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Whitehead Group of the Iwasawa algebra of GL2(Zp)

Solanki, Vishal

Awarding institution:King's College London

Download date: 22. Jun. 2020

Whitehead Group of the Iwasawa algebra of GL2(Zp)

Vishal Solanki

Doctor Of Philosophy in Mathematics

June 30, 2018

Abstract

Main conjectures in Iwasawa theory are interesting because they give a deep connection between

arithmetic and analytic objects in number theory. One of the most important recent developments

in Iwasawa theory is the formulation of non-commutative main conjectures by Coates, Fukaya, Kato,

Sujatha and Venjakob using K1 groups. Burns and Kato supplied a strategy to prove these non-

commutative main conjectures. After important special cases were proved by Kato and Hara, the

non-commutative main conjecture for totally real fields was proved by Kakde using this strategy (it

was proved independently by Ritter-Weiss). In this thesis we imitate Kakde’s computation of K1

groups in order to obtain a description of the K1 group of the Iwasawa algebra of GL2(Zp). While

we do not find an explicit description of this group, we do define another group which must contain

this K1 group.

Contents

1 Introduction 3

1.1 The strategy of Burns and Kato . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Our choice of F for GL2(Zp) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Application of our results to Iwasawa theory . . . . . . . . . . . . . . . . . . . . . . . 5

2 Preliminaries 6

2.1 Iwasawa algebras and some localisations . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 K-theory of Iwasawa algebras and localisations . . . . . . . . . . . . . . . . . . . . . 9

2.3 Classical Iwasawa theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4 Reformation using K-theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.5 The GL2 main conjecture for elliptic curves . . . . . . . . . . . . . . . . . . . . . . . 12

2.5.1 The Selmer group of E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.5.2 p-adic L-function of E, LE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.5.3 The main conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.6 The goal of this paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.6.1 The strategy of Kato . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.6.2 Our strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Choosing Fn, a suitable set of subgroups of Gn 18

3.1 Computing the p-part of the torsion subgroup of K1(Zp[Gn]) . . . . . . . . . . . . . 18

3.2 Conjugacy classes of GL2(Z/pnZ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.3 Elements of order prime to p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Image of ψn 41

4.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.2 Image of ψn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.3 Trace maps for Gn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5 Constructing map L 64

5.1 The explicit construction of the map f . . . . . . . . . . . . . . . . . . . . . . . . . . 65

5.2 Obtaining an explicit description of L . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5.3 Finding the group Θn,Zp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6 Whitehead group of the localised algebra

Λ(Gn)T ′ 78

6.1 Inspecting the twist τ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6.2 Verifying that we can take log . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

2

Chapter 1

Introduction

In this thesis we describe the Whitehead group or the first K-group, K1, of the Iwasawa algebra

of GL2(Zp) for odd prime p. We are interested in this K1 group because of its appearance in

non-commutative Iwasawa theory for elliptic curves without complex multiplication1. Let E be an

elliptic curve defined over Q. Put K∞ =⋃n≥1 Q(E[pn]) and G = Gal(K∞/Q). If E admits complex

multiplication by an order in an imaginary quadratic field F , then Gal(K∞/F ) is abelian. In this

case a main conjecture for E and the extension K∞/F can be formulated using the structure theory

for finitely generated modules over the Iwasawa algebra of K∞/F . Indeed this main conjecture was

proved in Rubin ([16] , [22]).

On the other hand, if E does not admit complex multiplication, then by a celebrated theorem of

Serre [18], G is an open subgroup of GL2(Zp). In fact, for all sufficiently large prime p, G is equal to

GL2(Zp). However, the whole approach of classical Iwasawa theory breaks down due to lack of good

structure theory for finitely generated modules over the Iwasawa algebra of GL2(Zp). Venjakob

[20] and Coates, Fukaya, Kato, Sujatha and Venjakob [4] bypassed structure theory by formulating

the main conjecture using algebraic K-theory. Thus it is essential to study K1 groups of Iwasawa

algebras of non-commutative p-adic Lie groups. Kato’s seminal paper [13] provided a strategy for

computing K1 of the Iwasawa algebra of a p-adic Lie group G using Iwasawa algebras of abelian

sub-quotients of G by working out a specific example of G ∼= Zp o Z×p . This result was generalised

independently by Kakde [11] and Ritter-Weiss ([15], [21]) after special cases were worked out by

Kato [13], Kakde [10] and Hara ([8], [9]). These computations of K1 groups of Iwasawa algebras and

certain localisations show that, in order to prove the non-commutative main conjecture, we must

prove “several” commutative main conjectures and prove certain congruences between commutative

p-adic L-functions (such a strategy usually assumes vanishing of certain µ-invariant but we will not

discuss this here). This is the so called Burns-Kato strategy for proving the non-commutative main

conjecture.

The computation of K1 by Ritter-Weiss requires working with all abelian sub-quotients. The

computation of K1 by Kato, generalized by Kakde, has an advantage that we do not necessarily

need all abelian sub-quotients. In this thesis we use this observation and describe K1 of the Iwasawa

algebra of GL2(Zp) by using abelian sub-quotients which come from “well-known” subgroups such

as Borel, Cartan, etc.

1An elliptic curve E admits complex multiplication if it has an endomorphism ring larger than the integers;

the endomorphism ring is a set of complex numbers which map the lattice, of the elliptic curve, to a subset of the

lattice. An elliptic curve with complex multiplication is one with endomorphism ring isomorphic to an imaginary

quadratic extension of the integers.

3

Let us now describe our results in more detail. From now on we assume p is odd.

1.1 The strategy of Burns and Kato

Let G be a p-adic Lie group. We put Λ(G) = Zp[[G]] := lim←−U

Zp[G/U ], for a pro-finite group G, where

U runs through open normal subgroups of G. If U is an open subgroup of G and V is a closed normal

subgroup of U such that U/V is abelian, then there are maps

θU,V : K1(Λ(G))→ K1(Λ(U))→ K1(Λ(U/V )) ∼= Λ(U/V )×

where the first map is the norm map2and the second map is induced by the natural projection

Λ(U))→ Λ(U/V ). Now let F be a collection of pairs (U, V ) as above. Then we have a map:

θF =∏

(U,V )∈F(θ(U,V ))(U,V )∈F : K1(Λ(G))→

∏(U,V )∈F

Λ(U/V )

The idea is to study the kernel and the image of θF .

1.2 Our choice of F for GL2(Zp)

We use the isomorphism

GL2(Zp) = lim←−n

GL2(Z/pnZ)

Using a result of Fukaya-Kato ([7], see ([11], Lemma 4.1)) we get

K1(Λ(GL2(Zp))) ∼= lim←−n

K1(Zp[GL2(Z/pnZ)])

We therefore study K1 of the group ring Zp[GL2(Z/pnZ)]. Consider the set of subgroups of

GL2(Z/pnZ) (these subgroups are defined in Chapter 3):

Fn = Zn, Cn, Tn,Kn, Nti , Nki |∀i = 1, 2, ..., n− 1

Consider the map

θn =∏

U∈Fn

(θ(U,[U,U ]))U∈Fn : K1(Zp[GL2(Z/pnZ)])→∏

U∈Fn

Zp[Uab]×

In fact, all subgroups in Fn are abelian already, so Zp[Uab]× = Zp[U ]× for each U ∈ Fn.

In Chapter 5 we prove that image of θn is contained in an explicitly defined subgroup Θn,Zpof∏U∈Fn Zp[U ]× (see Theorem 5.1). Unfortunately, we are unable to show that the image of θn

is exactly Θn,Zp . This is due to our lack of knowledge of the kernel and cokernel of the map Ldefined in Chapter 5. We also prove a similar result about twisted group rings R[GL2(Z/pnZ)]τ

for a specific ring R. This twisted group ring is the localisation of the Iwasawa algebra of a one

dimensional quotient of GL2(Zp) at the canonical Ore set of Coates, Fukaya, Kato, Sujatha and

Venjakob (for details see Section 2.1).

2To define this norm, we first notice that Λ(G) is a free Λ(U)-module of rank d = [G : U ], i.e. Λ(G) ∼= Λ(U)d.

So there is a map GLn(Λ(G))→ GLnd(Λ(U)) which induces the norm map K1(Λ(G))→ K1(Λ(U)).

4

1.3 Application of our results to Iwasawa theory

The main result (Theorem 5.1) states the following:

The image of θn is contained in Θn,Zp which is defined using the following conditions:

1. Θn,Zp ⊂∏U∈Fn Zp[U ]× such that xpZn

(λL,Zn ((xV )V ∈Fn )

)≡ ϕ(xZn ) (mod p3n−1)

2. For any (xV )V ∈Fn ∈ Θn,Zp , each xV is fixed by conjugation action of NGn (V )

3. For any (xV )V ∈Fn ∈ Θn,Zp , we have:

• NmV/Zn (xV ) = xZn for all V ∈ Fn

• 1plog

(NmU/Zm∩U

(xpUϕ(NU (xU ))p

2λL,U ((xV ))

ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )

))+TrU/Zm∩U

(µL,U (xU )

∑β∈(Z/piZ)×

rcn−i1,β

)= 1

plog

(NmCn/Zm∩Cn

(xpCn

NCn (ϕ(xCn ))pλL,Cn ((xV ))

ϕ(xCn )pϕ(NCn (xCn ))

))for U ∈ Nti , Nki and

m ≥ n− i

4. For any (xV )V ∈Fn ∈ Θn,Zp , we have:

• NmCn/Zm∩Cn(xpCnNCn (ϕ(xCn ))p

(λL,Cn ((xV )V ∈Fn )

))≡ NmCn/Zm∩Cn (ϕ(xCn )pϕ (NCn (xCn ))) (mod p3m+1)

• NmU/Zm∩U(xpUNU (ϕ(xU ))

(λL,U ((xV )V ∈Fn )

))≡ NmU/Zm∩U (ϕ(xU )ϕ(NU (xU ))) (mod p3m) for U ∈ Tn,Kn

• NmU/Zm∩U(xpUϕ(NU (xU ))p

2 (λL,U ((xV )V ∈Fn )

))≡ NmU/Zm∩U

(ϕ(xU )p

2ϕ(NU (xU ))

(νL,U (xCn )

))(mod p3m+1) for U ∈ Nti , Nki

For the definition of the maps λL,U , µL,N and νL,N , please refer to Definition 5.1.

Our algebraic result predicts certain congruences between abelian p-adic L-functions of ellip-

tic curves. Proving these congruences seems to be extremely hard at present. However, it may

be possible to numerically verify these and thus provide evidence for the non-commutative main

conjecture from [4].

5

Chapter 2

Preliminaries

2.1 Iwasawa algebras and some localisations

Let p be an odd prime number and G be a compact p-adic Lie group. We assume that G contains

a closed normal subgroup H such that G/H = Γ is isomorphic to Zp, the additive group of p-adic

integers. Define Λ(G) to be the Iwasawa algbra of G with coefficients in Zp:

Λ(G) = Zp[[G]] := lim←−U

Zp[G/U ]

where U runs through open normal subgroups of G.

We recall the canonical Ore set of [4]1:

T ′ := λ ∈ Λ(G) | Λ(G)/Λ(G)λ is a finitely generated Λ(H)- module

Following [4] put

T :=⋃i≥0

piT ′

It is proven in ([4], Theorem 2.4) that T ′ and T are multiplicatively closed subsets of Λ(G), do not

contain zero divisors and satisfies the Ore-conditions2 (both left and right). Consequently we can

localise Λ(G) with respect to T ′ and T and obtain inclusions:

Λ(G) → Λ(G)T ′ → Λ(G)T

Our aim is to study K1(Λ(G)), K1(Λ(G)T ′ ) and K1(Λ(G)T ) for G = GL2(Zp).

From now on we put G = GL2(Zp) and Gn = GL2(Z/pnZ). We also put H = SL2(Zp)

(Note that G/H ∼= Z×p ).

To study the localisation Λ(G)T ′ we write Λ(G) as an inverse limit of Iwasawa algebras of

one dimensional quotients of G and then show that the corresponding localisation of these Iwasawa

algebras is easy to study.

1In [4] the Ore set are denoted by S and S∗

2Ore-condition basically means that all right fractions with denominator in T can be written as left fractions

with denominator in T , and vice-versa.

6

Put Hn = SL2(Z/pnZ) and put Gn to be the quotient of G that makes the following dia-

gram commute: In other words Gn = G/ker(H → Hn).

1 → H → G → Z×p → 1

↓ ↓ =↓1 → Hn → Gn → Z×p → 1

Then G ∼= lim←−n

Gn.

Lemma 2.1 There exists an open central subgroup Γn in Gn such that Gn/Γn ∼= Gn.

Proof:

Let Kn = ker(G → Gn), and let Γn = Kn/ker(H → Hn). By the third isomorphism theorem,

Gn/Γn ∼= G/Kn ∼= Gn. Also notice that Γn is central in Gn since is it isomorphic to 1 + pnZp.

The lemma allows us to express the Iwasawa algebra Λ(Gn) as a twisted group ring. We recall the

definition of twisted group rings.

Definition 2.1 Let R be a ring and P be any finite group. Let

τ : P × P → R

be a 2-cocycle3. Then the twisted group ring, denoted by R[P ]τ , is a free R-module generated by

P . Denote the image of h ∈ P in R[P ]τ by h. Therefore every element in R[P ]τ can be written as∑h∈P rhh. The addition is component-wise and the multiplication has the following twist

h · h′ = τ(h, h′)hh′

Example: For us the most important example of twisted group rings comes as follows:

Let Q be a finite group with central subgroup Z such that Q/Z ∼= P . Choose a section s : P → Q

(this need not be a homomorphism but the identity must map to the identity). Then τ : P ×P → Z

defined by

τ(p1, p2) = s(p1)s(p2)s(p1p2)−1

is a 2-cocycle. Then for any ring A we have:

A[Z][P ]τ∼=−→ A[Q]∑

x∈Paxx 7→

∑x∈P

axs(x)

where ax lies in A[Z]. This map is an isomorphism.

Proof:

The map is clearly bijective but we must still prove that it is a homomorphism:∑x∈P axx 7→

∑x∈P axs(x) and

∑x∈P bxx 7→

∑x∈P bxs(x).

(∑x∈P axx)(

∑x∈P bxx)

=∑x∈P

( ∑x=yz

aybzτ(y, z)

)x 7→

∑x∈P

( ∑x=yz

aybzτ(y, z)

)s(x)

=∑x∈P

( ∑x=yz

aybzs(y)s(z)s(x)−1

)s(x))

= (∑x∈P axs(x))(

∑x∈P bxs(x))

3A map, τ , is a 2-cocycle if it satisfies the condition τ(p1, p2)τ(p1p2, p3) = (p1 ∗ τ(p2, p3)) τ(p1, p2p3)

7

Lemma 2.2 This is an isomorphism

Zp[[Gn]]∼=−→ Zp[[Γn]][Gn]τ

Furthermore, we may choose τ such that, for any A,B ∈ Gn, τ(A,A−1) = 12 = τ(A,12) and

τ(A,B) = τ(B,A).

Proof:

Since Gn/Γn ∼= Gn, we have the isomorphism. We define τ : Gn ×Gn → Γn in the following way

τ(X1, X2) = s(X1)s(X2)s(X1X2)−1

where s is any section from s : Gn → Gn. Any section is fine because we will get an element in Γn.

Let A,B ∈ Gτn, since A ·A−1 = 12 and A ·12 = A in Gn, by isomorphism, we also have A ·A−1 = 12

and A · 12 = A. Therefore τ(A,A−1) = 12 = τ(A,12).

By the way we have defined Γn, elements in Γn have unique determinants. Since det(AB) = det(BA)

in Gn and det(AB) = det(BA) in Gτn, we must also have det(τ(A,B)) = det(τ(B,A)) but τ maps

to Γn so we must have τ(A,B) = τ(B,A).

As before we define the canonical Ore set denoted again by T ′, in Λ(Gn) by

T ′ := λ ∈ Λ(Gn) | Λ(Gn)/Λ(Gn)λ is a finitely generated Zp-module

We have the following more convenient description of Λ(Gn)T ′ :

Lemma 2.3 ([10], Lemma 2.1) The set T = Λ(Γn) − pΛ(Γn) is a multiplicatively closed, left and

right Ore subset of Λ(Gn). The natural injection Λ(Gn)T → Λ(Gn)T ′ is an isomorphism.

Using this lemma we obtain the following result

Lemma 2.4 Let Λ(Gn)T ′ and Λ(Γn)T denote p-adic completions. Then the natural map

Λ(Γn)T [Gn]τ∼=−→ Λ(Gn)T ′

is an isomorphism.

Proof:

By using lemma 2.2 we have the following:

Zp[[Gn]] ∼= Zp[[Γn]][Gn]τ

By completing both sides and localizing we get:

( Λ(Γn)[Gn]τ )T ′ ∼= Λ(Gn)T ′

We use the lemma 2.3 and the fact the Gn is finite to get the following:

Λ(Γn)T [Gn]τ∼=−→ Λ(Gn)T ′

8

2.2 K-theory of Iwasawa algebras and localisations

In this section we will first define K0 for rings and categories of certain modules, and then define K1

for rings.

Definition 2.2 For any ring R, K0(R) is the Abelian group generated by elements [P ], where P is

a finitely generated projective R-module, with the following relations:

• if P ′ is isomorphic to P as R-modules, then [P ] = [P ′]

• if P = P ′ ⊕ P ′′ then [P ] = [P ′] + [P ′′]

Definition 2.3 For any ring homomorphism f : R → R′, K0(f) is the Abelian group generated

by elements [P, g,Q], where P and Q are a finitely generated projective R-module and g is any

isomorphism between R′ ⊗R P and R′ ⊗R Q as R′-modules. We have the following relations:

• if there exist hP : P∼=−→ P ′ and hQ : Q

∼=−→ Q′ such that g′ (idR′ ⊗ hP ) = (idR′ ⊗ hQ) g,

then [P, g,Q] = [P ′, g′, Q′]

• if g = g′ g′′ such that g′′ is any isomorphism between R′ ⊗R P and R′ ⊗R O and g′ is any

isomorphism between R′ ⊗R O and R′ ⊗R Q, then [P, g,Q] = [P, g′′, O] + [O, g′, Q]

• if there are three elements [P, g,Q], [P ′, g′, Q′] and [P ′′, g′′, Q′′] such that we have the following

short exact sequences compatible with g, g′ and g′′

0→ P ′ → P → P ′′ → 0 and 0→ Q′ → Q→ Q′′ → 0

then [P, g,Q] = [P ′, g′, Q′] + [P ′′, g′′, Q′′]

We write K0(R,R′) for K0(f) if f is a canonical injection from R to R′.

To define K1, we first need to define GL(R), the infinite general linear group over

R:

We define GL(R) as⋃n>0GLn(R) where we say GLn(R) ⊂ GLm(R) for n < m with the following

inclusion:

A ∈ GLn(R) =⇒(A 0

0 1m−n

)∈ GLm(R)

where 0 are zero matrices and 1n is the n by n identity matrix.

[GL(R), GL(R)] is called the commutator subgroup of GL(R), it is generated by the set

ABA−1B−1 | A,B ∈ GL(R).

Definition 2.4

K1(R) :=GL(R)

[GL(R), GL(R)]

In other words, K1(R) is the abelianization of the infinite general linear group.

In our case we have a surjective map (Λ(G))× K1(Λ(G)) ([4], Theorem 4.4).

Let I be an ideal of R, then GL(R, I) is the group of invertible matrices which are congru-

ent to the identity matrix modulo I. Now let E(R, I) denote the smallest normal subgroup

of GL(R) which contain all elementary matrices4 which are congruent to the identity mod-

ulo I. Set K1(R, I) := GL(R, I)/E(R, I). By Whitehead’s lemma ([14], Theorem 1.13)

E(R, I) = [GL(R), GL(R, I)], therefore K1(R, I) is abelian.

4Elementary matrices differ from the identity matrix by changing one of the zero entries to some r ∈ R

9

In the case that we have a finite group G, we can use the definition SK1(A[G]) := ker(K1(A[G])→K1(Q(A)[G])) where A is a Dedekind domain with Q(A) as its field of fractions. In the case that

the group G is pro-finite, G = lim←−U

U , we use the following:

SK1(A[[G]]) := lim←−U

SK1(A[U ])

Definition 2.5 For a Dedekind domain A and a pro-finite group G, we have

K′1(A[[G]]) := K1(A[[G]])/SK1(A[[G]])

Recall that G = GL2(Zp). From now on we put Λ = Λ(G).

Let ∂ be defined as the map in the following exact sequence ([19], Theorem 15.5):

K1(Λ)→ K1(ΛT ′ )∂−→ K0(Λ,ΛT ′ )→ K0(Λ)→ K0(ΛT ′ )→ 0

∂ : [f ] 7→ [Λn, f ,Λn]

where f ∈ GLn(Λ) such that f lifts to f ∈ GL(Λ).

It turns out that K0(Λ,ΛT ′ ) maps to 0 in K0(Λ) ([4], Proposition 3.4), thus, by exactness of the

sequence, ∂ is surjective. As G has no p-torsion, we have the following ([2], Proposition 3.4)

K1(ΛT ) ∼= K1(ΛT ′ )⊕K0(ΛT ′ ,ΛT )

and also that K0(ΛT ′ ,ΛT ) ∼= Zr for some r ≥ 0.

2.3 Classical Iwasawa theory

In this section we recall formulations of main conjectures using structure theory when G ∼= Zdp. For

the rest of this section we set G ∼= Zdp. Then Λ(G) is non-canonically isomorphic to the power series

ring in d variables over Zp. Fix d elements γ1, γ2, ..., γd ∈ G that topologically generate G. Then

Λ(G)∼=−→ Zp[[T1, T2, ..., Td]]

γi 7→ Ti + 1

The following is the structure theorem for finitely generated modules over Λ(G):

Theorem 2.1 [1] Let X be a finitely generated Λ(G)-module, then there is a map

X →⊕i

Λ(G)/(fi)⊕ Λ(G)r

where the kernel and cokernel are pseudo-null, i.e. they are annihilated by height two ideals of Λ(G).

Definition 2.6 A finitely generated Λ(G)-module X is pseudo-isomorphic to Y if there exists a

map

X → Y

with pseudo-null kernel and cokernel.

A finitely generated Λ(G)-module X is called torsion if for every x ∈ X, there exists f ∈ Λ(G)\0such that f · x = 0. In other words, X is a torsion Λ(G)-module if Q(Λ(G)) ⊗Λ(G) X = 0, where

Q(Λ(G)) is the total ring of fractions of Λ(G).

In the category of finitely generated torsion Λ(G)-modules, being pseudo-isomorphic is an

10

equivalence relation. If X is a finitely generated torsion Λ(G)-module, then there is a pseudo-

isomorphism

X →⊕i

Λ(G)/(fi)

for some fi in Λ(G).

Definition 2.7 Define the characteristic ideal

charΛ(G)(X) = (∏i

fi)Λ(G)

where the elements fi are defined as above.

In classical formulations of main conjectures in Iwasawa theory, one has interesting arithmetic objects

X (such as ideal class groups, or Selmer groups) which are torsion or are conjectured to be torsion

over Λ(G). Thus we can attach an arithmetic invariant charΛ(G)(X) to it. The main conjecture may

be stated as saying that there is a canonical generator L of the principal ideal charΛ(G)(X) whose

“evaluations” at various continuous representations of G are related to L-values. We will not make

this precise here. Let us only recall what evaluations mean. Let

ρ : G→ Qp×

be a continuous homomorphism. Then it extends to a map

ρ : Λ(G)→ Qp

This is classically denoted as µ 7→∫G ρ dµ.

We call it evaluation of µ at the continuous representation ρ.

More information about the classical case can be found in Washington [23].

2.4 Reformation using K-theory

In [20] and [4] main conjectures are reformulated without using structure theory but by using

algebraic K-groups, specifically K0 and K1 groups. We continue to assume G ∼= Zdp. Recall the

lower terms of K-theory localisation sequence ([19], Theorem 15.5):

K1(Λ(G))→ K1(Q(Λ(G)))∂−→ K0(Λ(G),Q(Λ(G)))→ 0

A finitely generated torsion Λ(G)-module gives a class [X] in K0(Λ(G),Q(Λ(G))). Any element

ξ ∈ K1(Q(Λ(G))) is a called a characteristic element of X if ∂(ξ) = [X]. This characteris-

tic element is well-defined up to multiplication by an element in Λ(G)×. By ([20], Remark 6.2),

we know that this definition of characteristic element is a generalized version of the one stated above.

Now let G be any p-adic Lie group. The localisation sequence for Λ(G) and Q(Λ(G)) exists

and one may use it to formulate main conjectures. For many technical reasons (as explained in

[20]) it is better to restrict to G having a closed normal subgroup H such that G/H = Γ ∼= Zp and

working with a smaller localisations Λ(G)T ′ or Λ(G)T . Here

T ′ := λ ∈ Λ(G) | Λ(G)/Λ(G)λ is a finitely generated Λ(H)- module

and T :=⋃i≥0 p

iT ′.As proven in [4], T ′ and T are multiplicatively closed subset of Λ(G), they do not contain zero divisors

and they satisfy the Ore condition (left and right). Furthermore, we have localisation sequences:

K1(Λ(G))→ K1(Λ(G)T ′ )∂−→ K0(Λ(G),Λ(G)T ′ )→ 1

11

K1(Λ(G))→ K1(Λ(G)T )∂−→ K0(Λ(G),Λ(G)T )→ 1

Let T ∈ T ′, T . If X is a finitely generated Λ(G)-module which is T -torsion5 then X gives a

class [X] in K0(Λ(G),Λ(G)T ). Any element ξ in K1(Λ(G)T ) such that ∂(ξ) = [X] is called a

characteristic element of X. In this setting the main conjecture simply gives a characteristic element

whose “evaluations” are related to L-values. Let us explain evaluations in this setting. Let

ρ : G→ GLn(O)

be a continuous homomorphism, for the ring of integers O in a finite extension L of Qp.

