Finite arithmetic subgroups of GL_{n}

Marcin Mazur

Department of Mathematics, the University of Chicago, 5734 S. University Avenue, Chicago, IL 60637 e-mail: mazur@math.uchicago.edu

Abstract We discuss the following conjecture of Kitaoka: if a finite subgroup G of GL_{n}(O_{K}) is invariant under the action of Gal(K/ℚ) then it is contained in GL_{n}(K^{ab}). Here O_{K} is the ring of integers in a finite, Galois extension K of ℚ and K^{ab} is the maximal, abelian subextension of K. Our main result reduces this conjecture to a special case of elementary abelian p−groups G. Also, we construct some new examples which negatively answer a question of Kitaoka.

1 Introduction.

About 20 years ago Yoshiyuki Kitaoka, motivated by some questions concerning quadratic forms, started the investigation of finite, stable under the action of Gal(K/ℚ) subgroups of GL_{n}(O_{K}), where O_{K} is the ring of integers in a finite, Galois extension K of ℚ. The subject seems to be important and interesting though we feel that it did not get sufficient attention. At present we have a very clear conjectural picture due to Kitaoka but the actual results are very fragmentary and the supporting evidence is rather poor. In the present article we recall the conjectures and reduce them to a very special case. Also, we describe some new examples which negatively answer one of Kitaoka’s questions.

For a finite, Galois extension K of the rationals let Γ = Gal(K/ℚ) and O_{K} be the ring of integers in K. In the case of totally real K the main conjecture is particularly easy to state:

Conjecture 1 Let G be a finite subgroup of GL_{n}(O_{K}) stable under the action of Γ. If K is totally real then G ⊆ GL_{n}(ℤ).

Conjecture 1 is equivalent to the following conjecture, important for the theory of arithmetic groups:

Conjecture 2 Let H be an algebraic group of compact type defined over ℚ. For any totally real number field K the equality H_{ℤ} = H_{O_{K}} holds.

In fact it would be enough to prove Conjecture 2 for orthogonal groups of positively defined quadratic forms over ℚ. As a corollary we would get the following, interesting result: Any two positively defined quadratic forms over ℤ, which are equivalent over a ring of integers of some totally real number field are already equivalent over ℤ.

For more details about the interrelation between the above conjectures and their consequences we refer to the beautiful book [6] (paragraph 4.8).

After stating Conjecture 1 it is natural to ask what happens for arbitrary Galois extensions of ℚ. To spell out a conjectural answer, due to Kitaoka, consider a free ℤ−module M of rank n with basis e_{1}, ..., e_{n}. The group GL_{n}(O_{K}) acts in a natural way on O_{K} ⊗ M = ⨁_{i=1}^{n} O_{K}e_{i}: (a_{i,j})e_{i} = ∑ a_{i,j}e_{j}. We say that a finite subgroup G of GL_{n}(O_{K}) if of A-type if there exists a decomposition M = ⨁_{i=1}^{k} M_{i} such that for every g ∈ G there are a permutation π(g) of {1, ..., k} and roots of unity ϵ_{i}(g) ∈ K such that ϵ_{i}(g)gM_{i} = M_{π(g)} for i = 1, ..., k. It is easy to see that π is a group homomorphism G ⟶ Σ_{k}.

Conjecture 3 Any finite subgroup of GL_{n}(O_{K}) stable under the action of Γ is of A-type.

Of course for totally real K Conjecture 3 reduces to Conjecture 1. The best result up to date toward Conjecture 3 is the following theorem due to Kitaoka and Suzuki ([4]):

Theorem 1 If the Galois group Γ is nilpotent then Conjecture 3 holds.

In the light of this result Conjecture 3 is easily seen to be equivalent to Conjecture 4 If a finite group G ⊆ GL_{n}(O_{K}) is Γ−stable then in fact G ⊆ GL_{n}(O_{K^{ab}}), where K^{ab} is the maximal abelian subextension of K.

Our main result reduces Conjecture 4 to the case of elementary abelian p−groups G.

