One-Cusped Congruence Groups of Bianchi Groups

One-Cusped Congruence Subgroups of Bianchi Groups

Kathleen L. Petersen

Abstract We show that there are only ﬁnitely many maximal congruence sub- 3 groups of the Bianchi groups such that the quotient by H has only one cusp.

1 Introduction

Let d√be a positive square free integer, and let Od denote the ring of integers of Q( −d). The groups PSL2(Od) are called the Bianchi groups and embed + 3 discretely in PSL2(C), which is isomorphic to Isom (H ), the orientation pre- 3 serving isometries of hyperbolic 3-space. The quotient Md = H /PSL2(Od) is a finite√ volume hyperbolic 3-orbifold√ with hd cusps where hd is the class number of 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 can 3 form the quotient MΥ = H /Υ, which is a finite volume hyperbolic 3-orbifold with finitely many cusps. We say that Υ has n cusps if this quotient has n cusps. The quotients Md are the prototypical examples of non-compact arithmetic 3-orbifolds, which are those orbifolds M such that M is commensurable with some Md. (Two orbifolds are commensurable if they share a finite sheeted cover.) For a non-zero ideal J ⊂ Od the principal congruence subgroup of level J is Γ(J ) = {A ∈ PSL2(Od): A ≡ I mod J} and has finite index in PSL2(Od). A (finite index) subgroup of PSL2(Od) is called 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 congruence subgroups of the Bianchi groups. Clearly, these must correspond to those Bianchi groups with associated class number equal to one, since if Υ < PSL2(Od), then MΥ covers Md. In fact, we will show that for d = 11, 19, 43, 67, or 163 that there are only finitely many one-cusped congruence subgroups of PSL2(Od). In addition, from our method of proof we will conclude that

1 Corollary 1.2. If d = 19, 43, 67, or 163 there are no torsion-free one-cusped congruence subgroups of PSL2(Od). A notable example of a one-cusped congruence subgroup is the fundamental group of the figure-eight knot complement, which injects into PSL2(O3) as an index 12 subgroup containing Γ(4) [7]. The fundamental group of the sister of the figure-eight knot complement, a knot in the lens space L(5, 1), also injects into PSL2(O3) as an index 12 subgroup containing Γ(2) [1]. Reid showed that the 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) onto Z, with a parabolic element generating the image. If d = 1 or 3 there are also infinitely many one-cusped subgroups, associated to torsion-free subgroups of finite index, e.g. subgroups of finite index in the fundamental group of the figure-eight knot complement [2]. There are examples of torsion-free one-cusped subgroups of PSL2(Od) corresponding to d = 1, 2, 3, 7, and 11, but historically such examples have proven difficult to find [2] [1]. In the setting of arithmetic manifolds and orbifolds, it has recently been shown that there is a finite number of commensurability classes of one-cusped orbifolds or manifolds of minimal volume [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). As PSL2(R) is isomorphic to the group of orientation preserving isometries of the 2 2 hyperbolic plane, H , we can form the quotient MQ = H /PSL2(Z). This quotient has finite volume as a hyperbolic 2-orbifold and has a single cusp. As in the

Bianchi group case, MQ is the prototype for non-compact arithmetic 2-orbifolds, which are defined as those orbifolds M commensurable with MQ. Rhode proved that there are at least two conjugacy classes of one-cusped subgroups of index n in the Modular group for every positive integer n [12]. One can define principal congruence subgroups and congruence subgroups of the Modular group analo- gous to the above Bianchi group definitions. Petersson proved that there are only finitely many one-cusped congruence subgroups of the Modular group, and that the index of any such group divides 55440 = 11 · 7 · 5 · 32 · 24 [13]. We now introduce a useful invariant of finite index subgroups of PSL2(Od), the level of the group, which has added meaning for congruence subgroups.

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

1 ` ± : ` ∈ L 0 1 is contained in Υ. Moreover, we say that Υ has Z-level l if l is the smallest positive integer such that the normal closure of the group generated by the

2 elements 1 lβ ± : β ∈ O 0 1 d is contained in Υ. We can now be more precise in regard to Theorem 1.1. We will show that any prime in Od that divides the Z-level of a one-cupsed congruence subgroup of PSL2(Od) has norm less than or equal to 11. In addition, if d = 1, 2, or 7, we show that there are only ﬁnitely many of odd Z-level, and if d = 3 we demonstrate 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 class number one, and let OK be the ring of integers of K. There are infinitely many maximal congruence subgroups of PSL2(OK ) that have two cusps. Moreover, for any even integer n, there are infinitely many primes P ⊂ OK such that there is an n-cusped congruence subgroup of OK -level P. The cases of the Modular group and the Bianchi groups differ vastly from those relating to other number fields. In general, let K be a number field with r real places and s complex places, and let OK be the ring of integers of r s K. The group PSL2(OK ) embeds discretely in PSL2(R) × PSL2(C) , which is isomorphic to the group of orientation preserving isometries of [H2]r × [H3]s. The quotient 2 r 3 s MK = [H ] × [H ] /PSL2(OK ) is equipped with a metric inherited from [H2]r × [H3]s and with respect to this 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 in other settings. For example, it is a famous conjecture that there are infinitely many real quadratics with class number equal to one. Principal congruence subgroups and congruence subgroups can be defined for any PSL2(OK ) similar to the imaginary quadratic case. We say that PSL2(OK ) has the Congruence Subgroup 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 or Q( −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 the topology of these quotients. By way of contrast to Theorem 1.1, if K is a number field with class number one, other than Q or an imaginary quadratic, and if i∈ / K, then assuming the Generalized Riemann Hypothesis, there are infinitely 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|. If p ∈ Z is a prime and P is a prime ideal in Od such that pOd ⊂ P, then we say that P lies over p, and p lies under P. Therefore p lies under P precisely when the norm of P, N(P), is a power of p. There are three possibilities for the decomposition of an ideal pOd when p is a rational prime

 2  P p is ramiﬁed pOd = P p is inert  P1P2 p is split.

This behavior can be completely classiﬁed. (See [8] for example.) If p is not ∼ inert and P lies over p, then Od/P = Fp and N(P) = p. If p is inert then ∼ 2 pOd = P, Od/P = Fp2 and N(P) = p .

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

d = 1 2 3 7 11 19 43 67 163 p = 2 R R I S I I I I I 3 I S R I S I I I I 5 S I I I S S I I I 7 I I S R I S I I I 11 I S I S R S S I I

Let K be a number field with r real places and s complex places. The ring of integers of a number field is a free abelian group of rank [K : Q]. There is always a Z-basis for OK , called an integral basis. The units in OK are precisely × those elements of norm ±1, and form a group, OK , under multiplication. The multiplicative group of units is isomorphic to a product of a free abelian group of rank r +s−1 and a finite cyclic group, consisting of roots of unity. Therefore, × |OK | is finite precisely when K is Q or an imaginary quadratic. This lies at the heart of the dichotomy√ between the associated groups, PSL2(OK ). Notice that when 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, and the non-zero principal ideals form a subgroup,√ Pd. The quotient, Id/Pd has finite√ 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 be convenient to use the notation J as well as (`). For example, we will use the notation Γ(`) = Γ(J ) for principal congruence subgroups. For the remainder of the section assume that J = (`) is a non-zero ideal with `1 and `2 ∈ Od such that (`) = (`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` as the reduction modulo J = (`) map from (P )SL2(Od) to (P )SL2(Od/(`)). The following 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/(ì)). ∼ 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) = n0 ns SL2(FN(P)). Since Od is a Dedekind domain we can write J = P0 ... Ps where the Pi are distinct prime ideals. We define Θ(J ) = [SL2(Od): G(J )]. Therefore [9] s Y 1 Θ(J ) = N(P )3ni 1 − . i N(P )2 i=0 i

Notice that for J = (`), and `1, and `2 as above, defining Θ(`) = Θ(J ) we have Θ(`) = Θ(`1)Θ(`2). We will commonly use Υ to denote a finite index (usually one-cusped and congruence) subgroup of PSL2(Od), and we define

G = φ`(Υ),Gi = φ`i (Υ).