Then this extends to a map from K1(Λ(G)T ′ ) to L⋃∞; we will state this map after we define

the augmentation map:

ϕ : ΛO(Γ)p → L

ϕ :∑

xgg 7→∑

xg

where ΛO(Γ) = O[[Γ]] and p is the kernel of the augmentation map from ΛO(Γ) to O. The map ρ

extends to the following map:

ξ(ρ) :=

ϕ(Φρ(ξ)) if Φρ(ξ) ∈ ΛO(Γ)p

∞ if Φρ(ξ) 6∈ ΛO(Γ)p

We will define Φρ now:

Let Q(ΛO(Γ)) be the field of fractions of ΛO(Γ), then by ([4], Lemma 3.3), we have a ring homo-

morphism Λ(G)T ′ → Mn(Q(ΛO(Γ))) defined by∑xgg 7→

∑xg(ρ(g) ⊗ g) where g is the image of

g ∈ G under the projection of G→ Γ. This homomorphism induces the following homomorphism:

Φρ : K1(Λ(G)T ′ )→ K1(Mn(Q(ΛO(Γ)))) = (Q(ΛO(Γ)))×

Following the classical notation we may denote the map as

µ 7→∫Gρ dµ = ξ(ρ)

Remark: Here we remark that if G ∼= Zp then Λ(G)T ′ = Q(Λ(G)). Furthermore any finitely

generated Λ(G)-module has a finite projective resolution.

Instead of X as above, we may even take a perfect complex C• of Λ(G)-modules whose co-

homologies are T -torsion i.e. Λ(G)T⊗L

Λ(G) C• is acyclic.

2.5 The GL2 main conjecture for elliptic curves

In this chapter we will state the GL2 main conjecture for elliptic curves, but first we need to define

many objects we will need:

Kn := Q(E[pn]), K∞ :=⋃n≥1

Kn, G = Gal(K∞/Q)

Λ := Λ(G) = Zp[[G]] := lim←−n

Zp[Gal(Kn/Q)]

5A module X is T -torsion if ∀x ∈ X, there exists t ∈ T such that t · x = 0.

12

2.5.1 The Selmer group of E

In this chapter we want to define Sel(E/K∞), the p-Selmer group of E over K∞. Let L be an

algebraic extension of Q, Lν is the completion of L at a place ν and Lν is an algebraic closure

of Lν . Also, E(Lν) are the Lν -rational points and Ep∞ =⋃n>0 E[pn]. Finally, H1(L,Ep∞ ) :=

H1(GL, Ep∞ ) is the 1st Galois cohomology group such that GL is the absolute Galois group of L;

the action of GL on Ep∞ is the action induced by the natural action of GL on Etors ∼= (Q/Z)2,

where Etors is the torsion subgroup of E(Q).

Sel(E/L) := ker((H1(L,Ep∞ )→∏ν

H1(Lν , E(Lν)))

where the product runs over all places of L. In our case, L = K∞ is an infinite extension of Q so

we define the completion to be Lν =⋃

(L′)ν where we union over the set of all finite extensions, L′

over Q, which are contained in L.

The Selmer group and the Tate-Shafarevich group6, Sha, can tell us about the correspond-

ing elliptic curve as one can see by the following exact sequence:

0→ E(L)⊗Z Qp/Zp → Sel(E/L)→ Sha(E/L)(p)→ 0

where Sha(E/L)(p) denotes the p-primary part of Sha(E/L).

X := X(E/K∞) = Sel(E/K∞)∨, this is the Pontryagin dual of the Selmer group

i.e. X(E/K∞) = Hom(Sel(E/K∞),Qp/Zp). It turns out that X(E/K∞) is finitely generated over

Λ ([5], Theorem 2.9).

2.5.2 p-adic L-function of E, LE

The p-adic L-function of E, LE , is a conjectural element of K1(ΛT ) and we expect it to be a

characteristic element of X. Before defining LE , we will define the L-function of E, L(E, ρ, s), where

ρ is an Artin representation7of G. Let A(G) be the set of Artin representations of G and let l and q

be two distinct prime numbers.

Let Il be the inertia group of GQl , the subgroup of GQl which fixes Zl/lZl, and Frobl be the Frobenius

automorphism of l in GQl/Il = Gal(Ql/Ql)/Il. Also let Kρ be a finite extension of Q such that we

have the vector space associated to ρ defined overKρ, and call this vector space Vρ. Also letH1q (E) :=

Hom(Tq(E)⊗Zq Qq ,Qq) such that Tq(E) := lim←−n E[qn] is the q-adic Tate module of E. Finally,

let λ be a prime of Kρ above q, then Pl(E, ρ, T ) := det(1−Frob−1l T ; (H1

q (E)⊗Qq (Vρ⊗K Kρ,λ))Il )

(we have used the following notation8; det(1− gT ; Vρ) = det(1− ρ(g)T ) where Vρ is the associated

vector space to the representation ρ). Then the L-function of E is defined by the following Euler

product:

L(E, ρ, s) =∏l

Pl(E, ρ, l−s)−1

where the product is over all primes l.

6Tate-Shafarevich group of E/L:

Sha(E/L) := ker((H1(L,E(L))→

∏ν

H1(Lν , E(Lν))))

7An Artin representation ρ : G → GLn(Zp) is a continuous representation such that ker(ρ) is open.8The reason we use the representation associated to a vector space which is fixed by Il, is because Frobl lives

in Gal(Qunraml /Ql) where Qunraml is the maximal unramified extension over Ql.

13

Let R = p⋃l prime | ordl(jE) < 0 where jE is the j-invariant9of E. We define LR(E, ρ, s) to

be the L-function with Euler factors removed of the primes in L, i.e:

LR(E, ρ, s) =∏l 6∈R

Pl(E, ρ, l−s)−1

Now we are ready to define LE :

Conjecture 2.1 ([2], Conjecture 7.3) Let M be the motive h1(E)(1). Assume that p ≥ 5 and that

E has good ordinary reduction at p. Then ∃LE ∈ K1(ΛT ) such that for all Artin representations of

G, the value at T = 0 of T−r(M)(ρ)Φρ(LE) is equal to

(−1)r(M)(ρ) LR(E, ρ∗, 1)

Ω∞(M(ρ∗))R∞(M(ρ∗))· Ωp(M(ρ∗))Rp(M(ρ∗)) ·

PL,p(W ∗ρ (1), 1)

PL,p(Wρ, 1)

For the purposes of this thesis, it is not important to define all of the notation in the above

conjecture, but it is all defined in Section 7 of [2].

2.5.3 The main conjecture

We can now state the Main Conjecture:

Conjecture 2.2 ([4], Conjecture 5.8) Assume that p ≥ 5, that E has good ordinary reduction at

p, and that X := X(E/K∞) is a finitely generated torsion Λ(G)-module. Granted Conjecture 2.1,

the p-adic L-function LE ∈ K1(ΛT ) is a characteristic element of X.

The important aspect to notice about Conjecture 2.2, is that we have an element that is related to

L-values when acted on by different Artin characters, and it is also a characteristic element of X.

If the main conjecture were proved, we would have the following:

Corollary 2.1 ([4], Corollary 5.9) Assume Conjecture 2.2. For any Artin representation of G,

ρ ∈ A(G), let ρ(g) = ρ(g−1)T where the T stands for transpose. ∀ρ ∈ A(G), such that ρ lands

in GLd(Zp), L(E, ρ, 1) 6= 0 ⇐⇒∏i≥0 #Hi(G, twρ(X))(−1)i is finite where twρ(X) = X ⊗Zp Zdp

such that we endow twρ(X) with the diagonal action. In this case, by ([4], Theorem 3.6), we have∏i≥0 #Hi(G, twρ(X))(−1)i = |LE(ρ)|dp.

2.6 The goal of this paper

2.6.1 The strategy of Kato

It is important for us to calculate K1(Λ). As we mentioned at the end of section 2.2, we have

K1(ΛT ) ∼= K1(ΛT ′ )⊕Zr for some r ≥ 0. Thus we only study K1(ΛT ′ ). Our methods use logarithms

and thus we need p-adically completed rings, hence we study K1

(ΛT ′

). As explained earlier, we

study K1 (Λ) and K1

(ΛT ′

)by studying K1 (Zp[Gn]) and K1

(Λ(Gn)T ′

). To do this we will imitate

the method used by Kakde in the proof of the non-commutative main conjecture for totally real fields.

Using the strategy of Burns and Kato, Kakde [11] proved the non-commutative main conjec-

ture for totally real fields under the µ = 0 condition ([11], Definition 2.8). In this proof, Kakde

9j-invariant: Let E be y2 = x3 + bx+ c, then the j-invariant of E is jE := 1728 b3

b3−27c2. ordl(a) is the integer

such that lordl(a)||a.

14

constructed the following commutative diagram to obtain a description of K′1 groups ([11], Proof of

Theorem 5.16):

1 → ker(Log) → K′1(Λ(G))Log−−−→ O[[Conj(G)]] → coker(Log) → 1

=↓ θ ↓ ψ ↓∼= =↓1 → ker(L) → Θ

L−→ Ψ → coker(L) → 1

Explaining the details of this diagram10is not necessary at this point, but it will be explained in

section 2.6.2.

The aim of this paper is to build on that work, and construct a similar diagram in the elliptic curve

case to obtain descriptions for K1 (Zp[Gn]) and K1

(Λ(Gn)T ′

)(see Section 2.6.2 for details).

2.6.2 Our strategy

Recall that G = GL2(Zp). In this section G is a finite group.

We will start this section by defining two important maps; Log, the integral logarithm

([14], Definition 6.1), and the map ψ. We will start with Log, which depends on the prime p from

the domain, Zp:

Definition 2.8 ([14], Definition 6.1) Let ϕ be the endomorphism of Qp ⊗Zp Zp[G]/[Zp[G],Zp[G]]

induced by the map∑xgg 7→

∑xggp. The integral logarithm is a map which is defined as follows:

Log : K1(Zp[G])→ Zp[G]/[Zp[G],Zp[G]]

Log : x 7→ log(x)−1

pϕ(log(x))

where [Zp[G],Zp[G]] denotes the additive group generated by elements [a1, a2] such that a1, a2 ∈Zp[G].

10One thing that should be noted is that the G here is the Galois group defined in a similar way to our G but for

totally real fields instead of elliptic curves.

15

The integral logarithm lands in Zp[G]/[Zp[G],Zp[G]] by ([14], Theorem 6.2):

Idea of the Proof:

For x in the Jacobson radical of Zp[G] one has:

Log(1− x) ≡ −∞∑k=1

1

pk(xpk − ϕ(xk)) (mod Zp[G]/[Zp[G],Zp[G]])

So one needs to show that pk|(xpk − ϕ(xk)), but in Zp, only p is non-invertible, so we can set

k = pn−1 where n is a positive integer. So we just need to show pn|(xpn − ϕ(xpn−1

)). To

show this, one expands x as∑agg, then one looks at the expansion of xp

nand then one finds

that this expansion is made up of terms in the form pn−tapt

g gpt for some t, 0 ≤ t ≤ n. By

definition of ϕ, we know p|(apg −ϕ(ag)), but when comparing xpn

with ϕ(xpn−1

)) we then find that

pn−tapt

g gpt ≡ pn−tϕ(ap

t−1

g )gpt

(mod pn), therefore pt|(apt

g − ϕ(apt−1

g )), so we have the desired

result.

Let Conj(G) be the set of all conjugacy classes [A]G of elements A ∈ G.

Proposition 2.1 ([17], Lemma 2.1) Let Zp[Conj(G)] denote the free Zp-module over the basis

Conj(G). Then we have isomorphism:

Zp[G]/[Zp[G],Zp[G]] ∼= Zp[Conj(G)]

as Zp-modules.

Before defining ψ, there is one last important property of Log we should state. Let

G = g ∈ G|g has order prime to p and let βn be the automorphism of Hn(G,Zp(G)) induced by

the map∑xgg 7→

∑xggp. If we look at Log as a map on the pro-p part of K1(Zp[G]), then by ([14],

Theorem 12.9(iii)), the kernel of Log is SK1(Zp[G]) ⊕ H1(G,Zp(G))β1 ⊕ H0(G, (Zp/2Zp)(G))β0

and the cokernel of Log is SK1(Zp[G]) ⊕ H1(G,Zp(G))β1 ⊕ H0(G, (Zp/2Zp)(G))β0 where

Hn(G,Zp(G))βn = ker(1 − βn) and Hn(G,Zp(G))βn = coker(1 − βn). Since we are taking p > 2,

we have that H0(G, (Zp/2Zp)(G))β0 is trivial.

Definition 2.9 Let U be a subgroup of G and let [A]U indicate the conjugacy class of A as an

element of U . Let TrG/U be the map:

Zp[Conj(G)] → Zp[Conj(U)]

[A]G 7→∑

X∈U\GX−1AX∈U

[X−1AX]U

and let projU be the natural projection from Zp[Conj(U)] to Zp[Conj(Uab)] = Zp[Uab], then:

ψ : Zp[Conj(G)] →∏U∈F Zp[Uab]

ψ : [A]G 7→∏U∈F projU TrG/U ([A]G)

As mentioned in the above section, we want to imitate the commutative diagram that was mentioned

in that section:

ker(Log) → K1(Zp[[G]])Log−−−→ Zp[[Conj(G)]] → coker(Log)

↓ θ ↓ ψ (1)

ker(L) 99K ΘL99K Ψ 99K coker(L)

Here Ψ ⊂∏U∈F Λ(Uab) ⊂ Qp ⊗Zp

∏U∈F Λ(Uab) is the image11of ψ and Θ ⊂

∏U∈F Λ(Uab)× is

a group that contains the image of θ. The bottom arrows are dotted to represent that they were

conjectural before they were constructed in this paper.

11The image of ψ lies in∏U∈F Zp[Uab] but we expect the image of

∏U∈F Λ(Uab)×, under the map L, to lie

outside of∏U∈F Zp[Uab]. This is why we are also interested in looking at Qp ⊗Zp

∏U∈F Zp[Uab]

16

To construct the map L, first we find Zp[[Conj(G))]]. Secondly, we want to show ψ is injective for

our choice of F and then describe its image, Ψ. A description of Ψ will help us define L by using

the following diagram ([14], Proof of Theorem 6.8):

K1(Zp[[G]])log−−→ Qp ⊗Zp Zp[[Conj(G)]]

1−ϕp−−−→ Qp ⊗Zp Zp[[Conj(G)]]

↓ θ ↓ ψ ↓ ψ∏U∈F Λ(Uab)×

log−−→ Qp ⊗Zp∏U∈F Zp[[Uab]] 99K Qp ⊗Zp

∏U∈F Zp[[Uab]]

Notice that the bottom maps combine to give us L. The first square is commutative so we can

concentrate on the second square. Since we will have a description for Zp[[Conj(G)]] and we will

understand ψ much better at this point, for any element x ∈ Zp[[Conj(G)]], we can compute ψ(x)

and ψ((1− ϕp

)(x) and compare these two elements to construct a map L. When we have a map L,

we can obtain a description of Θ by calculating a possible pre-image of Ψ under the map L.

If we could also prove that coker(Log) injects into coker(L) and that ker(Log) surjects onto ker(L),

then we would know that θ surjects onto Θ, but unfortunately we cannot do that in this paper.

Let Gn = GL2(Z/pnZ), by a result of Kakde ([11], Lemma 4.1)12, we can see that

K1(Zp[[G]]) ∼= lim←−nK1(Zp[Gn]) where the inverse limit is defined by projection. As a result

we can reduce diagram (1) and look at the following diagram:

ker(Log) → K1(Zp[Gn])Log−−−→ Zp[Conj(Gn)] → coker(Log)

↓ θ ↓ ψ

ker(L) 99K∏U∈Fn Λ(Uab)×

L99K Qp ⊗Zp

∏U∈Fn Zp[Uab] 99K coker(L)

where Fn is a subgroup of F such that F = ∪m≥1Fm.

So we can prove that ψ is injective by proving the restrictions of ψ to Gn is injective for all

n:

ψn : Zp[Conj(Gn)] −→∏

U∈Fn

Zp[Uab]

In this paper we calculate the image of ψn for general n and we call this image Ψn,Zp . We also

construct the map L and we construct a subgroup of∏U∈F Λ(Uab)× which, under the map L,

contains Ψn,Zp ; we call this group Θn,Zp and it contains the image of K1 (Zp[Gn]) under the map

θn. We also verify a similar result for K1

(Λ(Gn)T ′

). To carry on the work in this paper, the next

thing to do would be to calculate the kernel and cokernel of both Log and L. The work in this paper

is all done for a particular choice of Fn, which we stated in the next chapter.

12This result is a based on a result of Fukaya and Kato ([7], Proposition 1.5.1)

17

Chapter 3

Choosing Fn, a suitable set of subgroups of

Gn

In this chapter, we want to choose a set Fn of subgroups of Gn such that the kernel of θ would be

SK1(Zp[Gn]). We can achieve this result by using the set of all open subgroups [11] but then we

make it much harder to describe the image of ψn (Definition 2.9), and as a consequence, it becomes

harder to prove the main conjecture via the above strategy. Thus we aim to pick a Fn which is not

too large yet it has the desired kernel.

3.1 Computing the p-part of the torsion subgroup of K1(Zp[Gn])

To help us choose Fn which gives us the kernel SK1(Zp[Gn]), we look at the p-part of the torsion

subgroup of K1(Zp[Gn]). We do that in this chapter by using the following theorem:

Theorem 3.1 ([14], Theorem 12.5) Fix a prime p, let F be any finite extension of Qp, and let R ⊂F be the ring of integers. For any finite group G, let g1,...,gk be F -conjugacy class1representatives

for elements in G of order prime to p, and set

Ni = NFG (gi) = x ∈ G : xgix

−1 = gai , some a ∈ Gal(Fζni/F ) (ni = |gi|)

and Zi to be the centralizer group of gi as an element of G. Then

1. SK1(R[G]) ∼=⊕ki=1 H0(Ni/Zi;H2(Zi)/H

ab2 (Zi))(p)

2. tors(K′1(R[G]))(p)∼= [(µF )p]k ⊕

⊕ki=1 H

0(Ni/Zi;Zabi )(p)

H0 and H0 are the zeroth homology and cohomology2respectively. H2(Zi) = H2(Zi,Z) and Hab2 (Zi)

will be defined later. In our case [(µF )p] is trivial because we are taking F = Qp. When defining Fnit is important to include the subgroups U such that

∏U Zp[U ] contains tors(K′1(R[G]))(p). This

is because, by definition of Log, these subgroups lie in the kernel of Log and we need ker(Log) to

surject onto ker(L).

Here we will just state the full list of the representatives of conjugacy classes of Gn (see the

next section for the proof that this list is the full list):

1Let g and h be a group elements of order n in G. Since Gal(Fζn/F ) is a subgroup of (Z/nZ)×, Oliver writes

ga to denote the action of a on g for a ∈ Gal(Fζn/F ). We say g and h are F -conjugate if xhx−1 = ga.

18

AI AB AT AK ART,i ARK,i(x 0

0 x

) (x 0

1 x

) (w y2

1 w

) (z εy2

1 z

) (x piα2

1 x

) (x piεα2

1 x

)ARB,j ARBI,j,i ARBJ,j,i ARI,j ARJ,j(x 0

pj x

) (x piα2

pj x

) (x piεα2

pj x

) (x pjβ2

pj x

) (x pjεβ2

pj x

)

i, j = 1, 2, ..., n− 1 s.t j < i, x, y ∈ (Z/pnZ)× and w, z ∈ Z/pnZ s.t y 6≡ ±w (mod p)

α ∈ (Z/pn−iZ)× and β ∈ (Z/pn−jZ)×

The letter on the top represents the matrix under that letter.

In section 3.3 we calculate how many elements in Gn have order prime to p and what form these

matrices are in; there are only p(p− 1) distinct conjugacy classes with matrices of order prime to p,

and they all have a representation matrix in the form AI , AT or AK (defined in the table below).

The centralizer groups for these matrices are as follows:

Representatives Centralizers no. of prime to p classes

AI :=

(x 0

0 x

)GL2(Z/pnZ) p− 1

AT :=

(w y2

1 w

) (a by2

b a

):a ∈ Z/pnZb ∈ Z/pnZ

: p 6 |(a2 − b2y2)

(p−1)(p−2)

2

AK :=

(z εy2

1 z

) (a bεy2

b a


: p 6 |(a2 − b2εy2)

p(p−1)

2

We will refer to these centralizer groups as ZI , ZT , and ZK respectively. We will also denote the

normalizer groups in a similar way, i.e. NI , NT , and NK respectively. ZT and ZK are abelian3,

therefore we have Hab2 (ZT ) = H2(ZT ) and Hab

2 (ZK) = H2(ZK) (this will become clear later when

we define the second homology). Since the order of elements in the form AI and AT are either 1 or

p − 1 (see section 3.3) and ζp−1 ∈ Qp, we have NI = ZI and NT = ZT . So we only need to work

out NK :

AK :=

(z εy

y z

)=

(0 εy

1 0

)−1(z εy2

1 z

)(0 εy

1 0

)

Therefore we have [AK ] =[AK

]and we have AK ∈ Kn :=

(a εb

b a

):a, b ∈ Z/pnZs.t. p 6 |(a2 − εb2)

Since Kn is a group, any power of the matrix AK will also lie in Kn. By definition, if X ∈ NK then

we must have X−1AKX = (AK)m for some m, but any conjugate of AK lies in Kn in only two

cases: (a b

c d

)−1(z εy

y z

)(a b

c d

)=

(z εy

y z

)(

a b

c d

)−1(z εy

y z

)(a b

c d

)=

(z −εy−y z

)2H0(G;M) = MG = M/〈gm−m|g ∈ G,m ∈ M〉 and H0(G;M) = MG = m ∈ M|gm−m = 0 ∀g ∈ G.3Observe the following: (

a bu

b a

)−1 (c du

d c

)(a bu

b a

)=

(c du

d c

)

where a, b, c, d ∈ Z/pnZ such that p 6 |a2 − b2u and p 6 |c2 − d2u. Notice that u can be replaced with y2 or εy2 to

get ZT or ZK respectively. So we know the centralizers of both AT and AK are commutative, i.e. ZabT = ZT and

ZabK = ZK .

19

So we now know that NK/ZK is a group of order 1 or 2 depending on whether there exists an element

m in the Galois group, such that:(z εy

y z

)m=

(z −εy−y z

)

We can use this result in Theorem 3.1 to get the following:

tors(K′1(Zp[Gn]))(p)∼= [(µF )p]k ⊕

⊕ki=1H

0(Ni/Zi;Zabi )(p)

⊂⊕ki=1 H

0(1;Zabi )(p)

=⊕p−1(ZabI )(p) ⊕

⊕ (p−1)(p−2)2 (ZabT )(p) ⊕

⊕ p(p−1)2 (ZabK )(p)

=⊕p−1((Z/pnZ)×)(p) ⊕

⊕ (p−1)(p−2)2 (ZT )(p) ⊕

⊕ p(p−1)2 (ZK)(p)

=⊕ (p−1)(p−2)

2 (ZT )(p) ⊕⊕ p(p−1)

2 (ZK)(p)

To simplify the expression for SK1(Zp[Gn]), we need to defineH2(Zi) andHab2 (Zi). By ([6], Theorem

3.1) we know that H2(Zi) ∼= ker(Zi ∧ Zi[·,·]−−→ [Zi, Zi]) where Zi ∧ Zi is the exterior product and

[Zi, Zi] is the commutator subgroup. By ([14], Chapter 8a), we have Hab2 (Zi) = 〈g ∧ h ∈ H2(Zi) :

g, h ∈ Zi, gh = hg〉. Therefore we have Hab2 (ZT ) = H2(ZT ) and Hab

2 (ZK) = H2(ZK), therefore

H0(NT /ZT ;H2(ZT )/Hab2 (ZT )) = 1 = H0(NK/ZK ;H2(ZK)/Hab

2 (ZK)). Recall that NI = ZI :

SK1(Zp[GL2(Z/pnZ)]) ∼=⊕ki=1H0(Ni/Zi;H2(Zi)/H

ab2 (Zi))(p)

=⊕p−1H0(NI/ZI ;H2(ZI)/Hab

2 (ZI))(p)

=⊕p−1(H2(ZI)/Hab

2 (ZI))(p)

=⊕p−1(H2(Gn)/Hab

2 (Gn))(p)

Although SK1(Zp[Gn]) is not needed for the work done in this paper, it would be nice to compute

it explicitly but we do not know how to do it.

It turns out that we can take Fn = Zn, Cn, Tn,Kn, Nti , Nki |∀i = 1, 2, ..., n − 1 where

these subgroups are defined as follows:

Zn :=

(a 0

0 a

): a ∈ (Z/pnZ)×

Cn :=

(a 0

c a

):a ∈ (Z/pnZ)×

c ∈ Z/pnZ

Tn :=

(a 0

0 d

): a, d ∈ (Z/pnZ)×

Kn :=

(a εb

b a

):a, b ∈ Z/pnZs.t. p 6 |(a2 − εb2)

Nti :=

(a bpi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

Nki :=

(a bεpi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

Our calculation of tors(K′1(Zp[Gn]))(p) implies that we should included (ZT )(p) and (ZK)(p) in our

definition of Fn, but it turns out that we are better off using conjugates of AT and AK so we use

conjugates of ZT and ZK too. These groups are Tn and Kn respectively; we need to use these

replacement subgroups because we will conjugate4AT and AK to get the matrices

(w + y 0

0 w − y

)

20

and

(z εy

y z

).

