Tensor induction of relative syzygies

S. Bouc

Abstract: The main result of this paper states a formula expressing the effect of tensor induction on relative syzygies in the Dade group of a p-group. This provides tools to study the structure of this group, and a new proof of theorems A and B of [5]. Finally, it is possible to say when a relative syzygy is a torsion element of the Dade group.

1. Introduction

Let p be a prime number, and k be a field of characteristic p. If P is a finite p-group, a finitely generated kP -module M is called an endo-permutation module if End_{k}(M) is a permutation module, i.e. admits a P -invariant k-basis. Those modules appears in many different places in the modular representation theory of finite groups, and they have been studied intensively.

In [8],[9], E.C. Dade defined a group structure on the set of equivalence classes of endo-permutation kP -modules, which is now called the Dade group of P, and denoted by D_{k}(P). This group has natural functorial properties: when Q is a subgroup of P, there is an operation of restriction Res_{Q}^{P} from D_{k}(P) to D_{k}(Q). Similarly, if R is a quotient of P, there is an operation of inflation Inf_{R}^{P} from D_{k}(R) to D_{k}(P), and an operation Def_{R}^{P} from D_{k}(P) to D_{k}(R), called deflation in [5], which corresponds to Dade’s slash construction. These three operations are already defined and used in Dade’s papers.

It was observed by L. Puig ([13]) that when Q is a subgroup of P, tensor induction of modules gives an operation Ten_{Q}^{P} from D_{k}(Q) to D_{k}(P). There is moreover an obvious operation Iso_{P}^{P′} associated to a group isomorphism from P to a group P′. Recently, in a joint work with J. Thévenaz ([5]), we showed that these five operations can be described via a single formalism, using bisets.

Typical examples of endo-permutation modules are the syzygies of the trivial module. Following a suggestion of L. Puig, J.L. Alperin proved more generally that the syzygies of the trivial module relative to any permutation module are endo-permutation modules. These are defined as the kernel of the augmentation map kX → k sending each basis element x of the P -set X to 1 ∈ k, and denoted by Ω_{X}.

The aim of this paper is to describe the effect of the five functorial operations on these relative syzygies, i.e. to state formulas giving F_{P}^{P′}(Ω_{X}) in terms of elements of the Dade group of P′, for a P -set X, and one of the above operations F_{P}^{P′}. Actually, the only difficult case is the case of tensor induction, and this will be the central question of this work.

The paper is organized as follows: in sections 1 to 3, I recall some basic definitions and facts about relative projectivity and the Dade group. Section 4 describes the easy cases of the functorial aspect of relative syzygies. Section 5 is devoted to the proof of the formula for tensor induction.

This formula is used in section 6 to show that the subgroup of D_{k}(P) generated by relative syzygies is actually a subfunctor of D_{k}, i.e. that it is invariant by the five operations.

In section 7, some information on the structure of the Dade group is given: in particular, the subgroup generated by relative syzygies and by the intersection of the kernels of all restriction-deflation maps to elementary abelian sections has finite index in the whole Dade group, and the exponent of the quotient group divides the order of P. Roughly speaking, one can say that an element of D_{k}(P) can be recovered up to a power of p and up to a torsion element from its restriction-deflations to elementary abelian sections of P. This provides also an alternative proof of Theorem A and Theorem B of [5].

Finally in section 8, I state a theorem telling when a relative syzygy can be a torsion element in the Dade group. The case of Ω_{P/Q} was originally a question of J. Thévenaz.

2. Relative syzygies

When studying endo-permutation modules, one considers usually the case where the ground ring is a field of characteristic p (or a discrete valuation ring of characteristic 0 with residue field of characteristic p), and the group is a p-group. However, the first few results can be stated for an arbitrary commutative ring O, and arbitrary finite groups.

The symbol ⊗ denotes tensor product over O. If G is a group and M and N are OGmodules, their tensor product M ⊗ N is always viewed as an OG-module via diagonal action. Similarly the module Hom_{O}(M, N) is viewed as an OG-module with G-action given by (gφ)(m) = gφ(g^{−1}m) for g ∈ G, φ ∈ Hom_{O}(M, N), and m ∈ M. I denote by M^{∗} the O-dual of M, i.e. M^{∗} = Hom_{O}(M, O).

2.1. X-split morphisms

(2.1.1) Definition: Let G be a finite group.

• A morphism of OG-modules f : M → N is said to be split if there exists a morphism of OG-modules a : N → M such that f = f ∘ a ∘ f.

• Let X be a finite G-set. A morphism of OG-modules f : M → N is said to be X-split if the morphism OX ⊗ f : OX ⊗ M → OX ⊗ N is split.

• A short exact sequence of OG-modules

0 ⟶ L ⟶^{a} M ⟶^{b} N ⟶ 0

is said to be X-split if a is X-split, or equivalently, if b is X-split.

Note indeed that if f is a surjective (resp. injective) morphism of OG-modules, then f is split if and only if it admits a right (resp. left) inverse.

If f : M → N is a morphism of OG-modules, I denote by Sec(f) the set of O-linear morphisms a : N → M such that f = f ∘ a ∘ f. The group G acts on Sec(f) by conjugation: if g ∈ G and a ∈ Sec(f), the morphism g.a : N → M defined for n ∈ N by

(g.a)(n) = ga(g^{−1}n)

is clearly in Sec(f).

(2.1.2) Lemma: Let G be a finite group, let X be a finite G-set, and f : M → N be a morphism of OG-modules. Then the following are equivalent:

1. The morphism f is X-split.

2. For any x ∈ X, the stabilizer G_{x} of x in G admits a fixed point on Sec(f).

3. There exists a morphism of G-sets from X to Sec(f).

Proof: Let a : OX ⊗ N → OX ⊗ N be an O-linear morphism. Then a is defined for x ∈ X and n ∈ N by

(2.1.3)  a(x ⊗ n) = ∑_{y∈X} y ⊗ a_{y,x}(n)

where a_{y,x} ∈ Hom_{O}(N, M). Then a is a morphism of OG-modules if and only if for any g ∈ G, any x, y ∈ X and n ∈ N

a_{gy,gx}(n) = ga_{y,x}(g^{−1}n)

or equivalently g.a_{y,x} = a_{gy,gx}. Now for any x ∈ X and m ∈ M

(OX ⊗ f) ∘ a ∘ (OX ⊗ f)(x ⊗ n) = (OX ⊗ f) ∘ a(x ⊗ f(m))
= (OX ⊗ f)( ∑_{y∈X} y ⊗ a_{y,x}f(m))
= ∑_{y∈X} y ⊗ fa_{y,x}f(m)

The condition OX ⊗ f = (OX ⊗ f) ∘ a ∘ (OX ⊗ f) is equivalent to

(2.1.4)  fa_{y,x}f(m) = δ_{y,x}f(m)

for all x, y ∈ X and m ∈ M, the symbol δ_{y,x} being a Kronecker symbol.

In particular, if f is X-split, there exists such an a. Now for any x ∈ X, the element a_{x,x} is in Sec(f), and it is invariant by the action of G_{x}. Thus condition 2) holds.

Clearly condition 2) implies condition 3) (just choose a G_{x}-invariant element α_{x} of Sec(x) for a set of representatives [G\X] of G\X, and map g.x to g.α_{x} for any g ∈ G and x ∈ [G\X]).

Finally if 3) holds, let α be a morphism of G-sets from X to Sec(f). For x, y ∈ X, define a_{y,x} : N → M by

a_{y,x} = δ_{y,x}α(x)

This clearly defines (by equation 2.1.3) a morphism of OG-modules from OX ⊗ N to OX ⊗ M, which is in Sec(OX ⊗ f) by equation 2.1.4.

Assertion 3) of the lemma gives the following

(2.1.5) Corollary: With the same notation

1. If f : M → N is X-split, then so is f ⊗L : M ⊗L → N ⊗L, for any OG-module L.

2. Let Y be a finite G-set such that Hom_{G−set}(Y, X) ≠ ∅. Then if f : M → N is X-split, it is also Y -split.

3. If f : M → N is X-split, then f^{∗} : N^{∗} → M^{∗} is X-split.

Proof: If φ : X → Sec(f) is a morphism of G-sets, then the map x ↦ φ(x) ⊗ Id_{L} is a morphism of G-sets from X to Sec(f ⊗ Id_{L}). This proves 1). Assertion 2) is obvious. Assertion 3) uses the fact that since X is finite, the module OX is free and finitely generated over O. Hence (OX ⊗ M)^{∗} ≃ OX ⊗ M^{∗} (see [7] prop 2.29).

2.2. Relative projectivity

The notion of relative projectivity appears under various forms in representation theory and category theory, with successive generalizations: projectivity of an OGmodule with respect to a subgroup of the group G (see D.G. Higman [12], J.A. Green [11]), with respect to a G-set (see [1] def. 3.6.13), with respect to a module (see [6]), projectivity of an object relative to a functor (see [4] def 4.1), projectivity of an object with respect to a cotriple ([16] def. 8.6.5). The definition I will use here is the second one:

(2.2.1) Definition: Let G be a finite group, and X be a finite G-set. An OG-module M is projective relative to X, or X-projective, if it satisfies one of the following equivalent conditions:

1. There exists an OG-module N such that M is a direct summand of OX ⊗ N.

2. The morphism from End_{O}(OX) ⊗ M to M mapping f ⊗ m to tr(f)m is a split epimorphism of OG-modules.

3. The morphism from OX ⊗ M to M mapping x ⊗ m to m (for x ∈ X and m ∈ M) is a split epimorphism of OG-modules.

4. For any diagram

\begin{aligned}& & M \\ & & ↓ & α \\ L & ⟶^{β} & N \end{aligned}

such that the morphism OX ⊗ β is a split epimorphism, there exists a morphism of OG-modules γ : M → L such that β ∘ γ = α.

(2.2.2) Remarks: In assertion 2), I denote by tr(f) the trace of f, which is well defined since OX is O-free and finitely generated, since X is finite. The module OX is its own dual over O, and the functor M ↦ OX ⊗M from the category of OG-modules to itself, is its own left and right adjoint. The morphism in condition 2) is the counit of this adjunction. Now the equivalence of conditions 1), 2) and 4) is standard (see [16] for details). The equivalence of 2) and 3) is straightforward.

It leads to the following version of Higman’s criterion:

(2.2.3) Lemma: Let G be a finite group, and X be a finite G-set. If M is an OGmodule, the following are equivalent:

1. The module M is X-projective.

2. There exists a map of G-sets φ from X to End_{O}(M) such that

∑_{x∈X} φ(x) = Id_{M}

Proof: An O-linear map α : M → OX ⊗ M is defined by

α(m) = ∑_{x∈X} x ⊗ α_{x}(m)

where α_{x} ∈ End_{O}(M). This defines a morphism of OG-modules if and only if g.α_{x} = α_{gx} for all x ∈ X and g ∈ G. Moreover α is a section of ε if and only if ∑_{x} α_{x} is the identity of M.

(2.2.4) Corollary: With the same notation

1. If M is an X-projective OG-module, then so is M ⊗ N, for any OG-module N.

2. If Y is a finite G-set such that Hom_{G−set}(X, Y) ≠ ∅, and if M is an X-projective OG-module, then M is Y -projective.

3. If M is an X-projective OG-module, then so is M^{∗}.

Proof: Consider a map of G-sets from X to End_{O}(M) such that ∑_{x∈X} φ(x) = Id_{M}. For the first assertion, consider the map from X to End_{O}(M ⊗ N) sending x ∈ X to φ(x) ⊗ Id_{N}. For the second one, if f : X → Y is a map of G-sets, define a map from Y to End_{O}(M) by ψ(y) = ∑_{x∈f^{−1}(y)} φ(x). For the third assertion, note that if M is a direct summand of OX ⊗ N, then M^{∗} is a direct summand of (OX ⊗ N)^{∗}, which is isomorphic to OX ⊗ N^{∗} since X is finite.

2.3. Relative Shanuel’s lemma

The previous definitions lead to the following relative version of Shanuel’s lemma:

(2.3.1) Proposition: Let G be a finite group, and X be a finite G-set. If

0 → L → N → M → 0
0 → L′ → N′ → M → 0

are X-split exact sequences of OG-modules such that N and N′ are projective relative to X, then there is an isomorphism of OG-modules

L ⊕ N′ ≃ L′ ⊕ N

Proof: see [1], Lemma 3.9.1 or Lemma 1.5.3.

(2.3.2) Notation: If X is a G-set, then let Ω_{X} = Ω_{X}(O) denote the kernel of the O-linear augmentation map ε : OX → O sending each x ∈ X to 1 ∈ O.

More generally, if M is an OG-module, then Ω_{X}(M) denotes the kernel of the map ε ⊗ Id_{M} : OX ⊗ M → O ⊗ M ≃ M. The module Ω_{X}(M) is called the syzygy of M relative to X. Thus if M is O-flat, then Ω_{X}(M) is isomorphic to Ω_{X} ⊗ M.

Thus Ω_{X} is the set of linear combinations ∑_{x∈X} r_{x}x of elements of X with coefficients in O, such that ∑_{x} r_{x} = 0. If x_{0} ∈ X is given, then Ω_{X} admits a basis over O, formed of the elements x − x_{0}, for x ∈ X − {x_{0}}, hence it is a free O-module.

(2.3.3) Lemma: [Alperin] Let G be a finite group, and X be a non-empty finite G-set. Then there is an isomorphism of OG-modules

(Ω_{X} ⊗ Ω_{X}^{∗}) ⊕ OX ⊕ OX ≃ O ⊕ (OX ⊗ OX)

Proof: Clearly the module OX is X-projective, by condition 1) of definition 2.2.1. Moreover, the morphism ε : OX → O is X-split: the diagonal map x ↦ x ⊗ x is a section of OX ⊗ ε.

The module Ω_{X} is O-free and finitely generated, so Ω_{X}^{∗} is also O-free and finitely generated. Tensoring with the (O-free, hence O-flat) module Ω_{X}^{∗} the exact sequence

(2.3.4)  0 → Ω_{X} → OX → O → 0

gives the exact sequence

(2.3.5)  0 → Ω_{X} ⊗ Ω_{X}^{∗} → OX ⊗ Ω_{X}^{∗} → Ω_{X}^{∗} → 0

The module OX ⊗ Ω_{X}^{∗} is X-projective, and the map OX ⊗ Ω_{X}^{∗} → Ω_{X}^{∗} is X-split since ε is.

On the other hand, dualizing the sequence 2.3.4 gives the sequence

(2.3.6)  0 → O → OX → Ω_{X}^{∗} → 0

This sequence is exact since O is O-free. The morphism OX → Ω_{X}^{∗} is also X-split, by corollary 2.1.5. Now using Shanuel’s lemma and sequences 2.3.5 and 2.3.6 gives the isomorphism

(Ω_{X} ⊗ Ω_{X}^{∗}) ⊕ OX ≃ O ⊕ (OX ⊗ Ω_{X}^{∗})

Adding OX on both sides gives the isomorphism

(Ω_{X} ⊗ Ω_{X}^{∗}) ⊕ OX ⊕ OX ≃ O ⊕ (OX ⊗ Ω_{X}^{∗}) ⊕ OX

Since the sequence

0 → OX → OX ⊗ OX → OX ⊗ Ω_{X}^{∗} → 0

is split, the direct sum (OX ⊗ Ω_{X}^{∗}) ⊕ OX is isomorphic to OX ⊗ OX. Finally, I get the required isomorphism

(Ω_{X} ⊗ Ω_{X}^{∗}) ⊕ OX ⊕ OX ≃ O ⊕ (OX ⊗ OX)

This shows in particular that Ω_{X} ⊗ Ω_{X}^{∗}, which is isomorphic to End_{O}(Ω_{X}) since Ω_{X} is O-free and finitely generated, is a direct summand of a permutation OG-module.

One can also view this lemma from the point of view of Rickard’s endo-split permutation resolutions (see [14]): a complex C of OG-modules is an endo-split permutation resolution of the OG-module M if it is a bounded complex of permutation OG-modules with homology concentrated in one degree, isomorphic to M, and such that the complex End_{O}(C) is a split complex of OG-modules. With this definition:

(2.3.7) Lemma: The complex C : 0 → OX →^{ε} O → 0 is an endo-split permutation resolution of Ω_{X}. Proof: Clearly C has homology concentrated in one degree, and equal to Ω_{X}. Moreover, the complex End_{O}(C) is

0 → Hom_{O}(O, OX) → End_{O}(OX) ⊕ End_{O}(O) → Hom_{O}(OX, O) → 0

Hence it is isomorphic to the complex

0 → OX ⟶^{a} End_{O}(OX) ⊕ O ⟶^{b} OX → 0

where the map a is defined by a(x) = (c_{x}, 1) ∈ End_{O}(OX) ⊕ O for x ∈ X, the endomorphism c_{x} of OX being given by c_{x}(y) = x for y ∈ X. The map b is defined by b(f) = ∑_{x,y∈X} v_{x,y}x for f ∈ End_{O}(OX) with matrix f_{x,y} in basis X, and b(1) = − ∑_{x∈X} x.

Since C is a complex of free O-modules, the homology of End_{O}(C) is concentrated in the middle term, and isomorphic to End_{O}(Ω_{X}).

Now the map a has a section s, defined by s(f) = ∑_{x∈X} f_{x,x}x for f ∈ End_{O}(OX), and s = 0 on O. The map b has a section t defined by t(x) = (δ_{x}, 0) ∈ End_{O}(X) ⊕ O, where δ_{x} is the endomorphism of OX defined by δ_{x}(y) = δ_{x,y}x for y ∈ X. Hence C is an endo-split permutation resolution of Ω_{X}. (2.3.8) Corollary: Let M be an O-projective finitely generated OG-module. If End_{O}(M) is a direct summand of a permutation OG-module, then End_{O}(Ω_{X}(M)) is a direct summand of a permutation OG-module. Proof: As I already noted above Ω_{X}(M) ≃ Ω_{X} ⊗ M since M is flat over O. Moreover since Ω_{X} and M are O-projective and finitely generated, so is Ω_{X}(M). Now for any finitely generated projective O-module M, and any OG-module N, the linear map from M ⊗ N^{∗} to Hom_{O}(N, M) sending m ⊗ f, for m ∈ M and f ∈ N^{∗} to the homomorphism from N to M defined by u ↦ f(u)m, for u ∈ M, is an isomorphism of abelian groups (see [7] prop 2.29). In case M and N are OG-modules, this map is an isomorphism of OG-modules. It follows that Ω_{X}(M)^{∗} ≃ Ω_{X}^{∗} ⊗ M^{∗}, and also that

End_{O}(Ω_{X}(M)) ≃ Ω_{X}(M) ⊗ Ω_{X}(M)^{∗}
≃ Ω_{X} ⊗ Ω_{X}^{∗} ⊗ M ⊗ M^{∗}
≃ End_{O}(Ω_{X}) ⊗ End_{O}(M)

is the tensor product of two direct summands of permutation modules. Hence it is a direct summand of permutation module.

