New York Journal of Mathematics

New York J. Math. 12 (2006) 169–182.

Conjugacy classes of p-torsion in symplectic groups over S-integers

Cornelia Minette Busch

Abstract. For any odd prime p we consider representations of a group of order p in the symplectic group Sp(p−1, ℤ[1/n]) of (p−1)×(p−1)-matrices over the ring ℤ[1/n], 0 ≠ n ∈ ℕ. We construct a relation between the conjugacy classes of subgroups P of order p in the symplectic group and the ideal class group in the ring ℤ[1/n] and we use this relation for the study of these conjugacy classes. In particular we determine the centralizer C(P) and N(P)/C(P) where N(P) denotes the normalizer.

Contents

1. Introduction 169 2. A recall of algebraic number theory 171 3. Matrices of order p 172

3.1. A relation between matrices and ideal classes 172 3.2. The number of conjugacy classes 176

4. Subgroups of order p 178 4.1. The quotient of the normalizer by the centralizer 178 4.2. The centralizer of subgroups of order p 180 4.3. The action of the normalizer on the centralizer 182

References 182

1. Introduction

We define the group of symplectic matrices Sp(2n, R) over a ring R to be the subgroup of matrices M ∈ GL(2n, R) that satisfy

M^{T}JM = J = (\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix})

Received September 28, 2005. Mathematics Subject Classification. 20G05, 20G10. Key words and phrases. Representation theory, cohomology theory. Research supported by a grant “Estancias de jóvenes doctores y tecnólogos extranjeros en

España” (SB 2001-0138) from the Ministerio de Educación, Cultura y Deporte.

169^{ISSN 1076-9803/06}

where 1 ∈ M(n, R) denotes the identity. Our motivation for studying subgroups of odd prime order p in the symplectic group Sp(p − 1, ℤ[1/n]), 0 ≠ n ∈ ℕ, is given by the fact that the p-primary part of the Farrell cohomology of Sp(p − 1, ℤ[1/n]) is determined by the Farrell cohomology of the normalizer of subgroups of order p in Sp(p − 1, ℤ[1/n]) (see Brown [2]). First we consider the conjugacy classes of elements of order p in Sp(p − 1, ℤ[1/n]) and get the following result.

Theorem 3.14. There are

|C_{0}|2^{\frac{p−1}{2} +τ}

conjugacy classes of matrices of order p in Sp(p − 1, ℤ[1/n]), 0 ≠ n ∈ ℤ. Here C_{0} is the ideal class group of ℤ[1/n][ξ], ξ a primitive pth root of unity, and τ is the number of inert primes in ℤ[ξ + ξ^{−1}] that lie over primes in ℤ that divide n.

In order to prove this theorem we establish a relation between some ideal classes in ℤ[1/n][ξ] and the conjugacy classes of matrices of order p. We define equivalence classes [𝔞, a] of pairs (𝔞, a) where 𝔞 ⊆ ℤ[1/n][ξ] is an ideal with 𝔞\overline{𝔞} = (a) and the equivalence relation is

\begin{aligned}(𝔞, a) ∼ (𝔟, b) & ⇔ & ∃λ, μ ∈ ℤ[1/n][ξ] \ \{0\} \\ & & λ𝔞 = μ𝔟, λ\overline{λ}a = μ\overline{μ}b . \end{aligned}

We show that a bijection exists between the conjugacy classes of elements of order p in Sp(p − 1, ℤ[1/n]) and the set of equivalence classes [𝔞, a]. Sjerve and Yang (see [11]) construct an analogous bijection for Sp(p − 1, ℤ). We use the bijection described above in order to study the subgroups of order p in Sp(p − 1, ℤ[1/n]). We consider the case where n ∈ ℤ is such that ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}] are principal ideal domains because in this case the ideal class group of those rings is trivial. We get the following results.

Theorem 4.1. Let n ∈ ℤ be such that ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}] are principal ideal domains and moreover p | n. Let N(P) denote the normalizer and C(P) the centralizer of a subgroup P of order p in Sp(p − 1, ℤ[1/n]). Then

N(P)/C(P) ≅ ℤ/jℤ

where j | p − 1, j odd. For each j with j | p − 1, j odd, there exists a subgroup of order p in Sp(p − 1, ℤ[1/n]) with N(P)/C(P) ≅ ℤ/jℤ.

Theorem 4.2. Let n ∈ ℤ be such that ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}] are principal ideal domains. Then for a subgroup P of order p in Sp(p−1, ℤ[1/n]), the centralizer C(P) is

C(P) ≅ ℤ/2pℤ × ℤ^{σ^{+}}.

Here σ^{+} = σ if p ∤ n, σ^{+} = σ+1 if p | n and σ is the number of primes in ℤ[ξ +ξ^{−1}] that split in ℤ[ξ] and lie over primes in ℤ that divide n.

An application of these theorems is given in [5]; moreover they are a generalization of the results of Naffah [7] on the normalizer of SL(2, ℤ[1/n]).

Let U(\frac{p−1}{2}) ⊂ GL(\frac{p−1}{2}, ℂ) be the group of unitary matrices. We consider the homomorphism

\begin{matrix}U(\frac{p−1}{2}) & ⟶ & Sp(p − 1, ℝ) \\ X = A + iB & ⟼ & (\begin{matrix}A & B \\ −B & A \end{matrix}) \end{matrix}

p-torsion in symplectic groups 171

where A, B ∈ M(n, ℝ). In [3] a condition is given for the matrix X such that the image of X is conjugate to a matrix of order p in Sp(p − 1, ℤ). This is used in [4] to analyze the subgroups of order p in Sp(p − 1, ℤ) by considering the corresponding subgroups in U(\frac{p−1}{2}). Here we avoid the unitary group by taking an arithmetical approach.

2. A recall of algebraic number theory

For the convenience of the reader, we give a short introduction to algebraic number theory. More details and the proofs can be found in the books of Lang [6], Neukirch [8] and Washington [12].

Let p be an odd prime and let ξ be a primitive pth root of unity. Then ℤ[ξ] is the ring of integers of the cyclotomic field ℚ(ξ) and ℤ[ξ + ξ^{−1}] is the ring of integers of the maximal real subfield ℚ(ξ + ξ^{−1}) of ℚ(ξ). For an integer 0 ≠ n ∈ ℤ we consider the ring ℤ[1/n] and the extensions ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}]. It is well-known that ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}] are Dedekind rings. For j = 1, …, p − 1 let the Galois automorphism γ_{j} ∈ Gal(ℚ(ξ)/ℚ) be given by γ_{j}(ξ) = ξ^{j}. To simplify the notations, we define x^{(j)} := γ_{j}(x) for any x ∈ ℚ(ξ) and γ_{j} ∈ Gal(ℚ(ξ)/ℚ) as above. The Galois automorphism γ_{j} acts componentwise on a vector in ℚ(ξ)^{k}.