4X−1ATX =

(w + y 0

0 w − y

)and Y−1AKY =

(z εy

y z

)for the matrices X−1 =

(1 y

−1 y

)and Y =(

0 y

1 0

). So we need to conjugate each matrix in ZT by X−1 and each conjugate each matrix in ZK by Y .

Doing this gives us the subgroups Tn and Kn respectively.

21

3.2 Conjugacy classes of GL2(Z/pnZ)

In this section we will verify that the following table is the full table of representatives of all

conjugacy classes of GL2(Z/pnZ):

Table 1 ([3], Table 2 and 3):

Rep. no. of elts per class no. of classes(x 0

0 x

)1 pn − pn−1(

x 0

1 x

)p2n−2(p2 − 1) pn − pn−1(

w y2

1 w

)p2n−2(p2 + p)

p2n−2(p−1)(p−2)2(

z εy2

1 z

)p2n−2(p2 − p) p2n−1(p−1)

2(x piα2

1 x

)p2n−2(p2 − 1)

p2n−2−i(p−1)2

2(x piεα2

1 x

)p2n−2(p2 − 1)

p2n−2−i(p−1)2

2(x 0

pj x

)p2(n−1−j)(p2 − 1) pn−1(p− 1)(

x piα2

pj x

)p2(n−1−j)(p2 − 1)

p2n−2−i(p−1)2

2(x piεα2

pj x

)p2(n−1−j)(p2 − 1)

p2n−2−i(p−1)2

2(x pjβ2

pj x

)p2(n−1−j)(p2 + p)

p2n−2−j(p−1)2

2(x pjεβ2

pj x

)p2(n−1−j)(p2 − p) p2n−2−j(p−1)2

2



Here ε is a fixed non-square element in (Z/pnZ)×.

Theorem 3.2 Table 1 is an exhaustive list of all conjugacy classes of GL2(Z/pnZ) which tells us

the number of conjugacy classes of each type of matrix representative as well as the number of

elements in each class.

To prove this theorem we will verify the centralizers given in Table 2 below, and then use them to

verify the ‘number of elements per class’. Then we use simple counting arguments to verify the

‘number of classes’. We also need to verify that the representatives give distinct classes but in most

22

classes this is known because the ‘number of elements per class’ are different. Finally we use all

the information from Table 1 to confirm that this is the full list by checking the identity5∑

(no. of

elts per class)(no. of classes)= |GL2(Z/pnZ)|. This is a long process but the conjugacy classes are

essential to the work done in this paper so we show the full verification.

Table 2 ([3], Table 2):

Rep. Centralizers Size of Centralizers(x 0

0 x

)GL2(Z/pnZ) p4n−3(p2 − 1)(p− 1)(

x 0

1 x

) (a 0

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

p2n−1(p− 1)(

w y2

1 w

) (a by2

b a


: p 6 |(a2 − b2y2)

p2n−2(p− 1)2(

z εy2

1 z

) (a bεy2

b a


: p 6 |(a2 − b2εy2)

p2n−2(p2 − 1)(

x piα2

1 x

) (a bpiα2

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

p2n−1(p− 1)(

x piεα2

1 x

) (a bpiεα2

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

p2n−1(p− 1)(

x 0

pj x

) (a kpn−j

b a+ lpn−j

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

p2n+2j−1(p− 1)(

x piα2

pj x

) (a bpi−jα2 + kpn−j

b a+ lpn−j

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

p2n+2j−1(p− 1)(

x piεα2

pj x

) (a bpi−jεα2 + kpn−j

b a+ lpn−j

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

p2n+2j−1(p− 1)(

x pjβ2

pj x

) (a bβ2 + kpn−j

b a+ lpn−j


: p 6 |(a2 − b2β2)

p2(n+j−1)(p− 1)2(

x pjεβ2

pj x

) (a bεβ2 + kpn−j

b a+ lpn−j


: p 6 |(a2 − b2εβ2)

p2(n+j−1)(p2 − 1)


k, l ∈ Z/pjZ, α ∈ (Z/pn−iZ)× and β ∈ (Z/pn−jZ)×

Verifying Table 2

Proposition 3.1 In Table 2, the centralizers and their orders are correct.

Before we prove this proposition, let us set up some notation:


0 x

) (x 0

1 x

) (w y2

1 w

) (z εy2

1 z

) (x piα2

1 x

) (x piεα2

1 x


pj x

) (x piα2

pj x

) (x piεα2

pj x

) (x pjβ2

pj x

) (x pjεβ2

pj x

)The letters in top row represent the form6of the matrices under that letter.

We will conjugate each of these matrices by a generic element in GL2(Z/pnZ) and check that the

5This sum is a sum over each representation matrix.6At this point we are not interested in a notation which tells us the exact matrix; we only want to know the

form of matrix.

23

minimal conditions to fix these matrices are in fact the conditions which define the centralizer

groups in Table 2. Then we check the size of these centralizer groups and use these values

to verify the ‘number of elements per class’ stated in Table 1. This is done with a simple

formula: ‘number of elements per class’ is equal to the order of the group, |GL2(Z/pnZ)| in our

case, divided by the order of the centralizer group. Recall that |GL2(Z/pnZ)| = p4n−3(p2−1)(p−1).

Proof of Proposition 3.1:

We will verify each matrix representation case by case.

Matrix AI

This one is obvious since AI represents matrices in the centre of GL2(Z/pnZ).

Matrix AB

In this case, we want to show that the centralizer group is(a 0

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

and that this group has order p2n−1(p− 1):(a b

c d

)−1

.

(x 0

1 x

).

(a b

c d

)

=1

ad− bc

(−ab+ adx− bcx −b2

a2 ab+ adx− bcx

)To fix AB , we need b = 0 for the top right corner to be zero and then we get:(

x 0

a/d x

)

so we need just a = d and we have verified the centralizer group for AB .

In the centralizer group we have a choice of a ∈ (Z/pnZ)× and b ∈ Z/pnZ so |Centralizer Group| =|(Z/pnZ)×||Z/pnZ| = (pn − pn−1)(pn) = p2n−1(p− 1). This agrees with Table 2.

Therefore the number of elements per class=p4n−3(p2−1)(p−1)

p2n−1(p−1)= p2n−2(p2 − 1) which agrees with

Table 1.

Matrix AT

In this case, we want to show that the centralizer group is(a by2

b a


: p 6 |(a2 − b2y2)

and that this group has order p2n−2(p− 1)2:(a b

c d

)−1

.

(w y2

1 w

).

(a b

c d

)

=1

ad− bc

(−ab+ adw − bcw + cdy2 d2y2 − b2

a2 − c2y2 ab+ adw − bcw − cdy2

)

24

To fix AT , we need conditions:

w + cdy2−abad−bc = w

d2y2 − b2 = (ad− bc)y2

a2 − c2y2 = (ad− bc)w − cdy2−ab

ad−bc = w

so we need:

• d2y2 − b2 = a2y2 − c2y4 = (ad− bc)y2

• cdy2 − ab = 0

Let us split this problem into 2 different cases:

Case p|d

We need:

• d2y2 − b2 = a2y2 − c2y4 = (ad− bc)y2

• cdy2 − ab = 0

p cannot divide b or c so the second bullet point tells us that p|a. If we rearrange the first bullet

point as (d2 − a2)y2 = b2 − c2y4, then we can see that b ≡ ±cy2 (mod p). Set b = cy2 + kp such

that k ∈ Z/pnZ. Now put this into the second bullet point:

cdy2 − acy2 − akp = 0 ⇐⇒ cy2(d− a) = akp ⇐⇒ d =akp

cy2+ a

Now let’s put this in the first bullet point:

a2y2 − c2y4 =

(a2kp

cy2+ a2 − c2y2 − ckp

)y2

⇐⇒ 0 =a2kp

c− ckpy2

⇐⇒ c2kpy2 = a2kp

So we need kp = 0. This implies that we have b = cy2 and a = d and with these conditions we get:

1

a2 − c2y2

(w a2y2 − c2y4

a2 − c2y2 w

)=

(w y2

1 w

)

This centralizer consist of matrices with which agrees with Table 2.

Case p 6 |d

We need:

• d2y2 − b2 = a2y2 − c2y4 = (ad− bc)y2

• cdy2 − ab = 0

25

We can rearrange the bottom bullet point into the form c = abdy2

. We can use this in the first bullet

point:

d2y2 − b2 = a2y2 − ( abdy2

)2y4 = (ad− b( abdy2

))y2

⇐⇒ d4y2 − d2b2 = d2a2y2 − a2b2 = ad3y2 − ab2d⇐⇒ d2(d2y2 − b2) = a2(d2y2 − b2) = ad(d2y2 − b2)

So d = a or b = ±dy. If we set b = ±dy, we would have c = ±adydy2

= ±ay

so we would get a matrix

with determinant 0 therefore this condition gives us no matrices. If a = d then c = abay2

= by2

so we

have b = cy2 and we saw these conditions work in the last section and these are the conditions given

in Table 2.

In the centralizer group we have a choice of a ∈ Z/pnZ and b ∈ Z/pnZ but a restriction of p 6|(a2 − b2y2) but we can rewrite this as a choice of either:

• a ∈ p(Z/pnZ) and b ∈ (Z/pnZ)×

• OR a ∈ (Z/pnZ)× and b ∈ Z/pnZ such that a2 6≡ b2y2 (mod p)

So we get

|Centralizer Group|= |p(Z/pnZ)||Z/pnZ|+ |(Z/pnZ)×||(Z/pnZ)\b ∈ Z/pnZ : a2 ≡ b2y2 (p)|= (pn−1)(pn − pn−1) + (pn − pn−1)(pn − 2pn−1)

= (pn − pn−1)(pn − pn−1) = p2n−2(p− 1)2

This agrees with Table 2.


p2n−2(p−1)2= p2n−1(p + 1) = p2n−2(p2 + p)

which agrees with Table 1.

Matrix AK

In this case, we want to show that the centralizer group is(a bεy2

b a


: p 6 |(a2 − b2εy2)

and that this group has order p2n−2(p2 − 1):(a b

c d

)−1

.

(z εy2

1 z

).

(a b

c d

)

=1

ad− bc

(cdεy2 − ab− bcz + adz d2εy2 − b2

a2 − c2εy2 −cdεy2 + ab− bcz + adz

)This case is essentially the same as the AT case, just with y2 changed for εy2. This would give us the

same centralizer group as the centralizer group of AT but with y2 changed for εy2 and this agrees

with Table 2.

In the centralizer groups of matrices in the form AK we have a choice of a ∈ Z/pnZ and b ∈ Z/pnZbut a restriction of p 6 |(a2 − b2εy2) but we can rewrite this as a choice of either:


• OR a ∈ (Z/pnZ)× and b ∈ Z/pnZ such that a2 6≡ b2εy2 (mod p)

26

So we get

|Centralizer Group|= |p(Z/pnZ)||Z/pnZ|+ |(Z/pnZ)×||(Z/pnZ)\b ∈ Z/pnZ : a2 ≡ b2εy2 (p)|= (pn−1)(pn − pn−1) + (pn − pn−1)(pn)

= (pn − pn−1)(pn + pn−1) = p2n−2(p2 − 1)



p2n−2(p2−1)= p2n−1(p − 1) = p2n−2(p2 − p)

which agrees with Table 1.

Matrix ART,i

In this case, we want to show that the centralizer group is(a bpiα2

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

and that this group has order p2n−1(p− 1):(

a b

c d

)−1

.

(x piα2

1 x

).

(a b

c d

)

=1

bc− ad

(−cdα2pi + ab+ bcx− adx b2 − d2piα2

c2piα2 − a2 cdα2pi − ab+ bcx− adx

)Just like the previous section, we get the centralizer group by looking at the centralizer group of AT

and changing y2 for piα2, and this gives us the same centralizer group shown in Table 2.

In the centralizer group we have a choice of a ∈ (Z/pnZ)× and b ∈ Z/pnZ so |Centralizer Group| =|(Z/pnZ)×||Z/pnZ| = (pn − pn−1)(pn) = p2n−1(p− 1).



p2n−1(p−1)= p2n−2(p2 − 1) which agrees with

Table 1.

Matrix ARK,i

In this case, we want to show that the centralizer group is(a bpiεα2

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

and that this group has order p2n−1(p− 1):(

a b

c d

)−1

.

(x piεα2

1 x

).

(a b

c d

)

=1

bc− ad

(−cdεα2pi + ab+ bcx− adx b2 − d2piεα2

c2piεα2 − a2 cdεα2pi − ab+ bcx− adx

)Just like the previous section, we get the centralizer group by looking at the centralizer group of AT

and changing y2 for piεα2, and this gives us the same centralizer group shown in Table 2.

Also like the previous section we have a choice of a ∈ Z/pnZ and b ∈ Z/pnZ but a restriction of

p 6 |(a2 − b2εy2), therefore we know that

|Centralizer Group| = p2n−1(p− 1)

and the number of elements per class=p4n−3(p2−1)(p−1)

p2n−1(p−1)= p2n−2(p2 − 1) which agrees with Table

1 and Table 2.

27

Matrix ARB,j

In this case, we want to show that the centralizer group is(a kpn−j

b a+ lpn−j

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

and that this group has order p2n+2j−1(p− 1):(a b

c d

)−1

.

(x 0

pj x

).

(a b

c d

)

=1

ad− bc

(−abpj − bcx+ adx −b2pj

a2pj abpj − bcx+ adx

)To fix ARB,j , we need conditions:

x− abpj

ad−bc = x

−b2pj = 0

a2pj = (ad− bc)pj

x+ abpj

ad−bc = x

This means that we want pn−j |b2 and pn−j |ab but we cannot have p dividing both a and b therefore

pn−j |b. Set b = kpn−j . We also want a2pj = (ad − kpn−jc)pj = adpj therefore d = a + lpn−j .

These conditions give us:

1

a2 + alpn−j − ckpn−j

(a2x+ alpn−jx− ckpn−jx 0

a2pj a2x+ alpn−jx− ckpn−jx

)

=

(x 0

a2pj

a2+(al−ck)pn−jx

)On further inspection, we find that the bottom-left entry of this matrix is pj :

a2pj

a2 + (al − ck)pn−j=

a2pj + 0

a2 + (al − ck)pn−j=a2pj + (al − ck)pn

a2 + (al − ck)pn−j= pj

Thus we have verified Table 2 shows the correct centralizer group for ARB,j .

In the centralizer group we have a choice of k, l ∈ Z/pjZ, a ∈ (Z/pnZ)× and b ∈ Z/pnZ so

|Centralizer Group| = |Z/pjZ|2|(Z/pnZ)×||Z/pnZ| = (pj)2(pn−pn−1)(pn) = p2n+2j−1(p−1). This

agrees with Table 2.


p2n+2j−1(p−1)= p2(n−1−j)(p2 − 1) which agrees

with Table 1.

Matrix ARBI,j,i

In this case, we want to show that the centralizer group is(a bpi−jα2 + kpn−j

b a+ lpn−j

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ


c d

)−1

.

(x piα2

pj x

).

(a b

c d

)

=1

ad− bc

(−abpj + cdα2pi − bcx+ adx −b2pj + d2piα2

−c2piα2 + a2pj abpj − cdα2pi − bcx+ adx

)

28

To fix ARBI,j,i, we need conditions:

x− abpj−cdα2pi

ad−bc = x

d2piα2 − b2pj = (ad− bc)piα2

a2pj − c2piα2 = (ad− bc)pj

x+ abpj−cdα2pi

ad−bc = x

so we need:

• d2piα2 − b2pj = a2piα2 − c2p2i−jα4 = (ad− bc)piα2

• abpj − cdα2pi = 0

• a2pj − c2piα2 = (ad− bc)pj

Like before, we will split this into 2 cases:

Case p|a

We need:




p cannot divide b or c so we can rearrange the second bullet point as a = cdα2

bpi−j . Now we plug

that into the first bullet point:

d2piα2 − b2pj =c2d2α6

b2p3i−2j − c2p2i−jα4 =

(cd2α2

bpi−j − bc

)piα2

⇐⇒ b2d2piα2 − b4pj = c2d2α6p3i−2j − b2c2p2i−jα4 =(bcd2α2pi−j − b3c

)piα2

⇐⇒ b2pj(d2pi−jα2 − b2) = c2α4p2i−j(d2α2pi−j − b2) = bcα2pi(d2α2pi−j − b2

)So we have either d2α2pi−j = b2 or b2pj = c2α4p2i−j = bcα2pi but if we use the former condition, we

would get a matrix with determinant zero so this condition gives us no matrices. The latter condition

simplifies to b ≡ cα2pi−j (mod pn−j) and this gives us a = dbcα2pi−j ≡ d (mod pn−j). Note that

these conditions also satisfy the third bullet point. Set b = cα2pi−j + kpn−j and d = a + lpn−j ,

then we get: x −c2α4p2i−j+a2piα2

a2+alpn−j−c2α2pi−j−ckpn−j−c2α2pi+a2pj

a2+alpn−j−c2α2pi−j−ckpn−j x

=

x a2pj−c2α2pi

a2−c2α2pi−j+(al−ck)pn−jα2pi−j

a2pj−c2α2pi

a2−c2α2pi−j+(al−ck)pn−jx

Now the top-right and bottom-left entries simplify to α2pi and pj respectively:

a2pj − c2α2pi

a2 − c2α2pi−j + (al − ck)pn−j=

a2pj − c2α2pi + (al − ck)pn

a2 − c2α2pi−j + (al − ck)pn−j= pj

Therefore we get: (x α2pi

pj x

)So these conditions give us the same centralizer group as shown in Table 2.

29

Case p 6 |a

We need:




We can rearrange the second bullet point as b = cdα2

api−j . Now we plug this into the first bullet

point and get:

d2piα2 −c2d2α4

a2p2i−j = a2piα2 − c2p2i−jα4 =

(ad−

c2dα2

api−j

)piα2

⇐⇒ a2d2piα2 − c2d2α4p2i−j = a4piα2 − a2c2p2i−jα4 =(a3d− ac2dα2pi−j

)piα2

⇐⇒ d2piα2(a2 − c2pi−jα2) = a2piα2(a2 − c2pi−jα2) = adpiα2(a2 − c2α2pi−j

)So either a ≡ d (mod pn−i) or a2 = c2pi−jα2 but the latter condition gives us no invertible matrices.

By the third bullet point, we see that we must use the stricter condition of a ≡ d (mod pn−j). With

these conditions we get b = cdα2

api−j ≡ cpi−jα2 (mod pn−j), so these are the same conditions we

obtained in the last section so we know they give matrices in the centralizer group and they are the

same conditions given in Table 2.

Like the previous case we have a choice of k, l ∈ Z/pjZ, a ∈ (Z/pnZ)× and b ∈ Z/pnZ so

|Centralizer Group| = p2n+2j−1(p− 1). This agrees with Table 2.


p2n+2j−1(p−1)= p2(n−1−j)(p2 − 1) which agrees

with Table 1.

Matrix ARBJ,j,i

In this case, we want to show that the centralizer group is(a bpi−jεα2 + kpn−j

b a+ lpn−j

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ


c d

)−1

.

(x piεα2

pj x

).

(a b

c d

)

=1

bc− ad

(abpj − cdεα2pi + bcx− adx b2pj − d2piεα2

c2piεα2 − a2pj −abpj + cdεα2pi + bcx− adx

)This case is essentially the same as the ARBI,j,i case, just with piα2 changed for piεα2. This would

give us the same centralizer group as the centralizer group of ARBI,j,i but with piα2 changed for

piεα2 and this agrees with Table 2.

Also like the previous case we have a choice of k, l ∈ Z/pjZ, a ∈ (Z/pnZ)× and b ∈ Z/pnZ, therefore

we know that

|Centralizer Group| = p2n+2j−1(p− 1)

and the number of elements per class=p4n−3(p2−1)(p−1)

p2n+2j−1(p−1)= p2(n−1−j)(p2 − 1) which agrees with

Table 1 and Table 2.

30

Matrix ARI,j

In this case, we want to show that the centralizer group is(a bβ2 + kpn−j

b a+ lpn−j


: p 6 |(a2 − b2β2)

and that this group has order p2(n+j−1)(p− 1)2:(a b

c d

)−1

.

(x pjβ2

pj x

).

(a b

c d

)

=1

bc− ad

(−cdβ2pj + abpj + bcx− adx b2pj − d2pjβ2

c2pjβ2 − a2pj cdβ2pj − abpj + bcx− adx

)Just like the previous section, we get the centralizer group by looking at the centralizer group of

ARBI,j,i and changing piα2 for pjβ2, and this gives us the same centralizer group shown in Table 2.

In the centralizer groups of matrices in the form ARI,j we have a choice of k, l ∈ Z/pjZ, a ∈ Z/pnZand b ∈ Z/pnZ but a restriction of p 6 |(a2− b2β2) but we can rewrite this as a choice of k, l ∈ Z/pjZ,

and either:


• OR a ∈ (Z/pnZ)× and b ∈ Z/pnZ such that a2 6≡ b2β2 (mod p)

So we get

|Centralizer Group|= |Z/pjZ|2(|p(Z/pnZ)||Z/pnZ|+ |(Z/pnZ)×||(Z/pnZ)\b ∈ Z/pnZ : a2 ≡ b2β2 (p)|)= (pj)2((pn−1)(pn − pn−1) + (pn − pn−1)(pn − 2pn−1))

= (p2j)(pn − pn−1)(pn − pn−1) = p2(n+j−1)(p− 1)2



p2(n+j−1)(p−1)2= p2(n−1−j)(p2 + p) which agrees

with Table 1.

Matrix ARJ,j

In this case, we want to show that the centralizer group is(a bεβ2 + kpn−j

b a+ lpn−j


: p 6 |(a2 − b2εβ2)

and that this group has order p2(n+j−1)(p2 − 1):(a b

c d

)−1

.

(x pjεβ2

pj x

).

(a b

c d

)

=1

bc− ad

(−cdεβ2pj + abpj + bcx− adx b2pj − d2pjεβ2

c2pjεβ2 − a2pj cdεβ2pj − abpj + bcx− adx

)Just like the previous section, we get the centralizer group by looking at the centralizer group of

ARBI,j,i and changing piα2 for pjεβ2, and this gives us the same centralizer group shown in Table

2.

In the centralizer groups of matrices in the form ARJ,j we have a choice of k, l ∈ Z/pjZ, a ∈ Z/pnZand b ∈ Z/pnZ but a restriction of p 6 |(a2−b2εβ2) but we can rewrite this as a choice of k, l ∈ Z/pjZ,

and either:


31

• OR a ∈ (Z/pnZ)× and b ∈ Z/pnZ such that a2 6≡ b2εβ2 (mod p)

So we get

|Centralizer Group|= |Z/pjZ|2(|p(Z/pnZ)||Z/pnZ|+ |(Z/pnZ)×||(Z/pnZ)\b ∈ Z/pnZ : a2 ≡ b2εβ2 (p)|)= (pj)2((pn−1)(pn − pn−1) + (pn − pn−1)(pn))

= (p2j)(pn − pn−1)(pn + pn−1) = p2(n+j−1)(p2 − 1)



p2(n+j−1)(p2−1)= p2(n−1−j)(p2 − p) which agrees

with Table 1.

This concludes the proof of the proposition and we have also verified the second column in

Table 1 which states the number of elements in each class.

Verifying all representations generate distinct classes

Proposition 3.2 No representation matrix in Table 1 represent the same conjugacy classes as

another representation matrix of a different form.

Proof:

AI is clearly distinct from the rest. We know that most matrices do not represent the same class

because they have a different “number of elements per class”. So the only matrices that could give

the same class are AB , ART,i with ARK,i and ARB,j0 , ARBI,j0,i with ARBJ,j0,i for fixed j0. All

these matrices are in the form

(t s

pm t

)for s, t ∈ Z/pnZ and some fixed m = 0, 1, 2, ..., n − 1

with pm|s such that p 6 |(t2 − spm).(a b

c d

)−1

.

(t s

pm t

).

(a b

c d

)=

(t′ s′

pm t′

)

⇐⇒1

ad− bc

(−abpm + cds− bct+ adt d2s− b2pm

a2pm − c2s abpm − cds− bct+ adt

)=

(t′ s′

pm t′

)This would mean that we would have the following equations:

t− abpm−cdsad−bc = t′

d2s− b2pm = (ad− bc)s′

a2pm − c2s = (ad− bc)pm

t+ abpm−cdsad−bc = t′

which can be rewritten as:

• d2s− b2pm = a2s′ − c2 ss′

pm= (ad− bc)s′

• abpm − cds = 0

• a2pm − c2s = (ad− bc)pm

The second bullet point implies that t′ = t. We will spilt this problem into 2 cases:

32

Case p|aWe have:

• d2s− b2pm = a2s′ − c2 ss′

pm= (ad− bc)s′



p does not divide b and c so the second bullet point can be rearranged to a = cdsbpm

. Now plug this

into the first bullet point:

d2s− b2pm =c2d2s2s′

b2p2m− c2

ss′

pm= (

cd2s

bpm− bc)s′

⇐⇒ pm(d2 s

pm− b2) =

c2ss′

b2pm(d2 s

pm− b2) =

cs′

b(d2 s

pm− b2)

So we have d2 spm

= b2 or pm = c2ss′

b2pm= cs′

b. The former condition would give a zero determinant,

so we cannot conjugate by any matrix with this condition. The latter condition would give the

condition s = s′ meaning that any matrix with this condition would be in a centralizer group and

thus this would show that all representatives on Table 1 generate distinct conjugacy classes. Note

that the latter condition also satisfies the third bullet point.

