^{3})a + (X + Y + XY)b = 0.

Notice that X +Y +XY = 1+xy. With this information we can prove the following.

Lemma 6.1. The module V = Ω_{Q_{8}}^{2}(k) ⊕ k has a filtration

0 ⊆ V_{4} ⊆ V_{3} ⊆ V_{2} ⊆ V_{1} ⊆ V

where V/V_{1} ≅ k↑_{⟨xy⟩}^{Q_{8}}, where V_{i}/V_{i+1} ≅ k↑_{⟨x⟩}^{Q_{8}}, for i = 1, 2, 3 and V_{4} ≅ k↑_{⟨y⟩}^{Q_{8}}.

Proof. Let m denote the generator of the direct summand k of V. So we can assume that V is generated by a and b in Ω_{Q_{8}}^{2}(k) and by m. Let u = Xa + Y b ∈ Ω_{Q_{8}}^{2}(k). Then it can be checked that u = (0, X^{2}Y + X^{3}Y) ∈ kQ_{8}^{2} and in particular Y u = 0. Let V_{4} be the submodule of V generated by m + u. It is a module of dimension 2 which has one generator and on which y acts trivially. Hence it is isomorphic to k↑_{⟨y⟩}^{Q_{8}} as asserted.

Then V_{4} ∩ Ω_{Q_{8}}^{2}(k) has dimension 1 and is generated by Xu = X^{3}a. So we have that V/V_{4} ≅ Ω_{Q_{8}}^{2}(k)/⟨X^{3}a⟩. For the rest let V_{1} = ⟨ a, m ⟩, V_{2} = ⟨ Xa, m ⟩ and V_{3} = ⟨ X^{2}a, m ⟩. Then V/V_{1} ≅ Ω_{Q_{8}}^{2}(k)/⟨a⟩ ≅ k↑_{⟨xy⟩}^{Q_{8}} as desired and V_{1}/V_{4} ≅ ⟨a⟩/⟨X^{3}a⟩ ≅ Ω_{Q_{8}}^{1}(k↑_{⟨x⟩}^{Q_{8}}) ≅ Ω_{⟨x⟩}^{1}(k) ↑_{⟨x⟩}^{Q_{8}}.

Lemma 6.2. Let M be an indecomposable kQ_{8}-module such that M↓_{H}^{Q_{8}} ≅ k ⊕(free) for every proper subgroup H of Q_{8}. Then dim(M) + dim(Ω_{Q_{8}}^{2}(M)) ≤ 10. Moreover, Ω_{Q_{8}}^{4}(k) ≅ k.

Proof. We only sketch the argument, which is similar to the one used in the proof of Theorem 5.3. We tensor M with the module V = Ω_{Q_{8}}^{2}(k) ⊕ k of Lemma 6.1 and split off all the free modules appearing in the quotients of the filtration. We are left with a 10-dimensional module having a direct summand isomorphic to Ω_{Q_{8}}^{2}(M)⊕M. For the second statement, we choose M = Ω_{Q_{8}}^{2}(k), which is 9-dimensional, and we conclude that dim(Ω_{Q_{8}}^{2}(M)) ≤ 1.

Now we construct a 3-dimensional endo-trivial module L, assuming that the base field k contains a cubic root of unity ω. The action of the two generators x and y are given by the matrices

x = ( \begin{matrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix} ) and y = ( \begin{matrix}1 & 0 & 0 \\ ω & 1 & 0 \\ 0 & ω^{2} & 1 \end{matrix} )

respectively. It is clear that L↓_{⟨x⟩}^{Q_{8}} ≅ Ω_{⟨x⟩}^{1}(k), and therefore L is endo-trivial by Lemma 2.9, since the unique elementary abelian subgroup is the center ⟨x^{2}⟩.

Theorem 6.3. (Dade [Da1]) Let Q_{8} be the quaternion group of order 8.

(a) If the base field k does not contain cubic roots of unity, then T_{k}(Q_{8}) ≅ ℤ/4ℤ, generated by Ω_{Q_{8}}.

(b) If the base field k contains cubic roots of unity, the group T_{k}(Q_{8}) is isomorphic to ℤ/4ℤ⊕ℤ/2ℤ, generated by Ω_{Q_{8}} and the class of the 5-dimensional module Ω_{Q_{8}}^{1}(L), which has order 2.

Proof. We first note that Ω_{Q_{8}} has order 4, by Lemma 6.2 and the fact that Ω_{Q_{8}}^{2}(k) is non-trivial (it has dimension 9). Consider the cyclic subgroup H = ⟨x⟩ of order 4. The restriction map

Res_{H}^{Q_{8}} : T (Q_{8}) ⟶ T (H) ≅ ℤ/2ℤ

is surjective and we know that the class of Ω_{Q_{8}}^{2}(k) is in the kernel. We need to compute this kernel. If the class of M lies in the kernel, then dim(M) ≡ 1 (mod 4) and therefore the only possibility is that M↓_{K}^{Q_{8}} ≅ k⊕(free) for every proper subgroup K of Q_{8}. Thus Lemma 6.2 applies and we have dim(M) + dim(Ω_{Q_{8}}^{2}(M)) ≤ 10, assuming that M is indecomposable. If dim(M) = 1, then M ≅ k. If dim(M) = 9, then Ω_{Q_{8}}^{2}(M) ≅ k and so M ≅ Ω_{Q_{8}}^{2}(k). So we are left with the case dim(M) = 5.

We claim that M has a simple top. If M had a 2-dimensional top, then there would be a non-zero homomorphism ϕ : M → k which would split on restriction to H, because we know that M↓_{H}^{Q_{8}} ≅ k ⊕kH has a 2-dimensional top. Then Ker(ϕ) would be free on restriction to H, hence free by Chouinard’s theorem (which, in this case, is an easy consequence of Lemma 6.1). But this is impossible since dim(Ker(ϕ)) = 4. Thus M has a simple top, so its projective cover is kQ_{8} and N = Ω_{Q_{8}}^{1}(M) has dimension 3.

