Genus of Congruence Subgroups of the Modular Group

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GENUS OF CONGRUENCE SUBGROUPS OF THE MODULAR GROUP

YAP HUI HUI (B.Sc.(Hons), NUS)

A THESIS SUBMITTED FOR THE DEGREE OF MASTERS OF SCIENCE DEPARTMENT OF MATHEMATICS NATIONAL UNIVERSITY OF SINGAPORE 2003 Contents

Acknowledgements ii

Summary iii

1 Congruence Subgroups of SL2(Z) 1 1.1 Γˆ0(N), Γˆ1(N) and Γ(ˆ N)...... 1 1.2 Cusps of Γˆ0(N), Γˆ1(N) and Γ(ˆ N)...... 8 1.3 Cusp Widths ...... 23

2 The Modular Group PSL2(Z) 25 2.1 Γ0(N), Γ1(N) and Γ(N)...... 25 2.2 Indices of Subgroups of PSL2(Z)...... 27 2.3 Cusps of Γ0(N), Γ1(N) and Γ(N)...... 31 2.4 Cusp Widths ...... 36 2.5 The Genus Formula ...... 37

sqf 3 Genus of Γτ (m; m/d, ε, χ) 38 3.1 Larcher Congruence Subgroups ...... 38 sqf 3.2 Index of Γτ (m; m/d, ε, χ) in PSL2(Z) ...... 40 3.3 Number of Inequivalent Cusps ...... 42 3.4 Number of Elliptic Subgroups ...... 50 sqf 3.5 Genus Formula of Γτ (m; m/d, ε, χ) ...... 54

4 Genus of some Congruence Subgroups 55 4.1 Genus Formula of Γ1(M) ∩ Γ(N)...... 55 4.2 Genus Formula of Γ1(m; 2, 1, 2) ...... 57

Bibliography 60

i Acknowledgements

I sincerely thank my supervisor, A/P Lang Mong Lung, for his guidance and patience for the past one year of my canditure. Without him, this thesis would be impossible.

And also a very big thank you to all the lecturers whom have taught me, my family and friends.

ii Summary

The main objective of this thesis is to determine the genus formula of some Larcher congruence subgroups.

∗ Let G be a subgroup of ﬁnite index in PSL2(Z) and H = H ∪Q∪{∞}, where H is the upper half of the complex plane. Then the genus, g, of H∗/G (also referred to as the genus of G) is given by µ v v v g = 1 + − 2 − 3 − ∞ , 12 4 3 2 where µ = the index of G in PSL2(Z), v2 = the number of inequivalent elliptic subgroups of order 2 of G, v3 = the number of inequivalent elliptic subgroups of order 3 of G, v∞ = the number of inequivalent cusps of G.

Hence the study of v2, v3, v∞, indices, cusps and cusp widths are essential considerations in this thesis and compose the content of the four chapters of the thesis.

Chapter 1 begins by establishing auxillary results about the classical conguence subgroups of SL2(Z), the ﬁnite orders of elements in SL2(Z), and, cusps and cusp widths.

In Chapter 2, with the SL2(Z) case as a stepping stone, we are able to draw similar results for the modular group PSL2(Z). This treatment is neater and more systematic than a head-on approach with PSL2(Z). Chap- ter 2 also deals with the indices of subgroups of PSL2(Z) and introduces the genus formula.

iii Chapter 3 introduces the notions of “Larcher congruence subgroups” sqf and “Γτ (m; m/d, ε, χ)”. Here, the description for the equivalence of cusps sqf in Γτ (m; m/d, ε, χ) would be given, and, we will determine its genus via identiﬁcation with the congruence subgroup, Γ0(md)∩Γ1(m). The method- ology for ﬁnding the genus is in fact motivated by [L2].

Finally, in Chapter 4, we will extend the approach employed in Chapter 3 to procure the genus formula of other Larcher congruence subgroups.

iv Chapter 1

Congruence Subgroups of SL2(Z)

In this chapter, we will study the group, SL2(Z), and its classical con- ˆ ˆ ˆ gruence subgroups, Γ0(N), Γ1(N) and Γ(N).