Case p 6 |aWe have:

• d2s− b2pm = a2s′ − c2 ss′

pm= (ad− bc)s′



The second bullet point can be rearranged to b = cdsapm

. Now plug this into the first bullet point:

d2s−c2d2s2

a2pm= a2s′ − c2

ss′

pm= (ad−

c2ds

apm)s′

⇐⇒d2s

a2(a2 − c2

s

pm) = s′(a2 − c2

s

pm) =

ds′

a(a2 − c2

s

pm)

So we have c2 spm

= a2 or d2sa2

= s′ = ds′

a. The former condition would give a zero determinant, so we

cannot conjugate by any matrix with this condition. The latter condition would give the condition

s = s′ meaning that any matrix with this condition would be in a centralizer group and thus this

would show that all representatives on Table 1 do in fact generate distinct conjugacy classes. Note

that the latter condition also satisfies the third bullet point.

33

Verifying the “Number of Classes”

Recall Table 1:Rep. no. of elts per class no. of classes(x 0

0 x

)1 pn − pn−1(

x 0

1 x

)p2n−2(p2 − 1) pn − pn−1(

w y2

1 w

)p2n−2(p2 + p)

p2n−2(p−1)(p−2)2(

z εy2

1 z

)p2n−2(p2 − p) p2n−1(p−1)

2(x piα2

1 x

)p2n−2(p2 − 1)

p2n−2−i(p−1)2

2(x piεα2

1 x

)p2n−2(p2 − 1)

p2n−2−i(p−1)2

2(x 0

pj x

)p2(n−1−j)(p2 − 1) pn−1(p− 1)(

x piα2

pj x

)p2(n−1−j)(p2 − 1)

p2n−2−i(p−1)2

2(x piεα2

pj x

)p2(n−1−j)(p2 − 1)

p2n−2−i(p−1)2

2(x pjβ2

pj x

)p2(n−1−j)(p2 + p)

p2n−2−j(p−1)2

2(x pjεβ2

pj x

)p2(n−1−j)(p2 − p) p2n−2−j(p−1)2

2



Proposition 3.3 In Table 1, the third column is correct which tells us the number of classes for

each type of representation matrix.

Proof:

To count the ‘number of classes’, one simply counts the different values that can be taken for each

representative while taking into consideration if any class contains more than one representative, for

example we look at AI : (x 0

0 x

)We can pick x as any value in (Z/pnZ)× and clearly each of these representatives gives a unique

class, so the number of classes is |(Z/pnZ)×| = pn − pn−1. If fact, due to the proof in the previous

section we do not need to worry too much about any class containing more than one or two

representative; the proof in that section is general enough to be applicable to all representatives,

but in that proof we did replace squared terms for s ∈ Z/pnZ, so that proof does not take into

account that in AT , for example, −y and y give the same class. So the previous section tells us that

no class contains more than one or two representative, depending on whether there is a squared

term in the matrix or not. We will now verify, case by case, the number of classes for each matrix type:

Matrix AI

We can pick x as any value in (Z/pnZ)×, so the number of classes is |(Z/pnZ)×| = pn − pn−1.

34

Matrix AB

We can pick x ∈ (Z/pnZ)×, so the number of classes is |(Z/pnZ)×| = pn − pn−1.

Matrix AT

We can pick y ∈ (Z/pnZ)× and w ∈ Z/pnZ as long as y 6≡ ±w (mod p). Also, the class given

by y is also given by −y, so the number of classes is 12|(Z/pnZ)×| · |(Z/pnZ)\w ∈ Z/pnZ : y ≡

±w (mod p)| = 12

(pn − pn−1)(pn − 2pn−1) =p2n−2(p−1)(p−2)

2.

Matrix AK

We can pick y ∈ (Z/pnZ)× and z ∈ Z/pnZ. Also, the class given by y is also given by −y, so the

number of classes is 12|(Z/pnZ)×| · |Z/pnZ| = 1

2(pn − pn−1)(pn) =

p2n−1(p−1)2

.

Matrix ART,i

We can pick x ∈ (Z/pnZ)× and α ∈ (Z/pn−iZ)×. Also, the class given by α is also given by−α, so the

number of classes is 12|(Z/pnZ)×| · |(Z/pn−iZ)×| = 1

2(pn− pn−1)(pn−i− pn−i−1) =

p2n−2−i(p−1)2

2.

Matrix ARK,i



2(pn− pn−1)(pn−i− pn−i−1) =

p2n−2−i(p−1)2

2.

Matrix ARB,j

We can pick x ∈ (Z/pnZ)×, so the number of classes is |(Z/pnZ)×| = pn − pn−1.

Matrix ARBI,j,i or ARBJ,j,i



2(pn− pn−1)(pn−i− pn−i−1) =

p2n−2−i(p−1)2

2.

Matrix ARI,j or ARJ,j

We can pick x ∈ (Z/pnZ)× and β ∈ (Z/pn−jZ)×. Also, the class given by β is also given by−β, so the

number of classes is 12|(Z/pnZ)×| · |(Z/pn−jZ)×| = 1

2(pn−pn−1)(pn−j−pn−j−1) =

p2n−2−j(p−1)2

2.

Verifying Table 1 is exhaustive

Proposition 3.4 Table 1 gives the complete list of conjugacy classes of GL2(Z/pnZ).

Proof:

Now we must finally check the identity∑

(no. of elts per class)(no. of classes)= |GL2(Z/pnZ)| =

p4n−3(p2 − 1)(p− 1).

35

Recall that this sum is a sum over each representation matrix:∑(no. of elts per class)(no. of classes)

= pn − pn−1 + p2n−2(p2 − 1)(pn − pn−1) + p2n−2(p2 + p)(p2n−2(p−1)(p−2)

2

)+p2n−2(p2 − p)

(p2n−1(p−1)

2

)+∑ip2n−2(p2 − 1)

(p2n−2−i(p−1)2

2

)+∑ip2n−2(p2 − 1)

(p2n−2−i(p−1)2

2

)+∑jp2(n−1−j)(p2 − 1)

(pn−1(p− 1)

)+∑j,ip2(n−1−j)(p2 − 1)

(p2n−2−i(p−1)2

2

)+∑j,ip2(n−1−j)(p2 − 1)

(p2n−2−i(p−1)2

2

)+∑jp2(n−1−j)(p2 + p)

(p2n−2−j(p−1)2

2

)+∑jp2(n−1−j)(p2 − p)

(p2n−2−j(p−1)2

2

)

= pn−1(p− 1) + p3n−3(p2 − 1)(p− 1) + p4n−3(p2 − 1)( p−22

) + p4n−3(p− 1)2( p2

)

+∑ip4n−4−i(p2 − 1)(p− 1)2 +

∑jp3n−3−2j(p2 − 1)(p− 1)

+∑j,ip4n−4−2j−i(p2 − 1)(p− 1)2 +

∑jp4n−2−3j(p− 1)2

This is very messy so we will make things clearer by simplifying all the sums separately:

n−1∑i=1

p4n−4−i(p2 − 1)(p− 1)2 = p4n−5(p2 − 1)(p− 1)2n−2∑i=0

p−i

= p4n−5(p2 − 1)(p− 1)2(

1−p1−n1−p−1

)= p4n−4(p2 − 1)(p− 1)

−p3n−3(p2 − 1)(p− 1)

∑jp3n−3−2j(p2 − 1)(p− 1) = p3n−5(p2 − 1)(p− 1)

(1−p2−2n

1−p−2

)= p3n−3(p− 1)− pn−1(p− 1)

∑j,ip4n−4−2j−i(p2 − 1)(p− 1)2 = p4n−7(p2 − 1)(p− 1)2

n−2∑i=0

p−ii−1∑j=0

p−2j

= p4n−7(p2 − 1)(p− 1)2n−2∑i=0

p−i(

1−p−2i

1−p−2

)= p4n−5(p− 1)2

(n−2∑i=0

p−i −n−2∑i=0

p−3i

)= p4n−4(p− 1)− p3n−3(p− 1)

− p4n−5(p−1)2(p3−p6−3n)

p3−1∑jp4n−2−3j(p− 1)2 =

p4n−5(p−1)2(p3−p6−3n)

p3−1

36

Now that we have done that, we return to verifying the identity:

∴∑

(no. of elts per class)(no. of classes)

= pn−1(p− 1) + p3n−3(p2 − 1)(p− 1) +p4n−3(p−1)

2((p+ 1)(p− 2) + (p− 1)p)

+p4n−4(p2 − 1)(p− 1)− p3n−3(p2 − 1)(p− 1) + p3n−3(p− 1)− pn−1(p− 1)

+p4n−4(p− 1)− p3n−3(p− 1)− p4n−5(p−1)2(p3−p6−3n)

p3−1+p4n−5(p−1)2(p3−p6−3n)

p3−1

= (pn−1(p− 1)− pn−1(p− 1)) + (p3n−3(p2 − 1)(p− 1)− p3n−3(p2 − 1)(p− 1))

+p4n−3(p− 1)(p2 − p− 1) + p4n−4(p2 − 1)(p− 1) + p4n−4(p− 1)

+(p3n−3(p− 1)− p3n−3(p− 1)) + (p4n−5(p−1)2(p3−p6−3n)

p3−1− p4n−5(p−1)2(p3−p6−3n)

p3−1)

= 0 + 0 + p4n−3(p− 1)(p2 − p− 1) + p4n−4(p2 − 1)(p− 1) + p4n−4(p− 1) + 0 + 0

= p4n−3(p2 − 1)(p− 1)− p4n−2(p− 1) + p4n−2(p− 1)− p4n−4(p− 1)

+p4n−4(p− 1)

= p4n−3(p2 − 1)(p− 1) = |GL2(Z/pnZ)|

With this proposition we have also proved Theorem 3.2.

37

3.3 Elements of order prime to p

In this section we work out the elements in GL2(Z/pnZ) which have order prime to p.

Clearly if a matrix has order prime to p then any conjugate matrices also have order prime to p. So

we inspect conjugacy classes of GL2(Z/pnZ):


0 x

) (x 0

1 x

) (w y2

1 w

) (z εy2

1 z

) (x piα2

1 x

) (x piεα2

1 x


pj x

) (x piα2

pj x

) (x piεα2

pj x

) (x pjβ2

pj x

) (x pjεβ2

pj x

)



Proposition 3.5 There are only (p−1)(p2n+1−p2n−p2n−1 +1) matrices which have order prime

to p. To be precise there are p − 1 matrices in the form AI ,p2n−1(p2−1)(p−2)

2matrices conjugate

to matrices in the form AT andp2n(p−1)2

2matrices conjugate to matrices in the form AK .

Proof:

We will inspect each matrix representation case by case.

Matrix AI :

Clearly the order of AI =

(x 0

0 x

)is the same as x, so the order of AI divides pn − pn−1. So AI is

only prime to p if the order of x divides p− 1, therefore there are precisely p− 1 values that x can

take such that AI has order prime to p.

The identity matrix has order 1 and the rest of the matrices have order p− 1.

Matrix AB:

AmB =

(x 0

1 x

)m=

(xm 0

mxm−1 xm

)If AmB = I2 then xm = 1 and mxm−1 = 0, but x ∈ (Z/pnZ)× so mxm−1 = 0 if and only if pn|mtherefore the order of AB is never prime to p.

Matrix AT :

AmT =

(w y2

1 w

)m=

(y y2

1 −y

)−1(w + y 0

0 w − y

)m(y y2

1 −y

)

38

If AmT = I2 then (w + y)m = 1 and (w − y)m = 1. Set w + y = a and w + y = d then we get

a, d ∈ (Z/pnZ)× such that a 6= d. We know there are precisely p − 1 values x ∈ (Z/pnZ)× such

that the order of x is prime to p, therefore there are precisely p − 1 values that a can take and

p − 2 values that d can take such that AT is prime to p. So there are (p − 1)(p − 2) matrices in

the form AT which have order prime to p. However, these values do not always give us distinct

conjugacy classes; by proposition 3.3 we know that

(w y2

1 w

)is conjugate to both

(a 0

0 d

)and(

d 0

0 a

)but no other matrices in this form. Therefore there are

(p−1)(p−2)2

conjugacy classes with

representation matrices in the form AT which have order prime to p.

Any matrix in the form AT has order p − 1 since at least one value, w − y or w + y, would

have order p− 1 while the other would have order 1 or p− 1.

Matrix AK :

Consider the projection map GL2(Z/pnZ) → GL2(Z/pZ). The kernal is a p-group of order p4n−4

so any elements with order prime to p in the GL2(Z/pZ) will have a unique lift to an element in

GL2(Z/pnZ) which also has order prime to p.

AK =

(z εy2

1 z

)

Matrices in the form AK can be looked at as elements7in F×p2

and we clearly have an injection from

F×p2

to Z×p2

. We know Z×p2∼= F×

p2× P where P is some pro-p group, i.e. some group which is the

inverse limit of p-groups so every element in P has order p. AK maps to x + εy ∈ Z×p2

but for AK

to have order prime to p, we would need x+ εy ∈ F×p2× 1.

There are p choices for z and p − 1 choices for y that give us x + εy ∈ F×p2× 1. Therefore there

are p(p− 1) matrices in the form AK which have order prime to p. For a similar reason to the AT

case, we find that there arep(p−1)

2conjugacy classes with representation matrices in the form AK

which have order prime to p.

All matrices in this form have order which divides p2 − 1.

Matrix ART,i:

If we project ART,i ontoGL2(Z/pZ), we get the same matrix as the projection of AB ontoGL2(Z/pZ)

and we know that the projection of AB has order divisible by p thus the projection of ART,i has

order divisible by p. The order of ART,i is divisible by the order of its projection, so ART,i never

has order prime to p.

Matrix ARK,i:

This is exactly the same strategy as ART,i. So ARK,i never has order prime to p.

7Fp2

is the finite field of p2 elements and Zp2

is as unramified extension of Zp of degree 2. Fp2

can also be

thought of as a degree 2 extension of Fp and this would make it clear how we get a natural injection from F×p2

to

Z×p2

39

Matrix ARB,j:

Consider the projection map GL2(Z/pnZ) → GL2(Z/pZ). The kernal is a p-group of order p4n−4

and any element of order prime to p would project to an element of order prime to p. Therefore

any elements with order prime to p in the group GL2(Z/pZ) will have a unique lift to an element in

GL2(Z/pnZ) which also has order prime to p.

When projecting ARB,j to GL2(Z/pZ), we get the same elements when projecting AI to

GL2(Z/pZ). So we project to matrices in the form AI which lie in GL2(Z/pZ), but any matrix of

this form which also has order prime to p already lifts to a matrix in the form AI in GL2(Z/pnZ)

which has order prime to p. As a result, we know that there are no elements in the form ARB,j

which have order prime to p.

Matrix ARBI,j,i, ARBJ,j,i, ARI,j or ARJ,j:

This is exactly the same strategy as ARB,j . So ARBI,j,i, ARBJ,j,i, ARI,j and ARJ,j never have

order prime to p.

Therefore the total number of matrices which have order prime to p is∑

(matrix reps. which have

order prime to p)(no. of elts per conjugacy class of the matrix rep.):

p− 1 +(p−1)(p−2)

2× p2n−2(p2 + p) +

p(p−1)2× p2n−2(p2 − p)

= p− 1 +p2n−1(p2−1)(p−2)

2+p2n(p−1)2

2

= (p− 1)(2+p2n−1(p+1)(p−2)+p2n(p−1)

2)

= (p− 1)(2+p2n−1(p2−p−2+p2−p)

2)

= (p− 1)(1 + p2n−1(p2 − p− 1))

= (p− 1)(p2n+1 − p2n − p2n−1 + 1)

40

Chapter 4

Image of ψn

Recall diagram (1) from Chapter 2.6.2:


↓ θn ↓ ψnker(L) 99K

∏U∈Fn Λ(Uab)×

L99K Qp ⊗Zp


In the previous chapter we stated a suitable set of subgroups of Gn, called Fn. In this chapter we

verify that ψn (Definition 4.1) is injective with this choice of Fn.

After proving that ψn is injective, we provide the image of ψn which we call Ψn,Zp (defined in

Theorem 4.2). This will allow us to construct the map L in the next chapter.

4.1 Preliminaries

First recall Fn = Zn, Cn, Tn,Kn, Nti , Nki |∀i = 1, 2, ..., n − 1 where these subgroups are defined

as follows:

Zn :=

(a 0

0 a

): a ∈ (Z/pnZ)×

Cn :=

(a 0

c a

):a ∈ (Z/pnZ)×

c ∈ Z/pnZ

Tn :=

(a 0

0 d

): a, d ∈ (Z/pnZ)×

Kn :=

(a εb

b a

):a, b ∈ Z/pnZs.t. p 6 |(a2 − εb2)

Nti :=

(a bpi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

Nki :=

(a bεpi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

Next we recall the definition of ψn:

41

Definition 4.1 Let R be any Zp-algebra of characteristic 0. For U ∈ Fn, let TrGn/U be the map:

R[Conj(Gn)] → R[Conj(U)]

[A]Gn 7→∑

X∈U\GnX−1AX∈U

[X−1AX]U

and let projU be the natural projection from R[Conj(U)] to R[Conj(Uab)] = R[Uab], then:

ψn : R[Conj(Gn)] →∏U∈Fn R[Uab]

ψn : [A]Gn 7→∏U∈Fn projU TrGn/U ([A]Gn )

Remark: Although this chapter concentrates on finding the image of the map ψn in the case

R = Zp, the results in this chapter extend to any R i.e. for any Zp-algebra with characteristic 0, the

image of the map ψn is Ψn,R (defined in Theorem 4.2).

Throughout this chapter ψn will refer to the case R = Zp. By [12] we know that, when

n = 1, we can choose F1 = C1, Z1, T1,K1, so ψ1 is the following map:

ψ1 : Zp[Conj(G1)] −→ Zp[C1]× Zp[Z1]× Zp[T1]× Zp[K1]

Remark: If we compare this choice of F1 and our above choice of Fn, we notice that they coincide

for n = 1. We also notice that the subgroups Nti and Nki do not appear. This is because we

defined i to only takes values 1, 2, ..., n− 1. In this case n = 1, so i ∈ ∅.

For each individual subgroup, we use this notation: ψU := projU TrG/U ([A]G). This means we

can write ψn =∏U∈Fn ψU .

It will be useful if we now set up a more detailed notation to refer to the matrix representa-

tives of the conjugacy classes of GL2(Z/pnZ) from section 3.2:

ix c0x,1 AT AK ART,i ARK,i(x 0

0 x

) (x 0

1 x

) (w y2

1 w

) (z εy2

1 z

) (x piα2

1 x

) (x piεα2

1 x

)rcjx,1 ARBI,j,i ARBJ,j,i ARI,j ARJ,j(x 0

pj x

) (x piα2

pj x

) (x piεα2

pj x

) (x pjβ2

pj x

) (x pjεβ2

pj x

)



Where ε is a fixed non-square element in (Z/pnZ)×. Set c0x,y :=

(x 0

y x

)and

rcjx,β :=

(x 0

βpj x

). Also, note that we have not set up different notation for

AT , AK , ART,i, ARK,i, ARBI,j,i, ARBJ,j,iARI,j and ARJ,j . This is because, from now on, it is

better to use their respective conjugates

t0w,y :=

(w + y 0

0 w − y

)=

(1 y

−1 y

)(w y2

1 w

)(1 y

−1 y

)−1

k0z,y :=

(z εy

y z

)=

(0 y

1 0

)−1(z εy2

1 z

)(0 y

1 0

)

42

rtix,α :=

(x piα

α x

)=

(α 0

0 1

)−1(x piα2

1 x

)(α 0

0 1

)

rkix,α :=

(x piεα

α x

)=

(α 0

0 1

)−1(x piεα2

1 x

)(α 0

0 1

)

rcij,ix,α :=

(x piα

pjα x

)=

(α 0

0 1

)−1(x piα2

pj x

)(α 0

0 1

)

rcjj,ix,α :=

(x piεα

pjα x

)=

(α 0

0 1

)−1(x piεα2

pj x

)(α 0

0 1

)

rijx,β :=

(x+ βpj 0

0 x− βpj

)=

(1 β

−1 β

)(x pjβ2

pj x

)(1 β

−1 β

)−1

rjjx,β :=

(x pjεβ

pjβ x

)=

(0 β

1 0

)−1(x pjεβ2

pj x

)(0 β

1 0

)

Remark: The notation we use is referencing the paper [3] where we found the list of conjugacy

classes of Gn.

4.2 Image of ψn

To calculate the output of each matrix under the map ψn, we need to calculate the trace maps and

the projection maps. The trace maps are calculation heavy so we will leave full calculations of the

trace maps in section 4.3. Since every subgroup in Fn is already Abelian, these projection maps are

just the identity.

Here is a table of each map ψU applied to each conjugacy class for all U ∈ FnTable 3:

ψZn ψCn ψTn ψKn

ix [Gn : Zn]ix [Gn : Cn]ix [Gn : Tn]ix [Gn : Kn]ix

c0x,1 0∑

y∈(Z/pnZ)×

c0x,y 0 0

t0w,y 0 0 t0w,y + t0w,−y 0

k0z,y 0 0 0 k0

z,y + k0z,−y

rtix,α 0 0 0 0

rkix,α 0 0 0 0

rcjx,1 0 p2j∑

β∈(Z/pn−jZ)×

rcjx,β 0 0

rcij,ix,α 0 0 0 0

rcjj,ix,α 0 0 0 0

rijx,β 0 0 p2j(rijx,β + rijx,−β) 0

rjjx,β 0 0 0 p2j(rjjx,β + rjjx,−β)

43

ψNtu ψNku

ix [Gn : Ntu ]ix [Gn : Nku ]ix

c0x,1 0 0

t0w,y 0 0

k0z,y 0 0

rtix,α δiu(rtix,α + rtix,−α) 0

rkix,α 0 δiu(rkix,α + rkix,−α)

rcjx,1 p2ju+j∑v=1

δvn

∑β∈(Z/pn−jZ)×

rcjx,β

p2ju+j∑v=1

δvn


rcjx,β

rcij,ix,α δ(i−j)up

2j(rcij,ix,α + rcij,ix,−α) 0

rcjj,ix,α 0 δ(i−j)up2j(rcjj,ix,α + rcjj,ix,−α)

rijx,β 0 0

rjjx,β 0 0

We want to prove that ψn is injective and most effective and efficient way to do this seems

to be constructing a left inverse for ψn(Conj(Gn)). We will call this map δn. This map is

constructed by inspection of the above table, but here we will state δn and then prove it is the left

inverses via verification:

δn :∏

U∈Fn

Zp[U ]→ Qp[Conj(Gn)]

When defining δn, we write U ∩ Zm with U = Cn, Tn,Kn, Nti or Nki . We are using Zm to denote

the pre-image of Zm from Gm to Gn, i.e.

Zm :=

(a pmb

pmc a+ pmd

):a ∈ (Z/pnZ)×

b, c, d ∈ Z/pn−mZ

Also note that Zn ⊂ U for any U ∈ Fn, so any element aZn ∈ Zp[Zn] can also be thought of as an

element in Zp[U ].

Let δn =∑U∈Fn δU such that for any (aV )V ∈Fn ∈

∏U∈Fn Zp[U ], we have:

δU ((aV )) :=

aZn[Gn:Zn]

if U = Zn

1[NGn (U):U ]

(aU −

TrU/U∩Z1(aU )

[U :U∩Z1]

)+∑

1≤m<n1

[NGn (U∩Zm):U∩Zm]

×(TrU/U∩Zm (aU )−

TrU/U∩Zm+1(aU )

p

)if U = Tn,Kn or Cn

1[NGn (U):U ]

(aU −

TrU/U∩Z1(aU )

[U :U∩Z1]

)+∑

1≤m<n−i1

[NGn (U∩Zm):U∩Zm]

×(TrU/U∩Zm (aU )−

TrU/U∩Zm+1(aU )

p

)if U = Nti or Nki

Such that, whenever we have groups U ⊂ V , we say NV (U) denotes the normalizer of U as a subgroup

of V .

Theorem 4.1 δn ψn = 1Zp[Conj(Gn)]

In the definition of δn, we work out the traces of certain elements. So for each conjugacy class

[A] ∈ Conj(Gn), it will be useful to the proof of Theorem 4.1 if we know TrU/U∩Zm (ψU ([A])) for

44

U = Cn, Tn,Kn, Nti or Nki :

We know that the image of ψCn only contains terms with matrices in the form ix, rcjx,β and c0x,y .

So we only need to apply the trace map onto matrices in the aforementioned forms:

TrCn/(Zm∩Cn)(A) =

[Cn : Zm ∩ Cn]ix if A = ix

pmrcjx,β if A = rcjx,β and j ≥ m

0 otherwise

We know that the image of ψTn only contains terms with matrices in the form ix, rijx,β and t0w,y :

TrTn/(Zm∩Tn)(A) =

[Tn : Zm ∩ Tn]ix if A = ix

pm−1(p− 1)rijx,β if A = rijx,β and j ≥ m

0 otherwise

We know that the image of ψKn only contains terms with matrices in the form ix, rjjx,β and k0z,y :

TrKn/(Zm∩Kn)(A) =

[Kn : Zm ∩Kn]ix if A = ix

pm−1(p+ 1)rjjx,β if A = rjjx,β and j ≥ m

0 otherwise

We know that the image of ψNti

only contains terms with matrices in the form ix, rtix,α, rcij,ix,α and

rcjx,β :

TrNti/(Zm∩Nti )

(A) =

[Nti : Zm ∩Nti ]ix if A = ix

pmrcij,i+jx,α if A = rcij,i+jx,α and j ≥ m

pmrcjx,β if A = rcjx,β , i+ j ≥ n and j ≥ m

0 otherwise

We know that the image of ψNki

only contains terms with matrices in the form ix, rkix,α, rcjj,ix,α

and rcjx,β :

TrNki/(Zm∩Nki )

(A) =

[Nki : Zm ∩Nki ]ix if A = ix

pmrcjj,i+jx,α if A = rcjj,i+jx,α and j ≥ m

pmrcjx,β if A = rcjx,β , i+ j ≥ n and j ≥ m

0 otherwise

In fact, we have [Cn : Zm∩Cn] = pm, [Tn : Zm∩Tn] = pm−1(p−1), [Kn : Zm∩Kn] = pm−1(p+1),

[Nti : Zm ∩Nti ] = pm and [Nki : Zm ∩Nki ] = pm but we have chosen to separate the case A = ix

in this way because it makes the proof easier to understand. Details of these calculations can be

found in section 4.3.