Let A be a Dedekind ring and K the quotient field of A. Let L be a finite separable extension of K and B the integral closure of A in L. Let 𝔞 be an additive subgroup of L. The complementary set 𝔞′ of 𝔞 is the set of x ∈ L such that tr_{L/K}(x𝔞) ⊆ A. The different of the extension B/A is defined to be

D_{B/A} := B′_{L/K}^{−1}.

In ℤ[ξ] the different is generated by D = pξ^{(p+1)/2}/(ξ − 1). It is a principal ideal. This is also true for ℤ[1/n][ξ] (see Lang [6] or Serre [9]).

Let O be the ring of integers of a number field K. Let G = Gal(K/ℚ) be the Galois group of the extension and let 𝔮 be a prime ideal of O. The subgroup

G_{𝔮} = {σ ∈ G | σ𝔮 = 𝔮}

is called the decomposition group of 𝔮 over ℚ. The fixed field

Z_{𝔮} = {x ∈ K | σx = x for all σ ∈ G_{𝔮}}

is called the decomposition field of 𝔮 over ℚ. The decomposition group of a prime ideal σ𝔮 that is conjugate to 𝔮 is the conjugate subgroup G_{σ𝔮} = σG_{𝔮}σ^{−1}. Let 𝔮 ⊂ O be a prime ideal in O over the prime (q) in ℤ. Let κ(𝔮) := O/𝔮 and κ(q) := ℤ/qℤ. The degree f_{𝔮} of the extension of fields κ(𝔮)/κ(q) is called the residue class degree of 𝔮. We recall the following property. For any prime q ≠ p let f_{q} ∈ ℕ be the smallest positive integer such that

q^{f_{q}} ≡ 1 mod p.

Then (q) = (𝔮_{1} ⋯ 𝔮_{r}) where 𝔮_{1}, …, 𝔮_{r} are pairwise different prime ideals in ℚ(ξ) and all have residue class degree f_{q} (see Neukirch [8]).

Let p, q and ξ be as above. Let 𝔮^{+} ⊆ ℤ[ξ + ξ^{−1}] be a prime ideal that lies over q. We consider the ideal 𝔮^{+}ℤ[ξ] ⊂ ℤ[ξ] generated by 𝔮^{+}. Any prime q ≠ p is unramified and the prime p ramifies. Let σ ∈ G := Gal(ℚ(ξ)/ℚ) with σ(x) = \overline{x}. The Galois group G acts transitively on the set of prime ideals over q. It is known that f_{q} = |G_{𝔮}|. We have the following three cases:

(i) The prime 𝔮^{+} is inert: 𝔮^{+}ℤ[ξ] = 𝔮, a prime ideal in ℤ[ξ] that lies over q.

𝔮^{+}ℤ[ξ] = 𝔮 ⇔ 𝔮 = \overline{𝔮}
⇔ σ ∈ G_{𝔮}, i.e., G_{𝔮} contains an element of order 2,
⇔ f_{q} is even.

(ii) Primes that split in ℤ[ξ]: 𝔮^{+}ℤ[ξ] = 𝔮\overline{𝔮} where 𝔮 is a prime ideal in ℤ[ξ] that lies over q.

𝔮^{+}ℤ[ξ] = 𝔮\overline{𝔮} ⇔ 𝔮 ≠ \overline{𝔮}

⇔ σ ∉ G_{𝔮}, i.e., G_{𝔮} does not contain an element of order

2,

⇔ f_{q} is odd.

(iii) The ramified case: 𝔭^{+}ℤ[ξ] = 𝔭^{2} where 𝔭 := (1 − ξ) is the only prime ideal in ℤ[ξ] that lies over p. Moreover 𝔭^{+}ℤ[ξ] := ((1 − ξ)(1 − ξ^{−1})) = 𝔭\overline{𝔭} is the only prime ideal in ℤ[ξ + ξ^{−1}] that lies over p.

Let O_{K} be a Dedekind ring and let S be a finite set of prime ideals 𝔮 ⊆ O_{K}. We define

O_{K}^{S} := \{ \frac{f}{g} | f, g ∈ O, g ≢ 0 mod 𝔮 for 𝔮 ∉ S\} .

Let K be the quotient field of O_{K}. We call the group (O_{K}^{S})^{∗} the group of S-units of K. Let C(O_{K}), resp. C(O_{K}^{S}), denote the ideal class group of O_{K}, resp. O_{K}^{S}.

Proposition 2.1. For the group (O_{K}^{S})^{∗} defined above we have an isomorphism

(O_{K}^{S})^{∗} ≅ μ(K) × ℤ^{|S|+r+s−1}

where μ(K) denotes the group of roots of unity of K, r denotes the number of real embeddings of K and s denotes the number of conjugate pairs of complex embeddings of K.

Proof. See Neukirch [8]. □

Therefore

(O_{K}^{S})^{∗} ≅ μ(K) × ℤ^{r+s−1} × ℤ^{|S|}
≅ O_{K}^{∗} × ℤ^{|S|}.

3. Matrices of order p

3.1. A relation between matrices and ideal classes. The results obtained in this section are based on the bijection given by Proposition 3.3. Sjerve and Yang prove in [11] the analogous statement of this proposition for the group Sp(p − 1, ℤ). Since for our purpose it is important to understand the bijection and some proofs need a slightly different approach for the group Sp(p−1, ℤ[1/n]), we present in this subsection some of the proofs for the convenience of the reader.

Definition. Let I be the set of pairs (𝔞, a) where 𝔞 ⊆ ℤ[1/n][ξ] is a ℤ[1/n][ξ]-ideal and 0 ≠ a ∈ ℤ[1/n][ξ] is such that 𝔞\overline{𝔞} = (a) ⊆ ℤ[1/n][ξ]. Here \overline{𝔞} denotes the ideal generated by the complex conjugates of the elements of 𝔞. We define an equivalence relation on I.

\begin{aligned}(𝔞, a) ∼ (𝔟, b) & ⇔ & ∃λ, μ ∈ ℤ[1/n][ξ], λ, μ ≠ 0 \\ & & λ𝔞 = μ𝔟, λ\overline{λ}a = μ\overline{μ}b. \end{aligned}

Let [𝔞, a] denote the equivalence class of the pair (𝔞, a) and let I be the set of equivalence classes [𝔞, a].