1.1 Γˆ0(N), Γˆ1(N) and Γ(ˆ N)

Deﬁnition 1.1. Let SL2(Z) be the set of 2 × 2 matrices with each of its elements having integral entries and determinant 1, that is,

a b SL ( ) = a, b, c, d ∈ , ad − bc = 1 . 2 Z c d | Z

Deﬁnition 1.2. Let N∈N. We deﬁne the following subgroups of SL2(Z) to be, a b Γˆ (N) = ad − bc = 1, c ≡ 0 (mod N) , 0 c d | a b Γˆ (N) = ad − bc = 1, a ≡ d ≡ 1 (mod N), c ≡ 0 (mod N) , 1 c d | a b Γ(ˆ N) = ad − bc = 1, a ≡ d ≡ 1 (mod N), b ≡ c ≡ 0 (mod N) . c d |

Deﬁnition 1.3. Let G be a subgroup of SL2(Z). G is a congruence sub- ˆ group of SL2(Z) if there exists M ∈ N such that Γ(M) is a subgroup of

1 Chapter 1: Congruence Subgroups of SL2(Z) 2

ˆ ˆ ˆ G. Otherwise, G is a non-congruence subgroup. Thus, Γ0(N), Γ1(N), Γ(N) are all congruence subgroups of SL2(Z).

Theorem 1.4. The ﬁnite orders of elements in SL2(Z) are 1, 2, 3, 4 and 6. Proof. Let a b A = c d be an element of order n in SL2(Z). If A = ±I, then the order of A is either 1 or 2 respectively, which is trivial. So we may assume that A 6= ±I. Now, by direct calculation, the characteristic polynomial of A is,

4(x) = x2 − (a + d)x + 1.

Since A 6= ±I, 4(x) is the minimal polynomial of A and thus divides xn −1 i 2tπ which has roots e n , t = 0, 1, ..., n − 1. This means that the roots of i 2kπ i 2lπ 4(x) = 0 can be written as e n and e n for some 0 ≤ k, l ≤ n − 1 and k, l ∈ Z. Hence, 4(x) = x2 − (a + d)x + 1 i 2kπ i 2lπ = (x − e n )(x − e n ) 2 i 2kπ i 2lπ i 2kπ i 2lπ = x − (e n + e n )x + e n · e n .

By comparing coeﬃcients, we see that

i 2kπ i 2lπ e n · e n = 1, (1.1.1)

i 2kπ i 2lπ e n + e n = a + d ∈ Z. (1.1.2) i 2(k+l)π We can deduce from (1.1.1) that e n = 1 which implies n|(k + l) or l = −k + pn. Since 0 ≤ k, l ≤ n − 1, or, 0 ≤ k + l ≤ 2n − 2, we see that i 2lπ −i 2kπ p = 0 or 1. Substituting l = −k + pn, we get e n = e n . Thus (1.1.2) reduces to i 2kπ −i 2kπ 2kπ a + d = e n + e n = 2 cos . n As 2 cos(2kπ/n) = a + d ∈ Z and −1 ≤ cos(2kπ/n) ≤ 1, cos(2kπ/n) = −1, −1/2, 0, 1/2, 1. Note that gcd(k, n) = 1. Suppose not. Then gcd(k, n) = Chapter 1: Congruence Subgroups of SL2(Z) 3

±i 2kπ n n d > 1, and we have (e n ) d = 1. This implies that 4(x) divides x d − 1 which contradicts our choice of n. Consider cos(2kπ/n) = −1. Then 2kπ/n = π or n = 2k so k|n. But gcd(k, n) = 1, therefore k = 1 and n = 2. Similarly, for cos(2kπ/n) = −1/2, 0, 1/2, 1, we obtain n = 3, 4, 6 and 1 respectively. Thus, n = 1, 2, 3, 4 or 6.

Corollary 1.5. Let A ∈ SL2(Z) and A 6= ±I. If A is of ﬁnite order n, then

(i) n = 3 if and only if 4(x) = x2 + x + 1,

(ii) n = 4 if and only if 4(x) = x2 + 1,

(iii) n = 6 if and only if 4(x) = x2 − x + 1.

Proof. In the proof of Theorem 1.4, we have seen that 4(x) = x2 + x + 1, x2 +1, x2 −x+1 imply n = 3, 4 and 6 respectively. Conversely, suppose n = 3, 4 or 6. We ﬁrst compute the factorization of the following polynomials, namely, x3 − 1 = (x2 + x + 1)(x − 1), (1.1.3) x4 − 1 = (x2 + 1)(x − 1)(x + 1), (1.1.4) x6 − 1 = (x2 − x + 1)(x2 + x + 1)(x + 1)(x − 1). (1.1.5) Since A 6= ±I, 4(x) divides xn − 1. Clearly from (1.1.3), when n = 3, 4(x) = x2 + x + 1. Similarly, n = 4 implies 4(x) = x2 + 1. Let us now consider the case when n = 6. We notice that from (1.1.5),

4(x) = x2 − x + 1, or , x2 + x + 1.