2 Diagonal p−groups

Suppose that P is an abelian p−group sitting in GL_{n}(\overline{ℚ}) as a subgroup of diagonal matrices. In other words, there are abelian characters χ_{1}, ..., χ_{n} of G such that g = diag(χ_{1}(g), ..., χ_{n}(g)). After conjugating by a permutation matrix (which has entries in ℤ) we can and will assume that there are integers k_{0} = 0 < k_{1} < ... < k_{s} < k_{s+1} = n such that χ_{i} = χ_{j} iff k_{t} < i, j ≤ k_{t+1} for some 0 ≤ t ≤ s. We say in this case that P is in a strongly diagonal form. Set N(P) for the normalizer of P in GL_{n}(\overline{ℚ}) and C(P) for its centralizer. The following lemma is almost obvious and we omit a proof.

Lemma 1 The centralizer C(P) of P in GL_{n}(\overline{ℚ}) equals GL_{k_{1}−k_{0}}(\overline{ℚ})×⋯×GL_{k_{s+1}−k_{s}}(\overline{ℚ}). The group W (P) = N(P)/C(P) is finite.

Note that P is stable under the action of the absolute Galois group Γ_{ℚ} of ℚ. Thus both N(P) and C(P) are Γ_{ℚ}−stable. We denote by Π_{n} the group of permutation matrices in GL_{n}. If N ∈ N(P) and ϕ is the automorphism of P induced by conjugation with N then clearly the map χ ↦ χ ∘ ϕ permutes the characters χ_{1}, ..., χ_{n}. It is fairly obvious that we can choose a permutation π ∈ Σ_{n} such that χ_{i} ∘ ϕ = χ_{π(i)} and whenever i < j and χ_{π(i)} = χ_{π(j)} then π(i) < π(j). Moreover such permutation is unique and if we denote by T_{N} the corresponding permutation matrix then NT_{N}^{−1} ∈ C(P) and the association N ↦ T_{N} is a group homomorphism σ : N(P) ⟶ Π_{n} ∩ N(P). We denote by Π_{P} the image of σ. Thus we get the following

Lemma 2 The group N(P) is a semidirect product of C(P) and Π_{P}. In particular, the induced action of Γ_{ℚ} on W (P) is trivial.

Note that N(P) acts by conjugation on M_{k_{1}−k_{0}}(\overline{ℚ}) × ⋯ × M_{k_{s+1}−k_{s}}(\overline{ℚ}) which has a basis consisting of matrices with one entry equal to 1 and all others being 0. Withe respect to this basis the action of N(P) defines a representation ρ of N(P) in GL_{m_{1}^{2}+...+m_{s+1}^{2}}, where m_{i} = k_{i} − k_{i−1}. It is clear that this representation respects the action of Γ_{ℚ}. Also, if both a and a^{−1} have entries in O_{L} for some number field L, then so do ρ(a) and ρ(a^{−1}). The kernel of ρ equals to the product of centers of M_{k_{i}−k_{i−1}}(\overline{ℚ}). In particular, we get the following

Lemma 3 Suppose that G is a finite, Γ_{ℚ}−invariant subgroup of N(P) ∩ GL_{n}(O_{K}) for some Galois extension K of ℚ. Then ρ(G) is a finite, Γ_{ℚ}−invariant subgroup of GL_{m_{1}^{2}+...+m_{s+1}^{2}}(O_{K}) and the map ρ : G ⟶ ρ(G) commutes with the action of Γ_{ℚ}.

The last lemma will allow us to perform a “dirty trick”, i.e. apply Conjecture 3 to a quotient of G at the expense of raising n.