Proof of Theorem 4.1:

By definition δn ψn = 1Zp[Conj(Gn)] means that δn ψn acts like the identity on Conj(Gn). So

we will need to verify that δU ψn acts like the identity on each conjugacy class individually:

1. For ix, using the table we can easily see that δZn ψn([ix]) = [ix]. To prove that δnψn([ix]) =

[ix], we need to show that δU ψn([ix]) = 0 if U is not Zn. From the table we know that

ψn([ix]) = (aU )U∈Fn where aU = [Gn : U ][ix]. Notice that[U :U∩Zm+1]

p= [U : U ∩ Zm] for

1 ≤ m < n. We also have TrU/U∩Zm (aU ) = [U : U ∩ Zm]aU . Therefore we have:

TrU/U∩Zm (aU )−TrU/U∩Zm+1

(aU )

p= 0

45

and

aU −TrU/U∩Z1

(aU )

[U : U ∩ Z1]= 0

Therefore δU ψn([ix]) = 0 for all U ∈ Fn\Zn.

2. The cases for rcjx,1 and c0x,1 are similar so we will do c0x,1 and immediately do rcjx,1 after. For

c0x,1, using the table we know that δU ψn([c0x,1]) = 0 unless U is Cn, thus δn ψn([c0x,1]) =

δCn ψn([c0x,1]). Therefore we want to prove that δCn ψn([c0x,1]) = [c0x,1]:

In section 4.3 we found that:

NGn (Cn) =

(a 0

c d

):a, d ∈ (Z/pnZ)×

c ∈ Z/pnZ

and that:

NGn (Zm ∩ Cn) =

(a b0pn−m

c d

):a, d ∈ (Z/pnZ)×

c, b0 ∈ Z/pnZ

therefore [NGn (Cn) : Cn] = pn−1(p− 1) and [NGn (Zm ∩ Cn) : Zm ∩ Cn] = pn−1+2m(p− 1).

Now we can prove that δCn ψn([c0x,1]) = [c0x,1]:

δCn ψn([c0x,1]) = δCn

∑y∈(Z/pnZ)×

c0x,y

=

1

pn−1(p− 1)

∑y∈(Z/pnZ)×

[c0x,y ]− 0

+∑

1≤m<n

1

pn−1+2m(p− 1)(0− 0)

=∑

y∈(Z/pnZ)×

1

pn−1(p− 1)[c0x,1] + 0 = [c0x,1]

Now we work on the case with matrices in the form rcjx,1. Like before, by looking at the table

we find that δU ψn([rcjx,1]) = 0 if U is not Cn, Nti or Nki when i ≥ n− j. Thus we have

δn ψn([rcjx,1]) = δCn ψn([rcjx,1]) +

n−1∑i=n−j

(δNti ψn([rcjx,1]) + δN

ki ψn([rcjx,1]))

We want to show δn ψn([rcjx,1]) = [rcjx,1]. First we will calculate δCn ψn([rcjx,1]):

δCn ψn([rcjx,1])

= 1pn−1(p−1)

(p2j


[rcjx,β ]− p2j∑

β∈(Z/pn−jZ)×[rcjx,β ]

)

+∑

1≤m<j

1pn−1+2m(p−1)

(pm+2j


[rcjx,β ]− pm+2j∑


)

+ 1pn−1+2j(p−1)

(p3j


[rcjx,β ]− 0

)+

∑j<m<n

1pn−1+2m(p−1)

(0− 0)

=∑

β∈(Z/pn−jZ)×

1pn−j−1(p−1)

[rcjx,β ] + 0

= [rcjx,1]

46

Now we will calculate δNti ψn([rcjx,1]):

δNti ψn([rcjx,1])

= 1[NGn (N

ti):N

ti]

(p2j


[rcjx,β ]− p2j∑


)

+∑

1≤m<n−i

1[NGn (Zm∩Nti ):Zm∩Nti ]

(pm+2j


[rcjx,β ]− pm+2j∑


)

= 0

Note that j does not appear in the sum from m = 1 to n− i since i ≥ n− jThis would be the same for δN

ki ψn([rcjx,1]), so we will get

δn ψn([rcjx,1]) = δCn ψn([rcjx,1]) +n−1∑i=n−j

(δNti ψn([rcjx,1]) + δN

ki ψn([rcjx,1]))

= [rcjx,1] +n−1∑i=n−j

(0 + 0)

= [rcjx,1]

3. We will mirror the previous case; looking at t0w,y first and then rijx,β . For t0w,y , we know that

δU ψn([t0w,y ]) = 0 unless U is Tn, thus δn ψn([t0w,y ]) = δTn ψn([t0w,y ]). So we want to show

δTn ψn([t0w,y ]) = [t0w,y ]:


NGn (Tn) =

(a 0

0 d

),

(0 b

c 0

): a, b, c, d ∈ (Z/pnZ)×

and that NGn (Zm ∩ Tn) =(

a b0pn−m

c0pn−m d

),

(a0pn−m b

c d0pn−m

):a, b, c, d ∈ (Z/pnZ)×

a0, b0, c0, d0 ∈ Z/pnZ

therefore [NGn (Tn) : Tn] = 2 and [NGn (Zm ∩ Tn) : Zm ∩ Tn] = 2p3m−1(p − 1). Now we can

prove that δTn ψn([t0w,y ]) = [t0w,y ]:

δTn ψn([t0w,y ]) = δTn

(t0w,y + t0w,−y

)=

1

2

([t0w,y ] + [t0w,−y ]

)+

∑1≤m<n

1

2p3m−1(p− 1)(0)

= [t0w,y ]

For rijx,β , we also find δn ψn([rijx,β ]) = δTn ψn([rijx,β ]). So we want to show δTn ψn([rijx,β ]) = [rijx,β ]:

δTn ψn([rijx,β ]) =1

2(0) +

∑1≤m<j

1

2p3m−1(p− 1)(0)

+1

2p3j−1(p− 1)

(p3j−1(p− 1)([rijx,β ] + [rijx,−β ])

)

+∑

j<m<n

1

2p3m−1(p− 1)(0)

= [rijx,β ]

47

4. We will mirror the previous case; looking at k0z,y first and then rjjx,β . For k0

z,y , we know that

δU ψn([k0z,y ]) = 0 unless U is Kn, thus δn ψn([k0

z,y ]) = δKn ψn([k0z,y ]). So we want to

show δKn ψn([k0z,y ]) = [k0

z,y ]:


NGn (Kn) =

(a εb

b a

),

(a −εbb −a

):a, b ∈ Z/pnZa = 0 =⇒ b 6= 0

and that NGn (Zm ∩Kn) =(a εb+ kpn−m

b a+ lpn−m

),

(a −εb+ kpn−m

b −a+ lpn−m

):a, b, k, l ∈ Z/pnZa = 0 =⇒ b 6= 0

therefore [NGn (Kn) : Kn] = 2 and [NGn (Zm ∩Kn) : Zm ∩Kn] = 2p3m−1(p+ 1). Now we can

prove that δKn ψn([k0z,y ]) = [k0

z,y ]:

δKn ψn([k0z,y ]) = δKn

(k0z,y + k0

z,−y

)=

1

2

([k0z,y ] + [k0

z,−y ])

+∑

1≤m<n

1

2p3m−1(p+ 1)(0)

= [k0z,y ]

For rjjx,β , we also find δn ψn([rjjx,β ]) = δKn ψn([rjjx,β ]). So we want to show δKn ψn([rjjx,β ]) = [rjjx,β ]:

δKn ψn([rjjx,β ]) =1

2(0) +

∑1≤m<j

1

2p3m−1(p+ 1)(0)

+1

2p3j−1(p+ 1)

(p3j−1(p+ 1)([rjjx,β ] + [rjjx,−β ])

)

+∑

j<m<n

1

2p3m−1(p− 1)(0)

= [rjjx,β ]

5. We will now do the rest of the cases, rtix,α, rkix,α, rcij,ix,α and rcjj,ix,α. These cases are grouped

because these matrices are in the same form, A =

(x ε0piα

pjα x

)where 0 ≤ j < i < n

and ε0 is either ε, a fixed square-free element, or 1. Recall that rtix,α ∈ Nti , rcij,i+jx,α ∈ Nti ,

rkix,α ∈ Nki and rcjj,i+jx,α ∈ Nki . We will denote the groups, Nti and Nki as:

NA :=

(a bε0pi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

where i and ε0 are picked from the matrix A.

From the table, we can see that δn ψn([A]) = δNA ψn([A]), so we need to show that

δNA ψn([A]) = [A]. In section 4.3 we found:

NGn (NA) =

(a bε0pi

b a

),

(a −bε0pi

b −a

)and that

NGn (Zm ∩NA) =

(a bε0pi−m + kpn−m

b a+ lpn−m

),

(a −bε0pi−m + kpn−m

b −a+ lpn−m

)

48

where a ∈ (Z/pnZ)×, b ∈ Z/pnZ and k, l ∈ Z/pjZ such that α, i, j and ε0 are picked from the

matrix A and in the case that A = rtix,α or rkix,α we set j = 0. Therefore [NGn (NA) : NA] = 2

and [NGn (Zm ∩Nti ) : Zm ∩Nti ] = 2p3m. Now we can prove that δNA ψn([A]) = [A]:

In the case A = rtix,α or rkix,α:

δNA ψn([A]) = δNA

((x ε0αpi

α x

)+

(x −ε0αpi

−α x

))

=1

2

([(x ε0αpi

α x

)]+

[(x −ε0αpi

−α x

)])

+∑

1≤m<n−i

1

2p3m(0)

= [A]

In the case A = rcij,i+jx,α or rcjj,i+jx,α :

δNA ψn([A]) = δNA

(p2j

((x ε0αpi+j

αpj x

)+

(x −ε0αpi+j

−αpj x

)))

=1

2(0) +

∑1≤m<j

1

2p3m(0)

+1

2p3j

(p3j

([(x ε0αpi+j

αpj x

)]+

[(x −ε0αpi+j

−αpj x

)]))

+∑

j<m<n−i

1

2p3m(0)

= [A]

Note that j appear in the sum from m = 1 to n − i because if it did not, we would have

j ≥ n− i and this condition would give us a matrix in the form rcjx,β .

Thus Theorem 4.1 proves that ψn is injective for our choice of Fn.

As mentioned in chapter 2.6.2, we want to construct the map named L. To do this, it will

be useful to describe the image of ψn:

Theorem 4.2 Let R be a Zp-algebra of characteristic 0. We define Ψn,R in the following way:

1. Ψn,R ⊂ p3n−2R[Zn]×R[Cn]×R[Tn]×R[Kn]×n−1∏i=1

(R[Nti ]×R[Nki ])

2. For any (aV )V ∈Fn ∈ Ψn,R, each aV is fixed by conjugation action of NGn (V )

3. For any (aV )V ∈Fn ∈ Ψn,R, we have:

• TrV/Zn (aV ) = aZn for all V ∈ Fn

• TrNti/Zm∩Nti

(aNti

) = TrCn/Zm∩Cn (aCn ) for m ≥ n− i

49

• TrNki/Zm∩Nki

(aNki



• TrCn/Zm∩Cn (aCn ) ∈ p3mR[Cn]

• TrTn/Zm∩Tn (aTn ) ∈ p3m−1R[Tn]

• TrKn/Zm∩Kn (aKn ) ∈ p3m−1R[Kn]

• TrNti/Zm∩Nti

(aNti

) ∈ p3mR[Nti ]

• TrNki/Zm∩Nki

(aNki

) ∈ p3mR[Nki ]

Then im(ψn) = Ψn,R. So in our particular case, R = Zp, we have im(ψn) = Ψn,Zp .

Proof:

Looking at Table 3, we can clearly see that ψn satisfies conditions 1, 2 and 3 of Ψn,Zp . If we also

look at TrU/U∩Zm for U = Cn, Tn,Kn, Nti , or Nki (stated before the proof of Theorem 4.1) then

we also see that ψn satisfies condition 4 of Ψn,Zp therefore im(ψn) lies inside of Ψn,Zp . So to

prove this theorem, we will prove that δ is injective on Ψn,Zp . Let (aV )V ∈Fn ∈ Ψn,Zp such that

δn((aV )V ∈Fn ) = 0. By definition of δn, this means we have∑U∈Fn δU ((aV )) = 0. Our aim is to

prove that aU = 0 for each U ∈ Fn:

For all U ∈ Fn\Zn we will prove that δU ((aV )) = 0 and find an expression for aU in terms of

aZn . Then we will deduce that aZn is zero and as a result, we prove that aU = 0 for each U ∈ Fn.

Without loss of generality, set:

aCn =∑

x∈(Z/pnZ)×

∑y∈Z/pnZ

ax,yc0x,y

where ax,y is an element in Zp and c0x,y is the matrix

(x 0

y x

). By point (3), we know that

TrCn/Zn (aCn ) = aZn , therefore we get:

[Cn : Zm ∩ Cn]∑

x∈(Z/pnZ)×

∑y∈pm(Z/pnZ)

ax,yc0x,y = TrCn/Zm∩Cn (aCn )

Note that [Cn : Zn] = pn and [Cn : Zm ∩ Cn] = pm, thus we get two expressions:

aCn −TrCn/Z1∩Cn (aCn )

[Cn : Z1 ∩ Cn]=

∑x∈(Z/pnZ)×

∑y∈(Z/pnZ)×

ax,yc0x,y

and

TrCn/Zm∩Cn (aCn )−TrCn/Zm+1∩Cn (aCn )

p= pm

∑x∈(Z/pnZ)×

∑y∈pm(Z/pnZ)×

ax,yc0x,y

But we know that c0x,pmβ = rcmx,β . We also know that [c0x,y ] = [c0x,1] and [rcmx,y ] = [rcmx,1] for any

y ∈ (Z/p2Z)×, therefore we can use point (2), from the definition of Ψn,Zp , to say that the coefficients

ax,pmy are equal for different values of y, so we can simply write:

aCn −TrCn/Z1∩Cn (aCn )

[Cn : Z1 ∩ Cn]=

∑x∈(Z/pnZ)×

ax,1∑

y∈(Z/pnZ)×

c0x,y

and

TrCn/Zm∩Cn (aCn )−TrCn/Zm+1∩Cn (aCn )

p= pm

∑x∈(Z/pnZ)×

ax,pm∑

β∈(Z/pn−mZ)×

rcmx,β

50

Thus δCn ((aV )V ∈Fn ) =∑x∈(Z/pnZ)× ax,1[c0x,1] +

n−1∑m=1

1p2m

∑x∈(Z/pnZ)× ax,pm [rcmx,1]. Although

we are dividing by p2m, we remain in Zp[Cn] due to point (4) from the definition of Ψn,Zp . By the

definition of δn, we know that classes in the form [c0x,y ] and [rcmx,β ] can only appear in the image of

δCn (see the proof of Theorem 4.1 for details), thus, we must have δCn ((aV )V ∈Fn ) = 0. Therefore

we have:

aCn =TrCn/Z1∩Cn (aCn )

p

and

TrCn/Zm∩Cn (aCn ) =TrCn/Zm+1∩Cn (aCn )

p

By point (3) and the properties of the trace map, we have TrCn/Zn∩Cn (aCn ) = TrCn/Zn (aCn ) =

aZn . Therefore we get:

aCn =TrCn/Z1∩Cn (aCn )

p= · · · =

TrCn/Zn−1∩Cn (aCn )

pn−1=aZnpn

=aZn

[Cn : Zn]

So we can write aCn in terms of aZn :

aCn =aZn

[Cn : Zn]

We can do the exact same for Tn and Kn to obtain δTn ((aV )) = 0, δKn ((aV )) = 0, aTn =aZn

[Tn:Zn]

and aKn =aZn

[Kn:Zn].

We will now do the case aNti

. Without loss of generality, set:

aNti

=∑

x∈(Z/pnZ)×

∑α∈Z/pnZ

ax,αrtix,α

where ax,α is an element in Zp and rtix,α is the matrix

(x piα

α x

). By following the same steps

that we took for aCn , we get:

δNti

((aV )) = 0

and

aNti

=TrN

ti/Z1∩Nti

(aNti

)

p= · · · =

TrNti/Zn−i∩Nti

(aNti

)

pn−i

By point (3), we know that TrNti/Zn−i∩Nti

(aNti

) = TrCn/Zn−i∩Cn (aCn ), therefore we get:

aNti

=TrCn/Zn−i∩Cn (aCn )

pn−i=

aZn[Cn : Zn]

=aZn

[Nti : Zn]

We can do the same with aNki

to obtain δNki

((aV )V ∈Fn ) = 0 and aNki

=aZn

[Nki

:Zn].

Recall that we have∑U∈Fn δU ((aV )) = 0, but using the above results we can deduce that

δZn ((aV )) = 0, therefore aZn = 0. Throughout the proof, we have shown that for any U ∈ Fn, we

can write aU in terms of aZn , but aZn = 0, thus aU = 0 for all U ∈ Fn, therefore δ((aV )V ∈Fn ) = 0

if and only if (aV ) = 0, i.e. δ is injective. This proves that Ψn,Zp is the image of ψn.

51

4.3 Trace maps for Gn

Here we give details on how to calculate the trace maps of each conjugacy class over any subgroup

in Fn = Zn, Cn, Tn,Kn, Nti , Nki |∀i = 1, 2, ..., n−1. Each of these subgroups have been described

in chapter 4.1. We will also be calculating trace maps over the groups Zm ∩Cn, Zm ∩Tn, Zm ∩Kn,

Zm ∩Nti and Zm ∩Nki for m = 1, 2, ..., n− 1 where Zm denotes the pre-image of Zm from Gm to

Gn, i.e.

Zm :=

(a pmb

pmc a+ pmd

):a ∈ (Z/pnZ)×

b, c, d ∈ Z/pn−mZ

We define Cm, Tm, Km, N(m)

tiand N

(m)

kiin a similar way.

Throughout this section, we consider X =

(a b

c d

)to be any matrix in Gn and then look at

A′ := X−1AX where A is a representation matrix of a conjugacy class. Whichever group, U ∈ Fn,

that we are taking the trace over, we will want to find every matrix X such that A′ is in that group,

U . After we do that, we will want to see how many of these matrices are distinct inside of U\Gn,

then it will be easy to calculate the trace: TrGn/U ([A]Gn ) =∑

X∈U\GnX−1AX∈U

[X−1AX]U .

Whenever we have a group which contains no elements of a conjugacy class of a matrix, then the

trace of that matrix over that group is zero, i.e. if [A]Gn ∩ U = ∅ then TrGn/U ([A]Gn ) = 0.

In this chapter we set A′ = X−1AX and [A]U denotes the conjugacy class of A as an ele-

ment of U .

Case A = ix

Any matrix in the form ix is in the centre of Gn, so A′ = A and ix is in all the groups U ∈ Fn, so

for any U ∈ Fn we get the following:

TrGn/U ([A]Gn ) =∑

X∈U\GnX−1AX∈U

[X−1AX]U =∑

X∈U\Gn

[A]U = [Gn : U ][A]U

=|Gn||U |

[A]U

Case A = c0x,1 and A = rcjx,1

We are going to find the trace of c0x,1 and then find the trace of rcjx,1 because these cases are similar

and the work done for the A = c0x,1 case will help us with the A = rcjx,1 case. For the case A = c0x,1,

elements from [A]Gn are only contained in Cn and no other group in Fn, so we will have non-trivial

trace only over Cn: (a b

c d

)−1(x 0

1 x

)(a b

c d

)

52

=1

ad− bc

(−ab+ adx− bcx −b2

a2 ab+ adx− bcx

)To have this matrix, A′, in Cn, it is clear that −b2 and ab must be 0, so we must have b = 0. This

condition means we need X in the form

(a 0

c d

). But now we must find out how many of these

matrices are distinct in Cn\Gn, then we can apply the trace map. To do this, we are going to take

a generic element in Cn, multiply it by the matrix X and the result will be any matrix in Cn\Gnwhich is equivalent to the matrix X:(

x 0

y x

)(a 0

c d

)=

(ax 0

cx+ ay dx

)

Since b = 0 is divisible by p, we know that a and d must be invertible. If we set x = a−1 and

y = −ca−2 then we obtain the following:(a−1 0

−ca−2 a−1

)(a 0

c d

)=

(1 0

0 d′

)

where d′ = d/a. Therefore, in Cn\Gn, the matrices differ only by the values in the bottom right

entry of the above matrix, so the trace is only going to sum over different values of d:

TrGn/Cn ([A]Gn ) =∑

X∈Cn\GnX−1AX∈Cn

[X−1AX]Cn =∑d

[(x 0

d−1 x

)]Cn

=∑

y∈(Z/pnZ)×

[(x 0

y x

)]Cn

=∑

y∈(Z/pnZ)×

[c0x,y

]Cn

In the case A = rcjx,1, elements from [A]Gn are only contained in Cn, Nti and Nki for i ≥ n− j and

no other groups in Fn. Elements from [A]Gn are also contained in Zm∩Cn, Zm∩Nti and Zm∩Nkifor i ≥ n−j and m ≤ j but not Zm∩Tn or Zm∩Kn, so, out of all the groups we are interested in, we

will have non-trivial trace only over Cn, Nti , Nki , Zm∩Cn, Zm∩Nti and Zm∩Nki for i ≥ n−j and

m ≤ j. But (Zm∩Cn) ⊂ Cn, so we know that TrGn/(Zm∩Cn) = TrCn/(Zm∩Cn) TrGn/Cn , so once

we find TrGn/Cn ([A]) we can calculate TrGn/(Zm∩Cn) by just taking the trace of TrGn/Cn ([A]).

We also have (Zm ∩ Nti ) ⊂ Nti and (Zm ∩ Nki ) ⊂ Nki , so we can also find TrGn/(Zm∩Nti )and

TrGn/(Zm∩Nki )by first finding the trace of TrGn/Nti

([A]) and TrGn/Nki([A]) respectively.

First we will find the trace over Cn and Zm ∩ Cn then we will do the other cases. Now we

follow the same steps that we took for the case A = c0x,1:(a b

c d

)−1(x 0

βpj x

)(a b

c d

)

=1

ad− bc

(−abβpj + adx− bcx −b2βpj

a2βpj abβpj + adx− bcx

)So −b2βpj = 0 = abβpj , but p 6 |β therefore pn−j |b:(

a b0pn−j

c d

)−1(x 0

βpj x

)(a b0pn−j

c d

)=

(x 0

a2βad−b0cpn−j

pj x

)Like before, we must find out how many of these matrices are distinct in Cn\Gn:(

x 0

y x

)(a b0pn−j

c d

)=

(ax b0pn−jx

cx+ ay dx+ b0pn−jy

)

53

Since b0pn−j is divisible by p, we know that a and d must be invertible:(a−1 0

−ca−2 a−1

)(a b0pn−j

c d

)=

(1 b′pn−j

0 d′

)

where b′ = b0/a and d′ = d/a− b0ca−2pn−j :

TrGn/Cn ([A]Gn ) =∑

X∈Cn\GnX−1AX∈Cn

[X−1AX]Cn =∑b,d

[(x 0

d−1pj x

)]Cn

= p2j∑

β∈(Z/pn−jZ)×

[(x 0

βpj x

)]Cn

= p2j∑

β∈(Z/pn−jZ)×

[rcjx,β

]Cn

Now we can calculate the trace of A over Zm ∩Cn. We can use our previous calculation to see that,

for A ∈ (Zm ∩ Cn), we have X−1AX ∈ (Zm ∩ Cn) for any matrix X ∈ Cn.

We know (Zm ∩ Cn)\Cn = Zm\Cm, so in (Zm ∩ Cn)\Cn we have

[(a 0

c a

)]=[(

1 0

c′pn−m 1

)]. So now we are ready to work out the trace:

TrGn/(Zm∩Cn)([rcjx,β ]Gn ) = TrCn/(Zm∩Cn) TrGn/Cn ([rcjx,β ]Cn )

=∑

X∈(Zm∩Cn)\CnX−1AX∈(Zm∩Cn)

p2j∑

β∈(Z/pn−jZ)×

[X−1rcjx,βX](Zm∩Cn)

= p2j∑c


[(x 0

βpj x

)](Zm∩Cn)

= pm+2j∑

β∈(Z/pn−jZ)×

[rcjx,β

](Zm∩Cn)

Now we calculate the trace of A over Nti , Nki , Zm ∩Cn, Zm ∩Nti and Zm ∩Nki for i ≥ n− j and

m ≤ j. We will denote Nti and Nki as:

Ni :=

(a bε0pi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

where ε0 is either ε, a fixed square-free element, or 1. Now let’s follow the same steps we did in the

previous case: (a b

c d

)−1(x 0

βpj x

)(a b

c d

)

=1

ad− bc

(−abβpj + adx− bcx −b2βpj

a2βpj abβpj + adx− bcx

)For this matrix to belong to Ni, we need −b2βpj = a2βpi+jε0 and abβpj = 0. Since i ≥ n − j, we

need −b2βpj = 0 = abβpj but p 6 |β therefore pn−j |b:(a b0pn−j

c d

)−1(x 0

βpj x

)(a b0pn−j

c d

)=

(x 0

a2βad−b0cpn−j

pj x

)Like before, we must find out how many of these matrices are distinct in Ni\Gn:(

x ε0ypi

y x

)(a b0pn−j

c d

)=

(ax+ piε0cy ε0dypi + b0xpn−j

ay + cx dx+ b0ypn−j

)

54

By choosing x = a−1(1− cε0ypi) and y = −c(a2 − c2ε0pi)−1 we get the following:(x ε0ypi

y x

)(a b0pn−j

c d

)=

(1 b′pn−j

0 d′

)

where b′ ∈ Z/pjZ and d′ ∈ (Z/pnZ)×. This is the same as the last case:

TrGn/Ni ([A]Gn ) = p2j∑

β∈(Z/pn−jZ)×

[rcjx,β

]Ni

Now we can calculate the trace of A over Zm ∩Ni. We can use our previous calculation to see that,

for A ∈ (Zm ∩Ni), we have X−1AX ∈ (Zm ∩Ni) for any matrix X ∈ Ni since i ≥ n− j.

