Klaus Hoechsmann, April 3, 1998.

The Ferguson Relation II.

1. With notation as in my pq-paper, the aim of this paragraph is to compute the factors of the chain

Ẏ (ξ) ⊆ Ω̇^{∘}(ξ) ⊆ Ω̇(ξ) .

Proposition 1: Ω̇(ξ)/Ω̇^{∘}(ξ) is cyclic of order m with ϑ(ξ)^{2} as generator. Ω̇^{∘}(ξ)/Ẏ (ξ) is isomorphic to

\frac{H_{n}}{H_{n}^{2m}} × \frac{H_{q} × H_{p}}{F} where F = ⟨ τ_{p}^{h_{q}/m} × τ_{q}^{−h_{p}/m} ⟩

is cyclic of order m/m′′, and the isomorphism is given via the standard identification H ≃ ΔH/Δ^{2}H.

Proof. Consider the H_{n}-modules

M = (2ZZ + ΔH_{n}) ⊕ ΔH_{q} ⊕ ΔH_{p} ,
N = (2mΔH_{n} + Δ^{2}H_{n}) ⊕ Δ^{2}H_{q} ⊕ Δ^{2}H_{p} , (1)

with H_{n} acting diagonally on the three direct summands in both cases. To explore the group Ω̇(ξ)/Ẏ (ξ), we shall find it expedient to work modulo N rather than the more natural N^{♮} = Δ^{2}H_{n} ⊕ Δ^{2}H_{q} ⊕ Δ^{2}H_{p}. It will presently be seen that ẇ(N) = ẇ(N^{♮}) = Ẏ (ξ).

Since n = pq has just 2 prime factors, Ennola assures us that the kernel R of the standard surjection ẇ : M → Ω̇(ξ) is generated over ZZH_{n} by

r_{q} = ΣG̃_{q} ⊕ 0 ⊕ (1 − τ_{q}) and r_{p} = ΣG̃_{p} ⊕ (1 − τ_{p}) ⊕ 0 .  (2)

To obtain a system of generators of R over ZZ, we note that τr_{q} = r_{q} for any τ ∈ G̃_{q}, because the latter acts trivially on ΔH_{p}. Hence we can write σr_{q} for any σ ∈ H_{p} ≃ H_{n}/G̃_{q} without risk of confusion. Upshot: {σr_{q} | σ ∈ H_{p}} ∪ {ρr_{p} | ρ ∈ H_{q}} is a basis-plus-one for R over ZZ, and so is

{(σ − 1)r_{q} | σ ≠ 1} ∪ {(ρ − 1)r_{p} | ρ ≠ 1} ∪ {r_{p}, r_{q}} .  (3)

The first two pieces of this union lie in N — one of the reasons for using the latter — and R+N = ⟨r_{p}, r_{q}⟩+N. We now perform a change of basis in the ZZ-module ZZr_{p} + ZZr_{q}. Defining f, g ∈ R by the matrix equation

[f, g] = [r_{p}, r_{q}] [ \begin{matrix}h_{q}/m & k_{q} \\ −h_{p}/m & k_{p} \end{matrix} ]  (4)

with h_{q} and h_{p} chosen such that h_{q}k_{p} + h_{p}k_{q} = m, we clearly get ZZf + ZZg = ZZr_{p} + ZZr_{q}. To see the advantage of these new generators, let us rewrite r_{p} and r_{q}. Since (τ^{−1} − 1) + (τ − 1) ∈ Δ^{2}H_{n} for all τ ∈ H_{n}, and since G̃_{q} has a unique element ι of order 2, we get r_{q} = [(q − 1) + (ι − 1) + δ_{q}] ⊕ 0 ⊕ (1 − τ_{q}) with δ_{q} ∈ Δ^{2}H_{n}. Analogously for r_{p}, with the same ι ∈ G̃_{q} ∩ G̃_{p}, whence

