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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
):a ∈ Z/pnZb ∈ Z/pnZ
: 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
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)×
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
):a ∈ Z/pnZb ∈ Z/pnZ
: p 6 |(a2 − b2y2)
p2n−2(p− 1)2(
z εy2
1 z
) (a bεy2
b a
):a ∈ Z/pnZb ∈ Z/pnZ
: 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
):a ∈ Z/pnZb ∈ Z/pnZ
: p 6 |(a2 − b2β2)
p2(n+j−1)(p− 1)2(
x pjεβ2
pj x
) (a bεβ2 + kpn−j
b a+ lpn−j
):a ∈ Z/pnZb ∈ Z/pnZ
: p 6 |(a2 − b2εβ2)
p2(n+j−1)(p2 − 1)
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)
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:
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
)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
):a ∈ Z/pnZb ∈ Z/pnZ
: 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.
Therefore the number of elements per class=p4n−3(p2−1)(p−1)
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
):a ∈ Z/pnZb ∈ Z/pnZ
: 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:
• a ∈ p(Z/pnZ) and b ∈ (Z/pnZ)×
• 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)
This agrees with Table 2.
Therefore the number of elements per class=p4n−3(p2−1)(p−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).
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 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.
Therefore 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.
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
and that this group has order p2n+2j−1(p− 1):(a b
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:
• d2piα2 − b2pj = a2piα2 − c2p2i−jα4 = (ad− bc)piα2
• abpj − cdα2pi = 0
• a2pj − c2piα2 = (ad− bc)pj
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:
• d2piα2 − b2pj = a2piα2 − c2p2i−jα4 = (ad− bc)piα2
• abpj − cdα2pi = 0
• a2pj − c2piα2 = (ad− bc)pj
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.
Therefore 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.
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
and that this group has order p2n+2j−1(p− 1):(a b
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
):a ∈ Z/pnZb ∈ Z/pnZ
: 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:
• a ∈ p(Z/pnZ) and b ∈ (Z/pnZ)×
• 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
This agrees with Table 2.
Therefore the number of elements per class=p4n−3(p2−1)(p−1)
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
):a ∈ Z/pnZb ∈ Z/pnZ
: 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:
• a ∈ p(Z/pnZ) and b ∈ (Z/pnZ)×
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)
This agrees with Table 2.
Therefore the number of elements per class=p4n−3(p2−1)(p−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′
• abpm − cds = 0
• a2pm − c2s = (ad− bc)pm
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′
• abpm − cds = 0
• a2pm − c2s = (ad− bc)pm
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
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)×
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
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 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
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 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):
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)×
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:
ker(Log) → K1(Zp[Gn])Log−−−→ Zp[Conj(Gn)] → coker(Log)
↓ θn ↓ ψnker(L) 99K
∏U∈Fn Λ(Uab)×
L99K Qp ⊗Zp
∏U∈Fn Zp[Uab] 99K coker(L)
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
)
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)×
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
∑β∈(Z/pn−jZ)×
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
∑β∈(Z/pn−jZ)×
[rcjx,β ]− p2j∑
β∈(Z/pn−jZ)×[rcjx,β ]
)
+∑
1≤m<j
1pn−1+2m(p−1)
(pm+2j
∑β∈(Z/pn−jZ)×
[rcjx,β ]− pm+2j∑
β∈(Z/pn−jZ)×[rcjx,β ]
)
+ 1pn−1+2j(p−1)
(p3j
∑β∈(Z/pn−jZ)×
[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
∑β∈(Z/pn−jZ)×
[rcjx,β ]− p2j∑
β∈(Z/pn−jZ)×[rcjx,β ]
)
+∑
1≤m<n−i
1[NGn (Zm∩Nti ):Zm∩Nti ]
(pm+2j
∑β∈(Z/pn−jZ)×
[rcjx,β ]− pm+2j∑
β∈(Z/pn−jZ)×[rcjx,β ]
)
= 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 ]:
In section 4.3 we found that:
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 ]:
In section 4.3 we found that:
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 ) for m ≥ n− i
4. For any (aV )V ∈Fn ∈ Ψn,R, we have:
• 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
∑β∈(Z/pn−jZ)×
[(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
∑β∈(Z/pn−jZ)×
[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
)(a bδε0pi−j + kpn−j
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
)(a bδε0pi−j + kpn−j
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:
ker(Log) → K1(Zp[Gn])Log−−−→ Zp[Conj(Gn)] → coker(Log)
↓ θn ↓ ψnker(L) 99K
∏U∈Fn Λ(Uab)×
L99K Qp ⊗Zp
∏U∈Fn Zp[Uab] 99K coker(L)
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
)
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)×