3. The Dade group

From now on, following [15], I will suppose that the ring O is a complete noetherian local ring with residue field k of characteristic p > 0. Any direct summand of an O-free finitely generated O-module is O-free. If G is a finite group, then an OG-lattice is by definition an O-free finitely generated OG-module. Krull-Schmidt theorem holds for OG-lattices. If M is an indecomposable OG-lattice, then a vertex of M is a minimal subgroup P of G such that M is G/P -projective. It is unique up to conjugation in G, and it is a p-group.

3.1. Capped modules and permutation algebras

I will recall first some definitions and facts from Dade’s paper [9]:

(3.1.1) Definition: Let P be a p-group. An OP -lattice M is called an endo-permu-tation lattice if End_{O}(M) is a permutation OP -module, i.e. if it admits a P -invariant O-basis.

An endo-permutation OP -lattice M is said to be capped if it has some non-zero indecomposable direct summand with vertex P. Such a direct summand is unique up to isomorphism, and is called a cap of M, and denoted by cap(M).

Two capped endo-permutation OG-lattices M and N are said to be equivalent if cap(M) ≃ cap(N). This is equivalent to saying that O is a direct summand of M ⊗ N^{∗}.

Tensor product over O induces a group structure on the set of equivalence classes of capped endo-permutation lattices called the Dade group of capped endo-permutation OP -lattices, denoted by D′_{O}(P) (see [15] Remark 5.29.6).

If M is an endo-permutation OP -lattice, then A = End_{O}(M) is an O-simple algebra, i.e. it is isomorphic to a matrix algebra M_{n}(O) for some n (see [15] 1.7). Thus it is a permutation O-simple P -algebra. Moreover M is capped if and only if the Brauer quotient A[P] is non-zero (see [15] 5.28). This leads to the following definition:

(3.1.2) Definition: [see [15] 5.29] An O-simple permutation P -algebra such that A[P] ≠ 0 is called a Dade P -algebra (over O).

A Dade P -algebra is called neutral if there exists a permutation OP -lattice M having O as a direct summand, such that A ≃ End_{O}(M) as P -algebra.

Two Dade P -algebras A and B are called similar if A ⊗ B^{op} is neutral.

Tensor product of algebras induces a group structure on the set of equivalence classes of Dade P -algebras for this relation. This group is the Dade group of P, and is denoted by D_{O}(P). When the residue field k is algebraically closed, the natural group homomorphism D′_{O}(P) → D_{O}(P) mapping a capped OP -lattice to its endomorphism algebra is surjective (and even split see [15] Remark 5.29.6). The kernel is isomorphic to the group of one-dimensional characters of P (i.e. group homomorphisms from P to the multiplicative group of units of O). In particular if O = k is algebraically closed, then the groups D′_{O}(P) and D_{O}(P) are isomorphic.

3.2. Relative syzygies in the Dade group

Let P be a p-group. If X is a non-empty finite P -set, lemma 2.3.3 shows that there is an isomorphism of OP -lattices

(Ω_{X} ⊗ Ω_{X}^{∗}) ⊕ OX ⊕ OX ≃ O ⊕ (OX ⊗ OX)

Taking Brauer quotients gives

(3.2.1)  (Ω_{X} ⊗ Ω_{X}^{∗})[P] ⊕ k(X^{P}) ⊕ k(X^{P}) ≃ k ⊕ (k(X^{P}) ⊗ k(X^{P}))

It follows that the dimension over k of (Ω_{X} ⊗ Ω_{X}^{∗})[P] is

|X^{P} |^{2} − 2|X^{P} | + 1 = (|X^{P} | − 1)^{2}

This formula leads to the following:

(3.2.2) Definition: I will say that a non-empty finite P -set X is a Dade P -set if End_{O}(Ω_{X}) is a Dade P -algebra, or equivalently, if P does not admit a single fixed point on X (i.e. if X^{P} is empty or if |X^{P} | > 1).

(3.2.3) Notation: If P is a p-group, and X is a Dade P -set, I will denote by Ω_{X} the image of End_{O}(Ω_{X}(O)) in the Dade group D_{O}(P).

The following is elementary:

(3.2.4) Lemma: Let P be a p-group, and X be a Dade P -set.

1. If X^{P} ≠ ∅, then Ω_{X} = 0 in D_{O}(P).

2. If p ≠ 2 and if Ω_{X} = 0 in D_{O}(P), then X^{P} ≠ ∅.

Proof: If P has a fixed point on X, then the sequence

0 → Ω_{X} → OX →^{ε} O → 0

is split (mapping 1 ∈ O to x ∈ OX, for x ∈ X^{P}, is a section of ε). Thus Ω_{X} is a direct summand of OX, hence it is a permutation module, since P is a p-group. Hence Ω_{X} = 0 in D_{O}(P).

Conversely, if Ω_{X} = 0 in D_{O}(P), then there is a finite P -set Y such that Y^{P} ≠ ∅ and Ω_{X}(O) ⊗ Ω_{X}(O)^{∗} ≃ OY ⊗ OY as P -algebras. It follows in particular that (|X| − 1)^{2} = |Y |^{2}, for dimension reasons. Thus |X| = 1 + |Y |, since X is non-empty. Now if X^{P} = ∅, equation 3.2.1 gives

k(Y^{P} × Y^{P}) ≃ k

hence |Y^{P} | = 1. It follows that |Y | is congruent to 1 modulo p, thus |X| is congruent to 2 modulo p. Thus 2 ≡ |X| ≡ |X^{P} | = 0 modulo p, hence p = 2.

(3.2.5) Remark: Assertion 2) is false if p = 2: the Dade group D_{O}(P) of the cyclic group P of order 2 is trivial, thus Ω_{P/1} = 0, but P has no fixed points on P/1.

The first assertion of the previous lemma leads to the following

(3.2.6) Convention: If X is a finite P -set such that X^{P} ≠ ∅, I will set Ω_{X} = 0 in D_{O}(P). With this convention, the element Ω_{X} of D_{O}(P) is defined for any non-empty finite P -set X.

One could also define Ω_{∅} = 0 in D_{O}(P), since it is the only reasonable value for it. However, to avoid too many special cases in the proofs, I will not define Ω_{∅} in this paper.

(3.2.7) Lemma: Let P be a p-group. If X and Y are non-empty finite P -sets such that for any subgroup Q of P, the set X^{Q} is non-empty if and only if Y^{Q} is non-empty, then Ω_{X} = Ω_{Y} in D_{O}(P). Proof: The hypothesis on X and Y is equivalent to say that there is a morphism of P -sets from X to Y and a morphism of P -sets from Y to X. Now if X^{P} ≠ ∅, then Y^{P} ≠ ∅, and then Ω_{X} = Ω_{Y} = 0 in D_{O}(P). And if X^{P} = Y^{P} = ∅, in the sequence

0 → Ω_{Y} → OY → O → 0

the middle term is OX-projective by corollary 2.2.4, because is it OY -projective, and the right hand side morphism is OX-split by corollary 2.1.5, because it is OY -split. By Shanuel’s lemma, there is an isomorphism of OP -modules

Ω_{X} ⊕ OY ≃ Ω_{Y} ⊕ OX

As X^{P} = Y^{P} = ∅, the cap of Ω_{X} must be a direct summand of Ω_{Y}, since OX has no direct summand with vertex P. Hence cap(Ω_{X}) = cap(Ω_{Y}) in this case, and Ω_{X} = Ω_{Y} in D_{O}(P).

(3.2.8) Lemma: Let P be a p-group, let X be a non-empty finite P -set, and

0 → W → OX → V → 0

be an X-split exact sequence of OP -lattices. Then:

1. The lattice V is an endo-permutation OP -lattice if and only if W is an endopermutation OP -lattice.

2. If X^{P} = ∅, then V is a capped endo-permutation OP -lattice if and only if W is a capped endo-permutation OP -lattice.

3. If V and W are capped endo-permutation OP -lattices, then

W = Ω_{X} + V in D_{O}(P)

Proof: Tensoring the exact sequence of the lemma with W^{∗} gives an exact sequence

0 → W ⊗ W^{∗} → OX ⊗ W^{∗} → V ⊗ W^{∗} → 0

Dualizing the sequence of the lemma, and tensoring next with V, gives the exact sequence

0 → V ⊗ V^{∗} → V ⊗ OX → V ⊗ W^{∗} → 0

Now by Shanuel’s lemma

(3.2.9)  (W ⊗ W^{∗}) ⊕ (V ⊗ OX) ≃ (V ⊗ V^{∗}) ⊕ (OX ⊗ W^{∗})

Since the sequence of the lemma is X-split, there is moreover an isomorphism

OX ⊗ OX ≃ (OX ⊗ W) ⊕ (OX ⊗ V)

which gives also, since (OX ⊗ W)^{∗} ≃ OX ⊗ W^{∗}

OX ⊗ OX ≃ (OX ⊗ W^{∗}) ⊕ (OX ⊗ V^{∗})

It follows that

(W ⊗ W^{∗}) ⊕ (OX ⊗ V) ⊕ (OX ⊗ V^{∗}) ≃ (V ⊗ V^{∗}) ⊕ (OX ⊗ OX)

Hence W ⊗ W^{∗} is a permutation lattice if V ⊗ V^{∗} is. Similarly

(W ⊗ W^{∗}) ⊕ (OX ⊗ OX) ≃ (V ⊗ V^{∗}) ⊕ (OX ⊗ W^{∗}) ⊕ (OX ⊗ W)

and V ⊗ V^{∗} is a permutation lattice if W ⊗ W^{∗} is. This proves assertion 1).

If X^{P} = ∅, then the lattices OX, OX ⊗ W^{∗} and OX ⊗ V have no direct summand with vertex P. Equation 3.2.9 shows that

(W ⊗ W^{∗})[P] ≃ (V ⊗ V^{∗})[P]

Thus V has a direct summand of vertex P if and only if W does. This proves 2).

If V and W are capped, there are two cases: suppose first that X^{P} ≠ ∅. In this case V is a direct summand of OX ⊗ V, hence of OX ⊗ OX. If V is capped, then cap(V) is a permutation lattice of vertex P, hence isomorphic to O. Now W is also a direct summand of OX ⊗ W, hence of OX ⊗ OX. It follows that V = W = Ω_{X} = 0 in D_{O}(P), and the formula of assertion 3) holds.

And if X^{P} = ∅, the exact sequence

0 → Ω_{X} ⊗ V → OX ⊗ V → V → 0

and Shanuel’s lemma give

W ⊕ (OX ⊗ V) ≃ (Ω_{X} ⊗ V) ⊕ OX

This shows that the cap of W is a direct summand of Ω_{X} ⊗ V, since OX has no direct summand with vertex P. Hence W = Ω_{X} ⊗ V in the Dade group of P, as was to be shown.

4. Functorial properties of relative syzygies

4.1. Restriction, inflation, isomorphisms

(4.1.1) Lemma: Let P and Q be p-groups, and let f : P → Q be a group homomorphism. If M is an OQ-lattice (resp. if X is a Q-set), denote by Res_{f}M (resp. Res_{f}X) the OP -lattice obtained by restriction along f. Then if X is a non-empty finite Q-set, there is an isomorphism of OP -lattices

Res_{f}Ω_{X} ≃ Ω_{Res_{f} X}

Proof: Since Res_{f}OX ≃ ORes_{f}X, the restriction along f of the exact sequence

0 → Ω_{X} → OX →^{ε} O → 0

is the exact sequence

0 → Res_{f}Ω_{X} → ORes_{f}X →^{ε} O → 0

and the lemma follows. (4.1.2) Corollary:

1. Let Q ⊆ P be p-groups. If X is a non-empty finite P -set, then there is an isomorphism of OQ-lattices

Res_{Q}^{P}Ω_{X} ≃ Ω_{Res_{Q}^{P}X}

2. Let Q \underline{◁} P be p-groups. If X is a non-empty finite P/Q-set, then there is an isomorphism of OP -lattices

Inf_{P/Q}^{P}Ω_{X} ≃ Ω_{Inf_{P/Q}^{P}X}

3. Let φ : P → Q be a group isomorphism. If X is a non-empty finite P -set, then there is an isomorphism of OQ-modules

(Ω_{X})_{φ} = Ω_{X_{φ}}

where (Ω_{X})_{φ} (resp. X_{φ}) denotes the module Ω_{X} (resp. the set X) on which Q acts via φ^{−1}.

4.2. Deflation

Let Q \underline{◁} P be p-groups. One can define a deflation operation Def_{P/Q}^{P} : D_{O}(P) → D_{k}(P/Q) (called slash construction by Dade), in the following way: if M is a capped endo-permutation OP -module, then the Brauer quotient End_{O}(M)[Q] is a Dade P/Qalgebra over k, associated to an endo-permutation k(P/Q)-module Def_{P/Q}^{P}M. This construction is compatible with the equivalence relation defining the Dade group.

Comparison of equality in lemma 2.3.3 and equation 3.2.1 suggests the following: (4.2.1) Lemma: Let Q \underline{◁} P be p-groups, and X be a non-empty finite P -set.

1. If X^{Q} = ∅, then Def_{P/Q}^{P}Ω_{X} = 0 in D_{k}(P/Q).

2. If X^{Q} ≠ ∅, then Def_{P/Q}^{P}Ω_{X} = Ω_{X^{Q}} in D_{k}(P/Q)

Proof: Note first that Ω_{X} in the left hand side is in D_{O}(P), whereas Ω_{X^{Q}} in the right hand side is in D_{k}(P/Q). Actually for this lemma, I can suppose O = k.

If X^{P} ≠ ∅, then (X^{Q})^{P/Q} ≠ ∅, and there is nothing to prove. Now if X^{Q} = ∅, lemma 2.3.3 shows that (Ω_{X} ⊗ Ω_{X}^{∗})[Q] ≃ k as P/Q-algebras, and assertion 1) follows.

Finally if X^{P} = ∅, by lemma 2.3.7, the complex

(4.2.2)  0 → kX ⟶^{a} End_{k}(kX) ⊕ k ⟶^{b} kX → 0

is a split complex of kP -modules. Now (kX)[Q] ≃ k(X^{Q}) and since kX is a permutation module

(End_{k}(kX))[Q] ≃ End_{k}(k(X^{Q}))

The Brauer quotient at Q of 4.2.2 is the split complex of k(P/Q)-modules

0 → k(X^{Q}) ⟶^{a[Q]} End_{k}(k(X^{Q})) ⊕ k ⟶^{b[Q]} k(X^{Q}) → 0

with homology concentrated in the middle term. This homology can be viewed either as (End_{k}(Ω_{X}))[Q] or as End_{k}(Ω_{X^{Q}}). Hence those two algebras are isomorphic as (P/Q)-algebras, and the lemma follows.

5. Tensor induction

5.1. The formula for tensor induction

The previous lemmas describe the action of four of the five functorial operations on the Dade group. The only missing one is tensor induction, and this section will be devoted to the proof of the corresponding theorem. First I need some notation: (5.1.1) Notation: Let P be a group. I denote by s_{P} the set of conjugacy classes of subgroups of P. It is ordered by the following relation: if C and C′ are elements of s_{P}, then C ≤_{P} C′ if there exists Q ∈ C and Q′ ∈ C′ such that Q ⊆ Q′. I denote by μ_{P} the Möbius function of the poset s_{P}.

If Q is a subgroup of P, I denote by \overline{Q} its conjugacy class in P, viewed as an element of s_{P}. If Q and Q′ are subgroups of P, the notation Q ≤_{P} Q′ means that some conjugate of Q in P is contained in Q′, or equivalently, that \overline{Q} ≤_{P} \overline{Q′}. In this case, I will write μ_{P} (Q, Q′) instead of μ_{P} (\overline{Q}, \overline{Q′}).

Finally, I denote by [s_{P}] a set of representatives of s_{P}.

(5.1.2) Theorem: Let Q ⊆ P be p-groups. Let X be a non-empty finite Q-set. Then in the Dade group D_{O}(P)

Ten_{Q}^{P}Ω_{X} = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |\{a ∈ V \P/Q | X^{V^{a}∩Q} ≠ ∅\}| Ω_{P/U}

I will prove this theorem by induction on |P : Q|. There will be three steps: first the case P = Q, then the inductive step when |P : Q| ≥ p^{2}, and finally the case |P : Q| = p.

5.2. The case P = Q

This case is an easy consequence of the following lemma, due to J. Thévenaz:

(5.2.1) Lemma: [Thévenaz] Let P be a p-group. If X and Y are non-empty finite P -sets, then

Ω_{X⊔Y} + Ω_{X×Y} = Ω_{X} + Ω_{Y} in D_{O}(P)

Proof: Note first that the formula holds if X^{P} ≠ ∅: in this case Ω_{X} = 0, and Ω_{X⊔Y} = 0. Moreover there exists a morphism of P -sets from X ×Y to Y (the projection), and from Y to X × Y (mapping y ∈ Y to (x, y), for a given x ∈ X^{P}). Thus Ω_{Y} = Ω_{Y ×X}, and the formula holds. Hence I can assume X^{P} = Y^{P} = ∅ by symmetry.

Let C denote the complex 0 → OX → O → 0, and similarly let D denote the complex 0 → OY → O → 0. The tensor product of those complexes is the complex

0 → OX ⊗ OY → OX ⊕ OY → O → 0

which is isomorphic to the complex

0 ⟶ O(X × Y) ⟶^{a} O(X ⊔ Y) ⟶^{b} O ⟶ 0

where a is defined by a(x, y) = x − y ∈ O(X ⊔ Y) for x ∈ X and y ∈ Y, and b is defined by b(x) = b(y) = 1 for x ∈ X and y ∈ Y.

Since C and D are complexes of free O-modules, with homology concentrated in one degree, equal to Ω_{X} and Ω_{Y} respectively, the only non-zero homology group of C ⊗ D is the kernel of a, and it is isomorphic to Ω_{X} ⊗ Ω_{Y}. Now the kernel of b is Ω_{X⊔Y}, and there is an exact sequence

(5.2.2)  0 ⟶ Ω_{X} ⊗ Ω_{Y} ⟶ O(X × Y) ⟶^{a} Ω_{X⊔Y} ⟶ 0

Now the map a is (X × Y)-split: if (x_{0}, y_{0}) is given in X × Y, define a map σ_{x_{0},y_{0}} from Ω_{X⊔Y} to O(X × Y) by

σ_{x_{0},y_{0}}(∑_{x∈X} α_{x}x + ∑_{y∈Y} β_{y}y) = ∑_{x∈X} α_{x}(x, y_{0}) − ∑_{y∈Y} β_{y}(x_{0}, y) − ∑_{x∈X} α_{x}(x_{0}, y_{0})

The element ∑_{x∈X} α_{x}x+∑_{y∈Y} β_{y}y is in Ω_{X⊔Y} if and only if ∑_{x∈X} α_{x} +∑_{y∈Y} β_{y} = 0 Now

a∘σ_{x_{0},y_{0}}(∑_{x∈X} α_{x}x+ ∑_{y∈Y} β_{y}y) = ∑_{x∈X} α_{x}x+∑_{y∈Y} β_{y}y−∑_{x∈X} α_{x}y_{0}−∑_{y∈Y} β_{y}x_{0}−∑_{x∈X} α_{x}(x_{0}−y_{0})

Hence σ_{x_{0},y_{0}} is a section of a, and the map (x_{0}, y_{0}) ↦ σ_{x_{0},y_{0}} is clearly a map of P -sets from X × Y to Sec(a).

