One-Cusped Congruence Subgroups of Bianchi

Groups

Kathleen L. Petersen

Abstract

We show that there are only finitely many maximal congruence sub-groups of the Bianchi groups such that the quotient by H3 has only onecusp.

1 Introduction

Let d be a positive square free integer, and let Od denote the ring of integersof Q(

√−d). The groups PSL2(Od) are called the Bianchi groups and embed

discretely in PSL2(C), which is isomorphic to Isom+(H3), the orientation pre-serving isometries of hyperbolic 3-space. The quotient Md = H3/PSL2(Od) is afinite volume hyperbolic 3-orbifold with hd cusps where hd is the class numberof Q(

√−d). Famously, Q(

√−d) has class number one only when d = 1, 2, 3,

7, 11, 19, 43, 67, or 163. If Υ is a finite index subgroup of PSL2(Od), we canform the quotient MΥ = H3/Υ, which is a finite volume hyperbolic 3-orbifoldwith finitely many cusps. We say that Υ has n cusps if this quotient has ncusps. The quotients Md are the prototypical examples of non-compact arith-metic 3-orbifolds, which are those orbifolds M such that M is commensurablewith some Md. (Two orbifolds are commensurable if they share a finite sheetedcover.) For a non-zero ideal J ⊂ Od the principal congruence subgroup of levelJ is

Γ(J ) = {A ∈ PSL2(Od) : A ≡ I mod J }

and has finite index in PSL2(Od). A (finite index) subgroup of PSL2(Od) iscalled a congruence subgroup if it contains a principal congruence subgroup.Our first result is

Theorem 1.1. There are only finitely many maximal one-cusped congruencesubgroups of the Bianchi groups.

Clearly, these must correspond to those Bianchi groups with associated classnumber equal to one, since if Υ < PSL2(Od), then MΥ covers Md. In fact, wewill show that for d = 11, 19, 43, 67, or 163 that there are only finitely manyone-cusped congruence subgroups of PSL2(Od). In addition, from our methodof proof we will conclude that

1

Corollary 1.2. If d = 19, 43, 67, or 163 there are no torsion-free one-cuspedcongruence subgroups of PSL2(Od).

A notable example of a one-cusped congruence subgroup is the fundamentalgroup of the figure-eight knot complement, which injects into PSL2(O3) as anindex 12 subgroup containing Γ(4) [7]. The fundamental group of the sister ofthe figure-eight knot complement, a knot in the lens space L(5, 1), also injectsinto PSL2(O3) as an index 12 subgroup containing Γ(2) [1]. Reid showed thatthe figure-eight knot is the only arithmetic knot complement in S3 [14]. If d =2, 7, 11, 19, 43, 67, or 163, there are infinitely many one-cusped subgroups(not necessarily torsion-free) since there is a surjection from PSL2(Od) ontoZ, with a parabolic element generating the image. If d = 1 or 3 there arealso infinitely many one-cusped subgroups, associated to torsion-free subgroupsof finite index, e.g. subgroups of finite index in the fundamental group of thefigure-eight knot complement [2]. There are examples of torsion-free one-cuspedsubgroups of PSL2(Od) corresponding to d = 1, 2, 3, 7, and 11, but historicallysuch examples have proven difficult to find [2] [1]. In the setting of arithmeticmanifolds and orbifolds, it has recently been shown that there is a finite numberof commensurability classes of one-cusped orbifolds or manifolds of minimalvolume [4].

The behavior of the Bianchi groups is similar to the classical case of the Mod-ular group, PSL2(Z). The Modular group embeds discretely in PSL2(R). AsPSL2(R) is isomorphic to the group of orientation preserving isometries of thehyperbolic plane, H2, we can form the quotient MQ = H2/PSL2(Z). This quo-tient has finite volume as a hyperbolic 2-orbifold and has a single cusp. As in theBianchi group case, MQ is the prototype for non-compact arithmetic 2-orbifolds,which are defined as those orbifolds M commensurable with MQ. Rhode provedthat there are at least two conjugacy classes of one-cusped subgroups of index nin the Modular group for every positive integer n [12]. One can define principalcongruence subgroups and congruence subgroups of the Modular group analo-gous to the above Bianchi group definitions. Petersson proved that there areonly finitely many one-cusped congruence subgroups of the Modular group, andthat the index of any such group divides 55440 = 11 · 7 · 5 · 32 · 24 [13].

We now introduce a useful invariant of finite index subgroups ofPSL2(Od), the level of the group, which has added meaning for congruencesubgroups.

Definition 1.1. Let Υ < PSL2(Od). We say that Υ has Od-level L if L is anideal in Od maximal with respect to the property that the normal closure of thegroup generated by the elements{

±(

1 `0 1

): ` ∈ L

}is contained in Υ. Moreover, we say that Υ has Z-level l if l is the smallestpositive integer such that the normal closure of the group generated by the

2

elements {±

(1 lβ0 1

): β ∈ Od

}is contained in Υ.

We can now be more precise in regard to Theorem 1.1. We will show thatany prime in Od that divides the Z-level of a one-cupsed congruence subgroupof PSL2(Od) has norm less than or equal to 11. In addition, if d = 1, 2, or7, we show that there are only finitely many of odd Z-level, and if d = 3 wedemonstrate that there are only finitely many of Z-level relatively prime to 21.

In contrast to Theorem 1.1 we show

Theorem 1.3. Let K be Q or an imaginary quadratic number field with classnumber one, and let OK be the ring of integers of K. There are infinitely manymaximal congruence subgroups of PSL2(OK) that have two cusps. Moreover, forany even integer n, there are infinitely many primes P ⊂ OK such that there isan n-cusped congruence subgroup of OK-level P.

The cases of the Modular group and the Bianchi groups differ vastly fromthose relating to other number fields. In general, let K be a number fieldwith r real places and s complex places, and let OK be the ring of integers ofK. The group PSL2(OK) embeds discretely in PSL2(R)r × PSL2(C)s, whichis isomorphic to the group of orientation preserving isometries of [H2]r × [H3]s.The quotient

MK = [H2]r × [H3]s/PSL2(OK)

is equipped with a metric inherited from [H2]r × [H3]s and with respect tothis metric MK has finite volume. The orbifold MK has hK cusps, where hK

is the class number of K [19]. As opposed to the imaginary quadratic case,less is known about the distribution of number fields of class number one inother settings. For example, it is a famous conjecture that there are infinitelymany real quadratics with class number equal to one. Principal congruencesubgroups and congruence subgroups can be defined for any PSL2(OK) similarto the imaginary quadratic case. We say that PSL2(OK) has the CongruenceSubgroup Property (CSP) if all finite index subgroups are congruence subgroups.It has been shown that PSL2(OK) fails to have the CSP precisely when K = Q orQ(√−d) [15]. This is due to the fact that OK has infinitely many units precisely

when K is not Q or an imaginary quadratic. This dichotomy is reflected in thetopology of these quotients. By way of contrast to Theorem 1.1, if K is anumber field with class number one, other than Q or an imaginary quadratic,and if i /∈ K, then assuming the Generalized Riemann Hypothesis, there areinfinitely many maximal one-cusped (congruence) subgroups of PSL2(OK) [10].

3

2 Background

2.1 Number Theory

Recall that if L is a non-zero ideal in Od, the norm of L, N(L) = |Od/L|. Ifp ∈ Z is a prime and P is a prime ideal in Od such that pOd ⊂ P, then wesay that P lies over p, and p lies under P. Therefore p lies under P preciselywhen the norm of P, N(P), is a power of p. There are three possibilities for thedecomposition of an ideal pOd when p is a rational prime

pOd =

P2 p is ramifiedP p is inertP1P2 p is split .

This behavior can be completely classified. (See [8] for example.) If p is notinert and P lies over p, then Od/P ∼= Fp and N(P) = p. If p is inert thenpOd = P, Od/P ∼= Fp2 and N(P) = p2.

Table 1:Splitting Types of Small Primes in Od, R=Ramified, S=Split and I=Inert

d = 1 2 3 7 11 19 43 67 163p = 2 R R I S I I I I I3 I S R I S I I I I5 S I I I S S I I I7 I I S R I S I I I11 I S I S R S S I I

Let K be a number field with r real places and s complex places. The ringof integers of a number field is a free abelian group of rank [K : Q]. There isalways a Z-basis for OK , called an integral basis. The units in OK are preciselythose elements of norm ±1, and form a group, O×

K , under multiplication. Themultiplicative group of units is isomorphic to a product of a free abelian groupof rank r+s−1 and a finite cyclic group, consisting of roots of unity. Therefore,|O×

K | is finite precisely when K is Q or an imaginary quadratic. This lies at theheart of the dichotomy between the associated groups, PSL2(OK). Notice thatwhen K = Q(

√−d) the multiplicative group of units is simply ±1 unless d = 1

or 3, when it also contains ±i and (±1±√−3)/2, respectively.

A fractional ideal in Od is a set of the form αJ for some α ∈ Q(√−d) and

an ideal J ∈ Od. The non-zero fractional ideals of Od form a group, Id, andthe non-zero principal ideals form a subgroup, Pd. The quotient, Id/Pd hasfinite order, called the class number of Q(

√−d). As stated, the class number of

Q(√−d) is one precisely when Od is a principal ideal domain, when d = 1, 2, 3,

7, 11, 19, 43, 67, or 163.

4

2.2 The Groups (P)SL2(Od)

We will consider PSL2(Od) where the class number of Q(√−d) is one. Since

Od is a PID, for a non-zero ideal J , there is an ` such that J = (`). It will beconvenient to use the notation J as well as (`). For example, we will use thenotation Γ(`) = Γ(J ) for principal congruence subgroups. For the remainder ofthe section assume that J = (`) is a non-zero ideal with `1 and `2 ∈ Od suchthat (`) = (`1) ∩ (`2) and Od = (`1, `2). Define

G(J ) = G(`) = {A ∈ SL2(Od) : A ≡ I mod J }.

We have PSL2(Od)/Γ(J ) ∼= PSL2(Od/J ) and SL2(Od)/G(J ) ∼= SL2(Od/J ),and will often use the isomorphic groups interchangeably. We define φJ = φ` asthe reduction modulo J = (`) map from (P )SL2(Od) to (P )SL2(Od/(`)). Thefollowing decomposition will be essential [9]

SL2(Od)/G(`) ∼= SL2(Od/(`1))× SL2(Od/(`2)).

We define ρi to be projection ρi : SL2(Od/(`)) → SL2(Od/(`i)).If P is a prime ideal then Od/P ∼= FN(P), the finite field with N(P) el-

ements, and therefore PSL2(Od)/Γ(P) ∼= PSL2(FN(P)) and SL2(Od)/G(P) ∼=SL2(FN(P)). Since Od is a Dedekind domain we can write J = Pn0

0 . . .Pnss

where the Pi are distinct prime ideals. We define Θ(J ) = [SL2(Od) : G(J )].Therefore [9]

Θ(J ) =s∏

i=0

N(Pi)3ni

(1− 1

N(Pi)2

).

Notice that for J = (`), and `1, and `2 as above, defining Θ(`) = Θ(J ) we haveΘ(`) = Θ(`1)Θ(`2).