Now N↓_{H}^{Q_{8}} ≅ Ω_{H}^{1}(k) is uniserial and so N is uniserial. Let u ∈ N be a generator, so that u, Xu, X^{2}u is as basis of N, where X = x + 1. Write the action of Y = y + 1 as

Y u = λXu + μX^{2}u , λ, μ ∈ k .

If we choose the new generator v = u + μXu, then an easy computation yields Y v = λXv and Y^{2}v = λ(λ + 1)X^{2}v. Since X and Y must satisfy the relation X^{2} = Y^{2}, we deduce that λ(λ + 1) = 1, so that λ is a cubic root of unity.

If k does not contain cubic roots of unity, this shows that N, and hence M, cannot exist. Therefore T_{k}(Q_{8}) ≅ ℤ/4ℤ, generated by Ω_{Q_{8}}. If k contains a cubic root of unity ω, then the value of λ is either ω or ω^{2} = ω + 1. Thus N is isomorphic to the module L in the first case and to its dual in the second, as one can check easily. Moreover L ≅ L̸^{∗} because the actions of x and y on L/ Soc(L) are given by

x = ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ) y = ( \begin{matrix}1 & 0 \\ ω & 1 \end{matrix} ) ,

while the actions on L^{∗}/ Soc(L^{∗}) are

x = ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ) y = ( \begin{matrix}1 & 0 \\ ω^{2} & 1 \end{matrix} ) ,

so that L/ Soc(L) ≅ L̸^{∗}/ Soc(L^{∗}).

Finally, apart from the group of order 4 generated by Ω_{Q_{8}}, the group T_{k}(Q_{8}) contains 4 other elements, namely the classes of L, L^{∗}, M = Ω_{Q_{8}}^{−1}(L) and M′ = Ω_{Q_{8}}^{−1}(L^{∗}). Our analysis also shows that the kernel of the restriction to H consists of the classes [k], [Ω_{Q_{8}}^{2}(k)] = 2Ω_{Q_{8}}, [M] and [M′]. If this kernel was cyclic, then [M] would be a generator (because 2Ω_{Q_{8}} has order 2), so we would have that 2[M] = 2Ω_{Q_{8}}, hence

2[L] = 2[Ω_{Q_{8}}^{1}(M)] = 2Ω_{Q_{8}} + 2[M] = 0 = [L] + [L^{∗}] ,

contrary to the fact that [L] ≠ [L^{∗}]. So this kernel is a Klein four group and the classes of M and M′ ≅ Ω_{Q_{8}}^{1}(L) have order 2.

Now we turn to the generalized quaternion group of order 2^{n+1} (n ≥ 3),

Q_{2^{n+1}} = ⟨ x, y | x^{2^{n}} = 1, y^{2} = x^{2^{n−1}}, yxy^{−1} = x^{−1} ⟩ .

We consider the cyclic subgroup H =⟨x⟩ and we first construct a (2^{n}−1)-dimensional module L by extending to Q_{2^{n+1}} the module Ω_{H}^{1}(k).

Lemma 6.4. There exists a kQ_{2^{n+1}}-module L such that L↓_{H}^{Q_{2^{n+1}}} ≅ Ω_{H}^{1}(k). Moreover L is endo-trivial.

Proof. We have kH ≅ k[x]/(x − 1)^{2^{n}} and Ω_{H}^{1}(k) ≅ V = k[x]/(f) where

f = (x − 1)^{2^{n}−1} = ∑_{i=0}^{2^{n}−1} x^{i} .

Hence we have End_{kH}(V) ≅ V, via multiplication. Moreover, y acts on H by inversion and this action extends linearly to an action on kH, hence also on its quotient V. We write σ for this automorphism of the ring V = k[x]/(f). We claim that there exists an invertible element a ∈ V such that a σ(a) = x^{2^{n−1}}.

Postponing the proof of this claim, we deduce that V extends to Q_{2^{n+1}}. Indeed, consider the automorphism τ = m_{a}σ, where m_{a} denotes multiplication by a. Then we make y act on V via τ and this defines the extension, because

τm_{x} = m_{a}σm_{x} = m_{a}m_{σ(x)}σ = m_{σ(x)}τ = m_{x^{−1}} τ ,
τ^{2} = m_{a}σm_{a}σ = m_{a}m_{σ(a)}σ^{2} = m_{aσ(a)} = m_{x^{2^{n−1}}} .

The fact that the extended module is endo-trivial follows from Lemma 2.9, because its restriction to H is endo-trivial, hence also its restriction to the unique elementary abelian subgroup ⟨x^{2^{n−1}}⟩.

For the proof of the claim, we define

a = 1 + (x + x^{2})(1 + x^{2^{n−1}+2})( ∑_{i=0}^{2^{n−3}−1} x^{4i}) .

A tedious computation, using x^{2^{n}} = 1, shows that

a σ(a) = (∑_{k=0}^{2^{n}−1} x^{k}) − x^{2^{n−1}} = f + x^{2^{n−1}} = x^{2^{n−1}} ,

20 JON CARLSON AND JACQUES THÉVENAZ

since f = 0 in V. In order to facilitate the computation, it is useful to write a = 1 + b and to note that b σ(b) turns out to be equal to the sum of all odd powers of x, while 1 + b + σ(b) is the sum of all even powers of x except x^{2^{n−1}}.

Theorem 6.5. (Dade [Da1]) Let Q_{2^{n+1}} be the generalized quaternion group of order 2^{n+1} with n ≥ 3. The group T (Q_{2^{n+1}}) is isomorphic to ℤ/4ℤ ⊕ ℤ/2ℤ, generated by Ω_{Q_{2^{n+1}}} and the class of the (2^{n} + 1)-dimensional module Ω_{Q_{2^{n+1}}}^{1}(L), which has order 2.