But we have earlier just shown that 4(x) = x2 + x + 1 implies n = 3. Thus 4(x) = x2 − x + 1 if n = 6. Hence the result is proved.

Corollary 1.6. Let A ∈ SL2(Z). If A is of ﬁnite order (excluding orders of 1 and 2), then its characteristic polynomial, 4(x), is either x2 + x + 1, x2 + 1,or x2 − x + 1.

Proof. This follows immediately from Theorem 1.4. and Corollary 1.5. Chapter 1: Congruence Subgroups of SL2(Z) 4

−1 0 Theorem 1.7. is the only element of order 2 in SL ( ). 0 −1 2 Z

a b Proof. Suppose g = c d is an element of order 2 in SL2(Z), and g 6= −I. Then since g 6= ±I, the characteristic polynomial of g, 4(x), is the minimal polynomial of g which divides x2 − 1. But 4(x) = x2 − (a + d)x + 1 and so, x2 −1 = x2 −(a+d)x+1 which is obviously a contradiction. So g = −I.

Corollary 1.8. To determine whether a group, G, of SL2(Z) contains any −1 0 ˆ element of order 2, it suﬃces to check whether 0 −1 ∈ G. So, Γ(N) and ˆ −1 0 Γ1(N) do not contain any element of order 2 if and only if N ≥ 3 as 0 −1 ˆ ˆ is not contained in Γ(N) and Γ1(N) if and only if N ≥ 3. On the other −1 0 ˆ hand, 0 −1 ∈ Γ0(N) for all N ∈ N.

Theorem 1.9. Let N ∈ N. Γ(ˆ N) has no elements of order 3, 4 or 6 if and only if N ≥ 2. ˆ Proof. Suppose N = 1. Then Γ(1) = SL2(Z) which clearly possesses elements of order 3, 4, and 6. So, N ≥ 2. Conversely, suppose that N ≥ 2. Let 1 + aN bN A = ∈ Γ(ˆ N). cN 1 + dN By direct calculation, the characteristic polynomial of A is, 4(x) = x2 − (2 + N(a + d))x + 1. Now, by Corollary 1.6, we see that there are only three choices for 4(x), that is, x2 + x + 1, x2 + 1, or x2 − x + 1. In other words, if N ≥ 4, then Γ(ˆ N) will not contain any elements of order 3, 4 and 6. For N = 3, we have for any A ∈ Γ(3),ˆ

1 + 3a 3b A = , and 4(x) = x2 − (2 + 3(a + d))x + 1. 3c 1 + 3d Suppose A is of order 3, 4, or 6. Then by Corollary 1.6,

4(x) = x2 + x + 1, x2 + 1, or x2 − x + 1.

Let us now look at the coeﬃcient of x in 4(x). Note that 2 + 3(a + d) 6= 0 because a, d ∈ Z, and also, 2 + 3(a + d) = 1 implies 3(a + d) = −1 which Chapter 1: Congruence Subgroups of SL2(Z) 5 is impossible too. So, 2 + 3(a + d) = −1 which leads to a = −d − 1. Consequently, −2 − 3d 3b A = and |A| = (−2 − 3d)(1 + 3d) − 9bc = 1, 3c 1 + 3d which reduces to −2 − 9d − 9d2 − 9bc = 1, or −2 − 9(bc + d + d2) = 1, and we get a contradiction. For N = 2 and any A ∈ Γ(2),ˆ

1 + 2a 2b A = and 4(x) = x2 − (2 + 2(a + d))x + 1. 2c 1 + 2d

Similar to the case for Γ(3),ˆ by supposing A is of ﬁnite order 3, 4, or 6, we have 2 + 2(a + d) = 0 (not possible for 2 + 2(a + d) = ±1). As a result,

1 + 2a 2b A = . 2c 1 + 2d

Eventually, we obtain −2(bc + d + d2) = 1 which is again a contradiction. This means that Γ(ˆ N) has no elements of order 3, 4 and 6 if N ≥ 2.