Notice that Gi = φì (ΥΓ(ì)) (where ΥΓ(ì) is the composite of Υ and Γ(ì)) as Γ(ì) = kerφì . ∗ ∗ ∗ Let Υ be the SL2(Od) pull-back of Υ. We now define, G = φ`(Υ ) and ∗ ∗ ∗ ∗ ∗ ∗ Gi = φì (Υ ) = φì (Υ G(ì)) = ρi(G ). Define N1 < G1 as

∗ ∗ ∗ N1 = {g1 ∈ G1 :(g1, 1) ∈ G }

∗ ∗ ∗ and similarly for N2 . Let Ni = PNi . Notice that Ni is not deﬁned as the pull-back of Ni. It is an elementary exercise in group theory to show that ∗ ∗ ∼ ∗ ∗ ∗ ∗ ∗ ∗ ∗ G1/N1 = G2/N2 , |G | = |G1||N2 | = |G2||N1 | and

∗ ∗ ∗ [SL2(Od/`): G ] = [SL2(Od/(`1)) : G1][SL2(Od/(`2)) : N2 ]. Let x = [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(ì)] = [PSL2(Od/(ì)) : Gi] = [SL2(Od/(ì)) : Gi ]

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

∗ ∗ |G | x2Θ(`1) ∗ x1Θ(`2) |N1 | = ∗ = , |N2 | = . |G2| x 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 x Θ(` ) x Θ(` ) |N ∗| = 2 1 ∈ , |N ∗| = 1 2 ∈ . 1 x Z 2 x Z Additionally,

∗ x ∗ x [SL2(Od/(`1)) : N1 ] = and [SL2(Od/(`2)) : N2 ] = . x2 x1 Therefore ∗ ∗ ∗ ∗ x [G1 : N1 ] = [G2 : N2 ] = . x1x2

Notice that for some z dividing the order of the center of SL2(Od/(ì)) ∗ n |Ni|z = |Ni |. In particular, if (ì) = P where P is a prime ideal that does not 1 ∗ ∗ lie over 2, then the order of the center is 2 and |Ni| = 2 |Ni | or |Ni |, depending ∗ ∗ ∗ on whether ±I ∈ Ni or not. (Ni is defined with respect to Gi , not as the pull-back of Ni.) In this case 1 x Θ(` ) 1 x Θ(` ) |N | = 1 or 2 1 , |N | = 1 or 1 2 . 1 2 x 2 2 x

2.3 Peripheral Subgroups

As stated previously, Md = PSL2(Od) is a finite volume√ hyperbolic 3-orbifold with hd cusps, where hd is the class number of Q( −d). If Υ is a finite index 3 subgroup of PSL2(Od) then MΥ = H /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 a parabolic fixed point of Υ, that is if there is some parabolic ±A ∈ Υ such that

α β αC + β ±A = ± and ± A ·C := = C. γ δ γC + δ

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 therefore consists solely of parabolic elements unless d = 1 or 3. Two cusps are equivalent in MΥ if they are in the same Υ orbit. As such, the number of equivalence classes is the number of conjugacy classes of maximal peripheral subgroups of Υ, subgroups of the form StabC(Υ). This is the number of topological ends of MΥ.

Lemma 2.2.√Let Υ be a ﬁnite index subgroup of PSL2(Od) where the class number 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 of 0 0 truncated compact orbifolds Md and MΥ. This cover restricts to a cover of 0 0 0 0 ∂Md by ∂MΥ. The degree of the cover of Md by MΥ is the degree of the cover 0 0 of the cusp at infinity of Md by those cusps of ∂MΥ covering it. If MΥ has 0 one cusp, then as Md also has cusp, the cusp at infinity of MΥ is the only cusp 0 covering the cusp at infinity of Md. 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], 0 0 is the degree of the cover of the cusp at inﬁnity of Md by those cusps of ∂MΥ covering it. If

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

0 0 then only the cusp at inﬁnity√ of MΥ can cover the cusp at inﬁnity of Md. Since the 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

1 ` Λ(Υ) = ` ∈ : ± ∈ Υ . C 0 1

Moreover, let Λd = Λ(PSL2(Od)), Λ(J ) = Λ(Γ(J )), and Λ(`) = Λ(Γ(`)). Given a non-zero β ∈ Od, let Λβ(Υ) be the smallest positive integer b such that bβ ∈ 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 unity in Od are ±1. The fourth roots of unity are√ in O1, generated by i, and the sixth roots of unity are in O3 generated by (1 + −3)/2. If d 6= 1 or 3,

1 1 1 ω Stab (PSL (O )) = ± , ± ∞ 2 d 0 1 0 1 and geometrically, a cusp of Md is T × [0, ∞) where T is a torus. If d = 2, 7, 3 11, 19, 43, 67, or 163, and MΥ = H /Υ has one cusp, the degree of the covering of 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 S is the 2-sphere with four cone points with cone angles π. And

1 1 1 i −i 0 Stab (M ) = ± , ± , ± . ∞ 1 0 1 0 1 0 i

−i β Elements of the form ± for any β ∈ O , have order two. If Υ < 0 i 1 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 cone points each of cone angle 2π/3. And

1 1 1 ω ω−1 0 Stab (M ) = ± , ± , ± ∞ 3 0 1 0 1 0 ω where √ −1 + −3 ω = . 2 ω−1 β Elements of the form ± have order 3. As in the O case, if Υ < 0 ω 1 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 following deﬁnition.

Definition 2.2. Let Υ be a finite index one-cusped subgroup of PSL2(Od) then T (Υ) ∈ {1, 2, 3} is the number such that [PSL2(Od) : Υ] = T (Υ)[Λd : Λ(Υ)]. Specifically,   1 if d = 2, 3, 7, 11, 43, 67, 163  1 if d = 1, 3 and Υ has peripheral torsion T (Υ) =  2 if d = 1 and Υ has no peripheral torsion  3 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 is similar to that of the classical Modular group case and is omitted. The corollary follows directly. √ Lemma 3.1. (Wohlfahrt’s Lemma)Let Q( −d) have class number one, and 0 let Υ < PSL2(Od) be a congruence subgroup of Od-level (`). Then (` ) ⊆ (`) if and only if Υ contains Γ(`0). √ 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 congruence subgroup contained in Υ. 2. Υ has Z-level l if and only if Γ(l) is the maximal principal congruence subgroup 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 integer such that (l) ⊂ (`). The following is a generalization of the Modular group case [13]. The proof is omitted. Lemma 3.3. (The Ladder Lemma) Let Υ be a one-cusped congruence sub- m group of PSL2(Od). If Υ has Z-level p l where p is a prime, p 6 |l and 3 if p ≤ 3 m ≥ 2 if p > 3

m−1 m−1 n then ΥΓ(p l) has Z-level p l. If Υ has Od-level P where p is ramiﬁed and lies under P, and   5 if p = 2 n ≥ 4 if p = 3  3 if p > 5