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∑
β∈(Z/pn−1Z)×rc1xp,β
)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∑
β∈(Z/pn−1Z)×rc1xp,β ,
−p∑
β∈(Z/pn−1Z)×rc1xp,β
)Nkn−1 ,Cn×Ntn−1
rcjx,1 for
j < n− 1rcjx,1 −
rcj+1
xp,xp−1
p
(p2j
∑β∈(Z/pn−jZ)×
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
∑β∈(Z/pn−j−1Z)×
rcj+1xp,β ,
−p2j+1∑
β∈(Z/pn−j−1Z)×rcj+1xp,β
)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
∑β∈(Z/pn−j−1Z)×
rcj+1xp,β ,
−p2j+1∑
β∈(Z/pn−j−1Z)×rcj+1xp,β
)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
∑β∈(Z/pn−jZ)×
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∑
β∈(Z/pn−1Z)×rc1xp,β
)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∑
β∈(Z/pn−1Z)×rc1xp,β ,
−p∑
β∈(Z/pn−1Z)×rc1xp,β
)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∑
β∈(Z/pn−1Z)×rc1xp,β ,
−p∑
β∈(Z/pn−1Z)×rc1xp,β
)Nkn−1 ,Cn×Ntn−1
rcjx,1 for
j < n− 1
(p2j
∑β∈(Z/pn−jZ)×
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
(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
∑β∈(Z/pn−j−1Z)×
rcj+1xp,β ,
−p2j+1∑
β∈(Z/pn−j−1Z)×rcj+1xp,β
)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
∑β∈(Z/pn−j−1Z)×
rcj+1xp,β ,
−p2j+1∑
β∈(Z/pn−j−1Z)×rcj+1xp,β
)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
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)
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
(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 ))
)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
NmN/(Zn−i∩N)(xN )
) 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 ))
ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )
)−
∑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
NCn (ϕ(xCn ))pλL,Cn ((xV ))
ϕ(xCn )pϕ(NCn (xCn ))
)−
∑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 ))
ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )
)−
∑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
NCn (ϕ(xCn ))pλL,Cn ((xV ))
ϕ(xCn )pϕ(NCn (xCn ))
)−
∑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
(xpUNU (ϕ(xU ))λL,U ((xV ))
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
NmN/(Zn−i∩N)(xN )
) 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 )
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
• TrNki/Zm∩Nki
(aNki
) = TrCn/Zm∩Cn (aCn ) for m ≥ n− i
4. For any (aV )V ∈Fn ∈ Ψn,R, we have:
• 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
ϕ(NmN/(Zn−i−1∩N)(xN )p
NmN/(Zn−i∩N)(xN )
) p2[N:Zn−i∩N]
νL,N (xCn ) = NmCn/(Zn−i∩Cn)
(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
(λL,Zn ((xV )V ∈Fn )
)≡ 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:
• NmV/Zn (xV ) = xZn for all V ∈ Fn
74
• 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,R, 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
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:
• TrV/Zn (aV ) = aZn for all V ∈ Fn
• TrNti/Zm∩Nti
(aNti
) = TrCn/Zm∩Cn (aCn ) for m ≥ n− i
• TrNki/Zm∩Nki
(aNki
) = TrCn/Zm∩Cn (aCn ) for m ≥ n− i
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
(TrN/(Zn−i−1∩N)(aN )
[N : Zn−i−1 ∩N ]−TrN/(Zn−i∩N)(aN )
[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
pϕ (TCn (bCn )) + TCn (ϕ(bCn ))− [Gn : Cn]λf ((bV ))
)−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
pϕ (TCn (bCn )) + TCn (ϕ(bCn ))− [Gn : Cn]λf ((bV ))
)
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 ))
ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )
)−
∑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
NCn (ϕ(xCn ))pλL,Cn ((xV ))
ϕ(xCn )pϕ(NCn (xCn ))
)−
∑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
(xpUNU (ϕ(xU ))λL,U ((xV ))
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
NmN/(Zn−i∩N)(xN )
) p2[N:Zn−i∩N]
νL,N (xCn ) = NmCn/(Zn−i∩Cn)
(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
(λL,Zn ((xV )V ∈Fn )
)≡
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 )
3. For any (xV )V ∈Fn ∈ Θn, Zp[[Γn]](p)
, 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[[Γn]](p)
, 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)
80
• 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
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
NCn (ϕ(xCn ))pλL,Cn ((xV ))
ϕ(xCn )pϕ(NCn (xCn ))
))− 1plog
(NmU/Zm∩U
(xpUϕ(NU (xU ))p
2λL,U ((xV ))
ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )
))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 ))
ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )
≡ 1 (mod p) if U = Nti or Nki
xpCn
NCn (ϕ(xCn ))pλL,Cn ((xV ))
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
(λL,Cn ((xV )V ∈Fn )
))≡ NmCn/Zm∩Cn (ϕ(xCn )pϕ (NCn (xCn ))) (mod p3m+1)
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
ϕ(xCn )pϕ(NCn (xCn ))
)[Cn:Z1∩Cn]≡ 1 (mod p)
⇐⇒(xpCn
NCn (ϕ(xCn ))pλL,Cn ((xV )V∈Fn )
ϕ(xCn )pϕ(NCn (xCn ))
)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
ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )
)[U :Z1∩U ]
≡ 1 (mod p)
⇐⇒(
xpUϕ(NU (xU ))p
2λL,U ((xV ))
ϕ(xU )p2ϕ(NU (xU ))νL,U (xCn )
)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
Bibliography
[1] N. Bourbaki. Commutative Algebra. Hermann, Paris, 1965.