Proof. The group Q_{2^{n+1}} has no elementary subgroups of rank 2 and it has two conjugacy classes of quaternion subgroups of order 8, represented by H = ⟨ x^{2^{n−2}}, y ⟩ and H′ = ⟨ x^{2^{n−2}}, xy ⟩. Therefore, by Theorem 2.8, the restriction map

Res : T (Q_{2^{n+1}}) ⟶ T (H) ⊕ T (H′)

is injective. By Lemma 2.10, no kH-module of dimension ≡ 3, 5 (mod 8) can be in the image of the restriction map and so, by Theorem 6.3, the image of the restriction map to H is contained in the cyclic subgroup ⟨Ω_{H}⟩ of order 4, and similarly for H′. Thus we obtain an injective map

Res : T (Q_{2^{n+1}}) ⟶ ⟨Ω_{H}⟩ ⊕ ⟨Ω_{H′}⟩ ≅ ℤ/4ℤ ⊕ ℤ/4ℤ

and its image is contained in the even part, generated by (Ω_{H}, Ω_{H′}) and (Ω_{H}, −Ω_{H′}) (by the argument already used in the proof of Theorem 5.4).

Now Ω_{Q_{2^{n+1}}} restricts to (Ω_{H}, Ω_{H′}) and we claim that [L] restricts to ±(Ω_{H}, −Ω_{H′}) (we do not need to know the value of the sign). Since we have

dim(L) = 2^{n} − 1 ≡ −1 (mod 8) ,

the restriction of [L] to H must be either Ω_{H} or −Ω_{H}, and similarly with H′. By injectivity of the restriction map and the fact that L ≅ Ω̸_{Q_{2^{n+1}}}^{±1}(k) (because they do not have the same dimension), [L] cannot restrict to ±(Ω_{H}, Ω_{H′}), so it restricts to ±(Ω_{H}, −Ω_{H′}). This shows that the restriction map is surjective onto the even part of ⟨Ω_{H}⟩ ⊕ ⟨Ω_{H′}⟩, which is isomorphic to ℤ/4ℤ ⊕ ℤ/2ℤ. Finally, the class of Ω_{Q_{2^{n+1}}}^{1}(L) restricts to either (2Ω_{H}, 0) or (0, 2Ω_{H′}), which is a generator for the factor ℤ/2ℤ.

7. Semi-Dihedral 2-Groups

We continue our study of some specific 2-groups and describe in this section the group of endo-trivial modules for the semi-dihedral groups. The description of the whole Dade group will be given in Section 10.

Consider the semi-dihedral group

SD_{2^{n}} = ⟨ x, y | x^{2^{n−1}} = y^{2} = 1 , yxy = x^{2^{n−2}−1} ⟩ (n ≥ 4) .

We proceed in the same way as for dihedral groups by first constructing L as the relative syzygy of the trivial module with respect to the subgroup ⟨ y ⟩, that is,

0 ⟶ L ⟶ W ⟶ k ⟶ 0

where W = k↑_{⟨ y ⟩}^{SD_{2^{n}}} ≅ k[SD_{2^{n}}/⟨ y ⟩]. The subgroup E = ⟨ y, z ⟩ is the only elementary abelian subgroups of rank 2 up to conjugacy, where z = x^{2^{n−2}} is the central involution. Moreover, there is a unique conjugacy class of quaternion subgroups of order 8, represented by H = ⟨ yx, x^{2^{n−3}} ⟩. Exactly the same argument as for D_{2^{n}} shows that

L↓_{H}^{SD_{2^{n}}} ≅ Ω_{H}^{1}(k) ⊕ (free) and L↓_{E}^{SD_{2^{n}}} ≅ Ω_{E}^{−1}(k) ⊕ (free) .

By Lemma 2.9, the latter isomorphism suffices to show that L is endo-trivial.

Theorem 7.1. The group T (SD_{2^{n}}) is isomorphic to ℤ ⊕ ℤ/2ℤ, generated by Ω_{SD_{2^{n}}} and the class of the module Ω_{SD_{2^{n}}}^{1}(L), which has order 2.

Proof. By Theorem 2.8, the restriction map

T (SD_{2^{n}}) ⟶ T (E) ⊕ T (H)

is injective and, by the argument already used in the proof of Theorem 6.5, its image is contained in the even part of

T (E) ⊕ ⟨Ω_{H}⟩ ≅ ℤ ⊕ ℤ/4ℤ ,

generated by (Ω_{E}, Ω_{H}) and (−Ω_{E}, Ω_{H}). As these generators are the restrictions of Ω_{SD_{2^{n}}} and [L] respectively, it follows that the restriction map is an isomorphism onto this even part. The element [Ω_{SD_{2^{n}}}^{1}(L)] = Ω_{SD_{2^{n}}} + [L] restricts to (0, 2Ω_{H}), so it has order 2, and we see that T (SD_{2^{n}}) is isomorphic to ℤ ⊕ ℤ/2ℤ.

As in the case of dihedral groups, the result above can also be obtained using another construction of an endo-trivial kSD_{2^{n}}-module instead of L. By the method of Theorem 5.2 of [BoTh], one can prove that the tensor-induced module Ten_{E}^{SD_{2^{n}}}(Ω_{E}^{1}(k)) is an endo-permutation which belongs to the class of an indecomposable endo-trivial module M, and the restrictions of [M] to E and H are 2Ω_{E} and 0 respectively. Thus T (SD_{2^{n}}) is also generated by Ω_{SD_{2^{n}}} and [M], and it is [Ω_{SD_{2^{n}}}^{−2}(M)] which is the generator of order 2.