Suppose now that K is a number field and that H ⊆ GL_{n}(\overline{ℚ}) is an elementary abelian p−group stable under the action of the absolute Galois group Γ_{K} of K. There exists a matrix E ∈ GL_{n}(\overline{ℚ}) such that G = E^{−1}HE is in a strongly diagonal form. Since H is Γ_{K}−stable, we have E^{−1}E^{τ} ∈ N(G) for all τ ∈ Γ_{K}. In other words, the function τ^{−1} ↦ E^{−1}E^{τ} is a 1-cocycle of Γ_{K} with values in N(G). Conversely, suppose that f : Γ_{K} ⟶ N(G) is a one cocycle. Since H^{1}(Γ_{K}, GL_{n}) = 1, there exists a matrix E ∈ GL_{n}(\overline{ℚ}) such that f(τ^{−1}) = E^{−1}E^{τ} and the group EGE^{−1} is clearly Γ_{K}−stable. We say that L is the field of definition of EGE^{−1} if it is the smallest Galois extension of K such that EGE^{−1} ⊆ GL_{n}(L). Note that if F is another matrix representing f then EF^{−1} ∈ GL_{n}(K), so that EGE^{−1} and F GF^{−1} are conjugate over K. Clearly homologically equivalent cocycles lead to the same GL_{n}(K)−conjugacy classes of finite subgroups. Moreover, the field of definition for EGE^{−1} depends only on the homomorphism \overline{f} : Γ_{K} ⟶ W (G) obtained by composing f with the natural projection of N(G) onto W (G). In fact, for τ ∈ Γ_{K} and g ∈ G we have (EgE^{−1})^{τ} = E(E^{−1}E^{τ})g^{τ}(E^{−1}E^{τ})^{−1}E^{−1} and the action of E^{−1}E^{τ} = f(τ) ∈ N(G) clearly depends only on its image in N(G)/C(G). Thus for the determination of possible fields of definition we can restrict our attention to homomorphisms Γ_{K} ⟶ Π_{G}. But when we would like to study the fields of definition of subgroups in GL_{n}(O\overline{ℚ}) (O\overline{ℚ} is the ring of algebraic integers) the problem is much more subtle. It can a priori happen that some cocycle f gives rise to E such that EGE^{−1} is in GL_{n}(O\overline{ℚ}) but the corresponding \overline{f} does not have this property. We can not therefore restrict only to Hom(Γ_{K}, Π_{G}). It would be very interesting to get a nice description of those homomorphisms \overline{f} which come from a cocycle giving rise to a subgroup consisting of matrices with entries in algebraic integers. It is easy to see that Conjecture 4 for elementary abelian p−groups G is equivalent to saying that any such homomorphism (when K = ℚ) is trivial on the commutator subgroup of Γ_{ℚ}. We will see in the next section that Conjecture 4 reduces to this special case.

We will end this section by discussing some examples. From the very early days of the conjectures there was a desire to extend them to the relative case. In other words, we would like to know which fields can be the fields of definition for finite, Γ_{K}−stable subgroups of GL_{n}(\overline{ℚ}) with entries in algebraic integers. On page 260 of [3] (see also [6], paragraph 4.8) a totally real number field K is constructed for which there is a finite group with the field of definition being a totally real, unramified non-trivial extension with cyclic Galois group. There was a hope for a while that any such group has its field of definition contained in the cyclotomic extension of K but counterexamples have been produced in [4]. In the same paper the authors rise the question of whether the field of definition is always an abelian extension of K. We show that the answer is negative and the Galois group of the field of definition can be a nonabelian simple group. The idea of our construction is very simple. Consider a finite extension L of K of degree n and let τ_{1}, ..., τ_{n} be the embeddings L ↪ \overline{ℚ} over K. Let u_{1}, ..., u_{n} be a basis of L/K. Define the matrix U = (u_{i,j}) by u_{i,j} = u_{i}^{τ_{j}}. Plainly U is nonsingular and for any τ ∈ Γ_{K} we have U^{τ}U^{−1} = A_{τ} ∈ Π_{n}. This defines a homomorphism ψ : Γ_{K} ⟶ Π_{n} whose kernel is Γ_{M}, where M is the Galois closure of L/K. Fix a prime p and let i : Γ_{K} ⟶ ℤ/(p − 1)ℤ be the cyclotomic character which is given by the action of Γ_{K} on p−th roots of 1. Denote by P the maximal, elementary abelian p−subgroup of the diagonal matrices in GL_{n}(\overline{K}). The group P_{U} = U^{−1}P U is Γ_{K}−stable and for τ ∈ Γ_{K} and w ∈ P we have (U^{−1}wU)^{τ} = U^{−1}ψ(τ)^{−1}w^{i(τ)}ψ(τ)U. It is easy to see that τ acts trivially on P_{U} iff both ψ(τ) and i(τ) are trivial. In other words, the field of definition for P_{U} equals N = M(ξ_{p}), where ξ_{p} is a primitive p−th root of 1.

