TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 352, Number 1, Pages 457–475 S 0002-9947(99)02319-3 Article electronically published on July 26, 1999

FINITE GROUPS OF MATRICES OVER GROUP RINGS

GERALD CLIFF AND ALFRED WEISS

Abstract. We investigate certain finite subgroups Γ of GL_{n}(ZΠ), where Π is a finite nilpotent group. Such a group Γ gives rise to a Z[Γ × Π]-module; we study the characters of these modules to limit the structure of Γ. We also exhibit some exotic subgroups Γ.

1. Introduction

Let Π be a finite group. We set

SGL_{n}(ZΠ) = ker aug : GL_{n}(ZΠ) → GL_{n}(Z),

where aug is the usual augmentation map applied to each entry of GL_{n}(ZΠ).

Suppose that α is a homomorphism from a finite group Γ to SGL_{n}(ZΠ). We shall investigate the following problem.

Problem 0. Do there exist group homomorphisms σ_{i} : Γ → Π, i = 1, 2, …, n, and an element x ∈ GL_{n}(QΠ) such that x^{−1}α(γ)x = diag(σ_{i}(γ)), γ ∈ Γ?

This is analogous to a conjecture of Zassenhaus, who was interested in units of ZΠ of augmentation 1, i.e. SGL_{1}(ZΠ). Problem 0 is related to results on units of group rings, as shown in a special case in [MRSW]. There is a positive answer to Problem 0 if Π is a p-group [WAn] or if n = 1 and Π is nilpotent [WCr].

Given finite groups Γ and Π, set

G = Γ × Π, N = 1 × Π.

A homomorphism α : Γ → SGL_{n}(ZΠ) gives rise to a double action ZG-module M(α), defined as follows: as abelian group, M(α) is equal to the column vectors ZΠ^{n}, and the G-action is given by

m · (γ, π) = α(γ^{−1})mπ, (γ, π) ∈ G, m ∈ M.

There is a bijection between GL_{n}(ZΠ)-conjugacy classes of homomorphisms α and isomorphism classes of ZG-lattices M which satisfy

(a) G/N acts trivially on the N-fixed points M^{N}, and

(b) res_{N} M is a free ZN-module. (See [S, §38.6] for details.)

\begin{matrix}c \\ ◯ \end{matrix}1999 American Mathematical Society

Received by the editors November 1, 1997.

1991 Mathematics Subject Classification. Primary 20C10, 20C05; Secondary 16S34, 20H25.

Key words and phrases. Finite group, module, character.

This research was supported by grants from the Natural Sciences and Engineering Research Council of Canada.

Definition. Let D(Γ, Π) (D for “double-action”) be the set of characters of ZGmodules M which satisfy (a) and (b).

If σ : Γ → Π is a homomorphism, it is easy to show (Lemma 2.1 below) that the double action module M(σ) is isomorphic to the permutation module ind_{[σ]}^{G} Z, where [σ] denotes the subgroup {(γ, σ(γ)) : γ ∈ Γ} of G. Define

R(G) = the virtual complex characters of G,
R^{+}(G) = the proper characters of G,
P(Γ, Π) = Z-span of {ind_{[σ]}^{G} 1 : σ ∈ hom(Γ, Π)} ⊂ R(G),
P^{+}(Γ, Π) = Z_{≥0}-span of {ind_{[σ]}^{G} 1 : σ ∈ hom(Γ, Π)} ⊂ R^{+}(G).

Here P is for “permutation”. Then P^{+} is contained in D; it follows from [WAn] that if Π is a p-group, then D = P^{+}. Indeed, it turns out (Proposition 2.3 below) that there is a positive answer to Problem 0 for finite groups Γ, Π if and only if D = P^{+}. Note that D is closed under addition, and is a sub-semigroup of R^{+}(G). Problem 0 can be reformulated as

Problem 1. Is D = P^{+}?

In §6 we give some conditions under which Problem 1 has a positive solution. In §5 we give examples of groups Γ and Π for which D ≠ P^{+}. A counterexample to Problem 1 for n = 1 has been constructed by Roggenkamp and Scott [RS] (see [K]); there are also such examples in [B], but with no general principles of construction for them. We will show that for n > 1 counterexamples are so plentiful that our focus shifts to describing all of them. The feasibility of doing so is addressed by

Problem 2. Is the semigroup D finitely generated?

We will show that this is indeed true if Γ and Π are nilpotent; this is in §7. If we know that D is finitely generated, we are also interested in explicitly finding a generating set.

Analysis of D(Γ, Π) is complicated by questions about locally free class groups, so we are led to replace (b) in the definition of D by

(b′) res_{N} M is a locally free ZN-module.

Definition. Let D′(Γ, Π) be the set of characters of ZG-modules M which satisfy (a) and (b′).

The semigroup D′ is a good approximation to D, in the sense that rD′ ⊆ D ⊆ D′ for some positive integer r (Lemma 2.5 below); in other words, if M is a ZG-module satisfying (a) and (b′), then the direct sum M^{r} of r copies of M satisfies (a) and (b). We get first approximations to results about D by considering D′ instead. All known cases of the equality D = P^{+} actually have D′ = P^{+}.

One of our main results, Theorem 3.3 below, is a complete character-theoretic description of D′ in the case that Π is nilpotent. To explain this we introduce some notation. For a finite nilpotent group N, let N_{p} be its Sylow p-subgroup and N_{p′} its Sylow p-complement. Then we have R(G) = R(G_{p′}) ⊗ R(G_{p}), and we use this to define

Q_{p}(Γ, Π) = R(G_{p′}) ⊗ P(Γ_{p}, Π_{p}),
Q_{p}^{+}(Γ, Π) = {∑ η ⊗ λ : η ∈ R^{+}(G_{p′}), λ ∈ P^{+}(Γ_{p}, Π_{p})} .

We show in Theorem 3.3 below that if Π is nilpotent, then

D′(Γ, Π) = ⋂_{p} Q_{p}^{+}(Γ, Π).

Theorem 3.3 is crucial to most of the results in the rest of the paper.

We are also interested in knowing if Problem 1 has a “virtual” answer, that is, is the Z-span of D equal to P? This is discussed in §4.

Although the above definitions and problems are sensible for arbitrary finite groups Γ and Π, we shall need to assume that these groups are nilpotent for most of our results. This is because we rely on the local results of [WAn], which at this point have no analogues in the general case. From §3 through to the end of the paper, we will assume that Γ and Π are nilpotent.

2. Preliminaries

As in the introduction, let G = Γ × Π and N = 1 × Π. In this section, Π and Γ can be arbitrary finite groups.

Lemma 2.1. For σ ∈ hom(Γ, Π), the double action module M(σ) is isomorphic to ind_{[σ]}^{G} Z as ZG-modules.

Proof. By definition, M(σ) has Z-basis Π. Consider elements of ind_{[σ]}^{G} Z as linear combinations of the cosets of [σ] in G; use 1 × Π as coset representatives. Define

f : M(σ) → ind_{[σ]}^{G} Z, f(π) = [σ](1, π), π ∈ Π.

To check that f is a ZG-homomorphism, for (γ, π′) ∈ G,

f(π · (γ, π′)) = f(σ(γ^{−1})ππ′) = [σ](1, σ(γ^{−1})ππ′)
= [σ](γ, σ(γ))(1, σ(γ^{−1})ππ′)) = [σ](γ, ππ′) = f(π) · (γ, π′).

This proves the lemma.

Corollary 2.2. P^{+} ⊆ D.

Proposition 2.3. For a homomorphism α : Γ → SGL_{n}(ZΠ), the character χ of the double action module M(α) is in P^{+}(Γ, Π) if and only if there exist group homomorphisms σ_{i} ∈ hom(Γ, Π), 1 ≤ i ≤ n, and an element u ∈ GL_{n}(QΠ) such that uα(γ)u^{−1} = diag(σ_{1}(γ), …, σ_{n}(γ)) for all γ ∈ Γ.

Proof. Suppose that χ is in P^{+}(Γ, Π). By Lemma 2.1 there are homomorphisms σ_{i} : Γ → Π, 1 ≤ i ≤ k, such that there is a QG-isomorphism

f : Q ⊗ M(α) → Q ⊗ (⨁_{i=1}^{k} M(σ_{i})).

Comparing dimensions over Q gives k = n. Let {e_{j} : 1 ≤ j ≤ n} be the standard basis of the QΠ-column vectors Q ⊗ M(α), and write elements of Q ⊗ M(σ_{i}) as ⟨x⟩_{i}, x ∈ QΠ. Then Q ⊗ (⨁_{i=1}^{k} M(σ_{i})) has QN-basis {⟨1⟩_{i} : 1 ≤ i ≤ n} and, since f is a QN-homomorphism, we have

f(e_{j}) = ∑_{i}⟨u_{ij}⟩_{i} where u ∈ GL_{n}(QΠ).