Lemma 3.1. Let (𝔞, a) be a pair consisting of a ℤ[1/n][ξ]-ideal 𝔞 ⊆ ℤ[1/n][ξ] and 0 ≠ a ∈ ℤ[1/n][ξ]. Then (𝔞, a) ∈ I if and only if a ℤ[1/n]-basis α_{1}, …, α_{p−1} of 𝔞 exists such that

α^{T}J\overline{α}^{(i)} = δ_{1i}aD

where D = pξ^{(p+1)/2}/(ξ − 1) and α = (α_{1}, …, α_{p−1})^{T}.

Proof. The proof is analogous to the proof of Lemma 2.3 in [11]. □

Lemma 3.2. Let M, N be two (p − 1)×(p − 1)-matrices over ℤ[1/n] and let

α = (α_{1}, …, α_{p−1})^{T} ∈ ℤ[1/n][ξ]^{p−1}

where α_{1}, …, α_{p−1} are ℤ[1/n]-linear independent. If for i = 1, …, p − 1

α^{T}M\overline{α}^{(i)} = α^{T}N\overline{α}^{(i)}

then we have M = N.

Proof. It suffices to prove the case N = 0 because

α^{T}M\overline{α}^{(i)} = α^{T}N\overline{α}^{(i)} ⇔ α^{T}(M − N)\overline{α}^{(i)} = 0 = α^{T}0\overline{α}^{(i)}.

Let a_{i} = α^{T}M\overline{α}^{(i)}, then a_{i}^{(k)} = α^{(k)T}M(\overline{α}^{(i)})^{(k)}. For all k, l with 1 ⩽ k, l ⩽ p−1 let i be such that 1 ⩽ i ⩽ p − 1 and ki ≡ l mod p. Then (\overline{α}^{(i)})^{(k)} = \overline{α}^{(l)} and therefore α^{(k)T}M\overline{α}^{(l)} = 0 for k, l = 1, …, p − 1. This implies A^{T}MB = 0 where

A := (α_{i}^{(j)}) and B := (\overline{α}_{i}^{(j)})

are (p − 1)×(p − 1)-matrices. Since α_{1}, …, α_{p−1} are ℤ[1/n]-linear independent we have det A ≠ 0 and det B ≠ 0. But this yields M = 0. □

Proposition 3.3. A bijection ψ exists between the set of conjugacy classes of elements of order p in Sp(p − 1, ℤ[1/n]) and the set of equivalence classes of pairs [𝔞, a] ∈ I.

In order to prove this proposition, we first construct the bijection and then we show that the mapping we constructed is a bijection (Lemma 3.5, Lemma 3.6).

Let Y ∈ Sp(p − 1, ℤ[1/n]) be of order p. The eigenvalues of Y are the primitive pth roots of unity. An eigenvector

α = (α_{1}, …, α_{p−1})^{T} ∈ (ℤ[1/n][ξ])^{p−1}

exists for the eigenvalue ξ = e^{i2π/p} (i.e., Y α = ξα). The α_{1}, …, α_{p−1} are ℤ[1/n]- linear independent. Let 𝔞 be the ℤ[1/n]-module generated by α_{1}, …, α_{p−1}. Let a = D^{−1}α^{T}J\overline{α}. Then 𝔞 ⊆ ℤ[1/n][ξ] is a ℤ[1/n][ξ]-ideal and a = \overline{a}.

Lemma 3.4. The pair (𝔞, a) we construct above is an element of I.

Proof. By Lemma 3.1 it suffices to show that α^{T}J\overline{α}^{(i)} = 0 for i = 2, …, p − 1. Since Y α = ξα we have

Y α^{(i)} = ξ^{i}α^{(i)} and Y \overline{α}^{(i)} = \frac{1}{ξ^{i}} \overline{α}^{(i)},

2 ⩽ i ⩽ p − 1. Therefore

α^{T}J\overline{α}^{(i)} = \frac{ξ^{i}}{ξ} α^{T}Y^{T}JY \overline{α}^{(i)} = \frac{ξ^{i}}{ξ} α^{T}J\overline{α}^{(i)}

where the last equation follows from the fact that Y ∈ Sp(p − 1, ℤ[1/n]). Since ξ ≠ ξ^{j} we get α^{T}J\overline{α}^{(i)} = 0. □

Let Y, Ỹ ∈ Sp(p − 1, ℤ[1/n]) be matrices of odd prime order p. Let α ∈ (ℤ[1/n][ξ])^{p−1}, resp. β ∈ (ℤ[1/n][ξ])^{p−1} be an eigenvector of Y, resp. Ỹ, to the eigenvalue ξ, i.e., Y α = ξα and Ỹβ = ξβ. Let α = (α_{1}, …, α_{p−1})^{T}, β = (β_{1}, …, β_{p−1})^{T}. Let 𝔞 ⊆ ℤ[1/n][ξ], resp. 𝔟 ⊆ ℤ[1/n][ξ], be the ideal with ℤ[1/n]-basis α_{1}, …, α_{p−1}, resp. β_{1}, …, β_{p−1}. We define a = D^{−1}α^{T}J\overline{α} and b = D^{−1}β^{T}J\overline{β}. We show the injectivity of ψ.

Lemma 3.5. Let Y, Ỹ ∈ Sp(p − 1, ℤ[1/n]) be matrices of odd prime order p. Then Y and Ỹ are conjugate if and only if [𝔞, a] = [𝔟, b].

Proof. Let Y and Ỹ be conjugate. Then Q ∈ Sp(p − 1, ℤ[1/n]) exists such that Ỹ = Q^{−1}Y Q. Then QỸ = Y Q and for the eigenvector β to the eigenvalue ξ of Ỹ we get

Y Qβ = QỸβ = ξQβ

and therefore Qβ is an eigenvector of Y. But α is also an eigenvector to the eigenvalue ξ of Y. So λ, μ ∈ ℤ[1/n][ξ], λ, μ ≠ 0, exist such that

λα = μQβ = Qμβ.

Then λ𝔞 = μ𝔟 and for a = D^{−1}α^{T}J\overline{α}, b = D^{−1}β^{T}J\overline{β} we get λ\overline{λ}a = μ\overline{μ} b. This shows that [𝔞, a] = [𝔟, b].

In order to show the other direction we assume that λ, μ ∈ ℤ[1/n][ξ], λ, μ ≠ 0, exist such that λ𝔞 = μ𝔟 and λ\overline{λ}a = μ\overline{μ}b. Then a matrix Q ∈ GL(p − 1, ℤ[1/n]) exists such that λα = μQβ. We have

μQỸβ = μQξβ = ξμQβ = ξλα = λY α = μY Qβ

and therefore

QỸβ = Y Qβ.