The only problem with the above construction is that the group P_{U} does not consists of matrices with integral entries in general. Suppose however that O_{L} is a free O_{K}−module and that L/K is unramified. If we take for the u_{i}’s a basis for O_{L} over O_{K} then U ∈ GL_{n}(O_{M}) and consequently P_{U} ⊆ GL_{n}(O_{N}). For an explicit example we can use the results of [7]. Let K = ℚ(\sqrt{36497}). This field has class number 1 and an everywhere unramified, Galois, totally real extension L with Galois group A_{5} (the alternating group of degree 5). In particular, O_{L} is a free O_{K}−module. Our construction in this case gives a finite, Γ_{K}−stable subgroup of GL_{60}(O_{L(ξ_{p})}) with field of definition L(ξ_{p}), where p is arbitrary prime number. Extending our base field to K(ξ_{p}) we get an example with Galois group of the field of definition equal to A_{5}.

3 Reduction

We start by recalling Minkowski’s Lemma. Let K be an algebraic number field. For an ideal I of the ring of integers O_{K} we denote by R_{I} the natural map GL_{n}(O_{K}) ⟶ GL_{n}(O_{K}/I) and by GL_{n}(O_{K}, I) its kernel. Let β be a prime ideal of O_{K} lying over the rational prime p. By f we denote the ramification index of β.

Proposition 1 Any torsion element of GL_{n}(O_{K}, β) has p−power order. If Gl_{n}(O_{K}, β^{k}) contains an element of order p^{s} then k ≤ fp^{1−s}/(p − 1).

This result is classical but we include a proof for the readers convenience. Proof: Let x ∈ GL_{n}(O_{K}, β) be an element of prime order q. Let t be maximal such that x ∈ GL_{n}(O_{K}, β^{t}). Writing x = 1 + y we get 1 = (1 + y)^{q} and equivalently

−qy = ∑_{i=2}^{q} \binom{q}{i}y^{i}

If q ≠ p then the left hand side has entries in β^{t} but not all of them are in β^{t+1}. On the other hand, the right hand side has entries in β^{2t} which leads to a contradiction. Thus q = p and the left hand side of our equality has entries in β^{f+t} but not all of them are in β^{f+t+1}. The right hand side has entries in β^{min(f+2t,pt)}. Thus f + t ≥ min(f + 2t, pt) and therefore f + t ≥ pt, i.e. t ≤ f/(p − 1). This proves the lemma for s = 1. Also, since k ≤ t, we get that f + k ≥ pk. This implies that GL_{n}(O_{K}, β^{pk}) contains an element of order p^{s−1} and the result follows by induction. □

Corollary 1 If p ≠ 2 then GL_{n}(ℤ, p) is torsionfree. For p = 2 the group GL_{n}(ℤ, 4) is torsionfree and torsion elements of GL_{n}(ℤ, 2) are of order at most 2.

Suppose now that K is Galois over ℚ with Galois group Γ. Let G ⊆ GL_{n}(O_{K}) be a finite, Γ−stable group. For a prime β of O_{K} let G(β) be the kernel of R_{β} restricted to G. We will need the following lemma due to Kitaoka and Suzuki:

Lemma 4 If K is an abelian extension of ℚ then G(β) can be conjugated to diagonal matrices by an element of GL_{n}(ℤ).

For a proof see [4], Lemma 1.

Suppose that Conjecture 4 is false and let G be a counterexample of minimal order. We also assume that n is minimal possible for G and K is minimal. In particular, no element of Γ acts trivially on G. By Lemma 3 of [2] we may assume that there is a prime p such that K is unramified outside p. We denote by I_{β} and D_{β} the inertia and decomposition groups of a prime ideal β of O_{K} respectively. We have the following easy lemma:

Lemma 5 The Galois group Γ is generated by the inertia subgroups I_{β} for primes β lying over p.

Proof: Any subfield of K fixed by all the inertia of primes above p is unramified over ℚ hence coincides with ℚ. □

Clearly G(β) is not trivial for any prime β over p. Otherwise I_{β} would act trivially on G. Since the inertia subgroups of different primes over p are conjugate, this contradicts Lemma 5.

Lemma 6 G is generated by the subgroups G(β), where β runs over all primes above p.