Act by (γ^{−1}, 1), giving

f(e_{j})(γ^{−1}, 1) = ∑_{i}⟨u_{ij}⟩_{i}(γ^{−1}, 1) = ∑_{i}⟨σ_{i}(γ)u_{ij}⟩_{i}.

Since f(e_{j})(γ^{−1}, 1) = f(e_{j}(γ^{−1}, 1)), this equals

f(α(γ)e_{j}) = f(∑_{k} e_{k}α(γ)_{kj}) = ∑_{k} f(e_{k})(1, α(γ)_{kj})

= ∑_{k,i}⟨u_{ik}⟩_{i}(1, α(γ)_{kj}) = ∑_{i}⟨∑_{k} u_{ik}α(γ)_{kj}⟩_{i}.

Therefore for all i, j we have

σ_{i}(γ)u_{ij} = ∑_{k} u_{ik}α(γ)_{kj} so, as matrix equation, diag(σ_{i}(γ))u = uα(γ).

This implies that uα(γ)u^{−1} = diag(σ_{i}(γ)).

For the converse, given u ∈ GL_{n}(QΠ) such that uα(γ)u^{−1} = diag(σ_{i}(γ)), define

f : Q ⊗ M(α) → Q ⊗ (⊕_{i}M(σ_{i}))

by f(e_{j}) = ∑_{i}⟨u_{ij}⟩_{i}. It follows as above that f is a QG-isomorphism.

Lemma 2.4. Suppose that Π′ is a subgroup of Π, and set G′ = Γ×Π′, N′ = 1×Π′. Then ind_{G′}^{G} D(Γ, Π′) ⊆ D(Γ, Π).

Proof. We have G = NG′ and N ∩ G′ = N′. Let χ′ ∈ D(Γ, Π′) be the character of a ZG′-lattice M′ which satisfies (a) and (b) for G′, N in the definition of D in §1. We must check that M = ind_{G′}^{G} M′ satisfies (a) and (b) for G, N.

For (a) identify G/N and G′/N′ with Γ; there are Γ-isomorphisms

M^{N} ≅ M ⊗_{ZG} ZΓ ≅ M′ ⊗_{ZG′} ZG ⊗_{ZG} ZΓ ≅ M′^{N′},

so Γ acts trivially.

For (b), use Mackey decomposition to get

res_{N}^{G} M ≅ res_{N}^{G} ind_{G′}^{G} M′ ≅ ind_{N′}^{N} res_{N′}^{G′} M′,

because G = NG′ and N ∩ G′ = N′.

Lemma 2.5. rD′ ⊆ D ⊆ D′ for some positive integer r.

Proof. By definition, we have D ⊆ D′. We will find a positive integer r so that for any locally free ZN-lattice X, X^{r} is a free ZN-lattice. Let e be the exponent of the locally free class group of ZN, as defined in [CR, 49.10]. Then Y = X^{e} is stably free. From the Bass Cancellation Theorem [CR, 41.20] Y ⊕ Y is free, so X^{2e} is free, and the lemma is proved, with r = 2e.

To proceed further, we will apply the p-group results of [WAn]. In order to do this for all primes dividing |Π|, we will assume that Π is nilpotent. We next show that under this assumption, α(Γ) is also nilpotent.

Lemma 2.6. Suppose that Π is nilpotent; let ϕ_{p} denote the natural map

ϕ_{p} : GL_{n}(ZΠ) → GL_{n}(Z[Π/Π_{p}]).

Let H be a finite subgroup of SGL_{n}(ZΠ). Then the following hold:

1. ker ϕ_{p} ∩ H is a p-group.

2. If x ∈ SGL_{n}(ZΠ) has prime order r, then r divides |Π|.

3. ker ϕ_{p} ∩ H is a normal Sylow p-subgroup of H.

Proof. 1. Suppose that x ∈ ker ϕ_{p} ∩ H. Then x = 1 + δ, where all the entries of δ are in ZΠΔ(Π_{p}), where Δ(Π_{p}) is the augmentation ideal of ZΠ_{p}. Since ZΠΔ(Π_{p}) is a nilpotent ideal mod p, it follows that for a suitable positive integer m we have x^{p^{m}} = 1 + py, for some element y of the matrix ring M_{n}(ZΠ). Then, raising to p-powers, we get

x^{p^{m+i}} = 1 + p^{i}y_{i}, y_{i} ∈ M_{n}(ZΠ).

Now x has finite order, since it is in the finite group H; so there are only finitely many possible values of x^{p^{m+i}} as i varies. Then there is a subsequence of integers i for which x^{p^{m+i}} is constant, say z. We see from the last equation that z − 1 has coefficients divisible by arbitrarily high powers of p. This forces z = 1, and x has p-power order.

2. Use induction on Π. If |Π| = 1, then |SGL_{n}(ZΠ)| = 1. Suppose that |Π| > 1. Let p be a prime dividing |Π|. If x ∈ ker ϕ_{p}, then x has order p by 1, and r = p. If x ∉ ker ϕ_{p}, then ϕ_{p}(x) ∈ SGL_{n}(Z[Π/Π_{p}]) has order r dividing |Π/Π_{p}|, by induction.

3. Let Φ_{p} = ker ϕ_{p} ∩ H. From 1, Φ_{p} is a p-group. Suppose that y is an element of p-power order in H. From 2, SGL_{n}(Z[Π/Π_{p}]) has no element of order p; therefore y ∈ ker ϕ_{p}. So Φ_{p} contains all Sylow p-subgroups of H. This completes the proof.

Corollary 2.7. If Π is nilpotent then D(Γ, Π) = D(Γ_{nil}, Π), where Γ_{nil} is the largest nilpotent quotient of Γ. The same holds for D′.

Proof. If χ ∈ D(Γ, Π) is the character of M(α) then α(Γ) is nilpotent by Lemma 2.6, since each Sylow subgroup is normal. Thus χ ∈ D(Γ_{nil}, Π). The same assertion for D′ follows from Lemma 2.5.

3. Character-theoretic Description of D′

For the rest of the paper, we assume that Π is nilpotent; then by Corollary 2.7, it is no loss of generality to assume that Γ is nilpotent. Thus we shall always assume that Γ is nilpotent.

In this section we give a character-theoretic description of D′(Γ, Π).

Lemma 3.1. Let χ ∈ R(G) = R(G_{p′}) ⊗ R(G_{p}) be written uniquely in the form

χ = ∑_{η∈irr(G_{p′})} η ⊗ λ_{η}, λ_{η} ∈ R(G_{p}).

Then χ ∈ Q_{p}^{+}(Γ, Π) if and only if λ_{η} ∈ P^{+}(Γ_{p}, Π_{p}) for all λ.

Proof. Since R(G_{p′}) has Z-basis the irreducible complex characters irr(G_{p′}), and R^{+}(G_{p′}) is the non-negative linear combinations of irr(G_{p′}), the result follows.

Lemma 3.2. ⋂_{p} Q_{p} ⊆ Q ⊗ P.

Proof. Take χ ∈ ⋂_{p} Q_{p}. Let B_{p} be a maximal linearly independent subset of {ind_{[σ_{p}]}^{G_{p}} 1 : σ_{p} ∈ hom(Γ_{p}, Π_{p})}; extend this to a basis B̂_{p} of Q⊗R(G_{p}). Then ⨂_{p} B̂_{p} is a Q-basis of ⨂_{p}(Q⊗R(G_{p})) = Q⊗R(G), so we can write χ uniquely as a Q-linear combination in this basis. For a fixed p, χ is a linear combination of elements in (⨂_{l≠p} B̂_{l})⊗B_{p}. Varying over p shows that χ is a Q-linear combination of elements of ⨂ B_{p}. A collection {σ_{p} : Γ_{p} → Π_{p}}, one for each prime, corresponds to a single homomorphism σ : Γ → Π, whose restriction to Γ_{p} is σ_{p}. So χ ∈ Q ⊗ P(Γ, Π).

Theorem 3.3. Suppose that Π is nilpotent. Then D′ = ⋂_{p} Q_{p}^{+}.

Proof. We first show that D′ ⊆ ⋂_{p} Q_{p}^{+}. Suppose that χ ∈ D′, and let M be a ZG-lattice affording χ. Fix a prime p dividing |Π|. We want to apply Theorem 2 of [WAn] to the Z_{p}G_{p}-module M_{p} = res_{G_{p}}(Z_{p} ⊗_{Z} M), relative to the normal subgroup N_{p} = 1 × Π_{p} of G_{p}. We need to verify that

(a) the N_{p}-fixed points M_{p}^{N_{p}} have trivial G_{p}/N_{p}-action, and

(b) res_{N_{p}} M_{p} is a free Z_{p}N_{p}-module. Now (b) holds because res_{N} M is locally free. To prove (a), it is no loss to replace M by a direct sum of copies of M, and by Lemma 2.5, we may assume that res_{N} M is free. Then M ≅ M(α) for some α : Γ → GL_{n}(ZΠ); hence M^{N_{p}} ≅ M(ϕ_{p}α), where ϕ_{p} is as in Lemma 2.6. Now Γ_{p} is in the kernel of ϕ_{p}α by Lemma 2.6, so G_{p}/N_{p} acts trivially on M^{N_{p}}.