We will commonly use Υ to denote a finite index (usually one-cusped andcongruence) subgroup of PSL2(Od), and we define

G = φ`(Υ), Gi = φ`i(Υ).

Notice that Gi = φ`i(ΥΓ(`i)) (where ΥΓ(`i) is the composite of Υ and Γ(`i))as Γ(`i) = kerφ`i

.Let Υ∗ be the SL2(Od) pull-back of Υ. We now define, G∗ = φ`(Υ∗) and

G∗i = φ`i

(Υ∗) = φ`i(Υ∗G(`i)) = ρi(G∗). Define N∗

1 < G∗1 as

N∗1 = {g1 ∈ G∗

1 : (g1, 1) ∈ G∗}

and similarly for N∗2 . Let Ni = PN∗

i . Notice that N∗i is not defined as the

pull-back of Ni. It is an elementary exercise in group theory to show thatG∗

1/N∗1∼= G∗

2/N∗2 , |G∗| = |G∗

1||N∗2 | = |G∗

2||N∗1 | and

[SL2(Od/`) : G∗] = [SL2(Od/(`1)) : G∗1][SL2(Od/(`2)) : N∗

2 ].

Letx = [PSL2(Od) : Υ], xi = [PSL2(Od) : ΥΓ(`i)].

5

Notice that if Γ(`) < Υ then

x = [SL2(Od) : Υ∗] = [PSL2(Od/(`)) : G] = [SL2(Od/(`)) : G∗]

and

xi = [SL2(Od) : Υ∗G(`i)] = [PSL2(Od/(`i)) : Gi] = [SL2(Od/(`i)) : G∗i ]

Since Θ(`) = Θ(`1)Θ(`2), we conclude that

|N∗1 | =

|G∗||G∗

2|=

x2Θ(`1)x

, |N∗2 | =

x1Θ(`2)x

.

Therefore

Lemma 2.1. Let Υ be a subgroup of PSL2(Od) containing Γ(`). If (`) =(`1)∩(`2) and Od = (`1, `2) then with x = [PSL2(Od) : Υ] and xi = [PSL2(Od) :ΥΓ(`i)] we have

|N∗1 | =

x2Θ(`1)x

∈ Z, |N∗2 | =

x1Θ(`2)x

∈ Z.

Additionally,

[SL2(Od/(`1)) : N∗1 ] =

x

x2and [SL2(Od/(`2)) : N∗

2 ] =x

x1.

Therefore[G∗

1 : N∗1 ] = [G∗

2 : N∗2 ] =

x

x1x2.

Notice that for some z dividing the order of the center of SL2(Od/(`i))|Ni|z = |N∗

i |. In particular, if (`i) = Pn where P is a prime ideal that does notlie over 2, then the order of the center is 2 and |Ni| = 1

2 |N∗i | or |N∗

i |, dependingon whether ±I ∈ N∗

i or not. (N∗i is defined with respect to G∗

i , not as thepull-back of Ni.) In this case

|N1| =(

1 or12

)x2Θ(`1)

x, |N2| =

(1 or

12

)x1Θ(`2)

x.

2.3 Peripheral Subgroups

As stated previously, Md = PSL2(Od) is a finite volume hyperbolic 3-orbifoldwith hd cusps, where hd is the class number of Q(

√−d). If Υ is a finite index

subgroup of PSL2(Od) then MΥ = H3/Υ is non-compact, has finite volume,and has a finite number of cusps. Recall that ±A ∈ PSL2(C) is parabolic if±A 6= ±I and |trace(A)| = 2. We define C ∈ C ∪∞ to be a cusp of Υ if C is aparabolic fixed point of Υ, that is if there is some parabolic ±A ∈ Υ such that

±A = ±(

α βγ δ

)and ±A · C :=

αC + β

γC + δ= C.

6

For a cusp C we define the stabilizer of C in Υ as

StabC(Υ) = {±A ∈ Υ : ±A · C = C} .

In particular, Stab∞(Υ) consists of upper triangular matrices and thereforeconsists solely of parabolic elements unless d = 1 or 3. Two cusps are equivalentin MΥ if they are in the same Υ orbit. As such, the number of equivalenceclasses is the number of conjugacy classes of maximal peripheral subgroups ofΥ, subgroups of the form StabC(Υ). This is the number of topological ends ofMΥ.

Lemma 2.2. Let Υ be a finite index subgroup of PSL2(Od) where the classnumber of Q(

√−d) is one. Then Υ has one equivalence class of cusps if and

only if[PSL2(Od) : Υ] = [Stab∞(PSL2(Od)) : Stab∞(Υ)].

Proof. The covering of Md by MΥ induces a covering of the same degree oftruncated compact orbifolds M ′

d and M ′Υ. This cover restricts to a cover of

∂M ′d by ∂M ′

Υ. The degree of the cover of M ′d by M ′

Υ is the degree of the coverof the cusp at infinity of M ′

d by those cusps of ∂M ′Υ covering it. If MΥ has

one cusp, then as Md also has cusp, the cusp at infinity of M ′Υ is the only cusp

covering the cusp at infinity of M ′d. Therefore this covering degree,

[Stab∞(PSL2(Od)) : Stab∞(Υ)]

is the degree of the cover of MΥ to Md which is [PSL2(Od) : Υ].As remarked above, the degree of the cover of Md by MΥ, [PSL2(Od) : Υ],

is the degree of the cover of the cusp at infinity of M ′d by those cusps of ∂M ′

Υ

covering it. If

[PSL2(Od) : Υ] = [Stab∞(PSL2(Od)) : Stab∞(Υ)]

then only the cusp at infinity of M ′Υ can cover the cusp at infinity of M ′

d. Sincethe class number of Q(

√−d) is one, Md only has one cusp and consequently

MΥ has only one cusp as well.

Definition 2.1. For Υ < PSL2(Od) let

Λ(Υ) ={

` ∈ C : ±(

1 `0 1

)∈ Υ

}.

Moreover, let Λd = Λ(PSL2(Od)), Λ(J ) = Λ(Γ(J )), and Λ(`) = Λ(Γ(`)). Givena non-zero β ∈ Od, let Λβ(Υ) be the smallest positive integer b such thatbβ ∈ Λ(Υ).

Notice that Λ(Υ) is an abelian group, and Λ(Υ) is a subgroup of Λd = Od.Therefore if Υ is a finite index subgroup, Λ(Υ) is a full rank lattice in C.

7

Let {1, ω} be an integral basis for Od. If d 6= 1 or 3 the only roots of unityin Od are ±1. The fourth roots of unity are in O1, generated by i, and the sixthroots of unity are in O3 generated by (1 +

√−3)/2. If d 6= 1 or 3,

Stab∞(PSL2(Od)) =⟨±

(1 10 1

), ±

(1 ω0 1

)⟩and geometrically, a cusp of Md is T × [0,∞) where T is a torus. If d = 2, 7,11, 19, 43, 67, or 163, and MΥ = H3/Υ has one cusp, the degree of the coveringof Md by MΥ, which is [PSL2(Od) : Υ] by Lemma 2.2, equals [Λd : Λ(Υ)].

The cusp of M1 is a pillow cusp, which geometrically is S × [0,∞) where Sis the 2-sphere with four cone points with cone angles π. And

Stab∞(M1) =⟨±

(1 10 1

),±

(1 i0 1

), ±

(−i 00 i

)⟩.

Elements of the form ±(−i β0 i

)for any β ∈ O1, have order two. If Υ <

PSL2(O1) has one cusp, and if Stab∞(Υ) contains torsion, then [PSL2(O1) :Υ] = [Λ1 : Λ(Υ)]. If Υ has one cusp and a torsion-free stabilizer at infinity, then[PSL2(O1) : Υ] = 2[Λ1 : Λ(Υ)].

The cusp of M3 is geometrically S × [0,∞) where S is a sphere with 3 conepoints each of cone angle 2π/3. And

Stab∞(M3) =⟨±

(1 10 1

),±

(1 ω0 1

), ±

(ω−1 00 ω

)⟩where

ω =−1 +

√−3

2.

Elements of the form ±(

ω−1 β0 ω

)have order 3. As in the O1 case, if Υ <

PSL2(O3) has one cusp, and if Stab∞(Υ) contains torsion, then [PSL2(O3) :Υ] = [Λ3 : Λ(Υ)]. If Υ has a torsion-free stabilizer at infinity, then [PSL2(O3) :Υ] = 3[Λ3 : Λ(Υ)]. To assist with these extra factors, we have the followingdefinition.

Definition 2.2. Let Υ be a finite index one-cusped subgroup of PSL2(Od) thenT (Υ) ∈ {1, 2, 3} is the number such that [PSL2(Od) : Υ] = T (Υ)[Λd : Λ(Υ)].

Specifically,

T (Υ) =

1 if d = 2, 3, 7, 11, 43, 67, 1631 if d = 1, 3 and Υ has peripheral torsion2 if d = 1 and Υ has no peripheral torsion3 if d = 3 and Υ has no peripheral torsion.

8

3 Key Lemmas

In this section we collect several key lemmas used in the proof of Theorem 1.1.The following is a generalization of a theorem of Wohlfahrt [20]. The proof issimilar to that of the classical Modular group case and is omitted. The corollaryfollows directly.

Lemma 3.1. (Wohlfahrt’s Lemma)Let Q(√−d) have class number one, and

let Υ < PSL2(Od) be a congruence subgroup of Od-level (`). Then (`′) ⊆ (`) ifand only if Υ contains Γ(`′).

Corollary 3.2. Let Q(√−d) have class number one, and let Υ < PSL2(Od) be

a congruence subgroup.

1. Υ has Od-level (`) if and only if Γ(`) is the maximal principal congruencesubgroup contained in Υ.

2. Υ has Z-level l if and only if Γ(l) is the maximal principal congruence sub-group contained in Υ whose level can be generated by a rational integer.

3. Let Υ have Od-level (`) and Z-level l. Then l is the smallest positive integersuch that (l) ⊂ (`).

The following is a generalization of the Modular group case [13]. The proofis omitted.

Lemma 3.3. (The Ladder Lemma) Let Υ be a one-cusped congruence sub-group of PSL2(Od). If Υ has Z-level pml where p is a prime, p 6 |l and

m ≥{

3 if p ≤ 32 if p > 3

then ΥΓ(pm−1l) has Z-level pm−1l. If Υ has Od-level Pn where p is ramifiedand lies under P, and

n ≥

5 if p = 24 if p = 33 if p > 5

then ΥΓ(Pn−1) has Od-level Pn−1 or Pn−2.

Lemma 3.4. (The Index Lemma) Let Υ be a one-cusped congruence sub-group of PSL2(Od) of Z-level pnl where p is a prime and p 6 |l. If p > 3 then pn

divides [PSL2(Od) : Υ]. If p ≤ 3, then pn′ divides [PSL2(Od) : Υ] where n′ = nif p divides the Z-level of ΥΓ(pl) and n′ = n− 1 otherwise.

Moreover,

[PSL2(Od) : Υ] = [Stab∞(PSL2(Od)) : Stab∞(Υ)] = T (Υ)[Λd : Λ(Υ)].