[2] D. Burns and O. Venjakob. On descent theory and main conjectures in non-commutative
iwasawa theory. Journal of the Institute of Mathematics of Jussieu, 10(1):59118, 2011.
[3] I. Chen. Jacobians of modular curves associated to normalizers of cartan subgroups of level pn.
C. R. Math. Acad. Sci. Paris, 339:187–192, 2004.
[4] J. Coates, T. Fukaya, K. Kato, R. Sujatha, and O. Venjakob. The GL2 Main Conjecture
for Elliptic Curves without Complex Multiplication. Publications mathematiques de l’IHES,
101(1):163–208, 2005.
[5] J. Coates and S. Howson. Euler characteristics and elliptic curves ii. J. Math. Soc. Japan,
53(1):175–235, 01 2001.
[6] G. Ellis. The schur multiplier of a pair of groups. Applied Categorical Structures, 6(3):355–371,
1998.
[7] T. Fukaya and K. Kato. A formulation of conjectures on p-adic zeta functions in non-
commutative iwasawa theory, preprint, 2003.
[8] T. Hara. Iwasawa theory of totally real fields for certain non-commutative p-extensions. J.
Number Theory, 130:1068–1097, 2010.
[9] T. Hara. Inductive construction of the p -adic zeta functions for noncommutative p -extensions
of exponent p of totally real fields. Duke Math. J., 158(2):247–305, 06 2011.
[10] M. Kakde. Proof of the Main Conjecture of Noncommutative Iwasawa Theory for Totally Real
Number Fields in Certain Cases. J. Algebraic Geometry, 20:631–683, 2011.
[11] M. Kakde. The main conjecture of iwasawa theory for totally real fields. Invent. Math, 193:539–
626, 2013.
[12] M. Kakde. Some congruences for non-cm elliptic curves. In D. Loeffler and S. Zerbes, editors,
Elliptic Curves, Modular Forms and Iwasawa Theory, volume 188 of Springer Proceedings in
Mathematics & Statistics. Springer International Publishing, 2015.
[13] K. Kato. Iwasawa theory of totally real fields for galois extensions of heisenberg type. preprint.
[14] R. Oliver. Whitehead Groups of Finite Groups. Cambridge University Press, 1988.
[15] J. Ritter and A. Weiss. On the “main conjecture” of equivariant iwasawa theory. J. Amer.
Math. Soc., 2011.
84
[16] K. Rubin. The main conjectures of iwasawa theory for imaginary quadratic fields. Inventiones
mathematicae, 103(1):25–68, 1991.
[17] P. Schneider and O. Venjakob. K1 of certain iwasawa algebras, after kakde. 29, 10 2011.
[18] J.-P. Serre. Proprits galoisiennes des points d’ordre fini des courbes elliptiques. Inventiones
mathematicae, 15(4):259–331, 1971.
[19] R. Swan. Algebraic K-Theory. Springer, 1968.
[20] O. Venjakob. Characteristic elements in noncommutative Iwasawa theory. J. reine angew.
Math., 583:193–236, 2005.
[21] O. Venjakob. On the Work of Ritter and Weiss in Comparison with Kakde’s Approach, pages
159–182. Springer Berlin Heidelberg, Berlin, Heidelberg, 2013.
[22] X. Wan. Introduction to Skinner-Urbans Work on the Iwasawa Main Conjecture for GL2. In
Thanasis Bouganis and Otmar Venjakob, editors, Iwasawa Theory 2012, volume 7 of Contribu-
tions in Mathematical and Computational Sciences, pages 35–61. Springer Berlin Heidelberg,
2014.
[23] L. Washington. Introduction to Cyclotomic Fields. Springer-Verlag New York, 1997.
85