8. A Restrictive Condition on Relative Traces

In order to prove Conjecture 2.6, one has to consider the following question. Given a p-group G and a kG-module M such that M↓_{H}^{G} ≅ k ⊕ (free) for every proper subgroup H, is M isomorphic to k ⊕ (free) ? In view of Theorem 2.8, one has to consider this question only when G is extraspecial of exponent p if p is odd and when G is (almost) extraspecial if p = 2 (but G ≅ D̸_{8}, Q_{8}). The purpose of this section is to give a positive answer to the question above under one additional assumption. So one is reduced to the case where this assumption is not satisfied and it turns out that this provides a very restrictive condition on the module M. We shall have to assume that our base field k is perfect, but this is not a real restriction since we can always replace k by its algebraic closure.

Recall that, if M is a kG-module and H is a subgroup of G, then M^{H} denotes the set of H-fixed points in M. Moreover the relative trace map t_{H}^{G} is the k-linear

22 JON CARLSON AND JACQUES THÉVENAZ

map

t_{H}^{G} : M^{H} ⟶ M^{G} , m ↦ ∑_{g∈[G/H]} g · m ,

where [G/H] denotes a complete set of coset representatives. The image of t_{H}^{G} is written M_{H}^{G}. We are interested in the case where H is a normal subgroup of index p. In this case, the condition t_{H}^{G}(m) ≠ 0 means that the kG-submodule generated by m is a free k[G/H]-module (note that H acts trivially). The main result of this section is the following.

Theorem 8.1. Suppose that k is a perfect field. Let G be a p-group which is not elementary abelian and let M be a kG-module such that M↓_{J}^{G} ≅ k ⊕ (free) for every proper subgroup J of G. Assume that there exists a maximal subgroup H and a generator x of a one-dimensional trivial direct summand of M↓_{H}^{G} such that t_{H}^{G}(x) = 0. Then M ≅ k ⊕ (free).

It follows from this theorem that if M is a counter-example to Conjecture 2.6, then M must satisfy the following very restrictive condition: For every maximal subgroup H of G and for every one-dimensional trivial direct summand k·x of M↓_{H}^{G}, we must have t_{H}^{G}(x) ≠ 0, so that x generates a free k[G/H]-module. This condition is so restrictive that one expects that M does not exist, except in the case of the quaternion group of order 8.

The proof of Theorem 8.1 will be based on the following three lemmas.

Lemma 8.2. Let P be a non-trivial p-group and let N be a kP -module such that

N = T ⊕ F = T′ ⊕ F′ ,

where F and F′ are free and T ≅ T′ ≅ k. Then N = T ⊕ F′.

Proof. We have N^{P} = T ⊕ F^{P} and F^{P} = F_{1}^{P} = N_{1}^{P} = (F′)_{1}^{P} = (F′)^{P}. If x is a generator of T, then x ∈ N^{P} − N_{1}^{P} and so with respect to the decomposition N^{P} = T′ ⊕ (F′)^{P} = T′ ⊕ N_{1}^{P}, we have that x = y + z, y ∈ T′, z ∈ (F′)^{P}, and y ≠ 0. It follows that N = ⟨x⟩ ⊕ F′ = T ⊕ F′.

Lemma 8.3. Let G be a p-group which is not elementary abelian and let M be a kG-module such that M↓_{J}^{G} ≅ k ⊕ (free) for every proper subgroup J of G. Assume that there exists a maximal subgroup H and a generator x of a one-dimensional trivial direct summand of M↓_{H}^{G} such that x is fixed under G. Then M ≅ k ⊕ (free).

Proof. By assumption, T = k·x is a kG-submodule of M and M↓_{H}^{G} = T ↓_{H}^{G} ⊕ F where F is a free kH-module. Let J be any maximal subgroup of G with J ≠ H. We have M↓_{J}^{G} = T′ ⊕ F′ where T′ is a trivial and F′ a free kJ-module. Thus we obtain

(T′)↓_{J∩H}^{J} ⊕ (F′)↓_{J∩H}^{J} = M↓_{J∩H}^{G} = T ↓_{J∩H}^{G} ⊕ F ↓_{J∩H}^{H} ,

and so M↓_{J∩H}^{G} = T ↓_{J∩H}^{G} ⊕ (F′)↓_{J∩H}^{J} by Lemma 8.2, which applies since J ∩ H is non-trivial (otherwise G would be elementary abelian of rank 2). It follows that

M↓_{J}^{G} = T ↓_{J}^{G} ⊕ F′ .

This shows that the kG-module M/T is free on restriction to every maximal subgroup of G. Since G is not elementary abelian, M/T is free by Chouinard’s theorem and so M ≅ T ⊕ (free).

Let O be a complete discrete valuation ring of characteristic zero with residue field k. We also assume that O is unramified, that is, the prime number p is a generator of the maximal ideal of O. It is well-known that such a ring exists (and is unique up to isomorphism) if k is a perfect field (see [Se]).

Lemma 8.4. Let p be an odd prime and suppose that C is a cyclic group of order p acting O-linearly on O/p^{m}O where m ≥ 2. If x ∈ O/p^{m}O, then t_{1}^{C}(x) ≡ px (mod p^{2}).

Proof. Let u be a generator of C. Since C necessarily acts trivially on O/pO, we have ux = x + py for some y ∈ O/p^{m}O. Similarly uy ≡ y (mod p) and therefore u^{i}x ≡ x + ipy (mod p^{2}) for 0 ≤ i ≤ p − 1. Hence

t_{1}^{C}(x) = ∑_{i=0}^{p−1} u^{i}x ≡ px + ∑_{i=0}^{p−1} ipy ≡ px (mod p^{2}) ,

because p divides ∑_{i=0}^{p−1} i = p(p − 1)/2 (since p is odd).

Remark. We have in fact an equality t_{1}^{C}(x) = px because it follows from Lemma 2.4 in [Th2] that we necessarily have ux ≡ x (mod p^{m−1}). Note also that the lemma does not hold if p = 2 since we may have (u + 1)x = 0.