If Υ contains Γ(J ) then [PSL2(Od) : Υ] divides T (Υ)N(J ).

9

Proof. We will prove the second portion of the lemma first. From Lemma 2.2and the discussion following

[PSL2(Od) : Υ] = [Stab∞(PSL2(Od)) : Stab∞(Υ)] = T (Υ)[Λd : Λ(Υ)].

Since Γ(J ) < Υ, the cusp at infinity of MΓ(J ) covers the cusp at infinity of MΥ

which in turn covers the cusp at infinity of Md. Therefore, [Λd : Λ(Υ)] divides[Λd : Λ(J )] and hence [PSL2(Od) : Υ] divides T (Υ)[Λd : Λ(J )]. Let ` be suchthat (`) = J . Fixing an integral basis {1, ω} for Od, Λ(`) is generated by `and ω`, and Λd is generated by 1 and ω. Therefore [Λd : Λ(`)] = N(`) and weconclude that [PSL2(Od) : Υ] divides T (Υ)N(`).

We will prove the remainder of the lemma in the case where p > 3. Thecase p ≤ 3 is similar. Let Υ be as above and fix an integral basis {1, ω} forOd. For Υ1 < PSL2(Od), Λ(Υ1) is an abelian group, and if Υ2 < Υ1 thenΛ(Υ2) < Λ(Υ1). By above, [PSL2(Od) : Υ] = T (Υ)[Λd : Λ(Υ)]. It suffices toshow that pn divides [Λd : Λ(Υ)].

Fix l and p as in the statement of the lemma. We first consider the case wheren = 1, and we will induct on n. Assume that p does not divide [Λd : Λ(Υ)].Therefore p does not divide [Λ(Υ)Λ(l) : Λ(Υ)] = [Λ(l) : Λ(Υ) ∩ Λ(l)]. Sincep ∈ Z, [Λ(l) : Λ(pl)] = p2; we conclude that Λ(Υ) ∩ Λ(l) = Λ(l), and thereforeΛ(l) < Λ(Υ). This implies that

±(

1 l0 1

)and ±

(1 lω0 1

)∈ Υ.

By Wohlfahrt’s Lemma, the Z-level of Υ divides l contradicting the hypothesisthat it is pl.

Now assume that for all m such that 1 ≤ m < n, that for all one-cusped Υ ofZ-level pml, pm divides [Λd : Λ(Υ)]. Assume that Υ is a one-cusped congruencesubgroup of Z-level pnl and pn does not divide [Λd : Λ(Υ)]. By the LadderLemma and the inductive hypothesis, pn−1 divides [Λd : Λ(ΥΓ(pn−1l))]. SinceΛ(Υ)Λ(pn−1l) < Λ(ΥΓ(pn−1l)), p does not divide

[Λ(Υ)Λ(pn−1l) : Λ(Υ)] = [Λ(pn−1l) : Λ(Υ) ∩ Λ(pn−1l)]

but [Λ(pn−1l) : Λ(pnl)] = p2 and so Λ(pn−1l) = Λ(Υ) ∩ Λ(pn−1l) implying thatΛ(pn−1l) < Λ(Υ). Therefore

±(

1 pn−1l1 1

)and ±

(1 pn−1lω1 1

)∈ Υ.

This contradicts Wohlfahrt’s Lemma as the Z-level of Υ is pnl. We concludethat pn divides [Λd : Λ(Υ)], proving the lemma.

3.1 Vector Spaces

In this section we will describe a technique that will be used to encode infor-mation about one-cusped congruence subgroups, specifically their Od-levels, in

10

terms of a vector subspace. (We will use this in the proofs of Propositions 4.1and 4.2 which which are stated in the next section.) This will be used as aconvenient way to show that there cannot be one-cusped congruence subgroupsin certain situations.

Let Υ be a one-cusped congruence subgroup of PSL2(Od) of Od-level Pn

where n ≥ 2 and P is a prime lying over the rational prime p. Let q ∈ Od besuch that (q) = P. If p 6= 2 the quotient Γ(Pn−1)/Γ(Pn) is a three-dimensionalvector space, V, over FN(P). The correspondence is given by

±(

1 + qn−1α qn−1βqn−1γ 1 + qn−1δ

)∈ Γ(Pn−1) ↔ (α, β, γ) ∈ V

for α, β, γ, and δ in Od. This correspondence is well-defined modulo P. Thesubgroup Υ ∩ Γ(Pn−1) corresponds to a subspace, F , of V. Notice that ifA ∈ ΥΓ(Pn−1) then A ≡ B modulo Pn−1 for some B ∈ Υ. One can use thisfact to show that the action induced on V by conjugation by an element inΥΓ(Pn−1) preserves F . (The proof of this is similar to the classical case. See[13])

We will show this action explicitly. Given ±B ∈ Υ ∩ Γ(Pn−1) there is a

V =(

v1 v2

v3 v4

)in M2(Od) such that

B = I + qn−1V.

Let A = ±(

α βγ δ

)∈ PSL2(Od). The action of conjugating B by A, B →

ABA−1, interpreted in V sends the vector v = (v1, v2, v3) to the vector

A · v = ([αδ +βγ]v1−αγv2 +βδv3,−2αβv1 +α2v2−β2v3, 2γδv1−γ2v2 + δ2v3).

By the above discussion, if f ∈ F then A · f ∈ F when A ∈ ΥΓ(Pn−1). Wewill be explicitly making use of the action A · f for a few specific matrices. Letα, β ∈ Od and define

Aα,β = ±(

1 α0 1

) (0 −11 0

) (1 β0 1

).

Conjugating B by Aα,β takes v = (v1, v2, v3) to Aα,β · v =([2αβ − 1] v1 − αv2 +

[αβ2 − β

]v3,[

−2α2β + 2α]v1 + α2v2 +

[−α2β22 + 2αβ − 1

]v3,

2βv1 − v2 + β2v3

).

Lemma 3.5. (The Vector Space Lemma) Let Υ be a one-cusped congru-ence subgroup of PSL2(Od) of Od-level Pn. If Λ(Υ ∩ Γ(Pn−1)) = Λ(Pn) then(b1b2, b

22,−b2

1) ∈ F if and only if b1 = b2 = 0. Moreover, if p lies under P andis not inert and Λ(Υ ∩ Γ(Pn−1)) = Λ(Pn)

i) There are b1, b3 ∈ Fp such that (1, b1, 0) and (0, b3, 1) form a basis for F .

11

ii) If f = (f1, f2, f3) ∈ F then f2 = f1b1 + f3b3.

iii) If p > 2 then b21 − 4b3 is not a square in Fp.

In particular, neither (0, 1, 0) nor (0, 0, 1) are in F . The proof follows from thefact that such elements correspond to parabolic matrices, and is similar to [13].

4 Proof of Theorem 1.1

Theorem 1.1 and the discussion following it will be proven from a series ofpropositions. The proofs of the following three propositions will comprise thebulk of the manuscript. The proof of Proposition 4.2 will rely upon Proposition4.1, and the proof of Proposition 4.3 will rely upon Propositions 4.1 and 4.2.They will be proven in sections 4.1, 4.2, and 4.3 after we complete the proof ofTheorem 1.1.

Proposition 4.1. (Prime Z-Levels) Let Υ be a one-cusped congruence sub-group of Z-level p where p is the rational prime lying under a prime P ⊂ Od.Then N(P) ≤ 11.

Proposition 4.2. (Prime Power Z-Levels) Let Υ be a one-cusped congru-ence subgroup of Z-level pn where p is the rational prime lying under a primeP ⊂ Od. Then N(P) ≤ 11. Moreover, unless p = 2 and is not inert, p = 3 isramified or p = 7 and T (Υ) 6= 1, then

n ≤

0 if N(P) > 11, or p = 3 is inert and T (Υ) = 11 if p = 7, 11 is ramified

or p = 3 is inert and T (Υ) 6= 1or p = 5, 11 is splitor p = 7 is split and T (Υ) = 1

2 if p = 3 is splitor p = 2 is inert.

Proposition 4.3. Let Υ be a one-cusped congruence subgroup of Z-level l. ThenN(P) ≤ 11 for all primes P dividing (l).

Now it suffices to prove the following, which implies Theorem 1.1 and thediscussion following it.

Proposition 4.4. There are finitely many maximal one-cusped congruence sub-groups of the Bianchi groups. Furthermore, any prime in Od dividing the Z-levelof such a group has norm at most 11.

If d = 1, 2, or 7 there are finitely many one-cusped congruence subgroupsof odd Z-level, and if d = 3 there are finitely many one-cusped congruencesubgroups of Z-level relatively prime to 21.

If d = 11, 19, 43, 67, or 163 there are finitely many one-cusped congruencesubgroups of PSL2(Od).

12

Unless d = 3 and p = 3, or 7, or d = 1, 2, or 7 and p = 2, the proof will produceexplicit bounds on the maximal power of any prime p dividing the Z-level of aone-cusped congruence subgroup.

Proof. (Proof of Proposition 4.4)Let Υ be a one-cusped congruence subgroup of PSL2(Od) of Z-level l. Since

Υ has one cusp, Q(√−d) must have class number one, and therefore d =1 , 2,

3, 7, 11, 19, 43, 67, or 163. By Propostion 4.3, if P is a prime in Od dividing(l), then N(P) ≤ 11. Therefore, there is a finite set of primes, depending ond, such that the Z-level of any one-cusped congruence subgroup of PSL2(Od) isonly divisible by these primes. We now prove the following lemma.

Lemma 4.5. Let Q(√−d) have class number one, let P be a non-zero prime

in Od with N(P) ≤ 11, and let p ∈ Z lie under P. Then

a) If p is not a split or ramified 2, ramified 3, or split 7 in O3, there is aconstant c1 = c1(p, d) such that pc1 does not divide the Z-level of any one-cuspedcongruence subgroup of PSL2(Od).

b) If p is a split or ramified 2, ramified 3, or split 7 in O3, there is a constantc1 = c1(p, d) such that if pc1 divides the Z-level of a one-cusped congruencesubgroup, Υ, then if p = 7, Υ < ΥΓ(p) is a one-cusped congruence subgroup ofZ-level 7, and if p ≤ 3 then Υ < ΥΓ(p2) is a one-cusped congruence subgroupof Z-level p2.

Proof. (proof of Lemma 4.5) Let p be a prime such that N(P) ≤ 11 for allprimes P in Od lying over p. Let r ∈ Z be a prime distinct from p. Recall that

Θ(rm) = |SL2(Od/(rm))| =

r6m−2(r2 − 1) if r is ramifiedr6m−4(r4 − 1) if r is inertr6m−4(r2 − 1)2 if r is split.

Hence the power of p dividing Θ(rm) is less than a constant c(p, r), which isindependent of m. Let

c1(p, d) = 8 +∑

c(p, r)

where the sum is taken over all primes r ∈ Z distinct from p such that N(Q) ≤ 11for a prime Q lying over r.