ˆ Theorem 1.10. Let N ∈ N. Γ1(N) has no elements of order 3, 4 or 6 if and only if N ≥ 4. Proof. Let 1 + aN b A = cN 1 + dN ˆ be an element in Γ1(N). It is obvious that the characteristic polynomial of A, 4(x) = x2 − (2 + N(a + d))x + 1. Suppose N ≥ 4. Then similar to the proof of Theorem 1.9, if N ≥ 4, then 4(x) cannot be any of the 3 polynomials in Corollary 1.6, therefore A is not of order 3, 4, and 6. ˆ Conversely, Γ1(N) has no elements of order 3, 4 or 6 implies N ≥ 4 since 3 −1 ˆ 4 −7 10 −3 is an element of order 4 in Γ1(2) while 3 −5 is an element of order ˆ ˆ 3 in Γ1(3). For Γ1(1) = SL2(Z), it is obvious that it contains elements of order 3, 4, and 6, and so we must have N ≥ 4.

We state the following well known results from [Sh] before proceeding to Theorem 1.13. Chapter 1: Congruence Subgroups of SL2(Z) 6

Lemma 1.11. In SL2(Z), −1 −1 (i) All cyclic subgroups of order 3 are conjugate to . 1 0

0 −1 (ii) All cyclic subgroups of order 4 are conjugate to . 1 0

1 −1 (iii) All cyclic subgroups of order 6 are conjugate to . 1 0

Lemma 1.12. Let p be an odd prime. Then (i) −1 is a quadratic residue of p if and only if p ≡ 1 (mod 4), and

(ii) −3 is a quadratic residue of p if and only if p ≡ 1 (mod 3) for p > 3.

ˆ Theorem 1.13. Let N ∈ N and p be an odd prime. Γ0(N) has no elements of order 3, 4 or 6 if and only if (i) 4|N, or ∃ p|N such that p is of the form 4k + 3, and

(ii) 9|N, or ∃ p|N such that p is of the form 3k + 2. ˆ Proof. Suppose that Γ0(N) admits an element of order 4. In view of Lemma 1.11.(ii), a b 0 −1 d −b A = ∈ Γˆ (N), c d 1 0 −c a 0 a b for some ∈ SL ( ) with gcd(c, d) = 1 (because ad − bc = 1). c d 2 Z This implies that

b −a d −b ac + bd −a2 − b2 A = = ∈ Γˆ (N). d −c −c a c2 + d2 −ac − bd 0

So, c2 + d2 ≡ 0 (mod N). Since gcd(c, d) = 1, c2 + d2 ≡ 0 (mod N) is not solvable if 4|N. Suppose 4 is not a divisor of N. Then if 2|N, all the Chapter 1: Congruence Subgroups of SL2(Z) 7 remaining prime divisors of N must be odd. Moreover, as gcd(c, d) = 1, both c and d are odd. Clearly, c2 + d2 ≡ 0 (mod 2) is always admissible. This implies that we need only consider the odd prime divisors, pi’s, of N regardless of the parity of N. Note that if there exist some pi such that pi|N and pi|c, then pi is also a divisor of d which contradicts the fact that 2 2 gcd(c, d) = 1. Now, for all the odd prime divisors of N, c +d ≡ 0(mod pi), −1 and d (mod pi) exists as d is relatively prime to pi. Therefore, −1 2 (c·d ) ≡ −1 (mod pi). In consideration of Lemma 1.12.(i), the preceding congruence equation is solvable if and only if all the odd prime divisors of N are of the form 4k + 1, or equivalently, the congruence equation is not solvable if and only if there exists a prime divisor of N which is of the form 4k + 3. Now, by applying ˆ Lemma 1.11.(iii), if Γ0(N) admits an element of order 6, then a b 1 −1 d −b A = c d 1 0 −c a a + b −a d −b = c + d −c −c a ad + ac + bd −a2 − ab − b2 = ∈ Γˆ (N), c2 + cd + d2 −ac − bc − bd 0 and we get c2 + cd + d2 ≡ 0 (mod N). c2 + cd + d2 ≡ 0 (mod 9) and c2 + cd + d2 ≡ 0 (mod 2) are not solvable as gcd(c, d) =1 , Suppose that both 2 and 9 are not divisors of N. Since 2−1(mod N) and 4−1(mod N) exist as 2 is not a divisor of N, c2 + cd + d2 ≡ (c + 2−1 · d)2 + 3 · 4−1 · d2 ≡ 0 (mod N). This implies that (2c + d)2 + 3d2 ≡ 0 (mod N). By similar reasoning mentioned above, c, d and N are relatively prime to one another, so d−1 (mod N) exists and thus [d−1(3c + d)]2 ≡ −3 (mod N)