n−1 n−1 n−2 then ΥΓ(P ) has Od-level P or P . Lemma 3.4. (The Index Lemma) Let Υ be a one-cusped congruence sub- n n group of PSL2(Od) of Z-level p l where p is a prime and p 6 |l. If p > 3 then p n0 0 divides [PSL2(Od) : Υ]. If p ≤ 3, then p divides [PSL2(Od) : Υ] where n = n if p divides the Z-level of ΥΓ(pl) and n0 = 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 2.2 and 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 such that (`) = 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 we conclude that [PSL2(Od) : Υ] divides T (Υ)N(`). We will prove the remainder of the lemma in the case where p > 3. The case p ≤ 3 is similar. Let Υ be as above and fix an integral basis {1, ω} for Od. For Υ1 < PSL2(Od), Λ(Υ1) is an abelian group, and if Υ2 < Υ1 then Λ(Υ2) < Λ(Υ1). By above, [PSL2(Od) : Υ] = T (Υ)[Λd : Λ(Υ)]. It suffices to n show that p divides [Λd : Λ(Υ)]. Fix l and p as in the statement of the lemma. We first consider the case where n = 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)]. Since p ∈ Z, [Λ(l) : Λ(pl)] = p2; we conclude that Λ(Υ) ∩ Λ(l) = Λ(l), and therefore Λ(l) < Λ(Υ). This implies that 1 l 1 lω ± and ± ∈ Υ. 0 1 0 1

By Wohlfahrt’s Lemma, the Z-level of Υ divides l contradicting the hypothesis that it is pl. Now assume that for all m such that 1 ≤ m < n, that for all one-cusped Υ of m m Z-level p l, p divides [Λd : Λ(Υ)]. Assume that Υ is a one-cusped congruence n n subgroup of Z-level p l and p does not divide [Λd : Λ(Υ)]. By the Ladder n−1 n−1 Lemma and the inductive hypothesis, p divides [Λd : Λ(ΥΓ(p l))]. Since Λ(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 that Λ(pn−1l) < Λ(Υ). Therefore 1 pn−1l 1 pn−1lω ± and ± ∈ Υ. 1 1 1 1

This contradicts Wohlfahrt’s Lemma as the Z-level of Υ is pnl. We conclude n that p 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 4.1 and 4.2 which which are stated in the next section.) This will be used as a convenient way to show that there cannot be one-cusped congruence subgroups in certain situations. n Let Υ be a one-cusped congruence subgroup of PSL2(Od) of Od-level P where n ≥ 2 and P is a prime lying over the rational prime p. Let q ∈ Od be such that (q) = P. If p 6= 2 the quotient Γ(Pn−1)/Γ(Pn) is a three-dimensional vector space, V, over FN(P). The correspondence is given by 1 + qn−1α qn−1β ± ∈ Γ(Pn−1) ↔ (α, β, γ) ∈ V qn−1γ 1 + qn−1δ for α, β, γ, and δ in Od. This correspondence is well-deﬁned modulo P. The subgroup Υ ∩ Γ(Pn−1) corresponds to a subspace, F, of V. Notice that if A ∈ ΥΓ(Pn−1) then A ≡ B modulo Pn−1 for some B ∈ Υ. One can use this fact to show that the action induced on V by conjugation by an element in ΥΓ(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 v1 v2 V = in M2(Od) such that v3 v4

B = I + qn−1V.

α β Let A = ± ∈ PSL (O ). The action of conjugating B by A, B → γ δ 2 d −1 ABA , interpreted in V sends the vector v = (v1, v2, v3) to the vector

2 2 2 2 A · v = ([αδ + βγ]v1 − αγv2 + βδv3, −2αβv1 + α v2 − β v3, 2γδv1 − γ v2 + δ v3). By the above discussion, if f ∈ F then A · f ∈ F when A ∈ ΥΓ(Pn−1). We will be explicitly making use of the action A · f for a few speciﬁc matrices. Let α, β ∈ Od and deﬁne 1 α 0 −1 1 β A = ± . α,β 0 1 1 0 0 1

Conjugating B by Aα,β takes v = (v1, v2, v3) to Aα,β · v =

2 [2αβ − 1] v1 − αv2 + αβ − β v3, 2 2 2 2 −2α β + 2α v1 + α v2 + −α β 2 + 2αβ − 1 v3, 2 2βv1 − v2 + β v3 . Lemma 3.5. (The Vector Space Lemma) Let Υ be a one-cusped congru- n n−1 n ence subgroup of PSL2(Od) of Od-level P . If Λ(Υ ∩ Γ(P )) = Λ(P ) then 2 2 (b1b2, b2, −b1) ∈ F if and only if b1 = b2 = 0. Moreover, if p lies under P and is 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. 2 iii) If p > 2 then b1 − 4b3 is not a square in Fp. In particular, neither (0, 1, 0) nor (0, 0, 1) are in F. The proof follows from the fact 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 of propositions. The proofs of the following three propositions will comprise the bulk of the manuscript. The proof of Proposition 4.2 will rely upon Proposition 4.1, 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 of Theorem 1.1.

Proposition 4.1. (Prime Z-Levels) Let Υ be a one-cusped congruence subgroup 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 congruence subgroup of Z-level pn where p is the rational prime lying under a prime P ⊂ Od. Then N(P) ≤ 11. Moreover, unless p = 2 and is not inert, p = 3 is ramiﬁed or p = 7 and T (Υ) 6= 1, then   0 if N(P) > 11, or p = 3 is inert and T (Υ) = 1   1 if p = 7, 11 is ramiﬁed   or p = 3 is inert and T (Υ) 6= 1 n ≤ or p = 5, 11 is split   or p = 7 is split and T (Υ) = 1   2 if p = 3 is split  or p = 2 is inert.

Proposition 4.3. Let Υ be a one-cusped congruence subgroup of Z-level l. Then N(P) ≤ 11 for all primes P dividing (l). Now it suffices to prove the following, which implies Theorem 1.1 and the discussion following it. Proposition 4.4. There are finitely many maximal one-cusped congruence subgroups of the Bianchi groups. Furthermore, any prime in Od dividing the Z-level of such a group has norm at most 11. If d = 1, 2, or 7 there are finitely many one-cusped congruence subgroups of odd Z-level, and if d = 3 there are finitely many one-cusped congruence subgroups of Z-level relatively prime to 21. If d = 11, 19, 43, 67, or 163 there are finitely many one-cusped congruence subgroups of PSL2(Od).

12 Unless d = 3 and p = 3, or 7, or d = 1, 2, or 7 and p = 2, the proof will produce explicit bounds on the maximal power of any prime p dividing the Z-level of a one-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 on d, such that the Z-level of any one-cusped congruence subgroup of PSL2(Od) is only 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 a c1 constant c1 = c1(p, d) such that p does not divide the Z-level of any one-cusped congruence subgroup of PSL2(Od). b) If p is a split or ramified 2, ramified 3, or split 7 in O3, there is a constant c1 c1 = c1(p, d) such that if p divides the Z-level of a one-cusped congruence subgroup, Υ, then if p = 7, Υ < ΥΓ(p) is a one-cusped congruence subgroup of Z-level 7, and if p ≤ 3 then Υ < ΥΓ(p2) is a one-cusped congruence subgroup of Z-level p2. Proof. (proof of Lemma 4.5) Let p be a prime such that N(P) ≤ 11 for all primes P in Od lying over p. Let r ∈ Z be a prime distinct from p. Recall that  r6m−2(r2 − 1) if r is ramified m m  6m−4 4 Θ(r ) = |SL2(Od/(r ))| = r (r − 1) if r is inert  r6m−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 is independent of m. Let X c1(p, d) = 8 + c(p, r) where the sum is taken over all primes r ∈ Z distinct from p such that N(Q) ≤ 11 for a prime Q lying over r. Now assume that Υ is a one-cusped congruence subgroup of Z-level l. If l is a 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 4.5 follows from the Ladder Lemma. Therefore, we may assume that l is divisible by at least two distinct primes. If p is a prime dividing l and pn is the maximal n power of p dividing l, set l1 = p and l2 = l/l1. Let x = [PSL2(Od) : Υ] and x1 = [PSL2(Od) : ΥΓ(l1)]. By Lemma 2.1

