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Maths PhD thesis.
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Normal congruence subgroups of the
Bianchi groups and related groups
by
Robert M Scarth
A thesis submitted to
the Faculty of Science
at the University of Glasgow
for the degree of
Doctor of Philosophy
July 13, 2003
c©Robert M Scarth 1998
2
Summary
Let O be an order in an imaginary quadratic number field. This thesis is mainly concerned
with normal subgroups of SL2(O) and of PSL2(O). Suppose that O is a maximal order then
O is the ring of integers in the number field and the group PSL2(O) is a Bianchi group.
In chapter one we discuss the geometric background of these groups and introduce some
fundamental algebraic concepts; those of order and level. We also discuss the Congruence
subgroup problem. Chapter two is a discussion of the fundamental theorem of Zimmert
[93]. In chapter three we discuss PSL2(O) where O is not a maximal order. We derive a
formula for their index in the Bianchi groups and presentations for some of these groups.
In particular we derive a presentation for PSL2(Z[√−3]) and using this presentation get a
partial classification of the normal subgroups of PSL2(Z[√−3]).
Chapter four generalizes a result of Mason and Pride [62] about SL2(Z) to all but finitely
many SL2(O). This result shows that for an arbitrary normal subgroup of N SL2(O)
there is no relationship between the order and level of N . This is in distinction to the groups
SLn(O), n > 3, where the order and level of a normal subgroup coincide. This answers a
question of Lubotzky’s.
Let O be an order in an imaginary quadratic number field. Then O is a Noetherian
domain of Krull dimension one and has characteristic zero. Chapter five discusses SL2 over
the class of all Noetherian domains of Krull dimension one, including those of non-zero
characteristic. In particular we generalize the work of Mason [58] and derive a relationship
between the order and level of a normal congruence subgroup of SL2(K) for any Noetherian
domain of Krull dimension one, K. In chapter six we apply this work to SL2(O) and
i
construct a new and vast class of normal non-congruence subgroups of SL2(O). Finally we
take a closer look at some particular PSL2(O).
ii
“Results! Why man, I have gotten a lot of results. I know several thousand things that
don’t work.”
- Thomas Edison [50] p.121.
iii
List of Notation
R A Ring.
R∗ The group of units in the ring R.
charR The characteristic of R.
L A Local Ring.
u A unit in a ring R.
Fq The field of q elements.
d A positive square-free integer. p.4
m A positive integer. p.4
Od The ring of integers in the imaginary quadratic number field Q(√−d). p.4
Od,m The Order of index m in Od. p.4
ω
1+i√
d2
if d ≡ 3 (mod 4)
i√d else
p.4
D The discriminant of Q(√−d). p.21
q An ideal in a ring R.
eij The matrix with a 1 in the (i, j)th position and zero’s elsewhere. p.9
Eij(r) Denotes I + reij.
Eij An elementary matrix; Eij(1).
E(x) Denotes the matrix
x 1
−1 0
. p.43
D(µ) Denotes the matrix
µ 0
0 µ−1
. p.43
iv
SLn(R) The group of n× n matrices over R with determinant 1.
PSLn(R) SLn(R) with the centre factored out.
GLn(R) The group of n× n matrices over R with non-zero determinant.
PSL2(Z) The Modular group. p.2
PSL2(Od) The Bianchi groups. p.4
A The matrix
0 1
−1 0
∈ SL2(R). p.4
T The matrix
1 1
0 1
∈ SL2(R). p.4
U The matrix
1 mω
0 1
∈ SL2(Od,m). p.4
a, t, u The images of A, T , and U in PSL2(R), or PSL2(Od,m). p.4
M An arbitrary 2× 2 matrix
a b
c d
.
trM The trace of the matrix M , ie a+ d. p.5
detM The determinant of M , ie ad− bc.
En(R) The subgroup of SLn(R) generated by the elementary matrices. p.9
Un(R) The normal subgroup of SLn(R) generated by the unipotent matrices. p.24
En(R, q) The subgroup of En(R) generated by the q-elementary matrices. p.10
Hn(R, q) The subgroup of SLn(R) consisting of the q-scalar matrices. p.10
SLn(R, q) The principal congruence subgroup of level q. p.11
PSLn(R, q) The principal congruence subgroup of level q in PSL2(R). p.108
H(q) Alternative notation to Hn(R, q).
Γ(q) Alternative notation to SLn(R, q).
∆(q) Alternative notation to En(R, q). p.70
Un(R, q) The normal subgroup of SLn(R) generated by the unipotent matrices
in SLn(R, q). p.30
o(S) The order of the subgroup S. p.11
v
l(S) The level of the subgroup S. p.10
E0 (2, R; q) The set of normal subgroups of SL2(R) of level zero and order q. p.55
Z(d,m) The Zimmert Set. p.21
r(d,m) The cardinality of Z(d,m). p.21
ρ(d,m) The largest rank of a free quotient of Γd,m. p.32
Nd,m(n) The number of normal congruence subgroups of Γd,m of index exactly n. p.110
G An arbitrary group.
Fr The free group of rank r.
[x, y] The commutator of x and y, ie x−1y−1xy.
G′ The commutator subgroup of G.
Gab The abelianization of G, ie G/G′.
N An arbitrary normal subgroup of G.
N(x1, . . . ) The normal closure in G of the elements x1, . . . . p.97
an(G) The number of subgroups of G of index exactly n. p.97
gcd(x, y) The greatest common divisor of x and y.
H2 Hyperbolic 2-space. p.1
H3 Hyperbolic 3-space. p.3
vi
Contents
Summary i
Contents vii
Acknowledgements viii
Statement x
1 Introduction 1
1.1 Geometric Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Hyperbolic geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.2 Hyperbolic 3-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 The normal subgroups of SLn(R) . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2.1 SLn over a local ring . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.2.2 SLn over an arbitrary ring . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2.3 The Congruence Subgroup Problem . . . . . . . . . . . . . . . . . . . 15
1.3 Some Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2 Zimmert’s Theorem 21
2.1 Hyperbolic 3-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 Zimmert’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.3 Explicit result about the free generators . . . . . . . . . . . . . . . . . . . . 25
2.4 Unipotent matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
vii
2.5 Computation of Zimmert Sets . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3 The groups PSL2(Od,m) 39
3.1 Computation of |PSL2(Od) : PSL2(Od,m)| . . . . . . . . . . . . . . . . . . . 39
3.2 A miscellany of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.3 A presentation for SL2(Z[√−3]) . . . . . . . . . . . . . . . . . . . . . . . . 46
3.4 The Group PSL2(O3,2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.4.1 HNN and amalgam decomposition . . . . . . . . . . . . . . . . . . . . 50
3.4.2 Some Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.5 Presentations for some other PSL2(Od,m). . . . . . . . . . . . . . . . . . . . 58
4 Non-standard normal subgroups 60
4.1 Non-standard normal subgroups of SL2(Z) . . . . . . . . . . . . . . . . . . . 61
4.2 Non-standard normal subgroups of SL2(Od,m) . . . . . . . . . . . . . . . . . 62
5 Order and level... 72
5.1 Primary decomposition in Noetherian domains . . . . . . . . . . . . . . . . . 72
5.2 Wohlfahrt’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
5.3 Preliminary results about SL2(L) . . . . . . . . . . . . . . . . . . . . . . . . 79
5.4 Order and level of a normal congruence subgroup . . . . . . . . . . . . . . . 94
6 Non-congruence subgroups 106
6.1 Counting finite index subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 106
6.2 The growth of non-congruence subgroups in the groups SL2(Od,m) . . . . . . 109
6.3 SL2(L) where L is a local image of Od,m . . . . . . . . . . . . . . . . . . . 110
6.4 Non-congruence subgroups of small index in the Bianchi groups . . . . . . . 117
6.5 The groups PSL2(Od), d = 2, 7, 11 and PSL2(O3,2) and PSL2(O1,2) . . . . 124
6.5.1 Normal congruence subgroups of PSL2(O2) . . . . . . . . . . . . . . 127
131
Bibliography 134
viii
Acknowledgements
I would like to thank my supervisor, Dr. Alec Mason, for his patience, kindness and en-
couragement during the last three and a half years. I would also like to thank everyone at
the Department of Mathematics at the University of Glasgow for contributing to a friendly
and stimulating environment. Thanks also to the EPSRC for giving me financial support
between October 1995 and September 1998, the Department of Mathematics here at Glas-
gow for giving my funding to attend various conferences, and my parents for paying my rent
while I completed this work.
I would like to thank Andy, Iain, Mark Brightwell, Mark Ratter, Mohammad, and
Stephen for the many interesting and stimulating discussions we had about Politics, Foot-
ball, Philosophy, Theology, and even some Mathematics. I’d have learnt a lot less without
you.
Finally, I would like to thank my family,especially my parents, for their unfailing support
and belief throughout 21 years of education.
ix
Statement
This thesis is submitted in accordance with the regulations for the degree of Doctor of
Philosophy in the University of Glasgow.
Chapter one provides some motivation and introduces some fundamental concepts. Chapter
two is a discussion of Zimmert’s theorem [93]. Chapters three and six and the independ-
ent work of the author and chapters four and five are the joint work of the author and his
supervisor. References are given throughout.
x
Chapter 1
Introduction
1.1 Geometric Background
We are interested in groups acting on Hyperbolic space. Hyperbolic geometry is the best
known example of a non-Euclidean geometry. The importance of the parallel axiom and the
development of non-Euclidean geometry in the history of Mathematics, indeed the history
of Western thought cannot be overstated. See [41, 67, 82] for brief and accesible accounts.
1.1.1 Hyperbolic geometry
Let
H2 = z ∈ C : Im(z) > 0
and equip H2 with the following metric
ds =
√dx2 + dy2
y
We now have the Poincare half plane model for two dimensional hyperbolic geometry. The
geodesics in H2 are straight lines and semicircles orthogonal to the real axis. The historical
point is that given a geodesic L and a point P not on L there are infinitely many geodesics
passing through P which do not intersect L. That is Euclids parallel axiom does not hold.
1
We are interested in distance preserving maps, or rigid motions. The hyperbolic distance
between z1, z2 ∈ H2 shall be denoted ρ(z1, z2).
Definition. A function from H2 onto itself which preserves hyperbolic distance is called an
isometry. The group of isometries is denoted Isom(H2).
Consider the following set
M =
z 7→ az + b
cz + d: a, b, c, d ∈ R, ad− bc = 1
this is the set of Moebius transformations of C. They map H2 onto itself and are all isometries.
M can be identified with the group PSL2(R) via the obvious map.
Theorem. ([38] theorem 1.3.1) Isom(H2) is generated by PSL2(R) and the map z 7→ −z.
PSL2(R) is of index 2 in Isom(H2).
Let X be a metric space and G a group acting on X.
Definition. A family Mα : α ∈ A of subsets of X is called locally finite if for any compact
set K ⊂ X we have Mα ∩K 6= ∅ for only finitely many α ∈ A.
Definition. We say that a group G acts properly discontinuously on X if the G-orbit of any
point x ∈ X is locally finite.
Definition. A closed connected F ⊂ X, with int(F ) 6= ∅, is a fundamental region for G if
1. GF = X.
2. int(F ) ∩ g(int(F )) = ∅ ∀ 1 6= g ∈ G.
The existence of a fundamental region allows us, in particular to find a presentation for
the group (see [49]).
Definition. G 6 PSL2(C) is said to be discrete if it contains no sequence of matrices conver-
ging elementwise to the identity. Discrete subgroups of PSL2(R) are called
Fuchsian groups.
2
Example 1.1.1. PSL2(Z) is obviously discrete and therefore a Fuchsian group.
Suppose that G 6 PSL2(R) is not discrete. So G contains a sequence gn such that
gn → I. Suppose that G has a fundamental region F ⊂ H2 and let x ∈ int(F ). Then
gnx → x as n → ∞. So ∃N such that ∀ n > N int(F ) ∩ gn(int(F )) 6= ∅. Contradiction.
Hence G cannot have a fundamental region.
Let Γ be a Fuchsian group. Let p ∈ H2 be not fixed by any non-trivial element of Γ. The
Dirichlet fundamental polygon for Γ centred at p is
Dp(Γ) =z ∈ H2 : ρ(z, p) 6 ρ(Tz, p) ∀T ∈ Γ
Theorem. ([38] theorem 3.2.2) For every Fuchsian group Γ and every p ∈ H2 not fixed by
a non-trivial element of Γ, Dp(Γ) is a connected convex fundamental region for Γ.
Example 1.1.2. Let Γ = PSL2(Z). The set
F =
z ∈ H2 : |z| > 1, |Re(z)| 6 1
2
is a Dirichlet region for Γ centred at ki, some k > 1.
The theory of Fuchsian groups is of great interest and has been extensively studied. See
[6, 7, 38].
1.1.2 Hyperbolic 3-space
The upper half-space in Euclidean three-space gives a convenient model of 3-dimensional
hyperbolic space
H3 = (z, r) ∈ C× R : r > 0
which we equip with the hyperbolic metric
ds2 =dx2 + dy2 + dr2
r2
The group PSL2(C) acts on H3 in the following way. Let M ∈ PSL2(C) where
M =
α β
γ δ
3
Then
M(z, r) =
((δ − γz)(αz − β)− r2γα
τ,r
τ
)where
τ = |γz − δ|2 + r2|γ|2
Under this action the hyperbolic metric is PSL2(C)-invariant. SL2(C) acts on H3 in exactly
the same way. As above we want a class of discrete subgroups of PSL2(C). We make use of
the following
Proposition. [19] Let A be a discrete subring of C with a one then SL2(A) is a discrete
subgroup of SL2(C).
Proposition. [19] The discrete subrings of C with a one are
1. Z.
2. The ring of integers Od = Z + ωZ in an imaginary quadratic number field.
3. The orders Od,m = Z +mωZ in an imaginary quadratic number field.
where d is a positive square-free integer, m is a positive integer, and
ω =
1+i√
d2
if d ≡ 3 (mod 4)
i√d else
This gives us a class of discrete subgroups of SL2(C) and of PSL2(C). We have already
met the group PSL2(Z). It is known as the Modular group. It was Picard who, in 1883,
first studied PSL2(Z [i]), and this group is known as the Picard group [67, 73]. The groups
PSL2(Od) are called the Bianchi groups. They were first studied by Bianchi in the 1890s
[8, 9] as a natural class of discrete subgroups of PSL2(C). See [23] chapter 7 for a discussion
of their action on H3 and [25] for a discussion of their algebraic properties. The groups
PSL2(Od,m) are of finite index in the Bianchi groups. See [23] for a general treatment of
discrete subgroups of PSL2(C) acting on H3. The Modular group and the Bianchi groups
are the first arithmetic examples of such groups and are of great interest in number theory.
4
We take the opportunity here to introduce three matrices which will be very important in
what follows. Let R be any commutative ring with a one and let
A =
0 1
−1 0
, T =
1 1
0 1
∈ SL2(R)
We will also need the following
U =
1 mω
0 1
∈ SL2(Od,m).
We denote the corresponding matrices in PSL2 by a, t, u. This will be a general convention.
A description of a fundamental region for the Bianchi groups in H3 can be found in
[23] section 7.3. Swan [84] has used this to derive presentations for the group SL2(Od) for
d = 1, 2, 3, 7, 11, 5, 6, 15, 19. It is then easy to derive a presentation for PSL2(Od).
We now look at matrices of finite order in SL2(C) and SL2(Od,m). The results are well
known but the presentation is our own. Let
M =
a b
c d
∈ SL2(C)
and let tr = a+ d be the trace of M . Recall that conjugate matrices have the same trace.
Lemma 1.1.1. If tr = ±2 and M 6= I then M is conjugate to ±T and so is of infinite
order.
Proof. We can suppose that
M =
1 + α β
γ 1− α
Now a b
c d
1 1
0 1
d −b
−c a
=
1− ac a2
−c2 1 + ac
So letting a =
√β, and c = i
√γ and then choosing b, d, so that ad − bc = 1 we get the
result.
5
Lemma 1.1.2. If tr2 6= 4 (ie tr 6= ±2) then M is conjugate to α 0
0 α−1
for some α ∈ C.
Proof. Suppose that z ∈ C is a fixed point of M ie
az + b
cz + d= z
So cz2 + (d − a)z − b = 0. The discriminant of this quadratic is (d − a)2 + 4bc = (a +
d)2 − 4(ad− bc) = tr2 − 4. Now tr2 6= 4 so M has two distinct fixed points z1, and z2. Let
w = (z2 − z1)−1 and consider 1 −z2
w −z1w
a b
c d
1 −z2
w −z1w
−1
Using the fact that cz2i +(d−a)zi− b = 0, for i = 1, 2, we can show that this matrix is equal
to α 0
0 α−1
for some α ∈ C.
Lemma 1.1.3. If |tr| > 2 then M is of infinite order.
Proof. Clearly tr2 6= 4, so M is conjugate to α 0
0 α−1
for some α ∈ C. Suppose that |α| = 1, so |α−1| = 1 and |tr| = |α + α−1| 6 |α| + |α−1| = 2.
So |tr| 6 2. Contradiction. So |α| 6= 1, so |α| > 1, or |α−1| > 1. So |αn| → ∞, or |α−n| → ∞
as n→∞. Now α 0
0 α−1
n
=
αn 0
0 α−n
.
So Mn 6= I for every n ∈ Z.
6
Lemma 1.1.4. If |tr| = 2 and tr2 6= 4 then M is of infinite order.
Proof. Now tr2 6= 4 so M is conjugate to α 0
0 α−1
for some α ∈ C. As above if |α| 6= 1 then M is of infinite order. So suppose that |α| = 1, so
α−1 = α. So α + α−1 = α + α = 2Re(α). So 2 = |tr| = |α + α−1| = 2|Re(α)|, so Re(α) = 1.
Hence α = α−1 = 1 and M = I. Contradiction.
Lemma 1.1.5. Suppose that |tr| < 2. Then if tr /∈ R then M is of infinite order.
Proof. tr2 6= 4 so M is conjugate to α 0
0 α−1
for some α ∈ C and where α + α−1 = tr. Now if |α| 6= 1 then, as above, M is of infinite
order. So suppose that |α| = 1 then α−1 = α so tr = α+ α = 2Re(α) ∈ R.
Lemma 1.1.6. If tr = 0 then M2 = −I.
Proof. Now M is conjugate to
X =
α 0
0 α−1
some α ∈ C such that α+ α−1 = 0. So α2 + 1 = 0, so α = ±i. It is then easy to verify that
X2 = −I.
Lemma 1.1.7. If tr = 1 then M3 = −I and if tr = −1 then M3 = I.
Proof. Now M is conjugate to
X =
α 0
0 α−1
for some α ∈ C such that α + α−1 = tr. First suppose that tr = −1. Then α2 + α + 1 = 0,
so α(α2 + α + 1) = 0 and so α3 = 1. Thus X3 = I and so M3 = I. Similarly if tr = 1 then
α3 = −1 and so X3 = −I. Hence result.
7
Theorem 1.1.8. Let I 6= M ∈ SL2(Od,m). Then M is of finite order if and only if tr = 0,
or ±1.
Proof. Now tr ∈ Od,m so |tr|2 ∈ Z. First, if |tr|2 > 4 then M is of infinite order. Suppose
that |tr|2 < 4, so |tr|2 = 0, 1, 2, or 3, but if M is of finite order tr ∈ Od,m ∩ R = Z, so
|tr| = 0, or 1. Now if |tr| = 0 then tr = 0 and so M2 = −I. If |tr|2 = 1 then tr = ±1, so
M6 = I.
1.2 The normal subgroups of SLn(R)
Let R be a ring, with a one. Let n ∈ N, n > 2, and form the group SLn(R). It is natural to
ask the following
Question. What are the normal subgroups of SLn(R)?
The case where R is a field is simple; literally:
Theorem. [21] Let F be any field, n ∈ N, n > 2. Then PSLn(F) is simple with two
exceptions:
PSL2(F2) ∼= S3 and PSL2(F3) ∼= A4
where Fd denotes the field of d elements.
By another result of Dieudonne [21] SL2(F)′ = SL2(F), and so the only non-trivial
normal subgroup of SL2(F) is ±I. Note that the exceptions are both two dimensional
linear groups. We shall see that the two dimensional case is, in general, more complicated
than the higher dimensional cases. Further, when investigating normal subgroups of SL2(R)
one finds that ideals of index 2 or 3 in R play an important role. When we pass from fields
to rings we no longer have simplicity.
Example 1.2.1. Let m ∈ Z. Consider the following subgroup
Γ(m) = M ∈ SL2(L) : M ≡ I (mod m)
8
This is clearly a non-trivial normal subgroup of SL2(L). It is called the Principal congruence
subgroup of level m, and is a member of a very important class of normal subgroups, as we
shall see later on.
1.2.1 SLn over a local ring
The next easiest case after that of a field is a local ring,so let R = L, be a commutative
local ring, with maximal ideal m, and let Nm = |L : m|. We introduce two classes of normal
subgroup of SL2(L). Let q L be an ideal in L, then define
Γ(q) = M ∈ SL2(L) : M ≡ I (mod q)
this is the kernel of the natural map SLn(L) −→ SLn(L/q) and let
H(q) = M ∈ Γ : M ≡ kI (mod q), k ∈ L
H(q)/Γ(q) is the centre of SL2(L)/Γ(q). Let S 6 SL2(L), by the level of S, denoted l(S), we
mean the largest ideal, q of L such that Γ(q) 6 S, and by the order of S, denoted o(S), we
mean the smallest ideal, q of L such that S 6 H(q). Since Γ(q) 6 H(q) we have l(S) 6 o(S).
The first attempt to classify the Normal subgroups of SL2(L) was in 1961 by Klingenberg.
Theorem. [42] Let N 6 SL2(L) be of order q. Then N SL2(L) ⇔ l(N) = o(N) where,
for n = 2 we assume Nm 6= 3, and 2 is a unit.
Lacroix, in 1966 [43], dropped the condition that 2 was a unit and included the case
where Nm = 3 but was unable to deal with the case Nm = 2:
Theorem. Suppose that Nm > 2, and let N 6 SL2(L) be of order q. Then N SL2(L)⇔
o(N) = l(N) unless Nm = 3, and o(N) = L. If Nm = 3 and N SL2(L), o(N) = L then
N = SL2(L), or N = SL2(L)′.
However Lacroix provided examples of non-normal subgroups of SL2(L) of order L.
The case where Nm = 2 appears to be very complicated in general (see [43]). Mason [57]
has investigated the case where Nm = 2, m is principal and every ideal of L is a power of
9
m. In section (5.3) we investigate the normal subgroups of SL2(L) and introduce techniques
which allow us to deal with the case Nm = 2 and m nilpotent.
Klingenberg showed (roughly) that N SLn(L) ⇔ l(N) = o(N). This leads to the
following
Definition. Let S 6 SLn(L). If l(S) = o(S) we say that S is standard.
This gives rise to the standard criterion:
N SLn(L)⇔ N is standard
With slight modification these concepts carry over to the case of an arbitrary ring. We
remark that Costa and Keller [18] have characterized the normal subgroups of GL2(A) for an
arbitrary commutative local ring A in terms of certain commutator groups. Their solution
reduces to that of Klingenberg and Lacroix in the relevant cases. We mention it only in
passing here because we are mainly interested in the standard criterion, or, where that fails
the relationship between the order and level of a normal subgroup.
1.2.2 SLn over an arbitrary ring
We introduce the following subgroup of SLn(R):
En(R) =< I + reij : r ∈ R, i 6= j >
It is well known that when R = F is a field, SLn(F ) = En(F ). This remains true for some
rings. Firstly, in light of the previous section
Proposition. ([3] corollary 5.9.2) Let L be a semilocal ring. Then ∀n > 2 En(L) = SLn(L).
Proposition. ([33] proposition 2.4) Let R be a euclidean ring. Then ∀n > 2 En(R) =
SLn(R).
As Z is euclidean we have
Proposition. ([51] lemma 3.1) SL2(Z) = E2(Z).
10
Let Od be the ring of integers of Q(√−d). The Bianchi groups are the groups PSL2(Od),
we have already mentioned them in section (1.1.2). Now Od has a euclidean algorithm
⇔ d = 1, 2, 3, 7, 11 ([13] p.21), so if d = 1, 2, 3, 7, 11 then SL2(Od) = E2(Od). In fact
Proposition. ([14] Theorem 6.1)
SL2(Od) = E2(Od)⇔ d = 1, 2, 3, 7, 11
The groups PSL2(Od), d = 1, 2, 3, 7, 11 are called the Euclidean Bianchi groups. We can
now ask the following which is obviously related to our original question
Question. What are the En(R)-normalized subgroups of SLn(R)?
We now introduce an important concept, the SRn-condition. Let R be a ring. If
a1, . . . , an ∈ R such that∑Rai = R then ∃b1, . . . , bn−1 ∈ R such that
∑R(ai + bian) = R
then we say that R has stable range n, and we write SRn(R) or say R has SRn. The SR2-
condition is particularly important so we describe it separately: If Ra1 + Ra2 = R then
∃t ∈ R such that a1 + ta2 is a unit. See [16, 29, 85, 87, 88] for examples of rings with SR2.
Proposition. ([3] Proposition 5.3.4) Semilocal rings are SR2-rings.
Proposition. [3] If R is an SR2-ring then SL2(R) = E2(R).
We remark that Dedekind domains have SR3 ([3] theorem 3.5 page 239). We must also
modify our concept of level because, famously, in general not every normal subgroup of
SLn(R) contains a principal congruence subgroup. Let q R. Let
En(R, q) =< I + αeij : α ∈ q, 1 6 i, j 6 n, i 6= j >En(R)
Hn(R, q) = M ∈ SLn(R) : M ≡ kI (mod q), k ∈ R
Let S 6 SLn(R). By the level of S, denoted l(S), we mean the largest q R such that
En(R, q) 6 S. This is well defined because En(R, q1)En(R, q2) = En(R, q1 + q2). By the
order of S, denoted o(S) we mean the smallest q R such that S 6 Hn(R, q). As before,
we say S is standard if l(S) = o(S). We define a Principal congruence subgroup as before
SLn(R, q) = M ∈ SLn(R) : M ≡ I (mod q)
11
clearly En(R, q) 6 SLn(R, q). As in the case of a local ring Hn(R, q)/SLn(R, q) is the centre
of SLn(R)/SLn(R, q). In the case of a local ring our two concepts of level are the same
because:
Proposition. ([3] corollary 5.9.2) Let L be a local ring. Then ∀n > 2 En(L, q) = SLn(L, q).
Any subgroup of SLn(R) which contains a principal congruence subgroup is known as a
congruence subgroup. The question of whether every subgroup of finite index in SLn(R) is
a congruence subgroup or not is of great interest and is known as the Congruence Subgroup
Problem. We discuss it in the next section.
Most attempts to understand the normal subgroups of SLn(R) are centred round the
standard criterion, and we now describe some of these attempts.
Theorem. Let H 6 GLn(A). Then for n > 3, if A has SR2 [3], or is commutative [86], or
is a Banach algebra [88], or is von-Neumann regular [89] then
H is En(A)-normalized⇔ H is standard
We remark that there exist examples of rings for which the standard criterion fails for
n > 3 (See [28, 90] ). We now focus exclusively on commutative rings and ask how are
the E2(R)-normalized subgroups and the standard subgroups related? It turns out that the
answer depends very much on R.
Theorem. [3] IF R has SR2 and S 6 SL2(R) then S standard ⇒ S is E2(R)-normalized.
Costa and Keller ([17] theorem 2.6) have provided a partial converse
Theorem. Let R be an SR2 ring with 6 ∈ R∗. Then N SL2(R)⇒ N is standard.
The cases R = F2, or F3 show that 6 ∈ R∗ is necessary. Suppose now that A is a
Dedekind domain of arithmetic type (see the section at the end of this chapter on number
theory) so A has SR3, and suppose that A has infinitely many units. Serre has shown ([80]
Prop 2 p. 492)
12
Theorem. Let A be a Dedekind domain of arithmetic type and suppose that A∗ is infinite.
Then SL2(A) has no E2(A)-normalized subgroups of level zero and non-zero order.
Mason has proved the following
Theorem. [55] Let A be a Dedekind domain of arithmetic type with infinitely many units
then every standard subgroup of SL2(A) is E2(A)-normalized.
we also have
Theorem. [75] Let A be a Dedekind domain of arithmetic type with infinitely many units.
Then the E2(A)-normalized subgroups of SL2(A) are precisley the standard subgroups if and
only if the following three conditions hold for A:
1. All residue class fields of A have more than 3 elements.
2. 2 ∈ A∗ or 2 is unramified in A.
3. E2(A, a) = [E2(A), E2(A, a)] for every a A.
However Mason has shown that when A∗ is infinite the order and level of an E2(A)-
normalized subgroup are closely related.
Theorem. ([58]) Let N 6 SL2(A) be E2(A)-normalized. Let q = o(N), and q∗ = l(N).
Then
1. If A is contained in a number field and is not totally imaginary then 12q 6 q∗.
2. If A is contained in a number field and is totally imaginary then 12u0q 6 q∗.
3. If A is contained in a function field in one variable over a finite field then q3 6 q∗.
where u0 is defined as follows. Let m be the total number of roots of unity in A, and let u be
the A-ideal generated by u2 − 1 where u ∈ A∗. If m = 2 Let u0 = u + 2A and if m > 2 then
let u0 = u.
13
So we can see that in the two dimensional case the unit structure becomes important.
What happens to the standard criterion when the unit group is finite? Consider first the
group SL2(Z), here we see that not only does the standard criterion fail but it fails badly:
Theorem. [56, 62] The group SL2(Z) has 2ℵ0 non-normal standard subgroups and 2ℵ0 non-
standard normal subgroups.
In fact a more precise result is achieved. Let let E0(2,Z; q) denote the set of normal
subgroups of SL2(Z) of level zero, and order q.
Theorem. [62] Let 0 6= q Z then |E0(2,Z; q)| = 2ℵ0.
The situation in the groups SL2(Od) is similar
Theorem. [60] For every positive square-free integer d, the group SL2(Od) has 2ℵ0 non-
normal standard subgroups.