Now it follows from lemma 3.2.8 that in D_{O}(P)

Ω_{X} + Ω_{Y} = Ω_{X×Y} + Ω_{X⊔Y}

as was to be shown.

In the case P = Q, the equality to prove in theorem 5.1.2 takes the form of the following lemma:

(5.2.3) Lemma: Let P be a p-group and X be a non-empty finite P -set. Then in D_{O}(P)

(5.2.4)  Ω_{X} = ∑_{U,V ∈[s_{P}] ; {U≤_{P} V \atop X^{V} ≠∅}} μ_{P} (U, V)Ω_{P/U}

Proof: Suppose first that X is isomorphic to a disjoint union of copies of a single transitive P -set, isomorphic to P/S for some subgroup S of P. Then X^{V} is non-empty if and only if V ≤_{P} S. The sum in the right hand side of 5.2.4 is

∑_{{U,V ∈[s_{P}] \atop U≤_{P} V ≤_{P} S}} μ_{P} (U, V)Ω_{P/U}

Now for a given \overline{U} ∈ s_{P}, the sum for \overline{V} in the “interval” [\overline{U}, \overline{S}] of s_{P} of μ_{P} (U, V) is zero if \overline{U} ≠ \overline{S}, by definition of the Möbius function on s_{P}, and it is equal to 1 otherwise. The only term left in the previous sum is obtained for \overline{U} = \overline{S}, hence the sum is equal to Ω_{P/S}. On the other hand Ω_{X} = Ω_{P/S} in this case, by lemma 3.2.7. Thus formula 5.2.4 holds in this case.

For the general case, denote by Ω′_{X} the right hand side of 5.2.4. Clearly Ω′_{X} only depends on the set

F (X) = {Q ∈ [s_{P}] | X^{Q} ≠ ∅}

and by lemma 3.2.7 the same is true for Ω_{X}. I will prove that Ω_{X} = Ω′_{X} by induction on the cardinality of F (X).

If this cardinality is equal to 1, then P acts freely on X, hence X is a union of copies of P/1, and Ω_{X} = Ω′_{X} in this case.

If |F (X)| > 1, choose a maximal element Q of F (X). Then up to isomorphism, the set X can be written as Y ⊔ Z, where Z is a non-empty union of copies of P/Q, and Y is a set such that Y^{Q} = ∅.

If Y = ∅, then X = Z is a union of copies of P/Q, and the formula holds in this case.

Now if Y ≠ ∅, it is clear that |F (Y)| < |F (X)| since F (Y) ⊆ F (X) and Q ∈ F (X)−F (Y). Similarly F (Y ×Z) ⊆ F (Y) and |F (Y ×Z)| < |F (X)|. Hence formula 5.2.4 holds for Y and Y ×Z by induction hypothesis. It also holds for Z since Z is isomorphic to a union of copies of P/Q. Now by lemma 5.2.1

Ω_{X} = Ω_{Y} + Ω_{Z} − Ω_{Y ×Z}

It follows that

(5.2.5)  Ω_{X} = Ω′_{Y} + Ω′_{Z} − Ω′_{Y ×Z}

For a subgroup V of P set

Σ_{V} = ∑_{{U∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V)Ω_{P/U}

With this notation

Ω′_{X} = ∑_{V ∈F (X)} Σ_{V}

and equality 5.2.5 becomes

Ω_{X} = ∑_{V ∈F (Y)} Σ_{V} + ∑_{V ∈F (Z)} Σ_{V} − ∑_{V ∈F (Y ×Z)} Σ_{V}

Since F (Y × Z) = F (Y) ∩ F (Z), this is also

Ω_{X} = ∑_{V ∈F (Y)∪F (Z)} Σ_{V}

and since F (Y) ∪ F (Z) = F (Y ⊔ Z) = F (X), this gives finally

Ω_{X} = ∑_{V ∈F (X)} Σ_{V} = Ω′_{X}

This completes the proof of lemma 5.2.3.

5.3. The inductive step |P : Q| ≥ p^{2}

Suppose that formula of theorem 5.1.2 holds for all p-groups Q′ ⊆ P′ with index |P′ : Q′| < |P : Q|. Choose a subgroup S such that Q ⊂ S ⊂ P, the inclusions being proper ones. This is possible if |P : Q| ≥ p^{2}.

By induction hypothesis

Ten_{Q}^{S}Ω_{X} = ∑_{{U,V ∈[s_{S}] \atop U≤_{S}V}} μ_{S}(U, V) |\{a ∈ V \S/Q | X^{V^{a}∩Q} ≠ ∅\}| Ω_{S/U}

in D_{O}(S). Taking tensor induction up to P gives

Ten_{Q}^{P}Ω_{X} = ∑_{{U,V ∈[s_{S}] \atop U≤_{S}V}} μ_{S}(U, V) |\{a ∈ V \S/Q | X^{V^{a}∩Q} ≠ ∅\}| Ten_{S}^{P} Ω_{S/U}

since tensor induction is transitive. Since |P : S| < |P : Q|, by induction hypothesis

Ten_{S}^{P} Ω_{S/U} = ∑_{{A,B∈[s_{P}] \atop A≤_{P} B}} μ_{P} (A, B) |\{b ∈ B\P/S | (S/U)^{B^{b}∩S} ≠ ∅\}| Ω_{P/A}

Thus

Ten_{Q}^{P}Ω_{X} = ∑_{{U,V ∈[s_{S}] \atop U≤_{S}V}} μ_{S}(U, V) ∑_{{a∈V \S/Q \atop X^{V ^{a}∩Q}≠∅}} ∑_{{A,B∈[s_{P}] \atop A≤_{P} B}} μ_{P} (A, B) ∑_{{b∈B\P/S \atop B^{b}∩S≤_{S}U}} Ω_{P/A}

Changing the order of summations gives

Ten_{Q}^{P}Ω_{X} = ∑_{{A,B∈[s_{P}] \atop A≤_{P} B}}μ_{P} (A, B)( ∑_{b∈B\P/S} ∑_{V ∈[s_{S}]} ∑_{{a∈V \S/Q \atop X^{V ^{a}∩Q}≠∅}} ∑_{{U∈[s_{S}] \atop B^{b}∩S≤_{S}U≤_{S}V}} μ_{S}(U, V))Ω_{P/A}

The inner summation on U is zero unless V =_{S} B^{b} ∩ S. It follows that

Ten_{Q}^{P}Ω_{X} = ∑_{{A,B∈[s_{P}] \atop A≤_{P} B}} μ_{P} (A, B) ( ∑_{b∈B\P/S} ∑_{{a∈B^{b}∩S\S/Q \atop X^{(B^{b}∩S)^{a}∩Q}≠∅}} 1) Ω_{P/A}

Now (B^{b} ∩ S)^{a} ∩ Q = B^{ba} ∩ S ∩ Q = B^{ba} ∩ Q. Moreover, when b runs through B\P/S and a runs through B^{b} ∩ S\S/Q, the element c = ba runs through B\P/Q. Hence

Ten_{Q}^{P}Ω_{X} = ∑_{{A,B∈[s_{P}] \atop A≤_{P} B}} μ_{P} (A, B)( ∑_{{c∈B\P/Q \atop X^{B^{c}∩Q}≠∅}} 1)Ω_{P/A}

and finally

Ten_{Q}^{P}Ω_{X} = ∑_{{A,B∈[s_{P}] \atop A≤_{P} B}} μ_{P} (A, B) |\{c ∈ B\P/Q | X^{B^{c}∩Q} ≠ ∅\}| Ω_{P/A}

which completes the proof of the inductive step. 5.4. The case |P : Q| = p

(a) The case X^{Q} ≠ ∅. If X^{Q} ≠ ∅, the formula to check is the following

0 = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |P : V Q| Ω_{P/U}

The right hand side is also

S = ∑_{{U,V ∈[s_{P}] \atop U≤V ≤Q}} μ_{P} (U, V)pΩ_{P/U} + ∑_{{U,V ∈[s_{P}] \atop U≤V ≰Q}} μ_{P} (U, V)Ω_{P/U}

since either V ⊆ Q or V Q = P. As Ω_{P/P} = 0, I can fix a subgroup U ≠ P and look at the sum on V

∑_{{V ∈[s_{P}] \atop U≤_{P} V ≤P}} μ_{P} (U, V) = 0 = ∑_{{V ∈[s_{P}] \atop U≤_{P} V ≤Q}} μ_{P} (U, V) + ∑_{{V ∈[s_{P}] \atop U≤_{P} V ≰Q}} μ_{P} (U, V)

It follows that

S = ∑_{{U∈[s_{P}] \atop U≠P}} ∑_{{V ∈[s_{P}] \atop U≤_{P} V ≤Q}} (p − 1)μ_{P} (U, V)Ω_{P/U}

The sum on V is zero if U ≠ Q. Finally

S = (p − 1)Ω_{P/Q}

Then either p is odd, and this is zero since Ω_{P/Q} = Inf_{P/Q}^{P}Ω_{(P/Q)/(Q/Q)} has order 2 in D_{O}(P), or p = 2 and Ω_{P/Q} = 0, since it is inflated from D_{O}(P/Q) = 0. In both cases, the formula holds.

(b) The case X^{Q} = ∅. I denote by C the complex 0 → OX → O → 0, where O is in degree 0 and OX is in degree 1. It is a complex of O-free OQ-modules. I consider next the tensor induced complex T = Ten_{Q}^{P}C (see [2] section 4.1): for this, I choose a total ordering of a set [P/Q] of representatives of P/Q. Now the module T_{j} is isomorphic to

T_{j} = ⨁_{{A⊆[P/Q] \atop |A|=j}}(OX)^{⊗j}

It can also be viewed as the free O-module with basis the set of pairs (A, φ), where A is a subset of cardinality j of [P/Q] and φ is a map from A to X, or equivalently as the free O-module with basis the pairs (A, φ), where A is a subset of P such that AQ = A and |A/Q| = j, and φ is a Q-invariant map from A to X.

The action of g ∈ P on T_{j} is given by

g.(A, φ) = ε(A, g)(gA, gφ)

where gφ is the function from gA to X defined by

(gφ)(a) = φ(g^{−1}a)

and ε(A, g) is a sign, defined by

ε(A, g) = (−1)^{|{(a,b)∈[A/Q]^{2}|a<b, ga>gb}|}

where [A/Q] is the set of elements of [P/Q] representing elements of A, and the notation ga > gb means that this inequality holds between the representatives of ga and gb in [P/Q].

Therefore the module T_{j} is isomorphic as an OP -module to

⨁_{A⊆P ; {AQ=A,|A/Q|=j \atop A mod. P}} Ind_{P_{A}}^{P}O^{ε}Hom_{Q}(A^{op}, X)

where P_{A} denotes the stabilizer of A in P, and O^{ε}Hom_{Q}(A^{op}, X) is the free O-module with basis the set of Q-invariant maps from A to X, on which the element g ∈ P_{A} acts by

g.(A, φ) = ε(A, g)(A, gφ)

and ε(A, g) is the signature of g acting on A/Q. Note that if p is odd, this signature is always +1, since cycles of odd length are even. If p = 2 and O has characteristic 2, this sign is also +1=-1. Finally, if j ≠ p, then P_{A} is a proper subgroup of P, containing Q, hence equal to Q, and P_{A} acts trivially on A/Q since Q \underline{◁} P. The only case where signs do appear is the case j = p = 2 and O has characteristic different from 2. All other modules T_{j} are permutation modules.

The differential d_{j} : T_{j} → T_{j−1} is given on the element (A, φ), for A = a_{1}Q⊔…⊔a_{j}Q with a_{i} ∈ [P/Q] for 1 ≤ i ≤ j such that a_{1} < a_{2} < … < a_{j}, by

d_{j}((A, φ)) = ∑_{l=1}^{j}(−1)^{l−1}(A−a_{l}Q, φ_{|A−a_{l}Q})

Since C is a complex of free O-modules, with homology concentrated in degree one, isomorphic to Ω_{X}, the complex T has its homology concentrated in degree p. This homology group is ‘almost’ isomorphic to Ten_{Q}^{P}Ω_{X}: if p = 2 and O has characteristic different from 2, one has to take the sign into account: the element g of P acts on the module T_{p} = (OX)^{⊗p} as it acts on Ten_{Q}^{P}OX, up to a sign which is the signature of the permutation induced by g on the cosets P/Q, equal to 1 if g ∈ Q and to -1 if g ∉ Q.

Let j be an integer, with 1 ≤ j ≤ p. I will denote by Z_{j} the set of pairs (A, φ), where A is a subset of P such that AQ = A and |A/Q| = j, and φ ∈ Hom_{Q}(A^{op}, X). The set Z_{j} is a P -set, and the previous remarks show that T_{i} is isomorphic to the permutation module OZ_{i} if i < p. In particular the module T_{i} is Z_{i} projective.

(5.4.1) Lemma: The differential d_{i} : T_{i} → T_{i−1} is Z_{j}-split if i < p and i ≤ j. Proof: I will fix (A, φ) ∈ Z_{j}, where the set A is written as

A = a_{1}Q ⊔ … ⊔ a_{j}Q with ∀l, 1 ≤ l ≤ j, a_{l} ∈ [P/Q], a_{1} < a_{2} < … < a_{j}

I define a map θ_{i} from T_{i−1} to T_{i} in the following way: if (B, ψ) ∈ Z_{i−1}, where B can be written as

B = b_{1}Q ⊔ … ⊔ b_{i−1}Q with ∀l, 1 ≤ l ≤ i − 1, b_{l} ∈ [P/Q], b_{1} < b_{2} < … < b_{i−1}

denote by l_{B} the least integer l such that a_{l} ∉ B. Such an integer exists since i − 1 < j. Set I(B) = {a_{1}, …, a_{l_{B}−1}}, and denote by r_{B} the least integer r such that a_{l_{B}} < b_{r} (or r_{B} = i if a_{l_{B}} > b_{r} for all r). Now I set

θ_{i}((B, ψ)) = ∑_{U⊆I(B)}(−1)^{|U|+r_{B}−1}(B ⊔ a_{l_{B}}Q, [φ, ψ]_{U})

where [φ, ψ]_{U} is the map from B ⊔ a_{l_{B}}Q to X defined by

[φ, ψ]_{U}(c) = \{ \begin{matrix}ψ(c) & if c ∈ B−UQ \\ φ(c) & if c ∈ UQ ⊔ a_{l_{B}}Q \end{matrix}

The map θ_{i} in invariant under the action of the stabilizer P_{(A,φ)} of (A, φ) in P: indeed, since i < p, the stabilizer of A is equal to Q, and P_{(A,φ)} is the set of elements q of Q such that

φ(a) = φ(q^{−1}a) = φ(a(q^{−1})^{a}) = q^{a}φ(a)

for all a ∈ A. In other words

P_{(A,φ)} = ∩_{a∈A}^{a}Q_{φ(a)}

It acts trivially on P/Q. Thus if q ∈ P_{(A,φ)}, then q^{−1}C = C for any subset C of P such that CQ = C, and

qθ_{i}((q^{−1}B, q^{−1}ψ)) = ∑_{U⊆I(B)}(−1)^{|U|+r_{B}−1}q(B ⊔ a_{l_{B}}Q, [φ, q^{−1}ψ]_{U})
= ∑_{U⊆I(B)}(−1)^{|U|+r_{B}−1}(B ⊔ a_{l_{B}}Q, q[φ, q^{−1}ψ]_{U})

Moreover

q[φ, q^{−1}ψ]_{U}(c) = [φ, q^{−1}ψ]_{U}(q^{−1}c) = \{ \begin{matrix}(q^{−1}ψ)(q^{−1}c) = ψ(c) & if c ∈ B−UQ \\ φ(q^{−1}c) = φ(c) & if c ∈ UQ ⊔ a_{l_{B}}Q \end{matrix}

It follows that θ_{i} is P_{(A,φ)}-invariant.

I will show that d_{i}θ_{i} + θ_{i−1}d_{i−1} = Id_{T_{i−1}}. First

d_{i}θ_{i}((B, ψ)) = ∑_{U⊆I(B)}(−1)^{|U|+r_{B}−1}d_{i}(B ⊔ a_{l_{B}}Q, [φ, ψ]_{U})

and moreover

d_{i}(B ⊔ a_{l_{B}}Q, [φ, ψ]_{U}) = ∑_{m=1}^{r_{B}−1}(−1)^{m−1}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})+

+(−1)^{r_{B}−1}(B, f_{U})+

+ ∑_{m=r_{B}}^{i−1}(−1)^{m−1+1}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

where f_{m,U} denotes the restriction of [φ, ψ]_{U} to B ⊔ a_{l_{B}}Q−b_{m}Q, and f_{U} is the map equal to ψ on B−UQ and to φ on UQ.

Finally this gives

d_{i}θ_{i}((B, ψ)) = S_{1} + S_{2} + S_{3}

where S_{1}, S_{2} and S_{3} are given by

S_{1} = ∑_{U⊆I(B)} ∑_{m=1}^{r_{B}−1}(−1)^{|U|+r_{B}+m}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

S_{2} = ∑_{U⊆I(B)}(−1)^{|U|+r_{B}+r_{B}}(B, f_{U})

S_{3} = ∑_{U⊆I(B)} ∑_{m=r_{B}}^{i−1}(−1)^{|U|+r_{B}+m−1}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

On the other hand

θ_{i−1}d_{i−1}((B, ψ)) = θ_{i−1}( ∑_{m=1}^{i−1}(−1)^{m−1}(B−b_{m}Q, ψ_{|B−b_{m}Q}))

To compute θ_{i−1}((B −b_{m}Q, ψ_{|B−b_{m}Q})), I must distinguish the case b_{m} ∈ A and m ≤ r_{B} − 1, the case b_{m} ∉ A and m ≤ r_{B} − 1, and the case m ≥ r_{B}.