\begin{aligned}f & = [\frac{h_{q} − h_{p}}{m} (ι − 1) + δ_{∗}] ⊕ \frac{h_{q}}{m} (1 − τ_{p}) ⊕ \frac{h_{p}}{m} (τ_{q} − 1) , \\ g & = [2m + k(ι − 1) + ε_{∗}] ⊕ k_{q}(1 − τ_{p}) ⊕ k_{p}(1 − τ_{q}) , \end{aligned}  (5)

where k = k_{q} + k_{p} and δ_{∗}, ε_{∗} ∈ Δ^{2}H_{n}. What’s the point? For one thing, since gΔH_{n} ≡ 2mΔH_{n} modulo N^{♮}, we see that ẇ indeed maps N to Ẏ (ξ). Secondly, f lies in M^{∘} = ΔH_{n} ⊕ ΔH_{q} ⊕ ΔH_{p}, and even generates R^{∘} = R ∩ M^{∘} modulo N. In the exact sequence

0 → \frac{R^{∘} + N}{N} → \frac{M^{∘}}{N} → \frac{M}{R + N} → \frac{M}{R + M^{∘}} → 0 ,  (6)

whose third (non-zero) term is isomorphic to Ω̇(ξ)/Ẏ (ξ), while the second term is the easily recognized

\frac{M^{∘}}{N} ≃ \frac{H_{n}}{H_{n}^{2m}} × H_{q} × H_{p} ,  (7)

the two outer terms are now easily computable. The one on the right, for instance, is cyclic of order m (generated by 2 ⊕ 0 ⊕ 0 ∈ M), because R + M^{∘} = ⟨g⟩ + M^{∘} = ⟨2m⟩ + M^{∘}.

The first term on the left of (6) is also cyclic since R^{∘} + N = ⟨f⟩ + N. Its generator can be simplified to

f_{0} = 0 ⊕ \frac{h_{q}}{m} (1 − τ_{p}) ⊕ \frac{h_{p}}{m} (τ_{q} − 1) ,  (8)

(a non-relation) by showing that f − f_{0} ∈ N. Since 2(ι − 1) ∈ Δ^{2}H_{n}, the only interesting case is where exactly one of h_{q}/m and h_{p}/m — say, the latter — is even. Then 2m divides h_{p}, and ι ∈ G̃_{p}^{2m} implies (ι − 1) ∈ 2mΔH_{n} modulo Δ^{2}H_{n}. Hence R^{∘} + N = ⟨f_{0}⟩ + N, and the left hand term of (6) is mapped by the isomorphism (7) onto the cyclic group F ⊆ H_{q} × H_{p} described in the proposition.

To compute the order of F, note that (τ_{p}^{h_{q}/m})^{m/μ} = τ_{p}^{h_{q}/μ} is trivial in H_{q} if and only if μ is a divisor of [H_{q} : ⟨τ_{p}⟩], since the order of τ_{q} equals h_{q} divided by that index. Hence the smallest m/μ killing F is the one with μ = m′′, the greatest common divisor of [H_{q} : ⟨τ_{p}⟩] and [H_{p} : ⟨τ_{q}⟩]. End of proof.

To round out the picture, let us note that H_{n}/H_{n}^{2m} has order 2m^{2}, i.e., half the order of G_{n}/G_{n}^{2m}. This is because the involution ∗ is not in G_{n}^{2m}, since −1 cannot be a (2m)-th power in both F_{q}^{×} and F_{p}^{×}, as 4m cannot divide both q − 1 and p − 1.

2. According to Proposition 1, the index [Ω̇(ξ) : Ẏ (ξ)] equals 2m^{2}m′′h_{q}h_{p}, and of course so does [Ω(ξ) : Y (ξ)] if Y (ξ) contains the torsion element −1. Otherwise the undotted index is twice as big as the dotted one. Since −1 lies in Ω^{∘}(ξ), we may restrict our attention to [Ω^{∘}(ξ) : Y (ξ)].