Since β_{1}, …, β_{p−1} are ℤ[1/n]-linear independent, we have QỸ = Y Q and herewith

Ỹ = Q^{−1}Y Q.

It remains to show that Q ∈ Sp(p − 1, ℤ[1/n]). For i = 2, …, p − 1 we have

β^{T}Q^{T}JQ \overline{β}^{(i)} = \frac{λ\overline{λ}^{(i)}}{μ\overline{μ}^{(i)}} α^{T}J \overline{α}^{(i)} = 0 = β^{T}J \overline{β}^{(i)}

and for i = 1 we have

β^{T}Q^{T}JQ \overline{β} = \frac{λ\overline{λ}}{μ\overline{μ}} α^{T}J \overline{α} = \frac{b}{a} α^{T}J \overline{α} = β^{T}J \overline{β}

because λ\overline{λ} a = μ\overline{μ} b implies that \frac{λ\overline{λ}}{μ\overline{μ}} = \frac{b}{a}. Now it follows from Lemma 3.2 that Q^{T}JQ = J and this means that Q ∈ Sp(p − 1, ℤ[1/n]). □

Lemma 3.6. The mapping ψ is surjective.

Proof. If (𝔞, a) and α = (α_{1}, …, α_{p−1})^{T} are as in Lemma 3.1, then ξα_{1}, …, ξα_{p−1} is a new basis of 𝔞. Therefore X ∈ GL(p − 1, ℤ[1/n]) exists with Xα = ξα. It is evident that the order of X is p. We show that X ∈ Sp(p − 1, ℤ[1/n]). We have

α^{T}X^{T}JX\overline{α}^{(i)} = \frac{ξ}{ξ^{i}} α^{T}J\overline{α}^{(i)} = δ_{1i}α^{T}J \overline{α}

hence

α^{T}X^{T}JX \overline{α}^{(i)} = α^{T}J \overline{α}^{(i)}.

The last equation and Lemma 3.2 imply that X^{T}JX = J and therefore X ∈ Sp(p − 1, ℤ[1/n]). □

Let I be the set of equivalence classes of pairs (𝔞, a) ∈ I defined above. We define a multiplication on I by

[𝔞, a] · [𝔟, b] = [𝔞𝔟, ab].

The unit is [ℤ[1/n][ξ], 1] and the inverse of [𝔞, a] is [\overline{𝔞}, a] since

[𝔞, a] · [\overline{𝔞}, a] = [(a), a^{2}] = [ℤ[1/n][ξ], 1].

Lemma 3.7. Let (𝔞, a) ∈ I, λ ∈ ℤ[1/n][ξ], λ ≠ 0. Then:

(i) (λ𝔞, λ\overline{λ}a) ∈ I. (ii) (𝔞, λa) ∈ I if and only if λ ∈ ℤ[1/n][ξ + ξ^{−1}]^{∗}.

Proof. Trivial. □

Let

N : ℚ(ξ) ⟶ ℚ(ξ + ξ^{−1})

be the norm mapping, i.e., N(x) = x\overline{x} for x ∈ ℚ(ξ). Then

N(ℤ[1/n][ξ]^{∗}) ⊆ ℤ[1/n][ξ + ξ^{−1}]^{∗}.

Lemma 3.8. Let (𝔞, a), (𝔞, b) ∈ I. Then [𝔞, a] = [𝔞, b] if and only if

\frac{a}{b} ∈ N(ℤ[1/n][ξ]^{∗}).

Proof. Suppose that [𝔞, a] = [𝔞, b]. Then λ, μ ∈ ℤ[1/n][ξ], λ, μ ≠ 0, exist such that λ𝔞 = μ𝔞 and λ\overline{λ} a = μ\overline{μ} b. Let u = μ/λ, then u ∈ ℤ[1/n][ξ]^{∗} (since 𝔞 = (μ/λ) 𝔞) and a/b = μ\overline{μ}/λ\overline{λ} = u\overline{u}. This shows that a/b ∈ N(ℤ[1/n][ξ]^{∗}). Now let a/b = u\overline{u} for some u ∈ ℤ[1/n][ξ]^{∗}. Then [𝔞, a] = [𝔞, u\overline{u} b] = [u𝔞, u\overline{u} b] = [𝔞, b]. □

Lemma 3.9. Let (𝔞, a), (𝔟, b) ∈ I and λ𝔞 = μ𝔟 for some λ, μ ∈ ℤ[1/n][ξ], λ, μ ≠ 0. Then u ∈ ℤ[1/n][ξ + ξ^{−1}]^{∗} exists such that [𝔞, a] = [𝔟, ub].

Proof. If λ𝔞 = μ𝔟, then \overline{λ}\overline{𝔞} = \overline{μ}\overline{𝔟} and herewith

(λ\overline{λ} a) = λ𝔞\overline{λ}\overline{𝔞} = μ𝔟\overline{μ}\overline{𝔟} = (μ\overline{μ} b).

But then a unit u ∈ ℤ[1/n][ξ + ξ^{−1}]^{∗} exists with λ\overline{λ} a = μ\overline{μ} ub. Herewith

[𝔞, a] = [λ𝔞, λ\overline{λ} a] = [μ𝔟, μ\overline{μ} ub] = [𝔟, ub].  □

Proposition 3.10. Let C_{0} be the ideal class group of ℤ[1/n][ξ]. Then the sequence

1 ⟶ ℤ[1/n][ξ + ξ^{−1}]^{∗}/N(ℤ[1/n][ξ]^{∗}) ⟶^{δ} I ⟶^{η} C_{0} ⟶ 1

where δ([u]) = [ℤ[1/n][ξ], u], η([𝔞, a]) = [𝔞], is a short exact sequence.

Proof. Lemma 3.8 implies that δ is injective and η is well-defined and surjective. Moreover

η(δ([u])) = η([ℤ[1/n][ξ], u]) = [ℤ[1/n][ξ]]

and Lemma 3.9 implies that the kernel of η is equal to the image of δ. □

Corollary 3.11. There are

|C_{0}| · [ℤ[1/n][ξ + ξ^{−1}]^{∗} : N(ℤ[1/n][ξ]^{∗})]

conjugacy classes of matrices of order p in Sp(p − 1, ℤ[1/n]).