x Θ(l ) 1 2 ∈ x Z

13 n where Θ(l2) = |SL2(Od/(l2))|. By the Index Lemma, p divides x if p > 3 and n−1 c1−1 p divides x if p ≤ 3, so if n ≥ c1 then p divides x. By the above discus- c1−8 7 sion, p does not divide Θ(l2). We conclude that p divides x1. Therefore, n 7 ΥΓ(l1) is a one-cusped congruence subgroup whose Z-level divides p . Since p divides the index we conclude that the Z-level is a positive power of p, say ps. s We have Γ(p ) < ΥΓ(l1), and since ΥΓ(l1) has one cusp, by the Index Lemma x1 s 7 s divides T (ΥΓ(l1))N(p ). Therefore, p divides T (ΥΓ(l1))N(p ). The maximum s possible power of p dividing T (ΥΓ(l1))N(p ) is 2s + 1. Therefore s ≥ 3, and 3 thus the Z-level of ΥΓ(l1) is at least p . In case a) this contradicts Proposition 4.2 and we conclude that the pc1 cannot divide the Z-level of a one-cusped congruence subgroup. In case b), if p = 7, by repeated applications of the Ladder Lemma we conclude that ΥΓ(l1) is a subgroup of ΥΓ(p) which has Z-level p. If 2 p = 2 or 3, we conclude that ΥΓ(l1) is a subgroup of ΥΓ(p ) which has Z-level p2. 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) and Proposition 4.3 the Z-level of any one-cusped congruence subgroup divides Y J = pc1(p,d) where the product is taken over the finite number of primes where N(P) ≤ 11 for all P in Od lying over p. Therefore Γ(J) is contained in all one-cusped congruence subgroups. As a result, there are only finitely many one-cusped congruence subgroups in PSL2(Od) for these values of d. In O1, O2, and O7, the above argument shows that there are only finitely many 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 n depending on Υ such that the Z-level divides J2n where Y J = pc1(p,d) and the product is taken over the finite number of odd primes where N(P) ≤ 11 for all P ∈ Od lying over p. There are only finitely many of Z-level dividing c1(2,d) 2 J and by Lemma 4.5 b) if n > c1(2, d) then such a subgroup is contained in a one-cusped congruence subgroup of Z-level 4. This proves Proposition 4.4 for 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 congruence subgroup of Z-level p where p is a rational prime lying under P ⊂ Od then N(P) ≤ 11. We will do so with three lemmas. By Corollary 3.2 (see section 3), 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 lying 2 over p, then the Od-level is either P or P . 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 a prime 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-cusped congruence subgroup of Od-level P1P2. If p > 3 then

2 [PSL2(Od) : Υ] = T (Υ)p

2 and ΥΓ(Pi) has Od-level Pi. Moreover, if p ≤ 3 and [PSL2(Od) : Υ] = T (Υ)p 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 under P with N(P) > 11. There are no one-cusped congruence subgroups of Od-level Pn. This proves Proposition 4.1 for ramiﬁed primes. Together, Lemmas 4.6, 4.7 and 4.8 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 rational prime lying under P, and let {1, ω} be an integral basis for Od. Since Od is a principal ideal domain, P = (q) for some q ∈ Od. Recall that by the Index Lemma 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 that Λ(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 subgroup of PSL2(Fr) is r + 1 except for r = 2, 3, 5, 7, 9, or 11 [17]. Therefore, we have 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 is less 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] we conclude that if T (Υ) = 2, then as d = 1, N(P) = 2, 5, or 9. If T (Υ) = 3 then as 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 peripheral torsion, 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 generated 2 by p and pω, [Λd : Λ(p)] = p . 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 2 [Λd : Λ(Υ)] 6= 1 and we conclude that x = T (Υ)p or T (Υ)p . Notice that if ΥΓ(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 proper 2 divisor of [Λd : Λ(Υ)] so [Λd : Λ(ΥΓ(Pi))] = p and |Λd : Λ(Υ)] = p implying that x = T (Υ)p2. As a result, if p > 3 it suﬃces to show that the Od-level of ΥΓ(Pi) is Pi. Assume otherwise, that ΥΓ(P1) = PSL2(Od). We will use the notation ∼ from Section 2.2. We have G1 = ρ1(ΥΓ(P1)) = PSL2(Fp). We will identify ρ1(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 ∗ |N1 | = 1, 2, Θ(P1)/2 or Θ(P1). Since x Θ(P ) |N ∗| = 2 1 1 x 2 we conclude that x = x2Θ(P1), x2Θ(P1)/2, 2x2, or x2. Since x divides T (Υ)p 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 shows that 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-cusped congruence subgroups of Od-level P by Lemma 4.6. It suﬃces to show that there 2 are none of Od-level P as by the Ladder Lemma, this implies that there are no n one-cusped congruence subgroups of Od-level P for n > 0. Therefore, assume 2 that Υ is a one-cusped congruence subgroup of Od-level P 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 or 2 p . Notice that [Λd : Λ(Υ)] could equal p, for example when Λ(Υ) is generated by {1, pω}.