Proof: By Proposition 1 we get that G(γ) is a normal p−subgroup of G for any prime ideal γ of K lying over p. Thus the composition H of all G(γ), where γ runs over the primes above p, is a normal, Γ−stable p−subgroup of G. Suppose that H ≠ G. Then H ⊆ GL_{n}(K^{ab}) by minimality of G. Since K^{ab} is an abelian, unramified outside of p extension of ℚ, it is contained in ℚ(μ_{p^{∞}}) by class field theory. In particular, there is a unique prime π of O_{K^{ab}} over p. Clearly H(π) = H(β) = G(β) for all primes β over p, since π = β ∩ O_{K^{ab}}. In other words, G(β) does not depend on the prime β above p. Thus we have H = G(β) for all β over p and therefore all the inertia groups of primes over p act trivially on G/H. Consequently Γ acts trivially on G/H by Lemma 5. In particular, for any A ∈ G and any τ ∈ Γ there is an h ∈ H such that A^{τ} = hA. Note that Lemma 4 allows us to assume that H consists of diagonal matrices (after conjugating by a matrix with entries in ℤ, if necessary). By the following Lemma 7 we get that A ∈ GL_{n}(K^{ab}) and therefore that G ⊆ GL_{n}(K^{ab}), which contradicts our assumption. □

Lemma 7 Suppose that K is a Galois extension of ℚ unramified outside a prime p. Let A ∈ GL_{n}(O_{K}) has the property that for any τ ∈ Gal(K/ℚ) there is a diagonal matrix D_{τ} = diag(ξ_{1}(τ), ..., ξ_{n}(τ)) such that ξ_{i}(τ)’s are roots of unity of p−power order and A^{τ} = D_{τ}A. Then A ∈ GL_{n}(K^{ab}).

Proof: Let A = (a_{i,j}). Let β be a prime of O_{K} over p. For every i there exists k such that a_{i} = a_{i,k} is not in β. We have a_{i,j}^{τ} = ξ_{i}a_{i,j} for all j. In particular, a_{i}a_{i,j}^{−1} is stable under the action of Gal(K/ℚ) for all j. Thus there are rational numbers q_{i,j} such that a_{i,j} = a_{i}q_{i,j}.

Since a_{i}^{τ} = ξ_{i}(τ)a_{i}, there is an integer n such that (a_{i}^{p^{n}})^{τ} = a_{i}^{p^{n}} for all τ ∈ Gal(K/ℚ). Thus a_{i}^{p^{n}} = l_{i} ∈ ℚ and consequently they are integers prime to p. We get that \sqrt[p^{n}]{l_{i}} ∈ K. Since K is unramified outside p and l_{i} is prime to p, we get that l_{i} = ±t_{i}^{p^{n}} for some integers t_{i}. This implies that a_{i} = t_{i}ζ_{i} for some roots of unity ζ_{i} in K. It follows that A = diag(ζ_{1}, ..., ζ_{n})Q, where Q = (t_{i}q_{i,j}) ∈ GL_{n}(ℤ). □

Proposition 2 G is an elementary abelian p−group.

Proof: It follows from Lemma 6 that G is a p−group generated by the subgroups G(β) for primes β over p. Let W = G ∩ GL_{n}(O_{K^{ab}}). Every proper, Γ−stable subgroup of G is contained in W by minimality of G. In particular, we have the inclusion [G, G] ⊆ W and therefore W is a normal subgroup of G. Since there is a unique prime of O_{K^{ab}} over p, the subgroup H = G(β) ∩ W is independent on the prime β above p. Thus it is a Γ−stable, normal subgroup of G and by Lemma 4 we can assume that it consists of diagonal matrices. For a prime β over p let G′(β) be the image of G(β) in G/H. Since [G(β), G(β_{1})] ⊆ G(β) ∩ G(β_{1}) ∩ [G, G] ⊆ H, we get that [G′(β), G′(β_{1})] = 1. In other words, these groups pairwise commute and are commutative. Let F be the Frattini subgroup of G. Since F is characteristic, it is Γ−stable and consequently F ⊆ W. Note that F ∩ G(β) ⊆ W ∩ G(β) = H. Thus the groups G′(β) have trivial intersection with the image of F in G/H. It follows that they are elementary abelian p−groups. Consequently, these subgroups generate elementary abelian p−group. On the other hand they generate whole G/H, so G/H is an elementary abelian p−group generated by G′(β), where β runs over primes above p. It remains to show that H = 1.