The obvious question now is
Question. Does every group SL2(Od) have 2ℵ0 non-standard normal subgroups?
Mason has shown
Theorem. [59] For every positive square-free integer d, the group SL2(Od) has 2ℵ0 normal
subgroups of level zero.
The only normal subgroup of SL2(Od) with order 0 is the trivial subgroup. Thus there
are uncountably many non-standard normal subgroups in the groups SL2(Od). Later we ex-
tend this to show that for all but finitely many (d,m), and all 0 6= qOd,m, |E0(2,Od,m; q)| =
2ℵ0 . The exceptions are almost certainly due to an inadequacy in our proof. Mason has ob-
tained similar results in the final case. Let C be a Dedekind domain of arithmetic type
contained in a function field and with finitely many units. Let Γ = SL2(C).
Theorem. ([53] theorem 3.1) There exist 2ℵ0 normal subgroups of finite index in Γ which
have level zero.
Theorem. ([53] theorem 3.2) Let q C be such that Nq > c0, some constant c0. Then Γ
contains 2ℵ0 non-normal standard subgroups of level q.
14
1.2.3 The Congruence Subgroup Problem
Let R be a commutative ring with a one.
Congruence Subgroup Problem. Does every subgroup of SLn(R) of finite index contain
a principal congruence subgroup?
This question can, with some care, be made to make sense in PSLn(R), the details can
be found in section (6.4). The Congruence Subgroup Problem has a long history and has its
origins in the work of Fricke and Klein, and Pick in the 19th century (see [34] section 4.3 and
[11] chapter I.6 section B). Klein [39] pointed out, at a meeting of the Munich Academy on
6th December 1879, that the Modular Group, PSL2(Z), contains subgroups of finite index
which do not contain a principal congruence subgroup. This was later proven simultaneously
and independently by Fricke [27] and Pick [74]. We outline the proof, which can be found
in [51]. Chandler and Magnus [11] assert, without any evidence, that it may be pre-1914.
Lemma.PSL2(Z)
PSL2(Z, nZ)∼= PSL2(Zn)
Where Zn denotes the ring of integers mod n.
Lemma. The only non-abelian quotient groups that can appear in a composition series of
PSL2(Zn) are the groups PSL2(Zp), where p is prime.
Lemma. A11 is a quotient of PSL2(Z), and A11 is not isomorphic to any of the groups
PSL2(Zp).
The kernel of the map PSL2(Z) A11 is then a normal non-congruence subgroup of
PSL2(Z). It is then a simple matter to see that SL2(Z) must also contain non-congruence
subgroups. The positive solution of the problem in the n > 3 case was proved simultaneously
in 1965 by Mennicke [66] and Bass-Lazard-Serre [4] in the context of SL2(Z):
Theorem. Let n > 3. If H 6 SLn(Z) is of finite index then SL2(Z, nZ) 6 H for some
n 6= 0.
15
Again we see that in the two dimensional case the normal subgroup structure of SLn(Z)
is much more complicated than the higher dimensional cases. In fact the situation is a lot
more complicated than may at first appear because “most” subgroups of the Modular group
are non-congruence subgroups. We now outline two different ways in which this idea is made
precise.
The Modular group has the following presentation PSL2(Z) =< a, t; a2, (at)3 >∼= C2 ∗C3
[25, 51, 70, 81]. Newman [71] has derived an asymptotic formula for the number of subgroups
of a given index in the free product of finitely many cyclic groups. Applied to the Modular
group we get
Theorem. [71] Let an(G) denote the number of subgroups of G of index n. Then
an(PSL2(Z)) ∼ (12πe1/2)−1/2 exp
(n log n
6− n
6+ n
12 + n
13 +
log n
2
)Now let γn(PSL2(Z)) denote the number of congruence subgroups of PSL2(Z) of index
at most n. A special case of a theorem of Lubotzky [47] is
Theorem. There exists positive constants C1, C2 such that
nC1 log n/ log log n 6 γn(PSL2(Z)) 6 nC2 log n/ log log n.
So it can be seen that
γn(PSL2(Z))
an(PSL2(Z))−→ 0 as n −→∞
and in this sense most subgroups of the modular group are non-congruence subgroups. Again
it is now simple to see that in SL2(Z) most subgroups are non-congruence. We now outline
another interpretation. Let
F = S 6 SL2(Z) : |SL2(Z) : S| <∞
and
C = C 6 SL2(Z) : C a congruence subgroup
16
These constitute bases for neighbourhoods of the identity for two topologies on SL2(Q). Let
SL2(Q) be the completion relative to F and let SL2(Q) be the completion relative to C.
Since C ⊆ F we have a natural surjection
Π : SL2(Q) −→ SL2(Q)
We denote the kernel of this map by C(SL2,Z) and call it the non-congruence kernel. The
Congruence Subgroup Problem now becomes the following
Question. Is C(SL2,Z) trivial?
Theorem. [45]
C(SL2,Z) = Fω
where Fω is the free profinite group of countable rank.
That is the congruence kernel is enormous, and so again most subgroups of SL2(Z) are
non-congruence. It is simple to see that the same is true in the modular group PSL2(Z).
So far we have looked at the congruence subgroup problem in SLn(Z). What about more
general rings?
Let K be a global field (see section on number theory at the end of this chapter), S∞ ⊆
S ⊆ Ω, S 6= ∅ and O = OS the ring of S-integers. Form the group SLn(O). Clearly the
congruence subgroup problem makes sense in this group and we can construct the congruence
kernel C(SLn,O) in exactly the same way as C(SL2,Z). The congruence subgroup problem
then becomes to determine the structure of C(SLn,O). Bass-Milnor-Serre [5] proved the
following
Theorem. With the above notation suppose that O is a Dedekind domain of arithmetic type,
and n > 3. Then C(SLn,O) = 1 unless K is a number field which is totally complex and
O is the ring of integers, in which case C(SLn,O) ∼= µ(K), the (finite cyclic) group of all
roots of unity in K.
The n = 2 case was dealt with by Serre [80].
17
Theorem. With the above notation and n = 2 then
1. If |S| > 2 and it is not the case that K is a totally complex number field and O its ring
of integers then C(SL2,O) = 1.
2. If |S| > 2 and K is a totally complex number field and O is its ring of integers then
C(SL2,O) ∼= µ(K).
3. If |S| = 1 then C(SL2,O) is infinite.
Thus the Congruence Subgroup Problem fails completely only in the case where n = 2
and |S| = 1. So again we see that the two dimensional case is more complicated than the
higher dimensional cases, and in the two dimensional case the unit structure of the ring
becomes important. There are three families of rings O for which |S| = 1: Z , Od, and C (see
section on number theory). We have already seen Lubotzky’s characterization of C(SL2,Z).
He also proved
Theorem. [45] With the above notation let O = Od, or C. Then C(SL2,O) contains Fω,
the free profinite group of countable rank, as a closed subgroup.
So the Modular group and the Bianchi groups contain a great many non-congruence
subgroups. We have already mentioned that Fricke [27] and Pick [74] gave examples of non-
congruence subgroups of the modular group (see also [40]). Reiner [77] generalized their
construction and many authors have produced classes of non-congruence subgroups (see [37]
and [70]). Stothers has shown [83] that the minimal index of a non-congruence subgroup in
the Modular group is 7. McQuillan has classified the normal congruence subgroups of the
Modular group [65]. Drillick [22] has adapted the approach in Magnus [51] outlined earlier to
the Bianchi group PSL2(O1), known as the Picard Group. Britto [10] generalized Drillicks
arguments to construct an infinite family of non-congruence subgroups in PSL2(Od) for
d = 1, 2, 3, 7, 11, 5, 6, 15. Since the basis of this construction is a surjection of PSL2(Od)
onto An, n > 7, all of the normal non-congruence subgroups constructed by these methods
are of index 6k, for some k ∈ N. It is also the case that these normal non-congruence
subgroups are torsion free.
18
In chapter five we extend the work of Mason in [58] and derive a relationship between
the order and level of a normal congruence subgroup of PSL2(Od,m). We then use this
relationship in chapter six to show that nearly every normal subgroup in PSL2(Od,m) of
index not divisible by 6 is a non-congruence subgroup. Further they all have torsion. Thus
our normal non-congruence subgroups are all different from those constructed by Drillick
and Britto.
1.3 Some Number Theory
For more information see [34] section 2.2E and the references therein.
Let K be any field. Let
v : K −→ R
Consider the following four conditions:
1. v(a) > 0, and v(a) = 0⇔ a = 0.
2. v(ab) = v(a)v(b).
3. v(a+ b) 6 v(a) + v(b).
4. v(a+ b) 6 max(v(a), v(b)).
Clearly 1 and 4⇒ 3. If v satisfies 1, 2, and 3 we say v is a valuation on K. If v also satisfies
4, v is non-Archimedean, and if not it is Archimedean. If v(a) = 1 ∀a ∈ K − 0 then v is
the trivial valuation. We say that a field is global if it is a finite separable extension of Q or
of the quotient field of a polynomial ring Fd [X], where Fd is a finite field of order d. In the
first case K is a number field, in the second a function field. We say that two valuations v1,
and v2 are equivalent if for every a ∈ K we have v1(a) < 1⇔ v2(a) < 1. Let K be a global
field. Let Ω be a complete set of inequivalent non-trivial valuations on K and let S∞ ⊆ Ω
be the set of Archimedean valuations. Suppose S∞ ⊆ S ⊆ Ω and S 6= ∅ then
OS = x ∈ K : v(x) 6 1 ∀v /∈ S
19
is called the ring of S-integers of K. OS is a Dedekind domain. If S is finite then we say
OS is a Dedekind domain of arithmetic type. If K is a number field and S = S∞ then OS is
the ring of integers of K.
The completion of K with respect to an Archimedean valuation, v, is isomorphic (as a
topological field) to R, or C and we say v is real or complex accordingly. The number field
K is totally real if all valuations in S∞ are real, and totally complex if they are all complex.
Theorem. Let O = OS be a Dedekind domain of arithmetic type. Suppose |S| = 1 then one
of the following is the case
1. O = Z.
2. O = Od the ring of integers in Q(√−d), d a positive square free integer.
3. O = C, the coordinate ring of an affine curve obtained by removing a point from a
projective curve defined over a finite field.
farther these are precisely the Dedekind domains of arithmetic type with finitely many units.
Example 1.3.1. Let d ∈ N be square free. Let O be the ring of integers in Q(√d). Then
O is a Dedekind domain of arithmetic type, and by [13] theorem 11.4, O∗ is infinite.
20
Chapter 2
Zimmert’s Theorem
In this chapter we describe a topological method invented by Zimmert [93] and its extension
due to Grunewald and Schwermer [32]. It concerns the action of SL2(R) on hyperbolic 3-
space H3, where R is an order in an imaginary quadratic number field. See also [23] chapter
7 section 5 for a discussion of this method.
Recall that the Bianchi groups are PSL2(Od) where Od is the ring of integers in the
imaginary quadratic number field Q(√−d), and recall that
Od = Z + ωZ where ω =
1+i√
d2
if d ≡ 3 (mod 4)
i√d else
and where d is a positive square-free integer. Now let m ∈ N and let
Od,m = Z +mωZ
where ω is as above. The Od,m are the orders in the imaginary quadratic number field
Q(√−d). It is clear that Od = Od,1 and that |Od : Od,m| = m. We can form the groups
SL2(Od,m). Since mOd ⊆ Od,m we have SL2(Od,mOd) 6 SL2(Od,m) so that SL2(Od,m) is
of finite index in SL2(Od). Similar comments can be made to show that PSL2(Od,m) is of
finite index in PSL2(Od). The aim of Zimmert’s method is to prove the following
Theorem. For every d there exists m such that SL2(Od,m) has a free non-abelian quotient.
As PSL2(Od,m) = SL2(Od,m)/ < −I > the following theorem is clear
21
Theorem. For every d there exists an m such that PSL2(Od,m) has a free non-abelian
quotient.
We say that a group G is SQ-Universal if every countable group can be embedded in a
quotient of G. It follows from a result of P. M. Neumann [69] and the above theorem that
Theorem. [32] Every Bianchi group PSL2(Od) is SQ-universal.
Thus the Bianchi groups may be considered “large” [76]. It follows from SQ-Universality
that every Bianchi group has 2ℵ0 normal subgroups [69] and it is this that is the source of
the extremely complicated normal subgroup structure described in the previous chapter.
2.1 Hyperbolic 3-space
The upper half-space in Euclidean three-space gives a convenient model of 3-dimensional
hyperbolic space
H3 = (z, r) ∈ C× R : r > 0
which we equip with the hyperbolic metric
ds2 =dx2 + dy2 + dr2
r2
The group SL2(C) acts on H3 in the following way. Let M ∈ SL2(C) where
M =
α β
γ δ
Then
M(z, r) =
((δ − γz)(αz − β)− r2γα
τ,r
τ
)where
τ = |γz − δ|2 + r2|γ|2
Under this action the hyperbolic metric is SL2(C)-invariant. The group SL2(C) is generated
by the matrices 1 a
0 1
,
0 −1
1 0
22
where a ∈ C. These generators operate on H3 in the following way 1 a
0 1
(z, r) = (z + a, r) ,
0 −1
1 0
(z, r) =
(−z
|z|2 + r2,
r
|z|2 + r2
).
2.2 Zimmert’s Method
Let D be the discriminant of Q(√−d). It is well known that [52]
D =
−d if d ≡ 3 (mod 4)
−4d else
Definition. The Zimmert Set Z(d,m) is the set of all n ∈ N such that
1. 4n2 6 m2|D| − 3.
2. D is a quadratic non-residue modulo all the odd prime divisors of n, and if D 6≡ 5
(mod 8) then n is odd.
3. n > 0, gcd(n,m) = 1 and n 6= 2.
It is easy to prove that Z(d,m) = ∅ ⇔ (d,m) = (1, 1), or (3, 1), and if Z(d,m) 6= ∅ then
1 ∈ Z(d,m). We let r(d,m) = |Z(d,m)|. Zimmert’s theorem is
Theorem. [32] SL2(Od,m) has a free quotient of rank r(d,m).
Zimmert [93] proved the m = 1 case. We now outline the proof. Let
Bd,m =(z, r) ∈ H3 : τ > 1 ∀ coprime γ, δ ∈ Od,m
where, γ, δ coprime means that γOd,m + δOd,m = Od,m, and, as before
τ = |γz − δ|2 + r2|γ|2
Every point of H3 is equivalent to a point of Bd,m under the action of SL2(Od,m) and so the
natural map
Bd,m −→ SL2(Od,m)\H3
23
is surjective. Let
D = (s1 + s2mω, r) ∈ Bd,m : s1, s2 ∈ [0, 1]
D is a fundamental domain for SL2(Od,m) [84, 93].
Proposition 2.2.1. ([84] Proposition 3.9) Every h ∈ H3 has a neighbourhood U such that
σU ∩D 6= ∅ for only finitely many σ ∈ SL2(Od,m).
Let n ∈ Z(d,m), t ∈ Z such that (n, t) = 1. Let
Fn,t =
(z, r) ∈ Bd,m :
∣∣∣∣Im(z − tmω
n
)∣∣∣∣ 6 1
m4|D|2
By condition 1 in the definition of the Zimmert set we have Fn1,t2 ∩ Fn2,t2 6= ∅ ⇔ n1 = n2,
and t1 = t2.
Lemma 2.2.2. ([93] Hilfssatz 1) Let (z, r) ∈ Fn,t, let σ ∈ SL2(Od,m) such that σ(z, r) =
(z′, r′) ∈ Bd,m. Then ∃t′ ∈ Z such that gcd(n, t′) = 1 and
Im
(z − tmω
n
)= Im
(z′ − t′mω
n
).
Further
r >5
2m2|D|
Now for each n ∈ Z(d,m) define ϕn : Bd,m → S1, where S1 = z ∈ C : |z| = 1, by
ϕn(z, r) =
1 if (z, r) /∈⋃
(n,t)=1 Fn,t
exp2πi(
12
+ m4|D|22
Im(z − tmω
n
))if (z, r) ∈ Fn,t
There is a unique factorization of ϕn over SL2(Od,m)\H3 by a continuous map
fn : SL2(Od,m)\H3 −→ S1
This is well defined by lemma (2.2.2) and continuous by proposition (2.2.1). Suppose that
Z(d,m) = n1, . . . , nr. Let Y denote the one point union of r(d,m) copies of S1 with base
point 1 ie
Y =(z1, . . . , zr) ∈ S1 × · · · × S1 : zi 6= 1 for at most one i
24
Now define
f : SL2(Od,m)\H3 −→ Y
by
(z, r) 7−→ (f1(z, r), . . . , fr(z, r))
where fi denotes fni. Now f induces a homomorphism
f∗ : π1
(SL2(Od,m)\H3
)−→ π1 (Y, 1)
Now let g ∈ SL2(Od,m) and let h0 ∈ H3. Let P be any path from h0 to gh0. The image of P
in SL2(Od,m)\H3 is a loop and therefore represents some αg ∈ π1 (SL2(Od,m)\H3). Define
θ : SL2(Od,m) −→ π1
(SL2(Od,m)\H3
)by g 7→ αg. This map is well defined. We now have
σ : SL2(Od,m) −→ π1
(SL2(Od,m)\H3
)−→ π1(Y ) ∼= Fr
where r = r(d,m). Now consider the natural map
ψ : Fr Zr
Zimmert shows ([93] Satz 2(i)) that σψ is surjective. So 1 6= imσ 6 Fr, also SL2(Od,m) is
finitely generated, so imσ is a free group of rank s, where r 6 s < ∞ and so maps onto a
free group of rank r. Thus
Theorem 2.2.3. SL2(Od,m) maps onto a free group of rank r(d,m).
so clearly
Theorem 2.2.4. PSL2(Od,m) maps onto a free group of rank r(d,m).
2.3 Explicit result about the free generators
We have just seen that
σ : SL2(Od,m) π1(Y ) ∼= Fr
where r is the order of the Zimmert Set.
25
Lemma 2.3.1. The image, under σ, of the matrix
U =
1 mω
0 1
can be taken as a free generator of π1(Y ).
Proof. Let z0 = −im4|D|2 , z1 = z0 + mω, and let hi = (zi, 1). So uh0 = h1. Now define a
path P from h0 to h1 by P = (z(s), 1) : s ∈ [0, 1], where z(s) = (1−s)z0 +sz1 = z0 +smω.
Now
P ∩
(⋃t∈Z
F1,t
)= (P ∩ F1,0) ∪ h1
and h1 ∈ F1,1. So f1(P ) = S1. Now recall that
Y =(z1, . . . , zr) ∈ S1 × · · · × S1 : zi 6= 1 for at most one i
Let S1
i denote the ith circle of Y . Then S1i gives rise to gi ∈ π1(Y ) and g1, . . . , gr is a set
of free generators of π1(Y ) ∼= Fr. So
f∗(σ(u)) = g1g0 = x
where g0 ∈< g2, . . . , gr >. By Tietze transformations we can take x, g2, . . . , gr to be a set
of free generators of π1(Y ), as required.
Lemma 2.3.2. σ(u) is a free generator of imσ.
Proof. Clearly σ(u) ∈ imσ. We can then apply Proposition 2.10 on page 8 of [48] to get the
result.
2.4 Unipotent matrices
Let R be any commutative ring with a one. Recall that a matrix M ∈ GL2(R) is unipotent if
(M − I)2 = 0. Let U2(R) denote the normal subgroup of SL2(R) generated by the unipotent
matrices. Clearly E2(R) 6 U2(R) and so NE2(R), the normal subgroup generated by the
elementary matrices is contained in U2(R).
26
Lemma 2.4.1. Let M ∈ GL2(R). Then
M is unipotent ⇔ detM = 1 and trM = 2
⇔M =
1 + α γ
β 1− α
where α2 + βγ = 0, α, β, γ ∈ R.
Proof. Let
I 6= M =
a b
c d
∈ GL2(R)
be unipotent. Let t =trM = a+ d. Then
0 = (M − I)2 =
(a− 1)2 + bc b(t− 2)
c(t− 2) (d− 1)2 + bc
Now suppose that t 6= 2, so b = c = 0, so (a − 1)2 + bc = (a − 1)2 = 0, so a = 1, similarly
d = 1, so M = I. Contradiction. So t = 2, and d = 2 − a, so detM = a(2 − a) − bc =
−(a2 − 2a+ 1) + 1− bc = −((a− 1)2 + bc) + 1 = 1.
Conversely suppose that detM = 1 and trM = 2. So
M =
a b
c 2− a
so
(M − I)2 =
(a− 1)2 + bc 0
0 (1− a)2 + bc
and (a− 1)2 + bc = a2− 2a+1− bc = −(a(2−a)− bc)+1 = −1+1 = 0. So M is unipotent.
Now suppose that detM = 1 and tr M = 2. It is clear that M can be written in the
form 1 + α γ
β 1− α
where α2 + βγ = 0, some α, β, γ ∈ R. So conversely suppose that M is of this form. Then
(M − I)2 =
α2 + βγ αγ − αγ
αβ − αβ α2 + βγ
= 0
so M is unipotent.
27
Lemma 2.4.2. Let δ ∈ R. Then every conjugate of 1 δ
0 1
in SL2(R) is unipotent.
Proof. a b
c d
1 δ
0 1
d −b
−c a
=
1− acδ a2δ
−c2δ 1 + acδ
Lemma 2.4.3. Suppose that R is an integral domain and let α, β ∈ R then αR + βR is
principal if and only if ∃α′, β′ ∈ R such that α′, and β′ are coprime and αα′ + ββ′ = 0.
Proof. Suppose first that αR+βR = δR. So α = δr1, β = δr2, δ = αr3 +βr4, where ri ∈ R,
i = 1, 2, 3, 4. So δ = δ(r1r3 + r2r4), and because R is an integral domain and δ 6= 0, we have
r1r3 + r2r4 = 1. Let α′ = r2, and β′ = −r1. So α′, and β′ are coprime, and αα′ + ββ′ = 0.
Conversely suppose that ∃ coprime α′, β′ ∈ R such that αα′+ββ′ = 0. Now αα′ = −ββ′,
and α′ and β′ are coprime, so α = β′β′′, and β = α′α′′. So 0 = αα′ + ββ′ = α′β′(α′′ + β′′),
and as R is an integral domain and α′ 6= 0 6= β′ we have α′′ = −β′′. So αR + βR =
β′β′′R + α′α′′R = α′′(α′R + β′R) = α′′R.
Lemma 2.4.4. Suppose that R is an integral domain and let
M =
1 + α γ
β 1− α
be unipotent. Then M is conjugate in SL2(R) to 1 δ
0 1
if and only if αR + βR is principal.
28
Proof. Suppose that 1 + acδ a2δ
−c2δ 1− acδ
=
1 + α γ
β 1− α
.
So α = acδ, β = −c2δ, γ = a2δ. Clearly αR + βR ⊆ cδR. Now let x ∈ R and consider cδx.
cδx = cδx(ad− bc) as ad− bc = 1
= acδxd− c2δbx
= αxd+ βbx as α = acδ, β = −c2δ
∈ αR + βR
So αR + βR = δcR, ie is a principal ideal.
Conversely suppose that αR + βR is principal. Now a b
c d
1 + α γ
β 1− α
d −b
−c a
=
1 + 1β
(aα+ bβ) (cα + dβ) −1β
(aα+ bβ)2
1β
(cα + dβ)2 −1β
(aα+ bβ) (cα + dβ)
= Y
and αR + βR is principal so, by (2.4.3), ∃α′, β′ ∈ R such that α′, and β′ are coprime and
α′α+ β′β = 0. Let c = α′, and d = β′ so ∃a, b such that ad− bc = 1 and cα + dβ = 0, so
Y =
1 δ
0 1
where δ = −1
β(aα+ bβ)2.
Having set out the basics that we require in the context of an arbitrary commutative
ring we now turn to the case where R = Od,m is an order in an imaginary quadratic number
field. Let r be the order of the Zimmert Set and
σ : SL2(Od,m) −→ Fr
be the map given in the previous section. Let K = kerσ.
29
Lemma 2.4.5. ([93] Hilfssatz 2) Let O = Od,m and let M ∈ SL2(O), and α, β ∈ O such
that αO + βO is not a principal ideal. Let ε, t ∈ R such that 0 < ε 6 1, and |t| 6 ε. Let
h′ = (z, r) =
(−α+ t
β,
1
|β|√
2ε− ε2)
and suppose that Mh′ = (z′, r′). Then
r′ 6 |β|√
2ε− ε2
Proof. Suppose that
M =
a b
c d
so r′ = r/τ , where τ = |cz − d|2 + r2|c|2. Now
τ =
∣∣∣∣c−α+ t
β− d∣∣∣∣2 +
(1
|β|√
2ε− ε2)2
|c|2 >1
|β|2
For equivalently
|(c− α− βd) + ct|2 +(2ε− ε2
)|c|2 > 1
Now c and d are coprime and βO+αO is not a principal ideal so, by (2.4.3), 0 6= −cα−βd ∈
O, so | − cα− βd| > 1. Thus
|(−cα− βd) + ct|2 +(2ε− ε2
)|c|2 > (1− |ct|)2 +
(2ε− ε2
)|c|2
Suppose that |ct| > 1. Clearly (1 − |ct|)2 > 0. Also |c| > 1/|t| > 1/ε and so (2ε −
ε2)|c|2 =(
2ε− 1)|c|2ε2 > 2
ε− 1 > 1. Now suppose that |ct| 6 1. First of all observe that
2ε − ε2 > 2|t| − |t|2 ⇔ (ε − |t|)(2 − ε − |t|) > 0, and as 0 < |t| 6 ε 6 1, this is true, so
2ε− ε2 > 2|t| − |t|2. Now
(1− |ct|)2 + (2ε− ε2)|c|2 = 1− 2|ct|+ |ct|2 + (2ε− ε2)|c|2 > 1
⇔ |ct|2 − 2|ct|+ (2ε− ε2)|c|2 > 0
30
Now if |c| = 0 then this is certainly true, so suppose that |c| 6= 0, so as c ∈ O we have |c| > 1.
Now
|ct|2 − 2|ct|+ (2ε− ε2)|c|2
=|c|(|c||t|2 − 2|t|+ (2ε− ε2)|c|
)>|c|
(|c||t|2 − 2|t|+ (2|t| − |t|2)|c|
)=2|c||t|(|c| − 1) > 0.
So r′ = r/τ 6 |β|2(1/|β|)√
2ε− ε2 = |β|√
2ε− ε2.
Theorem 2.4.6. ([93] Satz 2(ii)) Let α, β, γ ∈ Od,m such that α2 + βγ = 0 and
αOd,m + βOd,m is not principal. Then
M =
1 + α γ
β 1 + α
∈ KProof. Let ε ∈ R and let
h1 = (z1, r1) =
(−α+ ε
β,
1
|β|√
2ε− ε2)
and suppose that Mh1 = h2 = (z2, r2). Now
τ =
∣∣∣∣β−α+ ε
β− (1− α)
∣∣∣∣2 +1
|β|2(2ε− ε2
)|β|2
= | − α+ ε− 1 + α|2 + 2ε− ε2
= (ε− 1)2 + 2ε− ε2
= 1
So r2 = r1, and
z2 =
(1− α− β−α+ ε
β
)((1 + α)
−α+ ε
β+α2
β
)− 1
|β|(2ε− ε2
)β(1 + α)
After some algebraic manipulation we see that
z2 =−α− εβ
31
so
h2 =
(−α− εβ
,1
|β|√
2ε− ε2)
Now let
P =
h′ =
(−α+ t
β,
1
|β|√
2ε− ε2)∈ H3 : t ∈ R,−ε 6 t 6 ε
so P is a path in H3 from h1 to h2. Let h′ ∈ P and M2 ∈ SL2(O). Then, by (2.4.5), if
M2h′ = (z′, r′) we have r′ 6 |β|
√2ε− ε2. Now choose ε so that |β|
√2ε− ε2 < 5/(2m2|D|).
So, by (2.2.2), M2h′ /∈ Fn,t for every n ∈ Z(d,m) and t ∈ Z coprime to n. Thus ϕn(h′) = 1
and it follows that M is in the kernel of the Zimmert map.
Theorem 2.4.7. With the above notation, there exists an epimorphism
τ :SL2(Od,m)
U2(Od,m)−→ Fs
where s = r − 1.
Proof. We have already seen, in the previous section that SL2(Od,m) has a free quotient of
rank s+ 1 where
U =
1 mω
0 1
can be taken as a free generator. Now consider a unipotent matrix
M =
1 + α γ
β 1− α
where α2 + βγ = 0. Now either αR + βR is principal or it isn’t. If it is not principal then,
by theorem (2.4.6), M ∈ K. If αR+βR is principal then, by lemma (2.4.4), M is conjugate,
in SL2(Od,m) to 1 δ
0 1
for some δ ∈ Od,m. Nowδ = z1 +mωz2, zi ∈ Z, but T ∈ SL2(Z) 6 K, so we can suppose that
M is conjugate to some power of U . So either M is in the kernel of the map SL2(Od,m) Fr
or it is conjugate to a power of U , which can be taken as a free generator of Fr. So clearly
SL2(Od,m)
U2(Od,m)
< u, x2, . . . , xr;>
< u >∼=< x2, . . . xr;>= Fs
32
Corollary 2.4.8.SL2(Od,m)
NE2(Od,m)
has a free quotient of rank s = r − 1.
Let R be any commutative ring and let q R. Then let
SLn(R, q) = ker(SLn(R)→ SLn(R/q))
and let Un(R, q) be the normal subgroup of SLn(R) generated by all unipotent matrices in
SLn(R, q), let En(R, q) denote the normal subgroup of En(R) generated by all q elementary
matrices and NEn(R, q) denote the normal subgroup of SLn(R) generated by q elementary
matrices.
Corollary 2.4.9. Let 0 6= q Od,m. Then
SL2(Od,m, q)
U2(Od,m, q)
has a free quotient of rank s = r − 1.