If m ≤ r_{B} − 1 and b_{m} = a_{n}, then l_{B−b_{m}Q} = n and I(B −b_{m}Q) = {a_{1}, …, a_{n−1}}. Moreover r_{B−b_{m}Q} = m since b_{m+1} is the least element of (B−b_{m}Q)/Q greater than a_{n}, and the m^{th} element of (B−b_{m}Q)/Q . In this case (B−b_{m}Q) ⊔ a_{n}Q = B and

θ_{i−1}((B−b_{m}Q, ψ_{|B−b_{m}Q})) = ∑_{U⊆{a_{1},...,a_{n−1}}}(−1)^{|U|+m−1}(B, g_{n,U})

where g_{n,U} is equal to ψ on B−(UQ ⊔ a_{n}Q) and to φ on UQ ⊔ a_{n}Q.

If m ≤ r_{B} − 1 and b_{m} ∉ A, then l_{B−b_{m}Q} = l_{B} and I(B −b_{m}Q) = I(B). Moreover r_{B−b_{m}Q} = r_{B} − 1. In this case

θ_{i−1}((B−b_{m}Q, ψ_{|B−b_{m}Q})) = ∑_{U⊆I(B)}(−1)^{|U|+r_{B}}((B ⊔ a_{l_{B}}Q)−b_{m}Q, h_{m,U})

where h_{m,U} is the map equal to ψ on (B ⊔ a_{l_{B}}Q)−(b_{m}Q ⊔ UQ) and to φ on a_{l_{B}}Q ⊔ UQ.

If m ≥ r_{B}, then again l_{B−b_{m}Q} = l_{B} and I(B−b_{m}Q) = I(B). Moreover r_{B−b_{m}Q} = r_{B}, and

θ_{i−1}((B−b_{m}Q, ψ_{|B−b_{m}Q})) = ∑_{U⊆I(B)}(−1)^{|U|+r_{B}−1}((B ⊔ a_{l_{B}}Q)−b_{m}Q, h_{m,U})

with the same notation.

Finally, this gives

θ_{i−1}d_{i−1}((B, ψ)) = S′_{1} + S′_{2} + S′_{3}

where S′_{1}, S′_{2} and S′_{3} are given by

S′_{1} = ∑_{{1≤m≤r_{B}−1 \atop b_{m}=a_{n}∈A}} ∑_{U⊆\{a_{1},...,a_{n−1}\}}(−1)^{m−1+|U|+m−1}(B, g_{n,U})

S′_{2} = ∑_{{1≤m≤r_{B}−1 \atop b_{m}∉A}} ∑_{U⊆I(B)}(−1)^{m−1+|U|+r_{B}}((B ⊔ a_{l_{B}}Q)−b_{m}Q, h_{m,U})

S′_{3} = ∑_{r_{B}≤m≤i−1} ∑_{U⊆I(B)}(−1)^{m−1+|U|+r_{B}−1}((B ⊔ a_{l_{B}}Q)−b_{m}Q, h_{m,U})

Replacing U by U ⊔ {a_{n}}, the double summation in S′_{1} can also be written

S′_{1} = ∑_{∅≠U⊆I(B)}(−1)^{|U|−1}(B, f_{U})

It follows in particular that the only non-zero term in S_{2} + S′_{1} is obtained in S_{2} for U = ∅, thus

S_{2} + S′_{1} = (−1)^{0}(B, f_{∅}) = (B, ψ)

Now the double summation in S_{1} looks like the double summation in S′_{2}: both are indexed by a subset U of I(B) and an integer m such that 1 ≤ m ≤ r_{B} − 1. In S′_{2}, I require moreover that b_{m} ∉ A. Apart from this, the general term of S_{1} is

(−1)^{|U|+r_{B}+m}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

whereas the general term of S′_{2} is

(−1)^{m−1+|U|+r_{B}}((B ⊔ a_{l_{B}}Q)−b_{m}Q, h_{m,U})

Both maps f_{m,U} and h_{m,U} are equal to ψ on the set (B ⊔ a_{l_{B}}Q)−(b_{m}Q ⊔ UQ) and to φ on the set (a_{l_{B}}Q ⊔ UQ)−b_{m}Q. Hence the corresponding terms vanish. The only left terms in S_{1} + S′_{2} are the terms of S_{1} for which b_{m} ∈ A. Thus

S_{1} + S′_{2} = ∑_{U⊆I(B)} ∑_{{1≤m≤r_{B}−1 \atop b_{m}∈A}}(−1)^{|U|+r_{B}+m}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

Switching the summation gives

S_{1} + S′_{2} = ∑_{{1≤m≤r_{B}−1 \atop b_{m}∈A}} ∑_{U⊆I(B)}(−1)^{|U|+r_{B}+m}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

Now for a given m and a given b_{m} ∈ A, the summation on U can be split as

∑_{{U⊆I(B) \atop b_{m}∈U}}(−1)^{|U|+r_{B}+m}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})+

+ ∑_{{U⊆I(B) \atop b_{m}∉U}}(−1)^{|U|+r_{B}+m}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

Now every U ∋ b_{m} can be written as U′ ⊔ {b_{m}}, where U′ ∌ b_{m}. Moreover the map f_{m,U′} is equal to ψ on

(B ⊔ a_{l_{B}}Q)−(b_{m}Q ∪ UQ) = (B ⊔ a_{l_{B}}Q)−(b_{m}Q ⊔ U′Q)

and to φ on (a_{l_{B}}Q ⊔ U′Q)−b_{m}Q = (a_{l_{B}}Q ⊔ UQ)−b_{m}Q. Since moreover |U| = |U′| + 1, the signs are opposite, and S_{1} + S′_{2} = 0.

Finally the double summation in S_{3} looks like the summation in S′_{3}: both are indexed by a subset U of I(B) and an integer m ≥ r_{B}. The general term of S_{3} is

(−1)^{|U|+r_{B}+m−1}((B ⊔ a_{l_{B}}Q)−b_{m}Q, f_{m,U})

whereas the general term of S′_{3} is

(−1)^{m−1+|U|+r_{B}−1}((B ⊔ a_{l_{B}}Q)−b_{m}Q, h_{m,U})

Here again, the maps f_{m,U} and h_{m,U} are the same. As the signs are opposite, the sum S_{3} + S′_{3} is zero.

This shows finally that

(d_{i}θ_{i} + θ_{i−1}d_{i−1})((B, ψ)) = S_{1} + S_{2} + S_{3} + S′_{1} + S′_{2} + S′_{3} = (B, ψ)

and d_{i}θ_{i} + θ_{i−1}d_{i−1} is the identity map of T_{i−1}. It follows that d_{i} = d_{i} ∘ θ_{i} ∘ d_{i}, as claimed.

Now for i < p, there is an exact sequence

0 ⟶ Kerd_{i} ⟶ T_{i} ≃ OZ_{i} ⟶^{d_{i}} Kerd_{i−1} ⟶ 0

and for every (A, φ) ∈ Z_{j}, for j ≥ i, the restriction of θ_{i} to Kerd_{i−1} is a section of d_{i}, which is invariant under the action of P_{(A,φ)}. By lemma 2.1.2, this means that this sequence is Z_{j}-split.

Moreover T_{i} ≃ OZ_{i} is Z_{i}-projective. Since the stabilizer of (A, φ) is contained in Q, the group P has no fixed points on Z_{i}. An easy induction using lemma 3.2.8, starting from the sequence

0 → Kerd_{1} → T_{1} → T_{0} ≃ O → 0

shows that Kerd_{i} is a capped endo-permutation OP -lattice, and that

(5.4.2)  Kerd_{i} = Ω_{Z_{i}} + Kerd_{i−1}

in D_{O}(P). At the last stage i = p, there is an exact sequence

(5.4.3)  0 → O^{ε} ⊗ Ten_{Q}^{P}Ω_{X} → T_{p} ≃ O^{ε} ⊗ Ten_{Q}^{P}OX → Kerd_{p−1} → 0

where O^{ε} is the module O on which g ∈ P acts by the sign of the permutation it induces on P/Q. The module Ten_{Q}^{P}OX is the permutation module OZ_{p}, and Z_{p} identifies with the set of Q-equivariant maps from P to X. In particular T_{p} is Z_{p}-projective.

The submodule T′_{P} = O^{ε} ⊗ Ten_{Q}^{P}Ω_{X} of T_{p} can be viewed as the set of linear combinations

∑_{φ∈Z_{p}} λ_{φ}φ

such that for all a ∈ [P/Q] and all Q-invariant maps ρ from P −aQ to X, the sum of the coefficients λ_{φ} corresponding to those φ’s for which φ_{|P−aQ} = ρ is zero.

Now the inclusion T′_{p} → T_{p} is Z_{p}-split: if φ ∈ Z_{p} is given, define a map θ from T_{p} to T′_{p} by

θ(ψ) = ∑_{U⊆[P/Q]}(−1)^{|U|}[φ, ψ]_{U}

where [φ, ψ]_{U} is the map equal to ψ on P −UQ, and to φ on UQ. The element θ(ψ) is in T′_{p}, for if a ∈ [P/Q] is given, and if a ∉ U, the maps [φ, ψ]_{U} and [φ, ψ]_{U⊔{a}} have the same restriction to P −aQ, and their coefficients in the expression of θ(ψ) are opposite.

Let g be an element of the stabilizer of φ in P, and let ε_{g} be the signature of the permutation it induces on P/Q. Then g maps the element ψ of T_{p} to ε_{g}(g.ψ), and

θ(ε_{g}gψ) = ∑_{U⊆[P/Q]}(−1)^{|U|}ε_{g}[φ, gψ]_{U}

It is clear moreover that

[φ, g.ψ]_{U} = [gφ, g.ψ]_{U} = g([φ, ψ]_{g^{−1}U})

so that replacing U by g^{−1}U in the summation gives

θ(ε_{g}gψ) = ∑_{U⊆[P/Q]}(−1)^{|U|}ε_{g}g([φ, ψ]_{g^{−1}U}) = g ∑_{U⊆[P/Q]}(−1)^{|U|}[φ, ψ]_{U} = gθ(ψ)

Hence θ is invariant by the stabilizer of φ. Moreover, it is a section of the inclusion of T′_{p} into T_{p}: indeed if f = ∑_{ψ∈Z_{p}} λ_{ψ}ψ is an element of T′_{p}, then

θ(f) = ∑_{ψ∈Z_{p}} λ_{ψ} ∑_{U⊆[P/Q]}(−1)^{|U|}[φ, ψ]_{U}

The coefficient μ_{h} of a given h ∈ Z_{p} in this summation is equal to

μ_{h} = ∑_{U⊆S}(−1)^{|U|} ∑_{ψ∈R_{U}} λ_{ψ}

where S is the set of elements a of [P/Q] for which φ(a) = h(a), and R_{U} is the set of elements ψ of Z_{p} which coincide with h on P −UQ.

Now if U ≠ ∅, the sum of coefficients λ_{ψ} for ψ in R_{U} is zero: indeed, if a ∈ U is given, this sum is equal to

∑_{g∈E_{a,U,h}} ∑_{{ψ∈Z_{p} \atop ψ_{|P−aQ}=g_{|P−aQ}}} λ_{ψ}

where E_{a,U,h} is the set of Q-equivariant maps from P −aQ to X which coincide with h on P −UQ. The inner sum is zero if f ∈ T′_{p}. Thus in the expression of μ_{h}, the only terms left correspond to U = ∅, and R_{∅} = {h}. Hence μ_{h} = λ_{h} for all h, and θ(f) = f if f ∈ T′_{p}.

This shows that the inclusion of T′_{p} into T_{p} is Z_{p}-split, and completes the proof of lemma 5.4.1.

(c) End of the proof of theorem 5.1.2 in the case |P : Q| = p. An element ψ ∈ Z_{p} is fixed by P if

ψ(gah) = h^{−1}ψ(a)

for all g, a ∈ P and h ∈ Q. Since P is transitive on P, this implies ψ(a) = ψ(1) for all a ∈ P, and finding such an element is equivalent to choosing ψ(1) ∈ X^{Q}. As this set is empty by assumption, it follows that Z_{p}^{P} = ∅.

Now tensoring sequence 5.4.3 with O^{ε} gives the sequence

0 → Ten_{Q}^{P}Ω_{X} → Ten_{Q}^{P}OZ_{p} → O^{ε} ⊗ Kerd_{p−1} → 0

This is still a Z_{p}-split exact sequence. Lemma 3.2.8 shows that Ten_{Q}^{P}Ω_{X} is a capped OP -lattice, and that

(5.4.4)  Ten_{Q}^{P}Ω_{X} = Ω_{Z_{p}} + Kerd_{p−1}

in D_{O}(P), since Kerd_{p−1} and O^{ε} ⊗ Kerd_{p−1} have the same image in that group.

Summarizing the successive equalities 5.4.2 and 5.4.4 gives finally the following expression for Ten_{Q}^{P}Ω_{X} in D_{O}(P)

Ten_{Q}^{P}Ω_{X} = Ω_{Z_{p}} + Ω_{Z_{p−1}} + … + Ω_{Z_{1}}

Now by lemma 5.2.3, for each i ∈ {1, …, p}

Ω_{Z_{i}} = ∑_{U,V ∈[s_{P}] ; {U≤_{P} V \atop Z_{i}^{V} ≠∅}} μ_{P} (U, V)Ω_{P/U}

The set Z_{i}^{V} is non-empty if and only if there exists a subset A of P such that V AQ = A and |A/Q| = i, and a map ψ : A → X such that

ψ(vaq) = q^{−1}ψ(a)

for all v ∈ V, a ∈ A, and q ∈ Q. If i < p, this implies V ⊆ Q (otherwise V Q = P, and A = V AQ = AV Q = AP = P), and it is equivalent to find an element ψ(a) ∈ X^{V^{a}} for each a ∈ A/Q. Thus Z_{i}^{V} is non-empty if V ⊆ Q and if there are at least i cosets aQ in P/Q such that X^{V^{a}} ≠ ∅. Note that aQ = V aQ and P/Q = V \P/Q in this case. Similarly V^{a} = V^{a} ∩ Q. Hence there are at least i double cosets V aQ such that X^{V^{a}∩Q} ≠ ∅.

If i = p, then Z_{p}^{V} is non-empty if there is a map ψ from P to X such that ψ(vaq) = q^{−1}ψ(a) for v ∈ V, a ∈ P, and q ∈ Q. If V ⊆ Q, the result is the same as before: such an element ψ exists if and only if for all a ∈ P, the set X^{V^{a}} is non-empty, or equivalently, if there are p double cosets V aQ such that X^{V^{a}∩Q} ≠ ∅.

And if V ⊈ Q, then V Q = P, and such a ψ is defined by ψ(1), which must be such that q^{−1}ψ(1) = ψ(1) for q ∈ Q whenever there is an element v ∈ V with vq = 1. In other words ψ(1) ∈ X^{V ∩Q}, and the set Z_{p}^{V} is non-empty if and only if X^{V ∩Q} is. Note that |V \P/Q| = 1 in this case.

Those remarks show that in D_{O}(P)

Ten_{Q}^{P}Ω_{X} = ∑_{i=1}^{p} ∑_{U,V ∈[s_{P}] ; {U≤_{P} V ≤Q \atop |E(V,Q,X)|≥i}} μ_{P} (U, V)Ω_{P/U} + ∑_{U,V ∈[s_{P}] ; {U≤_{P} V ≰Q \atop |E(V,Q,X)|=1}} μ_{P} (U, V)Ω_{P/U}

where E(V, Q, X) is the set of double cosets V aQ in P such that X^{V^{a}∩Q} ≠ ∅. Switching the summations on i and U, V gives

Ten_{Q}^{P}Ω_{X} = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V ≤Q}} |E(V, Q, X)|μ_{P} (U, V)Ω_{P/U} + ∑_{U,V ∈[s_{P}] ; {U≤_{P} V ≰Q \atop |E(V,Q,X)|=1}} μ_{P} (U, V)Ω_{P/U}

This second sum is also equal to

∑_{{U,V ∈[s_{P}] \atop U≤_{P} V ≰Q}} |E(V, Q, X)|μ_{P} (U, V)Ω_{P/U}

since when V ≰ Q, there is only one double coset V aQ. Finally, this gives

Ten_{Q}^{P}Ω_{X} = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} |E(V, Q, X)|μ_{P} (U, V)Ω_{P/U}

and this completes the proof of the case |P : Q| = p, and the proof of theorem 5.1.2.

6. Some elements in the Dade group

(6.0.1) Notation: I denote by D_{O}^{Ω}(P) the subgroup of D_{O}(P) generated by the relative syzygies Ω_{X}, for non-empty finite P -sets X.

Note that the reduction morphism D_{O}(P) → D_{k}(P) restricts to an isomorphism D_{O}^{Ω}(P) ≃ D_{k}^{Ω}(P), since it is always injective, and since the element Ω_{X} of D_{k}^{Ω}(P) is the reduction of the element Ω_{X} of D_{O}^{Ω}(P). Moreover corollary 4.1.2 and theorem 5.1.2 show that the operations of restriction, inflation, group isomorphism, and tensor induction preserve D_{O}^{Ω}. Lemma 4.2.1 shows that deflation maps D_{O}^{Ω} to D_{k}^{Ω}. This could be used to define a deflation map from D_{O}^{Ω}(P) to D_{O}^{Ω}(P/Q)).

6.1. Linear relations in the Dade group

(6.1.1) Proposition: Let P be a p-group. Then for any subgroup Q of P, for any finite P -set X, and any finite Q-set Y, the following equalities hold in D_{O}(P)

(6.1.2)  0 = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |V \P/Q| Ω_{P/U}

(6.1.3)  Ten_{Q}^{P}Ω_{Y} = − ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |\{a ∈ V \P/Q | Y^{V^{a}∩Q} = ∅\}| Ω_{P/U}

(6.1.4)  Ω_{X} = − ∑_{U,V ∈[s_{P}] ; {U≤_{P} V \atop X^{V} =∅}} μ_{P} (U, V)Ω_{P/U}

Proof: Equation 6.1.2 follows from the special case X^{Q} ≠ ∅ of theorem 5.1.2, which gives

Ten_{Q}^{P}Ω_{X} = 0 = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |V \P/Q| Ω_{P/U}

Now if Y is any Q-set, this formula and theorem 5.1.2 show that Ten_{Q}^{P}Ω_{Y} can also be written as

Ten_{Q}^{P}Ω_{Y} = − ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |\{a ∈ V \P/Q | Y^{V^{a}∩Q} = ∅\}| Ω_{P/U}

and in the special case P = Q, this gives for any P -set X

Ω_{X} = − ∑_{U,V ∈[s_{P}] ; {U≤_{P} V \atop X^{V} =∅}} μ_{P} (U, V)Ω_{P/U}

completing the proof.

6.2. The element Δ(P)

(6.2.1) Notation: If P is a p-group, I denote by M(P) the disjoint union of the P -sets P/Q, when Q runs through the set of maximal proper subgroups of P, i.e. the set of subgroups of P of index p. I denote by

Δ(P) = Ω_{M(P)}

the corresponding relative syzygy in D_{O}(P).