Suppose that H is not trivial. Thus it is a p−subgroup of the diagonal matrices of GL_{n} and there is no harm to assume that it is in a strongly diagonal form. We can associate to H its normalizer N(H), centralizer C(H) and representation ρ, as discussed in Section 2. Clearly G ⊆ N(H). Note that the kernel of ρ is contained in the group of diagonal matrices and the subgroup of diagonal matrices in G is exactly H (such matrices are clearly in the congruence subgroups and have entries in K^{ab}). Thus ρ(G) and G/H are isomorphic. By Lemma 3 the image ρ(G) is a finite, Γ−stable subgroup of GL_{m}(O_{K}) for some m ≥ n, Moreover, the action of Γ on ρ(G) coincides with its action on G/H induced from the action on G, i.e G/H and ρ(G) are isomorphic as Γ−modules. Directly from the definition of ρ we get that ρ(G(β)) ⊆ ρ(G)(β) for any prime β of O_{K}. In particular, ρ(G) is generated by the subgroups ρ(G)(β) for primes β over p. By minimality of our counterexample G we get that the commutator of Γ acts trivially on ρ(G). In other words, ρ(G) is contained in GL_{m}(K^{ab}). Since there is a unique prime π in K^{ab} over p, we have the equality ρ(G) = ρ(G)(π). Thus ρ(G) can be conjugated to diagonal matrices over ℤ by Lemma 4.

Now we will analyze the action of Γ on H and G/H

First note that p is odd. In fact, we already proved that [Γ, Γ] acts trivially on both H and G/H. It follows that the commutator [Γ, Γ] is an abelian p−group. In fact, for any g ∈ G the map τ ↦ g^{τ}g^{−1} is a homomorphism from [Γ, Γ] to H, and all these homomorphisms separate elements of [Γ, Γ]. If p = 2 then the abelianization Γ/[Γ, Γ] is a 2−group (K^{ab} is an abelian, unramified outside 2 extension of ℚ) and therefore Γ itself is a 2−group. Since 2−groups are nilpotent, we get a contradiction by Theorem 1. Thus in the case of p = 2 the group H has to be trivial.

Assume now that p is odd. Let μ be the subgroup of K^{×} consisting of roots of unity of p−power order. Γ acts on μ via the cyclotomic character i. To be more precise, i(τ) is the unique integer (mod |μ|) such that ξ^{τ} = ξ^{i(τ)} for all ξ ∈ μ . Since H consists of diagonal matrices, Γ acts on H via i, i.e. h^{τ} = h^{i(τ)} for all h ∈ H. The same is true for ρ(G), since it is conjugated to diagonal matrices over ℤ. Thus, b^{τ} = b^{i(τ)} for all b ∈ G/H = ρ(G) and τ ∈ Γ. As a consequence we get that for any g ∈ G the subgroup of G generated by g and H is Γ−stable. Let g ∈ G be any element which is not fixed by [Γ, Γ]. The subgroup < H, g > of G generated by g and H is Γ−stable and the commutator of G acts nontrivially on it. By minimality of G we get that G =< H, g >. In particular, G/H is a cyclic group of order p. We need to consider two cases:

\underline{case 1:} H is central in GL_{n}. Thus H is cyclic, G is abelian and either G is cyclic or G = H × ℤ/p. The former case implies that [Γ, Γ] acts trivially on G, which is impossible. Thus G = H × F, where F is cyclic of order p. We can assume that g is a generator of F. For any τ ∈ Γ we have τ(g) = u_{τ}g^{i(τ)} for some u_{τ} ∈ H. Clearly u_{τ} is of order p. By minimality of G it follows that H is generated by the u_{τ}’s. In particular, H has exponent p. Since H is cyclic, it is of order p. Let S be the subgroup of Γ which acts trivially on both H and G/H. Note that τ^{p−1} ∈ S for any τ ∈ Γ, because both H and G/H are cyclic of order p. Since the commutator of Γ is contained in S, the group Γ/S is cyclic of order dividing p − 1. The order has to be exactly p − 1 by looking at the action on p−th roots of unity. On the other hand, the map s^{−1} ↦ g^{s}g^{−1} is an injective homomorphism of S into H. Thus S is of order p. Putting these together we conclude that S = [Γ, Γ] is cyclic of order p and Γ/S =Galℚ(ζ)/ℚ is cyclic of order p − 1, where ζ is a primitive p−th root of 1.