Proof. This corollary is a direct consequence of Proposition 3.10 because the number of conjugacy classes of matrices of order p in Sp(p − 1, ℤ[1/n]) is equal to the cardinality of I. □

If ℤ[1/n][ξ] is a principal ideal domain the cardinality of C_{0} is 1 and the number of conjugacy classes of matrices of order p in Sp(p − 1, ℤ[1/n]) is given only by the index defined above. In fact we can choose n ∈ ℤ such that ℤ[1/n][ξ] is a principal ideal domain. Indeed let 𝔞_{1}, …, 𝔞_{h} be representatives of the ideal classes of ℚ(ξ). For j = 1, …, h choose n_{j} ∈ 𝔞_{j} with n_{j} ∈ ℤ[1/n][ξ]. It is possible to choose the n_{j} such that n_{j} ∈ ℤ. Then n = ∏_{j=1}^{h} n_{j} ∈ 𝔞_{k} for any k with 1 ⩽ k ⩽ h. For more details see Lang [6] and Neukirch [8].

3.2. The number of conjugacy classes. Let N : ℚ(ξ) ⟶ ℚ(ξ + ξ^{−1}) be the norm mapping defined above. Let n ∈ ℤ and ξ a primitive pth root of unity. The aim of this section is to compute the number of conjugacy classes of elements of order p in Sp(p − 1, ℤ[1/n]). Therefore we use Corollary 3.11.

Kummer proved that ℤ[1/n][ξ]^{∗} = ℤ[1/n][ξ + ξ^{−1}]^{∗} × ⟨−ξ⟩ where ⟨−ξ⟩ is the group of roots of unity in ℚ(ξ). This implies that

[ℤ[ξ + ξ^{−1}]^{∗} : N(ℤ[ξ]^{∗})] = [ℤ[ξ + ξ^{−1}]^{∗} : (ℤ[ξ + ξ^{−1}]^{∗})^{2}].

Moreover ℤ[ξ + ξ^{−1}]^{∗} ≅ ℤ^{(p−3)/2} × ℤ/2ℤ because of the Dirichlet unit theorem. Therefore

[ℤ[ξ + ξ^{−1}]^{∗} : N(ℤ[ξ]^{∗})] = 2^{\frac{p−1}{2}} .

Since the prime above p in ℤ[ξ] is principal, generated by 1 − ξ, and the prime above p in ℤ[ξ + ξ^{−1}] is principal, generated by N(1 − ξ) = (1 − ξ)(1 − ξ^{−1}), we get

[ℤ[1/p][ξ + ξ^{−1}]^{∗} : N(ℤ[1/p][ξ]^{∗})] = 2^{\frac{p−1}{2}} .

Proposition 3.12. Let p be an odd prime and let ξ be a primitive pth root of unity. Let S^{+} be a finite set of prime ideals in ℤ[ξ +ξ^{−1}], and let S be the set of the prime ideals in ℤ[ξ] that lie over those in S^{+}. Then

[(ℤ[ξ + ξ^{−1}]^{S^{+}})^{∗} : N ((ℤ[ξ]^{S})^{∗})] = 2^{\frac{p−1}{2} +τ}

where τ is the number of inert primes in S^{+}.

Proof. Let S := {𝔮_{1}, …, 𝔮_{k}} be a set of prime ideals in ℤ[ξ]. Then the isomorphism given by the generalization of the Dirichlet unit theorem implies that for each prime ideal 𝔮_{j} ∈ S, j = 1, …, k, g_{j} ∈ 𝔮_{j} exists such that each unit u ∈ (ℤ[ξ]^{S})^{∗} can be written

u = u′g_{1}^{n_{1}} ⋯ g_{k}^{n_{k}}

where u′ ∈ ℤ[ξ]^{∗}, n_{j} ∈ ℤ, j = 1, …, k. We compute the index we want to know by induction on the number of primes in S^{+}. Let T^{+} be a finite set of prime ideals in ℤ[ξ + ξ^{−1}]. Let T be the set of those prime ideals in ℤ[ξ] that lie over the prime ideals in T^{+}. Define S^{+} := T^{+} ∪ {𝔮^{+}} where 𝔮^{+} ⊂ ℤ[ξ + ξ^{−1}], 𝔮^{+} ∉ T^{+}, is a prime ideal. Let S be the set of the prime ideals in ℤ[ξ] that lie over the prime ideals in S^{+}. We have the following possibilities:

(i) The prime 𝔮^{+} is inert. Then S = T ∪ {𝔮} where 𝔮 is the prime that lies over 𝔮^{+}.

(ii) The prime 𝔮^{+} splits in ℤ[ξ]. Then S = T ∪ \{𝔮, \overline{𝔮}\} where 𝔮, \overline{𝔮} are the primes

that lie over 𝔮^{+}. (iii) The prime 𝔮^{+} lies over p. Then S = T ∪ {𝔭} where 𝔭 = (1 − ξ), the prime

over p. We have

(ℤ[ξ + ξ^{−1}]^{S^{+}})^{∗} ≅ (ℤ[ξ + ξ^{−1}]^{T^{+}})^{∗} × ℤ .

If the prime 𝔮^{+} is inert or if it lies over p, cases (i) and (iii) above, then

(ℤ[ξ]^{S})^{∗} ≅ ℤ[ξ]^{∗} × ℤ^{|S|} ≅ ℤ[ξ]^{∗} × ℤ^{|T |} × ℤ
≅ (ℤ[ξ]^{T})^{∗} × ℤ

and if the prime 𝔮^{+} splits in ℤ[ξ], case (ii) above, then

(ℤ[ξ]^{S})^{∗} ≅ (ℤ[ξ]^{T})^{∗} × ℤ^{2}.

We give a formula for the index

[(ℤ[ξ + ξ^{−1}]^{S^{+}})^{∗} : N ((ℤ[ξ]^{S})^{∗})]

in relation to the index

[(ℤ[ξ + ξ^{−1}]^{T^{+}})^{∗} : N ((ℤ[ξ]^{T} )^{∗})] .

If the prime 𝔮^{+} is inert, then

[(ℤ[ξ + ξ^{−1}]^{S^{+}})^{∗} : N ((ℤ[ξ]^{S})^{∗})] = 2 [(ℤ[ξ + ξ^{−1}]^{T^{+}})^{∗} : N ((ℤ[ξ]^{T} )^{∗})] .

If the prime 𝔮^{+} splits in ℤ[ξ] or if it lies over p, then

[(ℤ[ξ + ξ^{−1}]^{S^{+}})^{∗} : N ((ℤ[ξ]^{S})^{∗})] = [(ℤ[ξ + ξ^{−1}]^{T^{+}})^{∗} : N ((ℤ[ξ]^{T} )^{∗})] .