From Theorem 2 of [WAn], M_{p} is a permutation Z_{p}G_{p}-lattice; moreover, the permuted basis is a disjoint union of orbits whose point stabilizers are of the form [σ_{p}] = {(γ, σ_{p}(γ)) : γ ∈ Γ_{p}}, where σ_{p} ∈ hom(Γ_{p}, Π_{p}). Since Z_{p} ⊗ M is a summand of ind_{G_{p}}^{G} M_{p}, then Z_{p} ⊗ M is a summand of a permutation Z_{p}G-lattice.

Let L be an indecomposable summand of Z_{p} ⊗ M. Denote its vertex by 1 × D ⊆ G_{p′} ×G_{p}; then L is isomorphic to a summand of ind_{1×D}^{G_{p′} ×G_{p}} Z_{p} ≅ Z_{p}G_{p′} ⊗ind_{D}^{G_{p}} Z_{p}. Write Z_{p}G_{p′} = ⨁ X_{i} as a direct sum of (projective) indecomposables. Then L is a summand of X_{i} ⊗ ind_{D}^{G_{p}} Z_{p} for some i. We claim that Y = X_{i} ⊗ ind_{D}^{G_{p}} Z_{p} is indecomposable. The G_{p}-fixed points of Y are X_{i} ⊗ Z_{p} ≅ X_{i}, which is irreducible mod p as an F_{p}G_{p′}-module, since p does not divide |G_{p′}|. If Y were decomposable, it would be decomposable mod p, say as Z_{1} ⊕ Z_{2}, where Z_{1}, Z_{2} are nonzero F_{p}Gmodules. The G_{p}-fixed points of each of Z_{1}, Z_{2} are non-zero, since G_{p} is a p-group. This contradicts the irreduciblity of (Y/pY)^{G_{p}} as F_{p}G_{p′}-module. So Y is indeed indecomposable, and therefore L, which is a summand of Y, is Y itself. Thus Z_{p} ⊗ M is isomorphic to a sum of modules of the form X ⊗ ind_{D}^{G_{p}} Z_{p}, where X is a Z_{p}G_{p′}-module. Since each D is of the form [σ_{p}], it follows that

(3.1)  Z_{p} ⊗ M ≅ ⨁_{σ_{p}∈hom(Γ_{p},Π_{p})} X_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} Z_{p},

where each X_{σ_{p}} is a Z_{p}G_{p′}-module. Let ξ_{σ_{p}} be the character of X_{σ_{p}}. Then χ has the form

χ(g) = ∑_{σ_{p}} ξ_{σ_{p}}(g_{p′}) ind_{[σ_{p}]}^{G_{p}} 1(g_{p}).

In other words, writing R(G) = R(G_{p′}) ⊗ R(G_{p}), we have

χ = ∑_{σ_{p}∈Σ_{p}} ξ_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} 1, ξ_{σ_{p}} ∈ R^{+}(G_{p′}). (3.2)

Write ξ_{σ_{p}} as a sum of irreducible characters of G_{p′}:

(3.3) ξ_{σ_{p}} = ∑_{η∈irr(G_{p′})} b(σ_{p}, η)η, with unique integers b(σ_{p}, η) ≥ 0.

Changing the order of summation, we have χ = ∑_{η} η ⊗ λ_{η}, where

λ_{η} = ∑_{σ_{p}} b(σ_{p}, η) ind_{[σ_{p}]}^{G_{p}} 1 ∈ P^{+}(Γ_{p}, Π_{p}). (3.4)

It follows from Lemma 3.1 that χ ∈ ⋂_{p} Q_{p}^{+}.

Now suppose that χ ∈ ⋂_{p} Q_{p}^{+}; we will show that χ ∈ D′. From Lemma 3.2, χ ∈ Q ⊗ P; since characters in P are characters of permutation modules, we see that χ is rational valued.

Let p be a prime dividing |G|. Since χ is in Q_{p}^{+}, write χ as

χ = ∑_{η∈irr(G_{p′})} η ⊗ λ_{η}, λ_{η} ∈ P^{+}(Γ_{p}, Π_{p}).

Let ζ be a primitive |G|-th root of unity; let G denote the Galois group of Q(ζ) over Q, and let G_{p} denote the Galois group of Q_{p}(ζ) over Q_{p}, where Q_{p} is the p-adic rationals. Since χ and λ_{η} are rational valued, we have

χ = χ^{ω} = ∑_{η} η^{ω} ⊗ λ_{η}, ω ∈ G_{p}.

By uniqueness of the representation χ = ∑_{η} η ⊗ λ_{η}, it follows that λ_{η} = λ_{η^{ω}}, ω ∈ G_{p}. Partition irr(G_{p′}) into G_{p}-orbits. For an orbit O, λ_{η} is the same for all η in O; call this common value λ_{O}. Let τ_{O} denote ∑_{η∈O} η, which takes values in Q_{p}. Then

χ = ∑_{O} ∑_{η∈O} η ⊗ λ_{η} = ∑_{O} ∑_{η∈O} η ⊗ λ_{O} = ∑_{O} τ_{O} ⊗ λ_{O}. (3.5)

We have each λ_{O} ∈ P^{+}(Γ_{p}, Π_{p}), so we may choose non-negative integers b(σ_{p}, O) such that λ_{O} = ∑_{σ_{p}} b(σ_{p}, O) ind_{[σ_{p}]}^{G_{p}} 1. Let

ξ_{σ_{p}} = ∑_{O} b(σ_{p}, O)τ_{O}. (3.6)

Then, changing the order of summation, we have

χ = ∑_{O} τ_{O} ⊗ λ_{O} = ∑_{O} ∑_{σ_{p}} τ_{O} ⊗ b(σ_{p}, O) ind_{[σ_{p}]}^{G_{p}} 1 = ∑_{σ_{p}} ξ_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} 1.

Now ξ_{σ_{p}} is a Z_{≥0}-linear combination of {τ_{O}}, and each τ_{O} = ∑_{η∈O} η. Fix η ∈ irr(G_{p′}) in an orbit O. Since the order of G_{p′} is not divisible by p, the Schur index of η over the field Q_{p}(η) is 1, by [F, IV.9.5]. Let K = Q_{p}(η) and let X be a KG_{p′}-module affording η. By restriction of scalars, X is a Q_{p}G_{p′}-module whose character is the sum of all the algebraic conjugates of η over Q_{p}, namely τ_{O}. Thus τ_{O} is the character of a Q_{p}G_{p′}-module; hence so is ξ_{σ_{p}}. Choose a Z_{p}G_{p′}-lattice L_{σ_{p}} in this module, so ξ_{σ_{p}} is afforded by L_{σ_{p}}.

Next, define the Z_{p}G-lattice M(p) by

M(p) = ⨁_{σ_{p}} L_{σ_{p}} ⊗_{Z_{p}} ind_{[σ_{p}]}^{G_{p}} Z_{p}. (3.7)

We claim that (a) the N-fixed points M(p)^{N} have trivial G/N-action, and

(b) res_{N} M(p) is Z_{p}N-free.

To prove (a), note that M(p) has character χ which is in ⋂_{p} Q_{p}^{+}; by Lemma 3.2, χ ∈ Q ⊗ P. Since the N-fixed points of ind_{[σ]}^{G} Z have trivial G/N-action, (a) holds.

We now prove (b). Since G_{p} = N_{p}[σ_{p}] and N_{p} ∩ [σ_{p}] = 1, it follows that res_{N_{p}} ind_{[σ_{p}]}^{G_{p}} Z_{p} ≅ Z_{p}N_{p} by Mackey decomposition. Since

res_{N} M(p) ≅ ⨁_{σ_{p}}(res_{N_{p′}} L_{σ_{p}}) ⊗_{Z_{p}} Z_{p}N_{p} ≅ res_{N_{p′}} (⨁_{σ_{p}} L_{σ_{p}}) ⊗_{Z_{p}} Z_{p}N_{p},

it suffices to show that res_{N_{p′}} (⨁_{σ_{p}} L_{σ_{p}}) is free over Z_{p}N_{p′}. Since p does not divide |N_{p′}|, we need only show that the character χ′_{p} of ⨁_{σ_{p}} L_{σ_{p}} has the property that res_{N_{p′}} χ′_{p} is a multiple of the character ρ(N_{p′}) of the regular representation of N_{p′}.