We know (Zm ∩ Ni)\Ni = Zm\N(m)i , so in (Zm ∩ Ni)\Ni we have

[(a piε0b

b a

)]=[(

1 piε0b′

b′ 1

)]where b′ ∈ (Z/pmZ)×. This is also the same as the last case:

TrGn/(Zm∩Ni)([rcjx,β ]Gn ) = pm+2j


[rcjx,β

](Zm∩Ni)

Case A = t0w,y and A = rijx,β

Just like in the previous case, we find the trace of t0w,y first, because it assists us with finding the

trace of rijx,β . For the case A = t0w,y , elements from [A]Gn are only contained in Tn, and no other

group in Fn, so we will have non-trivial trace only over Tn:(a b

c d

)−1(w + y 0

0 w − y

)(a b

c d

)

=1

ad− bc

(adw − bcw + bcy + ady 2bdy

−2acy adw − bcw − bcy − ady

)To have A′ in Tn, we need 2bdy = 0 and −2acy = 0, but we also need X to have non-zero

determinant, so we have either both b and c are zero or both a and d are zero. This means that we

either need X =

(a 0

0 d

)or X =

(0 b

c 0

). Now let us find how many of these elements are

distinct in Tn\Gn: (x 0

0 y

)(a 0

0 d

)=

(ax 0

0 dy

)(

x 0

0 y

)(0 b

c 0

)=

(0 bx

cy 0

)It is clear that we can choose x and y in such a way that we can only pick X to be two distinct

matrices in Tn\Gn: (a−1 0

0 d−1

)(a 0

0 d

)=

(1 0

0 1

)(

b−1 0

0 c−1

)(0 b

c 0

)=

(0 1

1 0

)

55

It is also clear that these two matrices are distinct.

Now it is easy to calculate the trace:

TrGn/Tn ([A]Gn ) =∑

X∈Tn\GnX−1AX∈Tn

[X−1AX]Tn

=

[(w + y 0

0 w − y

)]Tn

+

[(w − y 0

0 w + y

)]Tn

=[t0w,y

]Tn

+[t0w,−y

]Tn

Now we look at the case A = rijx,β . Elements from [A]Gn are only contained in Tn and Zm ∩ Tnfor m ≤ j so we will have non-trivial trace only over Tn and Zm ∩ Tn for m ≤ j. Like in the

previous section, we find TrGn/Tn ([A]) and calculate the rest by using the fact that TrGn/(Zm∩Tn) =

TrTn/(Zm∩Tn) TrGn/Tn :(a b

c d

)−1(x+ βpj 0

0 x− βpj

)(a b

c d

)

=1

ad− bc

(adx− bcx+ adβpj + bcβpj 2bdβpj

−2acβpj adx− bcx− adβpj − bcβpj

)So we either have pn−j divides both b and c or pn−j divides both a and d:(

a b0pn−j

c0pn−j d

)−1(x+ βpj 0

0 x− βpj

)(a b0pn−j

c0pn−j d

)

=

(x+ adβ

ad−b0c0p2n−2j pj 0

0 x− adβad−b0c0p2n−2j p

j

)(

a0pn−j b

c d0pn−j

)−1(x+ βpj 0

0 x− βpj

)(a0pn−j b

c d0pn−j

)

=

(x+ bcβ

a0d0p2n−2j−bcpj 0

0 x− bcβa0d0p2n−2j−bcp

j

)Now we find out how many of these matrices are distinct in Tn\Gn:(

x 0

0 y

)(a b0pn−j

c0pn−j d

)=

(ax b0xpn−j

c0ypn−j dy

)(

x 0

0 y

)(a0pn−j b

c d0pn−j

)=

(a0xpn−j bx

cy d0ypn−j

)Again, it is clear that we can pick x and y in the following way:(

a−1 0

0 d−1

)(a b0pn−j

c0pn−j d

)=

(1 b′pn−j

c′pn−j 1

)(

b−1 0

0 c−1

)(a0pn−j b

c d0pn−j

)=

(a′pn−j 1

1 d′pn−j

)

56

Where a′, b′, c′, d′ ∈ Z/pjZ. So we get the following:

TrGn/Tn ([A]Gn ) =∑

X∈Tn\GnX−1AX∈Tn

[X−1AX]Tn

=∑b,c

[(x+ β

1−bcp2n−2j pj 0

0 x− β1−bcp2n−2j p

j

)]Tn

+∑a,d

[(x+ β

adp2n−2j−1pj 0

0 x− βadp2n−2j−1

pj

)]Tn

=∑b,c

[(x+ βpj 0

0 x− βpj

)]Tn

+∑a,d

[(x− βpj 0

0 x+ βpj

)]Tn

= p2j[rijx,β

]Tn

+ p2j[rijx,−β

]Tn

Now we can calculate the trace of A over Zm ∩ Tn. Like the previous section, we use our previous

calculation to see that, for A ∈ (Zm ∩ Tn), we always have X−1AX ∈ (Zm ∩ Tn) for any X ∈ Tn.

We know (Zm ∩ Tn)\Tn = Zm\Tm, so in (Zm ∩ Tn)\Tn we have

[(a 0

0 d

)]=

[(1 0

0 d′

)]where d′ ∈ (Z/pmZ)×. So now we are ready to work out the trace:

TrGn/(Zm∩Tn)([rijx,β ]Gn ) =

∑X∈(Zm∩Tn)\TnX−1AX∈(Zm∩Tn)

p2j [X−1rijx,βX](Zm∩Tn)

+∑

X∈(Zm∩Tn)\TnX−1AX∈(Zm∩Tn)

p2j [X−1rijx,−βX](Zm∩Tn)

=∑d

p2j

[(x+ βpj 0

0 x− βpj

)](Zm∩Tn)

+∑d

p2j

[(x− βpj 0

0 x+ βpj

)](Zm∩Tn)

= pm+2j−1(p− 1)

([rijx,β

](Zm∩Tn)

+[rijx,−β

](Zm∩Tn)

)

Case A = k0z,y and A = rjjx,β

Just like in the previous cases, we find the trace of k0z,y first. For the case A = k0

z,y , elements from

[A]Gn are only contained in Kn, and no other group in Fn, so we will have non-trivial trace only

over Kn: (a b

c d

)−1(z εy

y z

)(a b

c d

)

=1

ad− bc

(adx− bcx− aby + cdεy d2εy − b2y

a2y − c2εy adz − bcz + aby − cdεy

)To find the conditions on X such that A′ ∈ Kn, we need equate A′ to a generic term in Kn, i.e.(

w εx

x w

)where x,w ∈ Z/pnZ such that w2 − εx2 6= 0. So we have the following conditions:

• d2εy − b2y = (ad− bc)εx

• a2y − c2εy = (ad− bc)x

• aby − cdεy = 0

57

It turns out that these conditions imply that we have a = d and b = εc or that we have a = −d and

c = −εb. We are just stating this result for now but we will prove a more general result below. This

means that we either need X =

(a εb

b a

)or X =

(a −εbb −a

). Now let us find how many of

these elements are distinct in Kn\Gn:(z εy

y z

)(a εb

b a

)=

(az + εby ε(ay + bz)

ay + bz az + εby

)(

z εy

y z

)(a −εbb −a

)=

(az + εby −ε(ay + bz)

ay + bz −(az + εby)

)

Since

(a εb

b a

)is in Kn and

(a −εbb −a

)is not, it is clear that these two are distinct in Kn\Gn.

This also means that we can simply choose

(z εy

y z

)=

(a εb

b a

)−1

and this would give us a

z + εby = 1 and ay + bz = 0:(a εb

b a

)−1(a εb

b a

)=

(1 0

0 1

)(

a εb

b a

)−1(a −εbb −a

)=

(1 0

0 −1

)

Now it is easy to compute the trace:

TrGn/Kn ([A]Gn ) =∑

X∈Kn\GnX−1AX∈Kn

[X−1AX]Kn

=

[(z εy

y z

)]Kn

+

[(z −εy−y z

)]Kn

=[k0z,y

]Kn

+[k0z,−y

]Kn

Now we look at the case A = rjjx,β . Elements from [A]Gn are only contained in Kn and Zm∩Kn for

m ≤ j so we will have non-trivial trace only over Kn and Zm ∩Kn for m ≤ j. Like in the previous

section, we find TrGn/Kn ([A]) and calculate the rest by using the fact that TrGn/(Zm∩Kn) =

TrKn/(Zm∩Kn) TrGn/Kn :(a b

c d

)−1(x εβpj

βpj x

)(a b

c d

)

=1

ad− bc

(adx− bcx− abβpj + cdεβpj d2εβpj − b2βpj

a2βpj − c2εβpj adx− bcx+ abβpj − cdεβpj

)So we have the following conditions:

• (d2ε− b2)βpj = (ad− bc)εγpj

• (a2 − c2ε)βpj = (ad− bc)γpj

• pn−j |(ab− cdε)

where γ is any element in (Z/pn−jZ)×. Let us split this problem up into two cases:

58

Case p|aIn this case we would know that p does not divide either b or c and thus b and c would both be

invertible. So the third bullet point can be rearranged as a = cdε+lpn−j

bfor some l ∈ Z/pjZ. Since

p|a and p 6 |c, this equation tells us that p|d. Now we can plug this into the first bullet point and

then the second bullet point:

(d2ε− b2)βpj = ( cd2εb− bc)εγpj

⇐⇒ (d2ε− b2)βpj = (d2ε− b2) cbεγpj

Now we plug it into the second bullet point:

( c2d2ε2

b2− c2ε)βpj = ( cd

2εb− bc)γpj

⇐⇒ (d2ε− b2) c2

b2εβpj = (d2ε− b2) c

bγpj

Since p|d, we know that d2ε − b2 6= 0, therefore, these equations tell us that βpj = cbεγpj and

c2

b2εβpj = c

bγpj . This is only possible if b ≡ cε (mod pn−j) and β ≡ γ (mod pn−j) or if b ≡

−cε (mod pn−j) and β ≡ −γ (mod pn−j). In fact, all terms with γ and β are multiplied by pj so

any matrix with β ≡ γ (mod pn−j) is that same as having β = γ, and similar for β = −γ. Using the

equation a = cdε+lpn−j

b, we see now that we would have a ≡ d (mod pn−j) and b ≡ cε (mod pn−j)

when β = γ and we would have a ≡ −d (mod pn−j) and b ≡ −cε (mod pn−j) when β = −γ.

Case p 6 |aIn this case we would know that a is invertible. So the third bullet point can be rearranged as

b = cdε+kpn−j

afor some k ∈ Z/pjZ. Now we can plug this into the first bullet point and then the

second bullet point:

(d2ε− c2d2ε2

a2)βpj = (ad− c2dε

a)εγpj

⇐⇒ (a2 − c2ε) d2

a2εβpj = (a2 − c2ε) d

aεγpj

Now we can plug it into the second bullet point:

(a2 − c2ε)βpj = (ad− c2dεa

)γpj

⇐⇒ (a2 − c2ε)βpj = (a2 − c2ε) daγpj

These equations are satisfied only if either a2 ≡ c2ε (mod pn−j) or d2

a2εβpj = d

aεγpj and βpj = d

aγpj .

The former condition would imply that a ≡ c√ε (mod pn−j), but this is impossible since ε is square-

free by definition. The latter conditions are equivalent to a ≡ d (mod pn−j) and β = γ or if

a ≡ −d (mod pn−j) and β = −γ. Using the equation b = cdε+kpn−j

a, we see now that we would

have b ≡ cε (mod pn−j) and a ≡ d (mod pn−j) when β = γ and we would have b ≡ −cε (mod pn−j)

and a ≡ −d (mod pn−j) when β = −γ. These conditions agree with the previous case.

So we have either X =

(a εb+ kpn−j

b a+ lpn−j

)or X =

(a −εb+ kpn−j

b −a+ lpn−j

)where k, l ∈ Z/pjZ.

Now we find out how many of these matrices are distinct in Kn\Gn:(a εb

b a

)−1(a εb+ kpn−j

b a+ lpn−j

)=

(1 k′pn−j

0 1 + l′pn−j

)(

a εb

b a

)−1(a −εb+ kpn−j

b −a+ lpn−j

)=

(1 k′pn−j

0 −1 + l′pn−j

)

Where l = al′ + k′bε and k =

a−1(bl + k′(a2 − b2ε)) if p 6 |a

(bε)−1(al − l′(a2 − b2ε)) if p|a

59

Now we can compute the trace:

TrGn/Kn ([A]Gn ) =∑

X∈Kn\GnX−1AX∈Kn

[X−1AX]Kn

=∑k,l

[1

1 + lpn−j

(x+ xlpn−j εβpj

βpj x+ xlpn−j

)]Kn

+∑k,l

[1

−1 + lpn−j

(−x+ xlpn−j εβpj

βpj −x+ xlpn−j

)]Kn

=∑k,l

[(x εβpj

βpj x

)]Kn

+∑k,l

[(x −εβpj

−βpj x

)]Kn

= p2j

[(x εβpj

βpj x

)]Kn

+ p2j

[(x −εβpj

−βpj x

)]Kn

= p2j[rjjx,β

]Kn

+ p2j[rjjx,−β

]Kn

Now we can calculate the trace of A over Zm ∩Kn. Like the previous sections, we use our previous

calculation to see that, for A ∈ (Zm ∩Kn), we always have X−1AX ∈ (Zm ∩Kn) for any X ∈ Kn.

We know (Zm ∩Kn)\Kn = Zm\Km, so in (Zm ∩Kn)\Kn we have:

(a b

εb a

)∼

a′ 1

ε a′

if p 6 |b 1 pb′

pb′ 1

if p|b

where a′ ∈ Z/pmZ and b′ ∈ Z/pm−1Z . So now we are ready to work out the trace:

TrGn/(Zm∩Kn)([rjjx,β ]Gn ) =

∑X∈(Zm∩Kn)\KnX−1AX∈(Zm∩Kn)

p2j [X−1rjjx,βX](Zm∩Kn)

+∑

X∈(Zm∩Kn)\KnX−1AX∈(Zm∩Kn)

p2j [X−1rjjx,−βX](Zm∩Kn)

=∑a

p2j

[(x εβpj

βpj x

)](Zm∩Kn)

+∑b

p2j

[(x εβpj

βpj x

)](Zm∩Kn)

+∑a

p2j

[(x −εβpj

−βpj x

)](Zm∩Kn)

+∑b

p2j

[(x −εβpj

−βpj x

)](Zm∩Kn)

= pm+2j[rjjx,β

](Zm∩Kn)

+ pm+2j−1[rjjx,β

](Zm∩Kn)

+pm+2j[rjjx,β

](Zm∩Kn)

+ pm+2j−1[rjjx,−β

](Zm∩Kn)

= pm+2j−1(p+ 1)

([rjjx,β

](Zm∩Kn)

+[rjjx,−β

](Zm∩Kn)

)

Case A = rtix,α, A = rkix,α, A = rcij,ix,α and A = rcjj,ix,α

These cases are grouped because they are in the same form, A =

(x piε0α

pjα x

)where 0 ≤ j <

i < n and ε0 is either ε, a fixed square-free element, or 1. We will first do the cases A = rtix,α and

60

A = rkix,α. We will also denote the groups Nti and Nki as:

NA :=

(a bε0pi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

where i and ε0 are picked from the matrix A. For each matrix A in this form, elements from [A]Gnare only contained in NA, and no other group in Fn, so we will have non-trivial trace only over NA:(

a b

c d

)−1(x piε0α

α x

)(a b

c d

)

=1

ad− bc

(adx− bcx− abα+ cdε0piα d2ε0piα− b2α

a2α− c2ε0piα adx− bcx+ abα− cdε0piα

)To find the conditions on X such that A′ ∈ NA, we need equate A′ to a generic term in NA, i.e.(

y zε0pi

z y

)where y ∈ (Z/pnZ)× and z ∈ Z/pnZ. Now we have the following conditions:

• a2α− c2ε0piα = (ad− bc)z

• d2ε0piα− b2α = (ad− bc)zε0pi

• abα− cdε0piα = 0

It turns out that these conditions hold if and only if we have b = ±cε0pi and d = ±a. We are just

stating this result now because we prove a more general result below.

This means that we need X =

(a bε0pi

b a

)or X =

(a −bε0pi

b −a

)Now we need to find how many of these elements are distinct in NA\Gn:(

x ε0ypi

y x

)(a bε0pi

b a

)=

(bε0ypi + ax (aypj + bx)ε0pi

aypj + bx bε0ypi + ax

)(

x ε0ypi

y x

)(a −bε0pi

b −a

)=

(bε0ypi + ax −(aypj + bx)ε0pi

aypj + bx −(bε0ypi + ax)

)By choosing x = a−1(1− bε0ypi) and y = −b(a2 − b2ε0pi−j)−1 we get the following:(

x ε0ypi

y x

)(a bε0pi

b a

)=

(1 0

0 1

)(

x ε0ypi

y x

)(a −bε0pi

b −a

)=

(1 0

0 −1

)Now we can easily calculate the trace:

TrGn/NA ([A]Gn ) =∑

X∈NA\GnX−1AX∈NA

[X−1AX]NA

=

[(x ε0αpi

pjα x

)]NA

+

[(x −ε0αpi

−pjα x

)]NA

Now we will do the cases A = rcij,ix,α and A = rcjj,ix,α. When A = rcij,ix,α, A is only contained in

Nti−j and Zm ∩Nti−j for m ≤ j. When A = rcjj,ix,α, A is only contained in Nki−j and Zm ∩Nki−jfor m ≤ j. Now we will redefine NA as:

NA :=

(a bε0pi−j

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

61

where i,j and ε0 are picked from the matrix A. For each matrix A in this form, elements from

[A]Gn are only contained in NA and Zm ∩NA for m ≤ j, and no other group. Like in the previous

section, we find TrGn/NA ([A]) and calculate the rest by using the fact that TrGn/(Zm∩NA) =

TrNA/(Zm∩NA) TrGn/NA :(a b

c d

)−1(x piε0α

pjα x

)(a b

c d

)

=1

ad− bc

(adx− bcx− abpjα+ cdε0piα d2ε0piα− b2pjα

a2pjα− c2ε0piα adx− bcx+ abpjα− cdε0piα

)To find the conditions on X such that A′ ∈ NA, we need equate A′ to a generic term in NA, i.e.(

y z′ε0pi−j

z′ y

)where y ∈ (Z/pnZ)× and z′ ∈ Z/pnZ. Now we have the following conditions:

• a2pjα− c2ε0piα = (ad− bc)z′

• d2ε0piα− b2pjα = (ad− bc)z′ε0pi−j

• abpjα− cdε0piα = 0

We know that p cannot divide ad − bc because X is invertible, and we also know that i is strictly

greater than j so the first bullet point implies that we have pj |z′. Without loss of generality, we can

replace z′ with pjz and plug this into the three conditions:

• a2pjα− c2ε0piα = (ad− bc)zpj

• d2ε0piα− b2pjα = (ad− bc)zε0pi

• abpjα− cdε0piα = 0

Since i is strictly greater than j, the second bullet point implies p|b, therefore a and d are invertible.

Using this, we can rearrange the third bullet point to obtain b = cdε0pi−jα−kpn−jaα

where k ∈ Z/pjZ.

Plugging this into the first bullet point gives us the following:

a2pjα− c2ε0piα = adzpj − c2dzε0piα

aα

⇐⇒ pjα(a2 − c2ε0pi−j) = dazpj(a2 − c2ε0pi−j)

But a2 6= c2ε0pi−j because p 6 |a therefore pjα = dazpj thus aα ≡ dz (mod pn−j).

Plugging the expression for b into the second bullet point gives us the following:

d2ε0piα−c2d2ε20p

2i−jαa2

= adzpi − c2dzε0p2i−jε0a

⇐⇒ d2

a2ε0piα(a2 − c2ε0pi−j) = d

azε0pi(a2 − c2ε0pi−j)

Since a2 6= c2ε0pi−j we get d2

a2ε0piα = d

azε0pi. Now we can plug in aα ≡ dz (mod pn−j) which gives

us z2 ≡ α2 (mod pn−i). But if we look back at the equation pjα = dazpj , we see that we must use

the weaker condition z2 ≡ α2 (mod pn−j). So we now get X =

(a bδε0pi−j + kpn−j

b aδ + lpn−j

)where

δ2 ≡ 12 (mod pn−j):(a bδε0pi−j + kpn−j

b aδ + lpn−j

)−1(x piε0α

pjα x

)(a bδε0pi−j + kpn−j

b aδ + lpn−j

)

=

x+ 0a2ε0αp

i−b2ε20p2i−jα

a2δ−b2δε0pi−j+pn−j(al−bk)a2pjα−b2ε0piα

a2δ−b2δε0pi−j+pn−j(al−bk)x− 0

62

=

(x δpiε0α

δpjα x

)Finally, we find out how many of theses matrices are distinct in NA\Gn:(

x ε0ypi−j

y x


b aδ + lpn−j

)

=

(bε0ypi−j + ax (ay + bx)δε0pi−j + k′pn−j

ay + bx (bε0ypi−j + ax)δ + l′pn−j

)Where k′ = xk + ε0ylpi−j and l′ = yk + xl. By choosing x = a−1(1 − bε0ypi) and

y = −b(a2 − b2ε0pi)−1 we get the following:(x ε0ypi

ypj x


b aδ + lpn−j

)=

(1 k′pn−j

0 δ + l′pn−j

)so we effectively have pj choices for k′, l′ and 2 choices1for δ. Now we can calculate the trace:

TrGn/NA ([A]Gn ) =∑

X∈NA\GnX−1AX∈NA

[X−1AX]NA

= p2j

[( x ε0αpi

pjα x

)]NA

+

[(x −ε0αpi

−pjα x

)]NA

Now we calculate the trace of A over Zm ∩ NA. Like the previous sections, we use our previous

calculation to see that, for A ∈ (Zm ∩NA), we always have X−1AX ∈ (Zm ∩NA) for any X ∈ NA.

We know (Zm ∩ NA)\NA = Zm\N(m)A , so in (Zm ∩ NA)\NA we have

[(a pi−jε0b

b a

)]=[(

1 pi−jε0b′

b′ 1

)]where b′ ∈ Z/pmZ. So now we are ready to work out the trace:

TrGn/(Zm∩NA)([A]Gn ) =∑b′p2j

[(x ε0αpi

pjα x

)](Zm∩NA)

+∑b′p2j

[(x −ε0αpi

−pjα x

)](Zm∩NA)

= p2j+m

[( x ε0αpi

pjα x

)](Zm∩NA)

+

[(x −ε0αpi

−pjα x

)](Zm∩NA)

1Although our calculations show that δ2 ≡ 12 (mod pn−j), the terms kpn−j and lpn−j make it so that we can

treat δ as if it is equal to ±1.

63

Chapter 5

Constructing map L

In this chapter, we will construct a map L which makes the diagram below (diagram (1) from chapter

2.6.2) commute for general n:


↓ θn ↓ ψnker(L) 99K

∏U∈Fn Λ(Uab)×

L99K Qp ⊗Zp


where Fn = Zn, Cn, Tn,Kn, Nti , Nki |∀i = 1, 2, ..., n− 1. We will then find Θn,Zp (Definition 5.1),

a subgroup of∏U∈Fn Λ(Uab)× which, under the map L, contains Ψn,Zp (defined in Theorem 4.2).

The group Θn,Zp contains the image of θn. Recall the definition of ψn:

Definition 2.9 For U ∈ Fn, let TrGn/U be the map:

Zp[Conj(Gn)] → Zp[Conj(U)]

[A]Gn 7→∑

X∈U\GnX−1AX∈U

[X−1AX]U

and let projU be the natural projection from Zp[Conj(U)] to Zp[Conj(Uab)] = Zp[Uab], then:

ψn : Zp[Conj(Gn)] →∏U∈Fn Zp[Uab]

ψn : [A]Gn 7→∏U∈Fn projU TrGn/U ([A]Gn )

For each individual subgroup, we use this notation: ψU := projU TrG/U ([A]G). This means we

can write ψn =∏U∈Fn ψU .

We will use notation from chapter 4 for the conjugacy classes of GL2(Z/pnZ):

ix c0x,1 t0w,y k0z,y rtix,α rkix,α(

x 0

0 x

) (x 0

1 x

) (w + y 0

0 w − y

) (z εy

y z

) (x piα

α x

) (x piεα

α x

)rcjx,1 rcij,ix,α rcjj,ix,α rijx,β rjjx,β(x 0

pj x

) (x piα

pjα x

) (x piεα

pjα x

) (x+ βpj 0

0 x− βpj

) (x pjεβ

pjβ x

)



64

where ε is a fixed non-square element in (Z/pnZ)×.