This shows that if we add an inert prime to the set S the index is multiplied by 2, and if we add primes that split or the prime over p, then the index does not change. □

Corollary 3.13. Let n ∈ ℤ. Then

[ℤ[1/n][ξ + ξ^{−1}]^{∗} : N(ℤ[1/n][ξ]^{∗})] = 2^{\frac{p−1}{2} +τ}

where τ is the number of inert primes in ℤ[ξ + ξ^{−1}] that lie over primes in ℤ that divide n.

Proof. Let n ∈ ℤ and let S^{+}, resp. S, be the prime ideals in ℤ[ξ + ξ^{−1}], resp. ℤ[ξ], over the primes in ℤ that divide n. Then the assumption follows directly from Proposition 3.12. □

Now we have the main result of this section.

Theorem 3.14. There are

|C_{0}|2^{\frac{p−1}{2} +τ}

conjugacy classes of matrices of order p in Sp(p − 1, ℤ[1/n]), 0 ≠ n ∈ ℤ. Here C_{0} is the ideal class group of ℤ[1/n][ξ] and τ is the number of inert primes in ℤ[ξ + ξ^{−1}] that lie over primes in ℤ that divide n.

Proof. This follows directly from Corollary 3.11 and Corollary 3.13. □

4. Subgroups of order p

4.1. The quotient of the normalizer by the centralizer of subgroups of order p. The aim is to study the centralizers and normalizers of conjugacy classes of subgroups of order p in Sp(p − 1, ℤ[1/n]). We use the bijection between the set I of equivalence classes [𝔞, a] and the conjugacy classes of matrices of order p. Each conjugacy class of matrices generates a conjugacy class of subgroups of order p in Sp(p − 1, ℤ[1/n]). We determine the equivalence classes [𝔞, a] that correspond to the conjugacy classes of the elements of a subgroup.

Let Y ∈ Sp(p − 1, ℤ[1/n]) be of odd prime order p. We have seen that the conjugacy class of Y corresponds to an equivalence class [𝔞, a]. Let

α = (α_{1}, …, α_{p−1})^{T} ∈ (ℤ[1/n][ξ])^{p−1}

be an eigenvector of Y to the eigenvalue ξ = e^{i2π/p}. It is obvious that Y^{l} = ξ^{l}α. Let γ_{k} ∈ Gal(ℚ[ξ]/ℚ) be such that γ_{k}(ξ) = ξ^{k}. Then γ_{k}(ξ^{l}) = ξ^{kl}. If kl ≡ 1 mod p, then γ_{k}(ξ^{l}) = ξ and moreover

Y^{l}γ_{k}(α) = γ_{k}(Y^{l}α) = γ_{k}(ξ^{l}α) = γ_{k}(ξ^{l})γ_{k}(α) = ξ^{kl}γ_{k}(α) = ξγ_{k}(α).

So γ_{k}(α) is the eigenvector of Y^{l} to the eigenvalue ξ. Let 𝔟 be the ideal given by the ℤ[1/n]-basis γ_{k}(α_{1}), …, γ_{k}(α_{p−1}). Moreover let

b = D^{−1}(γ_{k}(α))^{T}Jγ_{k}(\overline{α}) = D^{−1}γ_{k}(α^{T}J\overline{α}).

So the conjugacy class of Y^{l} corresponds to the equivalence class [𝔟, b] with

𝔟 = γ_{k}(𝔞)
b = D^{−1}γ_{k}(Da) = D^{−1}γ_{k}(D)γ_{k}(a).

Let S be a multiplicative set such that S^{−1}ℤ = ℤ[1/n]. Then S^{−1}ℤ[ξ] = ℤ[1/n][ξ] and the different in ℤ[ξ] and in ℤ[1/n][ξ] are both principal ideals generated by D = p ξ^{(p+1)/2}/(ξ − 1). If p | n, then D is a unit in ℤ[1/n][ξ] since (ξ − 1) is a prime that divides p. If u, v ∈ ℤ[1/n][ξ]^{∗} are units with u = \overline{u}, v = −\overline{v}, then Du = −\overline{Du} and Dv = \overline{Dv}. This shows that the multiplication with D defines an isomorphism on ℤ[1/n][ξ]^{∗} that yields a bijection between the real and the purely imaginary units.

Theorem 4.1. Let n ∈ ℤ be such that ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}] are principal ideal domains and moreover p | n. Let N(P) denote the normalizer and C(P) the centralizer of a subgroup P of order p in Sp(p − 1, ℤ[1/n]). Then

N(P)/C(P) ≅ ℤ/jℤ

where j | p − 1, j odd. For each j with j | p − 1, j odd, there exists a subgroup of order p in Sp(p − 1, ℤ[1/n]) with N(P)/C(P) ≅ ℤ/jℤ.

Proof. Let n be such that ℤ[1/n][ξ] is a principal ideal domain. Then the ideal 𝔞 in the pair [𝔞, a] is a principal ideal. If 𝔞 = (x), then (x)\overline{(x)} = (x\overline{x}) = (a), i.e., a unit u exists such that a = ux\overline{x}. Then

[𝔞, a] = [(x), a] = [ℤ[1/n][ξ], u].

The conjugacy class of Y ∈ Sp(p − 1, ℤ[1/n]) corresponds to [ℤ[1/n][ξ], u]. We have seen that the conjugacy class of Y^{l}, 1 < l < p − 1, corresponds to the equivalence class [ℤ[1/n][ξ], D^{−1}γ_{k}(Du)] where γ_{k} ∈ Gal(ℚ[ξ]/ℚ) is defined so that γ_{k}(ξ^{l}) = ξ. The matrices Y and Y^{l} are conjugate if and only if

[ℤ[1/n][ξ], u] = [ℤ[1/n][ξ], D^{−1}γ_{k}(Du)].