Proof. Now
SL2(Od,m, q)U2(Od,m) > SL2(Od,m, q)E2(Od,m) = SL2(Od,m)
soSL2(Od,m, q)
U2(Od,m) ∩ SL2(Od,m, q)∼=SL2(Od,m)
U2(Od,m) Fs
thusSL2(Od,m, q)
U2(Od,m, q) Fs
33
2.5 Computation of Zimmert Sets
Given the definition of the Zimmert Set Z(d,m) it is a simple matter to write a computer
program to calculate any Zimmert Set. We have written such a program in GAP [24]. Since
we are interested in free non-abelian quotients it would be useful to know values of (d,m)
such that r(d,m) 6 1. Consider first the Zimmert sets Z(d, 1) which we denote Z(d), let
r(d) = r(d, 1). Mason, Odoni and Stothers [61] have proved
Theorem. For all but finitely many d r(d) > 2.
Obviously we would like a list of the d such that r(d) = 1. By means of a computer
search Mason et al [61] establish that up to 2× 1013 the only values of d for which r(d) = 1
are: 2, 5, 6, 7, 11, 14, 15, 17, 19, 21, 23, 26, 29, 30, 31, 35, 39, 41, 47, 51, 59, 66, 69, 71,
87, 89, 95, 101, 105, 110, 111, 119, 129, 159, 161, 191, 194, 209, 215, 230, 231, 255, 285,
311, 321, 335, 341, 374, 399, 426, 455, 479, 546, 591, 615, 671, 831, 959, 1095, 1119, 1239,
2415 and they conjecture that these are the only values of d for which r(d) = 1. Recall that
r(1) = r(3) = 0.
We extended this work by using our computer program to calculate pairs (d,m) such
that r(d,m) = 1. For every square free d ∈ N and every m ∈ N we checked every pair (d,m)
such that m2|D| 6 107 and found 104 pairs (d,m) such that r(d,m) = 1. Excluding those
mentioned above these are : (1, 2), (1, 3), (1, 6), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (3, 10),
(5, 2), (5, 3), (5, 4), (6, 2), (6, 3), (6, 4), (6, 5), (7, 2), (7, 3), (10, 3), (11, 2), (11, 4), (14, 2),
(15, 2), (15, 3), (21, 2), (23, 2), (26, 2), (31, 3), (35, 2), (35, 4), (35, 6), (39, 2), (41, 2), (55, 3),
(59, 2), (66, 2), (111, 2), (119, 2), (131, 2), (195, 2), (231, 2), (341, 2). Notice that, perhaps
surprisingly, r(d, 1) > 1 does not imply r(d,m) > 1, for m > 1; for example r(10, 1) = 2, yet
r(10, 3) = 1. It seems very reasonable to make the following
Conjecture. The only pairs (d,m) such that r(d,m) = 1 are those listed above.
It can be seen from the data above that for most d such that r(d) = 1, r(d, 2) > 1, so that
Zimmert’s theorem only just fails to give a free non-abelian quotient. However Zimmert’s
34
theorem is not best possible; for r(5) = 1 yet PSL2(O5) has a free quotient of rank 2. With
this in mind make the following
Definition. Let ρ(d,m) be the largest rank of a free quotient of SL2(Od,m).
ρ(d,m) is well defined as PSL2(Od,m) is finitely generated. Let ρ(d) = ρ(d, 1). Clearly
ρ(d,m) > r(d,m). If we have a presentation for PSL2(Od,m) we can compute ρ(d,m):
Theorem. ([61] theorem 6, and [78])
ρ(d) 0 1 2 3 5 7 8
d 1, 3 2, 7, 11, 19 5, 6, 15, 43 10, 13, 67 22 21, 163 37
Mason et al [61] have the following
Conjecture. ρ(d) > 1 for all d > 19.
We have no reason to doubt this. The only square free d 6 19 missing from the above
are 14, and 17, we partially close this gap
Proposition 2.5.1. 4 6 ρ(14) 6 5
Proof. The following presentation for PSL2(O14) can be found in [30] Proposition 3.5.
< g1, g2, g3, g4, g5, g6, g7 ; g21, (g1g2)
3, g1g3g−11 g−1
3 , g2g4g−12 g−1
4 ,
g5g−16 g7g
−13 g6g
−15 g3g
−17 , g6g4g1g
−13 g6g1g
−16 g3g
−14 g1g
−16 g1,
g1g6g1g2g−16 g1g
−16 g3g1g
−12 g−1
3 g6,
g−12 g−1
7 g6g−12 g1g
−16 g1g5g2g
−16 g7g2g1g
−13 g6g1g
−15 g3 >
From this it is easy to compute that PSL2(O14)ab ∼= Z6×Z5, and so ρ(14) 6 5. It is also easy
to see that by setting g1 = g2 = g5g−16 = 1 we get the free group of rank 4. So ρ(14) > 4.
Proposition 2.5.2. ρ(1, 2) = ρ(3, 2) = 1.
35
Proof. We have the following presentations from (3.5.1) and (3.3.1)
PSL2(O1,2) =< a, t, z, w; a2, z2, (at)3, (atz)2, [t, w] , (atw−1zw)2 >
PSL2(O3,2) =< a, t, w; a2, (at)3, (w−1awa)3, [t, w] >
from which it is easy to see PSL2(O1,2)ab = Z2 × Z, and PSL2(O3,2)
ab = Z6 × Z. So
ρ(1, 2) = ρ(3, 2) = 1.
Proposition 2.5.3. ρ(7, 2) = 2.
Proof. We have from section (3.5)
PSL2(O7,2) =< a, t, w, x, y ; a2, x2, [t, w] , (ax)3, (at)3, y = txyxt−1, (ytay−1x)2,
xyat−1y−1wt−1aw−1t >
Now suppose that PSL2(O7,2)/N is free then N contains the normal closure of all elements in
PSL2(O7,2) of finite order, soN(a, x, t) 6 N . It is easy to calculate that PSL2(O7,2)/N(a, x, t) ∼=
F2. So ρ(7, 2) = 2.
Proposition 2.5.4. 3 6 ρ(11, 2) 6 4.
Proof. We have from section (3.5)
PSL2(O11,2) =< a, t, w, k, l,m, n ; a2, k2, [m,n] , [t, w] , (at)3, tklkt−1l−1,
km−1ltat−1l−1m,nlat−1l−1n−1mwaw−1tm−1,
m−1tklt−1l−1mwt−1aw−1t >
so the kernel of a free quotient of PSL2(O11,2) must contain a, k, t. It is easy to compute
that PSL2(O11,2)/N(a, k, t) =< z, l,m, n; [m,n] >. This maps onto the free group of rank
3, so ρ(11, 2) > 3, and it abelianization is Z4 so ρ(11, 2) 6 4.
Given the evidence presented above it seems reasonable to make the following
Conjecture. The only values of (d,m) for which PSL2(Od,m) does not have a free non-
abelian quotient are: (1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (7, 1), (11, 1), (19, 1).
36
There are results later on in this thesis which depend on the fact that PSL2(Od,m) has
a free non-abelian quotient. Usually we then attempt to deal with the cases mentioned in
the conjecture. We are quite certain that the conjecture is true and so the idea is that
we have a proved a theorem (or a version of it) for all Bianchi groups. For example it is
not yet proven that PSL2(O2,2) has a free non-abelian quotient, so strictly speaking our
theorem which depends on PSL2(Od,m) having a free non-abelian quotient has not been
proved for PSL2(O2,2), although we are quite certain that it is true and all that is required
is a presentation of PSL2(O2,2) to verify it. Contrast this with PSL2(O2) where a slightly
different technique may be required or with PSL2(O3) where a radically different technique is
perhaps needed. For this reason we do not consider PSL2(O2,2) to be a “true” exception and
we refer to the pairs mentioned in the conjecture as the “true exceptions”. The conjecture
could (and probably will) be proved by means of a computer search. See the comments at
the end of section (3.5).
Recall theorem (2.4.7). To get a free non-abelian quotient we require that r(d,m) > 3.
In exactly the same way as above we have used our GAP [24] program to get a list of pairs
(d,m) such that r(d,m) = 2. Again for every square free d ∈ N and every m ∈ N we
checked every pair (d,m) such that m2|D| 6 107 and found 215 pairs (d,m) such that
r(d,m) = 2. These are: (1, 4), (1, 5), (1, 7), (1, 9), (2, 4), (2, 5), (2, 7), (3, 5), (3, 6), (3, 8),
(3, 12), (5, 5), (6, 6), (7, 5), (7, 6), (7, 9), (10, 1), (10, 2), (10, 6), (10, 9), (11, 3), (11, 6), (13, 1),
(13, 3), (13, 6), (14, 3), (14, 4), (15, 4), (15, 5), (17, 2), (17, 3), (17, 4), (17, 5), (19, 2), (19, 3),
(19, 4), (19, 6), (19, 12), (21, 3), (21, 4), (21, 5), (21, 6), (22, 1), (26, 3), (29, 2), (29, 3), (30, 2),
(30, 3), (30, 7), (31, 2), (33, 1), (33, 2), (33, 5), (34, 1), (38, 1), (38, 5), (39, 3), (39, 4), (39, 5),
(41, 1), (43, 1), (46, 1), (47, 2), (47, 3), (51, 2), (51, 4), (51, 6), (51, 8), (55, 1), (55, 2), (55, 6),
(61, 1), (62, 1), (62, 2), (65, 1), (65, 2), (66, 3), (69, 2), (69, 3), (70, 1), (71, 2), (74, 1), (77, 1),
(77, 2), (79, 1), (79, 3), (83, 1), (83, 2), (86, 1), (87, 2), (87, 3), (87, 4), (94, 3), (95, 2), (95, 3),
(101, 2), (105, 2), (110, 2), (114, 1), (129, 2), (131, 1), (134, 1), (138, 1), (143, 1), (143, 2),
(145, 3), (146, 1), (149, 1), (151, 1), (155, 1), (159, 2), (165, 1), (167, 1), (167, 2), (173, 1),
(179, 1), (182, 1), (183, 1), (185, 1), (186, 1), (195, 1), (195, 4), (199, 1), (206, 1), (215, 2),
37
(222, 1), (230, 2), (231, 3), (237, 1), (239, 1), (251, 1), (251, 2), (255, 2), (266, 1), (269, 1),
(271, 1), (285, 2), (287, 1), (290, 1), (299, 2), (314, 1), (327, 1), (329, 1), (339, 2), (359, 1),
(383, 1), (390, 1), (395, 2), (399, 2), (431, 1), (446, 1), (447, 1), (455, 2), (458, 1), (471, 1),
(494, 1), (497, 1), (503, 1), (506, 1), (519, 1), (545, 1), (551, 1), (563, 2), (569, 1), (623, 1),
(654, 1), (659, 2), (689, 1), (699, 2), (705, 1), (719, 1), (755, 2), (759, 1), (770, 1), (789, 1),
(791, 1), (815, 1), (831, 2), (854, 1), (887, 1), (935, 1), (1031, 1), (1055, 1), (1151, 1), (1169, 1),
(1190, 1), (1199, 1), (1209, 1), (1223, 1), (1271, 1), (1326, 1), (1335, 1), (1407, 1), (1511, 1),
(1551, 1), (1599, 1), (1751, 1), (1767, 1), (1823, 1), (1895, 1), (1959, 1), (1991, 1), (2015, 1),
(2159, 1), (2435, 2), (2639, 1), (2679, 1), (2735, 1), (3119, 1), (3311, 1), (3471, 1), (4479, 1),
(6215, 1), (6815, 1), (8655, 1). Again it seems reasonable to make the following
Conjecture. The only pairs (d,m) such that r(d,m) = 2 are those listed above.
38
Chapter 3
The groups PSL2(Od,m)
In chapter two we saw that almost all of the groups PSL2(Od,m) had a free non-abelian
quotient. In this chapter we take a closer look at these groups. For completeness we restate
the definition.
Let m, d ∈ N, d square-free then the orders in the imaginary quadratic number field
Q(√−d) are
Od,m = Z +mωZ where ω =
1+i√
d2
if d ≡ 3 (mod 4)
i√d else
The maximal order Od,1 is the ring of integers in Q(√−d) and is denoted Od.
| Od : Od,m | = m. We are considering the groups PSL2(Od,m). Since mOd ⊆ Od,m, we
have PSL2(Od,mOd) 6 PSL2(Od,m), so PSL2(Od,m) is of finite index in PSL2(Od). We
now compute this index.
3.1 Computation of |PSL2(Od) : PSL2(Od,m)|
First observe that
|PSL2(Od) : PSL2(Od,m)| = |PSL2(Od) : PSL2(Od,p1)| . . . |PSL2(Od,p1...pr−1) : PSL2(Od,m)|
where m = p1 . . . pr and pi are not necessarily distinct primes. So for m, p ∈ N, p prime, we
compute |PSL2(Od,m) : PSL2(Od,mp)|. We need the following well known lemma.
39
Lemma 3.1.1. Let R be any commutative ring. Let q R such that R/q is an SR2-ring.
Then the natural map
SL2(R) −→ SL2(R/q)
is onto.
Proof. R/q is SR2 so SL2(R/q) = E2(R/q) and clearly 1 α
0 1
7−→ 1 + q α+ q
q 1 + q
.
So the map is onto.
Lemma 3.1.2. For any d,m, p ∈ N such that d is square free, and p is prime
|SL2(Od,m) : SL2(Od,mp)| = |PSL2(Od,m) : PSL2(Od,mp)|
and so
|SL2(Od) : SL2(Od,m)| = |PSL2(Od) : PSL2(Od,m)|
Proof. First observe that
|PSL2(Od,m) : PSL2(Od,mp)| =|PSL2(Od,m) : PSL2(Od,m, pOd,m)||PSL2(Od,mp) : PSL2(Od,m, pOd,m)|
Now
|PSL2(Od,m) : PSL2(Od,m, pOd,m)| =∣∣∣∣PSL2
(Od,m
pOd.m
)∣∣∣∣Now, for any ring R and any ideal q of R such that R/q is finite, we have |PSL2(R/q)| =
ρ|SL2(R/q)|, where ρ = 1 if 2 ∈ q and ρ = 1/2 if 2 /∈ q. So∣∣∣∣PSL2
(Od,m
pOd,m
)∣∣∣∣ = ρ
∣∣∣∣SL2
(Od,m
pOd,m
)∣∣∣∣where ρ = 1 if p = 2 and ρ = 1/2 if p 6= 2. Also, as |Od,mp : pOd,m| = p,
|PSL2(Od,mp) : PSL2(Od,m, pOd,m)| = |PSL2(Fp)| = ρ|SL2(Fp)|
40
where ρ is as above (this follows from [79] theorem 8.14). Thus
|PSL2(Od,m) : PSL2(Od,m, pOd,m)||PSL2(Od,mp) : PSL2(Od,m, pOd,m)|
=ρ|SL2(Od,m/pOd,m)|
ρ|SL2(Fp)|
=|SL2(Od,m/pOd,m)|
|SL2(Fp)|
=|SL2(Od,m) : SL2(Od,m, pOd,m)||SL2(Od,mp) : SL2(Od,m, pOd,m)|
= |SL2(Od,m) : SL2(Od,mp)|
In what follows we work with SL2 as it makes the proofs a little simpler. Observe that
|SL2(Od,m) : SL2(Od,mp)| =|SL2(Od,m) : SL2(Od,m, pOd,m)||SL2(Od,mp) : SL2(Od,m, pOd,m)|
Lemma 3.1.3. |SL2(Od,mp) : SL2(Od,m, pOd,m)| = |SL2(Fp)| = p(p2 − 1).
Proof. Observe that |Od,mp : pOd,m| = p so
SL2(Od,mp)
SL2(Od,m, pOd,m)∼= SL2(Fp).
This group has the required order by [79] theorem 8.8 and 8.14
Now
|SL2(Od,m) : SL2(Od,m, pOd,m)| =∣∣∣∣SL2
(Od,m
pOd,m
)∣∣∣∣So we need to have some understanding of the structure of Od,m/pOd,m.
Lemma 3.1.4. If p | m thenOd,m
pOd,mis local with maximal ideal m = (mω).
Proof. Let R = Od,m/pOd,m. We compute R∗. Now
R = r +mωs : r, s = 0, 1, . . . , p− 1
Let ui = ri +mωsi ∈ R, i = 1, 2. Then
u1u2 = r1r2 + (r1s2 + r2s1)mω +m2ω2s1s2
41
suppose first that d ≡ 1, 2 (mod 4), so ω2 = −d. Then
u1u2 = r1r2 − dm2s1s2 + (r1s2 + r2s1)mω
= r1r2 + (r1s2 + r2s1)mω
since p | m. Now suppose that d ≡ 3 (mod 4), so ω2 = ω − (d+ 1)/4. Then
u1u2 = r1r2 −m2d+ 1
4s1s2 + (r1s2 + r2s1 +ms1s2)mω
= r1r2 + (r1s2 + r2s1)mω
again as p | m. Now suppose that r1 6= 0 and let r2 = r−11 , s2 = −r−2
1 s1, so u1u2 = 1. So
r1 6= 0⇒ u−11 exists. Hence r +mωs : r 6= 0 ⊆ R∗. But if r1 = 0 then u1u2 = r2s1mω 6= 1.
So r1 = 0⇒ u1 is not a unit. Hence
R∗ = r +mωs : r 6= 0
We now show that R − R∗ = mωs : s = 0, 1, . . . , p− 1 = (mω). Clearly R − R∗ ⊆ (mω).
Now if d ≡ 1, 2 (mod 4) then
mω(r +mωs) = rmω +m2ω2s
= rmω − dm2s
= rmω ∈ R−R∗
similarly if d ≡ 3 (mod 4) then
mω(r +mωs) = rmω +m2ω2s
= rmω +m2s
(ω − d+ 1
4
)= (r +ms)mω −m2s
d+ 1
4
= rmω ∈ R−R∗
Hence R−R∗ = (mω). So R is a local ring with maximal ideal m = (mω) of index p.
42
Lemma 3.1.5. If p - m thenOd,m
pOd,m
∼=Od
pOd
and the isomorphism is given by
r + (mω)s 7−→ r + (ms)ω
Proof. It is a trivial matter to verify that the map is a homomorphism. Now let r + sω ∈
Od/pOd. As p - m, m−1 exists and r + (m−1s)mω 7→ r + sω. So the map is onto. Now
suppose that r + mωs 7→ 1, so r + (ms)ω = 1 in Od/pOd, so r = 1 and ms ≡ 0 (mod p).
But p - m, so s = 0. So r +mωs = 1. So the map is injective. Hence result.
The structure of Od/pOd is well known and can be found in any book on algebraic number
theory. See, for example, [52] page 108. First recall the definition of the discriminant of
K = Q(√−d):
D =
−d if d ≡ 3 (mod 4),
−4d else.
Now χK , the quadratic character of K is defined as follows. If p | D then χK(p) = 0,
χK(2) =
1 if D ≡ 1 (mod 8),
−1 if D ≡ 5 (mod 8).
and if p 6= 2 then
χK(p) =
1 if D ≡ x2 (mod p),
−1 else.
Theorem. [52] Let p ∈ Z be prime then
pOd =
p1p2 for χK(p) = 1,
p for χK(p) = −1,
p2 for χK(p) = 0.
where p, pi are prime ideals of Od.
Lemma 3.1.6. If p | m then |SL2(Od,m/pOd,m)| = p3|SL2(Fp)|.
43
Proof. Now Od,m/pOd,m is local with maximal ideal m, of index p, so∣∣∣∣SL2
(Od,m
pOd,m
)∣∣∣∣ =
∣∣∣∣SL2
(Od,m
m
)∣∣∣∣ ∣∣∣∣SL2
(Od,m
pOd,m
,m
pOd,m
)∣∣∣∣ .and |SL2(Od,m/m)| = |SL2(Fp)|, and by lemma (4.2.9)
∣∣∣SL2
(Od,m
pOd,m, m
pOd,m
)∣∣∣ = p3.
Thus
Lemma 3.1.7. If p | m then
|SL2(Od,m) : SL2(Od,mp)| = p3
Lemma 3.1.8. If p - m then
∣∣∣∣SL2
(Od,m
pOd,m
)∣∣∣∣ =
p2(p2 − 1)2 if χK(p) = 1,
p4(p2 − 1) if χK(p) = 0,
p2(p4 − 1) if χK(p) = −1.
Proof. If χK(p) = −1 then Od,m/pOd,m∼= Fp2 and |SL2(Fp2)| = p2(p4 − 1), by [79] theorem
8.8 and the comments at the bottom of p.157. Now suppose that χK(p) = 1, so that R =
Od,m/pOd,m has two ideals m1, and m2 of index p. So, by (6.4.4), SL2(R) = SL2(R/m1) ×
SL2(R/m2) and |SL2(R)| = |SL2(Fp)|2 = p2(p2−1)2, by [79] theorem 8.8 and the comments
at the bottom of p.157. Finally suppose that χK(p) = 0, so L = Od,m/pOd,m is a local ring
and, as before |SL2(L)| = p3|SL2(Fp)| = p4(p2 − 1).
Thus
Lemma 3.1.9. Suppose that p - m then
|SL2(Od,m) : SL2(Od,mp)| =
p(p2 − 1) if χK(p) = 1,
p3 if χK(p) = 0,
p(p2 + 1) if χK(p) = −1.
Now observe that, from the above, if p - m
|SL2(Od,m) : SL2(Od,mpα)| = |SL2(Od : SL2(Od,pα)|
44
so that
|SL2(Od) : SL2(Od,m)| =r∏
i=1
|SL2(Od) : SL2(Od,pαii
)|
where m = pα11 . . . pαr
r and
|SL2(Od) : SL2(Od,pα)| = |SL2(Od) : SL2(Od,p)||SL2(Od,p) : SL2(Od,pα)|
= p3(α−1)|SL2(Od) : SL2(Od,p)|.
Thus
Theorem 3.1.10.
|PSL2(Od) : PSL2(Od,m)| = m3
r∏i=1
1
p3i
|PSL2(Od) : PSL2(Od,p)| ,
where m = pα11 . . . pαr
r , and
|PSL2(Od) : PSL2(Od,p)| =
p(p2 − 1) if χK(p) = 1,
p3 if χK(p) = 0,
p(p2 + 1) if χK(p) = −1.
3.2 A miscellany of results
Theorem 3.2.1. Let m1,m2 ∈ N be distinct, and suppose m2 | m1, so PSL2(Od,m1) 6
PSL2(Od,m2). Then PSL2(Od,m1) 6 PSL2(Od,m2).
Proof. Let
X =
α β
γ δ
∈ PSL2(Od,m2),
and let
M =
1 1 +m1ω
0 1
∈ PSL2(Od,m1)
so
XMX−1 ≡
1− αγ α2
−γ2 1 + αγ
( mod m1ωM2(Od)) .
45
Now suppose that d ≡ 1, 2 (mod 4). Let p be a rational prime such that p - d and p - m1,
so p - dm22. So ∃x, y ∈ Z such that px+ dm2
2y = 1 ie px+m2ω(−ωm2y) = 1. Let α = m2ω,
γ = p then
X =
m2ω −x
p −m2ωy
∈ PSL2(Od,m2)
and XMX−1 /∈ PSL2(Od,m1)
Now suppose that d ≡ 3 (mod 4), so ω2 = ω− (d+ 1)/4, so (d+ 1)/4 = ω(1− ω). Let p
be a rational prime such that p - (d+ 1)/4 and p - m1, so p - m22(d+ 1)/4. So ∃x, y ∈ Z such
that px+m22y(d+ 1)/4 = 1 ie px+m2ω(m2y(1− ω)) = 1. Then
X =
m2ω −x
p (1− ω)m2y
∈ PSL2(Od,m2)
and XMX−1 /∈ PSL2(Od,m1). Hence result.
Theorem 3.2.2. Let N PSL2(Od,m) be of index n. Then if 6 - n then o(N) = Od,m and
N has torsion.
Proof. Suppose that 6 - n, so 2 - n or 3 - n. If 2 - n then
a =
0 −1
1 0
∈ N,and a2 = 1 so N has torsion. Also a ≡ kI (mod q) ⇒ 1 ∈ q ie q = Od,m, so o(N) = Od,m.
If 3 - n then
at =
0 −1
1 1
∈ Nso, as above N has torsion and o(N) = Od,m.
3.3 A presentation for SL2(Z[√−3])
Dennis has shown [19]
Theorem. SL2(Z[√−3]) = E2(Z
[√−3]).
46
Following Fine [25] we apply the following, due to Cohn [15]
Theorem. Let R be a subring of C with the usual absolute valuation. Suppose that if α ∈ R,
and |α| 2 then α = 0, or |α| = 1, or |α| = √p, where p = 2, 3. Then E2(R) is generated
by E(x) with the following complete set of relations:
1. E(x)E(0)E(y) = −E(x+ y)
2. E(x)D(µ) = D(µ−1)E(µxµ)
3. (E(α)E(α))p = −I ∀α ∈ R such that |α| = √p, where p = 2, 3.
4. E(µ)E(µ−1)E(µ) = D(−µ)
Where x, y ∈ R, µ ∈ R∗, and
E(x) =
x 0
−1 0
, D(µ) =
µ 0
0 µ−1
.
Let O = Z + i√
3Z. Now O∗ = ±1, and the only elements of norm between 1 and 2
are ±i√
3. So we get for SL2(Z[√−3])
Generators:
E(x), where x ∈ R
Relators:
E(x)E(0)E(y) = −E(x+ y), (3.1)
−I central, (−I)2 = I, (3.2)
(E(i√
3)E(−i√
3))3 = −I, (3.3)
E(1)3 = −I, E(−1)3 = I. (3.4)
Using (3.1) we can reduce the generators to E(0), E(±1), E(±i√
3), as follows. Let z ∈ Z,
and suppose z > 0. Then
E(z) = −E(z − 1)E(0)E(1),
47
and
E(iz√
3) = −E(i(z − 1)√
3)E(0)E(i√
3).
We get similar formulas for z 6 0. Putting these together we get
E(z1 + iz2
√3) = −E(z1)E(0)E(iz2
√3).
If x, y = 0 then it is easy to see
E(O)2 = −I ⇔ E(x)E(0)E(y) = −E(x+ y) (3.5)
and (3.1) certainly implys the following
E(1)E(0)E(i√
3) = E(i√
3)E(0)E(1) (3.6)
E(1)E(0)E(−i√
3) = E(−i√
3)E(0)E(1) (3.7)
E(−1)E(0)E(i√
3) = E(i√
3)E(0)E(−1) (3.8)
E(−1)E(0)E(−i√
3) = E(−i√
3)E(0)E(−1) (3.9)
and we also get
E(−1) = −E(0)E(1)−1E(0)−1 (3.10)
E(−i√
3) = −E(0)E(i√
3)−1E(0)−1 (3.11)
Using (3.10), and (3.11) we can eliminate the generatorsE(−1), and E(−i√
3), so SL2(Z[√−3])
is generated by E(0), E(1), and E(i√
3). We can show that (3.7), (3.8), (3.9) can be deduced
from (3.6) as follows. RTP
E(1)E(0)E(−i√
3) = E(−i√
3)E(0)E(1)
now, using (3.10), and (3.11)
LHS = E(1)E(0)(−E(0)E(i√
3)−1E(0)−1)
= E(1)E(i√
3)−1E(0)−1
48
and
RHS = −E(0)E(i√
3)−1E(0)−1E(0)E(1)
= −E(0)E(i√
3)−1E(1)
now
E(1)E(i√
3)−1E(0)−1(−E(0)E(i√
3)−1E(1))−1
=− E(1)E(i√
3)−1E(0)−1E(1)−1E(i√
3)E(0)−1
=− E(1)E(1)−1E(0)−1E(i√
3)−1E(i√
3)E(0)−1
=− E(0)−2 = I
and so LHS = RHS. The others are done in a similar manner.
And so a presentation for SL2(Z[√−3]) is:
Generators:
E(0), E(1), E(i√
3),−I
Relators:
E(1)3 = −I
E(0)2 = −I
−I central, (−I)2 = I
(E(i√
3)E(0)E(i√
3)−1E(0)−1)3 = I
E(1)E(0)E(i√
3) = E(i√
3)E(0)E(1)
Letting
A = E(0) =
0 1
−1 0
, T = E(0)E(1)−1 =
1 1
0 1
,
U = −E(0)E(1)−1E(0)E(i√
3)−1 =
1 2ω
0 1
.
49
We get
Generators:
A, T, U, J
Relators:
A2 = (AT )3 = J,
J central, J2 = I
(U−1TAT−1UA−1)3 = [T, U ] = I
Let W = T−1U and set J = I to get
Theorem 3.3.1.
PSL2(O3,2) = PSL2(Z[√−3]) =< a, t, w; a2, (at)3, (w−1awa)3, [t, w] >
3.4 The Group PSL2(O3,2)
First of all observe
Theorem 3.4.1.
PSL2(O3,2)ab =< t,w; t6, [t, w] >∼= Z6 × Z
Letting N = N(w), the normal closure of w in PSL2(O3,2), we see that l(N) = 0 and
PSL2(O3,2)
N=< a, t; a2, (at)3 >= M.
We aim to decompose PSL2(O3,2) as a non-trivial free product with amalgamation and as
an HNN group. We do here for PSL2(O3,2) what Fine did in [25] for PSL2(O2), PSL2(O7),
PSL2(O11).
3.4.1 HNN and amalgam decomposition
Theorem 3.4.2. PSL2(O3,2) is an HNN extension of K3,2 with the Modular group asso-
ciated, where K3,2 = S3 ∗Z3 D(3, 3, 3), and D(3, 3, 3) =< x, y;x3, y3, (xy)3 > is one of von
Dyck’s groups (see [36]).