Since χ = ∑_{σ_{p}} ξ_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} 1 relative to R(G_{p′}) ⊗ R(G_{p}) and since the degree of

each ind_{[σ_{p}]}^{G_{p}} 1 is |G_{p} : [σ_{p}]| = |Π_{p}|, then res_{G_{p′}} χ = |Π_{p}|χ′_{p}; hence res_{N_{p′}} χ′_{p} is a Q-multiple of res_{N_{p′}} χ. Let y be a non-identity element of N_{p′}. Then y has the form (1, π), so no conjugate of y lies in [σ_{p}]. Since χ ∈ Q ⊗ P by Lemma 3.2, χ and therefore χ′_{p} both vanish on y. It follows that res_{N_{p′}} χ′_{p} is a multiple of ρ(N_{p′}), as desired. This proves (b).

We next show that χ is afforded by a QG-module. As in equation (3.5), we have

χ = ∑_{O′} τ_{O′} ⊗ λ_{O′},

where τ_{O′} is the orbit sum over an orbit of G acting on irr(G_{p′}), and λ_{O′} is the common value of λ_{η} for all η ∈ O′. Let R_{Q}^{+}(G) be the characters afforded by QG-modules.

Since τ_{O′} is a rational valued character of G_{p′}, then |G_{p′}|τ_{O′} ∈ R_{Q}^{+}(G_{p′}); since λ_{O′} is the character of a permutation module, then |G_{p′}|χ ∈ R_{Q}^{+}(G). Varying over p, since the greatest common divisor of the |G_{p′}| is 1, it follows that χ ∈ R_{Q}^{+}(G).

Let V be a QG-module affording χ. For each prime p we have an isomorphism

ϕ_{p} : Q_{p} ⊗_{Q} V → Q_{p} ⊗_{Z_{p}} M(p).

For each p let

V (p) = {v ∈ V : ϕ_{p}(1 ⊗ v) ∈ 1 ⊗ M(p)}.

Then let M = ⋂_{p} V (p). From [R, 5.3] we see that M is a ZG-lattice such that Z_{p} ⊗ M ≅ M(p). Then M affords χ, the fixed points M^{N} have trivial G/N-action, and res_{N}(Z_{p} ⊗ M) ≅ res_{N} M(p), so res_{N} M is locally free. Therefore χ is in D′, as desired. This completes the proof.

For later use, we record the following result, which is proved in the second paragraph of the proof of Theorem 3.3.

Proposition 3.4. If M satisfies (a) and (b′), then for each prime p, Z_{p} ⊗ M is a summand of a permutation lattice for Z_{p}G.

4. The Lattice Spanned by D′

In this section we show that the Z-span ZD′ of D′ is equal to ⋂_{p} Q_{p}. We also show that ZD′ is equal to P, if Γ is cyclic. We do not know if this result holds for arbitrary Γ.

Proposition 4.1. The Z-span of D′ is equal to ⋂_{p} Q_{p}.

Proof. It follows from Theorem 3.3 that

ZD′ ⊆ ⋂_{p} Q_{p}.

For the reverse inclusion, suppose that χ ∈ ⋂_{p} Q_{p}. For a fixed p dividing |G|, write

(4.1) χ = ∑_{η∈irr(G_{p′})} η ⊗ λ_{η} for unique λ_{η} ∈ P(Γ_{p}, Π_{p}).

By Lemma 3.2, χ ∈ Q ⊗ P. For a given η ∈ irr(G_{p′}), suppose that

(4.2) ⟨η, ind_{[σ_{p′}]}^{G_{p′}} 1⟩_{G_{p′}} = 0 for all σ_{p′} ∈ hom(Γ_{p′}, Π_{p′}).

Then η is orthogonal to P(Γ_{p′}, Π_{p′}), and hence

0 = ⟨χ, η ⊗ λ_{η}⟩_{G} = ⟨η ⊗ λ_{η}, η ⊗ λ_{η}⟩_{G} = ⟨η, η⟩_{G_{p′}} ⟨λ_{η}, λ_{η}⟩_{G_{p}}.

It follows that λ_{η} = 0. Thus in equation (4.1), we need only sum over η ∈ irr(G_{p′}) for which (4.2) does not hold. For such an η, if ⟨η, ind_{[σ_{p′}]}^{G_{p′}} 1⟩_{G_{p′}} ≠ 0, decompose

ind_{[σ_{p′}]}^{G_{p′}} 1, giving

−η = ∑ η̃ − ind_{[σ_{p′}]}^{G_{p′}} 1 (4.3)

for some η̃ ∈ irr(G_{p′}). Write λ_{η} = λ′_{η} − λ′′_{η} with λ′_{η}, λ′′_{η} ∈ P^{+}(Γ_{p}, Π_{p}); we get

χ = ∑ η ⊗ λ′_{η} + ∑(−η) ⊗ λ′′_{η}.

In this equation, replace −η using equation (4.3). We get χ = χ′_{p}−ξ_{p} with χ′_{p} ∈ Q_{p}^{+}, ξ_{p} ∈ P^{+}. Let ξ = ∑_{p||G|} ξ_{p} ∈ P^{+}; then χ + ξ = χ′_{p} + ∑_{l≠p} ξ_{l}, which is in Q_{p}^{+} for all p. Thus

χ = (χ + ξ) − ξ

with χ + ξ ∈ ⋂_{p} Q_{p}^{+} and ξ ∈ P^{+} ⊆ ⋂_{p} Q_{p}^{+}. Since ⋂_{p} Q_{p}^{+} = D′ by Theorem 3.3, the result is proved.

We next show that ZD′ = P if Γ is cyclic. Let Σ be a complete set of homomorphisms from Γ to Π up to conjugacy in Π. Our proof that ZD′ = P for cyclic Γ uses the next result, that {ind_{[σ]}^{G} 1 : σ ∈ Σ} is linearly independent over Q if Γ is cyclic. This lemma can fail if Γ is not cyclic, but it is possible that ZD′ = P can be proved by some other method in the non-cyclic case.

Lemma 4.2. The set {ind_{[σ]}^{G} 1 : σ ∈ Σ} is a basis of Q ⊗ P if Γ is cyclic.

Proof. For g = (γ, π) ∈ G and σ ∈ hom(Γ, Π), we have

ind_{[σ]}^{G} 1(g) = \begin{cases}|C_{Π}(π)|, & σ(γ) ∼ π, \\ 0, & else, \end{cases}

where ∼ denotes conjugacy in Π. If τ ∈ hom(Γ, Π) then τ is conjugate to some σ ∈ Σ. From the formula for ind_{[σ]}^{G} 1(g) above, then ind_{[τ]}^{G} 1 = ind_{[σ]}^{G} 1, so Q ⊗ P is spanned by {ind_{[σ]}^{G} 1 : σ ∈ Σ}.

Suppose that ∑_{σ∈Σ} a_{σ} ind_{[σ]}^{G} 1 = 0 with a_{σ} ∈ Q. Given τ ∈ Σ, let γ be a generator of Γ, and evaluate at (γ, τ(γ)). We get

ind_{[σ]}^{G} 1(γ, τ(γ)) = \begin{cases}|C_{Π}(τ(γ))|, & σ(γ) ∼ τ(γ), \\ 0, & else. \end{cases}

If τ and σ are distinct elements of Σ, then τ(γ) and σ(γ) are not conjugate in Π, so ind_{[σ]}^{G} 1(γ, τ(γ)) = 0. It follows that a_{τ} = 0, for all τ in Σ, and the result is proved.

Proposition 4.3. The Z-span of D′ is equal to P if Γ is cyclic.

Proof. Since P^{+} ⊆ D′ it suffices to show that D′ ⊆ P. Suppose that χ ∈ D′. Then χ ∈ Q ⊗ P from Theorem 3.3 and Lemma 3.2. Then χ = ∑_{σ∈Σ} a_{σ} ind_{[σ]}^{G} 1, where a_{σ} ∈ Q, and by Lemma 4.2, the a_{σ} ∈ Q are unique. We must show that each a_{σ} ∈ Z.

Since G is nilpotent, we pick Σ by picking complete sets Σ_{p} ⊆ hom(Γ_{p}, Π_{p}) up to conjugacy in Π_{p}, and then letting Σ be those homomorphisms whose restrictions to Γ_{p} are in Σ_{p}.

Fix a prime p dividing |G|. From Theorem 3.3, χ ∈ Q_{p}(Γ, Π) = R(G_{p′}) ⊗ P(Γ_{p}, Π_{p}). For σ ∈ Σ, we have ind_{[σ]}^{G} 1 = ind_{[σ_{p′}]}^{G_{p′}} 1 ⊗ ind_{[σ_{p}]}^{G_{p}} 1. Then

χ = ∑_{τ∈Σ_{p}}( ∑_{\begin{matrix}σ∈Σ \\ σ_{p}=τ \end{matrix}} a_{σ} ind_{[σ_{p′}]}^{G_{p′}} 1) ⊗ ind_{[τ]}^{G_{p}} 1.