Now assume that Υ is a one-cusped congruence subgroup of Z-level l. If l isa prime power, Proposition 4.2 implies Lemma 4.5 unless p = 2 and is not inert,p = 3 is ramified, or p = 7 and T (Υ) 6= 1. For these exceptions Lemma 4.5follows from the Ladder Lemma. Therefore, we may assume that l is divisibleby at least two distinct primes. If p is a prime dividing l and pn is the maximalpower of p dividing l, set l1 = pn and l2 = l/l1. Let x = [PSL2(Od) : Υ] andx1 = [PSL2(Od) : ΥΓ(l1)]. By Lemma 2.1

x1Θ(l2)x

∈ Z

13

where Θ(l2) = |SL2(Od/(l2))|. By the Index Lemma, pn divides x if p > 3 andpn−1 divides x if p ≤ 3, so if n ≥ c1 then pc1−1 divides x. By the above discus-sion, pc1−8 does not divide Θ(l2). We conclude that p7 divides x1. Therefore,ΥΓ(l1) is a one-cusped congruence subgroup whose Z-level divides pn. Since p7

divides the index we conclude that the Z-level is a positive power of p, say ps.We have Γ(ps) < ΥΓ(l1), and since ΥΓ(l1) has one cusp, by the Index Lemma x1

divides T (ΥΓ(l1))N(ps). Therefore, p7 divides T (ΥΓ(l1))N(ps). The maximumpossible power of p dividing T (ΥΓ(l1))N(ps) is 2s + 1. Therefore s ≥ 3, andthus the Z-level of ΥΓ(l1) is at least p3. In case a) this contradicts Proposition4.2 and we conclude that the pc1 cannot divide the Z-level of a one-cusped con-gruence subgroup. In case b), if p = 7, by repeated applications of the LadderLemma we conclude that ΥΓ(l1) is a subgroup of ΥΓ(p) which has Z-level p. Ifp = 2 or 3, we conclude that ΥΓ(l1) is a subgroup of ΥΓ(p2) which has Z-levelp2. This completes the proof of Lemma 4.5.

Now we continue the proof of Proposition 4.4. If d = 11, 19, 43, 67, or 163,the prime 2 is inert and 3 is unramified. Therefore, by Lemma 4.5 a) andProposition 4.3 the Z-level of any one-cusped congruence subgroup divides

J =∏

pc1(p,d)

where the product is taken over the finite number of primes where N(P) ≤ 11for all P in Od lying over p. Therefore Γ(J) is contained in all one-cuspedcongruence subgroups. As a result, there are only finitely many one-cuspedcongruence subgroups in PSL2(Od) for these values of d.

In O1, O2, and O7, the above argument shows that there are only finitelymany one-cusped congruence subgroups of odd Z-level. Moreover, if Υ is a one-cusped congruence subgroup and the Z-level is even, then there is a positive ndepending on Υ such that the Z-level divides J2n where

J =∏

pc1(p,d)

and the product is taken over the finite number of odd primes where N(P) ≤ 11for all P ∈ Od lying over p. There are only finitely many of Z-level dividing2c1(2,d)J and by Lemma 4.5 b) if n > c1(2, d) then such a subgroup is containedin a one-cusped congruence subgroup of Z-level 4. This proves Proposition 4.4for d = 1, 2 and 7. The proof in the case where d = 3 is similar.

4.1 Prime Z-Levels (Proof of Proposition 4.1)

In this section we prove Proposition 4.1, that if Υ is a one-cusped congruencesubgroup of Z-level p where p is a rational prime lying under P ⊂ Od thenN(P) ≤ 11. We will do so with three lemmas. By Corollary 3.2 (see section3), if the Z-level is p for an inert prime, then the Od-level is (p). If p is split as(p) = P1P2 then the Od-level is P1P2, P1 or P2. If p is ramified with P lyingover p, then the Od-level is either P or P2. First, in Section 4.1.1 we will show

14

Lemma 4.6. Let Υ be a one-cusped congruence subgroup of Od-level P for aprime P ⊂ Od, then N(P) ≤ 11.

This proves Proposition 4.1 for inert primes. Next, in Section 4.1.2 we prove

Lemma 4.7. Let p be a split prime with (p) = P1P2, and let Υ a one-cuspedcongruence subgroup of Od-level P1P2. If p > 3 then

[PSL2(Od) : Υ] = T (Υ)p2

and ΥΓ(Pi) has Od-level Pi. Moreover, if p ≤ 3 and [PSL2(Od) : Υ] = T (Υ)p2

then ΥΓ(Pi) has Od-level Pi for i = 1 and 2.

This proves Proposition 4.1 for split primes. Finally, in Section 4.1.3 we show

Lemma 4.8. Let n be a positive integer and p be a ramified prime lying underP with N(P) > 11. There are no one-cusped congruence subgroups of Od-levelPn.

This proves Proposition 4.1 for ramified primes. Together, Lemmas 4.6, 4.7 and4.8 prove Proposition 4.1.

4.1.1 Prime Od-levels (Proof of Lemma 4.6)

Let Υ be a one-cusped congruence subgroup of Od-level P, let p be the rationalprime lying under P, and let {1, ω} be an integral basis for Od. Since Od isa principal ideal domain, P = (q) for some q ∈ Od. Recall that by the IndexLemma

x = [PSL2(Od) : Υ] = T (Υ)[Λd : Λ(Υ)]

and x divides T (Υ)N(p). Notice that [Λd : Λ(Υ)] 6= 1 as this would imply thatΛ(Υ) is generated by 1 and ω and so by Wohlfahrt’s Lemma Γ(1) = PSL2(Od) <Υ.

Therefore if p is not inert x = T (Υ)p, and if p is inert x = T (Υ)p or T (Υ)p2.As Γ(P) < Υ, reduction modulo Γ(P) sends Υ to an index x subgroup of

PSL2(Od)/Γ(P) ∼= PSL2(FN(P)).

A famous theorem of Galois states that the smallest index of a proper subgroupof PSL2(Fr) is r + 1 except for r = 2, 3, 5, 7, 9, or 11 [17]. Therefore, wehave proven Lemma 4.6 in the case where T (Υ) = 1. Also, if p is inert and[Λd : Λ(Υ)] = p we conclude that p = 2 or 3 as otherwise x = T (Υ)p isless than N(P). So it remains to consider the case when T (Υ) = 2 or 3 and[Λd : Λ(Υ)] = N(P). By the classification of subgroups of PSL2(Fr) [17] weconclude that if T (Υ) = 2, then as d = 1, N(P) = 2, 5, or 9. If T (Υ) = 3 thenas d = 3, N(P) = 3, 4, 7, or 19. However, any index 57 subgroup of PSL2(O3)of O3-level P for a prime P lying over 19 can be shown to contain peripheraltorsion, hence T (Υ) = 1 for such a group Υ and and therefore Υ has 3 cusps.Hence N(P) ≤ 11.

15

4.1.2 Split Primes (Proof of Lemma 4.7)

Let Υ be as in the statement of the lemma, let q1 and q2 be such that (q1) = P1

and (q2) = P2 and let {1, ω} be an integral basis for Od. Since Λ(p) is generatedby p and pω, [Λd : Λ(p)] = p2. Since Γ(p) < Υ

x = [PSL2(Od) : Υ] = T (Υ)[Λd : Λ(Υ)]

and divides T (Υ)N(p) = T (Υ)p2 by the Index Lemma. By Wohlfahrt’s Lemma[Λd : Λ(Υ)] 6= 1 and we conclude that x = T (Υ)p or T (Υ)p2. Notice that ifΥΓ(Pi) has Od-level Pi the index

xi = [PSL2(Od) : ΥΓ(Pi)] = T (Υ)p or p

and a basis for Λ(ΥΓ(Pi)) is {qi, ωqi}. In this case, by Wohlfahrt’s Lemma Λ(Υ)is properly contained in Λ(ΥΓ(Pi)), and therefore [Λd : Λ(ΥΓ(Pi))] is a properdivisor of [Λd : Λ(Υ)] so [Λd : Λ(ΥΓ(Pi))] = p and |Λd : Λ(Υ)] = p2 implyingthat x = T (Υ)p2.

As a result, if p > 3 it suffices to show that the Od-level of ΥΓ(Pi) isPi. Assume otherwise, that ΥΓ(P1) = PSL2(Od). We will use the notationfrom Section 2.2. We have G1 = ρ1(ΥΓ(P1)) ∼= PSL2(Fp). We will identifyρ1(ΥΓ(P1)) with PSL2(Fp). Since p > 3, PSL2(Fp) is simple. As N1 / G1,N1 = {id} or PSL2(Fp) and so N1 has order equal to 1 or Θ(P1)/2. Therefore|N∗

1 | = 1, 2, Θ(P1)/2 or Θ(P1). Since

|N∗1 | =

x2Θ(P1)x

we conclude that x = x2Θ(P1), x2Θ(P1)/2, 2x2, or x2. Since x divides T (Υ)p2

by the Index Lemma, the first two are impossible. The second two imply that[Λd : Λ(Υ)] = [Λd : Λ(ΥΓ(P2))], contradicting Wohlfahrt’s Lemma. This showsthat ΥΓ(Pi) has Od-level Pi for i =1 and 2 by symmetry.

The case where p = 2 or 3 is similar.

4.1.3 Ramified Primes (Proof of Lemma 4.8)

Let q ∈ Od be such that (q) = P. If N(P) > 11 there are no one-cuspedcongruence subgroups of Od-level P by Lemma 4.6. It suffices to show that thereare none of Od-level P2 as by the Ladder Lemma, this implies that there are noone-cusped congruence subgroups of Od-level Pn for n > 0. Therefore, assumethat Υ is a one-cusped congruence subgroup of Od-level P2 with N(P) > 11,and let {1, ω} be an integral basis for Od. Since N(P) > 11 and p is ramified,T (Υ) = 1 as d 6= 1 or 3. By the Index Lemma, as Γ(P2) < Υ

x = [PSL2(Od) : Υ] = [Λd : Λ(Υ)]

and divides N(P2) = p2. By Wohlfahrt’s Lemma it is not 1, therefore x = p orp2. Notice that [Λd : Λ(Υ)] could equal p, for example when Λ(Υ) is generatedby {1, pω}.

16

First, assume that x = p2. We will use a vector space argument, and notationfrom Section 3.1. Here the vector space F corresponding to Υ ∩ Γ(P) moduloΓ(P2) is one-dimensional as

[Γ(P) : Υ ∩ Γ(P)] = [ΥΓ(P) : Υ] = p2.

Since Aα,β ∈ ΥΓ(P) for all α, β ∈ Od, Aα,β ·f ∈ F for all f ∈ F . By conjugatingwith A0,0 and A1,2 one can show this is not possible by implementing the VectorSpace Lemma.

Now it suffices to consider the x = p case. First, we will prove

Lemma 4.9. If x = p then PSL2(Z) < Υ.

Proof. Let φ be the reduction modulo P2 map and let

Υz = Υ ∩ PSL2(Z).

First, we will show that φ(Υz) = φ(Υ) ∩ φ(PSL2(Z)). To see this it suffices toshow that

φ(Υ) ∩ φ(PSL2(Z)) < φ(Υ ∩ PSL2(Z)).

If A ∈ φ(Υ) ∩ φ(PSL2(Z)) we have A = φ(C) = φ(B) for some C ∈ Υ andB ∈ PSL2(Z). Since φ(C) = φ(B), C ≡ B mod P2 and so B = CD for someD ∈ Γ(P2). Therefore B ∈ Υ since D ∈ Γ(P2) < Υ and therefore A = φ(B) forB ∈ Υ ∩ PSL2(Z).