16 First, assume that x = p2. We will use a vector space argument, and notation from Section 3.1. Here the vector space F corresponding to Υ ∩ Γ(P) modulo Γ(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 conjugating with A0,0 and A1,2 one can show this is not possible by implementing the Vector Space 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 to show 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 and 2 B ∈ PSL2(Z). Since φ(C) = φ(B), C ≡ B mod P and so B = CD for some D ∈ Γ(P2). Therefore B ∈ Υ since D ∈ Γ(P2) < Υ and therefore A = φ(B) for B ∈ 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 that Λ(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. 3 Therefore, Υ ∩ Γ(P) corresponds to a two-dimensional subspace, F, of (Fp) and satisﬁes the hypotheses of the Vector Space Lemma. In particular, there is a 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) = 2 2 −(0, 1, b3) ∈ F, so 1 = b3. As a result, b3 = ±1. Since b1 − 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 of Lemma 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 is a rational prime lying under a prime P ⊂ Od. If p = 2 then we assume that p is inert, if p = 3 we assume it is unramified, and if p = 7 we assume that T (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. We have shown in Lemma 4.8 that there are no one-cusped subgroups of Z-level any power of a ramified prime greater than 11. Therefore the following three lemmas will suffice. In Section 4.2.1 we will prove Lemma 4.10. (Powers of Ramified Primes) Let p be a ramified prime such that P lies over p and let Υ be a one-cusped congruence subgroup of Od-level Pn. Then n ≤ 1 if p = 7 or 11. With Lemma 4.8 this proves Proposition 4.2 for ramified primes. If p is an inert prime lying over P and p > 3, the Ladder Lemma implies that if Υ is a n one-cusped congruence subgroup of Od-level P , then ΥΓ(P) is a one-cusped congruence 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 4.2.2 will suffice. Lemma 4.11. (Powers of Inert Primes) Let p be an inert prime such that n P lies over p and let Υ be a one-cusped congruence subgroup of Od-level P . Then   1 or 2 if p = 2 and T (Υ) 6= 1  2 if p = 2 and T (Υ) = 1 n =  1 if p = 3 and T (Υ) 6= 1  0 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 n (p) = P1P2 and let Υ be a one-cusped congruence subgroup of Z-level p . As- sume p 6= 2 and if p = 7 that T (Υ) = 1. Then   2 if p = 3 n ≤ 1 if p = 5, 11 or p = 7 and T (Υ) = 1  0 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 there is a subgroup, H, of index p in PSL2(Fp) [17] and the pull-back of H necessarily has one cusp by Wohlfahrt’s Lemma. It is sufficient to show that there are no 2 3 one-cusped congruence subgroups of Od-level P or P . Let q ∈ Od be such that (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) = 1 and 2 x = [PSL2(Od) : Υ] = [Λd : Λ(Υ)] = p or p . To see this, recall that p divides x and x divides T (Υ)N(P2) = p2 by the Index Lemma. If x = p, ΥΓ(P) is necessarily PSL2(Od). We use a vector space argument as discussed in Section 3.1, applying the Vector Space Lemma. As ΥΓ(P) = PSL2(Od), A0,0 ∈ ΥΓ(P) and as in the proof of Lemma 4.8 we conclude that b1 = 0 and b3 = 1, or b3 = −1. Since A1,2 ∈ ΥΓ(P), one can derive a 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 of ΥΓ(P) is P, we use the Vector Space Lemma, where here F is a two-dimensional 3 subspace of (Fp) . When p = 7 and when p = 11, using the presentation for PSL2(Fp) in [16], and using [3] we find matrices of type Aα,β in PSL2(Fp) and similar 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) is 2 PSL2(Od) and as [ΥΓ(P) : Υ] = [Γ(P):Υ ∩ Γ(P)] = p under the vector space 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 results in a contradiction. We now show that there are no one-cusped congruence subgroups of Od- 3 2 3 level P . In this case [PSL2(Od) : Υ] = p or p by the Index Lemma, and the 3 2 2 fact that Λ(P ) < Λ(Υ). By the Ladder Lemma ΥΓ(P ) has Od-level P or P . Since we have established that there are no one-cusped congruence subgroups of 2 2 2 3 Od-level P , ΥΓ(P ) = ΥΓ(P) and has Od-level P. Therefore as Γ(P )/Γ(P ) 2 is a three-dimensional vector space over Fp, and Υ ∩ Γ(P ) corresponds to a one or two-dimensional subspace. In the two-dimensional case this results in a contradiction as above since Λ(Υ ∩ Γ(P2)) = Λ(P3). If F is one-dimensional, 2 as A0,0 and A1,−1 are both in ΥΓ(P ) = ΥΓ(P), conjugation of a possible generator 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. First we 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- n cusped congruence subgroup of Od-level P . 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 are no subgroups of index 2, 4 or 8 in PSL2(O3/(4)) [3]. There are subgroups of index 16, which can be seen to have one cusp by determining the corresponding lattices modulo (4) and hence determining the lattices of the pull-backs. For ∼ an inert 2, O3/(4) = Od/(4) for all other d with class number one, so the only one-cusped congruence subgroups of Od-level (2) or (4) are such index 16 subgroups. By the Ladder Lemma it suffices to show that there are no one-cusped con- 3 gruence subgroups of Od-level P . 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 index 16 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 contain Γ(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 conjugacy class of subgroups of index 6 and one of index 12 [3]. Both must have one cusp by index considerations, and each index 12 subgroup is contained in an index 6 subgroup [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 the fundamental group of the sister of the figure-eight knot complement, a knot in the lens space L(5, 1). Up to conjugation in PSL2(O3), Λ(ΥA) is generated by 1 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 index considerations ΥΓ(2) must be index 6 or 12 and therefore must be a subgroup of the index 6 subgroup. (Otherwise an isomorphism theorem implies that 3 divides [Γ(2) : Υ∩Γ(2)].) By searching low index subgroups of PSL2(O3/(4)) we see that there are 3 conjugacy classes of index 12 subgroups [3]. One corresponds to Υsister. The other two have O3-level (4), Υ8, and ΥB. The group Υ8 is isomorphic to the fundamental group of the figure-eight knot complement. Up to conjugation the lattice Λ(Υ8) is generated by 1 and 4ω and Λ(ΥB) is generated by 4 and ω − 1. There is one index 24 subgroup, ΥC , which is a subgroup of ΥB, and Λ(ΥC ) is generated by 4 and 2ω − 2. There are no index 48 subgroups contained 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 either has more than one cusp as seen by looking at the peripheral subgroup, cannot be a congruence subgroup by Wohlfahrt’s Lemma, or can be shown not to be congruence by explicitly computing the order of the group in MathematicaTM .

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

n Lemma 4.14. Let Υ be a one-cusped congruence subgroup of Od-level P where p = 3 is inert and lies under P. Then T (Υ) = 2, d = 1, and n = 1. Proof. Case I: T (Υ) = 1 As PSL2(F9) has no index 3 or 9 subgroups, there is no one-cusped subgroup of Od-level (3) [3] [17]. Assume Υ is a one-cusped congruence subgroup of Od-level (9). Then 2 3 4 x = [PSL2(Od) : Υ] = 3, 3 , 3 , or 3 by the Index Lemma. Using [3] we can see that there are no index 3, 9, or 27 subgroups of PSL2(O1)/N where N is the normal closure of the group generated by 1 9 1 9i ± and ± . 0 1 0 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). As 2 3 before since Γ(P ) < Υ ∩ Γ(P), Υ ∩ Γ(P) corresponds to a subspace F of (F9) . Since Λ(Υ) is generated by 9 and 9ω, (0, f, 0) or (0, 0, f) are in F only when f = 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= 1 Since T (Υ) 6= 1, T (Υ) = 2 and d = 1. Let Υ be a one-cusped congruence subgroup of O1-level (3). Then [PSL2(O1) : Υ] = 6 or 18. There is one conjugacy class 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-cusped congruence 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): 2 3 2 Γ(P )] = 9 . Let x = [PSL2(O1) : Υ]. Since 3 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 conjugacy class of index 3 subgroups but none have O1-level (9)[3]. We now consider the 3 case where x = 2 · 3 . The elements A0,0, A1,2, A1,0, A0,1, and A1,1 are all contained in the index 6 subgroup of PSL2(Od) which must be ΥΓ(3)[3]. Let 4 F correspond to Υ ∩ Γ(3), a 3 -dimensional vector space over F3. Notice that Λ(Υ ∩ Γ(3)) is generated by 9 and 3zi, or 9i and 3z, for z = 1 or 2. (We cannot say that (0, f, 0) or (0, 0, f) are in F only when f = 0. In fact, vectors of this form must be in F, but not both (0, 1, 0) and (0, i, 0), by Wohlfahrt’s Lemma.) 4 However, conjugation by A0,0 and A0,1 results in a contradiction. If x = 2 · 3 , 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 a contradiction.

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 be in 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 congruence subgroups of Z-level p for p ≤ 11. If Υ is a one-cusped congruence subgroup of Z-level pn, the Ladder Lemma implies that if p > 3, then ΥΓ(p) has Z-level p and hence p ≤ 11. (If p = 2 or 3, the Ladder Lemma implies that ΥΓ(p2) has 2 Z-level p if n ≥ 3.) Lemma 4.7 implies that if p > 3 and the Od-level of Υ is 2 p then [PSL2(Od) : Υ] = T (Υ)p and ΥΓ(Pi) has Od-level Pi for i = 1 and 2. 2 If p ≤ 3 and Υ has Od-level p then either the index is T (Υ)p and ΥΓ(Pi) has Od-level Pi or the index is T (Υ)p.