e1 e2 ei em Let N = p1 p2 ...pi ...pm . Then,

−1 2 [d (3c + d)] ≡ −3 (mod pi), for i = 1, 2, ..., m. So, from Lemma 1.12.(ii), we conclude that this is solvable if and only if all the prime divisors (pi > 3) of N are of the form 3k + 1. Equivalently, Chapter 1: Congruence Subgroups of SL2(Z) 8 c2 + cd + d2 ≡ 0 (mod N) is not solvable if and only if there exists a prime divisor of N which is of the form 3k +2. For A having order 3, using Lemma 1.11.(i), Lemma 1.12.(ii), and by a similar argument to the case when A is of order 6 produces the congruence equation c2 − cd + d2 ≡ 0 (mod N) which yields the same results as when A is of order 6. Hence the theorem holds.

1.2 Cusps of Γˆ0(N), Γˆ1(N) and Γ(ˆ N)

a b Deﬁnition 1.14. Let G be a subgroup of SL2(Z). g = c d ∈ G is said to be parabolic if trace of g, tr(g) = a + d = ±2.

Remark 1.15. Let a, b, c, d ∈ Z and z ∈ C ∪ {∞}. Then, for z 6= ∞, we deﬁne a b az + b z = , c d cz + d and for z = ∞, we deﬁne (a a b if c 6= 0 , ∞ = c c d ∞ if c = 0 .

Definition 1.16. Let G be a subgroup of SL2(Z). z ∈ C ∪ {∞} is a cusp of G if z is fixed by some non-trivial parabolic element g ∈ G, that is, z satisfies the condition gz = z.

Theorem 1.17. The set of cusps for SL2(Z) is Q ∪ {∞}. 1 1 Proof. Let the set of cusps for SL2(Z) be S. ∞ is a cusp of SL2(Z) as 0 1 1 1 is a parabolic element in SL2(Z), and 0 1 ∞ = ∞. So assume c 6= 0. For any a/c ∈ Q and gcd(a, c) = 1, we can ﬁnd inﬁnitely many b and d such a b that ad − bc = 1. As a consequence, c d ∈ SL2(Z). Now, a b a ∞ = , c d c Chapter 1: Congruence Subgroups of SL2(Z) 9

a b−1 a b a b−1 a ∞ = . c d c d c d c As a result, a b−1 a = ∞. c d c Note that a b 1 1 a b−1 c d 0 1 c d is a non-trivial parabolic element in SL2(Z). It can be easily checked with the above equalities that a b 1 1 a b−1 a a = . c d 0 1 c d c c Thus we have proved that Q ∪ {∞} ⊆ S. Conversely, let x ∈ S. Then for k l some A = m n ∈ SL2(Z), tr(A) = ±2, and A 6= ±I, we have Ax = x, kx + l = x, mx + n mx2 + (n − k)x − l = 0. (1.2.1) Case 1: m 6= 0. Therefore, the discriminant of the quadratic equation is, D = (n − k)2 + 4lm = (k + n)2 − 4(kn − lm) = (±2)2 − 4 (as A is parabolic and det(A) = 1.) = 0. Consequently, k − n x = ∈ . 2m Q Case 2: m = 0. Then det(A) = kn − lm = kn = 1, and k + n = ±2 force k = −1 and n = −1, or, k = 1 and n = 1. Thus, (1.2.1) reduces to −x = l − x or x = l + x. Suppose x 6= ∞, then we will obtain l = 0 from both of the previous equations. But A 6= ±I, so x = ∞. In other words, S ⊆ Q ∪ {∞}. This completes the proof of the theorem. Chapter 1: Congruence Subgroups of SL2(Z) 10

Theorem 1.18. The set of cusps for Γ(ˆ N) is Q ∪ {∞}. ˆ a b Proof. Let S denote the set of cusps for Γ(N), and c d ∈ SL2(Z). Since 1 N ˆ 0 1 is a parabolic element of Γ(N), and 1 N ∞ = ∞, 0 1 ∞ must be a cusp of Γ(ˆ N). Similar to the explanation of the preceding lemma, we have a b 1 N a b−1 a a = . c d 0 1 c d c c ˆ It can be easily checked that Γ(N) is a normal subgroup of SL2(Z). So, a b 1 N a b−1 ∈ Γ(ˆ N). c d 0 1 c d Furthermore, it is parabolic. This implies that a/c is a cusp of Γ(ˆ N). So Q ∪ {∞} ⊆ S. Conversely, we can prove that S ⊆ Q ∪ {∞} by a similar argument mentioned in Theorem 1.17. Thus S = Q ∪ {∞}.