I denote by D_{O}^{Δ}(P) the subgroup of D_{O}(P) generated by all the elements Ten_{Q}^{P}Δ(Q), when Q runs through the subgroups of P.

(6.2.2) Remark: It is clear by construction that every proper subgroup of P has fixed points on M(P). By lemma 3.2.4 and corollary 4.1.2, this shows that the restriction of Δ(P) to any proper subgroup of P is zero.

Similarly, if Q is a normal subgroup of P, then the set M(P)^{Q} identifies with the set M(P/Q). By lemma 4.2.1, this shows that Def_{P/Q}^{P}Δ(P) = Δ(P/Q).

Finally, it is clear that if ϕ : P → P′ is a group isomorphism, then the image of Δ(P) by ϕ in D_{O}(P) is equal to Δ(P′), since ϕ induces an isomorphism from M(P) to M(P′).

(6.2.3) Proposition: Let P be a p-group. Then

Δ(P) = − ∑_{U⊆P} μ(U, P)Ω_{P/U}

where μ(U, P) is the Möbius function of the poset of subgroups of P. Proof: Formula 6.1.4 gives

Δ(P) = − ∑_{U,V ∈[s_{P}] ; {U≤_{P} V \atop M(P)^{V} =∅}} μ_{P} (U, V)Ω_{P/U}

Now M(P)^{V} is empty if and only if V = P, and

Δ(P) = − ∑_{U∈[s_{P}]} μ_{P} (U, P)Ω_{P/U}

But μ_{P} (U, P) is the reduced Euler-Poincaré characteristic of the “interval” ]\overline{U}, \overline{P} [_{P} of classes of proper subgroups of P which are strictly greater than \overline{U} in s_{P}. If U does not contain the Frattini subgroup Φ(P) of P, this set is contractible, via the maps

\overline{V} ↦ \overline{V.Φ(P)} ↦ \overline{U.Φ(P)}

Now it suffices to sum over the classes of subgroups U containing Φ(P). Such subgroups are normal in P, and ]\overline{U}, \overline{P} [_{P} is isomorphic to the ordinary poset ]U, P [ of proper subgroups of P containing U as a proper subgroup. This poset has Euler-Poincaré characteristic μ(U, P) equal to (−1)^{n}p^{\binom{n}{2}} if P/U is of order p^{n}. If U ⊉ Φ(P), the poset ]U, P [ is contractible, by the same argument as above. Finally

Δ(P) = − ∑_{U⊆P} μ(U, P)Ω_{P/U}

as was to be shown.

(6.2.4) Remark: It is clear from the definition that the Frattini subgroup of P acts trivially on M(P), and the element Δ(P) should be inflated from the quotient P/Φ(P). This is indeed the case, since if μ(U, P) ≠ 0, then U ⊇ Φ(P), and Ω_{P/U} is inflated from P/Φ(P). This shows in particular that if p = 2 and P is cyclic, then Δ(P) = 0, since D_{O}(ℤ/2ℤ) = 0.

(6.2.5) Corollary: The element Δ(P) has infinite order in D_{O}(P) if P is not cyclic, it has order 2 if p is odd and P is cyclic and non-trivial, and it is zero if P is trivial or cyclic of even order. Proof: Since the natural reduction morphism D_{O}(P) → D_{k}(P) is injective (see [15] Proposition (29.4)), and since the reduction of Ω_{X} ∈ D_{O}(P) is Ω_{X} ∈ D_{k}(P), it suffices to consider the case O = k. Now Δ(P) = Inf_{P/Φ(P)}^{P}Δ(P/Φ(P)), and deflation is a section of inflation. In particular, inflation is injective, and it suffices to consider the case where P is elementary abelian.

In this case, the elements Ω_{P/Q}, for P/Q not cyclic, are linearly independent in D_{k}(P), by a theorem of Dade ([10]). In particular if P is not cyclic, the element Ω_{P/1} has infinite order, and it has order 2 if P is cyclic of order p > 2.

Now the element 2Δ(P) is a linear combination of elements Ω_{P/Q}, for P/Q noncyclic, and the coefficient of Ω_{P/1} is equal to 2μ(1, P), which is non-zero. Thus if P is not cyclic, the element Δ(P) has infinite order.

And if P is cyclic of order p ≥ 3, then Δ(P) = Ω_{P/1} has order 2. Finally, if |P | ≤ 2, then D_{k}(P) is the trivial group.

6.3. Characterization

The element Δ(P) is always inflated from P/Φ(P). When P is elementary abelian, it can be characterized as follows:

(6.3.1) Proposition: Let P be an elementary abelian p-group. Then:

1. There is a unique element Δ̃(P) in D_{O}^{Ω}(P) such that

Δ(P) = pΔ̃(P)

2. Let \underline{D}_{O}(P) denote the subgroup of D_{O}(P) defined by

\underline{D}_{O}(P) = ⋂_{Q⊂P} KerRes_{Q}^{P}

where Q runs through the proper subgroups of P. Then \underline{D}_{O}(P) is the subgroup of D_{O}(P) generated by Δ̃(P).

Proof: Assertion 1) is a consequence of the expression of Δ(P) given in proposition 6.2.3:

Δ(P) = − ∑_{U⊆P} μ(U, P)Ω_{P/U}

Since Ω_{P/P} is zero, the summation runs through proper subgroups U of P. If U has

index p^{n} in P, with n ≥ 2, then the coefficient μ(U, P) = (−1)^{n}p^{\binom{n}{2}} of Ω_{P/U} is a multiple of p. And if |P : U| = p, then the element Ω_{P/U} has order 2 in D_{O}(P) if p is odd, or it is zero if p = 2. In any case it is equal to pΩ_{P/U}. The existence of Δ̃(P) follows. The unicity is also clear, since D_{O}(P) has no p-torsion if P is elementary abelian (even for p = 2, since D_{O}(P) is torsion free in this case). Assertion 1) follows.

Now for assertion 2), first observe that Δ̃(P) is an element of \underline{D}_{O}(P): indeed, by remark 6.2.2, for any proper subgroup Q of P

pRes_{Q}^{P}Δ̃(P) = Res_{Q}^{P}Δ(P) = 0

Since D_{O}(Q) has no p-torsion by Dade’s theorem, it follows that

Res_{Q}^{P}Δ̃(P) = 0

as required.

I will prove that Δ̃(P) is a generator of \underline{D}_{O}(P) by induction on |P |. First observe that if Q and R are subgroups of P, then by corollary 4.1.2

Res_{R}^{P}Ω_{P/Q} = Ω_{Res_{R}^{P}P/Q}

Since moreover Res_{R}^{P}P/Q is a union of |P : QR| copies of R/R ∩ Q, it follows that

Res_{R}^{P}Ω_{P/Q} = Ω_{R/R∩Q}

Now by Dade’s theorem (see [10]), the elements Ω_{P/Q} for Q ⊂ P generate D_{O}(P). An element u = ∑_{Q⊂P} n_{Q}Ω_{P/Q} is in \underline{D}_{O}(P) if and only if for any proper subgroup R of P

0 = ∑_{Q⊂P} n_{Q}Ω_{R/R∩Q}

Moreover the elements Ω_{R/S}, for R/S non-cyclic, are linearly independent in D_{O}(P), and the elements Ω_{R/S}, for |R/S| = p, are elements of order 2 if p is odd, and independent over 𝔽_{2}. If p = 2, these elements are equal to zero.

It follows that u ∈ \underline{D}_{O}(P) if and only if for any pair of subgroups S ⊂ R ⊂ P (with proper inclusions)

(6.3.2)  ∑_{{Q⊂P \atop Q∩R=S}} n_{Q} \{ \begin{matrix}= 0 & if |R/S| ≥ p^{2} \\ ≡ 0 mod. 2 & if |R/S| = p > 2 \end{matrix}

If |P | ≤ 2, then D_{O}(P) = 0 and there is nothing to prove. If |P | = p > 2, then \underline{D}_{O}(P) = D_{O}(P) = ℤ/2ℤ, and there is nothing to prove. If |P | = 4, then \underline{D}_{O}(P) = D_{O}(P) is isomorphic to ℤ, generated by

Ω_{P/1} = −\frac{1}{2}Δ(P) = −Δ̃(P)

If |P | = p^{2} > 4, then condition 6.3.2 for a given subgroup R of index p in P and S = {1} gives

n_{1} + ∑_{{|Q|=p \atop Q≠R}} n_{Q} ≡ 0 mod. 2

which is equivalent to

n_{R} ≡ n_{1} + ∑_{|Q|=p} n_{Q} mod. 2

Hence n_{R} is independent of R modulo 2, equal to m, say. Therefore

m ≡ n_{1} + (p + 1)m ≡ n_{1} mod. 2

Finally u is in \underline{D}_{O}(P) if and only if it can be written

u = n_{1}(Ω_{P/1} + ∑_{|P:Q|=p} Ω_{P/Q}) = n_{1}(Ω_{P/1} − ∑_{|P:Q|=p} Ω_{P/Q})

But

Δ̃(P) = −Ω_{P/1} + ∑_{|P:Q|=p} Ω_{P/Q}

hence assertion 2) holds in this case.

The last special case is |P | = p^{3}. For any subgroup R of index p in P, condition 6.3.2 for S = 1 gives

(6.3.3)  n_{1} + ∑_{{|Q|=p \atop Q⊈R}} n_{Q} = 0

Summing these relations for subgroups R containing a given subgroup Q_{0} of order p of P gives

(p + 1)n_{1} + ∑_{{|Q|=p \atop Q≠Q_{0}}} pn_{Q} = 0

Adding n_{Q_{0}} to both sides gives

(p + 1)n_{1} + ∑_{|Q|=p} pn_{Q} = n_{Q_{0}}

The left hand side is independent of Q_{0}, thus n_{Q_{0}} is independent of Q_{0}, and equal to m, say. Equation 6.3.3 gives

n_{1} + p^{2}m = 0

Thus n_{1} = −p^{2}m. If p = 2, then

u = m(−p^{2}Ω_{P/1} + ∑_{|Q|=p} Ω_{P/Q}) = −mΔ̃(P)

since Ω_{P/R} = 0 for |P : R| = 2. Thus I can suppose p odd.

In this case for a subgroup R of index p in P, containing a subgroup S of order p, condition 6.3.2 gives

n_{S} + ∑_{|Q|=p^{2} ; {Q⊃S \atop Q≠R}} n_{Q} ≡ 0 mod. 2

Adding n_{R} to both sides gives also, since n_{S} = m

(6.3.4)  m + ∑_{{|Q|=p^{2} \atop Q⊃S}} n_{Q} ≡ n_{R} mod. 2

It follows that l_{S} = ∑_{{|Q|=p^{2} \atop Q⊃S}} n_{Q} is independent modulo 2 of S, since for two distinct

subgroups S and S′ of order p

l_{S} ≡ n_{S.S′} − m ≡ l′_{S} mod. 2

Let l denote this common value. Equation 6.3.4 shows then that n_{R} is independent of R modulo 2, and that

m + (p + 1)l ≡ m ≡ n_{R} mod. 2

thus also n_{R} ≡ −m modulo 2. Finally

u = −p^{2}mΩ_{P/1} + m ∑_{|Q|=p} Ω_{P/Q} − m ∑_{|R|=p^{2}} Ω_{P/R} = −mΔ̃(P)

and assertion 2) holds.

For the general case |P | = p^{d} ≥ p^{4}, I observe first that if Q and R are subgroups of P, then by lemma 4.2.1 Def_{P/R}^{P}Ω_{P/Q} is equal to zero if R ⊈ Q, and to Ω_{(P/R)/(Q/R)} if R ⊆ Q. Since moreover

Def_{Q/R}^{Q}Res_{Q}^{P} = Res_{Q/R}^{P/R}Def_{P/R}^{P}

it follows that Def_{P/R}^{P}u ∈ \underline{D}_{k}(P/R) if u ∈ \underline{D}_{O}(P). But if u = ∑_{Q⊂P} n_{Q}Ω_{P/Q}, then

Def_{P/R}^{P}u = ∑_{R⊆Q⊂P} n_{Q}Ω_{(P/R)/(Q/R)}

If R is non-trivial, then by induction hypothesis, there is an integer m_{R} depending only on R such that Def_{P/R}^{P}u = \frac{m_{R}}{p} Δ(P/R), i.e. after inflating to P

∑_{R⊆Q⊂P} n_{Q}Ω_{P/Q} = \frac{m_{R}}{p} ∑_{R⊆Q⊂P} μ(Q, P)Ω_{P/Q}

It follows that if |P/Q| > p^{2}, then n_{Q} = \frac{m_{R}}{p} μ(Q, P), and this equality is only a congruence modulo 2 if |P/Q| = p > 2. This must hold whenever R is a non-trivial subgroup of Q. Thus if R ⊆ R′ are non-trivial subgroups of index at least p^{2} in P, then

n_{R′} = \frac{m_{R}}{p} μ(R′, P) = \frac{m_{R′}}{p} μ(R′, P)

thus m_{R} = m_{R′} since μ(R′, P) ≠ 0. Since the poset of non-trivial subgroups of index at least p^{2} is connected if |P | ≥ p^{4}, the value of m_{R} is constant on this poset, equal to m, say. Moreover, if Q is a subgroup of index p > 2 in P, containing a non-trivial subgroup R of index at least p^{2}, then

n_{Q} ≡ \frac{m}{p}μ(Q, P) mod. 2

Hence there is an integer m such that for any non-trivial subgroup Q of P

n_{Q} = \frac{m}{p}μ(Q, P)

this equality being replaced by a congruence modulo 2 if |P : Q| = p > 2.

Now fix a subgroup R of index p in P. Condition 6.3.2 for S = {1} means that

n_{1} + ∑_{{|Q|=p \atop Q⊈R}} n_{Q} = 0

Thus

n_{1} = − ∑_{{|Q|=p \atop Q⊈R}} \frac{m}{p}μ(Q, P)

Now if P has order p^{d}, there are p^{d−1} subgroups Q of order p not contained in R, and for each of them μ(Q, P) = (−1)^{d−1}p^{\binom{d−1}{2}}. It follows that

n_{1} = −\frac{m}{p}(−1)^{d−1}p^{d−1}p^{\binom{d−1}{2}} = (−1)^{d} \frac{m}{p}p^{\binom{d}{2}} = \frac{m}{p}μ(1, P)

Thus n_{Q} = \frac{m}{p}μ(Q, P) for any subgroup Q of P, and

u = ∑_{Q⊂P} \frac{m}{p}μ(Q, P)Ω_{P/Q} = −mΔ̃(P)

This completes the proof.

It is convenient to extend the definition of Δ̃(P) to arbitrary p-groups, as follows:

(6.3.5) Notation: If P is a (non necessarily elementary abelian) p-group, I denote by Δ̃(P) the element of D_{O}^{Ω}(P) defined by

Δ̃(P) = Inf_{P/Φ(P)}^{P}Δ̃(P/Φ(P))

6.4. Linear independence

The elements Ten_{Q}^{P}Δ(Q) are almost linearly independent in the Dade group:

(6.4.1) Proposition: Let P be a p-group. For Q ∈ [s_{P}], let n_{Q} ∈ ℤ, such that

(6.4.2)  ∑_{Q∈[s_{P}]} n_{Q}Ten_{Q}^{P}Δ(Q) = 0

in D_{O}(P). Then

1. If Q is not cyclic, then n_{Q} = 0.

2. If Q is cyclic, non-trivial, and if p is odd, then n_{Q} ∈ 2ℤ.

Proof: Let R be a maximal element of [s_{P}] such that n_{R} ≠ 0. By Mackey’s formula, the restriction of 6.4.2 to R is

∑_{Q∈[s_{P}]} ∑_{x∈R\P/Q} n_{Q}Ten_{R∩^{x}Q}^{R}^{x}Res_{R^{x}∩Q}^{Q}Δ(Q) = 0

Since the restriction of Δ(Q) to any proper subgroup of Q is zero, the only terms left correspond to R^{x} ⊆ Q. But if Q is not equal to R^{x}, then n_{Q} = 0 by maximality of R. Finally, this gives

n_{R}|N_{P} (R) : R|Δ(R) = 0

Thus Δ(R) is a torsion element in D_{O}(R). This can only happen if R is cyclic, by corollary 6.2.5. This shows 1).

Now n_{Q} = 0 if Q is not cyclic. Let R be a maximal subgroup of P such that n_{R} ∉ 2ℤ. The same argument shows that

n_{R}|N_{P} (R) : R| ∈ 2ℤ

If p is odd, this gives n_{R} ∈ 2ℤ as required.

(6.4.3) Remark: Assertion 2) is definitely wrong if p = 2: recall that Δ(Q) = 0 if Q is cyclic and p = 2!

(6.4.4) Corollary: Let P be a p-group. Denote by nc(P) the number of conjugacy classes of non-cyclic subgroups of P, and by c̃(P) the number of conjugacy classes of non-trivial cyclic subgroups of P. Then:

1. If p is odd

D_{O}^{Δ}(P) ≃ ℤ^{nc(P)} ⊕ (ℤ/2ℤ)^{c̃(P)}

2. If p = 2

D_{O}^{Δ}(P) ≃ ℤ^{nc(P)}

6.5. Generation

The following proposition shows that up to a power of p, any relative syzygy is a linear combination of elements Ten_{R}^{P}Δ(R), for subgroups R of P:

(6.5.1) Proposition: Let P be a p-group, and X be a non-empty P -set. Then

|P |Ω_{X} = − ∑_{{R⊆Q⊆P \atop X^{R} ≠∅}} |R|μ(R, Q)Ten_{Q}^{P}Δ(Q)

Proof: Let E_{X} denote the right hand side of the equality of the proposition, i.e.