Notice that

[φ(PSL2(Z)) : φ(Υ) ∩ φ(PSL2(Z))] ≤ [φ(PSL2(Od)) : φ(Υ)] = p

and as p > 11 there are no subgroups of PSL2(Fp) of index less than p + 1. So

φ(PSL2(Z)) = φ(Υ) ∩ φ(PSL2(Z)) < φ(Υ)

Since φ(Υ) ∩ φ(PSL2(Z)) < φ(Υ) we conclude that

φ(Υ ∩ PSL2(Z)) = φ(PSL2(Z))

and since kerφ < Υ we conclude that PSL2(Z) < Υ, proving Lemma 4.9.

Continuing with the proof of Lemma 4.8, PSL2(Z) < Υ, and hence Λ(Υ)is generated by 1 and ωp. Since Λ(P) is generated by q and ωq, we see thatΛ(Γ(P) ∩ Υ) is generated by p and ωp. By assumption ΥΓ(P) = PSL2(Od),and hence

[Γ(P) : Υ ∩ Γ(P)] = [ΥΓ(P) : Υ] = p.

Therefore, Υ ∩ Γ(P) corresponds to a two-dimensional subspace, F , of (Fp)3

and satisfies the hypotheses of the Vector Space Lemma. In particular, there isa basis of the form {(1, b1, 0), (0, b3, 1)} for F .

As ΥΓ(P) = PSL2(Od), Aα,β ∈ ΥΓ(P) for all α, β ∈ Od. So, for all f ∈ F ,Aα,β · f ∈ F . Specifically, A0,0 · (1, b1, 0) = −(1, 0, b1) ∈ F . As f2 = f1b1 + f3b3,

17

we conclude that 0 = b1(b3 +1) and so b1 = 0 or b3 = −1. Now A0,0 · (0, b3, 1) =−(0, 1, b3) ∈ F , so 1 = b2

3. As a result, b3 = ±1. Since b21 − 4b3 is not a square,

b1 = 0 and b3 = 1, or b3 = −1.

Conjugating by A1,2 results in a contradiction. This completes the proof ofLemma 4.8.

4.2 Prime Power Z-Levels (Proof of Proposition 4.2)

Assume that Υ is a one-cusped congruence subgroup of Z-level pn where p isa rational prime lying under a prime P ⊂ Od. If p = 2 then we assume thatp is inert, if p = 3 we assume it is unramified, and if p = 7 we assume thatT (Υ) = 1. We will prove that N(P) ≤ 11 and there is a constant c1(d,P)such that n ≤ c1. We will deal with each prime splitting type separately. Wehave shown in Lemma 4.8 that there are no one-cusped subgroups of Z-levelany power of a ramified prime greater than 11. Therefore the following threelemmas will suffice. In Section 4.2.1 we will prove

Lemma 4.10. (Powers of Ramified Primes) Let p be a ramified prime suchthat P lies over p and let Υ be a one-cusped congruence subgroup of Od-levelPn. Then n ≤ 1 if p = 7 or 11.

With Lemma 4.8 this proves Proposition 4.2 for ramified primes. If p is aninert prime lying over P and p > 3, the Ladder Lemma implies that if Υ is aone-cusped congruence subgroup of Od-level Pn, then ΥΓ(P) is a one-cuspedcongruence subgroup of Od-level P. By Proposition 4.1, N(P) ≤ 11, so p ≤ 3.So for inert primes, the following lemma, which will be proven in Section 4.2.2will suffice.

Lemma 4.11. (Powers of Inert Primes) Let p be an inert prime such thatP lies over p and let Υ be a one-cusped congruence subgroup of Od-level Pn.Then

n =

1 or 2 if p = 2 and T (Υ) 6= 12 if p = 2 and T (Υ) = 11 if p = 3 and T (Υ) 6= 10 if p > 3 or p = 3 and T (Υ) = 1.

For a split prime, we prove the following in Section 4.2.3

Lemma 4.12. (Powers of Split Primes) Let p be a split prime such that(p) = P1P2 and let Υ be a one-cusped congruence subgroup of Z-level pn. As-sume p 6= 2 and if p = 7 that T (Υ) = 1. Then

n ≤

2 if p = 31 if p = 5, 11 or p = 7 and T (Υ) = 10 if p > 11.

18

4.2.1 Powers of Ramified Primes (Proof of Lemma 4.10)

By the classification theorem of subgroups of PSL2(Fp) for p = 7 and 11 thereis a subgroup, H, of index p in PSL2(Fp) [17] and the pull-back of H necessarilyhas one cusp by Wohlfahrt’s Lemma. It is sufficient to show that there are noone-cusped congruence subgroups of Od-level P2 or P3. Let q ∈ Od be suchthat (q) = N(P), and let {1, ω} be an integral basis for Od.

First, we prove that there are no one-cusped congruence subgroups of Od-level P2. Assume that Υ is such a group. Then as d = 7 or 11, T (Υ) = 1and

x = [PSL2(Od) : Υ] = [Λd : Λ(Υ)] = p or p2.

To see this, recall that p divides x and x divides T (Υ)N(P2) = p2 by the IndexLemma.

If x = p, ΥΓ(P) is necessarily PSL2(Od). We use a vector space ar-gument as discussed in Section 3.1, applying the Vector Space Lemma. AsΥΓ(P) = PSL2(Od), A0,0 ∈ ΥΓ(P) and as in the proof of Lemma 4.8 we con-clude that b1 = 0 and b3 = 1, or b3 = −1. Since A1,2 ∈ ΥΓ(P), one can derivea contradiction.

Therefore, we will assume that x = p2, and hence a basis for Λ(Υ) is {p, ωp}.We have two cases, either the Od-level of ΥΓ(P) is P or is 1. If the Od-level ofΥΓ(P) is P, we use the Vector Space Lemma, where here F is a two-dimensionalsubspace of (Fp)3. When p = 7 and when p = 11, using the presentation forPSL2(Fp) in [16], and using [3] we find matrices of type Aα,β in PSL2(Fp) andsimilar to the proof of Lemma 4.8 we obtain a contradiction.

We now assume that the Od-level of ΥΓ(P) is 1. Therefore ΥΓ(P) isPSL2(Od) and as [ΥΓ(P) : Υ] = [Γ(P) : Υ ∩ Γ(P)] = p2 under the vectorspace correspondence, Υ ∩ Γ(P) corresponds to a one - dimensional subspace,F . As before, Λ(Υ ∩ Γ(P)) is generated by p and ωp, and thus neither (0, 1, 0)nor (0, 0, 1) is in F . Conjugating a possible generator by A0,0 and A1,2 resultsin a contradiction.

We now show that there are no one-cusped congruence subgroups of Od-level P3. In this case [PSL2(Od) : Υ] = p2 or p3 by the Index Lemma, and thefact that Λ(P3) < Λ(Υ). By the Ladder Lemma ΥΓ(P2) has Od-level P or P2.Since we have established that there are no one-cusped congruence subgroups ofOd-level P2, ΥΓ(P2) = ΥΓ(P) and has Od-level P. Therefore as Γ(P2)/Γ(P3)is a three-dimensional vector space over Fp, and Υ ∩ Γ(P2) corresponds to aone or two-dimensional subspace. In the two-dimensional case this results in acontradiction as above since Λ(Υ ∩ Γ(P2)) = Λ(P3). If F is one-dimensional,as A0,0 and A1,−1 are both in ΥΓ(P2) = ΥΓ(P), conjugation of a possiblegenerator by A0,0 and A1,−1 results in a contradiction.

4.2.2 Powers of Inert Primes (Proof of Lemma 4.11)

By the Ladder Lemma and Proposition 4.1 we need only consider p ≤ 3. Firstwe will prove Lemma 4.11 for p = 2 and second for p = 3.

19

Lemma 4.13. Assume that 2 is inert and lies under P. Let Υ be a one-cusped congruence subgroup of Od-level Pn. If T (Υ) = 1, then n = 2 and[PSL2(Od) : Υ] = 16. If T (Υ) = 3, (d = 3) then n = 1 or 2.

Proof. Case I: T (Υ) = 1.As PSL2(Od)/Γ(P) ∼= PSL2(F4) which has no subgroups of index 2 or 4, [3]

there are no one-cusped subgroups of Od-level P. One can show that there areno subgroups of index 2, 4 or 8 in PSL2(O3/(4)) [3]. There are subgroups ofindex 16, which can be seen to have one cusp by determining the correspondinglattices modulo (4) and hence determining the lattices of the pull-backs. Foran inert 2, O3/(4) ∼= Od/(4) for all other d with class number one, so theonly one-cusped congruence subgroups of Od-level (2) or (4) are such index 16subgroups.

By the Ladder Lemma it suffices to show that there are no one-cusped con-gruence subgroups of Od-level P3. Let Υ be such a group, so [PSL2(Od) : Υ] =32 or 64, since Υ < ΥΓ(4) which by the Ladder Lemma is one of the index16 subgroups above. By looking at the peripheral subgroups, there are no one-cusped subgroups of index 2 or 4 in these index 16 subgroups that could containΓ(8).Case II: T (Υ) 6= 1.

Here d = 3 and T (Υ) = 3. First, we will consider subgroups of O3-level (2);such subgroups must have index 6 or 12. In PSL2(F4) there is one conjugacyclass of subgroups of index 6 and one of index 12 [3]. Both must have one cuspby index considerations, and each index 12 subgroup is contained in an index 6subgroup [3]. Therefore we have one-cusped congruence subgroups of O3-level(2) of index 6 and 12, ΥA and Υsister. The group Υsister is isomorphic to thefundamental group of the sister of the figure-eight knot complement, a knot inthe lens space L(5, 1). Up to conjugation in PSL2(O3), Λ(ΥA) is generated by1 and 2ω and Λ(Υsister) is generated by 2 and 2ω.

If Υ has O3-level (4) then [PSL2(O3) : Υ] = 6, 12, 24, or 48. By indexconsiderations ΥΓ(2) must be index 6 or 12 and therefore must be a subgroupof the index 6 subgroup. (Otherwise an isomorphism theorem implies that 3divides [Γ(2) : Υ∩Γ(2)].) By searching low index subgroups of PSL2(O3/(4)) wesee that there are 3 conjugacy classes of index 12 subgroups [3]. One correspondsto Υsister. The other two have O3-level (4), Υ8, and ΥB . The group Υ8 isisomorphic to the fundamental group of the figure-eight knot complement. Up toconjugation the lattice Λ(Υ8) is generated by 1 and 4ω and Λ(ΥB) is generatedby 4 and ω − 1. There is one index 24 subgroup, ΥC , which is a subgroup ofΥB , and Λ(ΥC) is generated by 4 and 2ω− 2. There are no index 48 subgroupscontained in the index 6 subgroup [3].

One can show that there is no one-cusped congruence subgroup of O3-level(8) by examining subgroups of the appropriate index in one of the aforemen-tioned groups modulo (8). For each subgroup of the appropriate index, it eitherhas more than one cusp as seen by looking at the peripheral subgroup, cannotbe a congruence subgroup by Wohlfahrt’s Lemma, or can be shown not to becongruence by explicitly computing the order of the group in MathematicaTM .