Lemma 3.8 shows that this equation is satisfied if and only if ω ∈ ℤ[1/n][ξ]^{∗} exists such that

D^{−1}γ_{k}(Du) = uω\overline{ω}. (1)

We know that u ∈ ℤ[1/n][ξ + ξ^{−1}]^{∗} and this implies that Du is purely imaginary. First we check if u ∈ ℤ[1/n][ξ + ξ^{−1}]^{∗} exists such that a special case of (1) holds, namely the case with ω = 1, i.e., we try to find γ_{k} and u such that γ_{k}(Du) = Du. The automorphism γ_{p−1}(= γ_{−1}) has order 2, i.e., γ_{p−1} yields the complex conjugation. Since u is real and therefore Du purely imaginary, we get γ_{p−1}(Du) = −Du. This proves that neither γ_{k}(Du) = Du nor (1) can be satisfied if k = p−1 (the image of ω\overline{ω} under any embedding of ℤ[1/n][ξ] in ℂ is a positive real number). Any automorphism γ_{k} ∈ Gal(ℚ[ξ]/ℚ) generates a subgroup ⟨γ_{k}⟩ ⊆ Gal(ℚ[ξ]/ℚ) and the order of this subgroup divides p−1, the order of Gal(ℚ[ξ]/ℚ). Let j = |⟨γ_{k}⟩| denote the order of γ_{k}. If j is even the order of γ_{k}^{j/2} is 2 and on the other hand γ_{k}^{r}(Du) = Du for any 1 < r < j. This yields a contradiction and therefore γ_{k}(Du) = Du cannot be satisfied if the order of γ_{k} is even. This implies that if γ_{k} and u exist with γ_{k}(Du) = Du, then the order of γ_{k} is odd.

The main theorem of Galois theory says that a subfield ℚ ⊆ K ⊆ ℚ(ξ) corresponds to the subgroup ⟨γ_{k}⟩ ⊆ Gal(ℚ[ξ]/ℚ) and that

K = {x ∈ ℚ(ξ) | ∀γ_{k^{r}} ∈ ⟨γ_{k}⟩, γ_{k^{r}}(x) = x}.

Let n ∈ ℤ with p | n, We have seen that in this case D = pξ^{(p+1)/2}/(ξ − 1) is a unit in ℤ[1/n][ξ]. We also know that D = −\overline{D}. Let γ_{k} ∈ Gal(ℚ[ξ]/ℚ) be of odd order j. Since complex conjugation commutes with the Galois automorphisms, we get for any r, 1 ⩽ r ⩽ j, γ_{k^{r}}(D) = −\overline{γ_{k^{r}}(D)}. Since j is odd,

∏_{r=1}^{j} γ_{k^{r}}(D) = (−1)^{j} ∏_{r=1}^{j} γ_{k^{r}}(\overline{D}) = − ∏_{r=1}^{j} γ_{k^{r}}(\overline{D}).

Moreover this product is invariant under γ_{k} since

γ_{k}(∏_{r=1}^{j} γ_{k^{r}}(D)) = ∏_{r=1}^{j} γ_{k}(γ_{k^{r}}(D)) = ∏_{r=1}^{j} γ_{k^{r}}(D).

Now consider the composition γ_{k} ∘ γ_{p−1} = γ_{−k} where the order of γ_{k} is odd. The order of γ_{−k} is even and ⟨γ_{k}⟩ is a subgroup of ⟨γ_{−k}⟩. Let L denote the subfield ℚ ⊆ L ⊆ K ⊆ ℚ(ξ) corresponding to ⟨γ_{−k}⟩. Sinnott constructs in [10] cyclotomic units in any subfield L of ℚ(ξ_{m}) where ξ_{m} is a mth root of unity. This means that units exist in L, that are contained in no subfield of L. Let v ∈ L be such a unit (v ∈ ℤ[ξ]). Then γ_{p−1}(v) = v, γ_{k}(v) = v since ⟨γ_{−k}⟩ fixes the elements of L. Let w := ∏_{r=1}^{j−1} γ_{k^{r}}(D) ∈ ℤ[1/n][ξ]. Then

w = ∏_{r=1}^{j−1} γ_{k^{r}}(D) = (−1)^{j−1} ∏_{r=1}^{j−1} \overline{γ_{k^{r}}(D)} = \overline{w}

since j is odd. Moreover Dw = ∏_{r=1}^{j} γ_{k^{r}}(D) and therefore γ_{k}(Dw) = Dw. We have Dw = −\overline{Dw} since w = \overline{w}. Now wv = \overline{wv} is a unit. Let u = wv, then the construction implies that

γ_{k}(Du) = γ_{k}(Dwv) = Dwv = Du.

So for any γ_{k} ∈ Gal(ℚ(ξ)/ℚ) of odd order, we found u ∈ ℤ[1/n][ξ + ξ^{−1}]^{∗} with γ_{k}(Du) = Du and such that

[ℤ[1/n][ξ], u] = [ℤ[1/n][ξ], D^{−1}γ_{k}(Du)].

If Y ∈ Sp(p − 1, ℤ[1/n]) is in the corresponding equivalence class then this is also true for Y^{l} with l such that γ_{k}(ξ^{l}) = ξ. If Y is conjugate to Y^{l} with γ_{k}(ξ^{l}) = ξ, then Y is also conjugate to Y^{l^{r}} where 1 ⩽ r ⩽ j and j is the order of γ_{k}. Indeed γ_{k^{r}}(ξ^{l^{r}}) = ξ for 1 ⩽ r ⩽ j and therefore l^{j} ≡ 1 mod p (since γ_{k^{j}} = id) and Y^{l^{j}} = Y because the order of Y is p. The l^{r} form a cyclic subgroup of ℤ/pℤ.

Let k, l ∈ ℤ be as above, i.e., γ_{k}(ξ^{l}) = ξ. Let γ_{l} ∈ Gal(ℚ(ξ)/ℚ) with γ_{l}(ξ) = ξ^{l}. Then γ_{l} = γ_{k}^{−1} and if j is the order of γ_{k}, then j is also the order of γ_{l}. Therefore l^{j} ≡ 1 mod p. This means that Y^{l^{j}} = Y and the Y^{l^{r}}, 1 ⩽ r ⩽ j are conjugate to Y. We know that j is odd and j | p − 1.

If j elements are conjugate in the subgroup generated by Y ∈ Sp(p − 1, ℤ[1/n]), and if j is maximal with this property, then for this subgroup N(P)/C(P) ≅ ℤ/jℤ since ⟨γ_{k}⟩ ≅ ℤ/jℤ. Since we showed that for any odd divisor j | p − 1 a matrix Y ∈ Sp(p − 1, ℤ[1/n]) exists for which j powers are conjugate, we showed that for any j | p − 1, j odd, a subgroup of order p exists in Sp(p − 1, ℤ[1/n]), for which N(P)/C(P) ≅ ℤ/jℤ. □

4.2. The centralizer of subgroups of order p.

Theorem 4.2. Let n ∈ ℤ be such that ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}] are principal ideal domains. Then for a subgroup P of order p in Sp(p−1, ℤ[1/n]), the centralizer C(P) is

C(P) ≅ ℤ/2pℤ × ℤ^{σ^{+}}.

Here σ^{+} = σ if p ∤ n, σ^{+} = σ+1 if p | n and σ is the number of primes in ℤ[ξ +ξ^{−1}] that split in ℤ[ξ] and lie over primes in ℤ that divide n.