Recall that c0x,y :=

(x 0

y x

)and rcjx,β :=

(x 0

βpj x

).

As explained in chapter 2.6.2, we will find L using the following diagram ([14], Proof of Theorem

6.8):

K1(Zp[Gn])log−−→ Qp ⊗Zp Zp[Conj(Gn)]

1−ϕp−−−→ Qp ⊗Zp Zp[Conj(Gn)]

↓ θ ↓ ψn ↓ ψn∏U∈Fn Zp[Uab]×

log−−→ Qp ⊗Zp∏U∈Fn Zp[Uab]

f99K Qp ⊗Zp

∏U∈Fn Zp[Uab]

Where ϕ is the endomorphism on Qp ⊗Zp Zp[Conj(Gn)], defined in the following way:

ϕ :∑

agg 7→∑

aggp

In the above diagram, the left square is commutative, so we just need to find a map f such that the

right square commutes and then we can just set L = f log.

5.1 The explicit construction of the map f

Recall Fn = Zn, Cn, Tn,Kn, Nti , Nki |∀i = 1, 2, ..., n − 1 where these subgroups are defined as

follows:

Zn :=

(a 0

0 a

): a ∈ (Z/pnZ)×

Cn :=

(a 0

c a

):a ∈ (Z/pnZ)×

c ∈ Z/pnZ

Tn :=

(a 0

0 d

): a, d ∈ (Z/pnZ)×

Kn :=

(a εb

b a

):a, b ∈ Z/pnZs.t. p 6 |(a2 − εb2)

Nti :=

(a bpi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

Nki :=

(a bεpi

b a

):a ∈ (Z/pnZ)×

b ∈ Z/pnZ

Proposition 5.1 f :=

∏U∈Fn fU such that:

fU ((aV )) :=

aU − pϕ(aU )− 1pϕ (TU (aU )) + pϕ (TU (aU ))− [Gn : U ]λf ((aV ))

−∑

N=Ntl,Nkl

l≤i

µf,N (aN )∑βrcn−l1,β + µf,U (aU )

∑βrcn−i1,β + νf,U (aCn ) if U = Nti or Nki

aCn − ϕ(aCn )− 1pϕ (TCn (aCn ))−

∑N=N

tl,Nkl

µf,N (aN )∑βrcn−l1,β

+TCn (ϕ(aCn ))− [Gn : Cn]λf ((aV )) if U = Cn

aZn −ϕ(aZn )

p− [Gn : Zn]λf ((aV )) if U = Zn

aU − ϕ(aU )p

+ 1p

(TU (ϕ(aU ))− ϕ(TU (aU ))

)− [Gn : U ]λf ((aV )) otherwise

65

where∑β is a sum of all β ∈ (Z/plZ)×, TU (aU ) =

TrU/Zn (aU )

[U :Zn], λf is a map from Qp ⊗Zp∏

U∈Fn Zp[U ] to Qp ⊗Zp Zp[Zn], µf,N is a map from Qp ⊗Zp Zp[N ] to Qp ⊗Zp Zp[Zn−i ∩ Cn] and

νf,N is a map from Qp ⊗Zp Zp[Cn] to Qp ⊗Zp Zp[N ] for N ∈ Nti , Nki |i = 1, 2, ..., n − 1. These

maps are defined1in the following way:

λf ((aV )) =1

p

∑W=Cn,Tn,Kn

TrW/Zn (ϕ (aW − TW (aW )))

[W : Zn]

µf,N (aN ) =p

2ϕ

(TrN/(Zn−i−1∩N)(aN )

[N : Zn−i−1 ∩N ]−TrN/(Zn−i∩N)(aN )

[N : Zn−i ∩N ]

)

νf,N (aCn ) = TrCn/(Zn−i∩Cn)

(aCn − ϕ(aCn )− TCn (aCn ) + TCn (ϕ(aCn ))

)Note that Zn is a subgroup of any U ∈ Fn, so any element aZn ∈ Qp⊗Zp Zp[Zn] can also be thought

as an element of Qp ⊗Zp Zp[U ]. In particular, for any U ∈ Fn, we have λf ((aV )) ∈ Qp ⊗Zp Zp[U ].

Also note that Zn−i ∩ Cn = Zn−i ∩ N for N ∈ Nti , Nki |i = 1, 2, ..., n − 1 so we have both

µf,N ∈ Qp ⊗Zp Zp[Zn−i ∩ Cn] and µf,N ∈ Qp ⊗Zp Zp[Zn−i ∩N ].

Proof:

We will verify that this map f makes the diagram commute by calculating ψn (1− ϕp

) and f ψnof each conjugacy class2 of Gn. In the previous chapter we calculated ψn(A) for each matrix

representation of each conjugacy class of Gn (Table 3). We will use this information to help us

calculate both ψn (1− ϕp

) and f ψn.

In the following tables we will use a condensed notion to save space when writing elements in

Qp ⊗Zp∏U∈Fn Zp[U ]:

Let (a1, a2, ..., am) ∈ A1 ×A2 × ...×Am

We will write (ak, al)Ak,Al to indicate that all entries are zero except for ak ∈ Ak and al ∈ Al. We

will also write (ai, a(U))Ai,An1×An2

×...Ani−1×Ani−1

...×Anm if the Athi entry is ai and the other

entries, U ∈ An1 ×An2 × ...Ani−1 ×Ani−1 ...×Anm , can be expressed as a function a(U).

1For more details on the trace maps used in λf , µf,N and νf,N , please refer to Chapter 4, before the proof of

Theorem 4.1, where we discuss TrU/(Zm∩U) for different values of m and groups U .2Since all maps f , ψn and ϕ act on Qp exactly like the identity map, we only need to calculate ψn (1 − ϕ

p)

and f ψn of each element in Conj(Gn) rather than the whole of Qp ⊗ Zp[Conj(Gn)].

66

A (1− ϕp

)(A) (ψn (1− ϕp

))(A)

ix ix − ixpp

([Gn : U ](ix − ixp/p))Fn

c0x,1 c0x,1 −rc1xp,xp−1

p

( ∑y∈(Z/pnZ)×

c0x,y − p∑

β∈(Z/pn−1Z)×rc1xp,β

)Cn

t0w,y t0w,y −(t0w,y)p

p

(t0w,y + t0w,−y −

ψn((t0w,y)p)

p

)Tn

k0z,y k0

z,y −(k0z,y)p

p

(k0z,y + k0

z,−y −ψn((k0z,y)p)

p

)Kn

rtix,α for

i < n− 1rtix,α −

(rtix,α)p

p

(rtix,α + rtix,−α −

ψn((rtix,α)p)

p

)Nti

rtn−1x,α rtn−1

x,α −rc1xp,αxp−1

p

(rtn−1x,α + rtn−1

x,−α − p∑

β∈(Z/pn−1Z)×rc1xp,β ,

−p∑


)Ntn−1 ,Cn×Nkn−1

rkix,α for

i < n− 1rkix,α −

(rkix,α)p

p

(rkix,α + rkix,−α −

ψn((rkix,α)p)

p

)Nki

rkn−1x,α rkn−1

x,α −rc1xp,αxp−1

p

(rkn−1x,α + rkn−1

x,−α − p∑


−p∑


)Nkn−1 ,Cn×Ntn−1

rcjx,1 for

j < n− 1rcjx,1 −

rcj+1

xp,xp−1

p

(p2j


rcjx,β − p2j+1

∑β∈(Z/pn−j−1Z)×

rcj+1xp,β

−p2j+1∑

β∈(Z/pn−j−1Z)×rcj+1xp,β

)Cn×

n−1∏i=n−j

(Nti×N

ki),N

tn−j−1×Nkn−j−1

rcn−1x,1 rcn−1

x,1 −ixpp

(p2n−2

∑β∈(Z/pZ)×

rcn−1x,β −

[Gn:Cn]p

ixp ,

− [Gn:U ]p

ixp)Cn×

n−1∏i=1

(Nti×N

ki),Fn\(Cn×

n−1∏i=1

(Nti×N

ki))

rcij,ix,α for

i < n− 1rcij,ix,α −

(rcij,ix,α)p

p

(p2j(rcij,ix,α + rcij,ix,−α)−

ψn((rcij,ix,α)p)

p

)Nti−j

rcij,n−1x,α rcij,n−1

x,α −rcj+1

xp,αxp−1

p

(p2j(rcij,n−1

x,α + rcij,n−1x,−α )− p2j+1


rcj+1xp,β ,

−p2j+1∑


)Ntn−1−j ,Cn×Nkn−1−j×

n−1∏i=n−j

(Nti×N

ki)

rcjj,ix,α for

i < n− 1rcjj,ix,α −

(rcjj,ix,α)p

p

(p2j(rcjj,ix,α + rcjj,ix,−α)−

ψn((rcjj,ix,α)p)

p

)Nki−j

rcjj,n−1x,α rcjj,n−1

x,α −rcj+1

xp,αxp−1

p

(p2j(rcjj,ix,α + rcjj,ix,−α))− p2j+1


rcj+1xp,β ,

−p2j+1∑


)Nkn−1−j ,Cn×Ntn−1−j×

n−1∏i=n−j

(Nti×N

ki)

rijx,β for

j < n− 1rijx,β −

(rijx,β

)p

p

(p2j(rijx,β + rijx,−β)−

ψn((rijx,β

)p)

p

)Tn

rin−1x,β rin−1

x,β −ixpp

(p2n−2(rin−1

x,β + rin−1x,−β)− [Gn:Tn]

pixp ,− [Gn:U ]

pixp)Tn,Fn\Tn

rjjx,β for

j < n− 1rjjx,β −

(rjjx,β

)p

p

(p2j(rjjx,β + rjjx,−β)−

ψn((rjjx,β

)p)

p

)Kn

rjn−1x,β rjn−1

x,β −ixpp

(p2n−2(rjn−1

x,β + rjn−1x,−β)− [Gn:Kn]

pixp ,− [Gn:U ]

pixp)Kn,Fn\Kn

Where i, j = 1, 2, ..., n− 1 s.t j < i, x, y ∈ (Z/pnZ)× and w, z ∈ Z/pnZ s.t y 6≡ ±w (mod p).

Also k, l ∈ Z/pjZ, α ∈ (Z/pn−iZ)× and β ∈ (Z/pn−jZ)×.

There was not enough space on the table to explicitly write out ϕ(t0w,y), ϕ(k0z,y), ϕ(rtix,α),

67

ϕ(rkix,α), ϕ(rcij,ix,α), ϕ(rcjj,ix,α), ϕ(rijx,β) and ϕ(rjjx,β). So we will describe them explicitly now:

(t0w,y)p = t0w′,y′ where

w′ =∑

0≤h≤p2|h

(p

h

)wp−hyh

and

y′ =∑

0≤h≤p2 6|h

(p

h

)wp−hyh

(k0z,y)p = k0

z′,y′′ where

z′ =∑

0≤h≤p2|h

(p

h

)zp−hyhε

h2

and

y′′ =∑

0≤h≤p2 6|h

(p

h

)zp−hyhε

h−12

So ψn((t0w,y)p) and ψn((k0z,y)p) are equal to ϕ(ψn(t0w,y)) and ϕ(ψn(k0

z,y)) respectively (this will be

useful when calculating (f ψn)(t0w,y) and (f ψn)(k0z,y) respectively).

(rcij,ix,α)p =

(x piα

pjα x

)p=

((x 0

0 x

)+

(0 piα

pjα 0

))p=

p∑h=0

(p

h

)(x 0

0 x

)p−h(0 piα

pjα 0

)h=

∑0≤h≤p

2|h

(p

h

)(xp−h(pi+jα2)

h2 0

0 xp−h(pi+jα2)h2

)

+∑

0≤h≤p2 6|h

(p

h

)(0 xp−h(pi+jα2)

h−12 piα

xp−h(pi+jα2)h−12 pjα 0

)

=

(xp +O(p) xp−1pi+1α+O(pi+2)

xp−1pj+1α+O(pj+2) xp +O(p)

)

Where O(q) stands for an element divisible by q. Since p does not divide x or α, we can conclude

that, for i = n − 1, (rcij,ix,α)p = rcj+1

xp,αxp−1 , and for i < n − 1, (rcij,ix,α)p = rcij+1,i+1x′,α′ , for some

x′ ∈ (Z/pnZ)× and α′ ∈ (Z/pn−iZ)×. So ψn((rcij,ix,α)p) is equal to p2ϕ(ψn(rcij,ix,α)) for i < n − 1.

This is similar3for rcjj,ix,α, rtix,α, rkix,α, rijx,β and rjjx,β .

3If i, j < n− 1 we have (rcjj,ix,α)p = rcjj+1,i+1

x′,α′ , (rtix,α)p = rci1,i+1

x′,α′ , (rkix,α)p = rcj1,i+1

x′,α′ , (rijx,β

)p = rij+1

x′,β′

and (rjjx,β

)p = rjj+1

x′,β′ .

Otherwise we have (rcjj,n−1x,α )p = rc

j+1

xp,αxp−1 , (rtn−1x,α )p = rc1

xp,αxp−1 , (rkn−1x,α )p = rc1

xp,αxp−1 , (rin−1x,β

)p =

ixp and (rjn−1x,β

)p = ixp .

68

A ψn(A)

ix ([Gn : U ]ix)Fn

c0x,1

( ∑y∈(Z/pnZ)×

c0x,y

)Cn

t0w,y

(t0w,y + t0w,−y

)Tn

k0z,y

(k0z,y + k0

z,−y

)Kn

rtix,α

(rtix,α + rtix,−α

)Nti

rkix,α

(rkix,α + rkix,−α

)Nki

rcjx,1

(p2j


rcjx,β

)Cn×

n−1∏i=n−j

(Nti×N

ki)

rcij,ix,α

(p2j(rcij,ix,α + rcij,ix,−α)

)Nti−j

rcjj,ix,α

(p2j(rcjj,ix,α + rcjj,ix,−α)

)Nki−j

rijx,β

(p2j(rijx,β + rijx,−β)

)Tn

rjjx,β

(p2j(rjjx,β + rjjx,−β)

)Kn

In most cases it is straight forward to calculate (f ψn)(A) since ψU (A) is non-zero for only one

subgroup U ∈ Fn. The greatest exception is A = ix, so we will explicitly do this case:

Set ψ(A) = (aU )U∈Fn then we have aU = [Gn : U ][ix] thus we have TrU/Zn (aU ) = [U : Zn][Gn :

U ][ix] therefore we have TU (aU ) = [Gn : U ][ix] = aU . Therefore λf ((aV )) = 0 and, for a similar

reason we get µf,N (aN ) = 0 = νf,N (aCn ) for all N ∈ Nti , Nki |i = 1, 2, ..., n − 1. So we clearly

get (fU ψn)(A) = [Gn : U ](ix − ixp/p) for the cases U = Zn, Tn,Kn. In the case U = Nti , Nki we

have:

(fU ψn)(A) = aU − pϕ(aU )−1

pϕ (TU (aU )) + pϕ (TU (aU ))

which also simplifies to [Gn : U ](ix − ixp/p). Finally, in the case U = Cn we have:

(fCn ψn)(A) = aCn − ϕ(aCn )−1

pϕ (TCn (aCn )) + TCn (ϕ(aCn ))

which also simplifies to [Gn : U ](ix − ixp/p), therefore (f ψn)(A) = ([Gn : U ](ix − ixp/p))Fn

The rest of the cases are in this table:A (f ψn)(A)

ix ([Gn : U ](ix − ixp/p))Fn

c0x,1

( ∑y∈(Z/pnZ)×

c0x,y − p∑


)Cn

t0w,y

(t0w,y + t0w,−y −

ψn((t0w,y)p)

p

)Tn

k0z,y

(k0z,y + k0

z,−y −ψn((k0z,y)p)

p

)Kn

rtix,α for

i < n− 1

(rtix,α + rtix,−α −

ψn((rtix,α)p)

p

)Nti

rtn−1x,α

(rtn−1x,α + rtn−1

x,−α − p∑


−p∑


)Ntn−1 ,Cn×Nkn−1

69

rkix,α for

i < n− 1

(rkix,α + rkix,−α −

ψn((rkix,α)p)

p

)Nki

rkn−1x,α

(rkn−1x,α + rkn−1

x,−α − p∑


−p∑


)Nkn−1 ,Cn×Ntn−1

rcjx,1 for

j < n− 1

(p2j


rcjx,β − p2j+1


rcj+1xp,β

−p2j+1∑


)Cn×

n−1∏i=n−j

(Nti×N

ki),N

tn−j−1×Nkn−j−1

rcn−1x,1

(p2n−2

∑β∈(Z/pZ)×

rcn−1x,β −

[Gn:Cn]p

ixp ,

− [Gn:U ]p

ixp)Cn×

n−1∏i=1

(Nti×N

ki),Fn\(Cn×

n−1∏i=1

(Nti×N

ki))

rcij,ix,α for

i < n− 1

(p2j(rcij,ix,α + rcij,ix,−α)−

ψn((rcij,ix,α)p)

p

)Nti−j

rcij,n−1x,α

(p2j(rcij,n−1

x,α + rcij,n−1x,−α )− p2j+1


rcj+1xp,β ,

−p2j+1∑


)Ntn−1−j ,Cn×Nkn−1−j×

n−1∏i=n−j

(Nti×N

ki)

rcjj,ix,α for

i < n− 1

(p2j(rcjj,ix,α + rcjj,ix,−α)−

ψn((rcjj,ix,α)p)

p

)Nki−j

rcjj,n−1x,α

(p2j(rcjj,ix,α + rcjj,ix,−α))− p2j+1


rcj+1xp,β ,

−p2j+1∑


)Nkn−1−j ,Cn×Ntn−1−j×

n−1∏i=n−j

(Nti×N

ki)

rijx,β for

j < n− 1

(p2j(rijx,β + rijx,−β)−

ψn((rijx,β

)p)

p

)Tn

rin−1x,β

(p2n−2(rin−1

x,β + rin−1x,−β)− [Gn:Tn]

pixp ,− [Gn:U ]

pixp)Tn,Fn\Tn

rjjx,β for

j < n− 1

(p2j(rjjx,β + rjjx,−β)−

ψn((rjjx,β

)p)

p

)Kn

rjn−1x,β

(p2n−2(rjn−1

x,β + rjn−1x,−β)− [Gn:Kn]

pixp ,− [Gn:U ]

pixp)Kn,Fn\Kn


k, l ∈ Z/pjZ, α ∈ (Z/pn−iZ)× and β ∈ (Z/pn−jZ)×

This table and the table for ψn (1− ϕp

) show that the above diagram is commutative for our choice

of f , and as a result diagram (1), from chapter 2.6.2, is also commutative for L = f log.

5.2 Obtaining an explicit description of L

We set L = f log so, to find an explicit description for L, we just write L =∏U∈F LU such that

LU = fU log i.e. we just replace aU with log(xU ) in the definition of fU :

70

Case U = Nti or Nki

In the case that U = Nti or Nki we have

fU = aU − pϕ(aU )− 1pϕ (TU (aU )) + pϕ (TU (aU ))− [Gn : U ]λf ((aV ))

−∑

N=Ntl,Nkl

l≤i

µf,N (aN )∑βrcn−l1,β + µf,U (aU )

∑βrcn−i1,β + νf,U (aCn )

where TU (aU ) =TrU/Zn (aU )

[W :Zn]and:

λf ((aV )) =1

p

∑W=Cn,Tn,Kn

TrW/Zn (ϕ (aW − TW (aW )))

[W : Zn]

µf,N (aN ) =p

2ϕ



[N : Zn−i ∩N ]

)

νf,N (aCn ) = TrCn/(Zn−i∩Cn)

(aCn − ϕ(aCn )− TCn (aCn ) + TCn (ϕ(aCn ))

)So LU would be defined as:

log(xU )− pϕ(log(xU ))− 1pϕ (TU (log(xU ))) + pϕ (TU (log(xU )))− [Gn : U ]λf (log((xV )))

−∑

N=Ntl,Nkl

l≤i

µf,N (log(xN ))∑βrcn−l1,β + µf,U (log(xU ))

∑βrcn−i1,β + νf,U (log(xCn ))

We can simplify TU (log(xU )) to log(NU (xU )) where NU (xU ) = NmU/Zn (xU )1

[U:Zn] .

We can also simplify λf (log((xV ))), µf,N (log(xN )) and νf,N (log(xCn )) but we will start with

λf (log((xV ))):

1p

∑W=Cn,Tn,Kn

TrW/Zn

(ϕ

(log(xW )−

TrW/Zn(log(xW ))

[W :Zn]

))[W :Zn]

= 1p

∑W=Cn,Tn,Kn

TrW/Zn

(1

[W :Zn] ([W :Zn]ϕ(log(xW ))−ϕ(log(NmW/Zn (xW )))))

[W :Zn]

= − 1p

∑W=Cn,Tn,Kn

1[W :Zn]2

TrW/Zn

(log(ϕ(NmW/Zn (xW ))

ϕ(xW )[W :Zn]

))= − 1

p

∑W=Cn,Tn,Kn

1[W :Zn]2

log(NmW/Zn

(ϕ(NmW/Zn (xW ))

ϕ(xW )[W :Zn]

))= − 1

[Gn:U ]plog(λL,U ((xV ))

)where λL,U ((xV )) =

∏W=Cn,Tn,Kn

NmW/Zn

(ϕ(NmW/Zn (xW ))

ϕ(xW )[W :Zn]

) [Gn:U]

[W :Zn]2 .

Now we will simplify our expression for µf,N (log(xN )):

p2ϕ

(TrN/(Zn−i−1∩N)(log(xN ))

[N :Zn−i−1∩N ]−TrN/(Zn−i∩N)(log(xN ))

[N :Zn−i∩N ]

)

= p2ϕ(

1[N :Zn−i∩N ]

(plog(NmN/(Zn−i−1∩N)(xN ))− log(NmN/(Zn−i∩N)(xN ))

))

= p2[N :Zn−i∩N ]

log

(ϕ

(NmN/(Zn−i−1∩N)(xN )p

NmN/(Zn−i∩N)(xN )

))

= µL,N (xN )

71

where µL,N (xN ) = log

(ϕ

(NmN/(Zn−i−1∩N)(xN )p


) p2[N:Zn−i∩N]

).

Finally we will simplify our expression for νf,N (log(xCn )):

TrCn/(Zn−i∩Cn)

(log(xCn )− ϕ(log(xCn ))− TCn (log(xCn )) + TCn (ϕ(log(xCn )))

)= −TrCn/(Zn−i∩Cn)

(− log(xCn ) + log(ϕ(xCn )) + log(NCn (xCn ))− log(NCn (ϕ(xCn )))

)= −TrCn/(Zn−i∩Cn)

(log(NCn (xCn )ϕ(xCn )

xCnNCn (ϕ(xCn ))

))= − 1

plog(νL,N (xCn ))

where νL,N (xCn ) = NmCn/(Zn−i∩Cn)

(NCn (xCn )ϕ(xCn )

xCnNCn (ϕ(xCn ))

)p.

Now we can simplify our expression for LU in the case that U = Nti or Nki :

log(xU )− pϕ(log(xU ))− 1pϕ (TU (log(xU ))) + pϕ (TU (log(xU )))− [Gn : U ]λf (log((xV )))

−∑

N=Ntl,Nkl

l≤i

µf,N (log(xN ))∑βrcn−l1,β + µf,U (log(xU ))

∑βrcn−i1,β + νf,U (log(xCn ))

= log(xU )− log((ϕ(xU ))p)− 1plog(ϕ(NU (xU ))) + log(ϕ(NU (xU ))p) + log(λL,U ((xV )))

−∑

N=Ntl,Nkl

l≤i

µL,N (xN )∑βrcn−l1,β + µL,U (xU )

∑β

rcn−i1,β −1plog(νL,U (xCn ))

= 1plog

(xpUϕ(NU (xU ))p

2λL,U ((xV ))


)−

∑N=N

tl,Nkl

l≤i

µL,N (xN )∑β

rcn−l1,β + µL,U (xU )∑β

rcn−i1,β

Case U = Cn

For the U = Cn case, we have

fCn = aCn − ϕ(aCn )− 1pϕ (TCn (aCn )) + TCn (ϕ(aCn ))

−[Gn : Cn]λf ((aV ))−∑

N=Nti,Nki

µf,N (aN )∑β

rcn−i1,β

So LCn is defined as:

log(xCn )− ϕ(log(xCn ))− 1pϕ (TCn (log(xCn ))) + TCn (ϕ(log(xCn ))

−[Gn : Cn]λf ((log(xV )))−∑

N=Nti,Nki

µf,N (log(xN ))∑βrcn−i1,β

= log(xCn )− log(ϕ(xCn ))− 1plog(ϕ (NCn (xCn ))) + log(NCn (ϕ(xCn )))

+ 1plog(λL,Cn ((xV )))−

∑N=N

ti,Nki

µL,N (xN )∑β

rcn−i1,β

= 1plog

(xpCn



)−

∑N=N

ti,Nki

µL,N (xN )∑βrcn−i1,β

where∑β is a sum of all β ∈ (Z/piZ)×.