50
Proof. We start with the presentation
< a, t, w; a2, (at)3, (w−1awa)3, [t, w] >
Letting v = w−1aw, the presentation becomes
< a, t, v, w; a2, (at)3, (av)3, v2, (tv)3, t = w−1tw, v = w−1aw >
Let
K3,2 =< a, t, v; a2, (at)3, (av)3, v2, (tv)3 >
We claim that PSL2(O3,2) is an HNN extension of K3,2 with < a, t >∼= M ∼=< v, t >
associated. First of all we show that < a, t >∼= M . Now as a2 = (at)3 = 1 it is clear that
M < a, t >. But < a, t >6 K3,2, and K3,2 M via the map t 7→ t, a 7→ a, and
v 7→ a. We see from this that < a, t > M , and so that M ∼=< a, t >. We can show that
< v, t >∼= M in exactly the same way. Let θ be the automorphism of K3,2 given by a 7→ v,
v 7→ a, and t 7→ t. We must show that θ(< a, t >) =< v, t >. Now θ(a) = v ∈< v, t >, and
θ(t) = t ∈< v, t >, so clearly θ(< a, t >) 6< v, t >. Similarly θ(< v, t >) 6< a, t >. Now
< v, t >= θ(θ(< v, t >)) 6 θ(< a, t >), and so θ(< a, t >) =< v, t >.
Now consider K3,2 and let s = at, m = av, so t = as, and v = am. Thus
K3,2 =< a, s,m; a2, s3,m3, (am)2, (sm2)3 >
Recall that
< a,m; a2,m3, (am)2 >∼= S3
and
< s,m; s3,m3, (sm−1)3 >∼= D(3, 3, 3)
So
K3,2 = S3 ∗m=m D(3, 3, 3)
We have the following theorem
51
Theorem 3.4.3. [48] If G maps onto a free product then G can be decomposed as an amal-
gamated free product. More precisely if θ : G A ∗ B and H = θ−1(A) ∩ θ−1(B), GA and
GB are disjoint copies of θ−1(A) and θ−1(B) respectively and HA, HB are copies of H in GA
and GB respectively Then G is isomorphic to the amalgamated free product GA ∗HA=HBGB.
As PSL2(O3,2) maps onto PSL2(Z) which is the free product of a 2-cycle and a 3-cycle,
we immediately get that PSL2(O3,2) is an amalgamated free product. However if we do
some direct calculations we get a nice decomposition.
Theorem 3.4.4.
PSL2(O3,2) = G1 ∗H G2
where G1 is an HNN extension of S3, G2 is an HNN extension of D(3, 3, 3), and
H = Z ∗ Z3.
Proof. We start with the presentation
< a, t, w; a2, (at)3, (w−1awa)3, [t, w] >
Letting t = as, x = w−1sw, m = w−1aw the presentation becomes
< a,w, s, x,m; a2, s3, (am)3, x = w−1sw,m = w−1aw, am = sx−1, x3,m2, (sx−1)3 >
Now let
G1 =< a,m,w; a2,m2, (am)3,m = w−1aw >
and
G2 =< w, s, x; s3, x3, (sx−1)3, x = w−1sw >
and
H1 =< w, am >6 G1
and
H2 =< w, sx−1 >6 G2
52
Using the normal form theorem for HNN extensions ([12] chapter 1 theorem 31) it can be
shown that H1∼= H2
∼= Z∗Z3. We can now combine the presentation for G1 and G2, adding
the relations w = w and am = sx−1, to get
< a,m,w, s, x, w; a2,m2, (am)3,m = w−1aw, s3, x3, (sx−1)3, x = w−1sw >
after applying Tietze transformations to eliminate w we get the earlier presentation for
PSL2(O3,2). Thus
PSL2(O3,2) = G1 ∗H G2
Now recall that
S3 =< x, y;x2, y2, (xy)3 >
It is clear that
< a,m; a2,m2, (am)3 >6 G1
and that < a >∼= Z2∼=< m >. It is very easy to see that under the automorphism, θ, of S3
where θ : a 7→ m and θ : m 7→ a, we have θ(< a >) =< m > and so G1 is an HNN extension
of S3 with two 2-cycles associated. Again it is easy to see that
D(3, 3, 3) =< s, x; s2, x3, (sx−1)3 >6 G2
and that under that automorphism of D(3, 3, 3) given by ϕ : s 7→ x, and ϕ : x 7→ s, we
have ϕ(< s >) =< x >∼= Z3 and so G2 is an HNN extension of D(3, 3, 3) with two 3-cycles
associated.
Now recall that PSL2(O3) does not decompose as a non-trivial free product with amal-
gamation or as an HNN group. However the above gives us a natural example of a subgroup
of finite index in PSL2(O3) which does. Hence
Theorem 3.4.5. PSL2(O3) is virtually a non-trivial product with amalgamation and vir-
tually an HNN group. More precisely |PSL2(O3) : PSL2(O3,2)| = 10 and PSL2(O3,2)
decomposes as a non-trivial free product with amalgamation and as an HNN group.
53
Contrast this with Fine’s [25] rather artificial construction of a subgroup of index 144
which decomposes as a non-trivial free product with amalgamation. We can pose the fol-
lowing
Question. Is 10 the least index of a subgroup of PSL2(O3) which decomposes as a free
product with amalgamation or as an HNN group?
3.4.2 Some Consequences
First recall that in an HNN group a torsion element must be conjugate to a torsion element
in the base ([12] chapter 1 exercise 22). We need the following
Lemma 3.4.6. Every element of finite order in D(3, 3, 3) is conjugate to one of:
x, x2, y, y2, xy, (xy)2
all of which have order 3.
Proof. This is easy to see by making D(3, 3, 3) act on the infinite lattice in Euclidean 2-space
made up of regular hexagons and triangles as given in [36] p.93.
Theorem 3.4.7. In PSL2(O3,2) every element of finite order is conjugate to one of:
a, at, (at)2, aw−1aw, (aw−1aw)2, w−1awt, w−1awt−1
where a has order 2 and the rest have order 3
Proof. Recall that PSL2(O3,2) is an HNN extension of K3,2 and K3,2 = S3 ∗Z3 D(3, 3, 3). Let
g ∈ PSL2(O3,2) be of finite order. So g is conjugate to an element of finite order in K3,2 and
so g is conjugate to an element of finite order in S3 or D(3, 3, 3). Now suppose that g has
order 2, then, as all the elements of finite order in D(3, 3, 3) have order 3, g is conjugate to
an element of order 2 in S3, recall
S3 =< a,m; a2,m3, (am)2 >
54
and S3 has exactly one conjugacy class of order 2. So g is conjugate to a.
Now suppose that g has order 3, so as S3 has only one conjugacy class of order 3 and it
is represented by m we have that g is conjugate to one of:
s, s2,m,m2, sm−1, s−1m
The result follows, as s = at, and m = aw−1aw.
Theorem 3.4.8. Let τ ∈ PSL2(O3,2) be a torsion element. Write x ∼ y if x and y are
conjugate. Then
1. If τ ∼ at, (at)2, w−1awt, w−1awt−1 then PSL2(O3,2)
N(τ)∼= Z2 × Z
2. If τ ∼ a then PSL2(O3,2)
N(τ)∼= Z3 × Z
3. If τ ∼ aw−1aw, (aw−1aw)2 then PSL2(O3,2)
N(τ)∼= Z×M
Proof. Recall
PSL2(O3,2) =< a, t, w; a2, (at)3, [t, w] , (w−1awa)3 >
set a = 1, so t3 = 1, so we get
PSL2(O3,2)
N(a)=< t,w; t3, [t, w] >∼= Z3 × Z
The others are similar
Theorem 3.4.9. Let T be the set of all torsion elements in PSL2(O3,2). Then N(T ) =
N(M).
Proof. N(T ) = N(a, at, aw−1aw,w−1awt, w−1awt−1), andN(M) = N(a, t). ClearlyN(M) ⊆
N(T ). Now suppose that a = t = 1. So a = at = aw−1aw = w−1awt = w−1awt−1 = 1. So
N(T ) ⊆ N(M).
Lemma 3.4.10. Let N Zp × Z =< x, y >, be of index n , where p is a rational prime.
Then if p - n then
N = N(x, yn)
55
if p | n then
N = N(x, yn), N(xmyk)
where n = pk, and m = 0, 1, . . . , p− 1.
Proof. Suppose p - n. Then x ∈ N and it is clear that N = N(x, yn).
Suppose that p | n, so n = pk, say. If, as above x ∈ N then N = N(x, yn) so suppose
that x /∈ N . Let l be the order of y mod N , so l | pk. Suppose that l | k, and l 6= k, so
yl = 1, and l < k, but then there are < pk distinct cosets of N . Contradiction. So if l | k
then l = k, and as l | pk we must have l = k, or pk.
Suppose that l = k. Then N(yk) 6 N . But N(yk) is of index pk = n, so N =
N(yk) = N(x0yk). Now suppose that l = pk. So y, y2, . . . , ypk are pk = n distinct cosets
representatives of N . Now x /∈ N , so x = yr some r, 1 6 r 6 pk, and 1 = xp = ypr, so
pk | pr so k | r, so r = k, 2k, . . . , pk. If r = pk then x = ypk = 1. Contradiction, as x /∈ N .
So r 6= pk, so x = yk, y2k, . . . , y(p−1)k. Hence, using the fact that Z/pZ is a field yk = xm
where m = 1, . . . , p− 1. So N(xmyk) 6 N . But N(xmyk) is of index n, so N = N(xmyk).
We now show that these groups are distinct. Suppose that
N(x, yn) = N(xmyk)
then yk ∈ N(x, ypk). Contradiction. Suppose that
N(xm1yk) = N(xm2yk) = N
where 0 6 m1,m2 6 p− 1. Then xm1yky−kx−m2 ∈ N , so xm1−m2 ∈ N , so m1 ≡ m2 (mod p)
so m1 = m2. Hence all the groups are distinct.
Theorem 3.4.11. Let N PSL2(O3,2) be of index n. Then if (n, 6) = 1
N = N(a, t, wn)
if 2 | n, and 3 - n then N is one of
N(a, t, wn), N(at, wn/2), N(at, awn/2)
56
if 2 - n, and 3 | n then N is one of
N(a, t, wn), N(a, wn/3), N(a, twn/3), N(a, t2wn/3)
Proof. Suppose that (n, 6) = 1, so working mod N , a = at = w−1awa = 1, so a = t = 1, so
N = N(a, t, wn). Now suppose that 2 - n, and 3 | n, so a = 1. So PSL2(O3,2)/N is a factor
of < t,w; t3, tw = wt >∼= Z3 × Z. So, by (3.4.10), N = N(a, t, wn), N(a, wn/3), N(a, twn/3),
N(a, t2wn/3). Now suppose 2 | n, and 3 - n, so at = w−1awa = 1. So PSL2(O3,2)/N a
factor of < a,w; a2, aw = wa >∼= Z2 × Z. So, by (3.4.10), N = N(a, t, wn), N(at, wn/2),
N(at, awn/2).
Theorem 3.4.12. Let n ∈ N. Let H = PSL2(O3,2)/PSL2(O3,2)n. Then
1. If (n, 6) = 1 then H ∼= Zn and PSL2(O3,2)n = N(a, t, wn).
2. If 3 - n and 2|n then H ∼= Z2 × Zn and PSL2(O3,2)n = N(at, wn).
3. If 2 - n and 3|n then H ∼= Z3 × Zn and PSL2(O3,2)n = N(a, wn).
Proof. Recall
PSL2(O3,2) =< a, t, w; a2, (at)3, [t, w] , (w−1awa)3 >
Suppose that (n, 6) = 1 then, working mod PSL2(O3,2)n, a = at = 1, so a = t = 1, and
wn = 1. So H ∼= Zn. Now suppose that 3 - n and 2 | n then at = 1, so a = t, and
(w−1awa) = 1 so [a, w] = 1, and wn = 1, so H ∼= Z2×Zn. Suppose that 2 - n and 3 | n then
a = 1 so t3 = 1, and wn = 1. So H ∼= Z3 × Zn.
Theorem 3.4.13.
PSL2(O3,2)′ < PSL2(O3,2)
2 ∩ PSL2(O3,2)3
Proof. Obviously PSL2(O3,2)′6 PSL2(O3,2)
2 ∩PSL2(O3,2)3. Consideration of index shows
that the inclusion is strict.
Remark. It is well known [70] that, in the Modular group PSL2(Z)′ = PSL2(Z)2 ∩
PSL2(Z)3. Fine ([25] Corollary 4.5.4.3) shows that if d = 2, 7, 11 then PSL2(Od)′ <
PSL2(Od)2 ∩ PSL2(Od)
3.
57
3.5 Presentations for some other PSL2(Od,m).
Using GAP [24] we find the following presentation for a subgroup of PSL2(O1) of index 8
< a, t, w, z; a2, z2, (at)3, (atz)2, [t, w] , (atw−1zw)2 >
where
a =
0 1
−1 0
, t =
1 1
0 1
, w =
1 2i
0 1
,
z = (uat)a(uat)−1 =
1− 2i 1− 2i
−2 2i− 1
.
Since all the generators lie in PSL2(O1,2), and, by (3.1.10), |PSL2(O1) : PSL2(O1,2)| = 8,
the group given by this presentation must be PSL2(O1,2). Hence
Theorem 3.5.1.
PSL2(O1,2) =< a, t, w, z; a2, z2, (at)3, (atz)2, [t, w] , (atw−1zw)2 >
We can show that PSL2(O1,2) decomposes as an HNN group and as a non-trivial amal-
gamated free product in the same way as we did for PSL2(O3,2).
Using GAP [24] we find the following presentation for a subgroup of PSL2(O11) of index
10
< a, t, w, k, l,m, n; a2, (at)3, k2, [m,n] , [t, w] , tklkt−1l−1, km−1ltat−1l−1m,
nlat−1l−1n−1mwaw−1tm−1,m−1tklt−1l−1mwt−1aw−1t >
Where
w =
1 2ω
0 1
, k = (uat)a(uat)−1 =
−1 + 2ω 5
2 1− 2ω
,
l = (ua)u2(ua)−1 =
−7 + 2ω 6 + 4ω
2ω 5− 2ω
,
m = (uaua)u(uaua)−1 =
−8 + 6ω 21 + 6ω
3 + 2ω 10− 6ω
,
58
n = (ua)uuatat−1au−1t−1(ua)−1 =
7 + 6ω 26− 14ω
6− 2ω −5− 6ω
.
Since all the generators lie in PSL2(O11,2), and, by (3.1.10), |PSL2(O11) : PSL2(O11,2)| =
10, the group given by this presentation must be PSL2(O11,2).
Again using GAP [24] wefind the following presentation for a subgroup of index 6 in
PSL2(O7):
< a, t, w, x, y ; a2, x2, [t, w] , (ax)3, (at)3, y = txyxt−1, (ytay−1x)2,
xyat−1y−1wt−1aw−1t >
where
w =
1 2ω
0 1
, x = (uat)a(uat)−1 =
1− 2ω −3
−2 −1 + 2ω
,
y = (ua)u2(ua)−1 =
−5 + 2ω −4− 2ω
2ω 3− 2ω
.
Since all the generators lie in PSL2(O7,2) and, by (3.1.10), |PSL2(O7) : PSL2(O7,2)| = 6
the group given by this presentation must be PSL2(O7,2).
The method of this section ie getting GAP [24] to list all subgroups of Γd of the relevent
index and then labouriously checking whether the generators for each lay in PSL2(Od,m) is
very unsophisticated and consequently of use in only very few cases. Swan [84] has developed
a method for computing presentations for PSL2(Od) which can easily be extended to the
PSL2(Od,m) ([78] page 628). However the computations rapidly become unwieldy as d
becomes large. Riley [78] developed a computer package (written in Fortran) called the
Poincare File which he used to find presentations for PSL2(Od) for 10 < d < 37 and
d = 43, 67, 163. Such an approach can be used to find presentations for PSL2(Od,m). The
implementation of this in GAP [24] could be an interesting future project.
59
Chapter 4
Non-standard normal subgroups of
SL2(Od,m)
Let R be a commutative ring with a one. Let n ∈ N such that n > 2. Let 1 6 i, j 6 n then
eij denotes the n× n matrix with a 0 in every position except (i, j), where it has a 1.
En(R) =< I + reij : 1 6 i, j 6 n, i 6= j >
Let q R
En(R, q) =< I + αeij : α ∈ q, 1 6 i, j 6 n, i 6= j >En(R)
the largest q such that En(R, q) 6 S is the level of S and is denoted l(S). This is well defined
because En(R, q1)En(R, q2) = En(R, q1 + q2).
Hn(R, q) = M ∈ SLn(R) : M ≡ kI (mod q), some k ∈ R
the smallest q such that S 6 Hn(R, q) is the order of S and is denoted o(S). S 6 SLn(R)
is said to be standard if l(S) = o(S).
E0(2, R; q) = N SL2(R) : o(N) = q, l(N) = 0
60
4.1 Non-standard normal subgroups of SL2(Z)
Here we describe the proof of the following theorem, mentioned in chapter one. This proof
appeared in [62].
Theorem. Let 0 6= q Z then |E0(2,Z; q)| = 2ℵ0.
First recall that the Modular group, PSL2(Z), has the presentation < x, y;x2, y3 > (see
[81]).
Theorem. [62] Every countable group can be embedded in an infinite simple group generated
by x, y of order 2, 3 respectively, and where xy has infinite order.
Corollary. [62] ∃2ℵ0 N SL2(Z) such that SL2(Z)/N is simple, and l(N) = 0.
Lemma 4.1.1. [62] Let X,Y be sets, X infinite. Let f : X −→ Y be a surjection. For
y ∈ Y let cy = |f−1(y)|. Then if ∃c0 such that cy 6 c0 |X| ∀y ∈ Y then |X| = |Y |.
Proof. Obviously |Y | 6 |X|. Now |X| =∑
y∈Y cy 6 |Y |cy 6 |Y |c0, and co |X| so
|X| 6 |Y |.
Theorem. [62] Let 0 6= q Z then |E0(2,Z; q)| = 2ℵ0.
Proof. Let
S = N SL2(Z) : l(N) = 0, SL2(Z)/N simple
so |S| = 2ℵ0 . We show that if N ∈ S then o(N) = Z and so |E0(2,Z,Z)| = |S| = 2ℵ0 . Let
N ∈ S, let q0 = o(N). So, since N 6 H(q0) SL2(Z) H(q0) = N or SL2(Z). Suppose
H(q0) = N then q0 = l(H(q0)) = l(N) = 0 so o(N) = 0, but o(N) = 0⇔ N = 1 or I,−I
and PSL2(Z) is not simple, so o(N) 6= 0. Hence H(q0) = SL2(Z) so q0 = l(H(q0)) =
l(SL2(Z)) = Z i.e. o(N) = Z.
Now suppose q 6= Z. Let S1 = H(q) ∩N : N ∈ S. Define ρ : S −→ S1 by ρ(N) = H(q)∩
N . Let X ∈ S1, and let S2 = ρ−1(X) ⊆ S. Suppose Y ∈ S2 so SL2(Z)/Y is simple and
61
o(Y ) = Z. Now Y H(q) = Y , or SL2(Z). Suppose that Y H(q) = Y , so H(q) 6 Y . So
q 6 l(Y ) = 0. Contradiction. Hence Y H(q) = SL2(Z), so
|Y : X| = |Y : H(q) ∩ Y | = |SL2(Z) : H(q)| ∞
so |S2| 6 ℵ0. Now S is infinite, ρ is a surjection and |S2| 6 ℵ0 2ℵ0 = |S|, so by the lemma
|S1| = |S| = 2ℵ0 .
We now show that if M ∈ S1 then l(M) = 0 and o(M) = q, and so |E0(2,Z; q)| = 2ℵ0 , as
required. Let M ∈ S1, so M = H(q) ∩N , some N ∈ S. Now M 6 N , so l(M) 6 l(N) = 0,
so l(M) = 0. Let q1 = o(M). Now M 6 H(q) so o(M) = q1 6 q. Let N be the image of N
in PSL2(Z), so, as −I ∈ NSL2(Z)
N∼=PSL2(Z)
N
since N is of infinite index in PSL2(Z), N is free and so contains elements of infinite order.
So N contains elements of infinite order. Now, as before
|N : M | = |N : H(q) ∩N | = |SL2(Z) : H(q)| <∞
so M contains elements of infinite order, so q1 = o(M) 6= 0. Now, as before, NH(q1) =
SL2(Z), so
H(q) = H(q) ∩NH(q1)
= H(q1)(N ∩H(q))
= H(q1)M
= H(q1).
so q1 = q. Hence result.
4.2 Non-standard normal subgroups of SL2(Od,m)
In this section we aim to extend the results proved above to the groups SL2(Od). This
answers a question of Lubotzky’s (MR92c : 20088). In fact we extend it to all SL2(Od,m)
62
with a free non-abelian quotient, more precisely what we require is that the group maps onto
the Modular group, M in a “nice” way. Recall from chapter two that it is conjectured that
the only SL2(Od,m) which do not have a free non-abelian quotient are SL2(O1), SL2(O1,2),
SL2(O2), SL2(O3), SL2(O3,2), SL2(O7), SL2(O11), and SL2(O19). We show that in fact
SL2(O1,2), SL2(O2), and SL2(O3,2) map onto the Modular group in a “nice” way and so
the result goes though for these groups as well, leaving only 5 true exceptions. We prove an
unsatisfactory version of the result for SL2(O7), SL2(O11), and SL2(O19) (which relies on
the fact that they have an infinite cyclic quotient) but we were unable to make any progress
with SL2(O1) and SL2(O3).
Lemma 4.2.1. Let N SL2(Od,m). Then
l(N) = 0⇔
1 1
0 1
or
1 mω
0 1
has infinite order mod N.
Proof. Suppose that 1 1
0 1
and
1 mω
0 1
have finite order mod N i.e. 1 s
0 1
,
1 tmω
0 1
∈ N, some s, t ∈ N.
Let M = st, let q = (M) Od,m, so q 6= 0. Let α ∈ q, so
α = (z1 +mωz2)M = z1M +mωz2M,
some z1, z2 ∈ Z. So 1 α
0 1
=
1 M
0 1
z1 1 mωM
0 1
z2
∈ N.
Hence E2(Od,m, q) 6 N , so l(N) 6= 0.
Conversely suppose l(N) 6= 0. So E2(Od,m, q) 6 N , some q 6= 0. Now |Od,m : q| ∞,
and so 1 + q has finite (additive) order i.e. s ∈ q, some s ∈ N. Similarly tmω ∈ q, some
63
t ∈ N. So 1 s
0 1
, and
1 tmω
0 1
∈ Ni.e. 1 1
0 1
, and
1 ω
0 1
have finite order mod N .
Recall the following result from lemma (2.3.2).
Theorem 4.2.2. Suppose r = r(d,m) > 1 then there exists a surjective homomorphism
ρ : SL2(Od,m) −→ Fr =< x1, . . . , xr >
such that
ρ
1 mω
0 1
= x1
Now consider the case r(d,m) > 1.
Lemma 4.2.3. If r(d,m) > 1 then ∃2ℵ0 N SL2(Od,m) such that SL2(Od,m)/N is simple,
l(N) = 0, and N contains elements of infinite order.
Proof. We have
ϕ : SL2(Od,m) Fr < x1, x2;x22, (x1x2)
3 >∼= PSL2(Z)
and
u =
1 mω
0 1
7→ x1 7→ t
Now by a result of Mason and Pride [62] mentioned in the previous section the Modular group
PSL2(Z) has 2ℵ0 normal subgroups N such that PSL2(Z)/N is simple and t has infinite
order mod N . Pulling these subgroups back to SL2(Od,m) we get 2ℵ0 normal subgroups N
of SL2(Od,m) such that SL2(Od,m)/N is simple and u has infinite order mod N . Further, as
kerϕ contains SL2(Z), N contains elements of infinite order. Hence result.
64
We can now generalize almost word for word the corresponding proof for the Modular
group.
Theorem 4.2.4. Suppose r(d,m) > 1 and let 0 6= q Od,m then |E0(2,Od,m; q)| = 2ℵ0.
Proof. Let
S = N SL2(Od,m) : l(N) = 0, SL2(Z) 6 N,SL2(Od,m)/N simple
so, by (4.2.3), |S| = 2ℵ0 . We show that ifN ∈ S then o(N) = Od,m and so |E0(2,Od,m,Od,m)| =
|S| = 2ℵ0 . Let N ∈ S, let q0 = o(N). So, since N 6 H(q0) SL2(Od,m) we have
H(q0) = N or SL2(Od,m). Suppose H(q0) = N then q0 = l(H(q0)) = l(N) = 0 so o(N) = 0,
but o(N) = 0 ⇔ N = 1 or I,−I and PSL2(Od,m) is not simple, so o(N) 6= 0. Hence
H(q0) = SL2(Od,m) so q0 = l(H(q0)) = l(SL2(Od,m)) = Od,m i.e. o(N) = Od,m.
Now suppose q 6= Od,m. Let S1 = H(q) ∩N : N ∈ S. Define ρ : S −→ S1 by
ρ(N) = H(q) ∩ N . Let X ∈ S1, and let S2 = ρ−1(X) ⊆ S. Suppose Y ∈ S2 so
SL2(Od,m)/Y is simple and o(Y ) = Od,m. So Y H(q) = Y , or SL2(Od,m). Suppose that
Y H(q) = Y , so H(q) 6 Y , so q 6 l(Y ) = 0. Contradiction. Hence Y H(q) = SL2(Od,m).
Now
|Y : X| = |Y : H(q) ∩ Y | = |SL2(Od,m) : H(q)| ∞
so |S2| 6 ℵ0. Now S is infinite, ρ is a surjection and |S2| 6 ℵ0 2ℵ0 = |S|, so by lemma
(4.1.1), |S1| = |S| = 2ℵ0 .
We now show that if M ∈ S1 then l(M) = 0 and o(M) = q, and so |E0(2,Od,m; q)| = 2ℵ0 ,
as required. Let M ∈ S1, so M = H(q)∩N , some N ∈ S. Now M 6 N , so l(M) 6 l(N) = 0,
so l(M) = 0. Let q1 = o(M). Now M 6 H(q) so o(M) = q1 6 q. Now N contains elements
of infinite order. So, as before,
|N : M | = |N : H(q) ∩N | = |SL2(Od,m) : H(q)| <∞
soM contains elements of infinite order, so q1 = o(M) 6= 0. As before, NH(q1) = SL2(Od,m),
65
so
H(q) = H(q) ∩NH(q1)
= H(q1)(N ∩H(q))
= H(q1)M
= H(q1).
so q1 = q. Hence result.
We have already seen in chapter two that when r(d,m) = 1 we may still have a free
non-abelian quotient. We now verify that in certain cases the mapping is “nice”
Lemma 4.2.5. For (d,m) = (5, 1), (6, 1), (15, 1), (14, 1), (7, 2), (11, 2), SL2(Od,m) has a
free quotient of rank 2 and the image of 1 mω
0 1
can be taken as a free generator. Further the kernel of this map contains an element of
infinite order.
Proof. Here we work with the presentations for PSL2(Od,m), as these are a homomorphic
images of SL2(Od,m) this is sufficient. In the following the presentations for d = 5, 6, 15 are
taken from [84], and for d = 14 from [30]. The presentations for PSL2(O7,2) and PSL2(O11,2)
can be found in (3.5). a, t, u have the usual meanings.
PSL2(O5) =< a, t, u, b, c; a2, (at)3, [t, u] , b2, (ab)2, (aubu−1)2, aca = tct−1, ubu−1cb = tct−1 >
so PSL2(O5)/N(a, t, b) =< u, c;>∼= F2.
PSL2(O6) =< a, t, u, b, c; a2, (at)3, [t, u] , b2, [a, c] , (atb)3, t−1ctubu−1 = bc, (atubu−1)3 >
so PSL2(O6)/N(a, t, b) =< u, c;>∼= F2.
PSL2(O15) =< a, t, u, c; a2, (at)3, [t, u] , [a, c] , ucuat = taucu >
66
so PSL2(O15)/N(a, t) =< u, c;>∼= F2.
PSL2(O14) =< a, t, u, b, c, d, e;a2, (at)3, [t, u] , [a, b] , cd−1eb−1dc−1be−1,
dau−1b−1dad−1baud−1a,
adtad−1ad−1bt−1ab−1d,
at−1ae−1dat−1d−1acatad−1eatb−1dac−1b >
after some calculation we see that PSL2(O14)/N(a, t, b, e, d) =< u, c;>∼= F2.
PSL2(O11,2) =< a, t, w, k, l,m, n; a2, (at)3, k2, [m,n] , [t, w] , tklkt−1l−1, km−1ltat−1l−1m,
nlat−1l−1n−1mwaw−1tm−1,m−1tklt−1l−1mwt−1aw−1t >
so PSL2(O11,2)/N(a, k, t,m, n) =< w, l;>∼= F2.
PSL2(O7,2) =< a, t, w, x, y ; a2, x2, [t, w] , (ax)3, (at)3, y = txyxt−1, (ytay−1x)2,
xyat−1y−1wt−1aw−1t >
so PSL2(O7,2)/N(a, x, t) =< w, y;>∼= F2.
It is clear from the proof of theorem (4.2.4) that it is sufficient for SL2(Od,m) to map
onto the Modular group with kernel of level zero. We remark that PSL2(O5)/N(b, c, u) ∼=
PSL2(O6)/N(b, c, u) =
< a, t; a2, (at)3 >= PSL2(Z).
Lemma 4.2.6.PSL2(O2)
N(u)=< a, t; a2, (at)3 >∼= PSL2(Z)
PSL2(O3,2)
N(w)=< a, t; a2, (at)3 >∼= PSL2(Z)
PSL2(O1,2)
N(a, t, (zw)3)=< z,w; z2, (zw)3 >∼= PSL2(Z)
Clearly the kernels contain elements of infinite order.
67
Proof. This is obvious given the following presentations
PSL2(O2) =< a, t, u; a2, (at)3, (u−1aua)2, [t, u] >
PSL2(O3,2) =< a, t, w; a2, (at)3, (w−1awa)3, [t, w] >
PSL2(O1,2) =< a, t, w, z; a2, z2, (at)3, (atz)2, [t, w] , (atw−1zw)2 >
The presentation for PSL2(O2) can be found in [25]. The presentation for PSL2(O1,2) can
be found in (3.5.1) and for PSL2(O3,2) in (3.3.1).