But {ind_{[τ]}^{G_{p}} 1} is a Z-basis of P(Γ_{p}, Π_{p}), from Lemma 4.2, so it follows that

ϕ_{τ} = ∑_{\begin{matrix}σ∈Σ \\ σ_{p}=τ \end{matrix}} a_{σ} ind_{[σ_{p′}]}^{G_{p′}} 1 ∈ R(G_{p′})

for all τ ∈ Σ_{p}. Let γ_{p′} be a generator of Γ_{p′}. Evaluate ϕ_{τ} at (γ_{p′}, σ_{p′}(γ′_{p})). For the unique σ ∈ Σ whose restriction to Γ_{p} is τ and whose restriction to Γ_{p′} is σ_{p′}, we get, as in the proof of Lemma 4.1,

|C_{Π_{p′}} (σ_{p′}(γ_{p′}))|a_{σ} ∈ Z

since the values of ϕ_{τ} are algebraic integers in Q. Thus

|Π_{p′}|a_{σ} ∈ Z, for all σ ∈ Σ.

This holds for all p. Since the greatest common divisor of |Π_{p′}| is 1, it follows that a_{σ} ∈ Z for all σ ∈ Σ, and the proof is complete.

Here is an example showing that Lemma 4.2 can fail if Γ is not cyclic. Let Γ and Π each be the direct product C_{p} × C_{p} of cyclic groups of prime order. Consider the p^{4} × p^{4} matrix whose rows are indexed by Σ = hom(Γ, Π) and columns by G = Γ × Π, with (σ, g)-entry given by ind_{[σ]}^{G} 1(g). Since ind_{[σ]}^{G} 1(1, x) = 0 if x is a nontrivial element of Π, then there is a column of zeros, and so the rows are linearly dependent.

5. Examples of D ≠ P^{+}

In this section we produce examples of groups Γ, Π and modules M having character which is in D(Γ, Π) but not in P^{+}(Γ, Π).

Lemma 5.1. For distinct primes p_{1} and p_{2}, let Γ = C_{p_{1}p_{2}}, the cyclic group of order p_{1}p_{2}, and let Π = C_{p_{1}p_{2}} × C_{p_{1}p_{2}}. Then D(Γ, Π) ≠ P^{+}(Γ, Π).

Proof. As before, let G = Γ × Π, N = 1 × Π. Choose sets Σ_{1}, Σ_{2} of homomorphisms σ : Γ → Π whose images σ(Γ) are precisely the subgroups of Π of order p_{1}, respectively, p_{2}. Let Σ = Σ_{1} ∪ Σ_{2}. Let τ : Γ → Π be the trivial map. The module M we will construct has character

χ = − ind_{[τ]}^{G} 1 + ∑_{σ∈Σ} ind_{[σ]}^{G} 1.

It will follow from our construction that χ ∈ D(Γ, Π). Let γ be a generator of Γ, and let g = (γ, 1) ∈ G. Then (γ, 1) ∈ [τ], so ind_{[τ]}^{G} 1(g) = |Π|, but (γ, 1) ∉ [σ] for all σ ∈ Σ, so ind_{[σ]}^{G} 1(g) = 0. Hence χ(g) < 0 and χ is not the character of a permutation module, so χ ∉ P^{+}(Γ, Π).

For σ ∈ Σ, M(σ) is the corresponding double-action module. Define

M(Σ) = ⨁_{σ∈Σ} M(σ).

For each σ ∈ Σ, let s(σ) = ∑_{x∈σ(Γ)} x ∈ ZΠ, and define the map

f_{σ} : M(σ) → M(τ), f_{σ}(m) = s(σ)m, m ∈ M(σ).

Define f : M(Σ) → M(τ) by f = ∑_{σ} f_{σ}. The key to the proof is the claim that f is an epimorphism of ZG-modules.

We now prove that the claim holds. Since s(σ)σ(γ^{−1}) = s(σ) = σ(γ^{−1})s(σ), then each f_{σ} is a ZG-homorphism, and so is f. To prove that f is surjective, it suffices to find v ∈ M(Σ) such that f(v) = 1 ∈ M(τ). Pick two distinct elements ϕ_{i}, ψ_{i} from Σ_{i}, i = 1, 2. Find integers n_{1}, n_{2} such that n_{1}p_{1} + n_{2}p_{2} = 1. Define v = ∑_{σ∈Σ} v_{σ} with v_{σ} ∈ M(σ) given by

v_{σ} = \begin{cases}n_{i} · 1, & σ ∈ Σ_{i}, σ ≠ ϕ_{i}, \\ n_{i}(1 − s(ψ_{i})), & σ = ϕ_{i}. \end{cases}

To compute f(v),

f(v) = n_{1} ∑_{\begin{matrix}σ∈Σ_{1} \\ σ≠ϕ_{1} \end{matrix}} s(σ) + n_{1}s(ϕ_{1})(1 − s(ψ_{1})) + n_{2} ∑_{\begin{matrix}σ∈Σ_{2} \\ σ≠ϕ_{2} \end{matrix}} s(σ) + n_{2}s(ϕ_{2})(1 − s(ψ_{2}).

Since

s(σ)s(σ̃) = Π̂_{p_{i}}, σ, σ̃ ∈ Σ_{i}, σ ≠ σ̃,

we get

f(v) = n_{1}( ∑_{σ∈Σ_{1}} s(σ) − Π̂_{p_{1}}) + n_{2}( ∑_{σ∈Σ_{2}} s(σ) − Π̂_{p_{2}}).

In the sum ∑_{σ∈Σ_{i}} s(σ) ∈ ZΠ_{p_{i}}, non-identity elements y ∈ Π_{p_{i}} occur exactly once, whereas 1 occurs p_{i} + 1 times and

∑_{σ∈Σ_{i}} s(σ) = p_{i} · 1 + Π̂_{p_{i}}.

We obtain

f(v) = n_{1}p_{1} · 1 + n_{2}p_{2} · 1 = 1.

Therefore f is indeed a ZG-epimorphism.

Now define M to be the kernel of f : M(Σ) → M(τ), so we have the exact sequence

0 → M → M(Σ) →^{f} M(τ) → 0.

Since res_{N} M(σ) is free for all σ, this sequence splits when restricted to N; hence res_{N} M is stably free. Since ZN satisfies the Eichler condition, it follows that res_{N} M is ZN-free by Jacobinski’s Cancellation Theorem [CR, 51.24]. (In fact, it can be shown directly that res_{N} M is ZN-free by exhibiting a basis; this is done in a special case below.) Since G/N acts trivially on M(Σ)^{N}, it does so on M^{N}. The character χ of M is therefore in D, and it follows from the exact sequence that χ = − ind_{[τ]}^{G} 1 + ∑_{σ∈Σ} ind_{[σ]}^{G} 1. This completes the proof.

From Proposition 2.4, there exists U ∈ SGL_{p_{1}p_{2}}(ZΠ) with U^{p_{1}p_{2}} = 1 and U not conjugate in GL(QΠ) to a diagonal matrix of group elements. We can exhibit such a matrix U by computing the action of a generator of Γ on an explicit ZN-basis of the free module res_{N} M. To simplify the exposition, we assume that

p_{1} = 2, p_{2} = 3.

Then pick

n_{1} = −1, n_{2} = 1.

Write Γ = ⟨c⟩ of order 6, Π = ⟨a⟩ × ⟨b⟩, where a and b each have order 6. Let

Σ_{1} = {σ_{1}, σ_{2}, σ_{3}}, σ_{1}(c) = a^{3}, σ_{2}(c) = b^{3}, σ_{3}(c) = a^{3}b^{3},

Σ_{2} = {σ_{4}, σ_{5}, σ_{6}, σ_{7}}, σ_{4}(c) = a^{2}, σ_{5}(c) = b^{2}, σ_{6}(c) = a^{2}b^{2}, σ_{7}(c) = a^{2}b^{4}.

Pick

ψ_{1} = σ_{1}, ϕ_{1} = σ_{3}, ψ_{2} = σ_{4}, ϕ_{2} = σ_{7}.

Define

s_{i} = s(σ_{i}), c_{i} = σ_{i}(c).

Identify M(Σ) with ZΠ^{7} and M(τ) with ZΠ, Then our map f takes ZΠ^{7} to ZΠ, given by

f(z_{1}, …, z_{7})^{T} = ∑_{i=1}^{7} s_{i}z_{i}.

The G-action on ZΠ^{7} becomes

(z_{1}, …, z_{7})^{T} (x, y) = (σ_{1}(x^{−1})z_{1}y ⋯, σ_{7}(x^{−1})z_{7}y)^{T} .

(Transposes are used because M(Σ) is a right ZΠ-module and endomorphisms are matrices over ZΠ acting on the left.) In this notation the element v in the proof of Lemma 5.1 is

v = (−1, −1, s_{1} − 1, 1, 1, 1, 1 − s_{4})^{T} .