20

The groups Υsister, Υ8 and their subgroups are the only one-cusped congru-ence subgroups that are known to be torsion-free.

Lemma 4.14. Let Υ be a one-cusped congruence subgroup of Od-level Pn wherep = 3 is inert and lies under P. Then T (Υ) = 2, d = 1, and n = 1.

Proof. Case I: T (Υ) = 1As PSL2(F9) has no index 3 or 9 subgroups, there is no one-cusped subgroup ofOd-level (3) [3] [17]. Assume Υ is a one-cusped congruence subgroup of Od-level(9). Then

x = [PSL2(Od) : Υ] = 3, 32, 33, or 34

by the Index Lemma. Using [3] we can see that there are no index 3, 9, or 27subgroups of PSL2(O1)/N where N is the normal closure of the group generatedby

±(

1 90 1

)and±

(1 9i0 1

).

Hence there are no index 3, 9, or 27 subgroups of PSL2(Od) that contain Γ(9)for any d such that 3 is inert. Therefore x = 81. Since ΥΓ(P) = PSL2(Od),[ΥΓ(P) : Υ] = 34, and therefore Υ ∩ Γ(P) is an index 34 subgroup of Γ(P). Asbefore since Γ(P2) < Υ∩Γ(P), Υ∩Γ(P) corresponds to a subspace F of (F9)3.Since Λ(Υ) is generated by 9 and 9ω, (0, f, 0) or (0, 0, f) are in F only whenf = 0, similar to the Vector Space Lemma. Conjugating by A0,1, A0,0 and A1,2,which are all in ΥΓ(P) = PSL2(Od), results in a contradiction.Case II: T (Υ) 6= 1Since T (Υ) 6= 1, T (Υ) = 2 and d = 1. Let Υ be a one-cusped congruencesubgroup ofO1-level (3). Then [PSL2(O1) : Υ] = 6 or 18. There is one conjugacyclass of index 6 one-cusped congruence subgroups and no index 18 subgroups[3]. By the Ladder Lemma, it suffices to show that there are no one-cuspedcongruence subgroups of O1-level (9). Assume that Υ is such a subgroup. First,notice that ΥΓ(3) 6= PSL2(O1). If this were the case, then as 2 divides [ΥΓ(3) :Υ] = [Γ(3) : Υ ∩ Γ(3)] implying that 2 divides [Γ(P) : Γ(P2)], but [Γ(P) :Γ(P2)] = 93. Let x = [PSL2(O1) : Υ]. Since 32 divides x and x divides 2N(9)by the Index Lemma, x = 2 · 32, 2 · 33, or 2 · 34.

The first case cannot occur because the index 6 subgroup has 1 conjugacyclass of index 3 subgroups but none have O1-level (9)[3]. We now consider thecase where x = 2 · 33. The elements A0,0, A1,2, A1,0, A0,1, and A1,1 are allcontained in the index 6 subgroup of PSL2(Od) which must be ΥΓ(3)[3]. LetF correspond to Υ ∩ Γ(3), a 34-dimensional vector space over F3. Notice thatΛ(Υ∩Γ(3)) is generated by 9 and 3zi, or 9i and 3z, for z = 1 or 2. (We cannotsay that (0, f, 0) or (0, 0, f) are in F only when f = 0. In fact, vectors of thisform must be in F , but not both (0, 1, 0) and (0, i, 0), by Wohlfahrt’s Lemma.)However, conjugation by A0,0 and A0,1 results in a contradiction. If x = 2 · 34,we have F corresponding to Υ ∩ Γ(P) as before. Here F is three-dimensional

21

and we can conclude that (0, 0, f) or (0, f, 0) are in F only when f = 0, as Λ(Υ)is generated by 9 and 9i. Conjugating by A1,2, A0,0, A1,1 and A0,1 result in acontradiction.

4.2.3 Powers of Split Primes (Proof of Lemma 4.12)

Let p be a rational prime such that P1 and P2 lie over p, and let q1 and q2 bein Od such that (qi) = Pi. We will now recall what we have proven thus far.We have shown in Proposition 4.1 that there are only one-cusped congruencesubgroups of Z-level p for p ≤ 11. If Υ is a one-cusped congruence subgroup ofZ-level pn, the Ladder Lemma implies that if p > 3, then ΥΓ(p) has Z-level pand hence p ≤ 11. (If p = 2 or 3, the Ladder Lemma implies that ΥΓ(p2) hasZ-level p2 if n ≥ 3.) Lemma 4.7 implies that if p > 3 and the Od-level of Υ isp then [PSL2(Od) : Υ] = T (Υ)p2 and ΥΓ(Pi) has Od-level Pi for i = 1 and 2.If p ≤ 3 and Υ has Od-level p then either the index is T (Υ)p2 and ΥΓ(Pi) hasOd-level Pi or the index is T (Υ)p.

Definition 4.1. Let Υ < PSL2(Od) be a congruence subgroup of Z-level l, andlet {1, ω} be an integral basis for Od. Then D is a proper diagonal of Λ(Υ) if Dis a primitive element of Λ(Υ) such that D = d1 + d2ω for 0 ≤ d1, d2 < l.

Lemma 4.15. (The Peripheral Lattice) Let Υ be a one-cusped congruencesubgroup of PSL2(Od) of Z-level pn where p is a prime. Then Λ(Υ) is generatedby Λ1(Υ) and Λω(Υ)ω if there are no proper diagonals. Otherwise, Λ(Υ) isgenerated by Λ1(Υ) and D (and also D′ and Λω(Υ)) for an appropriate diagonalD (and D′) in Λ(Υ).

Moreover, if D = pr(gpn−k+ω) then all other elements of the form pr(hpn−k+ωh′) are multiples of D modulo pn and ωpk, i.e. there are a, b, and c ∈ Z suchthat pr(hpn−k + ωh′) = aD + bpn + cωpk.

The proof of Lemma 4.15 follows from a theorem in the geometry of numberswhich states that if b1, b2 are two independent points of a lattice Λ in R2, andthe closed triangle with vertices 0, b1, b2 does not contain other points of Λ,then b1 and b2 form a basis for Λ [5]. Now we show

Lemma 4.16. Let p be split as (p) = P1P2 and let qi ∈ Od be such that(qi) = Pi. Let Υ be a one-cusped congruence subgroup of Od-level Pn1

1 Pn22 such

that n1 > n2 ≥ 0. Then [PSL2(Od) : Υ] = T (Υ)pn1+n2 and Λ(Υ) is generatedby qn1

1 qn22 and ωqn1

1 qn22 .

Proof. First, consider the case where n2 = 0. By the Index Lemma, since theZ-level is pn1 , [Λd : Λ(Υ)] divides pn1 . If p > 3, then pn1 divides [Λd : Λ(Υ)]and so Λ(Υ) = Λ(Pn1) and is generated by qn1

1 and ωqn11 . If p = 3 one can

check that since Υ has Od-level Pn11 that ΥΓ(P1) has Od-level P1. Therefore

Λ(Υ) is generated by qn11 and ωqn1

1 in this case as well. Hence [PSL2(Od) : Υ] =T (Υ)pn1 .

22

Now, assume both n1 and n2 are non-zero. Since p is a split prime with(p) = P1P2 there is an a ∈ Z such that P1 = (a + ω) and P2 = (a − ω).Therefore aP1 ≡ ωP1 mod p and −aP2 ≡ ωP2 mod p. It is a consequence ofthis and Wohlfahrt’s Lemma that Λ1(Υ) = Λω(Υ) = pn1 . (See Definition 2.1)From Lemma 4.15 we know that all elements in Λ(Υ) are of the form

AD + Bpn1 + ωCpn1

where D = pr(g + g′ω). From this and Wohlfahrt’s Lemma, we conclude thatgenerators of the lattice are qn1−n2

1 and pn1 , or qn1−n21 and ωpn1 . Therefore

[PSL2(Od) : Υ] = T (Υ)pn1+n2 .

The following three lemmas complete the proof of Proposition 4.2 as we havealready shown that p ≤ 11. The remainder of this section will consist of theirproofs.

Lemma 4.17. Let Υ be a one-cusped congruence subgroup of Od-level Pn whereP lies over p, which is split. Then unless p = 7 and T (Υ) 6= 1,

n ≤{

2 if p = 31 if 5 ≤ p ≤ 11.

Lemma 4.18. Let Υ be a one-cusped congruence subgroup of Od-level (pn)where p is split. Then unless p = 7 and T (Υ) 6= 1,

n ≤{

2 if p = 31 if 5 ≤ p ≤ 11.

Lemma 4.19. Let Υ be a one-cusped congruence subgroup of Od-level Pn11 Pn2

2

where n1 ≥ n2 and Pi lies over over p for i = 1, 2. Then unless p = 7 andT (Υ) 6= 1,

n1 ≤{

2 if p = 31 if 5 ≤ p ≤ 11.

Proof of Lemma 4.17Let Υ be as in the statement of the lemma. Let Υz denote Υ∩PSL2(Z) and

xz = [PSL2(Z) : Υz].

Notice that Υz is a congruence subgroup of PSL2(Z) containing Γz(pn), theprincipal congruence subgroup of PSL2(Z) of Z-level pn, that is, Γz(pn) =PSL2(Z) ∩ Γ(Pn). This implies that Υz has finite index in PSL2(Z). Onecannot a priori conclude that H2/Υz has one cusp, or that the Z-level of Υz ispn, only that pn divides the Z-level of Υz. Let φ denote the modulo Pn mapfrom PSL2(Od) to PSL2(Z/pnZ). (We will also use φ to denote the restriction

23

of this map to PSL2(Z).) As such, the kernel of the restriction to PSL2(Z) isΓz(pn). It is an elementary exercise to show that xz divides x.

Both PSL2(Od) and PSL2(Z) surject PSL2(Z/pnZ) via φ as φ(PSL2(Z))is isomorphic to PSL2(Z/pnZ) and |PSL2(Z/pnZ)| = |PSL2(Od/Pn)|. SinceΥz < Υ, x divides xz. We conclude that φ(Υz) = φ(Υ) and x = xz = T (Υ)pn.Let

Ab = ±(

1 b0 1

).

Notice that Ab is not contained in Υ for any 0 < b < pn by Lemma 4.16.Thus the Z-level of Υz is pn. Every cusp of Υz corresponds to the same one-dimensional lattice, as the stabilizer of every cusp is conjugate. Therefore,

T (Υ)pn = (the cusp width of Υz)× (the number of cusps of Υz)

where we say a cusp C of a subgroup Υ of PSL2(Z) has width w if after conju-gating Υ in PSL2(Z) such that the image of C is ∞, the stabilizer of infinity isgenerated by the positive integer w. As the cusp width of Υz is pn, if T (Υ) = 1,H2/Υz has one cusp. As the Z-level of Υz is pn, the lemma follows by Petersson’sresult [13].