Thus we have
Proposition 4.2.7. Let (d,m) = (1, 2), (2, 1), (3, 2), (5, 1), (6, 1), (14, 1), (15, 1). Let
0 6= q Od,m then |E0(2,Od,m; q)| = 2ℵ0.
The only “true exceptions” not covered above are SL2(O1), SL2(O3), SL2(O7), SL2(O11),
and SL2(O19). The results we can achieve are very unsatisfactory. When we have an infinite
cyclic quotient we get the following result
Theorem 4.2.8. Suppose SL2(Od,m)/K ∼= Z and l(K) = 0 then ∀ 0 6= q Od,m ∃N
SL2(Od,m) such that l(N) = 0 and o(N) = q.
The following lemma was proved in [63] as Theorem 4.1 for any n > 2 and any Dedekind
domain of arithmetic type, A.
Lemma 4.2.9. Let R be a commutative ring with a one, q1, q2 R such that R/q2 is an
SR2-ring q2 6 q1. ThenSL2(R, q1)
SL2(R, q2)is abelian ⇔ q2
1 6 q2
Furthermore, if SL2(R, q1)/SL2(R, q2) is abelian then SL2(R, q1)/SL2(R, q2) ∼= a3, where
a = q1/q2, and the isomorphism is given by: 1 + z x
y 1− z
←→ (x, y, z).
68
Proof. Denote SL2(R, qi) by Γ(qi). Now suppose that q21 6 q2 and let X, Y ∈ Γ(q1). Then
X − I, Y − I ≡ 0 (mod q1) and so we have
(X − I)(Y − I) ≡ (Y − I)(X − I) ≡ 0 (mod q21)
It follows that XY ≡ Y X (mod q2) and so Γ(q1)′ ⊆ Γ(q2). Conversely suppose that
Γ(q1) / Γ(q2) is abelian. Let x, y ∈ q1 then
[I + xe12, I + ye21] =
∗ ∗
∗ 1− xy
≡ I (mod q2)
and so xy ∈ q2 hence q21 ⊆ q2.
Recall that by lemma (3.1.1), ϕ : SL2(R) −→ SL2(R/q2) is surjective, and Γ(q2) =Kerϕ.
Let γ(q1) = ϕ(Γ(q1)) and observe that γ(q1) ∼= Γ(q1)/Γ(q2). We show that γ(q1) ∼= a3.
Define θ : a3 −→ SL2(R/q2) by
(x, y, z) 7−→
1 + z x
y 1− z
.
We first show that θ is a homomorphism. Let xi, yi, zi ∈ a where i = 1, 2. Then since q21 6 q2,
θ (x1, y1, z1) θ (x2, y2, z2) =
1 + z1 + z2 x1 + x2
y1 + y2 1− z1 − z2
= θ (x1 + x2, y1 + y2, z1 + z2)
So θ is a homomorphism.
We now show that θ is injective. Suppose that θ (x, y, z) = 1 i.e. 1 + z x
y 1− z
≡ I (mod q2)
so (x, y, z) = 0 and so θ is injective.
We now show that imθ = γ(q1). Let X ∈ imθ. Since X ∈ SL2(R/q2) and ϕ is surjective
∃Y ∈ SL2(R) such that ϕ (Y ) = X. We show Y ∈ Γ(q1) and so X = ϕ ( Y ) ∈ γ( q1 ).
Now X ∈ imθ so
X =
1 + z x
y 1− z
69
some x, y, z ∈ a. Let
Y =
a b
c d
where a, b, c, d ∈ R. So a ∈ 1 + z, b ∈ x, c ∈ y, d ∈ 1− z, so a, d ≡ 1 (mod q1) and b, c ≡ 0
(mod q1), so Y ≡ I (mod q1) i.e. Y ∈ Γ(q1). Hence imθ ⊆ γ(q1).
Now we show that γ(q1) ⊆ imθ. Let X ∈ γ(q1), so X = ϕ(Y ) some Y ∈ Γ(q1). RTP
X =
1 + z x
y 1− z
some x, y, z ∈ a. Let
Y =
a b
c d
where a, b, c, d ∈ R. Now Y ≡ I (mod q1) so a, d ≡ 1 (mod q1) and b, c ≡ 0 (mod q1). Let
x = b + q2, y = c + q2 so that x, y ∈ a. Now let z = ζ + q2 = (a − 1) + q2 and since a ≡ 1
(mod q1) we have z ∈ a. We show that d+ q2 = 1− z. Now (a− 1)(d− 1) ∈ q21 6 q2 and so
a + d ≡ 2 (mod q2). So now 1 − z = 1 − ζ + q2 = 1 − (a − 1) + q2 = 2 − a + q2 = d + q2.
Hence
X = ϕ(Y ) =
a+ q2 b+ q2
c+ q2 d+ q2
=
1 + z x
y 1− z
some x, y, z ∈ a. Hence γ(q1) ⊆ imθ. Hence result.
Theorem 4.2.10. Suppose that SL2(Od,m)/K ∼= Z and l(K) = 0. Then ∀ 0 6= q Od,m
∃N SL2(Od,m) such that l (N) = 0 and o (N) = q.
Proof. Let 0 6= q1 Od,m and consider H (q1) ∩K. Let q2 = o (H (q1) ∩K), so q2 6 q1.
Clearly l (H (q1) ∩K) = 0. We show q1 = q2. Suppose not. Now Γ/K ∼= Z so
H (q1) / H (q2) is cyclic, so Γ (q1) /Γ (q1) ∩ H (q2) is cyclic. Also Γ (q1) ∩ H (q2) /Γ (q2)
is central in Γ (q1) /Γ (q2). So Γ (q1) /Γ (q2) is central by cyclic, and so abelian. Hence
q21 6 q2. Let a = q1/q2. So Γ (q1) /Γ (q2) ∼= a3. Where 1 + z x
y 1− z
←→ (x, y, z)
70
We show a2 → Γ (q1) /Γ (q1) ∩H (q2). Define
ϕ : a2 −→ Γ (q1)
Γ (q1) ∩H (q2)
by
ϕ(x, y) =
1 x
y 1
.
ϕ is clearly a homomorphism. Now suppose that ϕ(x, y) = 1. So 1 x
y 1
∈ Γ (q1) ∩H (q2) ,
so x, y ∈ q2, so (x, y) = 0 in a2, hence ϕ is injective. So a2 is cyclic, so a is trivial i.e.
q1 = q2.
The only (d,m) excluded from the above theorem are (1, 1), and (3, 1). Neither SL2(O1)
nor SL2(O3) have infinite cyclic quotients and so using the above techniques we are unable
to say anything about E0(2,Od; q) for d = 1, 3.
However despite the difficulties mentioned above we feel confident in making the following
Conjecture. ∀(d,m) and ∀ 0 6= q Od,m, |E0(2,Od,m; q)| = 2ℵ0 .
71
Chapter 5
Order and level of a normal
congruence subgroup
We have seen that in general there is no relationship between the order and level of a normal
subgroup of SL2(Od,m). However if we restrict ourselves to normal congruence subgroups
we do find a nice relationship. Mason [58] has proved
Theorem. Let NSL2(Od) be a congruence subgroup of level q∗ and order q. Then 12q 6 q∗.
Suppose that N SL2(Od) is a congruence subgroup of order Od then SL2(Od, 12Od) 6
N . We use this fact to show that a large class of subgroups of the SL2(Od) are non-congruence
subgroups. First of all we aim to extend the above results to SL2(Od,m). In fact we extend
it to SL2 over a larger class of rings. The material in this chapter is the most technically
complex and perhaps the hardest to follow in this thesis, so we apologize to the reader for
the Cimmerian night which is about to descend.
5.1 Primary decomposition in Noetherian domains
The material in this section has been lifted from chapters four, seven, and eleven of Atiyah
and MacDonald [2]. Let A be any commutative ring.
Definition. q A is primary if q 6= A and xy ∈ q⇒ x ∈ q or yn ∈ q, some n > 0.
72
Definition. Let q A. Then the radical of q is
r(q) = x ∈ A : xn ∈ q, some n > 0
Proposition 5.1.1. ([2] Proposition 1.14) The radical of q A is the intersection of all
prime ideals which contain q.
Proposition 5.1.2. ([2] Exercise 1.13) Let a, b A. Then
r(ab) = r(a ∩ b) = r(a) ∩ r(b)
Proposition 5.1.3. ([2] Proposition 4.1) Let q A be primary. Then r(q) is the smallest
prime ideal containing q.
Definition. Let a A. A primary decomposition of a is
a =n⋂
i=1
qi
where qi A is primary.
Definition. If p = r(q) we say that q is p-primary.
Lemma 5.1.4. ([2] Lemma 3.1) Let qi A, 1 6 i 6 n be p-primary. Then q = ∩ni=1qi is
p-primary.
If in a primary decomposition, all the r(qi) are distinct and ∩i6=jqj * qi, i = 1, . . . , n
then we say that the decomposition is minimal. All primary decompositions discussed here
shall be minimal.
Recall that a ring A is Noetherian if it satisfies three equivalent conditions:
1. Every nonempty set of ideals in A has a maximal element.
2. Every ascending chain of ideals in A is stationary.
3. Every ideal in A is finitely generated.
From now on let A be Noetherian.
73
Definition. We say that a A is irreducible if a = b ∩ c⇒ a = b, or c.
Lemma 5.1.5. ([2] Lemma 7.11) Every ideal in A is a finite intersection of irreducible
ideals.
Definition. Let x ∈ A. Then the annihilator of x is Ann(x) = a ∈ A : ax = 0.
Lemma 5.1.6. ([2] Lemma 7.12) Every irreducible ideal in A is primary.
Thus we have proved the following
Theorem 5.1.7. ([2] Theorem 7.13) In a Noetherian ring, every ideal has a primary de-
composition.
Proposition 5.1.8. ([2] Proposition 7.14) In a Noetherian ring A, every ideal contains a
power of its radical.
Let R be a commutative ring. A proper chain of prime ideals
0 p1 · · · pn R
is of length n. The Krull dimension of R is the maximum length of chains of prime ideals
in R. Thus a domain is of Krull dimension one, if and only if all non-zero prime ideals are
maximal. From now on K shall denote a Noetherian domain of Krull dimension one.
Theorem 5.1.9. Let K be a Noetherian domain of Krull dimension one. Let 0 6= q K.
Then
q = p1 . . . pt
where pi is a primary ideal. Furthermore, for each i, r(pi) = mi, a maximal ideal; mnii 6 pi,
for some ni; and pi + pj = K, for i 6= j.
Proof. The radical of a primary ideal is prime, and so, as K is of Krull dimension one, is
maximal. Suppose that
q =t⋂
i=1
pi
74
is a minimal primary decomposition. Now r(pi) = mi is maximal and, as the decomposition
is minimal, mi 6= mj if i 6= j. Further mnii 6 pi, as every ideal contains a power of its radical.
Thus pi + pj = K if i 6= j. So q is an intersection of coprime ideals. Hence a product
q = pi . . . pt
Lemma 5.1.10. Let K be a Noetherian domain of Krull dimension one. Let q∗ 6 q be
ideals in K such that 0 6= q∗ 6= K. Suppose that q∗ = p∗1 . . . p∗t is a primary decomposition.
Then q = p1 . . . pt where pi = K or r(pi) = r(p∗i ) and p∗i 6 pi.
Proof. If q = K then the result is trivial, so suppose that K 6= q = p1 . . . ps is a primary
decomposition. We can suppose that r(p1) = r(p∗1) = m1. Let q∗0 = p∗2 . . . p∗t , q0 = p2 . . . ps.
So q∗ = p∗1q∗0, and q = p1q0. Now p∗1 + q∗0 = K = p1 + q0. Since r(p∗1) = r(p1) we have
p1 + q∗0 = K. So p∗1(p1 + q∗0) = p∗1, so p∗1p1 + p∗1q∗0 = p∗1. Now p∗1q
∗0 = q∗ 6 q 6 p1, and
p∗1p1 6 p1, so p∗1 6 p1. Now for those p∗i which have a radical distinct from all the r(pj) we
let pi = K.
Example 5.1.1. Let D be a Dedekind domain. Then by definition ([2] chapter 9) D is
a Noetherian domain of Krull dimension one. In this case every primary ideal is a prime
power.
Example 5.1.2. Let Od,m = Z +mωZ be an order in an imaginary quadratic number field.
Then Od,m is a Z-module of finite rank and so, by Hilbert’s basis theorem (see [2] Theorem
7.5) is Noetherian. Now Od,m is of finite index in Od and Od is a Dedekind domain and so a
Noetherian domain of Krull dimension one. Thus by theorem 20(1) on page 81 of [64], Od,m
is a Noetherian domain of Krull dimension one, it is clearly of characteristic zero. Note that
maximal orders are Dedekind domains. However in non-maximal orders not every primary
ideal is a power of a prime ideal.
For example consider the order O3,2 = Z + i√
3Z. Let m = (2, 1 + i√
3). Then m is a
maximal ideal of O3,2 of index 2 and m2 < 2O3,2 < m. So the ideal 2O3,2 is a primary ideal
which is not a power of a prime ideal.
75
Example 5.1.3. Let p ∈ Z be a rational prime and let K = Z/pZ. Let R = K [X] and let
q R be of finite index. R is Noetherian by Hilbert’s basis theorem. Consider the subring
S = K + q, of R. We show that S is a Noetherian domain of Krull dimension one. By
theorem 20(1) on page 81 of [64], if S is Noetherian then it is of Krull dimension one, so it
suffices to show that S is Noetherian. Now R is a PID so q = (f) where deg f = d. Let
g =∑aiX
i ∈ R and consider fg = b0f + · · ·+ btXtf , which is in the K-module generated
by fK,XfK, . . . , X tfK. Suppose that t > d, so degX tf > 2d = deg f 2. So, as R has a
Euclidean algorithm, X tf = hf2+r, where deg r < deg f 2 = 2d, and deg h 6 degX tf−2d <
degX tf . So that fg is in the K-module generated by fK,XfK, . . . , X t−1fK. Repeat the
above until we see that fg is in the K-module generated by fK,XfK, . . . , Xd−1fK. So
S = K + q is generated as a K-module by K, fK,XfK, . . . , Xd−1fK and so is Noetherian,
by Hilbert’s basis theorem. S is thus an example of a Noetherian domain of Krull dimension
one and non-zero characteristic.
5.2 Wohlfahrt’s Theorem
Wohlfahrt [91] showed that if S is a subgroup of SL2(Z), of level m then S is a congruence
subgroup if and only if SL2(Z,mZ) 6 S. That is he extended Klein’s concept of the level of
a congruence subgroup of the Modular group to an arbitrary subgroup of the Modular group.
This concept of level and Wohlfahrt’s theorem have been of great use in the construction of
non-congruence subgroups of the Modular group. We have seen how the concept of level can
be extended to SLn(R) where R is any ring with a one. In this section we extend Wohlfahrt’s
theorem to all Noetherian domains of Krull dimension one.
Definition. R is said to be an SR2-ring if a, b ∈ R such that gcd(a, b) = 1⇒ ∃t ∈ R such
that a+ tb ∈ R∗.
Lemma 5.2.1. Let A be any commutative ring with a one. Let a, b ∈ A such that gcd(a, b) =
1. Let q A such that A = A/q is an SR2-ring. Then ∃t ∈ A such that (a+ bt)A+ q = A.
76
Proof. As a, b are coprime ∃s, t ∈ A such that as + bt = 1. Now 1 = ϕ(1) = ϕ(as + bt) =
as + bt, where ϕ : A → A is the natural homomorphism. So a, b are coprime. So, as A is
an SR2-ring ∃t ∈ A such that (a + bt) = u, where u ∈ A∗, so (a + bt) ≡ u (mod q), so
u = (a+ bt)− q, some q ∈ q. Hence (a+ bt)A+ q = A.
Theorem 5.2.2. Let A be any commutative ring. Let qiA, i = 1, 2. Suppose that A = A/q1
is an SR2-ring. Then
SL2(A, q2) 6 E2(A, q2)SL2(A, q1)
Proof. Denote SL2(A, qi) by Γ(qi) and E2(A, qi) by ∆(qi). Let
X =
a b
c d
∈ Γ(q2).
Now ad− bc = 1 so a, b are coprime, and A is an SR2-ring, so by the lemma (5.2.1) ∃t ∈ A
such that (a+ bt)A is prime to q1. Let
T =
1 0
t 1
∈ ∆(A)
so
T−1XT =
a+ bt ∗
∗ ∗
∈ Γ(q2)
so wlog aA and q1 are coprime. Let x ∈ A such that ax ≡ 1 (mod q1). Now let
T1 =
1 0
1 1
, X1 =
1 x(1− a− b)
0 1
,
X2 =
1 a− 1
0 1
.
a ≡ 1 (mod q2), so X2 ∈ ∆(q2), and 1− a, b ∈ q2 so X1 ∈ ∆(q2). Now
T−11 XX1T1X2 ≡
1 0
q 1
(mod q1)
where q ∈ q2. So T−11 XX1T1X2 ∈ ∆(q2)Γ(q1). Hence X ∈ ∆(q2)Γ(q1), as X1, X2 ∈ ∆(q2),
and T1 ∈ ∆(A), and ∆(q2) ∆(A).
77
The following corollary is the classical form of Wohlfahrt’s theorem and is equivalent to
the theorem.
Corollary 5.2.3. Let G 6 SL2(A) have level q. Then G is a congruence subgroup if and
only if SL2(A, q) 6 G.
Corollary 5.2.4. Let A be an SR2-ring. Let q A. Then SL2(A, q) = E2(A, q). So in
particular SL2(A) = E2(A).
Proof. Take q1 = 0 in the theorem.
Corollary 5.2.5. Let A be any commutative ring. Let qi A, i = 1, 2. Suppose that
A = A/q2 is an SR2-ring. Then
SL2(A, q1 + q2) = E2(A, q1)SL2(A, q2)
Proof. Obviously Γ(q1)Γ(q2) 6 Γ(q1 + q2). Then
Γ(q1 + q2) 6 ∆(q1 + q2)Γ(q2) by theorem (5.2.2)
= ∆(q1)∆(q2)Γ(q2)
6 ∆(q1)Γ(q2)
Lemma 5.2.6. [3] Semilocal rings are SR2 rings.
Proof. Let R be a semilocal ring,and let m1, . . . ,mt be the maximal ideals of R. Let a, b ∈ R
such that gcd(a, b) = 1. RTP ∃t ∈ R such that a+ tb ∈ R∗.
∀i = 1, . . . , t suppose
∀s ∈ R, a+ bs ≡ 0 (mod mi)
Then, taking s = 0, we get a ∈ mi, and, taking s = 1, we get b ∈ mi. But a, b are coprime.
Contradiction. So ∀i ∃ti such that a + tib 6≡ 0 (mod mi). Then, by the Chinese Remainder
Theorem, ∃t ∈ R such that t ≡ ti (mod mi) ∀i. So
a+ tb ≡ a+ tib 6≡ 0 (mod mi) ∀i
So a+ tb ∈ R∗, as required.
78
Lemma 5.2.7. Let K be a Noetherian domain of Krull dimension one. Let 0 6= q K.
Then K/q is semilocal and therefore an SR2-ring.
Proof. AsK is Noetherian we have a primary decomposition q = ∩ti=1pi. AsK has dimension
one this is a product pi . . . pt and r(pi) is maximal. Let mi = r(pi). So ∀i ∃ni ∈ N such that
mnii 6 pi. So mn1
1 . . .mntt 6 q. Now suppose that m K/q is a maximal ideal. So m = m/q,
some maximal ideal m > q. Suppose that m 6= mi ∀i. Then m + mn11 . . .mnt
t = K, but
mn11 . . .mnt
t 6 q, so m + q = K. Contradiction. Hence K/q has only finitely many maximal
ideals and so is semilocal and therefore SR2.
Thus
Theorem 5.2.8. Let K be a Noetherian domain of Krull dimension one. Let qiK, i = 1, 2,
q1 6= 0. Then
SL2(K, q1 + q2) = E2(K, q2)SL2(K, q1)
SL2(K, q1 + q2) = SL2(K, q1)SL2(K, q2)
Let O be an order of an imaginary quadratic number field. So O is a Noetherian domain
of Krull dimension one, so we have a Wohlfahrt theorem in these rings. We remark that
maximal orders are Dedekind domains and in this case the result has already been proved
[63].
5.3 Preliminary results about SL2(L)
Let L be a commutative local ring with maximal ideal m. Let N SL2(L). Let o(N) = q,
and l(N) = q∗. We are interested in how q and q∗ are related. It will turn out that whether
2 is a unit or not is of critical importance. The case 2 ∈ L∗ is easier to handle than 2 /∈ L∗.
When 2 /∈ L∗ we have two cases, o(N) = L and o(N) 6= L both of which split into two
subcases, |L : m| > 2 and |L : m| = 2, with the latter case being the most difficult. Here we
present a complete account; the results up to lemma (5.3.15) are known.
79
Lemma 5.3.1. [3] Let L be a commutative local ring and let q L. Then
SL2(L, q) = E2(L, q)
and so, in particular
SL2(L) = E2(L)
Proof. This follows from (5.2.4) and (5.2.6)
Lemma 5.3.2. Let N SL2(L). Then o(N) is generated byc ∈ L :
∗ ∗c ∗
∈ N
Proof. o(N) is generated by b, c, a− d for all a b
c d
∈ N.The result follows as 0 −1
1 0
a b
c d
0 1
−1 0
=
d −c
−b a
∈ Nand
V T−1XTV −1 =
d −c
d− a+ c− b a− c
∈ Nwhere
T =
1 1
0 1
, V =
0 −1
1 0
Corollary 5.3.3. Let L be any local ring. Let N SL2(L) such that o(N) = L. Then
∃X =
∗ ∗c ∗
∈ N , such that c is a unit.
80
We make use of the following notation.
E12(x) =
1 x
0 1
, E21(y) =
1 0
y 1
The following lemma is a slight generalization of lemme 3.3 part (ii) of [44].
Lemma 5.3.4. Let A be a commutative ring. Let t ∈ A, and u ∈ A∗, and let
Y =
u t
0 u−1
∈ N SL2(A).
Then E12((u− u−1)α) ∈ N ∀α ∈ A.
Proof. 1 (u− u−1)α
0 1
=
1 −u−1α
0 1
u t
0 u−1
1 u−1α
0 1
u−1 −t
0 u
∈ N
Lemma 5.3.5. ( [54] lemma 1.1 ) Let A be a commutative ring. Let N SL2(A) and
suppose that
M =
a b
c d
∈ N.Then ∀u ∈ A∗ such that u2 ≡ 1( mod c), u4 − 1 ∈ l(N).
Proof. Let
M2 =
u t
0 u−1
where u ∈ A∗ such that u2 ≡ 1( mod c), and t ∈ A. Then
[M2,M ] = M−12 M−1M2M =
α β
γ δ
,
where
α = (u−1d+ ct)(ua+ tc)− u−1c(bu−1 + at),
81
γ = ac− u2ac− utc2,
δ = ad− u2bc− utcd.
Now if a−u2a−utc = 0 then γ = 0 and a−u2a−utc = 0⇔ a(1−u2) = utc. Since u2 ≡ 1(
mod c), u2 − 1 = kc, some k ∈ A. So let t = u−1ak, so γ = 0. Now tc = u−1a(1 − u2),
substituting this into the expressions for α and δ and simplifying we see that α− δ = u4− 1
and so by (5.3.4) we get u4 − 1 ∈ l(N).
Lemma 5.3.6. Let L be a commutative local ring with maximal ideal m. Let N SL2(L),
o(N) = q 6 m. Then 8q 6 l(N).
Proof. Let
M =
a b
c d
∈ Nby (5.3.2) it is enough to show that 8c ∈ l(N). Let u1 = 1 + c, u2 = 1 − c. As o(N) 6= L,
ui ∈ L∗, and u2i ≡ 1( mod c), so by (5.3.5) u4
i −1 ∈ l(N), so (u41−1)−(u4
2−1) = 8c(c2+1) ∈
l(N), and 1 + c2 ∈ L∗ so 8c ∈ l(N).
Lemma 5.3.7. Let L be a commutative local ring with maximal ideal m. Let N SL2(L).
Suppose that 2 ∈ L∗ and o(N) 6= L. Then o(N) = l(N).
Proof. As 2 ∈ L∗, 8 ∈ L∗ and so this follows from (5.3.6).
Lemma 5.3.8. Let L be a commutative local ring with maximal ideal m. Suppose that 2 = 0.
Let N SL2(L) and suppose that o(N) = q 6 m. Then
q(4) =< α4 : α ∈ q >6 l(N)
Proof. Consider ∗ ∗c ∗
∈ Nas o(N) 6= L, c ∈ m. Let u1 = 1 + xc, u2 = 1 + yc, some x, y ∈ L. Now, as 2 = 0,
c4(x4 − y4) = u41 − u4
2 ∈ l(N), by (5.3.5). Pick x, y ∈ L such that x = y + 1, so x4 − y4 = 1.
Hence result.
82
When |L : m| > 2 we can improve the above results. First we need Whitehead’s lemma
Lemma 5.3.9. Let R be any commutative ring with a one. Let u ∈ R∗. Then u 0
0 u−1
∈ E2(R).
Proof. Let u ∈ R∗. Then 1 u
0 1
1 0
−u−1 1
1 u
0 1
=
0 u
−u−1 0
∈ E2(R).
Now u2,−u ∈ R∗, so 0 u2
−u2 0
0 −u
u−1 0
=
u 0
0 u−1
∈ E2(R)
Lemma 5.3.10. ([43] Proposition 1.3.6 ) Let A be any commutative ring. Let q A. Let
Γ = SL2(A), ∆ = E2(A), ∆(q) = E2(A, q), Γ(q) = SL2(A, q), H(q) = H2(A, q). Then
[Γ, H(q)] 6 Γ(q)
[∆,∆(q)] 6 ∆(q)
and if ∃u ∈ A∗ such that u2 − 1 ∈ A∗ then
[∆,∆(q)] = ∆(q)
Proof. Let M1 ∈ Γ, M2 ∈ H(q). So [M1,M2] = M1M2M−11 M−1
2 ≡ M1kIM−11 k−1I ≡ I
mod q. So [Γ, H(q)] ∈ Γ(q). Now, as ∆,∆(q) are both normal in ∆, [∆,∆(q)] 6 ∆∩∆(q) =
∆(q).
Now let t ∈ A and suppose that u ∈ A∗ such that u2 − 1 ∈ A∗ and consider u 0
0 u−1
,
1 t
0 1
=
1 (u2 − 1)t
0 1
83
By Whitehead’s lemma (5.3.9) u 0
0 u−1
∈ ∆
Now let α ∈ q and let t = (u2 − 1)−1α ∈ q, so that 1 α
0 1
=
1 (u2 − 1)t
0 1
∈ [∆,∆(q)]
Lemma 5.3.11. ( [54] lemma 1.2 ) Let L be a commutative local ring with maximal ideal
m. Suppose that |L : m| > 2 and 2 /∈ L∗. Let N SL2(L) have order q 6 m. Let
M =
a b
c d
∈ N.Then 2c2, c4 ∈ l(N)⇒ 2c, c2 ∈ l(N).
Proof. Now
M2 = MT−1M−1T =
e f
g h
∈ Nwhere e = 1 + ac and g = c2. Note that 2g, and g2 ∈ l(N) = q. Now let
M3 = M2T (−t)M−12 T (t) =
1 + get t(1− e2 + get)
g2t 1− get+ g2t2
where t ∈ L. Let r = 1 + get and s = t(1− e2) + get2. So
D(r, r)T (r−1s) =
r 0
0 r
1 r−1s
0 1
=
r s
0 r
and g2, 2g ∈ q so
M3 ≡ D(r, r)T (r−1s)( mod q)
Let q = 1− r2 ∈ q and let
M4 =
r(1 + q) q
−q r
84
then detM4 = r2(1+q)+q2 = r2(2−r2)+(1−r2)2 = 1, so M4 ∈ SL2(L). Clearly M4 ∈ H(q).
Now
M4M3 ≡
r 0
0 r
r s
0 r
=
r2 rs
0 r2
( mod q)
and q = 1− r2 ∈ q, so r2 ≡ 1( mod q). Now rs = s+ tegs = s+ t2eg(1− e2) + t3e2g2, and
g2 ∈ q, so t3e2g2 ∈ q. Also 1− e2 = −(2ac+ a2g), so t2eg(1− e2) = −t2e(2gac+ a2g2), and
2g, g2 ∈ q, so rs ≡ s( mod q). Hence SR ≡ T (s)( mod q). So T (s) = T (t(1−e2))T (t2eg) ∈
NH(q). Now u−1 0
0 u
e f
g h
u 0
0 u−1
=
e u−2f
u2g h
∈ Nand repeat the above argument with this matrix to get T (t(1 − e2))T (t2eu2g) ∈ NH(q),
∀u ∈ L∗ ∀t ∈ L.
Now, as |L : m| > 2 we can find v ∈ L∗ such that v − 1 ∈ L∗. Now consider the above
with u = v, v − 1. So
T (t(1− e2))T (t2ev2g) ∈ NH(q)
and
T (t(1− e2))T (t2eg)T (t2egv2)T (−t2e2gv) ∈ NH(q)
Now 2g ∈ q, so T (−2t2e2gv) ∈ NH(q) and so it follows from the above that
T (t(1− e2)) ∈ NH(q)
∀t ∈ L. Hence e2−1 ∈ l(NH(q)). Now e2−1 = (1+ac)2−1 = a(2c+ac2), and ad− bc = 1,
c ∈ m, so a, d ∈ L∗, so 2c+ ac2 ∈ l(NH(q)). Now, again using the fact that |L : m| > 2, let
w ∈ L∗ such that w2 − 1 ∈ L∗, so w−1 0
0 w
a b
c d
w 0
0 w−1
=
a w−2b
w2c d
Repeating the above argument we see that 2w2c+ aw4c2 = w2(2c+ aw2c2) ∈ l(NH(q)). So,
as w2 ∈ L∗, 2c+ aw2c2 ∈ l(NH(q)). So 2c+ aw2c2 − (2c+ ac2) = ac2(w2 − 1) ∈ l(NH(q)),
and a, w2 − 1 ∈ L∗, so c2 ∈ l(NH(q)), and 2c ∈ l(NH(q)).