Deﬁnition 4.1. Let Υ < PSL2(Od) be a congruence subgroup of Z-level l, and let {1, ω} be an integral basis for Od. Then D is a proper diagonal of Λ(Υ) if D is 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 congruence n subgroup of PSL2(Od) of Z-level p where p is a prime. Then Λ(Υ) is generated by Λ1(Υ) and Λω(Υ)ω if there are no proper diagonals. Otherwise, Λ(Υ) is 0 generated by Λ1(Υ) and D (and also D and Λω(Υ)) for an appropriate diagonal D (and D0) in Λ(Υ). Moreover, if D = pr(gpn−k+ω) then all other elements of the form pr(hpn−k+ ωh0) are multiples of D modulo pn and ωpk, i.e. there are a, b, and c ∈ Z such that pr(hpn−k + ωh0) = aD + bpn + cωpk. The proof of Lemma 4.15 follows from a theorem in the geometry of numbers 2 which states that if b1, b2 are two independent points of a lattice Λ in R , and the 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 n1 n2 (qi) = Pi. Let Υ be a one-cusped congruence subgroup of Od-level P1 P2 such n1+n2 that n1 > n2 ≥ 0. Then [PSL2(Od) : Υ] = T (Υ)p and Λ(Υ) is generated n1 n2 n1 n2 by q1 q2 and ωq1 q2 .

Proof. First, consider the case where n2 = 0. By the Index Lemma, since the n1 n1 n1 Z-level is p , [Λd : Λ(Υ)] divides p . If p > 3, then p divides [Λd : Λ(Υ)] n1 n1 n1 and so Λ(Υ) = Λ(P ) and is generated by q1 and ωq1 . If p = 3 one can n1 check that since Υ has Od-level P1 that ΥΓ(P1) has Od-level P1. Therefore n1 n1 Λ(Υ) is generated by q1 and ωq1 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 of n1 this and Wohlfahrt’s Lemma that Λ1(Υ) = Λω(Υ) = p . (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 + g0ω). From this and Wohlfahrt’s Lemma, we conclude that n1−n2 n1 n1−n2 n1 generators of the lattice are q1 and p , or q1 and ωp . Therefore

n1+n2 [PSL2(Od) : Υ] = T (Υ)p .

The following three lemmas complete the proof of Proposition 4.2 as we have already shown that p ≤ 11. The remainder of this section will consist of their proofs.

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

2 if p = 3 n ≤ 1 if 5 ≤ p ≤ 11.

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

2 if p = 3 n ≤ 1 if 5 ≤ p ≤ 11.

n1 n2 Lemma 4.19. Let Υ be a one-cusped congruence subgroup of Od-level P1 P2 where n1 ≥ n2 and Pi lies over over p for i = 1, 2. Then unless p = 7 and T (Υ) 6= 1, 2 if p = 3 n ≤ 1 1 if 5 ≤ p ≤ 11. Proof of Lemma 4.17 Let Υ be as in the statement of the lemma. Let Υz denote Υ ∩ PSL2(Z) and

xz = [PSL2(Z):Υz].

n Notice that Υz is a congruence subgroup of PSL2(Z) containing Γz(p ), the n n principal congruence subgroup of PSL2(Z) of Z-level p , that is, Γz(p ) = n PSL2(Z) ∩ Γ(P ). This implies that Υz has ﬁnite index in PSL2(Z). One 2 cannot a priori conclude that H /Υz has one cusp, or that the Z-level of Υz is n n n p , only that p divides the Z-level of Υz. Let φ denote the modulo P map n from PSL2(Od) to PSL2(Z/p Z). (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) is n Γz(p ). It is an elementary exercise to show that xz divides x. n Both PSL2(Od) and PSL2(Z) surject PSL2(Z/p Z) via φ as φ(PSL2(Z)) n n n is isomorphic to PSL2(Z/p Z) and |PSL2(Z/p Z)| = |PSL2(Od/P )|. Since n Υz < Υ, x divides xz. We conclude that φ(Υz) = φ(Υ) and x = xz = T (Υ)p . Let 1 b A = ± . b 0 1 n Notice that Ab is not contained in Υ for any 0 < b < p by Lemma 4.16. n Thus the Z-level of Υz is p . Every cusp of Υz corresponds to the same one- dimensional lattice, as the stabilizer of every cusp is conjugate. Therefore,

n T (Υ)p = (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 conjugating Υ in PSL2(Z) such that the image of C is ∞, the stabilizer of infinity is n generated by the positive integer w. As the cusp width of Υz is p , if T (Υ) = 1, 2 n H /Υz has one cusp. As the Z-level of Υz is p , the lemma follows by Petersson’s result [13]. If T (Υ) 6= 1, necessarily d = 1 or 3. The only split prime less than or equal to 11 in O1 is 5, and in O3 is 7. The case where P lies over 7 and T (Υ) 6= 1 is excluded in our hypothesis. Assume that d = 1, p = 5 and T (Υ) = 2. First assume Υ is a one-cusped congruence subgroup of Od-level P. There is one conjugacy class of index 10 subgroups of PSL2(O1) of level P for any P lying over 5 and these all contain peripheral torsion and therefore have 2 cusps. If 2 2 Υ has Od-level P and index 50 modulo P , φ(Υ) is an index 10 subgroup of φ(ΥΓ(P)), an index 5 subgroup in PSL2(Z/25Z). Such a subgroup does not exist [3]. Proof of Lemma 4.18 n Let Υ be a one-cusped congruence subgroup of Od-level (p ). We will use the notation established in Section 2.2. For 5 ≤ p ≤ 11 it suffices to show that 2 there are no one-cusped congruence subgroups of Od-level (p ), establishing the result by the Ladder Lemma. As p > 3, we conclude by Lemma 4.7 that ΥΓ(p) 2 2 has Od-level (p), index p or T (Υ)p , and ΥΓ(P1) and ΥΓ(P2) have Od-levels k P1 and P2, respectively. By the Index Lemma, x = T (Υ)p where k = 2, 3 or 4, but the above shows that k 6= 2. First consider the case where T (Υ) = 1. Since there are no one-cusped 2 2 congruence subgroups of Od-level Pi for i = 1 or 2, ΥΓ(Pi ) = ΥΓ(Pi) and 2 ∗ 2 ∗ x1 = [SL2(Od/P1 ): G1] = p = [SL2(Od/P2 ): G2] = x2. ∗ ∗ With Ni a normal subgroup of Gi as defined as in Section 2.2,

∗ ∗ x 2 [G1 : N1 ] = = p or p . x1x2 ∗ As a result, the abelianization of G1 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 where p = 3. Proof of Lemma 4.19 n1 n2 Let Υ be a one-cusped congruence subgroup of Od-level P1 P2 with n1 ≥ n2. By the Ladder Lemma, if p > 3 and n2 > 1 then ΥΓ(P1P2) has Z-level p. 2 2 Therefore p ≤ 11 by Proposition 4.1. If 5 ≤ p ≤ 11 then ΥΓ(P1 P2 ) has Z-level 2 2 p . From Lemma 4.18, the Od-level cannot be (p ), and from Lemma 4.17 it 2 cannot be Pi . So it suﬃces to show that there are no one-cusped congruence 2 subgroups of Od-level P1 P2. (Similarly, if p = 3 it suﬃces to rule out the 3 3 2 existence of such groups of Od-level P1 P2 and P1 P2 .) 2 First, assume that Υ has Od-level P1 P2 and 5 ≤ p ≤ 11. By Lemma 4.16,

3 [PSL2(Od) : Υ] = T (Υ)p .