Let {e_{1}, e_{2}, …, e_{7}} be the standard basis of ZΠ^{7}. Extend the “unimodular column” v to a ZΠ-basis {v, e_{2}, …, e_{7}} of ZΠ^{7}. Then do “elementary operations” by setting

m_{i} = e_{i+1} − vf(e_{i+1}) = e_{i+1} − vs_{i+1}, 1 ≤ i ≤ 6,

and then {v, m_{1}, m_{2}, …, m_{6}} is a ZΠ-basis of ZΠ^{7}. The m_{i} are in M by construction, so {m_{1}, …, m_{6}} is a ZΠ-basis of M.

We want the action of a generator of Γ on M in this basis. The matrix of (c^{−1}, 1) in the standard basis of ZΠ^{7} is D = diag(c_{1}, …, c_{7}). So we need only see the effect of two changes of basis. Set

A = (\begin{matrix}−1 \\ −1 & 1 \\ s_{1} − 1 & & 1 \\ 1 & & & 1 \\ 1 & & & & 1 \\ 1 & & & & & 1 \\ 1 − s_{4} & & & & & & 1 \end{matrix}) , B = (\begin{matrix}1 & −s_{2} & −s_{3} & −s_{4} & −s_{5} & −s_{6} & −s_{7} \\ & 1 \\ & & 1 \\ & & & 1 \\ & & & & 1 \\ & & & & & 1 \\ & & & & & & 1 \end{matrix}) .

The action of (c^{−1}, 1) in the basis {v, m_{1}, m_{2}, …, m_{6}} is given by the ZΠ-matrix X = B^{−1}A^{−1}DAB. Since M is a G-submodule of ZΠ^{7}, this matrix has the form

(\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 \\ ∗ \\ ∗ \\ ∗ & & & U \\ ∗ \\ ∗ \\ ∗ \end{matrix})

for some U ∈ GL_{6}(ZΠ), and it is this U that we want. Using c_{i}s_{i} = s_{i}, and denoting c_{ij} = c_{i} − c_{j}, s_{1}^{∗} = s_{1} − 1, s_{4}^{∗} = s_{4} − 1, we get U =

(\begin{matrix}c_{2} + c_{21}s_{2} & c_{21}s_{3} & c_{21}s_{4} & c_{21}s_{5} & c_{21}s_{6} & c_{21}s_{7} \\ c_{13}s_{1}^{∗}s_{2} & c_{3} + c_{13}s_{1}^{∗}s_{3} & c_{13}s_{1}^{∗}s_{4} & c_{13}s_{1}^{∗}s_{5} & c_{13}s_{1}^{∗}s_{6} & c_{13}s_{1}^{∗}s_{7} \\ c_{14}s_{2} & c_{14}s_{3} & c_{4} + c_{14}s_{4} & c_{14}s_{5} & c_{14}s_{6} & c_{14}s_{7} \\ c_{15}s_{2} & c_{15}s_{3} & c_{15}s_{4} & c_{5} + c_{15}s_{5} & c_{15}s_{6} & c_{15}s_{7} \\ c_{16}s_{2} & c_{16}s_{3} & c_{16}s_{4} & c_{16}s_{5} & c_{6} + c_{16}s_{6} & c_{16}s_{7} \\ c_{71}s_{4}^{∗}s_{2} & c_{71}s_{4}^{∗}s_{3} & c_{71}s_{4}^{∗}s_{4} & c_{71}s_{4}^{∗}s_{5} & c_{71}s_{4}^{∗}s_{6} & c_{7} + c_{71}s_{4}^{∗}s_{7} \end{matrix}) .

Note that this matrix has trace −1 + ∑_{i=1}^{7} c_{i}, so it is a counterexample to the strategy of [MRSW]. Actually every χ ∈ D′ − P^{+} gives such a counterexample, by Lemma 1 of [WCr] generalized to matrices.

Lemma 5.2. Let p be an odd prime, and let Γ = C_{4} × C_{p} and Π = Q_{8} × C_{p} × C_{p}, where Q_{8} is the quaternion group of order 8. Then D(Γ, Π) ≠ P^{+}(Γ, Π).

Proof. Pick τ ∈ hom(Γ, Π) whose image is one of the cyclic subgroups of order 4 of Q_{8}. Choose Σ_{2} to consist of three homomorphisms Γ → Π whose images are the two other subgroups of order 4 in Q_{8} as well as the subgroup of order 2. Choose Σ_{p} to consist of p + 1 elements of hom(Γ, Π) so that im σ_{p} are the nontrivial cyclic p-subgroups and so that σ_{2} = τ_{2}. Set Σ = Σ_{2} ∪ Σ_{p}, and

χ′ = − ind_{[τ]}^{G} 1 + ∑_{σ∈Σ} ind_{[σ]}^{G} 1.

Relative to R(G) = R(G_{p}) ⊗ R(G_{2}),

χ′ = (− ind_{[1]}^{G_{p}} 1 + ∑_{σ∈Σ_{p}} ind_{[σ_{p}]}^{G_{p}} 1) ⊗ ind_{[τ_{2}]}^{G_{2}} 1 + ∑_{σ∈Σ_{2}} ind_{[1]}^{G_{p}} 1 ⊗ ind_{[σ_{2}]}^{G_{2}} 1,

where the expression in parentheses is in R^{+}(G_{p}). So χ ∈ Q_{2}^{+}. Relative to R(G) = R(G_{2}) ⊗ R(G_{p}) we have

χ′ = (− ind_{[τ_{2}]}^{G_{2}} 1 + ∑_{σ∈Σ_{2}} ind_{[σ_{2}]}^{G_{2}}) ⊗ ind_{[1]}^{G_{p}} 1 + ∑_{σ∈Σ_{p}} ind_{[σ_{p}]}^{G_{p}} 1 ⊗ ind_{[τ_{2}]}^{G_{2}} 1,

where the expression in parentheses is in R^{+}(G_{2}). Therefore χ ∈ Q_{p}^{+}.

By Theorem 3.3, χ′ ∈ D′; hence χ = rχ′ ∈ D by Lemma 2.5. We shall show that χ ∉ P^{+}. By Lemma 4.2, it is enough to check that τ is not Π-conjugate to any σ ∈ Σ. But the image of τ, which is normal in Π, is different from the images of all σ ∈ Σ. This completes the proof.

6. On Problem 1

In this section we prove that D′ = P^{+} if Π is nilpotent and Γ has prime-prower order. Since P^{+} ⊆ D ⊆ D′, then Problem 1 has a positive answer in this case. We also completely deal with Problem 1 if Γ is cyclic.

Theorem 6.1. Suppose that Γ is an l-group for some prime l and that Π is nilpotent. Then D′(Γ, Π) = P^{+}(Γ, Π).

Proof. Suppose that χ ∈ D′. For each prime p ≠ l, use Theorem 3.3 to write χ relative to R(G_{p′}) ⊗ R(G_{p}) as χ = ∑_{σ_{p}} ξ_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} 1. Since Γ_{p} = 1, the only

σ_{p} : Γ_{p} → Π_{p} in this sum is the trivial map; hence ind_{[σ_{p}]}^{G_{p}} 1 is the character ρ(G_{p}) of the regular representation and χ = ξ_{σ_{p}} ⊗ ρ(G_{p}). In particular, χ vanishes off G_{p′}. Varying p ≠ l, it follows that χ vanishes off G_{l}.

Define the class function λ on G_{l} by λ(g) = χ(g)/|G_{l′}|. Then relative to R(G) = R(G_{l′}) ⊗ R(G_{l}) (actually with scalars extended to Q) we have χ = ρ(G_{l′}) ⊗ λ.

At the prime l write χ = ∑_{η∈irr(G_{l′})} η⊗χ_{η} with χ_{η} ∈ P^{+}(Γ_{l}, Π_{l}). Since ρ(G_{l′}) = ∑_{η∈irr(G_{l′})} η(1)η, we get

χ = ρ(G_{l′}) ⊗ λ = ∑_{η} η ⊗ η(1)λ = ∑_{η} η ⊗ χ_{η},

so we deduce that χ_{η} = η(1)λ for all η ∈ irr(G_{l′}). Take η to be the trivial character; then λ = χ_{1} ∈ P^{+}(Γ_{l}, Π_{l}), and we can write λ = ∑_{σ_{l}} a_{σ_{l}} ind_{[σ_{l}]}^{G_{l}} 1, where each a_{σ_{l}} is a non-negative integer.

Since ρ(G_{l′}) ⊗ ind_{[σ_{l}]}^{G_{l}} 1 = ind_{[σ_{l}]}^{G} 1, we have

χ = ρ(G_{l′}) ⊗ λ = ∑_{σ_{l}} a_{σ_{l}}ρ(G_{l′}) ⊗ ind_{[σ_{l}]}^{G_{l}} 1 = ∑_{σ_{l}} a_{σ_{l}} ind_{[σ_{l}]}^{G} 1.

Thus χ ∈ P^{+}(Γ, Π), and the proof is complete.