85
So Γ(2c+ c2) 6 NH(q). Let q0 = (2c) + (c2). Now
Γ(q0) = [Γ,Γ(q0)] by (5.3.10) and (5.3.1)
6 [Γ, NH(q)] as Γ(q0) 6 NH(q0)
6 [Γ, N ] [Γ, H(q)]
6 NΓ(q) by (5.3.10)
6 N as Γ(q) 6 N
Lemma 5.3.12. ( [54] theorem 1.3 ) Let L be a commutative local ring with maximal ideal
m. Suppose that |L : m| > 2, 2 /∈ L∗, and 2 6= 0. Let N SL2(L), o(N) = q 6 m. Then
2q 6 l(N).
Proof. We show that 2c4, c8 ∈ l(N) and apply (5.3.11). Let
M =
a b
c d
∈ Nso c ∈ m. Let x, y ∈ L and consider 1 + xc, 1 + yc ∈ L∗, so by (5.3.5),
(1 + xc)4 − (1 + yc)4 = 4c(x− y) + 6c2(x2 − y2) + 4c3(x3 − y3) + c4(x4 − y4)
= (x− y)c (2 + (x+ y)c)(2 + 2(x+ y)c+ (x2 + y2)c
)∈ l(N)
Now, using the fact that |L : m| > 2 choose x = u, y = v such that u, v ∈ L∗, u + v = 1, so
u− v = 1− 2v ∈ L∗, as 2 /∈ L∗. So
c(2 + c)(2 + 2c+ (1− 2uv)c2) ∈ l(N)
Choosing x = 1, y = 0, we see that
c(2 + c)(2 + 2c+ c2) ∈ l(N)
86
so that
2c3(2 + c) ∈ l(N)
Now, again using the fact that |L : m| > 2 choose u ∈ L∗ such that u2−1 ∈ L∗. Conjugating
X by D(u, u−1) and repeating the above argument shows that
2c3(2 + u2c) ∈ l(N)
so, since u2 − 1 ∈ l(N), we see that
2c4 ∈ l(N)
Now, with x = 1, y = 0, we see that
c4 + 4c3 + 6c2 + 4c ∈ l(N)
so, multiplying by c4 we get
c8 + 4c7 + 6c6 + 4c5 = c8 + 2c4(2c3 + 3c2 + 2c) ∈ l(N)
so
c8 ∈ l(N)
Hence result.
Lemma 5.3.13. ( [54] theorem 1.3 ) Let L be a commutative local ring with maximal ideal
m. Suppose that |L : m| > 2, and 2 = 0. Let N SL2(L) and suppose that o(N) = q 6 m.
Then q(2) 6 l(N).
Proof. By (5.3.8) 2c4, c4 ∈ l(N). Hence by (5.3.11), 2c, c2 ∈ l(N). Clearly 2c = 0. Hence
result.
Lemma 5.3.14. Let L be a commutative local ring with maximal ideal m. Let N SL2(L).
Suppose that o(N) = L and |L : m| > 3. Then l(N) = L.
Proof. Let ∗ ∗c ∗
∈ N87
such that c ∈ L∗. Now let u ∈ L∗, so u2 − 1 = cc−1(u2 − 1), so u2 ≡ 1( mod c). So by
(5.3.5) u4 − 1 ∈ l(N). Now if ∀u ∈ L∗ u4 − 1 ∈ m, |L : m| 6 5. So if |L : m| > 5 then we are
done.
Suppose that |L : m| = 5. Let α ∈ m, and let u1 = 1 + α, u2 = 1 − α, so ui ∈ L∗ and
u2i ≡ 1 (mod c), so, by (5.3.5), 8α(1 + α2) = u4
1 − u42 ∈ l(N). Now |L : m| = 5 so 2 ∈ L∗, so
8 ∈ L∗, and 1+α2 ∈ L∗, so α ∈ l(N), hence m 6 l(N). Now let MSL2(F5) be the image of
N under the natural map SL2(L) SL2(F5), so o(M) = F5, M 6= ±I, and as PSL2(F5)
is simple, M ±I = SL2(F5). So M is of index 1 or 2 in SL2(F5) but SL2(F5)′ = SL2(F5)
[72], so M = SL2(F5). So N = SL2(L) and l(N) = L. Now suppose that |L : m| = 4. Let
u ∈ L∗ − 1. Then u4 − 1 = u− 1 /∈ m, so l(N) = L.
Lemma 5.3.15. Let L be a commutative local ring with maximal ideal m. Let N SL2(L).
Suppose that o(N) = L and |L : m| = 3. Then m 6 l(N).
Proof. As o(N) = L, by (5.3.3),
∃
∗ ∗c ∗
∈ Nsuch that c ∈ L∗. Letα ∈ m and let u1 = 1 +α, u2 = 1−α, so ui ∈ L∗ and u2
i ≡ 1( mod c).
So 8α(1 + α2) = u41 − u4
2 ∈ l(N) by (5.3.5). As |L : m| = 3, 2 is a unit and, as α ∈ m, 1 + α2
is also a unit. So α ∈ l(N). Hence m 6 l(N).
Lemma 5.3.16. Let L be a commutative local ring with nilpotent maximal ideal m of index
2. Then |SL2(L)| = 2α3.
Proof. Let n ∈ N be minimal such that mn = 0. Let Γ = SL2(L), and Γ(i) = SL2(L,mi).
Then
|Γ| = |Γ : Γ(1)||Γ(1) : Γ(2)| . . . |Γ(n− 2) : Γ(n− 1)||Γ(n− 1) : Γ(n)|
First, as m of index 2, Γ/Γ(1) ∼= PSL2(F2) ∼= S3, so |Γ : Γ(1)| = 6. Now, Γ(i− 1)/Γ(i) ∼= Z32
by lemma (4.2.9), so |Γ(i− 1) : Γ(i)| = 8. Hence |Γ| = 23(n−1)+13. Hence result.
88
Lemma 5.3.17. Let L be a commutative local ring with nilpotent maximal ideal m of index
2. Let N SL2(L). Then
o(N) = L⇔ |SL2(L) : N | = 2α
Proof. Let Γ denote SL2(L). Suppose first that o(N) = L. If Γ(m) 6 N then, as Γ/Γ(m) ∼=
S3, |Γ : N | = 1, or 2. So suppose that Γ(m) N and suppose that 3 | |Γ : N |. Now
NΓ(m) = Γ, or is of index 2 in Γ so 3 | |NΓ(m) : N | and 3 | |NΓ(m) : Γ(m)| so 9 | |Γ :
N ∩ Γ(m)|. So 9 | |Γ|, contradicting lemma (5.3.16). So 3 - |Γ : N | ie |Γ : N | = 2α.
Now suppose that o(N) 6= L. So N 6 H(m) = Γ(m) and |Γ : Γ(m)| = 6, so 3 | |Γ : N |.
Hence result.
Now observe the following consequence of (5.3.17)
Corollary 5.3.18. Let L be a commutative local ring with maximal nilpotent ideal m of
index 2. Let N SL2(L), o(N) = L and M ∈ SL2(L) of order 3. Then M ∈ N .
Lemma 5.3.19. Let L be a commutative local ring with nilpotent maximal ideal m of index
2. Let N SL2(L), o(N) = L. Then T 4 ∈ N , and T 2 ∈ N ⇔ −I ∈ N .
Proof. AT is of order 3 so AT ∈ N , so A ≡ T−1( mod N), and A4 = I, so T 4 ∈ N . Also
A2 = −I so T 2 ≡ −I( mod N).
Lemma 5.3.20. Let a ∈ L, u ∈ L∗. Then a u
−cu−1 −(1 + a)
is of order 3, where c = a2 + a+ 1 ∈ L∗.
Proof. Tedious calculation.
Lemma 5.3.21. Let L be a commutative local ring with nilpotent maximal ideal m, of index
2. Let α ∈ m and Let N SL2(L) be of order L. Then 1 4α
0 1
∈ N.89
Proof. By lemma (5.3.20) 1 u
−3u−1 −2
∈ Nfor every unit u. Let α ∈ m , u = 1 + α, and suppose that u−1 = 1 + β, some β ∈ m. Now 1 u
−3u−1 −2
=
1 u
0 1
A
1 u−1
0 1
A−1
1 3u
0 1
A
1 −u−1
0 1
A−1
=
1 1 + α
0 1
A
1 1 + β
0 1
A−1
1 3 + 3α
0 1
A
1 −1− β
0 1
A−1
= T
1 α
0 1
AT
1 β
0 1
A−1T 3
1 3α
0 1
AT−1
1 −β
0 1
A−1
and AT ∈ N , A−1T 3 ≡ A−1T−1 ≡ I( mod N), AT−1 ≡ ATT−2 ≡ T−2( mod N), so
T
1 4α+ β
0 1
T−2
1 −β
0 1
A−1 ∈ N.
so 1 4α
0 1
A−1T−1 ∈ N
and A−1T−1 ∈ N . Hence 1 4α
0 1
∈ Nas required.
Theorem 5.3.22. Let L be a commutative local ring with nilpotent maximal ideal m of index
2. Let N SL2(L) be of order L. Then 4L 6 l(N).
Proof. We show E2(L, 4L) 6 N . Recall that E2(L, 4L) is generated by conjugates of
E12(x) =
1 x
0 1
90
where x ∈ 4L. Let x ∈ 4L, so x = 4y, some y ∈ L. Now y ∈ L∗ or y ∈ m. Suppose that
y ∈ L∗, so y = 1 + α, where α ∈ m. So x = 4y = 4 + 4α, so
E12(x) =
1 x
0 1
=
1 4
0 1
1 4α
0 1
∈ N.If y ∈ m then similarly e12(x) ∈ N . So E2(L, 4L) 6 N , as required.
Since we are also interested in PSL2(R), for commutative rings R we consider normal
subgroups of SL2 which contain −I.
Lemma 5.3.23. Let L be a commutative local ring with nilpotent maximal ideal m of index
2. Let N SL2(L) have order L and let α ∈ m. Suppose that −I ∈ N . Then 1 2α
0 1
∈ Nand so 2L 6 l(N).
Proof. −I ∈ N so by (5.3.19), T 2 ∈ N . Now, by (5.3.20), ∀u ∈ L∗, 0 u
−u−1 −1
=
1 u
0 1
A
1 u−1
0 1
A−1
1 2u
0 1
∈ NLet α ∈ m, let u = 1 + α ∈ L∗ and suppose that u−1 = 1 + α′, α′ ∈ m. So that 1 1 + α
0 1
A
1 1 + α′
0 1
A−1
1 2 + 2α
0 1
∈ Nso
T
1 α
0 1
AT
1 α′
0 1
A−1T 2
1 2α
0 1
∈ Nbut, as AT , T 2 ∈ N , 1 α+ α′
0 1
1 2α
0 1
∈ N
91
Now, by (5.3.20), and since −I ∈ N , 1 u
−u−1 0
= A
1 u−1
0 1
A−1
1 u
0 1
∈ Nso
A
1 1 + α′
0 1
A−1
1 1 + α
0 1
∈ Nie
AT
1 α′
0 1
A−1T
1 α
0 1
∈ Nhence, as AT, T 2 ∈ N , we have 1 α+ α′
0 1
∈ N.Hence 1 2α
0 1
∈ N.Now using the fact that T 2 ∈ N and arguing in exactly the same way as in theorem (5.3.22),
we see that E2(L, 2L) 6 N , ie 2L 6 l(N).
Lemma 5.3.24. Let L be a commutative local ring with maximal nilpotent ideal m of index
2 and suppose that 2 = 0 in L. Let N SL2(L) have order L. Then
m(2) =< α2 : α ∈ m >6 l(N)
Proof. By lemma (5.3.20), and since 2 = 0, a u
−cu−1 1 + a
∈ Nand this matrix equals 1 u
0 1
A
1 u−1
0 1
A−1
1 au
0 1
A
1 −au−1
0 1
A−1
92
Now let α, β ∈ m, let u = 1 + α, a = 1 + β, and suppose that u−1 = 1 + α′, where α′ ∈ m.
Then au = 1 + α+ β + αβ, and au−1 = 1 + α′ + β + α′β, and so 1 1 + α
0 1
A
1 1 + α′
0 1
A−1
1 1 + α+ β + αβ
0 1
A
1 −(1 + α′ + β + α′β)
0 1
A−1
lies in N , and so, as AT ∈ N , and 2 = 0, we get 1 (α− α′)β
0 1
∈ NNow (1 +α)(1 +α′) = 1, so α+α′ +αα′ = 0, so, as 2 = 0, α−α′ = αα′, and α′ = α(1 +α′),
so α− α′ = α2u−1. So 1 α2u−1β
0 1
∈ NNow let x ∈ L, if x ∈ m then 1 α2(1 + α)−1x
0 1
∈ N.Now if x /∈ m then x = 1 + β, some β ∈ m, and 1 α2(1 + α)−1x
0 1
=
1 α2(1 + α)−1
0 1
1 α2(1 + α)−1β
0 1
Now 1 u
−u−1 0
= A
1 u−1
0 1
A−1
1 u
0 1
∈ Nby lemma (5.3.20). So using the fact that u = 1 + α, u−1 = 1 + α′, and AT ∈ N we get 1 α+ α′
0 1
∈ Nand so, as α+ α′ + αα′ = 0 we get 1 α2(1 + α)−1
0 1
∈ N93
and so 1 α2(1 + α)−1x
0 1
∈ N.Hence α2(1 + α)−1 ∈ l(N) ∀α ∈ m, and so α2 ∈ l(N) ∀α ∈ m. Hence result.
5.4 Order and level of a normal congruence subgroup
Throughout this section K shall be a Noetherian domain of Krull dimension one, so K has
primary decomposition and a Wohlfahrt theorem. Let N SL2(K), and suppose N is a
congruence subgroup. Let q∗ = l(N) 6= 0, and q = o(N). We ask how q∗ and q are related.
We are particularly interested in the case where q = K. The following 3 lemmas can be
proved using elementary group theory.
Lemma 5.4.1. Let G be any group and let A,B,C G. Then [BA,CA] 6 [B,C]A.
Lemma 5.4.2. Let A,B G. Then [A,B] 6 A ∩B.
Lemma 5.4.3. Let G be a group and let A,B,C G where C 6 A. Then A ∩ BC =
(A ∩B)(A ∩ C).
The following lemma is a generalization of [58] lemma 3.3.
Lemma 5.4.4. Let N SL2(K) = Γ, l(N) = q∗ = q∗1q∗2 6= 0, where q∗1 + q∗2 = K, and q1 is
primary. Then by (5.1.10), o(N) = q = q1q2, where q∗1 6 q1. Let N0 = (N ∩ Γ(q∗2))Γ(q∗1),
q′ = l(N0), and N = NΓ(q∗1). Then
1. o(N) = q1
2.[Γ, N
]6 N0
3. q′ = q∗1
94
Proof. Now
o(N) = o(NΓ(q∗1))
= o(N) + q∗1
= q + q∗1
= q1q2 + q∗1
= (q1 ∩ q2) + (q1 ∩ q∗1)
= q1 ∩ (q2 + q∗1) by the modular law (see [2] p.6)
= q1 ∩K = q1.
Now, as K has a Wohlfahrt theorem and q∗1 + q∗2 = K, we have, by (5.2.8), Γ = Γ(q∗1 + q∗2) =
Γ(q∗1)Γ(q∗2), so
[Γ, N
]= [Γ(q∗1)Γ(q∗2), NΓ(q∗1)]
6 [Γ(q∗2), N ] Γ(q∗1) by lemma (5.4.1)
6 N0 by lemma (5.4.2)
Now Γ(q∗1) 6 N0, so q∗1 6 l(N0) = q′. RTP q′ 6 q∗1. Now N0 is a congruence subgroup of
level q′, so by Wohlfahrt (5.2.3) Γ(q′) 6 N0, so
Γ(q′q∗2) = Γ(q′) ∩ Γ(q∗2) as q′ + q∗2 = K
6 N0 ∩ Γ(q∗2)
= ((N ∩ Γ(q∗2))Γ(q∗1)) ∩ Γ(q∗2)
= (N ∩ Γ(q∗2))Γ(q∗) by (5.4.3)
6 NΓ(q∗)
= N
So q′q∗2 6 l(N) = q∗1q∗2. Now q′ + q∗2 = q∗1 + q∗2 = K, so q′ ∩ q∗2 = q′q∗2 6 q∗1q
∗2 = q∗1 ∩ q∗2, and
95
q∗1 6 q′, so q∗1 ∩ q∗2 6 q′ ∩ q∗2. Hence q′ ∩ q∗2 = q∗1 ∩ q∗2. So
q′ = q′ ∩ (q∗1 + q∗2)
= (q′ ∩ q∗2) + q∗1
= (q∗1 ∩ q∗2) + q∗1
= q∗1.
The following lemma is a generalization of [58] lemma 3.4.
Lemma 5.4.5. Let NSL2(K), l(N) = q∗ = q∗0p∗ 6= 0, where p∗ is primary and q∗0+p∗ = K,
so by (5.1.10), o(N) = q = q0p, where p∗ 6 p. Let N0 = (N ∩Γ(q∗0))Γ(p∗), and N = NΓ(p∗).
Let L denote the local ring K/p∗. Let ϕ : SL2(K)→ SL2(L) be the natural homomorphism.
Let M0 = ϕ(N0), and M = ϕ(N). Then
1. M0 SL2(L)
2. l(M0) = 0
3. o(M) = p/p∗
4.[E2(L),M
]6 M0
Proof. 1 is obvious. For 4 recall that[SL2(K), N
]6 N0, on applying ϕ we see that[
SL2(L),M]
6 M0. Now as L is local so an SR2-ring we have E2(L) = SL2(L). For
3, as M = ϕ(N), and o(N) = p we get o(M) = p/p∗. Now suppose that l(M0) = a 6= 0. So
E2(L, a) = SL2(L, a) 6 M0, so E2(K, a) 6 N0. So, as l(N0) = p∗, we have a 6 p∗, so a = p,
so a = 0. Contradiction. Hence l(M0) = 0.
Lemma 5.4.6. ([57] lemma 2.1) Let L be any local ring. Let NSL2(L) such that o(N) = L.
Then ∃X ∈ N of the form ∗ ∗1 1
96
Proof. By (5.3.3)
∃M1 =
a b
c d
∈ Nsuch that c is a unit. Now 1 c−1d
0 1
a b
c d
1 −c−1d
0 1
=
∗ −c−1
c 0
= M2 ∈ N
Let
T =
1 −t
0 1
Then
[T,M2] = T−1M−12 TM2 =
1 + t2c2 t
tc2 1
∈ NChoose t = c−2 to get the result.
Let L be a local ring, with maximal ideal m, and residue field K = L/m. The following
theorem is a slight generalization of [58] theorem 2.2.
Theorem 5.4.7. Let L be a commutative local ring. Let N SL2(L),M 6 GL2(L) such
that N 6 M , and [E2(L),M ] 6 N . Let q = o(N), q∗ = l(N), q0 = o(M), so q∗ 6 q 6 q0.
Then q = q0, unless K = F2, and q0 6= L in which case mq0 6 q 6 q0.
Proof. Let
X =
a b
c d
∈MLet δ = ad−bc ∈ L∗. Then, as [E2(L),M ] 6 N , [T (r), X] ∈ N and so (d2−δ)rδ−1+r2cdδ−1 ∈
q ∀r ∈ L. So ∀u ∈ L∗, (d2−δ)δ−1+ucdδ−1 ∈ q, so (u−1)cd ∈ q. Suppose that q0 6 m. Then
c ∈ m, as o(M) = q0. Now ad − bc = δ ∈ L∗ and c ∈ m, so bc ∈ m. So if a or d ∈ m then
δ ∈ m. Contradiction. Hence a, d /∈ m ie a, d ∈ L∗. So, as (u− 1)cd ∈ q, (u− 1)c ∈ q. When
|K| > 2 we can choose u ∈ L∗ such that u − 1 ≡ 1 (mod q). So c ∈ q. Hence, by (5.3.2),
q0 = o(M) 6 q. So q0 = q. If |K| = 2 then u = 1 + α, where α ∈ m, so u − 1 = α ∈ m, so
αc ∈ q, so, by (5.3.2), mq0 6 q.
97
Now suppose that q0 = L, so ∃X0 ∈ N ⊆M of the form ∗ ∗1 1
So(u− 1)c = u− 1 ∈ q ∀u ∈ L∗. Thus if |K| > 2 then q = L = q0. Suppose that |K| = 2, let
α ∈ m, so u = 1 +α ∈ L∗, so α = u− 1 ∈ q, so m 6 q. Suppose that m = q and consider the
natural homomorphism f : SL2(L) → SL2(F2). Now o(N) = q so f(N) = 1, and X0 ∈ M
and [SL2(L),M ] 6 N , so f(X0) is central. But SL2(F2) has trivial centre, so f(X0) = I.
But X0 =
∗ ∗1 1
. So, under the map L→ F2, 1 7→ 0. Contradiction. Hence q = L.
We are now in a position to address the main problem of this chapter; to derive a
relationship between the order and level of a normal congruence subgroups of SL2 over a
Noetherian domain of Krull dimension one. We deal first with the case of a normal subgroup
of order K. This is for two reasons, first we are able to obtain a better bound in this case, in
fact our bound is best possible, and second, when we apply these results to the construction
of non-congruence subgroups of the Bianchi groups we do so by showing a large class of
normal subgroups has order Od,m.
Theorem 5.4.8. Let K be a Noetherian domain of Krull dimension one and suppose that
charK 6= 2, or 3. Let N SL2(K) be a congruence subgroup. Suppose that o(N) = K and
that l(N) = q∗. Then 12K 6 q∗.
Proof. Let q∗ = p∗1 . . . p∗t be a primary decomposition. Fix i ∈ 1, . . . , t and consider p∗i .
Let q∗0 = ∩j 6=ip∗j , p∗ = p∗i , and so we can apply (5.4.5) let q0 = p = K. Let Li = K/p∗i and
denote the maximal ideal of Li by mi. By an abuse of notation we also use mi to denote the
corresponding maximal ideal of K, ie r(p∗i ) = mi. Let ϕi : SL2(K)→ SL2(Li) be the natural
homomorphism. Let N0 = (N ∩ Γ(q∗0))Γ(p∗), N = NΓ(p∗). Let M0 = ϕi(N0), M = ϕi(N).
Then by lemma (5.4.5) l(M0) = 0, o(M) = p/p∗ = Li, and[E2(Li),M
]6 M0. Now by
theorem (5.4.7), as o(N) = K then o(M0) = o(M) = Li.
98
Suppose that 2 ∈ L∗i . Then, by (5.3.14) o(M0) = l(M0) unless |Li : mi| = 3, in which
case, by (5.3.15), mi 6 l(M0). If o(M0) = l(M0) then Li = 0 and so p∗i = K. If |Li : mi| = 3
and mi 6 l(M0) then Li is the field of 3 elements and 3K 6 p∗i .
Now suppose that 2 /∈ L∗i . If |Li : mi| > 3 then, by (5.3.14), l(M0) = o(M0) and so
p∗i = K. If |Li : mi| = 3 then, by (5.3.15), mi 6 l(M0) and so Li is the field of 3 elements
and 3K 6 p∗i . If |Li : mi| = 2 and 2 6= 0 in Li then, by (5.3.22), 4o(M0) 6 l(M0) and so
4K 6 p∗i . If |Li : mi| = 2 and 2 = 0 in Li then 2 ∈ p∗i so 2K 6 p∗i . So, in all cases 12K 6 p∗i .
Now q∗ = p∗1 . . . p∗t =
⋂ti=1 p∗i > 12K. Hence result.
Note from the proof above that the 12K comes from ideals of index 2 and ideals of index
3. When one or both of these are not present we get a better bound.
Corollary 5.4.9. Let K be a Noetherian domain of Krull dimension one. Let N SL2(K)
be a congruence subgroup. Suppose that o(N) = K and l(N) = q∗. Then
1. If K has no ideals of index 2, then 3K 6 q∗.
2. If K has no ideals of index 3 then 4K 6 q∗.
3. If K has no ideals of index 2 or 3 then q∗ = K.
Corollary 5.4.10. Let K be a Noetherian domain of Krull dimension one and suppose that
charK = 0. Let N SL2(K) be a congruence subgroup which contains −I. Suppose that
o(N) = K, and l(N) = q∗. Then 6K 6 q∗.
Proof. This follows from (5.4.8) and (5.3.23).
Lemma 5.4.11. Let K be a Noetherian domain of Krull dimension one. Let p ∈ Z be a
rational prime. Then K has only finitely many maximal ideals of index p.
Proof. Let m K be a maximal ideal of index p. Now ∀x ∈ K xp − x ∈ m. So consider the
ideal
q =∑x∈K
(xp − x)K
99
So q 6 m. Now, by (5.1.9), q = p1 . . . pt, where pi is primary, r(pi) = mi, a maximal
ideal; mnii 6 pi, for some ni; and pi + pj = K, for i 6= j. Suppose that m is distinct
from mi, i = 1, . . . , t. Now mnii 6 pi so K = mni
i + m 6 pi + m. So ∀i pi + m = K, so
m + q = m + (p1 . . . pt) = K. But q 6 m. Contradiction. So m = mi, some i = 1, . . . , t.
Hence result.
In light of this lemma make the following
Definition. Let K be a Noetherian domain of Krull dimension one. Let p ∈ Z be a rational
prime. Let
Mp =⋂m K : |K : m| = p
and if K has no ideals of index p letMp = K.
Lemma 5.4.12. Let R be a commutative ring of characteristic 2. Let q1, q2 R. Then
(q1q2)(4) = q
(4)1 q
(4)2 .
Proof. First recall that qi =< α4 : α ∈ qi > and q(4)1 q
(4)2 =< α4β4 : α ∈ q1, β ∈ q2 >. Let
α ∈ q1, and β ∈ q2 so α4β4 = (αβ)4 ∈ (q1q2)(4). So q
(4)1 q
(4)2 6 (q1q2)
(4).
Conversely let γ ∈ q1q2, so γ =∑αiβiri where αi ∈ q1, βi ∈ q2, ri ∈ R. So γ4 =
(∑αiβiri)
4 =∑α4
iβ4i r
4i as R is of characteristic 2. So γ4 ∈ q
(4)1 q
(4)2 . Hence result.
Recall that if charK = p, some prime p then any other rational prime, q is a unit in K.
Using this fact we get the following
Theorem 5.4.13. Let K be a Noetherian domain of Krull dimension one and suppose that
charK = p, for some prime p. Let N SL2(K) be a congruence subgroup. Suppose that
o(N) = K and l(N) = q∗. Then
1. If charK > 3 then q∗ = K.
2. If charK = 3 then M3 6 q∗ and if K has no ideals of index 3 then q∗ = K.
3. If charK = 2 then M22 6 q∗ and if K has no ideals of index 2 then q∗ = K.
100
Proof. Part 1 follows from (5.4.8). For part 2 consider the setup in the proof of theorem
(5.4.8). As charK = 3 then 2 is a unit in K and Li so, by (5.3.14), o(M0) = l(M0) unless
|Li : mi| = 3, in which case, by (5.3.15), mi 6 l(M0). If o(M0) = l(M0) then p∗i = K. If
|Li : mi| = 3 and mi 6 l(M0) then mi 6 p∗i . The result follows.
For part 3 we use the same setup as theorem (5.4.8) but in this case, as charK = 2 then
2 /∈ L∗i . Now if |Li : mi| > 3 then p∗i = K, as before. Also |Li : mi| 6= 3 for if |Li : mi| = 3
then 3 ∈ mi but as charK = 2 then 3 is a unit. Now suppose that |Li : mi| = 2, as 2 = 0 we
have, by (5.3.24) m(2)i 6 l(M0) and so m
(2)i 6 p∗i . The result follows.
We now deal with case of a normal congruence subgroup of order di stinct from K.
Theorem 5.4.14. Let K be a Noetherian domain of Krull dimension one and suppose that
charK 6= 2, or 3. Let N SL2(K) be a congruence subgroup. Suppose that o(N) = q 6= K,
and l(N) = q∗. Then 48q 6 q∗.
Proof. Let q∗ = p∗1 . . . p∗t be a primary decomposition. Then, by (5.1.10) q = p1 . . . pt, where
p∗i 6 pi. Fix i ∈ 1, . . . , t and consider p∗i . Let q∗0 = ∩j 6=ip∗j , p∗ = p∗i , q0 = ∩j 6=ipj, and
let p = pi. Let Li = K/p∗i and let ϕi : SL2(K) → SL2(Li) be the natural homomorphism.
Let N0 = (N ∩ Γ(q∗0))Γ(p∗), N = NΓ(p∗). Let M0 = ϕi(N0), M = ϕi(N). Then by lemma
(5.4.5) l(M0) = 0, o(M) = p/p∗, and[E2(Li),M
]6 M0.
First suppose that |Li : mi| > 2, so by theorem (5.4.7) o(M0) = o(M). Now suppose that
2 ∈ L∗i . So, by (5.3.7), and (5.3.14), o(M0) = l(M0) = 0 unless |Li : mi| = 3 and o(M0) = Li,
in which case, by (5.3.15), mi 6 l(M0). If o(M0) = l(M0) then pi = p∗i . If |Li : mi| = 3
and o(M0) = Li then mi 6 p∗i ie 3pi 6 p∗i . Now suppose that 2 /∈ L∗i . If o(M0) = Li
and |Li : mi| > 3 then, by (5.3.14), l(M0) = Li and so p∗i = K. Also |Li : mi| 6= 3 for if
|Li : mi| = 3 then 3 ∈ mi and, as 2 ∈ mi, 1 = 3−2 ∈ mi. Suppose that o(M0) 6= Li and 2 6= 0
in Li. Then as |Li : mi| > 2, by (5.3.12), 2o(M0) 6 l(M0) = 0, and so 2pi 6 p∗i . Finally if
2 = 0 in Li then 2 ∈ p∗i , so 2pi 6 p∗i .