If T (Υ) 6= 1, necessarily d = 1 or 3. The only split prime less than or equalto 11 in O1 is 5, and in O3 is 7. The case where P lies over 7 and T (Υ) 6= 1is excluded in our hypothesis. Assume that d = 1, p = 5 and T (Υ) = 2. Firstassume Υ is a one-cusped congruence subgroup of Od-level P. There is oneconjugacy class of index 10 subgroups of PSL2(O1) of level P for any P lyingover 5 and these all contain peripheral torsion and therefore have 2 cusps. IfΥ has Od-level P2 and index 50 modulo P2, φ(Υ) is an index 10 subgroup ofφ(ΥΓ(P)), an index 5 subgroup in PSL2(Z/25Z). Such a subgroup does notexist [3].Proof of Lemma 4.18

Let Υ be a one-cusped congruence subgroup of Od-level (pn). We will usethe notation established in Section 2.2. For 5 ≤ p ≤ 11 it suffices to show thatthere are no one-cusped congruence subgroups of Od-level (p2), establishing theresult by the Ladder Lemma. As p > 3, we conclude by Lemma 4.7 that ΥΓ(p)has Od-level (p), index p2 or T (Υ)p2, and ΥΓ(P1) and ΥΓ(P2) have Od-levelsP1 and P2, respectively. By the Index Lemma, x = T (Υ)pk where k = 2, 3 or4, but the above shows that k 6= 2.

First consider the case where T (Υ) = 1. Since there are no one-cuspedcongruence subgroups of Od-level P2

i for i = 1 or 2, ΥΓ(P2i ) = ΥΓ(Pi) and

x1 = [SL2(Od/P21 ) : G∗

1] = p = [SL2(Od/P22 ) : G∗

2] = x2.

With N∗i a normal subgroup of G∗

i as defined as in Section 2.2,

[G∗1 : N∗

1 ] =x

x1x2= p or p2.

As a result, the abelianization of G∗1 is divisible by p. One can check that this

is not the case [3] [17].

24

The case where d = 1, p = 5 and T (Υ) = 2 is similar, as is the case wherep = 3.Proof of Lemma 4.19

Let Υ be a one-cusped congruence subgroup of Od-level Pn11 Pn2

2 with n1 ≥n2. By the Ladder Lemma, if p > 3 and n2 > 1 then ΥΓ(P1P2) has Z-level p.Therefore p ≤ 11 by Proposition 4.1. If 5 ≤ p ≤ 11 then ΥΓ(P2

1P22 ) has Z-level

p2. From Lemma 4.18, the Od-level cannot be (p2), and from Lemma 4.17 itcannot be P2

i . So it suffices to show that there are no one-cusped congruencesubgroups of Od-level P2

1P2. (Similarly, if p = 3 it suffices to rule out theexistence of such groups of Od-level P3

1P2 and P31P2

2 .)First, assume that Υ has Od-level P2

1P2 and 5 ≤ p ≤ 11. By Lemma 4.16,

[PSL2(Od) : Υ] = T (Υ)p3.

Also with N∗1 defined in Section 2.2

[SL2(Z/p2Z) : N∗1 ] =

T (Υ)p3

x2

with x2 = 1 or p. But this implies that p2 divides |SL2(Z/pZ)| which is not thecase.

The proof in the case p = 3 is similar to the above and to the proof of Lemma4.17.

4.3 Composite Z-Levels (Proof of Proposition 4.3)

Recall that Proposition 4.3 states that if Υ is a one-cusped congruence subgroupof Z-level l then N(P) ≤ 11 for all P dividing (l). By Corollary 3.2 (see section3) it suffices to show that N(P) ≤ 11 for all primes P dividing the Od-level ofΥ. First, we will prove

Lemma 4.20. Let Υ be a one-cusped congruence subgroup of Z-level l. Let pbe the largest prime dividing l. Then N(P) ≤ 11 for all P in Od lying over p.

Proof. Let Υ be a one-cusped congruence subgroup of Od-level (`). As Od is aDedekind domain,

(`) = Pn00 Pn1

1 . . .Pnss

where Pi is a prime for 0 ≤ i ≤ s and Pi 6= Pj for all i 6= j. Let pi be therational prime lying under Pi, and order the Pi such that pi+1 ≤ pi. We willshow that N(P0) ≤ 11.

The Index Lemma implies that if pi is unramified, then pnii divides x =

[PSL2(Od) : Υ], and if pi is ramified then pdni

2 ei divides x, where dre is the

ceiling function. By Proposition 4.1, we may assume that s > 0, and if p0 = p1

then s > 1.We will use the notation established in Section 2.2. The approach will be to

choose non-zero `1 and `2 in Od with (`) = (`1) ∩ (`2) and (`1, `2) = Od such

25

that there is a rational prime p dividing x but not Θ(`2). Let l be the smallestrational integer such that (l) ⊂ (`), and let li be the smallest rational integersuch that (li) ⊂ (`i). Therefore, as

x1Θ(`2)x

∈ Z

by Lemma 2.1, p divides

x1 = [PSL2(Od) : ΥΓ(`1)]

and hence ΥΓ(`1) is a proper one-cusped congruence subgroup of PSL2(Od),whose Od-level divides (`1). Notice that if p0 is split with P0P1 = (p0) thenp0 = p1. If l2 is relatively prime to p0, then as p0 is chosen to be larger thanany pi dividing l2, and since

Θ(`2) =∏Pi|`2

N(Pi)3ni−2(N(Pi)2 − 1)

we conclude that unless p0 = 3 that p0 can only divide Θ(`2) if it divides thefactor p2

i +1 corresponding to an inert pi. We will assume that it is not the casethat p0 = 3 and ps = 2 as here N(P0) ≤ 11. We will break down the proof ofLemma 4.20 into four cases.

First, we prove a useful lemma.

Lemma 4.21. If Υ is a one-cusped congruence subgroup of Z-level 2a23a35a5

where ap = 0 unless p is inert, then a3 = a5 = 0 unless d = 1, in which caseneither 2 nor 5 is inert, and a3 may equal 1.

Proof. Let P2, P3, and P5 lie over 2, 3, and 5, respectively. We may assume thatd 6= 1, and hence there are no one-cusped congruence subgroups of Z-level anypower of 3 or 5. First, assume that a5 = 0. Let (`1) = Pa3

3 and (`2) = Pa22 . We

will be using subgroups N1 and N∗1 defined in Section 2.2. By Proposition 4.2,

x1 = [PSL2(Od) : ΥΓ(`1)] = 1 as there are no one-cusped congruence subgroupsof Z-level a power of 3, and hence none of Od-level a power of P3. Therefore byLemma 2.1

x1Θ(`2)x

=26a2−4 · 3 · 5

x∈ Z

and we conclude that 32 does not divide x. By the Index Lemma we see thata3 =1 or 2. If a3 = 1 then as PSL2(F9) is simple, under the isomorphismPSL2(Od/P3) ∼= PSL2(F9), N1 = {id} or PSL2(F9), as it is a normal subgroup.Since

[PSL2(F9) : N1] =(

1 or12

)[SL2(F9) : N∗

1 ]

and[SL2(F9) : N∗

1 ] =x

x2

26

we conclude that x = x2, 2x2, x2Θ(`1) or x2Θ(`1)/2. Recall that x2 = [PSL2(Od) :ΥΓ(`2)] and therefore x2 divides T (Υ)[Λd : Λ(`2)], hence 3 does not divide x2

as d 6= 3. But, 3 does divide x by the Index Lemma. Therefore x 6= x2 or2x2. But Θ(`1) cannot divide x as 5 divides Θ(`1) but not x, since x dividesT (Υ)N(2a23a3) = T (Υ)22a232a3 . Therefore a3 = 2 but in this case as 3 and 2are inert and 32 does not divide x, we see that 32 cannot divide either Λ1(Υ)or Λω(Υ) (see Definition 2.1). Therefore Λ(3 ·2a2) is contained in Λ(Υ), contra-dicting the fact that a3 = 2. The case where a2 or a3 is zero is similar.

Finally, assume that none of a5, a3 or a2 is zero. Let (`1) = Pa55 Pa3

3 and(`2) = Pa2

2 . From above we conclude that x1 = 1 and as

x1Θ(`2)x

=26a2−4 · 3 · 5

x∈ Z

we see that 52 does not divide x. Therefore a5 = 1 by the Index Lemma andsimilar to above we have a contradiction if we set (`1) = P5 and (`2) = Pa3

3 Pa22 .

Case 1: p0 6= p1 and p0 does not divide p2i + 1 for any inert pi dividing l .

Let(`1) = Pn0

0

and(`2) = Pn1

1 Pn22 . . .Pns

s .

Since p0 > pj for all 0 < j ≤ s, p0 does not divide Θ(`2), as

Θ(`2) =s∏

j=1

N(Pj)3nj−2(N(Pj)2 − 1)

but p0 does divide x by the Index Lemma. Therefore, p0 divides x1 sincex1Θ(`2)/x is an integer. So ΥΓ(Pn0

0 ) is a one-cusped subgroup of PSL2(Od)with Od- level a positive power of the prime P0, and therefore Z-level a positivepower of p0. Hence N(P0) ≤ 11 by Proposition 4.2.

Case 2: p0 6= p1 but p0 divides p2i + 1 for some inert pi dividing l.

One can check that if p0 > 11 is ramified, that p0 does not divide p2i +1 for any

inert pi. Therefore we may assume that p0 is unramified. One can show that ifpi and pj are inert, p0 > pi > pj , and p0 divides p2

i + 1, that either p0 does notdivide p2

j + 1 or p0 = 5 and pi = 3. Lemma 4.21 implies that in this case p0 issplit, and therefore N(P0) = 5.

If p20 divides x, let

(`1) = Pn00

and(`2) = Pn1

1 Pn22 . . .Pns

s .

27

Using the notation in Section 2.2 recall that

x1Θ(`2)x

∈ Z

by Lemma 2.1. Since p0 > pi, p20 does not divide p2

i + 1, and as p0 does notdivide p2

j + 1 for any other inert prime pj , p20 does not divide Θ(`2). As we are

assuming p20 divides x, we conclude that p0 divides x1 and ΥΓ(Pn0

0 ) has Od-level a positive power of P0, and therefore Z-level a positive power of p0. HenceN(P0) ≤ 11 by Proposition 4.2.

Therefore we may assume that p20 does not divide x. By the Index Lemma,

this implies that n0 = 1 since p0 > 3, and we let

(`1) = P0Pnii , and (`2) = Pn1

1 . . .Pni−1i−1 P

ni+1i+1 . . .Pn2

s .

Now p0 divides x but not Θ(`2), so p0 divides x1. Therefore ΥΓ(P0Pnii ) is a one-

cusped congruence subgroup of Od- level P0 or P0Pmii for some 1 ≤ mi ≤ ni.

In the first case, N(P0) ≤ 11 by Proposition 4.1.It now suffices to assume that ΥΓ(P0Pni

i ) has Od-level P0Pmii (and hence

ΥΓ(P0Pnii ) = ΥΓ(P0Pmi

i )) and p20 does not divide [PSL2(Od) : ΥΓ(P0Pmi

i )].Abusing notation, let

(`) = P0Pmii , (`1) = Pmi

i , (`2) = P0,

andx = [PSL2(Od) : ΥΓ(P0Pmi

i )].