Now let τ ∈ Γ maps to a generator of Γ/S. Thus τ acts on p − th roots of unity by raising to a power t, where t is a primitive root mod p. We have g^{τ} = ug^{t} and g^{τ^{i}} = u^{it^{i−1}}g^{t^{i}} for some u ∈ H and any i ∈ ℤ. If u ≠ 1 then τ has order p(p − 1) and Γ is cyclic- a contradiction. Thus u = 1 for any τ mapping to a generator of Γ/S. But such elements generate Γ, so Γ preserves F. But then the commutator of Γ acts trivially on F and H, hence on G, again a contradiction.

\underline{case 2:} H is not central in GL_{n}. Then the centralizer C(H) equals GL_{a_{1}} × ... × GL_{a_{s}}, where a_{i} < n for all i. Since Z = G ∩ C(H) is Γ−stable, we get that [Γ, Γ] acts trivially on Z by minimality of n. To be more precise, the projection Z_{i} of Z into GL_{a_{i}}(O_{K}) is Γ−stable and either Z_{i} has order smaller than the order of Z or Z_{i} is isomorphic to Z but sits in matrices of smaller dimension. In both cases [Γ, Γ] acts trivially on Z_{i} by minimality of G and n. Since this is true for all i, it follows that [Γ, Γ] acts trivially on Z. Clearly G ⊆ N(H). Thus we can write g = cw with c ∈ C(H) and w a permutation matrix. Since Γ acts on G/H via the cyclotomic character, there is τ ∈ Γ such that g^{τ} = ug^{2} for some u ∈ H. On the other hand, we have g^{τ} = c^{τ}w and consequently ucwcw = c^{τ}w, so w = (uc)^{−1}c^{τ}c^{−1} ∈ C(H). Thus g ∈ Z and therefore G = Z has trivial [Γ, Γ]−action contrary to our assumption.

Both cases have led to a contradiction and therefore our assumption that H is not trivial has to be wrong. □

We summarize our investigation in the following theorem:

Theorem 2 Suppose that Conjecture 4 is true in the following very special situation: — K is a Galois extension of ℚ unramified at all finite primes q ≠ p;

— G is an elementary abelian p−group contained in GL_{n}(O_{K}) and stable under the action of GalK/ℚ;

— G(β) ∩ GL_{n}(K^{ab}) = 1 for all primes β over p and these subgroups generate G. Then it is true in general.

Remark. Recently Kitaoka proved ([5]) that Conjecture 4 is true for n = 2. The main step in the proof is the reduction to abelian groups G. Note however that our reduction can not be used for this purpose. The point is that we used our ”dirty trick” of raising the dimension. It would be nice to have a proof of Theorem 2 which does not use any such methods. Unfortunately we were unable to produce any satisfactory argument in general. However for totally real fields one can avoid raising the dimension for groups G of odd order. The key point is that we have Corollary 1 which gives triviality of the congruence subgroups over ℤ. It is worth pointing out that in the totally real case (or more generally, when complex conjugation is central) it is known that Conjecture 4 holds for n < 43 (see [1]).

References

[1] Y. Kitaoka, Finite arithmetic subgroups of GL_{n}, II, Nagoya Math. J. 77 (1980), 137–143.

[2] Y. Kitaoka, Finite arithmetic subgroups of GL_{n}, III, Proc. Indian Acad. Sci. (Math. Sci.) 104 (1994), 201–206.

[3] Y. Kitaoka, Arithmetic of quadratic forms, Cambridge Tracts in Mathematics 106, Cambridge University Press, 1993.

[4] Y. Kitaoka and H. Suzuki, Finite arithmetic subgroups of GL_{n}, IV, Nagoya Math. J. 142 (1996), 183–188.

[5] Y. Kitaoka, Finite arithmetic subgroups of GL_{n}, V, Nagoya Math. J. 146 (1997), 131–148.

[6] V. P. Platonov and A. S. Rapinchuk, Algebraic Groups and Number Theory, Pure and Applied Math. 139. Academic Press, Inc., Boston Ma, 1994.

[7] K. Yamamura, On unramified Galois extensions of real quadratic number fields, Osaka J. Math. 23 (1986), 471–478.