∗ Also with N1 deﬁned in Section 2.2

3 2 ∗ T (Υ)p [SL2(Z/p Z): N1 ] = x2

2 with x2 = 1 or p. But this implies that p divides |SL2(Z/pZ)| which is not the case. The proof in the case p = 3 is similar to the above and to the proof of Lemma 4.17.

4.3 Composite Z-Levels (Proof of Proposition 4.3) Recall that Proposition 4.3 states that if Υ is a one-cusped congruence subgroup of Z-level l then N(P) ≤ 11 for all P dividing (l). By Corollary 3.2 (see section 3) it suﬃces to show that N(P) ≤ 11 for all primes P dividing the Od-level of Υ. First, we will prove

Lemma 4.20. Let Υ be a one-cusped congruence subgroup of Z-level l. Let p be 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 a Dedekind domain, n0 n1 ns (`) = P0 P1 ... Ps where Pi is a prime for 0 ≤ i ≤ s and Pi 6= Pj for all i 6= j. Let pi be the rational prime lying under Pi, and order the Pi such that pi+1 ≤ pi. We will show that N(P0) ≤ 11. ni The Index Lemma implies that if pi is unramiﬁed, then pi divides x = ni d 2 e [PSL2(Od) : Υ], and if pi is ramiﬁed then pi 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 smallest rational integer such that (l) ⊂ (`), and let li be the smallest rational integer such that (li) ⊂ (`i). Therefore, as

x Θ(` ) 1 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) then p0 = p1. If l2 is relatively prime to p0, then as p0 is chosen to be larger than any pi dividing l2, and since

Y 3ni−2 2 Θ(`2) = N(Pi) (N(Pi) − 1)

Pi|`2 we conclude that unless p0 = 3 that p0 can only divide Θ(`2) if it divides the 2 factor pi +1 corresponding to an inert pi. We will assume that it is not the case that p0 = 3 and ps = 2 as here N(P0) ≤ 11. We will break down the proof of Lemma 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 2a2 3a3 5a5 where ap = 0 unless p is inert, then a3 = a5 = 0 unless d = 1, in which case neither 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 that d 6= 1, and hence there are no one-cusped congruence subgroups of Z-level any a3 a2 power of 3 or 5. First, assume that a5 = 0. Let (`1) = P3 and (`2) = P2 . We ∗ will be using subgroups N1 and N1 deﬁned in Section 2.2. By Proposition 4.2, x1 = [PSL2(Od) : ΥΓ(`1)] = 1 as there are no one-cusped congruence subgroups of Z-level a power of 3, and hence none of Od-level a power of P3. Therefore by Lemma 2.1 x Θ(` ) 26a2−4 · 3 · 5 1 2 = ∈ x x Z and we conclude that 32 does not divide x. By the Index Lemma we see that a3 =1 or 2. If a3 = 1 then as PSL2(F9) is simple, under the isomorphism ∼ PSL2(Od/P3) = PSL2(F9), N1 = {id} or PSL2(F9), as it is a normal subgroup. Since 1 [PSL ( ): N ] = 1 or [SL ( ): N ∗] 2 F9 1 2 2 F9 1 and ∗ x [SL2(F9): N1 ] = 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 or 2x2. But Θ(`1) cannot divide x as 5 divides Θ(`1) but not x, since x divides a2 a3 2a2 2a3 T (Υ)N(2 3 ) = T (Υ)2 3 . Therefore a3 = 2 but in this case as 3 and 2 2 2 are inert and 3 does not divide x, we see that 3 cannot divide either Λ1(Υ) a2 or Λω(Υ) (see Deﬁnition 2.1). Therefore Λ(3 · 2 ) is contained in Λ(Υ), contradicting the fact that a3 = 2. The case where a2 or a3 is zero is similar. a5 a3 Finally, assume that none of a5, a3 or a2 is zero. Let (`1) = P5 P3 and a2 (`2) = P2 . From above we conclude that x1 = 1 and as

x Θ(` ) 26a2−4 · 3 · 5 1 2 = ∈ x x Z

2 we see that 5 does not divide x. Therefore a5 = 1 by the Index Lemma and a3 a2 similar to above we have a contradiction if we set (`1) = P5 and (`2) = P3 P2 .

2 Case 1: p0 6= p1 and p0 does not divide pi + 1 for any inert pi dividing l . Let n0 (`1) = P0 and n1 n2 ns (`2) = P1 P2 ... Ps .

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

s Y 3nj −2 2 Θ(`2) = N(Pj) (N(Pj) − 1) j=1 but p0 does divide x by the Index Lemma. Therefore, p0 divides x1 since n0 x1Θ(`2)/x is an integer. So ΥΓ(P0 ) is a one-cusped subgroup of PSL2(Od) with Od- level a positive power of the prime P0, and therefore Z-level a positive power of p0. Hence N(P0) ≤ 11 by Proposition 4.2.

2 Case 2: p0 6= p1 but p0 divides pi + 1 for some inert pi dividing l. 2 One can check that if p0 > 11 is ramiﬁed, that p0 does not divide pi + 1 for any inert pi. Therefore we may assume that p0 is unramiﬁed. One can show that if 2 pi and pj are inert, p0 > pi > pj, and p0 divides pi + 1, that either p0 does not 2 divide pj + 1 or p0 = 5 and pi = 3. Lemma 4.21 implies that in this case p0 is split, and therefore N(P0) = 5. 2 If p0 divides x, let n0 (`1) = P0 and n1 n2 ns (`2) = P1 P2 ... Ps .

27 Using the notation in Section 2.2 recall that

x Θ(` ) 1 2 ∈ x Z

2 2 by Lemma 2.1. Since p0 > pi, p0 does not divide pi + 1, and as p0 does not 2 2 divide pj + 1 for any other inert prime pj, p0 does not divide Θ(`2). As we are 2 n0 assuming p0 divides x, we conclude that p0 divides x1 and ΥΓ(P0 ) has Od- level a positive power of P0, and therefore Z-level a positive power of p0. Hence N(P0) ≤ 11 by Proposition 4.2. 2 Therefore we may assume that p0 does not divide x. By the Index Lemma, this implies that n0 = 1 since p0 > 3, and we let

ni n1 ni−1 ni+1 n2 (`1) = P0Pi , and (`2) = P1 ... Pi−1 Pi+1 ... Ps .

ni Now p0 divides x but not Θ(`2), so p0 divides x1. Therefore ΥΓ(P0Pi ) is a one- mi cusped congruence subgroup of Od- level P0 or P0Pi for some 1 ≤ mi ≤ ni. In the ﬁrst case, N(P0) ≤ 11 by Proposition 4.1. ni mi It now suﬃces to assume that ΥΓ(P0Pi ) has Od-level P0Pi (and hence ni mi 2 mi ΥΓ(P0Pi ) = ΥΓ(P0Pi )) and p0 does not divide [PSL2(Od) : ΥΓ(P0Pi )]. Abusing notation, let

mi mi (`) = P0Pi , (`1) = Pi , (`2) = P0, and mi x = [PSL2(Od) : ΥΓ(P0Pi )].