Proof. Let Y ∈ Sp(p−1, ℤ[1/n]) be of order p and let [𝔞, a] be the equivalence class corresponding to the conjugacy class of Y. Let P be the subgroup generated by Y. Let Z ∈ Sp(p − 1, ℤ[1/n]) be an element of the centralizer of Y, i.e., Z^{−1}Y Z = Y or Y Z = ZY. Then Z is an element of the centralizer of P. If α is an eigenvector of Y to the eigenvalue ξ, then so is Zα:

ξZα = Zξα = ZY α = Y Zα.

But this means that Zα = wα for some w ∈ ℤ[1/n][ξ] and w is a unit since Z is invertible. Therefore

(Zα)^{T}J\overline{Zα}^{(i)} = α^{T}Z^{T}JZ\overline{α}^{(i)} = wα^{T}J\overline{w}^{(i)}\overline{α}^{(i)}
= w\overline{w}^{(i)}α^{T}J\overline{α}^{(i)} = δ_{1i}aw\overline{w}^{(i)}D

and, since δ_{1i} = 0 for i ≠ 1, we get

(Zα)^{T}J\overline{Zα} = aw\overline{w}D.

But Z ∈ Sp(p − 1, ℤ[1/n]) and therefore

(Zα)^{T}J\overline{Zα} = α^{T}Z^{T}JZ\overline{α} = α^{T}J\overline{α} = aD.

This implies that w\overline{w} = 1. In order to determine the centralizer C(P) of a subgroup P ⊆ Sp(p−1, ℤ[1/n]) of order p, we have to find the units w ∈ ℤ[1/n][ξ]^{∗} that satisfy w\overline{w} = 1. This corresponds to the kernel of the norm mapping

\begin{aligned}N : & ℤ[1/n][ξ]^{∗} & ⟶ & ℤ[1/n][ξ + ξ^{−1}]^{∗} \\ & x & ⟼ & x\overline{x}. \end{aligned}

Brown [1] and Sjerve and Yang [11] showed that the kernel of the norm mapping

\begin{aligned}N′ : & ℤ[ξ]^{∗} & ⟶ & ℤ[ξ + ξ^{−1}]^{∗} \\ & x & ⟼ & x\overline{x} \end{aligned}

is the set of roots of unity

ker(N′) = {±ξ^{r} | ξ^{p} = 1, 1 ⩽ r ⩽ p}.

It is obvious that ker(N′) ⊆ ker(N). The prime ideals that lie over the primes in ℤ and divide n yield units in ℤ[1/n][ξ]^{∗} \ ℤ[ξ]^{∗}. Let 𝔮^{+} ⊆ ℤ[ξ + ξ^{−1}] be a prime over a prime q | n and let 𝔮 ⊆ ℤ[ξ] be a prime over 𝔮^{+}. If 𝔮^{+} is inert, then 𝔮 = \overline{𝔮} and if 𝔮^{+} splits, then 𝔮^{+}ℤ[ξ] = 𝔮\overline{𝔮}. A generalization for S-units of the Dirichlet unit theorem says that for each prime 𝔮_{j}, j = 1, …, k, over n a g_{j} ∈ 𝔮_{j} exists such that any unit u ∈ (ℤ[1/n][ξ])^{∗} can be written as

u = u′g_{1}^{n_{1}} ⋯ g_{k}^{n_{k}}

where u′ ∈ ℤ[ξ]^{∗}, n_{j} ∈ ℤ, j = 1, …, k. So the group of units ℤ[1/n][ξ + ξ^{−1}]^{∗} is generated by ℤ[ξ + ξ^{−1}]^{∗}, the inert primes over n, the primes over n that split and, if p | n, the prime over p. The inert primes yield nontrivial elements in ℤ[1/n][ξ +ξ^{−1}]^{∗}/N(ℤ[1/n][ξ]^{∗}) since for those holds w\overline{w} = w^{2} ≠ 1 for w ≠ ±1. The centralizer C(P) is a finitely generated group whose torsion subgroup is isomorphic to the group of roots of unity in ℚ(ξ) and whose rank is equal to σ if p ∤ n and to σ + 1 if p | n where

σ^{+} = rank(ℤ[1/n][ξ]^{∗}) − rank(ℤ[1/n][ξ + ξ^{−1}]^{∗}).

This difference is equal to the number of primes in ℤ[ξ + ξ^{−1}] that split or ramify in ℤ[ξ] and lie over primes in ℤ that divide n. This follows directly from a generalization of the Dirichlet unit theorem and proves our theorem. □

182 Cornelia Minette Busch

4.3. The action of the normalizer on the centralizer of subgroups of order p.

Theorem 4.3. Let N(P) be the normalizer and C(P) the centralizer of a subgroup P of order p in Sp(p − 1, ℤ[1/n]). Let p be an odd prime, ξ a primitive pth root of unity, n ∈ ℤ such that ℤ[1/n][ξ] and ℤ[1/n][ξ + ξ^{−1}] are principal ideal domains and moreover p | n. Then the action of N(P)/C(P) on C(P) is given by the action of the Galois group Gal(ℚ(ξ)/ℚ) on the group of units ℤ[1/n][ξ]^{∗}. Moreover N(P)/C(P) acts faithfully on C(P).

Proof. We have seen in the proof of Theorem 4.2 that the centralizer of a subgroup of order p in Sp(p − 1, ℤ[1/n]) is given by the kernel of the norm mapping ℤ[1/n][ξ]^{∗} ⟶ ℤ[1/n][ξ + ξ^{−1}]^{∗}, x ⟼ x\overline{x}. Herewith the centralizer is isomorphic to a subgroup of the group of units ℤ[1/n][ξ]^{∗}. In the proof of Theorem 4.1 we identify the quotient N(P)/C(P) with a subgroup of the Galois group Gal(ℚ(ξ)/ℚ). Herewith the action of the quotient N(P)/C(P) on the centralizer C(P) is given by the action of the subgroup of Gal(ℚ(ξ)/ℚ) corresponding to N(P)/C(P) on the kernel of the norm mapping ℤ[1/n][ξ]^{∗} ⟶ ℤ[1/n][ξ + ξ^{−1}]^{∗}. Since it is nontrivial, the action of N(P)/C(P) on C(P) is faithful. □

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Katholische Universität Eichstätt-Ingolstadt, MGF, D-85071 Eichstätt, Germany cornelia.busch@ku-eichstaett.de

This paper is available via http://nyjm.albany.edu/j/2006/12-10.html.