Now we can start the proof of Theorem 8.1. So we let M be a kG-module such that M↓_{J}^{G} ≅ k ⊕ (free) for every proper subgroup J of G and we suppose that there exists a maximal subgroup H and a generator x of a one-dimensional trivial direct summand of M↓_{H}^{G} such that t_{H}^{G}(x) = 0. If p = 2, the k[G/H]-module generated by x is not free, so it is trivial. Thus x is fixed under G and the result follows from Lemma 8.3. So from now on we assume that p is odd. If G is cyclic of order p^{2}, the proof is easy and is left to the reader. Since G is not elementary abelian, we can now assume that G has order p^{n} with n ≥ 3.

We let O be as in Lemma 8.4 (O exists because k is perfect). The main result of [Al2] asserts that every endo-trivial kG-module can be lifted to an endo-trivial OG-lattice. Therefore, since M is endo-trivial by Lemma 2.9, there exists an endotrivial OG-lattice L such that L/pL ≅ M. The decomposition M↓_{H}^{G} = k·x ⊕ (free) lifts to a decomposition L↓_{H}^{G} = O·x̃ ⊕ F where F is a free OH-lattice. The onedimensional lattice O·x̃ corresponds to a character λ : H → O^{∗}. But O is unramified and p is odd, so there are no p-th roots of unity in O because the cyclotomic polynomial is irreducible over O. Therefore λ has to be trivial and so x̃ ∈ L^{H}.

Now L^{H}/L_{1}^{H} ≅ O·x̃/p^{n−1}O·x̃ because F^{H} = F_{1}^{H} and t_{1}^{H} is multiplication by p^{n−1} on the trivial summand O·x̃. Clearly the cyclic group G/H acts on L^{H}/L_{1}^{H}, so by Lemma 8.4 (which applies because n − 1 ≥ 2), we have

t_{H}^{G}(x̃) ≡ px̃ (mod p^{2}L^{H} + L_{1}^{H}) .

24 JON CARLSON AND JACQUES THÉVENAZ

Because the image of px̃ in L^{H}/L_{1}^{H} is non-zero, t_{H}^{G}(x̃) ≠ 0, and we have constructed a non-zero G-fixed point in L.

But now we have t_{H}^{G}(x) = 0 in M by assumption, and t_{H}^{G}(x̃) = py for some y ∈ L with y ∈ L^{G}. Moreover y cannot be a multiple of p, because otherwise t_{H}^{G}(x̃) would be a multiple of p^{2}, and px̃ cannot be a multiple of p^{2} in L^{H}/L_{1}^{H}. Since the images of px̃ and py are equal in L^{H}/L_{1}^{H} ≅ O/p^{n−1}O, which is uniserial, the image of y must be a generator of L^{H}/L_{1}^{H}. Therefore there exists λ ∈ O^{∗} such that y ≡ λx̃ (mod L_{1}^{H}). But L_{1}^{H} = p^{n−1}O·x̃ ⊕ F_{1}^{H} and it follows that L is also the direct sum of O·y and F. This shows that y is a generator of a one-dimensional trivial direct summand of L↓_{H}^{G} which is fixed under G. After reducing modulo p, the same holds for the image of y in M, so Lemma 8.3 applies and the result follows.

9. Extraspecial Groups of Exponent p

The reader might ask the question of why the methods of Section 4 do not work for the extraspecial groups of exponent p (where p is odd). In one sentence, the answer is that there is no general analog of Theorem 4.4. The proof involves an analysis of the cohomology ring of the algebra defined similar to the definition of B in Section 4. The outcome was something of a surprise to us. It may be of some independent interest because very few examples of such cohomology rings have been computed and the examples show the extent to which the cohomology rings of groups can differ from the cohomology rings of very similar algebras.

The examples rely on some computer calculations that were performed only for the primes p = 3 and p = 5. It seems likely that with some effort it would be possible to produce a complete proof of the structure of the cohomology rings for all primes and without any of the ambiguities that we will leave. However with no motivation beyond general curiosity, it seems sufficient to stick to the restricted cases. We hope the reader will find this satisfactory.

Throughout the section we let G denote an extraspecial group of order p^{3} and let z denote a central element of order p in G. We will consider both cases, that G has exponent p and that G has exponent p^{2}. Consequently, the algebra B = kG/kG(z − 1)^{p−1} will depend upon which group is chosen. In either case we let Z = z − 1 and D ≅ k[Z]/(Z^{p−1}) denote the subalgebra of B generated by 1 and z. Also, as in Section 4, we let C = B/(B·Z) ≅ k(G/⟨z⟩). Because G/⟨z⟩ is elementary abelian of order p^{2}, Ext_{kG}^{∗}(k, k) ≅ k[x, w] ⊗ Λ(a, b) where Λ is the exterior algebra generated by a and b.

In the cases that p = 3 and p = 5, projective resolutions for the cohomology rings of the algebras B associated to the two groups were constructed on the computer. We also constructed the multiplicative generators of the cohomology ring. The projective resolutions were only constructed out to degree 10 but as we see below, this is more than sufficient to compute the cohomology rings. The computations were done on a Toshiba Tecra with 98 Mb. of RAM. The programs were run using the MAGMA computer algebra system [BoCa] using a new package for performing homological algebra calculations for modules over basic algebras. The package was written mostly by the first author, with generous help from John Cannon and Allan Steel, and is first included in the release of MAGMA Version 2.5 in June of 1999. The results of the calculation are presented in the following.

Proposition 9.1. Suppose that p = 3 or 5. In the case that G has exponent p, the dimensions of Ext_{B}^{n}(k, k) for n = 0, 1, …, 15 are

1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92

and the cohomology ring is minimally generated by two elements in degree 1 and five elements in degree 2. In the case that the exponent of G is p^{2}, the dimensions of Ext_{B}^{n}(k, k) for n = 0, 1, …, 15 are

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

and the cohomology ring is minimally generated by two elements in degree 1 and three elements in degree 2.