We now leave our neat torsion-free world and consider the surjection w : M̂ → Ω(ξ), where M̂ = (2ZZ + ΔG_{n}) ⊕ ΔG_{q} ⊕ ΔG_{p}. It yields an isomorphism onto Ω^{∘}(ξ)/Y (ξ) from the last term of the exact sequence

0 → \frac{R̂^{∘} + N̂}{R_{tor} + N̂} → \frac{M̂^{∘}}{R_{tor} + N̂} → \frac{M̂^{∘}}{R̂^{∘} + N̂} → 0 ,  (9)

with N̂ = (2mΔG_{n} + Δ^{2}G_{n}) ⊕ Δ^{2}G_{q} ⊕ Δ^{2}G_{p}, and with R_{tor} denoting the kernel of w restricted to

M_{tor} = (∗_{n} − 1)ZZG_{n} ⊕ (∗_{q} − 1)ZZG_{q} ⊕ (∗_{p} − 1)ZZG_{p} ,  (10)

the kernel of standard map M̂ → M. Note that M_{tor} is a submodule of M̂^{∘}, and that (9) is a lifting of part of (6). We shall prove that the middle term of (9) is twice as large as its counterpart in (6), and that the same holds for the first term if and only if a certain integer i_{p,q} is odd. Remember that τ_{p} now denotes an element of G_{q} instead of H_{q} — and so on. That is, we are abusing the same letters for elements of G_{n}, G_{q} and G_{p} which had formerly labelled their images in H_{n}, H_{q} and H_{p}.

Proposition 2: Y (ξ) contains −1 if and only if the integer

i_{p,q} = \frac{h_{q}h_{p}}{mm′′} + \frac{[G_{q} : ⟨τ_{p}⟩]}{m′′} + \frac{[G_{p} : ⟨τ_{q}⟩]}{m′′} is odd.

Proof. To compute the size of middle term in (9), we start by observing that M̂^{∘}/N̂ is isomorphic to (G_{n}/G_{n}^{2m}) × G_{q} × G_{p}, a group of order 16m^{2}4h_{q}h_{p}. Next we need a handle on R_{tor} + N̂. Since each of the generators (∗_{d} − 1) of M_{tor} goes to −1, and since the map w is G_{n}-equivariant, we get

w : (∗_{n} − 1)α ⊕ (∗_{q} − 1)β ⊕ (∗_{p} − 1)γ ⟼ (−1)^{a+b+c} ,  (11)

where a, b, c are the augmentations of α ∈ ZZG_{n}, β ∈ ZZG_{q}, and γ ∈ ZZG_{p}. Since 2(∗_{q} − 1) ∈ Δ^{2}G_{q} and 2(∗_{p} − 1) ∈ Δ^{2}G_{p}, this shows

R_{tor} + N̂ = ⟨s_{p}, s_{q}⟩ + N̂ ,  (12)

with s_{p} = (∗_{n} − 1) ⊕ 0 ⊕ (∗_{p} − 1) and s_{q} = (∗_{n} − 1) ⊕ (∗_{q} − 1) ⊕ 0. Since s_{p}, s_{q}, and s_{p} + s_{q} all lie outside N̂, they generate a group of order 4 in M̂^{∘}/N̂, whence

[M̂^{∘} : R_{tor} + N̂ ] = 4m^{2}h_{q}h_{p} .  (13)

Now we are ready to tackle the first term of (9). For the full kernel R̂ of the map w : M̂ → Ω(ξ), we have

R̂ + N̂ = ⟨r̂_{p}, r̂_{q}⟩ + R_{tor} + N̂ ,  (14)

where r̂_{q} = ΣG_{q} ⊕ 0 ⊕ (1 − τ_{q}) = [(q − 1) + (ι_{q} − 1) + δ_{q}] ⊕ 0 ⊕ (1 − τ_{q}) is the obvious lift of our old r_{q}, with G_{q} denoting G_{q} × 1 ⊆= G_{q} × G_{p} = G_{n}, and ι_{q} standing for the element ∗_{q} × 1 in that subgroup. Similarly — mutatis mutandis — for r̂_{p}. It is important that these are actual relations (not just modulo torsion) — as was verified is Eq.(14) of the pq-paper.