E_{X} = − ∑_{{R⊆Q⊆P \atop X^{R} ≠∅}} |R|μ(R, Q)Ten_{Q}^{P}Δ(Q)

Let Q be a subgroup of P. Then by formula 6.1.3

Ten_{Q}^{P}Δ(Q) = − ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |\{a ∈ V \P/Q | M(Q)^{V^{a}∩Q} = ∅\}| Ω_{P/U}

Now M(Q)^{V^{a}∩Q} = ∅ is equivalent to Q ⊆ V^{a}, and in this case V.a.Q = V.^{a}Q.a = V.a. Hence, summing now for a ∈ P/V instead of a ∈ V \P gives

(6.5.2)  Ten_{Q}^{P}Δ(Q) = − ∑_{U,V ∈[s_{P}] ; {U≤_{P} V \atop a∈P/V, Q^{a}⊆V}} μ_{P} (U, V)Ω_{P/U}

Suppose first that X is isomorphic to a disjoint union of a single transitive P -set P/S for some subgroup S of P. Then X^{R} is non-empty if and only if (P/S)^{R} is, and

E_{X} = E_{P/S} = − ∑_{{R⊆Q⊆P \atop (P/S)^{R} ≠∅}} |R|μ(R, Q)Ten_{Q}^{P}Δ(Q)

Now (P/S)^{R} ≠ ∅ is equivalent to R ≤_{P} S. Using equation 6.5.2

E_{X} = ∑_{{R⊆Q⊆P \atop R≤_{P} S}} |R|μ(R, Q) ∑_{U,V ∈[s_{P}] ; {U≤_{P} V \atop a∈P/V, Q⊆^{a}V}} μ_{P} (U, V)Ω_{P/U}

Changing the order of summations gives

E_{X} = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) ∑_{a∈P/V} ∑_{{R⊆P \atop R≤_{P} S}} |R| ( ∑_{R⊆Q⊆^{a}V} μ(R, Q)) Ω_{P/U}

The inner summation on Q is zero unless R = ^{a}V. Since R ≤_{P} S, this gives

E_{X} = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V ≤_{P} S}} μ_{P} (U, V)|P/V ||V |Ω_{P/U} = |P | ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V ≤_{P} S}} μ_{P} (U, V)Ω_{P/U}

Now the summation on V for a given U is zero, unless U =_{P} S, hence

E_{X} = |P |Ω_{P/S}

Moreover Ω_{X} = Ω_{P/S} in this case, by lemma 3.2.7. Hence the formula of the proposition holds if X is isomorphic to a union of copies of a single transitive P -set.

As in the proof of lemma 5.2.3, the proof of the general case of the proposition can be completed by induction on the cardinality of the set

F (X) = {Q ∈ [s_{P}] | X^{Q} ≠ ∅}

Clearly, both sides of the equality of the proposition depend only on F (X).

If |F (X)| = 1, then P acts freely on X, and X is a disjoint union of copies of P/1. The formula holds in this case. And if |F (X)| > 1, choose a maximal element S of F (X). Then X can be written as a disjoint union Y ⊔ Z, where Z is a disjoint union of copies of P/S, and Y^{S} = ∅.

If Y = ∅, then X = Z is a union of copies of P/S, and the formula holds. And if Y ≠ ∅, then by lemma 5.2.1

Ω_{X} = Ω_{Y} + Ω_{Z} − Ω_{Y ×Z}

But |F (Y)| < |F (X)| since F (Y) ⊆ F (X) and S ∈ F (X) − F (Y). Similarly since F (Y × Z) ⊆ F (Y), I have |F (Y × Z)| ≤ |F (Y)| < |F (X)|. By induction hypothesis, the proposition holds for Y and Y × Z. It also holds for Z since Z is a union of copies of P/Q. Thus

|P |Ω_{X} = E_{Y} + E_{Z} − E_{Y ×Z}

But as at the end of proof of lemma 5.2.3, the right hand side is equal to E_{Y ⊔Z} = E_{X}, and this completes the proof.

The following special case seems of interest (see remark (4.14) of [5]): (6.5.3) Corollary: Let P be a p-group. Then in D_{O}(P)

|P |Ω_{P/1} = − ∑_{V ∈a_{p}(P)} |V |μ(V)Ten_{V}^{P} Δ(V)

where a_{p}(P) is the set of non-trivial elementary abelian p-groups of P. Proof: This is just the case X = P/1 of the proposition. Then X^{U} ≠ ∅ implies U = {1}, and then μ(1, V) ≠ 0 implies that V is elementary abelian.

7. Structure of D_{O}(P): partial results

7.1. Elementary abelian sections

The elements Ten_{Q}^{P}Δ(Q) of section 6.2 provide tools to study the structure of D_{O}(P) as abelian group. Unfortunately, they are not enough to elucidate this structure completely. The main problem here is the same problem we already encountered with J. Thévenaz in [5]: we have almost no information on the subgroup K_{O}(P) of D_{O}(P) defined hereafter: (7.1.1) Notation: Let P be a p-group. If R \underline{◁} Q are subgroups of P, I will say that Q/R is a section of P. I will denote by Defres_{Q/R}^{P} the map from D_{O}(P) to D_{k}(Q/R) defined by

Defres_{Q/R}^{P} = Def_{Q/R}^{Q} ∘ Res_{Q}^{P}

Similarly, I will denote by Teninf_{Q/R}^{P} the map from D_{O}(Q/R) to D_{O}(P) defined by

Teninf_{Q/R}^{P} = Ten_{Q}^{R} ∘ Inf_{Q/R}^{Q}

I denote by K_{O}(P) the intersection of the kernel of the maps Defres_{E}^{P} associated to elementary abelian sections E of P.

Recall from [5] that K_{k}(P) is a finite subgroup of D_{k}(P). Clearly for any elementary abelian section E of P, the diagram

D_{O}(P) \underrightarrow{ρ} D_{k}(P)
Defres_{E}^{P} ↓  ↓ Defres_{E}^{P}
D_{k}(E) \overrightarrow{Id} D_{k}(E)

where the horizontal map ρ is the reduction morphism, is commutative. Hence ρ maps K_{O}(P) in K_{k}(P). Since ρ is injective, it follows that K_{O}(P) is a subgroup of K_{k}(P), hence it is a finite group. Actually K_{k}(P) is conjecturally zero if p is odd (and it is known to be non-zero in some cases if p = 2).

7.2. Connection with the Burnside ring

In [5], we showed that there is an exact sequence

0 ↦ ℚD_{k}(P) → ℚB(P) → ℚR_{ℚ}(P) → 0

where B(P) is the Burnside ring of P, and R_{ℚ}(P) the ring of rational representations of P. This sequence is moreover functorial in P, in the sense of [3]. In particular the inclusion ℚD_{k}(P) → ℚB(P) is compatible with restriction to subgroups, and changes tensor induction into ordinary induction of P -sets. The image of this inclusion is the subspace generated by the idempotents e_{Q}^{P} of B(P) indexed by non-cyclic subgroups Q of P.

It is easy to see that the subspace of elements u ∈ ℚB(P) such that Res_{Q}^{P}u = 0 for any proper subgroup Q of P is generated by the idempotent e_{P}^{P}. Thus the image of Δ(P) in ℚB(P) should be proportional to e_{P}^{P}.

Now the expression of e_{P}^{P} is

e_{P}^{P} = \frac{1}{|P |} ∑_{Q⊆P} |Q|μ(Q, P)P/Q

(7.2.1) Lemma: The element pe_{P}^{P} lies in B(P), i.e. for all Q ⊆ P, the expression

p\frac{μ(Q, P)}{|P : Q|} is an integer.

Proof: Indeed, in the expression of e_{P}^{P}, the only non-zero terms correspond to subgroups Q containing Φ(P). Now if Q ⊇ Φ(P) is such that |P : Q| = p^{k}

\frac{|Q|}{|P |}μ(Q, P) = \frac{μ(P, Q)}{|P : Q|} = (−1)^{k}p^{\binom{k}{2}−k}

If k = 0 or k ≥ 3, then \binom{k}{2} − k ≥ 0, and if k = 1 or k = 2, it is equal to -1. Thus

p\frac{|Q|}{|P |}μ(Q, P) is always an integer, and the lemma follows.

Since multiplication by P/Q in B(P) is restriction Res_{Q}^{P} followed by induction Ind_{Q}^{P}, this lemma shows that it should be of interest to consider the following operation:

(7.2.2) Notation: Let P be a p-group. I denote by τ_{P} the endomorphism of D_{O}(P) defined by

τ_{P} = ∑_{Q⊆P} \frac{pμ(Q, P)}{|P : Q|}Ten_{Q}^{P}Res_{Q}^{P}

If R is a subgroup of P, I denote by α(R, P) the integer defined by

α(R, P) = ∑_{S⊆R} \frac{pμ(S, P)}{|P : S|}

Note that α(R, P) = 0 if R ⊉ Φ(P), and that α(R, P) = α(R/Φ(P), P/Φ(P)) otherwise.

(7.2.3) Lemma: Let P be a p-group.

1. If R is a subgroup of P, then

τ_{P} (Ω_{P/R}) = −α(R, P)Δ(P)

2. The image of D_{O}^{Ω}(P) by τ_{P} is equal to the subgroup of D_{O}(P) generated by Δ(P).

Proof: If R is a subgroup of P, then

τ_{P} (Ω_{P/R}) = ∑_{Q⊆P} \frac{pμ(Q, P)}{|P : Q|}Ten_{Q}^{P}Res_{Q}^{P}Ω_{P/R}

By corollary 4.1.2 and theorem 5.1.2

Ten_{Q}^{P}Res_{Q}^{P}Ω_{P/R} = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V) |\{a ∈ V \P/Q | (P/R)^{V^{a}∩Q} ≠ ∅\}| Ω_{P/U}

Thus τ_{P} (Ω_{P/R}) is expressed as a double summation. Summing first on U and V, next on Q gives

τ_{P} (Ω_{P/R}) = ∑_{{U,V ∈[s_{P}] \atop U≤_{P} V}} μ_{P} (U, V)β(R, V)Ω_{P/U}

where β(R, V) is defined by

β(R, V) = ∑_{Q⊆P} \frac{pμ(Q, P)}{|P : Q|} |\{a ∈ V \P/Q | V^{a} ∩ Q ≤_{P} R\}|

This is also equal to

β(R, V) = ∑_{Q⊆P} \frac{pμ(Q, P)}{|P : Q|} ∑_{{a∈P \atop Q∩V^{a} ≤_{P} R}} \frac{|Q ∩ V^{a}|}{|Q||V |}

Thus

β(R, V) = ∑_{{a∈P \atop S≤_{P} R}} \frac{p|S|}{|P ||V |} ∑_{{Q⊆P \atop Q∩V^{a} =S}} μ(Q, P)

It is well known that the sum on Q is zero if V^{a} ≠ P, i.e. if V ≠ P. Moreover if V = P

β(R, P) = ∑_{{a∈P \atop S≤_{P} R}} \frac{p|S|}{|P |^{2}} μ(S, P) = ∑_{S≤_{P} R} \frac{pμ(S, P)}{|P : S|}

Note that μ(S, P) is zero if S ⊉ Φ(P). And if S ⊇ Φ(P), then S \underline{◁} P, and S ≤_{P} R is equivalent to S ⊆ R. This gives finally

β(R, P) = ∑_{S⊆R} \frac{pμ(S, P)}{|P : S|} = α(R, P)

Moreover

τ_{P} (Ω_{P/R}) = α(R, P) ∑_{U∈[s_{P}]} μ_{P} (U, P)Ω_{P/U}

But as I already mentioned in the proof of proposition 6.2.3, the coefficient μ_{P} (U, P) is equal to μ(U, P). Thus

τ_{P} (Ω_{P/R}) = −α(R, P)Δ(P)

and assertion 1) follows.

It remains to check that Δ(P) is in τ_{P} (D_{O}^{Ω}(P)): but the element Δ̃(P) is in D_{O}^{Ω}(P) and

(7.2.4)  τ_{P} (Δ̃(P)) = p ∑_{Q⊆P} \frac{μ(Q, P)}{|P : Q|}Ten_{Q}^{P}Res_{Q}^{P}Δ̃(P) = pΔ̃(P) = Δ(P)

since the restriction of Δ̃(P) to any proper subgroup of P is zero. Assertion 2) follows.

(7.2.5) Corollary: Let P be an elementary abelian p-group. There exists a well defined linear form λ_{P} from D_{O}(P) to ℤ if |P | ≥ p^{2}, or to ℤ/2ℤ if |P | = p > 2, such that for any element u of D_{O}(P)

τ_{P} (u) = λ_{P} (u)Δ(P)

Proof: This is clear since if P is abelian, then D_{O}(P) = D_{O}^{Ω}(P), and since Δ(P) has infinite order if P is non-cyclic, and order 2 if P is cyclic of order p > 2.

With this definition, lemma 7.2.3 shows that λ_{P} (Ω_{P/R}) = −α(R, P). More generally:

(7.2.6) Proposition: Let P be an elementary abelian p-group (of order at least 3) and X be a non-empty finite P -set. Then

λ_{P} (Ω_{X}) = − ∑_{{Q⊆P \atop X^{Q} ≠∅}} \frac{pμ(Q, P)}{|P : Q|}

Proof: By lemma 5.2.3, and since P is abelian

Ω_{X} = ∑_{{U⊆V ⊆P \atop X^{V} ≠∅}} μ(U, V)Ω_{P/U}

Thus

λ_{P} (Ω_{X}) = − ∑_{{U⊆V ⊆P \atop X^{V} ≠∅}} μ(U, V) ∑_{W ⊆U} \frac{pμ(W, P)}{|P : W |}

= − ∑_{{W ⊆V ⊆P \atop X^{V} ≠∅}} \frac{pμ(W, P)}{|P : W |} ( ∑_{W ⊆U⊆V} μ(U, V))

The sum on U is zero, unless W = V, and the proposition follows.

Though it is not really necessary, I would like to indicate how to compute the integers α(R, P). By the above remarks, it suffices to consider the case where P is elementary abelian:

(7.2.7) Lemma: Let P be an elementary abelian p-group, and R be a subgroup of P. Then

α(R, P) = \frac{pμ(R, P)}{|P : R|}m_{P,R}

where m_{P,R} is defined as in lemme 16 of [3] by

m_{P,R} = \frac{1}{|P |} ∑_{X.R=P} |X|μ(X, P)

Let |P | = p^{d} and |R| = p^{r}.

• If d = 1, then α(R, P) = −1 if r = 0 and α(R, P) = p − 1 if r = 1.

• If d ≥ 2, then α(R, P) = 0 if r ≥ d − 1 and

α(R, P) = (−1)^{d−r}p^{\binom{d−r−1}{2}}(1 − p^{d−2})(1 − p^{d−3}) … (1 − p^{d−r−1})

if r ≤ d − 2.

Proof: By definition

α(R, P) = \frac{p}{|P |} ∑_{S⊆R} |S|μ(S, P)

By classical formulae, if S ⊆ R, then μ(S, P) can be computed as

μ(S, P) = ∑_{{R′∩R=S \atop R′.R=P}} μ(S, R′)μ(R′, P)

This gives

α(R, P) = \frac{p}{|P |} ∑_{R′.R=P} μ(R, P)μ(R′, P)|R ∩ R′| = (\frac{pμ(R, P)}{|P : R|})\frac{1}{|P |} ∑_{R′.R=P} |R′|μ(R′, P)

And the first formula holds. The other ones are a consequence of the computation of m_{P,R} (see formula (4.8) of [5]):

• If r = 0, then m_{P,R} = 1.

• If d = r = 1, then m_{P,R} = 1 − 1/p.

• If d ≥ 2, then m_{P,R} = 0 if r ≥ d − 1, and

m_{P,R} = (1 − p^{d−2})(1 − p^{d−3}) … (1 − p^{d−r−1})

if r ≤ d − 2.

Moreover if |P | = p^{d} and |R| = p^{r}, then \frac{pμ(R,P)}{|P:R|} = (−1)^{d−r}p^{\binom{d−r−1}{2}}.

(7.2.8) Remark: Note that in the case d = 1 the value λ_{P} (Ω_{P/R}) is the image of α(R, P) in ℤ/2ℤ. Hence it is equal to 1 if R = 1, and to 0 if R = P since p is odd in this case. This is consistent with the equality Ω_{P/P} = 0.

(7.2.9) Lemma: An element u of D_{O}(P) lies in K_{O}(P) if and only if for any subgroup Q of P

τ_{Q/Φ(Q)}(Defres_{Q/Φ(Q)}^{P}u) = 0

Proof: One way is obvious. Now if τ_{Q/Φ(Q)}(Defres_{Q/Φ(Q)}^{P}u) = 0 for any subgroup Q of P, I will prove by induction on |Q| that Defres_{Q/Φ(Q)}^{P}u = 0, and the lemma will follow, since for any elementary abelian section E = Q/R of P

Defres_{E}^{P} = Def_{Q/R}^{Q/Φ(Q)} ∘ Defres_{Q/Φ(Q)}^{P}

The result is clear if |Q| = 1. If it holds for groups of order less than |Q|, set v = Defres_{Q/Φ(Q)}^{P}u. Let R be a subgroup of Q of index p. Then R ⊇ Φ(Q), and

Res_{R/Φ(Q)}^{Q/Φ(Q)}v = Def_{R/Φ(Q)}^{R/Φ(R)}Defres_{R/Φ(R)}^{P}u = 0

since |R| < |Q|. This shows that v ∈ \underline{D}_{O}(Q/Φ(Q)). Hence there exists an integer m

such that v = mΔ̃(Q/Φ(Q)). Now

0 = τ_{Q/Φ(Q)}(Defres_{Q/Φ(Q)}^{P}u) = τ_{Q/Φ(Q)}(mΔ̃(Q/Φ(Q))) = mΔ(Q/Φ(Q)) = pv

Thus v = 0 as required, since D_{O}(Q/Φ(Q)) has no p-torsion.

7.3. Tensor induction and τ_{P}

In the Burnside ring, there is a Frobenius formula

X.Ind_{Q}^{P}Y = Ind_{Q}^{P}((Res_{Q}^{P}X).Y )

In particular in ℚB(P), the product of e_{P}^{P} with any element induced from a proper subgroup of P is zero. Hence it should be possible to prove the following:

(7.3.1) Lemma: Let P be a p-group, and Q be a proper subgroup of P. Then

τ_{P} ∘ Ten_{Q}^{P} = 0

Proof: Let u ∈ D_{O}(Q). Then

τ_{P} (Ten_{Q}^{P}u) = ∑_{R⊆P} \frac{pμ(R, P)}{|P : R|}Ten_{R}^{P}Res_{R}^{P}Ten_{Q}^{P}u

Applying Mackey’s formula gives

τ_{P} (Ten_{Q}^{P}u) = ∑_{R⊆P} \frac{pμ(R, P)}{|P : R|} ∑_{x∈R\P/Q} Ten_{R∩^{x}Q}^{P}^{x}Res_{R^{x}∩Q}^{Q}u

Moreover

Ten_{R∩^{x}Q}^{P}^{x}Res_{R^{x}∩Q}^{Q}u = ^{x}Ten_{R^{x}∩Q}^{P}Res_{R^{x}∩Q}^{Q}u = Ten_{R^{x}∩Q}^{P}Res_{R^{x}∩Q}^{Q}u

since P acts trivially on D_{O}(P). Moreover if μ(R, P) ≠ 0 then R ⊇ P and R \underline{◁} P. Thus

τ_{P} (Ten_{Q}^{P}u) = ∑_{R⊆P} \frac{pμ(R, P)}{|P : R|}|P : R.Q|Ten_{R∩Q}^{P}Res_{R∩Q}^{Q}u

= ∑_{R⊆P} \frac{pμ(R, P)|R ∩ Q|}{|Q|}Ten_{R∩Q}^{P}Res_{R∩Q}^{Q}u

Grouping together the terms for which R ∩ Q is a given subgroup S of Q gives

τ_{P} (Ten_{Q}^{P}u) = ∑_{S⊆Q} \frac{p|S|}{|Q|}( ∑_{{R⊆P \atop R∩Q=S}} μ(R, P))Ten_{S}^{P} Res_{S}^{Q}u

If Q is a proper subgroup of P, then the sum on R is zero, and the lemma follows.