The important thing to note here is that in the first case the dimensions of the cohomology are growing at a quadratic rate (complexity 3) while in the second case they are only growing at a linear rate (complexity 2). This leads us directly to the following. For notation we let V(u, v, …) denote the linear k-subspace spanned by the elements u, v, . . ..

Theorem 9.2. Assume that p = 3 or 5 and |G| = p^{3}. If G is the extraspecial group of exponent p, then

Ext_{B}^{∗}(k, k) ≅ (k[a, b, z, y, x, w]/I) ⊗ k[ζ] (9.2.1)

where a and b have degree one, z, y, x, w and ζ have degree two, and I is an ideal generated by relations of the form

ab = 0 , az, ay, bz, by ∈ V(ax, aw, bx, bw) ,
z^{2}, zy, y^{2} ∈ V(zx, zw, yx, yw, x^{2}, xw, w^{2}) .

(The actual values of these products is not determined.) If G is the extraspecial group of exponent p^{2}, then

Ext_{B}^{∗}(k, k) ≅ (k[x, w] ⊗ Λ(a, b))/I ⊗ k[ζ] (9.2.2)

where a and b have degree 1, z, y, x, w and ζ have degree 2, and I is the ideal generated by the relation ab + w = 0.

Remark. The isomorphisms (9.2.1) and (9.2.2) are certainly isomorphisms of Ext_{C}^{∗}(k, k)-modules. We expect that they are also ring isomorphisms, but this is not so clear since we are not dealing with group algebras.

Proof. In both cases we have the spectral sequence

E_{2}^{r,s} = Ext_{C}^{r}(k, Ext_{D}^{s}(k, k)) ⇒ Ext_{B}^{r+s}(k, k)

as in Section 4. The action of C on Ext_{D}^{s}(k, k) is trivial, since Z is central. If p = 3, D is uniserial of length 2 and k has a free resolution of period 1, while if p > 3, D is uniserial of length p−1 and k has a free resolution of period 2. Therefore, as in the case of cyclic groups, we have Ext_{D}^{∗}(k, k) = k[t, u] with t^{2} = u if p = 3 and t^{2} = 0 if p > 3. Also Ext_{C}^{∗}(k, k) ≅ k[x, w] ⊗ Λ(a, b), so we actually have that

E_{2}^{∗,∗} ≅ Ext_{C}^{∗}(k, k) ⊗ Ext_{D}^{∗}(k, k)

as a ring.

In the case of the first group, the d_{2} differential on the spectral sequence sends the element t ∈ E_{2}^{0,1} to the element ab ∈ E_{2}^{2,0} and the element u ∈ E_{2}^{0,2} to zero. Note that d_{2}(t) is determined by the map on the spectral sequences (see Proposition 4.2) and our knowledge of what happens in the case of the group extension (see Section 10 of [BeCa]). The fact that d_{2}(u) = 0 can be deduced from knowing that d_{2} is a differential, if we assume that the multiplication is right. It can also be deduced from a count of the dimensions of the terms on the E_{2}-page of the spectral sequence. That is, the dimensions of E_{3}^{1,1} and E_{3}^{2,0} are at most 2 and hence, if d_{2}(u) were not zero, then the dimension of Ext_{B}^{r+s}(k, k) would be at most 4, contradicting the previous proposition. The E_{3} page of the spectral sequence is obtained by taking the homology with respect to this differential. In total degree 2, there are five linearly independent classes represented by the elements x, w in E_{3}^{2,0}, at, bt in E_{3}^{1,1}, and u in E_{3}^{2,0}. All of these are new generators of the ring structure. We let z and y be the classes of at and bt respectively, and let ζ be the class of u.

But now the computer calculation guarantees that there are five generators of degree 2 in Ext_{B}^{∗}(k, k). So every one of these generators must be in the kernel of the differential d_{3}. The statement of real importance is that d_{3}(u) = 0. Thus the E_{3} page of the spectral sequence is generated as a ring by the element on rows 0 and 1 and by u. The differential d_{3} and all higher differential must be zero because they are zero on these elements. So E_{3} = E_{∞}. This gives us the generators of the cohomology ring.

The actual relations are somewhat ambiguous, because the E_{∞} page of the spectral sequence is only a graded version of the cohomology. So for example, the classes of a · bt = ay and at · bt = zy are zero on the E_{∞} page of the spectral spectral but they may actually be equal to nonzero elements in E_{∞}^{3,0} = V(ax, aw, bx, bw) or respectively in E_{∞}^{3,1} ⊕ E_{∞}^{4,0}. More analysis would be needed to settle these cases.

Now suppose that G has exponent p^{2}. In this case, d_{2}(u) = 0 again, but d_{2}(t) = ab + w. These statements follow from a suitable modification of the same reasoning as in the case of the first group. The main difference from the other case is that this element is not a divisor of zero in Ext_{C}^{∗}(k, k). It follows that on the E_{3} page all of the odd numbered rows are zero, because the differential d_{2} : E_{2}^{2m+1,s} → E_{2}^{2m,s+2} takes the form

d_{2}(γ · u^{m}t) = γ · u^{m}(ab + w) ≠ 0

for all non-zero γ ∈ Ext_{C}^{∗}(k, k). Moreover there are three generators in degree two: the classes of x and w in E_{3}^{2,0} and u. Again the computer calculation tells us that the spectral sequence stops at the E_{3} page.

Corollary 9.3. Suppose that G has exponent p. Let H_{1}, …, H_{p+1} be the maximal subgroups of G, and for each i let U_{i} be the subalgebra U_{i} = kH_{i}/kH_{i} · Z^{p−1} ⊆ B.