We again apply the change of basis (4), to get ZZr̂_{p} + ZZr̂_{q} = ZZf̂+ ZZĝ. Since ĝ lies outside M̂^{∘}, we can concentrate on

f̂ = [ − \frac{h_{p}}{m} (ι_{q} − 1) + \frac{h_{q}}{m} (ι_{p} − 1) + δ_{∗}] ⊕ \frac{h_{q}}{m} (1 − τ_{p}) ⊕ \frac{h_{p}}{m} (τ_{q} − 1) ,  (15)

which we shall now simplify modulo N̂ by means of the congruence

[ − \frac{h_{p}}{m} (ι_{q} − 1) + \frac{h_{q}}{m} (ι_{p} − 1) + δ_{∗}] ≡ \frac{h_{p}h_{q}}{m^{2}} (∗_{n} − 1) (mod 2mΔG_{n} + Δ^{2}G_{n}) .  (16)

Let us take time out to establish this. Modulo Δ^{2}G_{n}, the δ_{∗} disappears and the coefficients of (ι_{q}−1), (ι_{p}−1), and (∗_{n} − 1) are reduced to 0 or 1 depending on parity. Since (ι_{q} − 1) + (ι_{p} − 1) = (∗_{n} − 1) modulo Δ^{2}G_{n}, all is well as long as h_{p}/m and h_{q}/m are either both even or both odd. If h_{p}/m is even and h_{q}/m is odd, the right hand side of (16) vanishes modulo Δ^{2}G_{n}, while the left is reduced to (ι_{p} − 1) ≡ h_{p}(σ − 1)2mΔG_{n}, where σ is a generator of G_{p} × 1 ⊆ G_{n}. This proves (16).

We are ready to pull the plug: R̂^{∘} + N̂ = ⟨f̂⟩ + R_{tor} + N̂ = ⟨f̂_{0}⟩ + R_{tor} + N̂, with
f̂_{0} = \frac{h_{p}h_{q}}{m^{2}} (∗_{n} − 1) ⊕ \frac{h_{p}}{m} (τ_{q} − 1) .  (17)

The first term in (9) is a cyclic group F̂ generated by the element f̂_{0}, which by Paragraph 1 has order m/m′′ modulo M_{tor}. But (m/m′′)f̂_{0} is congruent to

f̂_{!} = \frac{h_{p}h_{q}}{mm′′} (∗_{n} − 1) ⊕ \frac{[G_{q} : ⟨τ_{p}⟩]}{m′′}(1 − ∗_{q}) ⊕ \frac{[G_{p} : ⟨τ_{q}⟩]}{m′′}(∗_{p} − 1)  (18)

as one sees by writing τ_{p} as the [G_{q} : ⟨τ_{p}⟩]-th power of a suitable generator ρ of G_{q} and noting that ρ^{h_{q}} = ∗_{q}, with similar acrobatics in G_{p}. Hence the orders of F̂ and F are equal — and [Ω^{∘}(ξ) : Y (ξ)] = 2·[Ω̇^{∘}(ξ) : Ẏ (ξ)], i.e.,−1 ∉ Y (ξ) — if and only if f̂_{!} is trivial modulo R_{tor} + N̂. Otherwise F̂ has twice the order of F and −1 ∈ Y (ξ).

To finish the argument, we need the relation M_{tor} ∩ (R_{tor} + N̂) = R_{tor} + (M_{tor} ∩ N̂) = R_{tor}, which follows at once from M_{tor} ∩ N̂ ⊆ R_{tor}. Indeed an element a(∗_{n} − 1) ⊕ b(∗_{q} − 1) ⊕ c(∗_{p} − 1) with integer coefficients a, b, c cannot be in N̂ unless all these coefficients are even. This is obvious for b and c, but also holds for a because ∗_{n} ∉ G_{n}^{2m}, as noted at the end of Paragraph 1. Therefore f̂_{!} is trivial modulo R_{tor} + N̂ if and only if it is in R_{tor}, which means that w(f̂_{!}) = (−1)^{i_{p,q}} equals 1.