It is an elementary exercise to show that with p0 and pi as above, pi does notdivide p0 ± 1 unless pi = 2 and p0 = 5. Similar to the proof of Lemma 4.21 onecan show that p0 is not inert. Therefore, pi divides x by the Index Lemma, butnot

Θ(`2) = p0(p0 + 1)(p0 − 1)

since p0 is not inert. Since x1Θ(`2)/x ∈ Z we conclude that pi divides x1.Therefore ΥΓ(Pmi

i ) has Od-level a positive power of Pi and therefore Z-level apositive power of pi. Hence N(Pi) ≤ 11 by Proposition 4.2. As we are assumingthat pi is inert, we conclude that pi is 2 or 3 and therefore p0 = 5 which mustbe split.

Case 3: p0 = p1 and p0 does not divide p2i + 1 for any inert pi dividing l.

Let(`1) = Pn0

0 Pn11

and(`2) = Pn2

2 Pn33 . . .Pns

s .

As beforex1Θ(`2)

x∈ Z

28

and so p0 divides x but not Θ(`2). Thus p0 divides x1 and ΥΓ(Pn00 Pn1

1 ) hasOd- level Pm0

0 Pm11 or, say, Pm0

0 . In either case, the Z-level is a power of p0 andby Proposition 4.2, N(P0) ≤ 11.

Case 4: p0 = p1 and p0 divides p2i + 1 for some inert pi dividing l.

We may assume that p0 > 3. If p20 divides x, then let (`1) = Pn0

0 Pn11 and

(`2) = Pn22 . . .Pns

s . Sincex1Θ(`2)

x∈ Z

and as p20 does not divide Θ(`2), p0 divides x1 and therefore the Z- level of

ΥΓ(Pn00 Pn1

1 ) is a proper power of p0. By Proposition 4.2 we conclude thatN(P0) ≤ 11. So assume that p2

0 does not x and therefore n0 = n1 = 1 by theIndex Lemma. Let

(`1) = P0P1Pnii

and(`2) = Pn2

2 . . .Pni−1i−1 P

ni+1i+1 . . .Pns

s .

Therefore p0 divides x, but not Θ(`2).By previous work, we may assume ΥΓ(P0P1Pni

i ) has Od- level P0P1Pmii or

P0Pmii where 1 ≤ mi ≤ ni. Abusing notation let (`1) = Pmi

i , and (`2) = P0P1

or P0, respectively. Recall that

x1Θ(`2)x

∈ Z

and pi divides x by the Index Lemma. As commented before, pi does not divide

Θ(`2) = (p0(p0 + 1)(p0 − 1))2 or p0(p0 + 1)(p0 − 1)

unless p0 = 5 and pi = 2. If pi does divide Θ(`2), by Lemma 4.21, p0 is notinert, so N(P0) ≤ 11. If pi does not divide Θ(`2) we conclude that pi dividesx1, and hence ΥΓ(P0P1Pni

i )Γ(Pmii ) has Od-level a power of Pi and therefore

N(Pi) ≤ 11. Since pi is inert and N(Pi) ≤ 11 pi = 2 or 3, implying that p0 = 5and we conclude that N(P0) ≤ 11 by Lemma 4.21, completing the proof ofLemma 4.20

Table 2: Θ(P) for small primes P in Od

N(P) 2 3 4 5 7 9 11 25Θ(P) 2· 3 233 223· 5 233· 5 243· 7 24325 233· 5· 11 243· 5213

Now we complete the proof of Proposition 4.3. Let Υ be a one-cuspedcongruence subgroup of Od-level (`). It suffices to do a few calculations. As inthe previous proof, write

(`) = Pn00 Pn1

1 . . .Pnss

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with Pi lying over pi, and ordered so that pi ≥ pi+1. By Lemma 4.20 we knowthat N(P0) ≤ 11. Therefore either p0 is split or ramified and p0 ≤ 11 or p0 isinert and p0 = 2 or 3. It suffices to rule out the possibility that p0 is split orramified and there is an inert prime pi such that 11 > pi > 3. We will call suchan inert prime p, lying under P, with n the largest power of P dividing (`). Theonly possibilities are p = 5 or 7. If p = 7, notice that gcd(p, Θ(Q)) = 1 for anyother possible prime Q dividing (`). Let (`1) = Pn. Using the fact that

x1Θ(`2)x

∈ Z

we conclude that p divides x1, since p divides x by the Index Lemma but pdoes not divide Θ(`2). Therefore ΥΓ(Pn) is a one-cusped congruence subgroupof Z-level a proper power of p, which contradicts Proposition 4.2. Therefore itsuffices to show that p 6= 5, which can be shown in a manner similar to thatused in the proof of Lemma 4.21.

5 Proof of Corollary 1.2

To prove the corollary it is enough to show that 6 does not divide the index[PSL2(Od) : Υ] for any one-cusped congruence subgroup, Υ [6]. If p ∈ Z isprime, and inert or ramified, let Pp denote the prime lying over p. If p is split,then we will write (p) = PpQp. One can show the following

Proposition 5.1. Let Υ be a one-cusped congruence subgroup of PSL2(Od),where d =19, 43, 67, or 163. Then Υ contains torsion and

a) If d = 19 then the Z-level Υ divides 1127252225.

b) If d = 43 then the O43-level is one of the following:

P11,Q11,P11Q11,P22 ,P2

2P11,P22Q11,P2

2P11Q11.

c) If d = 67 or 163 then [PSL2(Od) : Υ] = 16 and the Od-level of Υ is (4).

The proof is similar to those of Proposition 4.3 and Lemma 4.21.

6 Proof of Theorem 1.3

First, we will prove the theorem for K = Q. We will say a cusp C of a subgroupΥ of PSL2(Z) has width w if after conjugating Υ in PSL2(Z) such that the imageof C is ∞, the stabilizer of infinity is generated by the positive integer w. Inother words, if the degree of the cover of the cusp of MZ by C is w. Let p be anodd prime and let φ denote the modulo p map from PSL2(Z) to PSL2(Fp). Bythe classification of subgroups of PSL2(Fp) there is a maximal subgroup, H, ofindex p + 1 [17]. Since p is odd,

Γ(p) = {A ∈ PSL2(Z) : A ≡ I mod p}

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has (p2 − 1)/2 cusps all of width p. By Wohlfahrt’s Lemma, Υ = φ−1(H) hastwo cusps, one of width one, and one of width p. Therefore, letting p vary overall odd primes, there are infinitely many maximal subgroups of PSL2(Z) withtwo cusps.

The group H has a normal Sylow p-subgroup, S, and the quotient of H byS is cyclic of order (p − 1)/2 [17]. As p does not divide (p − 1)/2, the cuspof width one in φ−1(H) must lift to (p − 1)/2 cusps of width one in φ−1(S).Therefore the cusp of width p must lift to (p− 1)/2 cusps of width p in φ−1(S).So φ−1(S) has p−1 cusps. If l divides (p−1)/2 there is a subgroup L such thatS < L < H and [H : L] = l. The preimage, φ−1(L) has l cusps of width oneand l of width p, for a total of 2l cusps. Therefore we have subgroups of Z-levelp with 2l cusps for all divisors, l, of (p− 1)/2. For a fixed l, there are infinitelymany primes p such that p ≡ 1 mod 2l, and as a result we have infinitely manymaximal 2l-cusped subgroups.

Let K = Q(√−d) for d ∈ {2, 7, 11, 19, 43, 67, 163}. Here, we say that a cusp

C of a subgroup Υ of PSL2(Od) has width w if after conjugating Υ in PSL2(Od)such that the image of C is ∞, we have [Λd : Λ(Υ)] = w. Let P be a prime inOd, such that q = N(P) is odd. If A < PSL2(Fq) and the pre-image of A hasn-cusps, we will say that A has n cusps. If p is split, lying under P then weanalyze the quotient PSL2(Od) by Γ(P) and as above conclude that H has twocusps and for all divisors, l, of (p−1)/2 there is a subgroup L of PSL2(Fp) suchthat L has l cusps of width one and l cusps of width p, for a total of 2l cusps.

Now consider the case where p is inert, so the quotient is PSL2(Fp2). Asabove there is an index p2 + 1 subgroup, H, and H has a normal Sylow p-subgroup, S. The quotient H/S is cyclic of order (p2− 1)/2 [17]. One can showthat H has one cusp of width one and one cusp of width p2. Since S / H, and[H : S] = (p2 − 1)/2 we conclude that the cusp of with one in H is coveredby (p2 − 1)/2 cusps of width one in S, as all of the covers must have the samewidth. Therefore the cusp of width p2 is covered by (p2 − 1)/2 cusps of widthp2. We conclude that S has (p2 − 1)/2 cusps of width one and (p2 − 1)/2 ofwidth p2. As H/S is cyclic of order (p2 − 1)/2, for any l dividing (p2 − 1)/2,there is a subgroup L of H of index l which has l cusps of width one and l ofwidth p2, for a total of 2l cusps.

Therefore, combining the split and inert cases, given a positive integer l, weneed only show that there are infinitely many primes P ⊂ Od with

N(P) ≡ 1 mod 2l.

By Dirichlet’s Theorem on primes in an arithmetic progression, there are in-finitely many primes p ∈ Z such that p ≡ 1 mod 2l [18]. For any P lying overp, N(P) = p or p2 and therefore N(P) ≡ 1 mod 2l as well.

Now consider the case where d = 1 or 3. Notice that in O1 = Z[i], a primep splits when p ≡ 1 mod 4 and is inert if p ≡ −1 mod 4. In O3 = Z[ 1+

√−3

2 ]a prime p splits when p ≡ 1 mod 6 and p is inert when p ≡ −1 mod 6. Thisis due to the fact that both are cyclotomic extensions. In O1, consider a splitprime, p, lying under P and the quotient PSL2(Fp). Here {id} has (p2 − 1)/4

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torsion-free cusps, all of width p. Since |S| = p, we conclude that all cusps of Sare torsion-free and S has (p−1)/4 cusps of width one and (p−1)/4 of width p.As before, we conclude that H has one cusp (with torsion) of width one and onecusp (with torsion) of width p. Since φ−1(S) is peripheral torsion-free, there isa representative of peripheral torsion in the cyclic quotient H/S. Let T be thegroup such that S < T < H which corresponds to that quotient, so [T : S] = 2.Therefore, T has (p − 1)/4 cusps of with one and (p − 1)/4 of width p, all ofwhich have torsion. As above, for any l dividing (p−1)/4 we have a subgroup ofH containing T with 2l cusps. Since for any l, there are infinitely many primesp ∈ Z with p ≡ 1 mod 4l, these primes split and the PSL2(Fp) have subgroupswith 2l cusps. So, there is a subgroup of PSL2(O1) containing Γ(P) with 2lcusps, l of width one and l of width p. Similarly, for O3 for a given l, as thereare infinitely many primes p ∈ Z with p ≡ 1 mod 6l, these primes split and thePSL2(Fp) have subgroups with 2l cusps.

7 acknowledgements

This paper is part of the author’s doctoral thesis [11]. The author would like tothank her advisor, Alan Reid, for his guidance and support, and the referee formany useful comments and suggestions.

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