Now suppose that |Li : mi| = 2. So if o(M) = Li ( ie pi = K ) then, by (5.4.7),
o(M0) = o(M) and if o(M) 6= Li ( ie pi 6= K ) then, by (5.4.7), mio(M) 6 o(M0). Now
|Li : mi| = 2 so 2 /∈ L∗i and 2 6= 0. If o(M) = Li then, by (5.4.7), o(M) = o(M0) and, by
101
(5.3.22), 4o(M0) 6 l(M0) so 4pi 6 p∗i . If o(M) 6= Li ( and so o(M0) 6= Li ) then, by (5.4.7),
mio(M) 6 o(M0) and, by (5.3.6), 8o(M0) 6 l(M0) so 16pi 6 p∗i . If 2 = 0 in Li then 2 ∈ p∗i ,
so 2pi 6 p∗i .
So in all cases 48pi 6 p∗i and so, as before 48q 6 q∗.
Again, when ideals of index 2 or 3 are not present we can improve the bound
Corollary 5.4.15. Let K be a Noetherian domain of Krull dimension one and suppose that
charK 6= 2, or 3. Let N SL2(K) be a congruence subgroup. Suppose that o(N) = q 6= K,
and l(N) = q∗. Then
1. If K has no ideals of index 2 then 6q 6 q∗. In particular if 6 is a unit then q∗ = q.
2. If K has no ideals of index 3 then 16q 6 q∗.
3. If K has no ideals of index 2 or 3 then 2q 6 q∗.
Example 5.4.1. Let K = Z[
16
]. So charK = 0, and 6 is a unit in K further, K is a
Noetherian domain of Krull dimension one, in fact K is a Dedekind domain of arithmetic
type. Recall the material in the section on number theory (1.3). Let p ∈ Z be prime and
define
vp : Q→ R
as follows. Let x ∈ Q, so x = pα ab
where p - a, and p - b. Then let vp(x) = p−α. Let v∞
denote the usual Archimedean valuation and let S = v∞, v2, v3 and form OS as indicated
in section (1.3). We claim that K = OS and so is a Dedekind domain of arithmetic type.
Let x ∈ OS, so x = a/b where a, b ∈ Z are coprime. Suppose that a prime p 6= 2, 3 divides b,
b = pγb′, γ > 0, p - b′ say. Now a and b are coprime so p - a, so x = p−γ ab′, so vp(x) = pγ > 1,
as γ > 0, so x /∈ OS. Contradiction. Hence OS ⊆ K. Now consider a2α3β ∈ K and let p 6= 2, 3
be prime. So vp(a
2α3β ) = vp(pγa′
2α3β ) = p−γ 6 1, as γ > 0, so a2α3β ∈ OS. So K ⊆ OS and so K
is a Dedekind domain of arithmetic type.
102
Theorem 5.4.16. Let K be a Noetherian domain of Krull dimension one and suppose that
charK = p, for some prime p. Let N SL2(K) be a congruence subgroup. Suppose that
o(N) = q 6= K, and l(N) = q∗. Then
1. If charK > 3 then q∗ = q.
2. If charK = 3 then qM3 6 q∗.
3. If charK = 2 then q(4)M(4)2 6 q∗.
Proof. Part 1 is obvious. For part 2 we use the same set up as before. As charK = 3
then 2 ∈ K∗ and so, by examining the above proofs it is obvious that pi = p∗i unless
|Li : mi| = 3 and o(M0) = Li, in which case, by (5.3.15), mi 6 l(M0). In this case mi 6 p∗i .
Hence q∗ = p∗1 . . . p∗t =
⋂p∗i > qM3.
Now suppose that charK = 2 so that 2 = 0 and 2 /∈ L∗i . Again we use the usual setup.
First suppose that |Li : mi| > 3 so, by theorem (5.4.7), o(M0) = o(M). If o(M0) = Li
( ie pi = K ) then, by (5.3.14), l(M0) = o(M0) and so p∗i = pi = K. If o(M0) 6= Li (ie
pi 6= K) then, by (5.3.13) o(M0)(2) 6 l(M0) and so p
(2)i 6 p∗i . Now |Li : mi| 6= 3 for if
|Li : mi| = 3 then 3 ∈ mi and 3 is a unit. Now suppose that |Li : mi| = 2. If o(M) = Li (ie
pi = K) then, by (5.4.7), o(M0) = o(M) = Li, so by (5.3.24), m(2)i 6 l(M0), so m
(2)i 6 p∗i .
If o(M) 6= Li (so o(M0) 6= Li) then, by (5.4.7), mio(M) 6 o(M0) and so, as o(M0) 6= Li,
by (5.3.8) o(M0)(4) 6 l(M0) so, applying (5.4.12), m
(4)i p
(4)i = (mipi)
(4) 6 p∗i . It follows that
q∗ =⋂
p∗i 6M(4)2
⋂p
(4)i = q(4)M(4)
2 (again using (5.4.12)).
In [57] Mason dealt with the case of a normal subgroup of SL2(L) where L is a com-
mutative local ring with principal maximal ideal m of index 2. This meant that in [58] he
was only able to derive a relationship between the order and level of a normal congruence
subgroup of SL2 over a Dedekind domain. By dropping this condition in section (5.3) we
were able to generalize Mason’s work to a normal congruence subgroup of SL2 over any
Noetherian domain of Krull dimension one. For comparison we state Mason’s results
103
Theorem. ([58] theroem 3.6) Let A be a Dedekind domain of arithmetic type (see section
(1.3)) and suppose that A is contained in a number field and is not totally imaginary. Let
N SL2(A) be a congruence subgroup. Suppose that o(N) = q, and l(N) = q∗ 6= 0. Then
1. If A has no ideals of index 2 or 3 and 2 is unramified in A then q = q∗.
2. If A has no ideals of index 2 and 2 is unramified in A then 3q 6 q∗.
3. If A has no ideals of index 2 or 3 then 2q 6 q∗.
4. If A has no ideals of index 2 then 6q 6 q∗.
5. If A has no ideals of index 3 then 4q 6 q∗.
6. Otherwise, 12q 6 q∗.
Theorem. ([58] theorem 3.14) Let A be a Dedekind domain of arithmetic type and suppose
that A is contained in a function field in one variable over a finite field k. Let N SL2(A)
be a congruence subgroup. Suppose that o(N) = q, and l(N) = q∗ 6= 0. Then
1. If charK > 3 or |k| = 3α, where α > 1 then q = q∗.
2. If |k| = 3 then qM3 6 q∗.
3. If |k| = 2α, where α > 1 then q2 6 q∗.
4. If |k| = 2 then q2M22 6 q∗.
So when N has order K we get the same results as Mason. However when N has order
distinct for K our results are not as good as Mason’s. It is clear from comparing the above
with our theorems that the problems arise when K has an ideal of index 2 and when the
characteristic is 2. It is not known whether our results are best possible or not. Our results
are definitely an extension of Mason’s for consider the following argument
Lemma 5.4.17. Let L be a commutative local ring with principal maximal ideal m. Then if
q L such that mi+1 6 q 6 mi then q = mi+1, or mi.
104
Proof. Suppose that m = αL, so mi = αiL. Let k = L/m, so mi+1/mi is a k-vector space.
Let x + mi+1 ∈ mi/mi+1, where x ∈ mi, so x = αiy, where y ∈ L. Let y = y + m ∈ k. So
y(αi + mi+1) = x + mi+1. So mi/mi+1 is generated by αi + mi+1 as a k-vector space and so
is one dimensional.
Suppose that mi+1 6 q 6 mi. So q/mi+1 is a subspace of mi/mi+1. So, as mi/mi+1 is one
dimensional q = mi, or mi+1.
Lemma 5.4.18. Let L be a commutative local ring with principal maximal ideal m. Suppose
that⋂∞
i=1 mi = 0. Then ∀ 0 6= q L, q = mα, some α > 1.
Proof. ∃α ∈ N such that q 6 mα, otherwise q = 0. Consider q + mα+1. By (5.4.17),
q + mα+1 = mα+1, or mα. If q + mα+1 = mα+1 then q 6 mα+1, contradicting the maximality
of α. So q + mα+1 = mα. Now let M = mα/q, so M is a finitely generated L-module. Then
mM = (q + mα+1)/q = mα/q = M . So, by Nakayama’s lemma ( see [2] page 21 ), M = 0, ie
q = mα.
Now, again consider O = O3,2 = Z + i√
3Z. Let m = (2, 1 + i√
3). Then, as in example
(5.1.2), m is a maximal ideal of O of index 2 and m2 < 2O < m. Let L = O/mn, any n > 2,
so in L the ideal 2L is not a power of the maximal ideal and so L has a maximal ideal of
index 2 which is not principal. In a Dedekind domain ideals are 32-generated. That is let q
be an ideal in a Dedekind domain and let x ∈ q, then ∃y ∈ q such that q is generated by
x and y. As a consequence of this every local image of a Dedekind domain has a principal
maximal ideal.
Example 5.4.2. In this chapter we have derived a relationship between the order and level
of a normal congruence subgroup. What about non-normal congruence subgroups? Here
unfortunately there is no relationship. Let d, p ∈ Z such that p is a rational prime and
d > 0 is square free. Consider the groups SL2(Od,p). By (3.2.1), SL2(Od,p) 6 SL2(Od),
and clearly SL2(Od, pOd) 6 SL2(Od,p) so SL2(Od,p) is a non-normal congruence subgroup
of SL2(Od). However A, T ∈ SL2(Od,p), so o(SL2(Od,p)) = Od and Od,p/pOd∼= Fp, so
l(SL2(Od,p)) = pOd.
105
Chapter 6
Non-congruence subgroups of the
Bianchi groups
We now apply the results of the previous chapter to the Bianchi groups PSL2(Od,m). We take
two view points. First we look at the growth of non-congruence subgroups. Then we look at
normal subgroups of small index, showing that all but finitely many normal congruence sub-
groups are of index divisible by 6. Further we classify the normal congruence subgroups with
torsion in PSL2(Od), d = 7, 11, 19, and PSL2(O1,2) and the normal congruence subgroups of
index not divisible by 6 in PSL2(O2) and PSL2(O3,2). Normal subgroups of PSL2(Od,m) of
index not divisible by 6 all have torsion. In [25] Fine shows that for d = 2, 7, 11 all principal
congruence subgroups of PSL2(Od) are torsion free with the exception of PSL2(O2, ωO2).
Let xi ∈ G, we use the notation N(xi) to denote the normal closure of the elements xi in G.
6.1 Counting finite index subgroups
In this section we introduce some basic notions about counting subgroups of finite index.
The material has been lifted from the survey article of Lubotzky [46]. Let G be any group.
Let an(G) denote the number of subgroups of G of index exactly n. Ideally we would like to
get a formula for an(G). The first attempt to do this was by M. Hall in 1949 [35] in which
106
a recursive formula for the number of subgroups of finite index in a free group of finite rank
was given. This method was extended and simplified by Dey [20] and Wohlfahrt [92].
Proposition 6.1.1. [46] Let tn(G) be the number of transitive permutation representations
of G on n objects. Then
an(G) =tn(G)
(n− 1)!
Proof. Let H 6 G be a subgroup of index n. G acts in an obvious way on G/H. Identify
G/H with the set 1, . . . , n such that H is identified with 1. There are (n−1)! ways of doing
this. Each of these gives rise to a homomorphism ϕ : G→ Sn such that ϕ(G) is transitive and
Stab(1) = γ ∈ G : ϕ(γ)(1) = 1 = H. Conversely every transitive permutation representa-
tion of G on n objects give a subgroup Stab(1) of index n. Thus (n− 1)!an(G) = tn(G).
Let hn(G) denote the number of homomorphisms from G to Sn.
Lemma 6.1.2. [46]
hn(G) =n∑
k=1
(n− 1
k − 1
)tk(G)hn−k(G)
Proof. Suppose that the orbit of 1 is of length k, 1 6 k 6 n. There are
(n− 1
k − 1
)ways
of choosing such an orbit, tk(G) ways of acting on it, and hn−k(G) ways of acting on its
complement.
Theorem 6.1.3. [46] Let G be any group then
an(G) =1
(n− 1)!hn(G)−
n−1∑k=1
1
(n− k)!hn−k(G)ak(G).
107
Proof.
hn(G) =n∑
k=1
(n− 1
k − 1
)tk(G)hn−k(G) by (6.1.2)
=n∑
k=1
(n− 1
k − 1
)(k − 1)!ak(G)hn−k(G) by (6.1.1)
=n∑
k=1
(n− 1)!
(n− k)!ak(G)hn−k(G)
= (n− 1)!an(G) +n−1∑k=1
(n− 1)!
(n− k)!ak(G)hn−k(G).
Therefore
an(G) =1
(n− 1)!hn(G)−
n−1∑k=1
1
(n− k)!hn−k(G)ak(G).
Now consider the free group on r generators. Since each generator of Fr can be mapped
to one of n! elements of Sn we see that hn(Fr) = (n!)r. Thus
Theorem 6.1.4. [35] Let Fr be the free group on r generators. Then
an(Fr) = n(n!)r−1 −n−1∑k=1
[(n− k)!]r−1 ak(Fr).
From this it follows [71] that
Theorem 6.1.5.
an(Fr) ∼ n(n!)r−1
In [20] Dey extended this work to free products
Theorem 6.1.6. Let G = ∗ni=1Ai be a free product of groups. Let hin = hn(Ai). Then
an(G) =1
(n− 1)!
(r∏
i=1
hin
)−
n−1∑k=1
1
(n− k)!ak(G)
(r∏
i=1
hin−k
).
108
Proof. This follows from (6.1.2) and the fact that hn(G) =∏n
i=1 hin.
Recall that the Modular group PSL2(Z) = Z2 ∗ Z3. So we can apply the above theorem
from which we get
Proposition 6.1.7. [71]
an(PSL2(Z)) ∼ (12πe1/2)−1/2 exp
(n log n
6− n
6+ n1/2 + n1/3 +
log n
2
).
6.2 The growth of non-congruence subgroups in the
groups SL2(Od,m)
Recall the following results
Theorem. (2.2.3) There exists a surjective homomorphism
SL2(Od,m) −→ Fr
where r = r(d,m) the rank of the Zimmert Set and a, t lie in the kernel.
Theorem. (5.4.8) Let N SL2(Od,m) be a congruence subgroup of order Od,m. Then
SL2(Od,m, 12Od,m) 6 N .
Recall that Zimmert’s theorem was not best possible; r(5) = 1, but SL2(O5) F2.
So, as before, let ρ(d,m) denote the largest rank of a free quotient of SL2(Od,m). Let FK
denote the kernel of the map from SL2(Od,m) onto Fρ(d,m). Let N SL2(Od,m) containing
FK. As t ∈ FK then N has order Od,m. Therefore if N is a congruence subgroup it must
contain SL2(Od,m, 12Od,m). Observe that S 6 SL2(Od,m) is a congruence subgroup if and
only if coreS is a congruence subgroup, where by the core of S we mean the largest normal
subgroup contained in S. Thus
Proposition 6.2.1. Let S 6 SL2(Od,m) and suppose that FK 6 S. Then with finitely many
exceptions S is a non-congruence subgroup.
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Using Newmans result (6.1.5) we get
Theorem 6.2.2. Asymptotically, the number of non-congruence subgroups in the group,
SL2(Od,m), of index precisely n is at least n(n!)ρ−1 where ρ = ρ(d,m) > r(d,m), the rank of
the Zimmert Set.
The vast number of non-congruence subgroups in the above theorem all come from the
largest free quotient, all have order Od,m and all contain elements of finite order. Are most
non-congruence subgroups like this? It would be interesting to know about the growth of
non-congruence subgroups of order q 6= Od,m, or of torsion free non-congruence subgroups.
We now take the opposite view and look at normal subgroups of the Bianchi groups
PSL2(Od,m) of small index with the aim of determining whether they are congruence or
non-congruence. First, we have to take a look at SL2(L) where L is a local image of Od,m.
6.3 SL2(L) where L is a local image of Od,m
Let L be a finite local homomorphic image of Od,m with maximal ideal of index 2. Then we
can apply (5.3.22) to get
Theorem 6.3.1. Let L be a finite local homomorphic image of Od,m with a maximal ideal
m of index 2. Let N SL2(L) be of order L. Then 4L 6 l(N).
We now classify all normal subgroups of SL2(L) with order L by means of the commutator
subgroup viz
Theorem 6.3.2. Let N SL2(L). Then o(N) = L⇔ SL2(L)′ 6 N .
Now o(SL2(L)′) = L, as
[A, T ] =
0 −1
1 0
,
1 1
0 1
=
1 −1
−1 2
so (⇐) is obvious. First we describe the ideals of Od,m of index 2.
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Theorem 6.3.3. If m ≡ 0 (mod 2) then Od,m has exactly one ideal of index 2, namely
(2,mω). If m ≡ 1 (mod 2) then the ideals of index 2 are:
(2, 1 +mω) if d ≡ 1 (mod 4)
(2,mω) if d ≡ 2 (mod 4)
(2,mω), (2, 1 +mω) if d ≡ 7 (mod 8)
none if d ≡ 3 (mod 8)
Proof. The m ≡ 0 (mod 2) case follows from lemma (3.1.4). If m ≡ 1 (mod 2) then we use
the isomorphism given in lemma (3.1.5) ie Od,m/2Od,m∼= Od/2Od where mω ↔ ω. So we
can assume that m = 1. The generators of the maximal ideals can now be found using the
theorem on page 107 of [52]. Let
fω(X) =
X2 + d if d ≡ 1, 2 (mod 4)
X2 −X + d+14
if d ≡ 3 (mod 4)
We decompose fω (mod 2) then if gi is a factor of fω (mod 2), an ideal of index 2 in Od
is given by (2, gi(ω)) (see p. 107 of [52]). First, if d ≡ 3 (mod 8) then D ≡ 5 (mod 8), so
χK(2) = −1 (see chapter 3 for the definition) so Od/2Od is a field and Od,m has no ideals of
index 2. Now suppose that d ≡ 1, 2 (mod 4) so
fω(X) = X2 + d ≡
(X + 1)2 (mod 2) if d ≡ 1 (mod 4)
X2 (mod 2) if d ≡ 2 (mod 4)
So if d ≡ 1 (mod 4) then (2, 1 + mω) is the only ideal of index 2 in Od,m and if d ≡ 2
(mod 4) then (2,mω) is the only ideal of index 2 in Od,m. Now suppose d ≡ 7 (mod 8), so
(d+ 1)/4 ≡ 0 (mod 2), so fω(X) ≡ X2 −X ≡ X(X + 1) (mod 2). So Od,m has exactly two
ideals of index 2, namely (2,mω), and (2, 1 +mω).
We compute SL2(L)ab using a method developed by P. M. Cohn in [15] which we now
describe.
Theorem 6.3.4. Let L be a commutative local ring. Then
SL2(L)ab ∼= (L/M)+
111
where M is the additive subgroup of L generated by (u2 − 1)x : x ∈ L, u ∈ L∗ and 3(u+ 1)(v + 1) : u, v ∈ L∗,
and (L/M)+ is understood to be an additive group. The isomorphism is given by
E(x) =
x 1
−1 0
7−→ x− 3 (mod M)
Proof. This follows from [15] theorem 2, [14] theorem 4.1, and the fact that SL2(L) = E2(L)
[3].
First, ignore the case d ≡ 7 (mod 8) and m ≡ 1 (mod 2) and let L denote the local ring
Od,m/4Od,m.
Lemma 6.3.5. If m ≡ 0 (mod 2) then SL2(L)ab ∼= Z2 × Z4.
Proof. L = r +mωs : r, s = 0, 1, 2, 3, and it is easy to show that
L∗ = r +mωs : r ≡ 1 (mod 2). Let u ∈ L∗ then u2 − 1 = 0, or 2mω. Now 2mω(r +
mωs) = 0, or 2mω. So (u2 − 1)x = 0, 2mω. Let u, v ∈ L∗, it is trvial to show that
3(u+ 1)(v + 1) ∈ 0, 2mω. So M = 0, 2mω, so (L/M)+ ∼= Z2 × Z4. Hence result.
Similarly we can prove:
Lemma 6.3.6. Suppose that d ≡ 1, 2 (mod 4) and m ≡ 1 (mod 2). Then
SL2(L)ab ∼= Z2 × Z2.
Consider the following two matrices
T =
1 1
0 1
, U =
1 mω
0 1
Under Cohn’s map SL2(L) SL2(L)ab they behave as follows
T = E(−1)E(0)−1 7→ (−1− 3)− (0− 3) = −1
U = E(−mω)E(0)−1 7→ (−mω − 3)− (0− 3) = −mω
and so, from the proofs of the above lemmata we see that SL2(L)ab is generated by the images
of T and U ie if m ≡ 0 (mod 2) then SL2(L)ab ∼= Z2×Z4 =< U, T ;U2, T 4, TU = UT > and
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if d ≡ 1, 2 (mod 4) and m ≡ 1 (mod 2) then SL2(L)ab ∼= Z2 × Z2 =< U, T ;U2, T 2, TU =
UT >. It follows from this that only one of the normal subgroups of SL2(L) of index 2
contains T . We will use this observation in the next section and we record it as a lemma.
Lemma 6.3.7. Suppose m ≡ 0 (mod 2) or d ≡ 1, 2 (mod 4) and m ≡ 1 (mod 2). Then
∃N SL2(L) of index 2 which does not contain T .
We now deal with the case d ≡ 7 (mod 8) and m ≡ 1 (mod 2). In this case Od,m has
two ideals of index 2, so Od,m/4Od,m is not local. Instead we let m be one of the ideals of
index 2 and L = Od,m/m3. We need to describe m3.
Lemma 6.3.8. If m = (2,mω) then m3 = (8,mω), (8, 4 +mω), or (8,±2 +mω).
Proof. First we calculate generators for m2. Clearly m2 = (4, 2mω, (mω)2). Now (mω)2 =
m(mω) −m2 d+14
. Now d ≡ 7 (mod 8) so (d + 1)/4 ≡ 0, 2 (mod 4) and m ≡ 1 (mod 2) so
m2 ≡ 1 (mod 4) so
(mω)2 =
mω (mod m2) if d ≡ 15 (mod 16)
2 +mω (mod m2) if d ≡ 7 (mod 16)
and 2(2 +mω) ≡ 2mω (mod 4) so m2 = (4,mω), or (4, 2 +mω).
Now m3 = mm2. First suppose that m2 = (4,mω), so d ≡ 15 (mod 16). So m3 =
(8, 2mω, (mω)2). Again (mω)2 = m(mω)−m2(d+ 1)/4. Now m ≡ 1 (mod 2) so m(mω) ≡
mω (mod m3). Since m2 ≡ 1 (mod 8) and (d + 1)/4 ≡ 0, 4 (mod 8), so m2(d + 1)/4 ≡
0, 4 (mod 8). So (mω)2 ≡ mω, or 4 + mω (mod m3). So m3 = (8,mω), or (8, 4 + mω).
Now suppose that m2 = (4, 2 + mω). So m3 = (8, 4mω, 2(2 + mω),mω(2 + mω)). Now
2×2(2+mω) ≡ 4mω, and we can show that (mω)2 ≡ ±2+mω. Also 2(±2+mω) ≡ 4+2mω.
So m3 = (8,±2 +mω). Hence result.
In a similar way we can prove
Lemma 6.3.9. If m = (2, 1 +mω) then m3 = (8,±1 +mω).
Lemma 6.3.10. Suppose that d ≡ 7 (mod 8) and m ≡ 1 (mod 2). Let mOd,m be an ideal
of index 2, so m = (2,mω), or (2, 1 +mω). Let L = Od,m/m3. Then SL2(L)ab ∼= Z4.
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Proof. Suppose first that m = (2,mω), so m3 = (8,mω), (8, 4 + mω), or (8,±2 + mω).
Suppose r+mωs ∈ L, we can assume that 0 6 r, s 6 7. Then r+mωs ≡ r+mωs−s(mω) = r
(mod (8,mω)), or r+mωs ≡ r+mωs−s(4+mω) = r−4s (mod (8, 4+mω)), or r+mωs ≡
r + mωs − s(±2 + mω) = r ± 2s (mod (8,±2 + mω)). So L = 0, 1, 2, 3, 4, 5, 6, 7 ∼= Z8.
Applying Cohn’s method we see that SL2(L)ab ∼= Z4.
Now suppose that m = (2, 1 +mωs), so m3 = (8,±1 +mω). As above we can show that
L = 0, 1, 2, 3, 4, 5, 6, 7 and so SL2(L)ab ∼= Z4.
We now present the proof of theorem (6.3.2) as a series of lemmas.
Lemma 6.3.11. Suppose that d ≡ 7 (mod 8) and m ≡ 1 (mod 2). Let N SL2(L). Then
o(N) = L⇔ SL2(L)′ 6 N .
Proof. Let Γ denote SL2(L). Now o(Γ′) = L, so (⇐) is obvious. Conversely suppose that
o(N) = L and Γ′ N . So, by (5.3.17), |Γ : N | = 2α and Γ/N is non-abelian, so α > 3. So
Γ/N is a 2-group of order > 8, so ∃MΓ such that N 6 M and |Γ : M | = 8, and o(M) = L.
Now Γab = Z4, so Γ/M is non-abelian, so Γ/M ∼= Q8, or D8, the only non-abelian groups
of order 8 [68]. Now Qab8∼= Dab
8∼= Z2 × Z2, so (Γ/M)ab ∼= Z2 × Z2. But Γ Γ/M , so
Γab (Γ/M)ab, so Z4 Z2 × Z2. Contradiction. Hence result.
Unfortunately this technique does not work in the other cases. Instead we use the ob-
servation (5.3.18) that if N SL2(L) is of order L then N contains every matrix in SL2(L)
of order 3. Let N(3) denote the normal closure of all elements of order 3. We attempt to
show that SL2(L)′ = N(3). Note that because we have shown (5.3.22) that if o(N) = L
then SL2(L, 4L) 6 N we can work mod 4. Recall that SL2(L) = E2(L) and let
E12(x) =
1 x
0 1
, E21(y) =
1 0
y 1
.
So SL2(L) is generated by E12(x) and E21(y) where x, y = 1,mω. It is easy to compute that
[E12(x), E21(y)] =
1 + xy + (xy)2 −x2y
xy2 1− xy
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we want to show that [E12(x), E21(y)] ∈ N(3) for (x, y) = (1, 1), (1,mω), (mω, 1), (mω,mω).
We make frequent use of lemma (5.3.20) without comment.
Lemma 6.3.12. [E12(1), E21(1)]3 = I.
Lemma 6.3.13. Suppose that d ≡ 1, 2 (mod 4) and m ≡ 0 (mod 2). Then SL2(L)′ = N(3).
Proof. In this case (mω)2 = −dm2 = 0 as m ≡ 0 (mod 2) and we are working mod 4. This
shows that [E12(mω), E21(mω)] = I. Now
[E12(1), E21(mω)] =
1 +mω −mω
0 1−mω
=
−1 1
−1 0
0 −(1−mω)
1 +mω −1
.
Similarly for [E12(mω), E21(1)].
Lemma 6.3.14. Suppose that d ≡ 3 (mod 4) and m ≡ 0 (mod 2). Then SL2(L)′ = N(3).
Proof. First note that if m ≡ 0 (mod 4) then (mω)2 = 0, and if m ≡ 2 (mod 4) then
(mω)2 = 2mω. This renders the m ≡ 0 (mod 4) case identical to the d ≡ 1, 2 (mod 4) cases
above. So suppose m ≡ 2 (mod 4). Then
[E12(1), E21(mω)] =
1−mω −mω
2mω 1−mω
and this equals −1 −1
1 0
2mω 1−mω
−(1 +mω) −1 + 2mω
∈ N(3)
Similarly [E12(mω), E21(1)] ∈ N(3). Now
[E12(mω), E21(mω)] =
1 + 2mω 0
0 1 + 2mω
and this equals −1 +mω −1
1 +mω −mω
−1 −1 +mω
1−mω 0
−1 −1 + 2mω
1 + 2mω 0
∈ N(3)
115
Lemma 6.3.15. Suppose that d ≡ 1 (mod 4) and m ≡ 1 (mod 2). Then SL2(L)′ = N(3).
Proof. In this case (mω)2 = −1, so
[E12(mω), E21(1)] =
mω 1
mω 1−mω
and this equals −1 1
−1 0
2 0 −mω
−mω −1
∈ N(3).
similarly [E12(mω), E21(1)] ∈ N(3). Also
[E12(mω), E21(mω)] =
1 −mω
mω 2
so [E12(mω), E21(mω)]3 = I. Hence result.
Lemma 6.3.16. Suppose that d ≡ 2 (mod 4) and m ≡ 1 (mod 2). Then T 2 ∈ N(3), and
so −I ∈ N(3).
Proof. By lemma (5.3.20) 0 1 +mω
3 +mω −1
∈ N(3)
and this equals 1 1 +mω
0 1
A
1 3 +mω
0 1
A−1
1 2 + 2mω
0 1
= T
1 mω
0 1
AT
1 2 +mω
0 1
A−1T−1
1 3 + 2mω
0 1
so, as AT,A−1T−1 ∈ N(3), and 4 = 0, we see that T 2 ∈ N(3).
Lemma 6.3.17. Suppose that d ≡ 2 (mod 4) and m ≡ 1 (mod 2). Then SL2(L)′ = N(3).
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Proof. In this case (mω)2 = 2, so
[E12(1), E21(mω)] =
−1 +mω −mω
2 1−mω
and this equals −1 1−mω
1 +mω 0
2 + 2mω 1− 2mω
1 1 + 2mω
∈ N(3)
Similarly [E12(mω), E21(1)] ∈ N(3). However
[E12(mω), E21(mω)] =
−1 2mω
2mω −1
and this equals
−I
2mω 1
−1 + 2mω −1 + 2mω
−1 −1 + 2mω
1 + 2mω 0
∈ N(3)
Hence result.
We have now proved theorem (6.3.2). From the above we can get an improved lower
bound for the level in some cases
Theorem 6.3.18. Suppose that d ≡ 1, 2 (mod 4) and m ≡ 1 (mod 2). Let L = Od,m/4Od,m.