7.4. Some subgroups of D_{O}(P)

Lemma 7.2.9 leads to the following:

(7.4.1) Lemma: Let P be a p-group and u be an element of D_{O}(P). Then u lies in D_{O}^{Δ}(P) + K_{O}(P) if and only if for any subgroup R of P there is an integer n_{R} such that

τ_{R/Φ(R)}(Defres_{R/Φ(R)}u) = n_{R} p |N_{P} (R) : R| Δ(R/Φ(R))

Proof: The element u lies in D_{O}^{Δ}(P) + K_{O}(P) if and only if there exists integers n_{Q} for subgroups Q of P such that

u − ∑_{Q∈[s_{P}]} n_{Q}Ten_{Q}^{P}Δ(Q) ∈ K_{O}(P)

It is equivalent to require that for any subgroup R of P

(7.4.2)  Defres_{R/Φ(R)}^{P}u = Defres_{R/Φ(R)}^{P} ( ∑_{Q∈[s_{P}]} n_{Q}Ten_{Q}^{P}Δ(Q) ∈ K_{O}(P))

Let S be any normal subgroup of R. Then by Mackey’s formula

Defres_{R/S}^{P}Ten_{Q}^{P}Δ(Q) = Def_{R/S}^{R}Res_{R}^{P}Ten_{Q}^{P}Δ(Q)
= ∑_{x∈R\P/Q} Def_{R/S}^{R}Ten_{R∩^{x}Q}^{R}^{x}Res_{R^{x}∩Q}^{Q}Δ(Q)

Since the restriction of Δ(Q) to any proper subgroup of Q is zero, this is also

Defres_{R/S}^{P}Ten_{Q}^{P}Δ(Q) = ∑_{{y∈P/R \atop Q^{y}⊆R}} Def_{R/S}^{R}Ten_{Q^{y}}^{R}Δ(Q^{y})

since obviously Δ(Q)^{y} = Δ(Q^{y}). Now the element Δ(Q^{y}) is inflated from Q^{y}/Φ(Q^{y}), thus by corollary (3.11) of [5]

Def_{R/S}^{R}Ten_{Q^{y}}^{R}Δ(Q^{y}) = Ten_{Q^{y}.S/S}^{R/S}Iso_{Q^{y}/Q^{y}∩S}^{Q^{y}.S/S}Def_{Q^{y}/Q^{y}∩S}^{Q^{y}}Δ(Q^{y})

(Actually corollary (3.11) of [5] only deals with the case O = k, but clearly one can reduce everything from O to k before taking deflations.) Finally by remark 6.2.2 this gives

Defres_{R/S}^{P}Ten_{Q}^{P}Δ(Q) = ∑_{{y∈P/R \atop Q^{y}⊆R}} Ten_{Q^{y}.S/S}^{R/S}Δ(Q^{y}.S/S)

Suppose now that R/S is elementary abelian. Applying τ_{R/S} gives by lemma 7.3.1 and equation 7.2.4

τ_{R/S}Defres_{R/S}^{P}Ten_{Q}^{P}Δ(Q) = ∑_{{y∈P/R \atop Q^{y}.S=R}} τ_{R/S}(Δ(R/S)) = p ∑_{{y∈P/R \atop Q^{y}.S=R}} Δ(R/S)

If moreover S = Φ(R), this is zero unless Q is a conjugate of R. Finally

(7.4.3)  τ_{R/Φ(R)}Defres_{R/Φ(R)}^{P}Ten_{R}^{P}Δ(R) = p |N_{P} (R):R| Δ(R/Φ(R))

Applying this in equation 7.4.2 gives

τ_{R/Φ(R)}Defres_{R/Φ(R)}^{P}(u) = n_{R} p |N_{P} (R) : R| Δ(R/Φ(R))

as was to be shown.

Of course, this condition is void if |R/Φ(R)| ≤ 2. Suppose now that u = |P |v, for v ∈ D_{O}(P). Since the image of τ_{R/Φ(R)} is generated by Δ(R/Φ(R)), and since

\frac{|P |}{p|N_{P} (R):R|} is an integer if R is non-trivial, there is an integer n_{R} such that

\frac{|P |}{p|N_{P} (R):R|}τ_{R/Φ(R)}Defres_{R/Φ(R)}^{P}(v) = n_{R}Δ(R/Φ(R))

More precisely, this integer is equal to

n_{R} = \frac{|P |}{p|N_{P} (R):R|}λ_{R/Φ(R)}(Defres_{R/Φ(R)}^{P}v)

(this equality being a congruence modulo 2 if R is non-trivial and cyclic of odd order). These remarks lead to the following notation:

(7.4.4) Notation: Let P be a p-group. I denote by s′_{P} the set of conjugacy classes of subgroups R of P such that |R/Φ(R)| ≥ 3, and by [s′_{P}] the set [s_{P}] ∩ s′_{P}. I denote by ψ_{P} the endomorphism of D_{O}(P) defined by

ψ_{P} (v) = ∑_{R∈[s′_{P}]} \frac{|P |}{p|N_{P} (R):R|}λ_{R/Φ(R)}(Defres_{R/Φ(R)}^{P}v)Ten_{R}^{P}Δ(R)

Note that by definition of Δ̃(Q), this is also

(7.4.5)  ψ_{P} (v) = ∑_{R∈[s′_{P}]} \frac{|P |}{|N_{P} (R):R|}λ_{R/Φ(R)}(Defres_{R/Φ(R)}^{P}v)Ten_{R}^{P}Δ̃(R)

Now by definition of the maps λ_{R/Φ(R)} and of the elements Δ(R), this is also

ψ_{P} (v) = ∑_{R∈[s_{P}]} \frac{|P |}{p|N_{P} (R):R|}Teninf_{R/Φ(R)}^{P}τ_{R/Φ(R)}Defres_{R/Φ(R)}^{P}(v)

(where the summation may run over the set [s_{P}] − {1}). Finally, the map ψ_{P} is equal to

(7.4.6)  ψ_{P} = ∑_{{R∈[s_{P}] \atop Φ(R)⊆Q⊆R}} \frac{|P |μ(Q, R)}{|N_{P} (R) : Q|}Teninf_{Q/Φ(R)}^{P}Defres_{Q/Φ(R)}^{P}

(7.4.7)  ψ_{P} = ∑_{Φ(R)⊆Q⊆R} |Q|μ(Q, R)Teninf_{Q/Φ(R)}^{P}Defres_{Q/Φ(R)}^{P}

(7.4.8) Proposition: Let P be a p-group. Then

1. The image of ψ_{P} is contained in D_{O}^{Δ}(P).

2. The kernel of ψ_{P} is equal to K_{O}(P).

3. The image of ψ_{P} − |P |Id is contained in K_{O}(P).

4. The kernel of ψ_{P} − |P |Id contains D_{O}^{Ω}(P).

Proof: Assertion 1) follows from the definition of ψ_{P}. Assertion 2) follows from lemma 7.2.9, and from proposition 6.4.1 and corollary 7.2.5. Assertion 3) follows from the proof of lemma 7.4.1 and from the construction of ψ_{P}.

For assertion 4), by linearity, it is enough to prove that if X is a non-empty P -set, then

ψ_{P} (Ω_{X}) = |P |Ω_{X}

Now by expression 7.4.5

ψ_{P} (Ω_{X}) = ∑_{R∈[s′_{P}]} \frac{|P |}{|N_{P} (R):R|}λ_{R/Φ(R)}(Defres_{R/Φ(R)}^{P}Ω_{X})Ten_{R}^{P}Δ̃(R)

By corollary 4.1.2 and proposition 7.2.6

λ_{R/Φ(R)}(Defres_{R/Φ(R)}^{P}Ω_{X}) = − ∑_{{Φ(R)⊆S⊆R \atop X^{S}≠∅}} \frac{pμ(S, R)}{|R : S|}

ψ_{P} (Ω_{X}) = − ∑_{R∈[s′_{P}]} \frac{|P |}{|N_{P} (R):R|} ∑_{{Φ(R)⊆S⊆R \atop X^{S}≠∅}} \frac{pμ(S, R)}{|R : S|}Ten_{R}^{P}Δ̃(R)

= − ∑_{R∈[s′_{P}] ; {Φ(R)⊆S⊆R \atop X^{S} ≠∅}} \frac{|P |}{|N_{P} (R)|}|S|pμ(S, R)Ten_{R}^{P}Δ̃(R)

= − ∑_{S⊆R⊆P ; {\overline{R}∈s′_{P} \atop X^{S}≠∅}} |S|μ(S, R)Ten_{R}^{P}Δ(R)

since pΔ̃(P) = Δ(P), and since μ(S, R) = 0 if S ⊉ Φ(R). Now assertion 4) follows from Proposition 6.5.1, and from the fact that Δ(R) = 0 if \overline{R} ∈ s_{P} − s′_{P}.

(7.4.9) Proposition: Let P be a p-group. Then:

1. The subgroup D_{O}^{Ω}(P) is generated by the elements Ω_{P/Q}, for Q ⊂ P.

2. One has |P |D_{O}^{Ω}(P) ⊆ D_{O}^{Δ}(P) ⊆ D_{O}^{Ω}(P). The quotient group D_{O}^{Ω}(P)/D_{O}^{Δ}(P) is finite, with exponent dividing the order of P.

3. One has |P |D_{O}(P) ⊆ D_{O}^{Δ}(P) + K_{O}(P) ⊆ D_{O}^{Ω}(P) + K_{O}(P) ⊆ D_{O}(P). The quotient groups D_{O}(P)/(D_{O}^{Δ}(P) + K_{O}(P)) and D_{O}(P)/(D_{O}^{Ω}(P) + K_{O}(P)) are finite, with exponent dividing |P |.

4. The intersection D_{O}^{Δ}(P) ∩ K_{O}(P) is zero.

5. The intersection D_{O}^{Ω}(P) ∩ K_{O}(P) is finite, with exponent dividing the order of P.

6. The torsion part of D_{O}(P) is equal to

D_{O}^{tors}(P) = (D_{O}^{Δ})^{tors}(P) ⊕ K_{O}(P)

7. The torsion part of D_{O}^{Ω}(P) has exponent dividing 2|P |.

8. There is an isomorphism

D_{O}(P) ≃ D_{O}^{Δ}(P) ⊕ K_{O}(P)

Proof: Assertion 1) is a trivial consequence of lemma 5.2.3. Assertion 2) follows from proposition 6.5.1, and from the fact that D_{O}^{Ω}(P) is finitely generated.

Assertion 3) follows from assertions 1) and 3) of proposition 7.4.8.

Now if u ∈ D_{O}^{Ω}(P) ∩ K_{O}(P), then ψ_{P} (u) = 0 = |P |u, by assertions 2) and 4) of proposition 7.4.8. This proves 5), and 4) follows, since moreover D_{O}^{Δ}(P) has no p-torsion by corollary 6.4.4.

The restriction of ψ_{P} to the torsion subgroup of D_{O}(P) gives an exact sequence

0 ⟶ K_{O}(P) ⟶ D_{O}^{tors}(P) ⟶^{ψ_{P}}(D_{O}^{Δ})^{tors}(P)

Moreover, the group (D_{O}^{Δ})^{tors}(P) is a 2-group, which is trivial if p = 2. Now the restriction of ψ_{P} to this subgroup is equal to |P |Id, and this is always the identity of (D_{O}^{Δ})^{tors}(P). Hence the previous exact sequence is split, and this proves 6).

If u is a torsion element of D_{O}(P), then 2u is an element of K_{O}(P), since 2D_{O}(Q) is a free group if Q is elementary abelian. Thus 2u ∈ D_{O}^{Ω}(P) ∩ K_{O}(P), and 2|P |u = 0.

Finally, since the restriction of ψ_{P} to (D_{O}^{Δ})^{tors}(P) is equal to the identity, the map ψ_{P} induces an injection

i_{P} : D_{O}(P)/ ((D_{O}^{Δ})^{tors}(P) ⊕ K_{O}(P)) ↪ D_{O}^{Δ}(P)/(D_{O}^{Δ})^{tors}(P)

and the quotient group is finite. This shows that the image of i_{P} is a free ℤ-module of rank equal to the number nc(P) of conjugacy classes of non-cyclic subgroups of P. Thus

D_{O}(P) ≃ ℤ^{nc(P)} ⊕ (D_{O}^{Δ})^{tors}(P) ⊕ K_{O}(P)

proving assertion 8).

Proposition 7.4.9 gives a new proof of theorems A and B of [5]:

(7.4.10) Corollary: Let P be a p-group. Then

1. The dimension over ℚ of ℚD_{O}(P) is equal to the number of conjugacy classes of non-cyclic subgroups of P.

2. If p is odd, the quotient of the torsion part of D_{O}(P) by K_{O}(P) is an 𝔽_{2}-vector space of dimension equal to the number of conjugacy classes of non-trivial cyclic subgroups of P.

Proof: Both assertions are clear, since tensoring by ℚ kills all finite groups.

(7.4.11) Remark: Actually, the previous results also provide bases for those vector spaces: the elements Ten_{Q}^{P}Δ(Q), for non-cyclic elements Q of [s_{P}], form a ℚ-basis of ℚD_{O}(P). Similarly, the elements Ten_{Q}^{P}Δ(Q), for non-trivial cyclic elements Q of [s_{P}], form a basis of the 𝔽_{2}-vector space of assertion 2).

7.5. Inverting p

(7.5.1) Notation: If R is a commutative ring, and P is a p-group, I denote by RD_{O}(P) the tensor product

RD_{O}(P) = R ⊗_{ℤ} D_{O}(P)

and similarly I set RD_{O}^{Ω}(P) = R ⊗_{ℤ} D_{O}^{Ω}(P), RD_{O}^{Δ}(P) = R ⊗_{ℤ} D_{O}^{Δ}(P), and RK_{O}(P) = R ⊗_{ℤ} K_{O}(P). If u ∈ D_{O}(P), I denote by u the element 1 ⊗ u of RD_{O}(P), and I still denote by Res_{Q}^{P}, Ten_{Q}^{P}, Inf_{P/R}^{P}, Def_{P/R}^{P}, Iso_{P}^{P′} the R-linear extensions to RD_{O} of the corresponding operations defined on D_{O}.

(7.5.2) Proposition: Let R be a commutative ring in which p is invertible. Let π_{P} = \frac{1}{|P |}ψ_{P}. Then for u ∈ D_{O}(P)

π_{P} (u) = ∑_{Q∈[s′_{P}]} \frac{1}{|N_{P} (Q) : Q|}λ_{Q/Φ(Q)}(Defres_{Q/Φ(Q)}^{P}u)Ten_{Q}^{P}Δ̃(Q)

The map π_{P} is an idempotent, with image RD_{O}^{Δ}(P) and kernel RK_{O}(P). Thus

RD_{O}(P) ≃ RD_{O}^{Δ}(P) ⊕ RK_{O}(P)

and this decomposition is functorial in P, in that π_{P} commutes with restriction, tensor induction, inflation, deflation, and group isomorphisms. Proof: This is just a reformulation of proposition 7.4.9: tensoring with R kills all finite p-groups since p in invertible. Thus

RD_{O}(P) = RD_{O}^{Δ}(P) ⊕ RK_{O}(P)

Now π_{P} = \frac{1}{|P |}ψ_{P}. Since the image of ψ_{P} is contained in the kernel of ψ_{P} − |P |Id by proposition 7.4.8, it follows that ψ_{P}^{2} = |P |ψ_{P}, hence π_{P} is an idempotent, with image RD_{O}^{Δ}(P) and kernel RK_{O}(P).

It it easy to see that if F : D_{O}(P) → D_{O′}(P′) is one of the operations of the proposition (note that the ring may change from O to O′ = k in the case of deflation), then

F (RK_{O}(P)) ⊆ RK_{O}(P′) and F (RD_{O}^{Δ}(P)) ⊆ RD_{O′}^{Δ}(P′)

this last inclusion coming from the equality RD_{O}^{Δ}(P) = RD_{O}^{Ω}(P), and from corollary 4.1.2 and from theorem 5.1.2. Now if u ∈ RD_{O}(P), then

u = π_{P} (u) + (1 − π_{P} (u))

is the decomposition of u as sum of an element of RD_{O}^{Δ}(P) and an element of RK_{O}(P). Thus

F (u) = F (π_{P} (u)) + F (1 − π_{P} (u))

This is the decomposition of F (u) in D_{O′}(P′) as sum of an element of RD_{O′}^{Δ}(P′) and an element of RK_{O′}(P′). Hence in particular

π_{P′}(F (u)) = F (π_{P} (u))

and F commutes with π_{P}.

So there should be formulas expressing the effect of the natural operations on the elements Δ(P). The effect of restriction, tensor induction, group isomorphism and deflation can be deduced from the previous computations. The effect of inflation follows from the following proposition:

(7.5.3) Proposition: Let P be a p-group and R be a normal subgroup of P. Then

|P |Inf_{P/R}^{P}Δ(P/R) = ∑_{{U⊆V ⊆P \atop U.R=P}} |U|μ(U, V)Ten_{V}^{P} Δ(V)

in D_{O}(P). Proof: Since

Inf_{P/R}^{P}Δ(P/R) = ∑_{R⊆Q⊆P} μ(Q, P)Ω_{P/Q}

it follows from proposition 6.5.1 that

|P |Inf_{P/R}^{P}Δ(P/R) = ∑_{R⊆Q⊆P} μ(Q, P) ∑_{{U⊆V ⊆P \atop U≤_{P} Q}} |U|μ(U, V)Ten_{V}^{P} Δ(V)

Now if μ(Q, P) ≠ 0, then Q ⊇ Φ(P), and Q is a normal subgroup of P. Thus U ≤_{P} Q is equivalent to U ⊆ Q, and

|P |Inf_{P/R}^{P}Δ(P/R) = ∑_{R⊆Q⊆P ; {U ⊆V ⊆P \atop U ⊆Q}} |U|μ(Q, P)μ(U, V)Ten_{V}^{P} Δ(V)

Now the sum of μ(Q, P) for U.R ⊆ Q ⊆ P is zero, unless U.R = P. Thus

|P |Inf_{P/R}^{P}Δ(P/R) = ∑_{{U⊆V ⊆P \atop U.R=P}} |U|μ(U, V)Ten_{V}^{P} Δ(V)

as was to be shown.