Corollary 6.2. Γ is a subgroup of SGL_{n}(ZΠ) if and only if Γ is isomorphic to a subgroup of Π^{n} (the direct product of n copies of Π.)

Proof. Suppose that Γ ⊆ SGL_{n}(ZΠ). From Corollary 2.7, Γ is nilpotent, so in order to prove that Γ is a subgroup of Π^{n} it suffices to prove that Γ_{l} ⊆ Π^{n} for each prime l dividing Γ. Hence we may assume that Γ is an l-group. Then Theorem 6.1 implies that D(Γ, Π) = P^{+}(Γ, Π), and then from Proposition 2.3, uΓu^{−1} ⊆ Π^{n}.

The converse is clear.

Theorem 6.3. Suppose that Π is nilpotent. Then D = P^{+} for all cyclic Γ if and only if Π has at most one non-cyclic Sylow p-subgroup.

Proof. Suppose that Π has at most one non-cyclic Sylow p-subgroup. We will show that D′ = P^{+}, and therefore that D = P^{+}. Fix a prime p, which exists by hypothesis, such that Π_{p′} is cyclic, and therefore has a faithful character λ of degree 1. Let χ be in D′. Choose Σ as in the proof of Proposition 4.3, namely

Σ = {σ ∈ hom(Γ, Π) : σ_{p} ∈ Σ_{p}},

where Σ_{p} ⊂ hom(Γ_{p}, Π_{p}) is a complete set of homomorphisms up to conjugacy in Π_{p}. By Proposition 4.3, χ ∈ P, and since Γ is cyclic, we may write, by Lemma 4.2,

χ = ∑_{σ∈Σ} a_{σ} ind_{[σ]}^{G} 1 for unique a_{σ} ∈ Z.

We must show that a_{σ} ≥ 0 for all σ.

Relative to R(G) = R(G_{p′}) ⊗ R(G_{p}), we have ind_{[σ]}^{G} 1 = ind_{[σ_{p′}]}^{G_{p′}} 1 ⊗ ind_{[σ_{p}]}^{G_{p}} 1, giving

χ = ∑_{σ} a_{σ} ind_{[σ_{p′}]}^{G_{p′}} 1 ⊗ ind_{[σ_{p}]}^{G_{p}} 1.

From equation (3.2), we have

χ = ∑_{σ_{p}∈Σ_{p}} ξ_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} 1, ξ_{σ_{p}} ∈ R^{+}(G_{p′}).

Comparing these equations, using linear independence of {ind_{[σ]}^{G} 1 : σ ∈ Σ} from Lemma 4.2, we get

ξ_{σ_{p}} = ∑_{τ} a_{τ} ind_{[τ_{p′}]}^{G_{p′}} 1 ∈ R^{+}(G_{p′})

where the sum is over τ ∈ Σ such that τ_{p} = σ_{p}. Then ⟨ξ_{σ_{p}}, η⟩_{G_{p′}} ≥ 0 for all irreducible characters η of G_{p′} = Γ_{p′} × Π_{p′}, in particular for η = λ^{∗}σ_{p′} ⊗ λ, where λ^{∗} is the contragredient of λ. But

⟨ξ_{σ_{p}}, λ^{∗}σ_{p′} ⊗ λ⟩ = ∑_{τ} a_{τ} ⟨1, res_{[τ_{p′}]}^{G_{p′}}(λ^{∗}σ_{p′} ⊗ λ)⟩_{[τ_{p′}]}

= ∑_{τ} a_{τ} /|Γ_{p′}| ∑_{γ∈Γ_{p′}} λ(σ_{p′}γ^{−1})λ(τ_{p′}γ) = ∑_{τ} a_{τ} ⟨λσ_{p′}, λτ_{p′}⟩_{Γ_{p′}} ,

and this equals a_{σ} since λσ_{p′} and λτ_{p′} are different irreducible characters of Γ_{p′} unless σ_{p′} = τ_{p′}, that is, τ = σ. Thus a_{σ} ≥ 0.

Conversely, suppose that Π has at least 2 non-cyclic Sylow p-subgroups. First suppose that Π has a subgroup of the form Π′ = C_{p_{1}} × C_{p_{1}} × C_{p_{2}} × C_{p_{2}} ≅ C_{p_{1}p_{2}} × C_{p_{1}p_{2}}, where p_{1}, p_{2} are distinct primes. Using the construction in Lemma 5.1 with Γ = C_{p_{1}p_{2}}, G′ = Γ×Π′, N′ = 1×Π′, there exists a ZG′-lattice M′ satisfying (a) and (b) whose character is χ′ = − ind_{[τ]}^{G′} 1+∑_{σ∈Σ′} ind_{[σ]}^{G′} 1. By Lemma 2.4, χ = ind_{G′}^{G} χ′ is in D(Γ, Π). But χ = − ind_{[τ]}^{G} 1+∑_{σ∈Σ′} ind_{[σ]}^{G} 1 is not in P^{+} by Lemma 4.2, since τ = 1 is not Π-conjugate to an element of Σ′. Thus D(Γ, Π) ≠ P^{+}(Γ, Π).

If Π has at least 2 non-cyclic Sylow p-subgroups but does not have a subgroup isomorphic to C_{p_{1}} × C_{p_{1}} × C_{p_{2}} × C_{p_{2}}, where p_{1}, p_{2} are distinct primes, then Π_{2} is a quaternion group, and Π has a subgroup of the form Π′ = C_{p} × C_{p} × Q_{8} where p is an odd prime. Apply the construction of Lemma 5.2, where Γ = C_{p} × C_{4} ≅ C_{4p}, G′ = Γ × Π′, N′ = 1 × Π′, to get

χ′ = r(− ind_{[τ]}^{G′} 1 + ∑_{σ∈Σ′} ind_{[σ]}^{G′} 1) ∈ D(Γ, Π′) for some r ≥ 1.

As above, χ = ind_{G′}^{G} χ′ is in D(Γ, Π) by Lemma 2.4, but to show that χ ∉ P^{+} we must be careful in our choice of τ. Let A be a cyclic normal subgroup of index 2 in Π_{2} and choose τ whose image is A ∩ Π′. Then the construction of Lemma 5.2 applies, since im τ, which is normal in Π, is not Π-conjugate to any im σ with σ ∈ Σ′. Applying Lemma 4.2 as before completes the proof.

7. Finite generation of D

Theorem 7.1. If Π is nilpotent, then D′ and D are finitely generated semigroups.

Proof. Set

X = ⨁_{H≤G} ind_{H}^{G} Z, where H varies over all subgroups of G.

For each prime p dividing |G|, enumerate the distinct non-isomorphic indecomposable summands of Z_{p} ⊗ X: suppose they are X(p, i), 1 ≤ i ≤ n_{p}, and suppose that X(p, i) affords the character χ(p, i) of G. Let n = ∑ n_{p}, summed over primes dividing |G|. We shall use some ideas in the proof of a result of Jones [CR, 33.2]. Following some of the notation of [CR], let C denote the additive semigroup of n-tuples of non-negative integers; partially order C by writing (a_{i}) ≤ (b_{i}) in C if a_{i} ≤ b_{i} for 1 ≤ i ≤ n. If the ZG-lattice M satisfies (a) and (b′), then by Proposition 3.4 and the Krull-Schmidt Theorem for Z_{p}G-lattices, Z_{p} ⊗ M can be written uniquely as a direct sum of modules X(p, i):

Z_{p} ⊗ M ≅ ⨁_{1≤i≤n_{p}} X(p, i)^{m(p,i)}, for unique non-negative integers m(p, i).

Let θ(M) denote the ordered n-tuple in C whose entries are the integers m(p, i). Let D̂′ be the set of n-tuples θ(M) ∈ C where M ranges over all ZG-lattices which satisfy (a) and (b′); similarly, let D̂ be the set of θ(M) ∈ C where M satisfies (a) and (b). Given θ(M) = (a(p, i)) in D̂, we associate to θ(M) the character ∑_{p} a(p, i)χ(p, i); this gives us a mapping of D̂ onto D. Similarly we have a mapping of D̂′ onto D′. Thus the theorem will follow if we can prove finite generation of D̂′ and D̂.

We first prove that D̂′ is finitely generated. From Step 3 in the proof of Jones’ Theorem [CR, p. 689], any subset of C has a finite set of minimal elements in the partial order we have given C. Let S be the finite set of minimal elements of D̂′ − {0}. We claim that this set S generates D̂′. To prove this, let θ(M) be an element of D̂′; we shall show that θ(M) is a sum of elements of S. This is true if θ(M) ∈ S, so assume that there is an element s ∈ S which is strictly smaller than θ(M). Also assume that if L is a ZG-lattice such that θ(L) ∈ D̂′ and which has smaller Z-rank than M, then θ(L) is a sum of elements of S. Suppose that s = θ(M′). Then locally at each prime, M′ is a direct summand of M, so [CR, 31.12], there is a lattice M′′ in the same genus as M′ such that M ≅ M′′ ⊕ M_{0} for some ZG-lattice M_{0}. Then θ(M′′) = θ(M′), and M_{0} satisfies (a) and (b′). Moreover, by our assumption on lattices with ranks smaller than that of M, θ(M_{0}) is a sum of elements of S. Since θ(M) = s + θ(M_{0}), then θ(M) is a sum of elements of S, as claimed.