There exists a B-module M such that every restriction M↓_{U_{i}}^{B} is a free U_{i}-module, but M is not a free B-module.

Proof. The proof is just an imitation of standard L_{ζ} technology for group algebras (see [Be] or [Ev]). The element ζ ∈ Ext_{B}^{2}(k, k) can be represented by a cocycle ζ̃: Ω_{B}^{2}(k) ⟶ k, where Ω_{B}^{2}(k) is the kernel of ∂_{1} in a minimal B-projective resolution (… → Q_{1} ⟶^{∂_{1}} Q_{0} → k → 0) of k. Because of the position of ζ on the edge of the spectral sequence, we know that the restriction of ζ to D is not zero and hence the exact sequence

0 ⟶ L_{ζ} ⟶ Ω_{B}^{2}(k) ⟶^{ζ̃} k ⟶ 0

splits as a sequence of D-modules, because Ω_{D}^{2}(k) ≅ k, so Ω_{B}^{2}(k)↓_{D}^{B} ≅ k ⊕ (free) and ζ̃ is non-zero on the trivial summand k. Consequently, the kernel L_{ζ} of ζ̃ is free as a D-module.

Now the trick is to choose a C-module N which is not projective but whose restriction to every k(H_{i}/⟨z⟩) is a free module. Take the tensor product M = N ⊗ L_{ζ}. Because D acts trivially on N and N ⊗ (L_{ζ}/ZL_{ζ}) is free on restriction to every H_{i}/⟨z⟩, it is straightforward to verify that the restriction of M to every U_{i} is a free module.

It remains only to show that M is not a free module. We know that multiplication by ζ is an injective map on the cohomology ring of B and that Ext_{B}^{n}(Ω_{B}^{2}(k), k) ≅ Ext_{B}^{n+2}(k, k). So from the long exact sequence on cohomology we get that

Ext_{B}^{n}(L_{ζ}, k) ≅ Ext_{B}^{n+2}(k, k)/ζ Ext_{B}^{n}(k, k).

Therefore we must have that no element of the subring k[x, w] of Ext_{B}^{∗}(k, k) can annihilate Ext_{B}^{n}(L_{ζ}/ZL_{ζ}, k). It follows that the variety V_{G/⟨z⟩}(L_{ζ}/ZL_{ζ}) has dimension 2 and must be all of V_{G/⟨z⟩}(k). By the tensor product theorem for varieties (see [Be, Ev]), N ⊗(L_{ζ}/ZL_{ζ}) is not a free C-module. Thus M is not a free B-module.

Finally, we end the section with a brief demonstration that the method of [Ca2] used in Theorem 4.4 allows us to bound dimensions but does not solve the problem. Still it may provide us with some information.

Proposition 9.4. Let G be an extraspecial group of order p^{3} and exponent p. Let M be a kG-module such that for every proper subgroup H of G, M↓_{H}^{G} ≅ k ⊕ Q for some free kH-module Q. Then dim(M ⊕ Ω^{2p+1}(M)) ≤ (p + 1)^{2}p^{3}.

Proof. We follow the proof of the main theorem of [Ca2] as outlined in the proof of Theorem 4.4. Let H_{1}, …, H_{p+1} be all the maximal subgroups of G and, for each i, let β_{i} ∈ Ext_{kG}^{2}(k, k) denote the Bockstein of the degree one element corresponding to H_{i}. We know that β_{1} ⋯ β_{p+1} = 0 [BeCa]. Also β_{i} is represented by the sequence

0 ⟶ k ⟶ k↑_{H_{i}}^{G} ⟶ k↑_{H_{i}}^{G} ⟶ k ⟶ 0

where the middle map is multiplication by x − 1 for some element x of G that is not in H_{i}. The product of the Bocksteins is the splice of the sequences. So the product (which is 0) is represented by a sequence of the form

E′ : 0 ⟶ k ⟶ V_{2p+1} ⟶ … ⟶ V_{0} ⟶ k ⟶ 0

28 JON CARLSON AND JACQUES THÉVENAZ

where each V_{i} is a trivial module induced from some maximal subgroup. Now we tensor the above sequence with M and take the complex W_{∗} that we get by truncating the ends of the sequence as in the proof of 4.4. Then for each i there is a j such that

W_{i} = M ⊗ V_{i} = M ⊗ k↑_{H_{j}}^{G} ≅ (M↓_{H_{j}}^{G} ⊗ k)↑_{H_{j}}^{G} ≅ k↑_{H_{j}}^{G} ⊕ Q_{i}

where Q_{i} is projective.

Again exactly as in the proof of 4.4, we take a projective resolution (P_{∗}, ϵ) of W_{∗} and let N denote the cokernel of ∂_{2p+2} in the projective resolution. The difference with the proof of 4.4 is that the module N is not projective though it is a direct sum of M ⊕ Ω^{2p+1}(M) ⊕ P for some projective module P. As in the proof of the main theorem in [Ca2] we can get that N has a filtration whose quotients are the induced modules Ω^{2p+1}(W_{0}), …, Ω(W_{2p}), Ω^{0}(W_{2p+1}) where Ω^{0} is the non-projective part of the module.

The conclusion of this is that

dim(M ⊕ Ω^{2p+1}(M)) ≤ dim(N) = ∑_{i=0}^{2p+1} dim(Ω^{i}(W_{2p+1−i})) .

But dim(Ω^{i}(W_{2p+1−i})) = p dim(Ω_{H}^{i}(k)) for the appropriate maximal subgroup H. Since H is an elementary abelian subgroup of order p^{2} we can check that

dim(Ω_{H}^{2i}(k)) + dim(Ω_{H}^{2i+1}(k)) = (2i + 1)p^{2} .

Hence the dimension of N is (1 + 3 + 5 + … + 2p + 1)p^{3} = (p + 1)^{2}p^{3} as asserted.