(7.5.4) Remark: Proposition 7.5.3 shows that the decomposition

D_{O}^{tors}(P) = (D_{O}^{Δ})^{tors}(P) ⊕ K_{O}(P)

of proposition 7.4.9 is functorial in P, i.e. that the torsion part of D_{O}^{Δ}(P) is mapped to itself by tensor induction, restriction, inflation, deflation, and group isomorphism. If p = 2, there is nothing to prove, since (D_{O}^{Δ})^{tors}(P) = 0 in this case. And if p is odd, the only non trivial fact to check here is that if P/R is cyclic (and non trivial), then Inf_{P/R}^{P}Δ(P/R) can be expressed as a linear combination of elements Ten_{V}^{P} Δ(V) corresponding to cyclic subgroups V of P.

Since Δ(P/R) has order 2 in this case, proposition 7.5.3 gives

Inf_{P/R}^{P}Δ(P/R) = ∑_{{U⊆V ⊆P \atop U.R=P}} |U|μ(U, V)Ten_{V}^{P} Δ(V)

The coefficient of Ten_{V}^{P} Δ(V) in the right hand side is zero if V.R ≠ P, and otherwise it is equal to

∑_{{U⊆V \atop U.(V ∩R)=P}} |U|μ(U, V) = |V |m_{V,V ∩R}

But if V.R = P, the group V/V ∩ R is cyclic, isomorphic to P/R. Then the constant m_{V,V ∩R} is zero if V is non cyclic, as I already recalled in the proof of lemma 7.2.7. And if V is cyclic (hence non trivial since V.R = P), then m_{V,V ∩R} = 1 if V ∩ R ≠ V, and m_{V,V} = 1 − 1/p. Since moreover Ten_{V}^{P} Δ(V) has order 2 in this case, this gives

Inf_{P/R}^{P}Δ(P/R) = ∑_{V ⊆P ; {V cyclic, V ≠P \atop V.R=P}} Ten_{V}^{P} Δ(V)

as claimed.

7.6. Back to the Burnside ring

The linear forms λ_{Q} lead to a more explicit version of the functorial morphism ℚD_{k} → ℚB. Recall the following notation from [3] page 704, or lemma 4.7 of [5]:

(7.6.1) Notation: If P is a finite group, there is a constant m(P) ∈ ℚ such that if N is a maximal normal subgroup such that m_{P,N} ≠ 0, then m_{P,N} = m(P). Moreover, if M is a normal subgroup of P such that m_{P,M} ≠ 0, then

m_{P,M} = \frac{m(P)}{m(P/M)}

In the case of a p-group P, the constant m(P) can be computed as follows:

• If P = {1}, then m(P) = 1.

• If P is cyclic and non-trivial, then m(P) = 1 − 1/p.

• If |P/Φ(P)| = p^{d}, with d ≥ 2, then

m(P) = (1 − p)(1 − p^{2}) … (1 − p^{d−2})

(7.6.2) Proposition: Let P be a p-group. Define a morphism α_{P} from ℚD_{k}(P) to ℚB(P) by

α_{P} (u) = ∑_{{Q∈[s_{P}] \atop Q non−cyclic}} λ_{Q/Φ(Q)}(Defres_{Q/Φ(Q)}^{P}u)\frac{1}{m(Q)}e_{Q}^{P}

Then there is an exact sequence

0 ⟶ ℚD_{k}(P) ⟶^{α_{P}} QB(P) ⟶^{χ_{P}} ℚR_{ℚ}(P) ⟶ 0

(where χ_{P} is the character map), which is functorial with respect to P, i.e. it is compatible with restriction, induction, deflation, inflation, and group isomorphisms. Proof: We showed in [5] (theorem D) that such an exact sequence of functors exists. Moreover

ℚD_{O}(P) ≃ ℚD_{O}^{Δ}(P) ≃ ℚD_{k}^{Δ}(P)

has a ℚ-basis consisting of the elements Ten_{Q}^{P}Δ(Q), for non-cyclic Q ∈ [s_{P}].

Since Δ(P) is the only element, up to a scalar, in the intersection of the kernels of the maps Res_{Q}^{P} for Q ⊂ P (Q ≠ P), the map α_{P} must send Δ(P) to some multiple γ_{P} e_{P}^{P} of e_{P}^{P}. Then by functoriality, it must send Ten_{Q}^{P}Δ(Q) to

γ_{Q}Ind_{Q}^{P}e_{Q}^{Q} = γ_{Q}|N_{P} (Q) : Q|e_{Q}^{P}

By functoriality again, it must map Def_{P/R}^{P}Δ(P) = Δ(P/R) to

Def_{P/R}^{P}(γ_{P} e_{P}^{P}) = m_{P,R}γ_{P} e_{P/R}^{P/R}

for any normal subgroup R of P. It follows that γ_{P/R} = m_{P,R}γ_{P}. Now if P is noncyclic the constant m_{P,R} is non-zero if and only if P/R is non-cyclic, and in this case m_{P,R} = m(P)/m(P/R). This shows that γ_{P} m(P) is independent of the non-cyclic group P, and I can suppose it is equal to p. Now the proposition follows from assertion 3) of proposition 7.4.8, since ℚK_{O}(P) = 0.

8. Some torsion elements in D_{O}^{Ω}(P)

8.1. Which relative syzygies are torsion elements

The following proposition characterizes the P -sets X such that Ω_{X} is a torsion element in D_{O}(P). In the case of a transitive P -set X, implication 1) ⇒ 2) was originally a question of J. Thévenaz:

(8.1.1) Proposition: Let P be a p-group, and X be a P -set. Then the following are equivalent:

1. The element Ω_{X} is a torsion element in D_{O}(P).

2. There exists a normal subgroup N of P such that P/N is cyclic or generalized quaternion, and such that N ⊇ P_{x} for all x ∈ X, with equality for some x ∈ X. Moreover in this case Ω_{X} = Ω_{P/N}.

Before proving this proposition, let me state an equivalent condition to assertion 1. First some notation:

(8.1.2) Notation: If X is a P -set, I denote by F_{X} the set of elements g ∈ P such that X^{g} ≠ ∅.

If A is a subset of P, I denote by ^{p} \sqrt{A} the set of elements g ∈ P such that g^{p} ∈ A.

(8.1.3) Lemma: Let P be a p-group, and X be a P -set. Then the following are equivalent:

1. The element Ω_{X} is a torsion element in D_{O}(P).

2. Let W be any subgroup of P. Then

• If W ⊆ F_{X}, the set X^{W} is non-empty.

• If W ⊈ F_{X}, then p|W ∩ F_{X}| = |W ∩ ^{p} \sqrt{F_{X}}|

Proof: By proposition 6.5.1

|P |Ω_{X} = − ∑_{{R⊆Q⊆P \atop X^{R} ≠∅}} |R|μ(R, Q)Ten_{Q}^{P}Δ(Q)

Now Ω_{X} is a torsion element if and only if |P |Ω_{X} is. Since Δ(Q) is a torsion element if Q is cyclic, this is equivalent to saying that there exists a positive integer n such that

0 = n ∑_{R⊆Q⊆P ; {X^{R} ≠∅ \atop Q non−cyclic}} |R|μ(R, Q)Ten_{Q}^{P}Δ(Q)

By proposition 6.4.1, this is equivalent to requiring that for any non-cyclic subgroup Q of P, the integer

n_{Q} = ∑_{{R⊆Q \atop X^{R} ≠∅}} |R|μ(R, Q)

is zero. On the other hand, if Q is cyclic, then the set of subgroups of Q is totally ordered, and there is a biggest subgroup R_{0} of Q such that X^{R_{0}} ≠ ∅. There are three cases:

• Either R_{0} = Q, or equivalently X^{Q} ≠ ∅, then

n_{Q} = ∑_{R⊆Q} |R|μ(R, Q) = ∑_{{x∈Q \atop <x>⊆R⊆Q}} μ(R, Q) = φ(|Q|)

where φ is the Euler function.

• Either R_{0} = Φ(Q) ≠ Q, and in this case

n_{Q} = |R_{0}|μ(R_{0}, Q) = −|R_{0}| = −|Q|/p

• Either R_{0} is a proper subgroup of Φ(Q), and n_{Q} = 0 in this case.

Let W be an arbitrary subgroup of P. Then, by Möbius inversion, requiring that the n_{Q}’s have some prescribed values, depending on Q, is equivalent to requiring that the sum of n_{Q}’s for subgroups Q of W have prescribed values, depending on W. Now summing the n_{Q}’s for Q ⊆ W gives

∑_{Q⊆W} n_{Q} = ∑_{{R⊆Q⊆W \atop X^{R} ≠∅}} |R|μ(R, Q)

and the sum on Q is equal to zero if R ≠ W, and equal to 1 otherwise. Thus

∑_{Q⊆W} n_{Q} = \{ \begin{matrix}0 & if X^{W} = ∅ \\ |W | & otherwise \end{matrix}

But this is also the sum of n_{Q}’s for cyclic subgroups Q of W, i.e.

∑_{Q⊆W ; {Q cyclic \atop X^{Q} ≠∅}} φ(|Q|) − ∑_{Q⊆W ; {Q cyclic \atop X^{Q} =∅≠X^{Φ(Q)}}} |Q|/p

Since φ(|Q|) is the number of generators of Q, the first sum is also

∑_{{g∈W \atop X^{g} ≠∅}} 1 = |W ∩ F_{X}|

where F_{X} is the set of elements g ∈ P such that X^{g} ≠ ∅.

Now a cyclic subgroup Q of W such that X^{Q} = ∅ ≠ X^{Φ(Q)} is a subgroup generated by an element g of W such that g ∉ F_{X}, but g^{p} ∈ F_{X}, and Q admits |g| − |g|/p such generators. Hence the second sum is equal to

∑_{g∈(W ∩ ^{p} \sqrt{F_{X}})−(W ∩F_{X})} \frac{|g|/p}{|g| − |g|/p} = \frac{|W ∩ ^{p} \sqrt{F_{X}}| − |W ∩ F_{X}|}{p − 1}

This gives

∑_{Q⊆W} n_{Q} = |W ∩ F_{X}| − \frac{|W ∩ ^{p} \sqrt{F_{X}}| − |W ∩ F_{X}|}{p − 1} = \frac{p|W ∩ F_{X}| − |W ∩ ^{p} \sqrt{F_{X}}|}{p − 1}

Finally, the element Ω_{X} is a torsion element if and only if for any subgroup W of P

(8.1.4)  \frac{p|W ∩ F_{X}| − |W ∩ ^{p} \sqrt{F_{X}}|}{p − 1} = \{ \begin{matrix}0 & if X^{W} = ∅ \\ |W | & otherwise \end{matrix}

If W ⊆ F_{X}, then W ⊆ ^{p} \sqrt{F_{X}}, and the left hand side of equation 8.1.4 is equal to |W |. In particular it is non-zero, and X^{W} must be non-empty. Thus

(8.1.5)  W ⊆ F_{X} ⇒ ∃x ∈ X, W ⊆ P_{x}

On the other hand, if W ⊈ F_{X}, then in particular X^{W} = ∅. Hence

(8.1.6)  W ⊈ F_{X} ⇒ p|W ∩ F_{X}| = |W ∩ ^{p} \sqrt{F_{X}}|

This completes the proof of the lemma.

Condition 8.1.6 alone implies that F_{X} is a normal subgroup of P:

(8.1.7) Lemma: Let P be a p-group. Let A be a normal subset of P, containing 1, such that for any subgroup W of P

W ⊈ A ⇒ |W ∩ ^{p} \sqrt{A}| = p|W ∩ A|

Then A is a normal subgroup of P. Proof: I first prove by induction on |W | that if W is a cyclic subgroup of P, then W ∩ A is a subgroup of W. The result is true if W = {1}, since 1 ∈ A. Thus I can suppose that W is a non-trivial cyclic subgroup of P, and by induction Φ(W) ∩ A is a subgroup of Φ(W). Moreover

W ∩ ^{p} \sqrt{A} = \{g ∈ W | g^{p} ∈ A\} = \{g ∈ W | g^{p} ∈ Φ(W) ∩ A\}

since g^{p} ∈ Φ(W) for all g ∈ P. But the set of elements g of W such that g^{p} belongs to the proper subgroup Φ(W) ∩ A of W is a subgroup of W, of order p|Φ(W) ∩ A|. Then either W ⊆ A, and there is nothing to prove, or by assumption

|W ∩ ^{p} \sqrt{A}| = p|W ∩ A| = p|Φ(W) ∩ A|

Since W ∩ A ⊇ Φ(W) ∩ A, it follows that W ∩ A = Φ(W) ∩ A is a subgroup of W, as required.

It follows that if a ∈ A, then the subgroup generated by a is contained in A. In particular if a ∈ A, then a^{p} ∈ A. Equivalently A ⊆ ^{p} \sqrt{A}.

Now I will prove the lemma by induction on |P |. If P = {1}, the result holds since 1 ∈ A. Suppose that the result holds for all proper subgroups of P. If P′ is such a subgroup, set A′ = P′ ∩ A. This is a normal subset of P′, containing 1. Moreover, if W′ is a subgroup of P′, not contained in A′, then W′ is a subgroup of P, not contained in A. Denote by ^{′p} \sqrt{A′} the set of elements g′ of P′ such that g′^{p} ∈ A′. Then

′p \sqrt{A′} = P′ ∩ ^{p} \sqrt{A}

Thus

|W′ ∩ ^{′p} \sqrt{A′}| = |W′ ∩ ^{p} \sqrt{A}| = p|W′ ∩ A| = p|W′ ∩ A′|

By induction hypothesis A′ = P′ ∩ A is a subgroup of P.

Let H be a subgroup of index p of P, such that N = H ∩A has maximal cardinality. Then N is a normal subgroup of P, since it is a subgroup by the previous argument, and it is also a normal subset of P.

Suppose that A ⊈ H, and choose a ∈ A−H. Then let W =<a, N >.

If W ≠ P, then W ∩ A is a subgroup of P. Since W is generated by elements of A, this shows that W ⊆ A. Let H′ be a subgroup of index p of P, containing W. Such a subgroup exists since W ≠ P. Then H′ ∩ A contains W, and W contains strictly N, since a ∉ H. This contradicts the assumption on H.

Thus W = < a, N > =P, and then P/N is cyclic, generated by the image of a. In particular it admits a unique subgroup of index p, generated by the image of a^{p}. Since H/N has index p in P/N, it follows that H = <a^{p}, N >. Now H is generated by elements of A, and H ∩ A is a subgroup of P. Thus H ⊆ A. This implies in particular that P = ^{p} \sqrt{A}, since g^{p} ∈ H for all g ∈ P. Hence either A = P, or |P | = p|P ∩A| = p|A|. In this case |A| = |H|, thus A = H. In both cases A is a subgroup of P.

Now if A ⊆ H, then H ∩ A = A is a subgroup of P, and the lemma follows.

8.2. Proof of proposition 8.1.1

Suppose that Ω_{X} is a torsion element. By the previous lemmas, the set N = F_{X} is a subgroup of P, contained in F_{X}. Thus X^{N} ≠ ∅, by 8.1.5, and there is an element x ∈ X such that N ⊆ P_{x}, hence N = P_{x}. It follows that N ⊇ P_{y} for all y ∈ X, and that Ω_{X} = Ω_{P/N} by lemma 3.2.7.

Now if W/N is a non-trivial elementary abelian subgroup of P/N

|W | = |W ∩ ^{p} \sqrt{N}| = p|W ∩ N| = p|N|

thus |W/N| = p. The group P/N has p-rank at most 1, hence it is cyclic or generalized quaternion. This shows that 1) implies 2).

Conversely if N is a normal subgroup of P such that P/N is cyclic or generalized quaternion, then it it well known that Ω_{P/N} has order

• 1 if P/N is trivial or of order 2.

• 2 if P/N is cyclic of order at least 3.

• 4 if P/N is generalized quaternion.

Thus 2) implies 1), and the proof is complete.

(8.2.1) Remark: One can check easily that the conditions of lemma 8.1.3 hold for X = P/N, if P/N is cyclic or generalized quaternion: indeed in this case, the group P/N has a unique subgroup M/N of order p, if it is non trivial. Then clearly ^{p} \sqrt{N} = M.

If W is a subgroup of P not contained in N, then W.N contains M, and M = N.(W ∩ M). Thus |M||W ∩ N| = |N||W ∩ M|, or

|W ∩ ^{p} \sqrt{N}| = |W ∩ M| = p|W ∩ N|

This can be viewed as a new proof of the fact that Ω_{P} is a torsion element if P is quaternion. This is not a new proof for the case of cyclic groups, since it requires Dade’s theorem.

8.3. Acknowledgements

I am grateful to J. Thévenaz for careful reading of a first version of this paper, for pointing out mistakes and misprints, and also for many useful suggestions.

References

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Contents

1 Introduction 1

2 Relative syzygies 2 2.1 X-split morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2.2 Relative projectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.3 Relative Shanuel’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 The Dade group 7 3.1 Capped modules and permutation algebras . . . . . . . . . . . . . . . . 8 3.2 Relative syzygies in the Dade group . . . . . . . . . . . . . . . . . . . . 8

4 Functorial properties of relative syzygies 11 4.1 Restriction, inflation, isomorphisms . . . . . . . . . . . . . . . . . . . . . 11 4.2 Deflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5 Tensor induction 12 5.1 The formula for tensor induction . . . . . . . . . . . . . . . . . . . . . . 12 5.2 The case P = Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 5.3 The inductive step |P : Q| ≥ p^{2} . . . . . . . . . . . . . . . . . . . . . . . 15 5.4 The case |P : Q| = p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6 Some elements in the Dade group 25 6.1 Linear relations in the Dade group . . . . . . . . . . . . . . . . . . . . . 25 6.2 The element Δ(P) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 6.3 Characterization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.4 Linear independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 6.5 Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

7 Structure of D_{O}(P): partial results 34 7.1 Elementary abelian sections . . . . . . . . . . . . . . . . . . . . . . . . . 34 7.2 Connection with the Burnside ring . . . . . . . . . . . . . . . . . . . . . 35 7.3 Tensor induction and τ_{P} . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 7.4 Some subgroups of D_{O}(P) . . . . . . . . . . . . . . . . . . . . . . . . . . 40 7.5 Inverting p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 7.6 Back to the Burnside ring . . . . . . . . . . . . . . . . . . . . . . . . . . 46

8 Some torsion elements in D_{O}^{Ω}(P) 48 8.1 Which relative syzygies are torsion elements . . . . . . . . . . . . . . . . 48 8.2 Proof of proposition 8.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 51 8.3 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Serge Bouc UFR de Mathématiques Université Paris 7-Denis Diderot 75251. Paris Cedex 05 , France email: bouc@math.jussieu.fr