It is an elementary exercise to show that with p0 and pi as above, pi does not divide p0 ± 1 unless pi = 2 and p0 = 5. Similar to the proof of Lemma 4.21 one can show that p0 is not inert. Therefore, pi divides x by the Index Lemma, but not Θ(`2) = p0(p0 + 1)(p0 − 1) since p0 is not inert. Since x1Θ(`2)/x ∈ Z we conclude that pi divides x1. mi Therefore ΥΓ(Pi ) has Od-level a positive power of Pi and therefore Z-level a positive power of pi. Hence N(Pi) ≤ 11 by Proposition 4.2. As we are assuming that pi is inert, we conclude that pi is 2 or 3 and therefore p0 = 5 which must be split.

2 Case 3: p0 = p1 and p0 does not divide pi + 1 for any inert pi dividing l. Let n0 n1 (`1) = P0 P1 and n2 n3 ns (`2) = P2 P3 ... Ps . As before x Θ(` ) 1 2 ∈ x Z

28 n0 n1 and so p0 divides x but not Θ(`2). Thus p0 divides x1 and ΥΓ(P0 P1 ) has m0 m1 m0 Od- level P0 P1 or, say, P0 . In either case, the Z-level is a power of p0 and by Proposition 4.2, N(P0) ≤ 11.

2 Case 4: p0 = p1 and p0 divides pi + 1 for some inert pi dividing l. 2 n0 n1 We may assume that p0 > 3. If p0 divides x, then let (`1) = P0 P1 and n2 ns (`2) = P2 ... Ps . Since x Θ(` ) 1 2 ∈ x Z 2 and as p0 does not divide Θ(`2), p0 divides x1 and therefore the Z- level of n0 n1 ΥΓ(P0 P1 ) is a proper power of p0. By Proposition 4.2 we conclude that 2 N(P0) ≤ 11. So assume that p0 does not x and therefore n0 = n1 = 1 by the Index Lemma. Let ni (`1) = P0P1Pi and n2 ni−1 ni+1 ns (`2) = P2 ... Pi−1 Pi+1 ... Ps .

Therefore p0 divides x, but not Θ(`2). ni mi By previous work, we may assume ΥΓ(P0P1Pi ) has Od- level P0P1Pi or mi mi P0Pi where 1 ≤ mi ≤ ni. Abusing notation let (`1) = Pi , and (`2) = P0P1 or P0, respectively. Recall that x Θ(` ) 1 2 ∈ x Z and pi divides x by the Index Lemma. As commented before, pi does not divide

2 Θ(`2) = (p0(p0 + 1)(p0 − 1)) or p0(p0 + 1)(p0 − 1) unless p0 = 5 and pi = 2. If pi does divide Θ(`2), by Lemma 4.21, p0 is not inert, so N(P0) ≤ 11. If pi does not divide Θ(`2) we conclude that pi divides ni mi x1, and hence ΥΓ(P0P1Pi )Γ(Pi ) 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 = 5 and we conclude that N(P0) ≤ 11 by Lemma 4.21, completing the proof of Lemma 4.20

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

N(P) 2 3 4 5 7 9 11 25 Θ(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-cusped congruence subgroup of Od-level (`). It suﬃces to do a few calculations. As in the previous proof, write

n0 n1 ns (`) = P0 P1 ... Ps

29 with Pi lying over pi, and ordered so that pi ≥ pi+1. By Lemma 4.20 we know that N(P0) ≤ 11. Therefore either p0 is split or ramified and p0 ≤ 11 or p0 is inert and p0 = 2 or 3. It suffices to rule out the possibility that p0 is split or ramified and there is an inert prime pi such that 11 > pi > 3. We will call such an inert prime p, lying under P, with n the largest power of P dividing (`). The only possibilities are p = 5 or 7. If p = 7, notice that gcd(p, Θ(Q)) = 1 for any n other possible prime Q dividing (`). Let (`1) = P . Using the fact that x Θ(` ) 1 2 ∈ x Z we conclude that p divides x1, since p divides x by the Index Lemma but p n does not divide Θ(`2). Therefore ΥΓ(P ) is a one-cusped congruence subgroup of Z-level a proper power of p, which contradicts Proposition 4.2. Therefore it suffices to show that p 6= 5, which can be shown in a manner similar to that used 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 is prime, 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:

2 2 2 2 P11, Q11, P11Q11, P2 , P2 P11, P2 Q11, P2 P11Q11. 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 subgroup Υ of PSL2(Z) has width w if after conjugating Υ in PSL2(Z) such that the image of C is ∞, the stabilizer of inﬁnity is generated by the positive integer w. In other words, if the degree of the cover of the cusp of MZ by C is w. Let p be an odd prime and let φ denote the modulo p map from PSL2(Z) to PSL2(Fp). By the classiﬁcation of subgroups of PSL2(Fp) there is a maximal subgroup, H, of index p + 1 [17]. Since p is odd,

Γ(p) = {A ∈ PSL2(Z): A ≡ I mod p}

30 has (p2 − 1)/2 cusps all of width p. By Wohlfahrt’s Lemma, Υ = φ−1(H) has two cusps, one of width one, and one of width p. Therefore, letting p vary over all odd primes, there are infinitely many maximal subgroups of PSL2(Z) with two cusps. The group H has a normal Sylow p-subgroup, S, and the quotient of H by S is cyclic of order (p − 1)/2 [17]. As p does not divide (p − 1)/2, the cusp of 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 that S < L < H and [H : L] = l. The preimage, φ−1(L) has l cusps of width one and l of width p, for a total of 2l cusps. Therefore we have subgroups of Z-level p with 2l cusps for all divisors, l, of (p − 1)/2. For a fixed l, there are infinitely many primes p such that p ≡ 1 mod 2l, and as a result we have infinitely many maximal 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 in Od, such that q = N(P) is odd. If A < PSL2(Fq) and the pre-image of A has n-cusps, we will say that A has n cusps. If p is split, lying under P then we analyze the quotient PSL2(Od) by Γ(P) and as above conclude that H has two cusps and for all divisors, l, of (p − 1)/2 there is a subgroup L of PSL2(Fp) such that 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 ). As above 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 show that 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 covered by (p2 − 1)/2 cusps of width one in S, as all of the covers must have the same width. Therefore the cusp of width p2 is covered by (p2 − 1)/2 cusps of width p2. We conclude that S has (p2 − 1)/2 cusps of width one and (p2 − 1)/2 of width 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 of width p2, for a total of 2l cusps. Therefore, combining the split and inert cases, given a positive integer l, we need 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 over p, 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 O = [i], a prime 1 Z √ 1+ −3 p splits when p ≡ 1 mod 4 and is inert if p ≡ −1 mod 4. In O3 = Z[ 2 ] a prime p splits when p ≡ 1 mod 6 and p is inert when p ≡ −1 mod 6. This is due to the fact that both are cyclotomic extensions. In O1, consider a split 2 prime, p, lying under P and the quotient PSL2(Fp). Here {id} has (p − 1)/4

31 torsion-free cusps, all of width p. Since |S| = p, we conclude that all cusps of S are 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 one cusp (with torsion) of width p. Since φ−1(S) is peripheral torsion-free, there is a representative of peripheral torsion in the cyclic quotient H/S. Let T be the group 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 of which have torsion. As above, for any l dividing (p−1)/4 we have a subgroup of H containing T with 2l cusps. Since for any l, there are inﬁnitely many primes p ∈ Z with p ≡ 1 mod 4l, these primes split and the PSL2(Fp) have subgroups with 2l cusps. So, there is a subgroup of PSL2(O1) containing Γ(P) with 2l cusps, l of width one and l of width p. Similarly, for O3 for a given l, as there are inﬁnitely many primes p ∈ Z with p ≡ 1 mod 6l, these primes split and the PSL2(Fp) have subgroups with 2l cusps.

7 acknowledgements

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

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