72

Case U = Zn, Tn or Kn

In the case U = Zn, fZn = aZn −ϕ(aZn )

p− [Gn : Zn]λf ((aV )), so LZn is defined as:

log(xZn )− ϕ(log(xZn ))

p− [Gn : Zn]λf (log((xV )))

= log(xZn )− 1plog(ϕ(xZn )) + 1

plog(λL,Zn ((xV )))

= 1plog

(xpZn

λL,Zn ((xV ))

ϕ(xZn )

)Finally, in the case U = Tn or Kn, fU = aU − ϕ(aU )

p+ 1

p

(TU (ϕ(aU )) − ϕ(TU (aU ))

)− [Gn :

U ]λf ((aV )), so LU is defined as:

log(xU )− ϕ(log(xU ))p

+ 1p

(TU (ϕ(log(xU )))− ϕ(TU (log(xU )))

)− [Gn : U ]λf (log((xV )))

= log(xU )− 1plog(ϕ(xU )) + 1

plog(NU (ϕ(xU )))− 1

plog(ϕ(NU (xU ))) + 1

plog(λL,U ((xV )))

= 1plog

(xpUNU (ϕ(xU ))λL,U ((xV ))

ϕ(xU )ϕ(NU (xU ))

)

Now it is very easy to state L:

L =∏U∈F LU such that

LU ((xV )) :=

1plog

(xpUϕ(NU (xU ))p

2λL,U ((xV ))


)−

∑N=N

tl,Nkl

l≤i

µL,N (xN )∑

β∈(Z/plZ)×rcn−l1,β

+µL,U (xU )∑

β∈(Z/piZ)×rcn−i1,β if U = Nti or Nki

1plog

(xpCn



)−

∑N=N

ti,Nki

µL,N (xN )∑

β∈(Z/piZ)×rcn−i1,β if U = Cn

1plog

(xpZn

λL,Zn ((xV ))

ϕ(xZn )

)if U = Zn

1plog


ϕ(xU )ϕ(NU (xU ))

)if U = Tn or Kn

Where NU (xU ) = NmU/Zn (xU )1

[U:Zn] , λL,U is a map from∏U∈Fn Zp[U ]× to Q×p ⊗Zp Zp[Zn]×,

µL,N is a map from Zp[N ]× to Q×p ⊗Zp Zp[Zn−i ∩ Cn]× and νL,N is a map from Zp[Cn]× to

Q×p ⊗Zp Zp[N ]× for N ∈ Nti , Nki |i = 1, 2, ..., n− 1. These maps are defined in the following way:

λL,U ((xV )) =∏

W=Cn,Tn,Kn

NmW/Zn

(ϕ(NmW/Zn (xW ))

ϕ(xW )[W :Zn]

) [Gn:U]

[W :Zn]2

µL,N (xN ) = log

ϕ(NmN/(Zn−i−1∩N)(xN )p


) p2[N:Zn−i∩N]

νL,N (xCn ) = NmCn/(Zn−i∩Cn)

(NCn (xCn )ϕ(xCn )

xCnNCn (ϕ(xCn ))

)p

73

Note that Zn is a subgroup of any U ∈ Fn, so any element xZn ∈ Zp[Zn]× can also be thought

as an element of Zp[U ]×. In particular, for any U ∈ Fn, we have λL,U ((xV )) ∈ Q×p ⊗Zp Zp[U ]×.

Also note that Zn−i ∩ Cn = Zn−i ∩ N for N ∈ Nti , Nki |i = 1, 2, ..., n − 1 so we have both

µL,N ∈ Q×p ⊗Zp Zp[Zn−i ∩ Cn]× and µL,N ∈ Q×p ⊗Zp Zp[Zn−i ∩N ]×.

5.3 Finding the group Θn,Zp

Now we can easily find conditions for a group Θn,Zp ⊂∏U∈Fn Λ(Uab)× such that Ψn,Zp ⊂ L(Θn,Zp )

Recall that NV (U) denotes the normalizer of U as a subgroup of V . Now recall the definition of

Ψn,R:

Let R be a Zp-algebra of characteristic 0. Ψn,R is defined using the following conditions:

1. Ψn,R ⊂ p3n−2R[Zn]×R[Cn]×R[Tn]×R[Kn]×n−1∏i=1

(R[Nti ]×R[Nki ])

2. For any (aV )V ∈Fn ∈ Ψn,R, each aV is fixed by conjugation action of NGn (V )



• TrNti/Zm∩Nti

(aNti


• TrNki/Zm∩Nki

(aNki



• TrCn/Zm∩Cn (aCn ) ∈ p3mR[Cn]

• TrTn/Zm∩Tn (aTn ) ∈ p3m−1R[Tn]

• TrKn/Zm∩Kn (aKn ) ∈ p3m−1R[Kn]

• TrNti/Zm∩Nti

(aNti

) ∈ p3mR[Nti ]

• TrNki/Zm∩Nki

(aNki

) ∈ p3mR[Nki ]

Now we define Θn,R:

Definition 5.1 First recall the maps:

λL,U ((xV )) =∏

W=Cn,Tn,Kn

NmW/Zn

(ϕ(NmW/Zn (xW ))

ϕ(xW )[W :Zn]

) [Gn:U]

[W :Zn]2

µL,N (xN ) = log



) p2[N:Zn−i∩N]


(NCn (xCn )ϕ(xCn )

xCnNCn (ϕ(xCn ))

)pLet R be a Zp-algebra of characteristic 0. We define Θn,R using the following conditions:

1. Θn,R ⊂∏U∈Fn R[U ]× such that xpZn


)≡ ϕ(xZn ) (mod p3n−1)

2. For any (xV )V ∈Fn ∈ Θn,R, each xV is fixed by conjugation action of NGn (V )

3. For any (xV )V ∈Fn ∈ Θn,R, we have:


74

• 1plog

(NmU/Zm∩U

(xpUϕ(NU (xU ))p

2λL,U ((xV ))


))+TrU/Zm∩U

(µL,U (xU )

∑β∈(Z/piZ)×

rcn−i1,β

)= 1

plog

(NmCn/Zm∩Cn

(xpCn




m ≥ n− i

4. For any (xV )V ∈Fn ∈ Θn,R, we have:








2 (λL,U ((xV )V ∈Fn )

))≡ NmU/Zm∩U

(ϕ(xU )p

2ϕ(NU (xU ))

(νL,U (xCn )


Theorem 5.1 The image of θn is contained in Θn,Zp .

Proof:

We prove this theorem by proving each individual condition of Ψn,Zp is satisfied in the image of

Θn,Zp under the map L. We start with conditions 2 and 3. Recall condition 2 and 3 from the

definition of Ψn,Zp :

2. For any (aV )V ∈Fn ∈ Ψn,Zp , each aV is fixed by conjugation action of NGn (V )

3. For any (aV )V ∈Fn ∈ Ψn,Zp , we have:


• TrNti/Zm∩Nti

(aNti


• TrNki/Zm∩Nki

(aNki


Due to the properties of logarithm and the definitions of norm4and trace, these conditions are

clearly satisfied in the image of L by the respective conditions 2 and 3 from Θn,Zp .

Now we prove condition 1. Condition 1 in the definition of Ψn,Zp says this:

1.Ψn,Zp ⊂ p3n−2Zp[Zn]× Zp[Cn]× Zp[Tn]× Zp[Kn]×n−1∏i=1

(Zp[Nti ]× Zp[Nki ]

)This is equivalent to the condition ∀(aV )V ∈Fn ∈ Ψn,Zp we have (aV )V ∈Fn ∈

∏U∈Fn Zp[U ]

such that p3n−2|aZn . Now we replace aU with LU ((xV )V ∈Fn ). By definition of L this condition is

equivalent5to:

p3n−2

∣∣∣∣1p log(xpZnλL,Zn ((xV ))

ϕ(xZn )

)By the properties of logarithm, one can see that these conditions are equivalent to condition 1 of

Θn,Zp .

4Let V be a group and U a subgroup of V. For any A ∈ V the norm is defined in the following way:

NmV/U (A) :=∏

X∈U\VX−1AX∈U

X−1AX

5At this point we are not concerned with proving L((xV )) ∈∏U∈Fn Zp[U ], we want to prove that image of θn

is contained in Θn,Zp so we are satisfied with L((xV )) ∈∏U∈Fn Qp ⊗Zp Zp[U ]

75

Finally, we prove condition 4. Condition 4 from the definition of Ψn,Zp states:

4. For any (aV )V ∈Fn ∈ Ψn,Zp , we have:

• TrCn/Zm∩Cn (aCn ) ∈ p3mZp[Cn]

• TrTn/Zm∩Tn (aTn ) ∈ p3m−1Zp[Tn]

• TrKn/Zm∩Kn (aKn ) ∈ p3m−1Zp[Kn]

• TrNti/Zm∩Nti

(aNti

) ∈ p3mZp[Nti ]

• TrNki/Zm∩Nki

(aNki

) ∈ p3mZp[Nki ]

We rewrite these conditions in the form p3m|TrCn/Zm∩Cn (aCn ), p3m−1|TrTn/Zm∩Tn (aTn ),

p3m−1|TrKn/Zm∩Kn (aKn ), p3m|TrNti/Zm∩Nti

(aNti

) and p3m|TrNki/Zm∩Nki

(aNki

).

Recall the definition of µf,N (aN ) =

p

2ϕ



[N : Zn−i ∩N ]

)

for N ∈ Nti , Nki |i = 1, ..., n− 1.By condition 4 from the definition of Ψn,Zp , we know that p3m|TrN/(Zm∩N)(aN ), but we also have

[N : Zm ∩N ] = pm, therefore p2(n−i)+1|µf,N (aN ).

Recall that µL,N ∈ Q×p ⊗Zp Zp[Zn−i ∩ Cn]×, thus µL,N will have the form:

µL,N =∑

j≥n−i

∑x,β

ax,β,jrcjx,β

where ax,β,j ∈ p2j+1Z×p . Therefore we have:

µf,N (aN )∑β

rcn−i1,β =∑

j≥n−i

∑x,β

ax,β,jrcjx,β

But we also know that

TrCn/Zm∩Cn

(rcjx,β

)=

pmrcjx,β if j ≥ m

0 otherwise

Therefore we can conclude that p3m+1|TrCn/Zm∩Cn

( ∑N=N

ti,Nki

µf,N (aN )∑βrcn−i1,β

).

Recall that LCn = fCn log, so we have TrCn/Zm∩Cn (aCn ) = TrCn/Zm∩Cn (fCn log(xCn )). Let

log(xCn ) = bCn then we have:

TrCn/Zm∩Cn (fCn (bCn )) =

TrCn/Zm∩Cn

(bCn − ϕ(bCn )− 1

pϕ (TCn (bCn )) + TCn (ϕ(bCn ))− [Gn : Cn]λf ((bV ))

−∑

N=Nti,Nki

µf,N (bN )∑βrcn−i1,β

)

= TrCn/Zm∩Cn

(bCn − ϕ(bCn )− 1


)−TrCn/Zm∩Cn

( ∑N=N

ti,Nki

µf,N (bN )∑βrcn−i1,β

)This means that the condition p3m|TrCn/Zm∩Cn (aCn ) implies that we need

p3m|TrCn/Zm∩Cn(bCn − ϕ(bCn )−

1


)

76

By the same reasoning, we also find that the conditions p3m|TrNti/Zm∩Nti

(aNti

) and

p3m|TrNki/Zm∩Nki

(aNki

) imply that we need:

p3m|TrU/Zm∩U(bU − pϕ(bU )−

1

pϕ (TU (bU )) + pϕ (TU (bU ))− [Gn : U ]λf ((bV )) + νf,U (bCn )

)for U = Nti or Nki

Now we can rewrite these five conditions in the following forms:

p3m

∣∣∣∣TrCn/Zm∩Cn(

1

plog

(xpCnNCn (ϕ(xCn ))pλL,Cn ((xV ))

ϕ(xCn )pϕ (NCn (xCn ))

))

p3m−1

∣∣∣∣TrTn/Zm∩Tn(

1

plog

(xpTnNTn (ϕ(xTn ))λL,Tn ((xV ))

ϕ(xTn )ϕ(NTn (xTn ))

))

p3m−1

∣∣∣∣TrKn/Zm∩Kn(

1

plog

(xpKnNKn (ϕ(xKn ))λL,Kn ((xV ))

ϕ(xKn )ϕ(NKn (xKn ))

))

p3m

∣∣∣∣TrNti/Zm∩Nti1

plog

xpNtiϕ(NN

ti(xN

ti))p

2λL,N

ti((xV ))

ϕ(xNti

)p2ϕ(NNti

(xNti

))νL,Nti

(xCn )

p3m

∣∣∣∣TrNki/Zm∩Nki1

plog

xpNkiϕ(NN

ki(xN

ki))p

2λL,N

ki((xV ))

ϕ(xNki

)p2ϕ(NNki

(xNki

))νL,Nki

(xCn )

Like before, by the properties of logarithm one can see that these conditions are equivalent to

condition 4 in Θn,Zp and thus the theorem is proved.

77

Chapter 6

Whitehead group of the localised algebra

Λ(Gn)T ′

In this chapter we will modify the work done in this paper to prove that θn

(K1

(

Λ(Gn)T ′

))lies

in the group Θτn, Zp[[Γn]](p)

which is defined below. Recall the isomorphism from lemma 2.4:

Zp[[Γn]]T [Gn]τ ∼= Λ(Gn)T ′

where T = Zp[[Γn]] − pZp[[Γn]]. But Zp[[Γn]] has inverse elements for all elements in T except for

powers of p, therefore Zp[[Γn]]T∼= Zp[[Γn]](p).

We expect to have the following commutative diagram:

ker(Log) → K1

(

Λ(Gn)T ′

)Log−−−→ Zp[[Γn]](p)[Conj(Gn)]τ → coker(Log)

↓ θn ↓ ψnker(L) → Θτ

n, Zp[[Γn]](p)

L−→ Qp ⊗Zp Ψτn, Zp[[Γn]](p)

→ coker(L)

where Θτn,R and Ψτn,R is the notation used to denote the analogue to Θn,R and Ψn,R which use

twisted group rings1. Also θn and ψn are the analogue maps of θn and ψn respectively. If this

diagram commutes then we have proved that θn

(K1

(

Λ(Gn)T ′

))is contained in Θτ

n, Zp[[Γn]](p)

.

This is very similar to the work we have already done in this paper so we only need to verify that

the twist does not change the image of the maps and that the maps L and Log are still well-defined

on Θτn, Zp[[Γn]](p)

and K1

(

Λ(Gn)T ′

)respectively. We need to check this is because both L and Log

are defined using logarithm.

6.1 Inspecting the twist τ

This section is focused on the map ψn and understanding the module2 Zp[[Γn]](p)[Conj(Gn)]τ .

Let A,X ∈ Gn, then their images in Zp[[Γn]](p)[Gn]τ are A and X respectively. Recall that, for

elements in Gn, multiplication in Zp[[Γn]](p)[Gn]τ has the twist

A ·X = τ(A,X)AX

1Ψτn,R and Θτn,R have the same definitions as Ψn,R and Θn,R except that instead of Ψn,R ⊂∏U∈Fn R[U ]

and Θn,R ⊂∏U∈Fn R[U ]× we have Ψτn,R ⊂

∏U∈Fn R[U ]τ and Θτn,R ⊂

∏U∈Fn (R[U ]τ )×.

2R[Conj(G)]τ is the R-module over the basis of all conjugacy classes of G in the twisted group ring R[G]τ i.e.

two elements g and h in G are conjugate in R[G]τ if there exists x ∈ G such that g = x−1 · h · x.

78

Lemma 6.1 For any A,X ∈ Gn, we have X−1 ·A ·X = X−1AX in [Gn]τ

Proof:

X−1 ·A ·X = τ(X−1, A)X−1A ·X = τ(X−1, A)τ(X−1A,X)X−1AX

So we just need to prove that τ(X−1, A)τ(X−1A,X) = 12

The 2-cocycle, τ , has the following properties τ(A,A−1) = 12 = τ(A,12) and τ(A,B) = τ(B,A)

(see lemma 2.1).

By definition of 2-cocycle we know that

τ(B,A)τ(BA,X) =(B ∗ τ(A,X)

)τ(B,AX)

where ∗ denotes conjugation. But τ(A,X) ∈ Γn so τ(A,X) is in the centre of Gn, therefore B ∗τ(A,X) = τ(A,X)

τ(X−1, A)τ(X−1A,X) =(X−1 ∗ τ(A,X)

)τ(X−1, AX)

= τ(A,X)τ(X−1, AX)

= τ(A,X)τ(AX,X−1)

=(A ∗ τ(X,X−1)

)τ(A,XX−1)

= 12 · τ(A,12)

= 12

With this lemma, we know that the conjugacy class basis of Zp[[Γn]](p)[Conj(Gn)]τ have a

one-to-one correspondence with the basis of Zp[Conj(Gn)]. We also know that ψn acts onZp[[Γn]](p)[Conj(Gn)]τ in same that ψn acts on Zp[Conj(Gn)], thus the image of ψn is in fact

Ψτn, Zp[[Γn]](p)

.

6.2 Verifying that we can take log

This section will focus on the maps Log and L.

By ([11], Section 5.5.2), we know that Log is well defined on K1

(

Λ(Gn)T ′

). Also due to ([11],

Section 5.5.2), we only need to show that L is well defined on Zp[[Γn]](p) in order to prove that L is

well defined on Θτn, Zp[[Γn]](p)

.

Recall the definition of L:

79

L =∏U∈F LU such that

LU ((xV )) :=

1plog

(xpUϕ(NU (xU ))p

2λL,U ((xV ))


)−

∑N=N

tl,Nkl

l≤i

µL,N (xN )∑

β∈(Z/piZ)×rcn−l1,β

+µL,U (xU )∑

β∈(Z/piZ)×rcn−i1,β if U = Nti or Nki

1plog

(xpCn



)−

∑N=N

ti,Nki

µL,N (xN )∑

β∈(Z/piZ)×rcn−i1,β if U = Cn

1plog

(xpZn

λL,Zn ((xV ))

ϕ(xZn )

)if U = Zn

1plog


ϕ(xU )ϕ(NU (xU ))

)if U = Tn or Kn

where NU (xU ) = NmU/Zn (xU )1

[U:Zn] , λL,U is a map from∏U∈Fn Zp[U ]× to Q×p ⊗Zp Zp[Zn]×,

µL,N is a map from Zp[N ]× to Q×p ⊗Zp Zp[Zn−i ∩ Cn]× and νL,N is a map from Zp[Cn]× to

Q×p ⊗Zp Zp[N ]× for N ∈ Nti , Nki |i = 1, 2, ..., n− 1. These maps are defined in the following way:

λL,U ((xV )) =∏

W=Cn,Tn,Kn

NmW/Zn

(ϕ(NmW/Zn (xW ))

ϕ(xW )[W :Zn]

) [Gn:U]

[W :Zn]2

µL,N (xN ) = log



) p2[N:Zn−i∩N]


(NCn (xCn )ϕ(xCn )

xCnNCn (ϕ(xCn ))

)pAlso recall the definition of Θτ

n, Zp[[Γn]](p)

:

1. Θn, Zp[[Γn]](p)

⊂∏U∈Fn ( Zp[[Γn]](p)[U ]τ )× such that xpZn


)≡

ϕ(xZn ) (mod p3n−1)

2. For any (xV )V ∈Fn ∈ Θn, Zp[[Γn]](p)

, each xV is fixed by conjugation action of NGn (V )


, we have:


• 1plog

(NmU/Zm∩U

(xpUϕ(NU (xU ))p

2λL,U ((xV ))


))+TrU/Zm∩U

(µL,U (xU )

∑β∈(Z/piZ)×

rcn−i1,β

)= 1

plog

(NmCn/Zm∩Cn

(xpCn




m ≥ n− i


, we have:




80





2 (λL,U ((xV )V ∈Fn )

))≡ NmU/Zm∩U

(ϕ(xU )p

2ϕ(NU (xU ))

(νL,U (xCn )


Lemma 6.2 L is well defined on Θτn, Zp[[Γn]](p)

Proof:

As mentioned above, we will only show that L is well defined on Zp[[Γn]](p).

The logarithm is only defined on elements in the form 1 + py where y ∈ Zp[[Γn]](p). By inspecting

L, we see that we need to prove the following

xpZn

λL,Zn ((xV ))

ϕ(xZn )≡ 1 (mod p) if U = Zn

xpUNU (ϕ(xU ))λL,U ((xV ))

ϕ(xU )ϕ(NU (xU ))≡ 1 (mod p) if U = Tn or Kn

In fact, the condition for U = Zn is already satisfied by condition 1 of Θτn, Zp[[Γn]](p)

.

We have left out the cases when U = Cn, Nti or Nki for now because first we want to inspect

µL,N (xN ). We can rearrange the second bullet point of condition 3 of the definition of Θτn, Zp[[Γn]](p)

to get the following:

TrU/Zm∩U

(µL,U (xU )

∑β∈(Z/piZ)×

rcn−i1,β

)=

1plog

(NmCn/Zm∩Cn

(xpCn



))− 1plog

(NmU/Zm∩U

(xpUϕ(NU (xU ))p

2λL,U ((xV ))


))for U ∈ Nti , Nki and m ≥ n − i. But by condition 4 of Θτ

n, Zp[[Γn]](p)

we know that the RHS

belongs to p3m Zp[[Γn]](p).

Also, if we take m = n− i then we get:

TrU/Zn−i∩U

µL,U (xU )∑

β∈(Z/piZ)×

rcn−i1,β

= pn−iµL,U (xU )∑

β∈(Z/piZ)×

rcn−i1,β

Thus we can conclude that µL,N (xN ) is divisible by p2(n−i).

So for U = Cn, Nti or Nki we need to prove the following

xpUϕ(NU (xU ))p

2λL,U ((xV ))


≡ 1 (mod p) if U = Nti or Nki

xpCn


ϕ(xCn )pϕ(NCn (xCn ))≡ 1 (mod p) if U = Cn

Let us inspect λL,U ((xV )):

λL,U ((xV )) =∏

W=Cn,Tn,Kn

NmW/Zn

(ϕ(NmW/Zn (xW ))

ϕ(xW )[W :Zn]

) [Gn:U]

[W :Zn]2

81

By condition 3 of Θτn, Zp[[Γn]](p)

we know that NmW/Zn (xW ) = xZ , so we have

λL,U ((xV )) =∏W=Cn,Tn,Kn

NmW/Zn

(ϕ(xZ)

ϕ(xW )[W :Zn]

) [Gn:U]

[W :Zn]2

=∏W=Cn,Tn,Kn

(NmW/Zn (ϕ(xZ))

NmW/Zn (ϕ(xW )[W :Zn])

) [Gn:U]

[W :Zn]2

=∏W=Cn,Tn,Kn

(ϕ(xZ)

NmW/Zn (ϕ(xW ))

) [Gn:U][W :Zn]

We know that ϕ(xW ) ≡ xpW (mod p), therefore NmW/Zn (ϕ(xW )) ≡ NmW/Zn (xW )p (mod p) and

then we can use condition 3 of Θτn, Zp[[Γn]](p)

again

λL,U ((xV )) ≡∏

W=Cn,Tn,Kn

(xpZxpZ

) [Gn:U][W :Zn]

≡ 1 (mod p)

So for the cases U = Tn or Kn, we need

xpUNU (ϕ(xU ))

ϕ(xU )ϕ(NU (xU ))≡ 1 (mod p)

But this is clearly satisfied since ϕ(xU ) ≡ xpU (mod p).

Now we will check the case U = Cn. Condition 4 of Θτn, Zp[[Γn]](p)

tells us

NmCn/Zm∩Cn

(xpCnNCn (ϕ(xCn ))p



But ϕ(xCn ) ∈ Z1 ∩ Cn (see proof of Proposition 5.1) and NCn (xCn ) ∈ Zn ⊂ Z1 ∩ Cn, thus we get

NmCn/Z1∩Cn

(xpCnNCn (ϕ(xCn ))p(λL,Cn ((xV )V ∈Fn ))

)≡ (ϕ(xCn )pϕ(NCn (xCn )))[Cn:Z1∩Cn] (mod p)

⇐⇒ NmCn/Z1∩Cn (ϕ(xCn )NCn (ϕ(xCn ))p) ≡ (ϕ(xCn )pϕ(NCn (xCn )))[Cn:Z1∩Cn] (mod p)

⇐⇒ (ϕ(xCn )NCn (ϕ(xCn ))p)[Cn:Z1∩Cn] ≡ (ϕ(xCn )pϕ(NCn (xCn )))[Cn:Z1∩Cn] (mod p)

⇐⇒(ϕ(xCn )NCn (ϕ(xCn ))p


)[Cn:Z1∩Cn]≡ 1 (mod p)

⇐⇒(xpCn

NCn (ϕ(xCn ))pλL,Cn ((xV )V∈Fn )


)p≡ 1 (mod p)

The exact same method can be used for the cases U = Nti or Nki :

We know that νL,U (xCn ), ϕ(xU ) ∈ Z1 ∩ U and NU (xU ) ∈ Zn ⊂ Z1 ∩ U , so we just use condition 4

of Θτn, Zp[[Γn]](p)

:

NmU/Z1∩U

(xpUϕ(NU (xU ))p

2)≡(ϕ(xU )p

2ϕ(NU (xU ))

(νL,U (xCn )

))[U :Z1∩U ](mod p)

⇐⇒(ϕ(xU )ϕ(NU (xU ))p

2)[U :Z1∩U ]

≡(ϕ(xU )p

2ϕ(NU (xU ))

(νL,U (xCn )

))[U :Z1∩U ](mod p)

⇐⇒(

ϕ(xU )ϕ(NU (xU ))p2


)[U :Z1∩U ]

≡ 1 (mod p)

⇐⇒(

xpUϕ(NU (xU ))p

2λL,U ((xV ))


)p≡ 1 (mod p)

82

With that last lemma we have proved the following

Theorem 6.1 The image of K1

(

Λ(Gn)T ′

)under the map θn is contained in Θτ

n, Zp[[Γn]](p)

.

83

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