A computer check shows that when p is 3 or 5, then dim(k⊕Ω^{2p+1}(k)) = (p+2)p^{3}. So there is potentially a lot of room for torsion endo-trivial modules here, although we conjecture that there are none.

10. The Dade Group

In this section, we first prove an injectivity result for the Dade group D(G) of a finite p-group G as a consequence of the main theorem of this paper. Then we compute the Dade group of dihedral, semi-dihedral and quaternion 2-groups, using the results of Sections 5, 6, and 7.

Let Y be the class of p-groups appearing in Theorems 2.7 and 2.8. In order to include cyclic groups in our discussion, we let X be the class of p-groups consisting of all groups in Y and of the cyclic groups of order p if p is odd and of order 4 if p = 2. If G = C_{p^{n}} is cyclic, then the restriction map T (C_{p^{n}}) → T (C_{p}), respectively T (C_{2^{n}}) → T (C_{4}), is injective (in fact bijective). Therefore Theorem 2.7 implies that, for any finite p-group G, the restriction map

Res : T (G) ⟶ ∏_{E} T (E)

is injective, where E runs over all subgroups of G belonging to X. Note that if there are no such subgroups (i.e. if |G| is 1 or 2), then the product is zero, but T (G) = 0 too. Recall that if H is a subgroup of G and K is a normal subgroup of H, then H/K is called a section of G.

Theorem 10.1. If G is a finite p-group, the product of all restriction-deflation maps

Def_{H/K}^{H} Res_{H}^{G} : D(G) ⟶ ∏_{H/K} D(H/K)

is injective, where H/K runs over all sections of G belonging to X.

Proof. The argument is the same as in the proof of Theorem 1.6 in [BoTh]. Let a ∈ D(G) be in the kernel of the map. If J is a non-trivial subgroup of G, then a_{J} = Def_{N_{G}(J)/J}^{N_{G}(J)} Res_{N_{G}(J)}^{G}(a) lies in the kernel of the restriction-deflation maps from D(N_{G}(J)/J) to all sections of N_{G}(J)/J belonging to X. By induction, a_{J} = 0. Since this holds for every J ≠ 1, we have that a ∈ T′(G) by definition of T′(G) (see Section 2). But T′(G) ≅ T (G) (by Lemma 2.1) and a lies in the kernel of the restriction to subgroups in X. As mentioned above, this restriction map is injective, so we have a = 0.

Now we give the structure of the Dade group of dihedral, semi-dihedral, and quaternion 2-groups. In the three cases, the result follows by induction from the following result.

Lemma 10.2. Let G = D_{2^{n}}, SD_{2^{n}}, or Q_{2^{n}}. Then D(G) ≅ T (G) ⊕ D(G/Z(G)). Moreover D(G/Z(G)) ≅ D(D_{2^{n−1}}).

Proof. Recall that Z = Z(G) has order 2 and that G/Z ≅ D_{2^{n−1}} in the three cases, so that the second statement is clear. Now the inflation map Inf_{G/Z}^{G} is a section of the deflation map Def_{G/Z}^{G} : D(G) → D(G/Z) and therefore

D(G) ≅ Ker(Def_{G/Z}^{G}) ⊕ D(G/Z) .

The result will follow if we prove that Ker(Def_{G/Z}^{G}) = T′(G). It is easy to see that every non-trivial subgroup H of G either contains Z or has order 2 (this can be proved using induction and the fact that a maximal subgroup of G is cyclic, dihedral, or quaternion). If H ≥ Z, we have that

Def_{N_{G}(H)/H}^{N_{G}(H)} Res_{N_{G}(H)}^{G} = Def_{N_{G}(H)/H}^{N_{G}(H)/Z} Def_{N_{G}(H)/Z}^{N_{G}(H)} Res_{N_{G}(H)}^{G}
= Def_{N_{G}(H)/H}^{N_{G}(H)/Z} Res_{N_{G}(H)/Z}^{G/Z} Def_{G/Z}^{G} ,

and therefore Ker(Def_{G/Z}^{G}) ⊆ Ker(Def_{N_{G}(H)/H}^{N_{G}(H)} Res_{N_{G}(H)}^{G}). If now H is a non-central subgroup of order 2, then N_{G}(H)/H has order 2, so D(N_{G}(H)/H) = 0. It follows that Ker(Def_{G/Z}^{G}) is contained in the kernel of the restriction-deflation maps to all sections N_{G}(H)/H, H ≠ 1. Since G/Z is among these sections, we have equality and Ker(Def_{G/Z}^{G}) = T′(G), as was to be shown.

In the following theorem, we use the standard convention that D_{4} is the Klein four-group and we note that D(D_{4}) ≅ ℤ by Theorem 2.2.

Theorem 10.3.

(a) D(D_{2^{n}}) ≅ ℤ^{2n−3}.

(b) D(SD_{2^{n}}) ≅ ℤ^{2n−4} ⊕ ℤ/2ℤ.

(c) D(Q_{2^{n}}) ≅ ℤ^{2n−5} ⊕ ℤ/4ℤ ⊕ ℤ/2ℤ if n ≥ 4.

(d) D(Q_{8}) ≅ ℤ ⊕ ℤ/4ℤ ⊕ ℤ/2ℤ if the base field k contains cubic roots of unity, and D(Q_{8}) ≅ ℤ ⊕ ℤ/4ℤ otherwise.

Proof. In each case, the result follows by induction and Lemma 10.2, using the structure of T (G) given in Theorems 5.4, 6.3, 6.5, and 7.1.

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J. F. C.: Department of Mathematics, University of Georgia, Athens, Georgia 30602, USA

  E-mail address: jfc@sloth.math.uga.edu

J. T.: Institut de Mathématiques, Université de Lausanne, CH-1015 Lausanne, Switzerland

  E-mail address: Jacques.Thevenaz@ima.unil.ch