Let N SL2(L), o(N) = L. Then 2L 6 l(N).
6.4 Non-congruence subgroups of small index in the
Bianchi groups
Here we need to be very clear about the differences between SL2 and PSL2 so we restate
the relevant definitions. Let O = Od,m be an order in an imaginary quadratic number field.
Let 0 6= q O.
SL2(O, q) = M ∈ SL2(O) : M ≡ I (mod q)
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SL2(O, q) is the principal congruence subgroup of level q. Let S 6 SL2(O). We say that S
is a congruence subgroup if SL2(O, q) 6 S for some q O. If q is the largest O-ideal such
that SL2(O, q) 6 S we say that S is of level q. Otherwise S is a non-congruence subgroup.
Let
ϕ : SL2(O) −→ PSL2(O)
be the natural homomorphism. Let q O, then
PSL2(O, q) = ϕ(SL2(O, q))
PSL2(O, q) is the principal congruence subgroup of level q. Let S 6 PSL2(O). We say that
S is a congruence subgroup if PSL2(O, q) 6 S, for some q O. If q is the largest O-ideal
such that PSL2(O, q) 6 S we say that S is of level q. Otherwise S is a non-congruence
subgroup.
Lemma 6.4.1. Let N SL2(O) such that −I ∈ N . Let N = ϕ(N). Then
N is a congruence subgroup ⇔ N is a congruence subgroup
Further N and N have the same level.
Proof. Suppose that N is a congruence subgroup. So SL2(O, q) 6 N , some 0 6= q O. So
PSL2(O, q) = ϕ(SL2(O, q)) 6 ϕ(N) = N ie N is a congruence subgroup.
Conversely suppose that N is a congruence subgroup. So PSL2(O, q) 6 N , some q O.
Suppose that SL2(O, q) N , so ∃M1 ∈ SL2(O, q) such that M1 /∈ N . But ϕ(M1) ∈
ϕ(SL2(O, q)) = PSL2(O, q) 6 N = ϕ(N). So ∃M2 ∈ N such that ϕ(M1) = ϕ(M2). So
M1M−12 ∈ kerϕ = ±I, so M1 = ±M2. Now M1 6= M2, as M1 /∈ N , and M2 ∈ N .
So M1 = −M2 = −IM2, but −I ∈ N and M2 ∈ N , so M1 ∈ N . Contradiction. So
SL2(O, q) 6 N ie N is a congruence subgroup.
So the normal congruence subgroups of SL2(O) of level q that contain −I are in one to
one correspondence with the normal congruence subgroups of PSL2(O) of level q and they
clearly have the same index. So we attempt to classify the normal congruence subgroups of
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PSL2(O) of index n by classifying the normal congruence subgroups of SL2(O) of index n
which contain −I.
Lemma 6.4.2. Let O = Od,m and suppose that N SL2(O). Let n ∈ N. Then if T n, and
Un ∈ N then E2(O, nO) 6 N .
Proof. Recall that
E2(O, nO) =< I + αeij : α ∈ nO, i 6= j >E2(O) .
Let α ∈ nO, so α = n(z1 +mωz2), where z1, z2 ∈ Z. Now
I + αe12 =
1 nz1 + nmωz2
0 1
= (T n)z1(Un)z2 ∈ N.
Similarly I + αe21 ∈ N . So, as N is normalized by E2(O) then E2(O, nO) 6 N .
Recall the following results
Theorem. (3.2.2) Let O = Od,m. Let N SL2(O) be of index n and suppose that 6 - n.
Then N is of order O.
Theorem. (5.4.10) Let O = Od,m. Let N SL2(O) be a congruence subgroup of order O
that contains −I. Then SL2(O, 6O) 6 N .
So that any normal congruence subgroup of SL2(Od,m) of index not divisible by 6 and
containing −I must contain SL2(Od,m, 6Od,m). This gives us a method to classify the normal
congruence subgroups of SL2(Od,m) of index not divisible by 6. In fact we can be more
precise.
Lemma 6.4.3. Let O = Od,m. Let N SL2(O) be of index n that contains −I and suppose
that 6 - n. Then
1. If 2 | n and 3 - n then N is a congruence subgroup ⇔ SL2(O, 2O) 6 N .
119
2. If 2 - n and 3 | n then N is a congruence subgroup ⇔ SL2(O, 3O) 6 N .
3. If (6, n) = 1 then N is a non-congruence subgroup.
Proof. Now, N is a normal congruence subgroup of index n, so using (6.4.2) and (5.2.5),
SL2(O, nO) 6 N . Applying Wohlfahrts theorem (5.2.3) we see that
SL2(O, nO)SL2(O, 6O) = SL2(O, gcd(6, n)O) 6 N from which the result follows.
Let Nd,m(n) denote the number of normal congruence subgroups of PSL2(Od,m) of index
precisely n. So by the above lemma if (6, n) = 1 then Nd,m(n) = 0. If SL2(Od,m) has
an infinite cyclic quotient then it has a normal subgroup of every index. In contrast the
Modular group has only two normal subgroups which are not of index 6k, some k ∈ N ([70]
theorems 8.6 and 8.7). However by (6.4.3), with only finitely many exceptions, every normal
congruence subgroup of SL2(Od,m) is of index 6k, some k ∈ N. We need the following lemma
Lemma 6.4.4. Let K be a Noetherian domain of Krull dimension one. Let q1, q2 K such
that q1 + q2 = K. Then
SL2(K/q1q2) ∼= SL2(K/q1)× SL2(K/q2)
Proof. By the Chinese remainder theorem K/q1q2∼= K/q1 ×K/q2 where the isomorphism
is given by x+ q1q2 7→ (x+ q1, x+ q2) = (x1, x2). Define ϕ : SL2(K/q1q2)→ SL2(K/q1)×
SL2(K/q2) by a b
c d
7→ a1 b1
c1 d1
,
a2 b2
c2 d2
where a = a+q1q2 7→ (a1, a2) etc. It is simple, if tedious to check that ϕ is an monomorphism
(indeed it is true for any commutative ring). To show that ϕ is onto we need Wohlfahrt’s
theorem (5.2.5). By Wohlfahrt SL2(K, q1)SL2(K, q2) = SL2(K, q1 + q2) = SL2(K). Also
SL2(K, q1) ∩ SL2(K, q2) = SL2(K, q1q2). So in the group SL2(K/q1q2) = G we have two
normal subgroups H, K, say such that HK = G and H ∩K = 1. So by [79] theorem 4.1 we
have G ∼= H ×K. Hence result.
Lemma 6.4.5. Let O = Od,m. Let n ∈ N and suppose that 2 | n and 3 - n. Then
120
1. If O has no ideals of index 2 then Nd,m(n) = 0.
2. If O has an ideal of index 2 then Nd,m(n) =
3 if n = 2
1 if n = 4
0 else
Proof. Let N SL2(O) be a congruence subgroup of index n that contains −I. Then by
(6.4.3) SL2(O, 2O) 6 N . There are 3 cases, depending on how many ideals of index two O
has. If O has no ideals of index 2 then SL2(O)/SL2(O, 2O) ∼= SL2(F4) but −I ∈ N and
PSL2(F4) is simple, so Nd,m(n) = 0. Now suppose that O has exactly one ideal of index 2.
Let L denote the local ring O/2O, so SL2(O)/SL2(O, 2O) ∼= SL2(L). Now corresponding
to N is N SL2(L), o(N) = L. So, by (6.3.2), SL2(L)′ 6 N . Now −I ∈ N so, by (5.3.19),
(6.3.5) and (6.3.6), SL2(O)/N is a factor of Z2 × Z2 which, by (3.4.10), has exactly 3 non-
trivial normal subgroups, all of index 2. The result follows in this case. Finally suppose
that O has two ideals of index 2. In this case, by (6.4.4), SL2(O)/SL2(O, 2O) ∼= S3 × S3,
so |SL2(O) : SL2(O, 2O)| = 2232. So N is of index 2 or 4, so SL2(O)/N is abelian. Now
(SL2(O)/SL2(O, 2O))ab ∼= Sab3 ×Sab
3∼= Z2×Z2 and so the answer is as the previous case.
Lemma 6.4.6. Let O = Od,m. Let n ∈ N and suppose that 2 - n and 3 | n. Then
1. If O has no ideals of index 3 then Nd,m(n) = 0.
2. If O has exactly one ideal of index 3 then Nd,m(n) =
1 if n = 3
0 else
3. If O has exactly two ideals of index 3 then Nd,m(n) =
4 if n = 3
1 if n = 9
0 else
Proof. Let N SL2(O) be a congruence subgroup of index n that contains −I. Then
by (6.4.3), SL2(O, 3O) 6 N . First suppose that O has no ideals of index 3. Then
SL2(O)/SL2(O, 3O) ∼= SL2(F9) but −I ∈ N and PSL2(F9) is simple, so Nd,m(n) = 0.
Now suppose that O has exactly one ideal of index three, m, say and let L denote the local
121
ring O/3O. So, by passing to SL2(L) and applying (5.3.15) we see that SL2(O,m) 6 N .
Now SL2(O)/SL2(O,m) = SL2(F3), −I ∈ N and, by [79] exercise 8.11, PSL2(F3) ∼= A4
which has exactly one normal subgroup, which is of index 3. Hence result in this case.
Finally suppose that O has exactly two ideals of index three, mi, i = 1, 2, say. Now, by
(6.4.4),SL2(O)
SL2(O, 3O)∼= SL2(F3)× SL2(F3)
and |SL2(F3)| = 24 (this follows from [79] theorem 8.8), so |SL2(O) : SL2(O, 3O)| = 2632.
So N is of index 3 or 9 and so SL2(O)/N is abelian. Now SL2(F3)ab ∼= Z3 so(
SL2(O)
SL2(O, 3O)
)ab
∼= Z3 × Z3
and, −I2 = I so −I ∈ SL2(O/3O)′. Now, by (3.4.10), Z3 × Z3 has exactly 4 non-trivial
normal subgroups, all of index 3. Hence result in this case.
In [31] Grunewald and Schwermer determine the minimum index of a non-congruence
subgroup of a Bianchi group. Let ncs(d) denote the minimum index of a non-congruence
subgroup of PSL2(Od). Their main result is
Theorem.
ncs(d) =
5 if d = 1
4 if d = 2
22 if d = 3
3 if d = 7
2 else
In the Modular group the least index of a non-congruence subgroup if 7 ([83] theorem
5.4). There is considerable overlap with Grunewald and Schwermer’s theorem and what
follows but our techniques are completely different. First recall the following
Theorem.
ϕ :SL2(Od,m)
U2(Od,m) Fs
where s = r − 1, and r = r(d,m), the rank of the Zimmert Set and U ∈ kerϕ.
122
Theorem 6.4.7. If r = r(d,m) > 2 then SL2(Od,m) has a normal non-congruence subgroup
containing −I of every index.
Proof. Since r > 2 we have
ϕ :SL2(Od,m)
U2(Od,m) Z
and clearly −I ∈ kerϕ. Suppose that N SL2(Od,m) such that kerϕ 6 N , so T, U ∈ N , so
l(N) = Od,m. So, if N is a congruence subgroup then N = SL2(Od,m). Thus SL2(Od,m) has
a normal non-congruence subgroup containing −I of every index.
Corollary 6.4.8. If r = r(d,m) > 2 then PSL2(Od,m) contains a normal non-congruence
subgroup of every index.
Not included in the above theorem are the cases d = 1, 2, 3, 7, 11, 19, and m = 1 and
(d,m) = (1, 2), (3, 2). We are unable to deal with the cases (d,m) = (1, 1), (3, 1) but we take
a closer look at the others later. Recall that it is conjectured that these are the only values
of (d,m) such that PSL(Od,m) does not have a free non-abelian quotient.
All the non-congruence subgroups produced above contain the Zimmert kernel and so
have torsion. We now show that in any Bianchi group the number of normal congruence
subgroups with torsion is finite. Recall
Theorem. (1.1.8) I 6= M ∈ SL2(Od,m) is of finite order if and only if trM = 0, or ±1.
We now show that the order of a subgroup with torsion is severely restricted. Let M ∈
SL2(Od,m) and recall that the order, o(M), of the matrix M =
a b
c d
is the ideal of
Od,m generated by b, c, a− d. The order of any subgroup containing M contains o(M).
Lemma 6.4.9. Let M ∈ SL2(Od,m). Then if trM = 0 then 2 ∈ o(M).
Proof. a + d = 0, so a = −d. Now a − d = 2a ∈ o(M). Also 1 = ad − bc = −a2 − bc, so
1 + a2 ∈ o(M). Thus 2a2 + 2 ∈ o(M) and, because 2a ∈ o(M) we have 2a2 ∈ o(M). Hence
2 ∈ o(M).
Lemma 6.4.10. Let M ∈ SL2(Od,m). Then if trM = ±1 then 3 ∈ o(M).
123
Proof. First suppose that trM = −1, so a + d = −1 and a = −1 − d. Now a − d =
2a + 1 ∈ o(M). Also 1 = ad − bc = −a − a2 − bc, so a2 + a + 1 = −bc ∈ o(M). So
2a2 + 1 = 2(a2 + a+ 1)− (2a+ 1) ∈ o(M). Further 1− a = 2a2 + 1− (2a+ 1)a ∈ o(M). So
3 = 2(1− a) + 2a+ 1 ∈ o(M). Similarly if trM = 1.
Theorem 6.4.11. There are only finitely many normal congruence subgroups with torsion
in SL2(Od,m).
Proof. Let N SL2(Od,m) and suppose that N has torsion and is a congruence subgroup.
So N contains a matrix M , say with trM = 0, or ±1. So 2, or 3 ∈ o(N), so 6 ∈ o(N). Now
N is a normal congruence subgroup so, by (5.4.14), SL2(Od,m, 288Od,m) 6 N and there are
only finitely many such subgroups.
Example 6.4.1. There are infinitely many non-normal congruence subgroups with torsion
in SL2(Od,m). The groups SL2(Od,mn), for n ∈ N, are an example.
Recall that it is conjectured that PSL2(Od,m) has a free non-abelian quotient for all values
of (d,m) except for (1, 1), (2, 1), (3, 1), (7, 1), (11, 1), (19, 1), (1, 2), and (3, 2). The normal
subgroups of PSL2(O1) were studied in [26] where, in particular, the normal subgroups of
index < 60 were classified and it was shown that for a wide collection of n, PSL2(O1) had no
normal subgroups of index n. The normal subgroups PSL2(O3) were studied in [1] where the
normal subgroups of index < 960 were classified and shown to all be congruence subgroups.
We now take a closer look at the others.
6.5 The groups PSL2(Od), d = 2, 7, 11 and PSL2(O3,2) and
PSL2(O1,2)
In this section we deal exclusively with PSL2(Od,m), which we denote by PSL2(Od,m), or
PSL2(Od), if m = 1. We look first at the groups PSL2(Od), for d = 2, 7, 11. In [25]
section 4.5.3, Fine attempts to classify the normal subgroups of PSL2(Od), d = 2, 7, 11
124
but his classification is incomplete. We first of all correct his errors. Recall the following
presentations
Theorem. ([25] theorem 4.3.1)
PSL2(O2) =< a, t, u; a2, (at)3, (u−1aua)2, [t, u] >
PSL2(O7) =< a, t, u; a2, (at)3, (u−1auat)2, [t, u] >
PSL2(O11) =< a, t, u; a2, (at)3, (u−1auat)3, [t, u] >
We now state and prove the correct version of Fine’s results
Theorem 6.5.1. Let NPSL2(O2) be of index n and suppose that 6 - n. Then if (n, 6) = 1
N = N(a, t, un)
if 2 | n and 3 - n then N is one of
N(a, t, un), N(at, un/2), N(at, aun/2)
if 2 - n and 3 | n then N is one of
N(a, t, un), N(a, un/3), N(a, tun/3), N(a, t2un/3)
Proof. Suppose that (n, 6) = 1 so a = at = u−1aua = 1, so a = t = 1, so PSL2(O2)/N is
a factor of Z. SoN = N(a, t, un). Suppose that 2 | n and 3 - n, so at = 1, so PSL2(O2)/N is a
factor of< a, u; a2, au = ua >∼= Z2×Z. So, by (3.4.10), N = N(a, t, un), N(at, un/2), N(at, aun/2).
Now suppose that 2 - n and 3 | n, so a = u−1aua = 1, so PSL2(O2)/N is a factor of <
t, u; t3, tu = ut >∼= Z3×Z. So, by (3.4.10), N = N(a, t, un), N(a, un/3), N(a, tun/3), N(a, t2un/3).
Fine ([25] section 4.5.3) correctly classified the normal subgroups of PSL2(O2) of index
coprime to 6, didn’t deal with the cases 2 | n and 3 - n or 2 - n and 3 | n, and erroneously
claimed to have classified the normal subgroups of PSL2(O2) with torsion of index 6k. As
can be seen above we have dealt with the case of a normal subgroup of index not divisible
125
by 6 but the case of an index divisible by 6 is very complicated and we have been unable
to deal with it. The complications arise because if N PSL2(O2) is of index 6k and has
torsion then PSL2(O2)/N is a factor of Z× PSL2(Z) ([25] theorem 4.5.3).
Lemma. ([25] theorem 4.5.3) If N PSL2(O7) has torsion then PSL2(O7)/N is a factor
of Z2 × Z.
Theorem 6.5.2. Let N PSL2(O7) be of index n. Then if 2 - n
N = N(a, t, un)
if 2 | n and N has torsion then N is one of
N(a, t, un), N(at, un/2), N(at, aun/2)
Proof. Suppose that 2 - n so a, u−1auat ∈ N , so a, t ∈ N , so PSL2(O7)/N is a factor of
< u >∼= Z. So N = N(a, t, u2).
Now suppose that 2 | n and N has torsion, so PSL2(O7)/N is a factor of Z2 × Z so, by
(3.4.10), N is one of
N(a, t, un), N(at, un/2), N(at, aun/2)
Missing from Fine’s classification were the groups N(at, aun/2).
Lemma. ([25] theorem 4.5.3) If N PSL2(O11) has torsion then PSL2(O11)/N is a factor
of Z3 × Z.
Theorem 6.5.3. Let N PSL2(O11) be of index n. Then if 3 - n then
N = N(a, t, un)
if 2 - n, and 3 | n, or 6 | n and N has torsion then N is one of
N(a, t, un), N(a, un/3), N(a, tun/3), N(a, t2un/3)
126
Proof. Suppose that 3 - n, so at = u−1auat = 1, so a = t = 1, so N = N(a, t, un). Suppose
that 2 - n, and 3 | n, so a = 1, so t3 = 1, so PSL2(O11)/N is a factor of < t, u; t3, [t, u] >∼=
Z3 × Z. Similarly, if 6 | n, and N has torsion, then PSL2(O11)/N is a factor of Z3 × Z. So
that, by (3.4.10), N = N(a, t, un), N(a, un/3), N(a, tun/3), N(a, t2un/3).
Missing from Fine’s classification were the groups N(a, tun/3) and N(a, t2un/3).
6.5.1 Normal congruence subgroups of PSL2(O2)
Now consider O2 and let ω = i√
2. By (6.3.3), O2 has one ideal of index 2, namely ωO2
and two ideals of index 3, namely (1 + ω)O2, and (1 − ω)O2. Thus, by (6.4.5) and (6.4.6),
if n ∈ N such that 6 - n then
N2(n) =
3 if n = 2
4 if n = 3
1 if n = 4, 9
0 else
and so from (6.5.1) all the normal subgroups of PSL2(O2) of index 2
N(a, t, u2), N(at, u), N(at, au)
are congruence subgroups, and all the normal subgroups
N(a, t, u3), N(a, u), N(a, tu), N(a, t2u)
of index 3 are congruence subgroups. Exactly one of the normal subgroups
N(a, t, u4), N(at, u2), N(at, au2)
of index 4 is a congruence subgroup and exactly one of the normal subgroups
N(a, t, u9), N(a, u3), N(a, tu3), N(a, t2u3)
of index 9 is a congruence subgroup. All other normal subgroups of PSL2(O2) of index not
divisible by 6 are non-congruence subgroups.
127
Lemma 6.5.4.
PSL2(O2, ωO2) 6 N(at, u)
Proof. PSL2(O2)/PSL2(O2, ωO2) ∼= S3, so PSL2(O2) has a normal congruence subgroup
of index 2. By (6.5.1), the normal subgroups of PSL2(O2) of index 2 are:
N(a, t, u2), N(at, u), N(at, au)
Now u ∈ PSL2(O2, ωO2), so u ∈ N . So, clearly, N 6= N(a, t, u2), N(at, au). Hence N =
N(at, u).
Lemma 6.5.5.
N(a, t, u4), N(at, au2)
are non-congruence subgroups, and
PSL2(O2, 2O2) 6 N(at, u2), N(a, t, u2), N(at, au)
Proof. Suppose N is one of these groups. Then PSL2(O2)/N is abelian. Suppose N is
a congruence subgroup, so because N is of index 2, or 4 we have PSL2(O2, 2O2) 6 N .
Now (PSL2(O2)/PSL2(O2, 2O2))ab ∼= Z2 × Z2. Suppose N = N(a, t, u4), N(at, au2). Then
PSL2(O2)/N ∼= Z4. So N is a non-congruence subgroup. By (6.5.1), the only other normal
subgroup of PSL2(O2) of index 4 is N(at, u2). Thus PSL2(O2, 2O2) 6 N(at, u2).
Now, by (3.4.10), Z2 × Z2 has 3 normal subgroups of index 2. By (6.5.1), the normal
subgroups of PSL2(O2) of index 2 are
N(a, t, u2), N(at, u), N(at, au)
and we have already seen that PSL2(O2, ωO2) 6 N(at, u).
Lemma 6.5.6.
PSL2(O2, (1 + ω)O2) 6 N(a, tu)
128
Proof. PSL2(O2)/PSL2(O2, (1 + ω)O2) ∼= A4, and A4 has a normal subgroup of index 3.
By (6.5.1), the normal subgroups of PSL2(O2) of index 3 are:
N(a, t, u3), N(a, u), N(a, tu), N(a, t2u)
Now tu ∈ PSL2(O2, (1 + ω)O2), so tu ∈ N . So, clearly N 6= N(a, t, u3), N(a, u). Suppose
that N = N(a, t2u). Then t = t2uu−1t−1 ∈ N , so a, t, u ∈ N . Contradiction. Hence
N = N(a, tu).
Lemma 6.5.7.
PSL2(O2, (1− ω)O2) 6 N(a, t2u)
Proof. PSL2(O2)/PSL2(O2, (1 − ω)O2) ∼= A4, so, as above, PSL2(O2) has a normal con-
gruence subgroup, N , of index 3. So, by (6.5.1), N is one of
N(a, t, u3), N(a, u), N(a, t2u)
Now tu−1 ∈ PSL2(O2, (1− ω)O2), so tu−1 ∈ N . So clearly N 6= N(a, t, u3), N(a, u). Hence
N = N(a, t2u).
Lemma 6.5.8.
N(a, t, u9), N(a, tu3), N(a, t2u3)
are non-congruence subgroups, and
PSL2(O2, 3O2) 6 N(a, u3), N(a, t, u3), N(a, u)
Proof. Let N be one of these groups and suppose that N is a congruence subgroup. Then be-
causeN is of index 3, or 9 we have PSL2(O2, 3O2) 6 N . Now ifN = N(a, t, u9), N(a, tu3), N(a, t2u3)
then PSL2(O2)/N ∼= Z9, so as (PSL2(O2)/PSL2(O2, 3O2))ab ∼= Z3 × Z3, N is a non-
congruence subgroup. The only other normal subgroup of index 9 is N(a, u3). Hence
PSL2(O2, 3O2) 6 N(a, u3).
Now, by (3.4.10), Z3 × Z3 has 4 normal subgroups of index 3. By (6.5.1), the normal
subgroups of PSL2(O2) of index 3 are
N(a, t, u3), N(a, u), N(a, tu), N(a, t2u)
129
and we have already seen that PSL2(O2, (1+ω)O2) 6 N(a, tu), and PSL2(O2, (1−ω)O2) 6
N(a, t2u). Hence result.
Hence
Theorem 6.5.9. The normal congruence subgroups of PSL2(O2) of index not divisible by
6 are precisely:
Group Level Index Abelianization
N(at, u) ω 2 Z23 × Z
N(a, t, u2) 2 2 Z32 × Z3 × Z
N(at, au) 2 2 Z3 × Z
N(a, tu) 1 + ω 3 Z32 × Z
N(a, t2u) 1− ω 3 Z22 × Z
N(a, u) 3 3 Z32 × Z
N(a, t, u3) 3 3 Z42 × Z3 × Z
N(at, u2) 2 4 Z23 × Z
N(a, u3) 3 9 Z72 × Z
Proof. Presentations for each of these groups were found using GAP [24]. These presentations
were then abelianized by hand.
Remark. So we have shown that every subgroup of PSL2(O2) of index 2 is a congruence
subgroup and we have an example of a (normal) non-congruence subgroup of PSL2(O2) of
index 4. Now PSL2(O2)/PSL2(O2, ωO2) ∼= S3 and S3 has a non-normal subgroup of index
3. Thus PSL2(O2) has a non-normal congruence subgroup of index 3. It is a simple matter
to use GAP [24] to show that PSL2(O2) has exactly one non-normal subgroup of index 3.
Thus all subgroups of PSL2(O2) of index 3 are congruence subgroups. So the least index
of a non-congruence subgroup of PSL2(O2) is 4. This replicates part of Grunewald and
Schwermer’s theorem ([31] proposition 3.1)
130
6.5.2 Normal congruence subgroups of PSL2(O7), PSL2(O11),
PSL2(O3,2), and PSL2(O1,2)
In exactly the same way as we did for PSL2(O2) we can classify Normal congruence sub-
groups of PSL2(O7), PSL2(O11), PSL2(O3,2), and PSL2(O1,2). However we do have
Lemma 6.5.10. Let N PSL2(O7) be of index n. Suppose that 6 | n and N has torsion.
Then N is a non-congruence subgroup.
Proof. Suppose that N is a congruence subgroup. By (6.5.2) a, or at ∈ N , so by (6.4.1)
there is a normal congruence subgroup of SL2(O7) corresponding to N , of order O7. By
applying (5.4.9), (5.3.19) and then (6.4.1) again we see that PSL2(O7, 2) 6 N . From (6.5.2)
we can see that PSL2(O7)/N is abelian. Now (PSL2(O7)/PSL2(O7, 2))ab ∼= Z2 × Z2, so
6 | 4. Contradiction. Hence N is a non-congruence subgroup.
and in a similar way we can prove
Lemma 6.5.11. Let N PSL2(O11) be of index n. Suppose that 6 | n and N has torsion.
Then N is a non-congruence subgroup.
Lemma 6.5.12. Let N PSL2(O1,2) be of index n. Suppose that N has torsion and 6 | n.
Then N is a non-congruence subgroup.
So that we get:
Theorem 6.5.13. The normal congruence subgroups of PSL2(O7) with torsion are precisely:
Group Level Index Abelianization
N(at, u) ω 2 Z3 × Z
N(at, au) 1− ω 2 Z3 × Z
N(a, t, u2) 2 2 Z2 × Z
N(at, u2) 2 4 Z23 × Z
131
Remark. So we have shown that all subgroups of PSL2(O7) of index 2 are congruence
subgroups and we have given an example of a (normal) non-congruence subgroup of index
3 in PSL2(O7). So 3 is the least index of a non-congruence subgroup of PSL2(O7). This
replicates a part of Grunewald and Schwermer’s theorem ([31] proposition 3.1).
Theorem 6.5.14. The normal congruence subgroups of PSL2(O11) with torsion are pre-
cisely:
Group Level Index Abelianization
N(a, u) ω 3 Z22 × Z
N(a, t2u) 1− ω 3 Z22 × Z
N(a, t, u3) 3 3 Z3 × Z
N(a, tu) 3 3 Z22 × Z
N(a, u3) 3 9 Z62 × Z
Remark. As stated earlier the unique subgroup of PSL2(O11) of index 2 is a non-congruence
subgroup so clearly 2 is the least index of a non-congruence subgroup. Again this replicates
a part of Grunewald and Schwermer’s theorem ([31] proposition 3.1).
Theorem 6.5.15. The normal congruence subgroups of PSL2(O3,2) of index not divisible
by 6 are precisely:
Group Level Index Abelianization
N(at, aw) m2 2 Z23 × Z
N(a, t, w2) 2 2 Z2 × Z3 × Z
N(at, w) 2 2 Z23 × Z
N(a, tw) m3 3 Z2 × Z
N(at, w2) 2 4 Z33 × Z
where m2 = (2, 1 + i√
3) is the ideal of index 2 in O3,2, and m3 = (3, 2 + 2i√
3) is the ideal
of index 3 in O3,2.
132
Remark. So we have shown that every subgroup of PSL2(O3,2) of index 2 is a congruence
subgroup and we have given an example of a (normal) non-congruence subgroup of index 3
in PSL2(O3,2). So 3 is the least index of a non-congruence subgroup of PSL2(O3,2). This is
an extension of Grunewald and Schwermer’s theorem ([31] proposition 3.1).
Theorem 6.5.16. The normal congruence subgroups of PSL2(O1,2) with torsion are pre-
cisely
Group Level Index Abelianization
N(at, tz, w) m2 2 Z3 × Z2
N(a, t, z, w2) 2 2 Z42 × Z
N(at, aw, tz) 2 2 Z2
N(at, tz, w2) 2 4 Z23 × Z
where m2 = (2, 2i) is the ideal of index 2 in O1,2.
Remark. So there are exactly three congruence subgroups of index 2 in PSL2(O1,2). It is
a simple matter to use GAP [24] to show that there are exactly 8 subgroups of PSL2(O1,2)
of index 2. So there are exactly 5 non-congruence subgroup of index 2 in PSL2(O1,2) and
2 is the least index of a non-congruence subgroup of PSL2(O1,2). This is an extension of
Grunewald and Schwermer’s theorem ([31] proposition 3.1).
133
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