We next show that D̂ is finitely generated. Let S_{0} be the set of s ∈ S so that s = θ(M) for some M such that res_{N} M is stably free but s = θ(M′) for no M′ such that res_{N} M′ is free. Let r be as in Lemma 2.5. Set

T = (D̂ ∩ {∑_{s∈S} a_{s}s : 0 ≤ a_{s} ≤ r, s ∈ S}) ∪ {(r + 1)s : s ∈ S_{0}}.

Note that if res_{N} M is stably free, then since r + 1 ≥ 2, res_{N} M^{r+1} is free by [CR, 41.20], so T ⊆ D̂. We claim that T generates D̂. Suppose that d = θ(D) ∈ D̂. As above, we assume that if θ(L) ∈ D̂ and L has smaller Z-rank than D, then θ(L) is a sum of elements of T. Since D̂′ is generated by S, we can write

d = ∑_{s∈I} a_{s}s for a subset I ⊆ S with a_{s} ≥ 1, s ∈ I.

Write a_{s} = b_{s} + rc_{s} where 1 ≤ b_{s} ≤ r and c_{s} ≥ 0, and set

e = ∑_{s∈I} b_{s}s, f = r ∑_{s∈I} c_{s}s,

so d = e + f. We have d ∈ D̂, and since rD′ ⊆ D, then f ∈ D̂, so we can find ZG lattices D and F satisfying (a) and (b) with d = θ(D) and f = θ(F). Also, e = θ(E′) for a lattice E′ satisfying (a) and (b′). Now θ(D) = θ(E′ ⊕ F), so D and E′ ⊕ F are in the same genus, and locally for all p, F is a direct summand of D.

We will apply a result of Roiter and Jacobinski [CR, 31.32]; we must check that every irreducible QG-composition factor of QF occurs more often as a composition factor of QD. This is so because d = e + f and e = ∑_{s∈I} b_{s}s, where b_{s} > 0 for s ∈ I. Thus D ≅ E ⊕ F for some ZG-lattice E. Restricting to N, we see that res_{N} E is stably free. If it is actually free (so, in particular, if we have the Eichler condition for QN), then e = θ(E) ∈ T, and we are done.

From [CR, 41.20], if ∑_{s∈I} b_{s} > 1, then res_{N} E is free; so we may assume that ∑_{s∈I} b_{s} = 1. Thus I contains a single element s_{0}, and e = s_{0}. If s_{0} ∉ S_{0}, then writing s_{0} = θ(E′) where E′ satisfies (a) and (b), it follows that D is in the same genus as E′ ⊕ F; we replace D by E′ ⊕ F and we are done as before. So we assume that s_{0} ∈ S_{0}.

Suppose that f = 0. Then d = s_{0} ∉ D̂. Hence f ≠ 0. Then c_{s} ≥ 1 for some s ∈ I; since I = {s_{0}}, then f = c_{s_{0}}rs_{0} with c_{s_{0}} ≥ 1. Then e = (r + 1)s_{0} + (c_{s_{0}} − 1)(rs_{0}) with (r + 1)s_{0} ∈ T and (c_{s_{0}} − 1)(rs_{0}) = θ(L) ∈ D̂, where L has smaller rank than D. Thus d is indeed a sum of elements of T, and the proof is complete.

8. Complements

Assume we have Γ, Π and G = Γ × Π as above. As in the proof of Theorem 3.3, G is the Galois group of Q(ζ) over Q, where ζ be a primitive |G|-th root of unity, and G_{p} denotes the Galois group of Q_{p}(ζ) over Q_{p}. We identify G_{p} as a subgroup of G, namely the decomposition group at any prime of Q(ζ) above p. Let Σ_{p} be a complete set of elements of hom(Γ_{p}, Π_{p}) up to conjugation by Π_{p}. As in [WCr], define a label for χ to be a collection b = {b_{p}} of functions

b_{p} : Σ_{p} × irr(G_{p′}) → Z_{≥0},

one for each prime p, so that on writing

χ = ∑_{η∈irr(G_{p′})} η ⊗ λ_{η} relative to R(G) = R(G_{p′}) ⊗ R(G_{p})

we have

(i)_{p} λ_{η} = ∑_{σ_{p}∈Σ_{p}} b_{p}(σ_{p}, η) ind_{[σ_{p}]}^{G_{p}} 1,

(ii)_{p} b_{p}(σ_{p}, η^{ω}) = b_{p}(σ_{p}, η) for all ω ∈ G_{p}.

Theorem 8.1. Suppose that Γ and Π are nilpotent. Then labels for χ are in bijection with genera of ZG-lattices with character χ which satisfy (a) and (b′).

Proof. The proof comes from a closer look at the proof of Theorem 3.3. Suppose that M is a ZG-lattice which satisfies (a) and (b′). From equation (3.1), we have

Z_{p} ⊗ M ≅ ⨁_{σ_{p}∈hom(Γ_{p},Π_{p})} X_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} Z_{p}.

Instead of summing over σ_{p} ∈ hom(Γ_{p}, Π_{p}), we may sum over Σ_{p}, because replacing σ_{p} by a Π_{p}-conjugate gives a G_{p}-conjugate of [σ_{p}], hence a module isomorphic to ind_{[σ_{p}]}^{G_{p}} Z_{p}. Then the different groups [σ_{p}], σ_{p} ∈ Σ_{p}, are the vertices of the summands of Z_{p} ⊗ M, so the modules X_{σ_{p}} are unique up to isomorphism, and their characters ξ_{σ_{p}} give the well-defined equation

χ = ∑_{σ_{p}∈Σ_{p}} ξ_{σ_{p}} ⊗ ind_{[σ_{p}]}^{G_{p}} 1, ξ_{σ_{p}} ∈ R^{+}(G_{p′}).

Then (ii)_{p} follows from equation (3.3) and (i)_{p} comes from equation (3.4). Since these depend only on Z_{p} ⊗ M, the same label would be attached to any lattice in the same genus as M.

Conversely, suppose that b is a label. From (ii)_{p}, b(σ_{p}, η) just depends on the G_{p}-orbit O containing η. Then as in equation (3.6), we let ξ_{σ_{p}} = ∑_{O} b(σ_{p}, O)τ_{O}, and ξ_{σ_{p}} ∈ R_{Q_{p}}^{+}(G_{p′}). The lattice

M(b_{p}) = ⨁_{σ_{p}} L_{σ_{p}} ⊗_{Z_{p}} ind_{[σ_{p}]}^{G_{p}} Z_{p}

in equation (3.7) satisfies the local versions (a_{p}) and (b_{p}) of (a) and (b), and has character χ by (i)_{p}. Then the ZG-lattice M = M(b) at the end of the proof has Z_{p} ⊗ M ≅ M(b_{p}) for all p, hence it satisfies (a) and (b′); the construction of M depends on the identifications ϕ_{p}, but the genus of M is well-defined.

Corollary 8.2. If Γ is cyclic, there is only one genus of ZG-modules having a given character in D′.

Proof. By Lemma 4.2, for each p the set {ind_{[σ_{p}]}^{G_{p}} 1 : σ_{p} ∈ Σ_{p}} is linearly independent; hence there is only one solution to (i)_{p}, and only one label for a character χ.

Remarks. Given χ ∈ D′, we want to decide whether χ ∈ D. We begin by determining all labels b for χ: this is a purely character-theoretic problem. For each b one then constructs a lattice M = M(b) in the genus of the label. Deciding whether the genus of M contains an M′ with res_{N} M′ stably free can then be approached by genus class group methods, generalizing Theorem 3 of [WCr]. Carrying this out is a long computation which will answer the existence question “Is χ in D?” (at least when we have the Eichler condition for ZΠ). However, the construction of α : Γ → SGL_{n}(ZΠ) with double-action character χ takes still more calculation.

Nevertheless this is how the first example of §5 was found. It is typical of such computations that once an M satisfying (a) and (b) is found, it is simpler to describe it directly, as we have done in §5.

Finally there is the issue of finding χ ∈ D′, no multiple of which is in P^{+}, in the first place. This amounts to finding generators of D′, and this again is a problem of character theory, by Theorem 3.3. In examples we have considered, D′ has many generators, even when Γ is cyclic. More exploration of this, perhaps by computer, is still needed.

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Department of Mathematical Sciences, University of Alberta, Edmonton, Alberta, Canada T6G 2G1

   E-mail address: gcliff@math.ualberta.ca
   E-mail address: aweiss@math.ualberta.ca