Lemma 1.19. Let G1 and G2 be subgroups of SL2(Z). Denote the set of cusps for G1 and G2 by S1 and S2 respectively. If G1 ⊆ G2, then S1 ⊆ S2.

Proof. The proof is straightforward. Let x ∈ S1. Then gx = x, where g is a non-trivial parabolic element of G1. But g ∈ G2. So x ∈ S2.

ˆ ˆ Corollary 1.20. The sets of cusps for Γ1(N) and Γ0(N) are the same, which is, Q ∪ {∞}. ˆ ˆ Proof. Let C1, C0, and C be the sets of cusps for Γ1(N), Γ0(N) and SL2(Z) respectively. Since ˆ ˆ ˆ Γ(N) ⊆ Γ1(N) ⊆ Γ0(N) ⊆ SL2(Z), and combining the results of Theorem 1.17, Theorem 1.18. and Lemma 1.19, we have the following,

C ⊆ C1 ⊆ C0 ⊆ C. Hence we are done. Chapter 1: Congruence Subgroups of SL2(Z) 11

Deﬁnition 1.21. Let G be a subgroup of SL2(Z), and x1 and x2 be cusps of G. Then x1 and x2 are G-equivalent (also described as equivalent in G and equivalent modulo G) if gx1 = x2 for some g ∈ G. Moreover, we denote the equivalence classes of x1 and x2 by [ x1] and [ x2] respectively. In other words, we write [ x1] = [ x2] for x1 and x2 being G-equivalent.

Theorem 1.22. All the cusps of SL2(Z) are equivalent to ∞, that is to say, SL2(Z) has only one equivalence class of cusps, that is, [∞].

Proof. We know from Theorem 1.17. that the set of cusps for SL2(Z) is Q ∪ {∞}. But we have also seen that for any a/c ∈ Q, c 6= 0, and gcd(a, c) = 1, there always exists a b ∈ SL ( ) c d 2 Z such that a b a ∞ = . c d c

This implies that Q ⊆ [∞] and therefore SL2(Z) has only one equivalence class of cusps, namely, [∞].

Remark 1.23. It has been a common practice to replace the term “equivalence class of cusps” by “cusp” itself.

ˆ Lemma 1.24. The set of inequivalent cusps of Γ0(N) is a subset of ˆ ˆ ˆ {Γ0(N)g1∞, Γ0(N)g2∞, ..., Γ0(N)gm∞},

m [ ˆ where SL2(Z) = Γ0(N)gi. i=1 ˆ Proof. Let a/c be a cusp of Γ0(N). Then hai a = Γˆ (N) . c 0 c Chapter 1: Congruence Subgroups of SL2(Z) 12

Since a/c is a cusp of SL2(Z), by Theorem 1.22, there exists a b ∈ SL ( ) c d 2 Z such that a b a ∞ = . c d c Then, hai a b = Γˆ (N) ∞. c 0 c d However, a b Γˆ (N) = Γˆ (N)g , for some integer i satisfying 0 ≤ i ≤ m. 0 c d 0 i Therefore, hai = Γˆ (N)g ∞, c 0 i and the lemma follows.

ˆ Theorem 1.25. The number of inequivalent cusps for Γ0(N) is equal to ˆ the number of double cosets of the form Γ0(N)\SL2(Z)/SL2(Z)∞, where ±1 m SL ( ) = {β ∈ SL ( )| β∞ = ∞} = | m ∈ . 2 Z ∞ 2 Z 0 ±1 Z

m [ ˆ Proof. Let SL2(Z) = Γ0(N)gi. Suppose that i 6= j and i=1 ˆ ˆ Γ0(N)gi∞ = Γ0(N)gj∞. ˆ Then for some γ ∈ Γ0(N),

gi∞ = γgj∞, or

−1 −1 gj γ gi∞ = ∞. −1 −1 This means that gj γ gi is an element in SL2(Z)∞, and

−1 −1 gj γ giSL2(Z)∞ = SL2(Z)∞, Chapter 1: Congruence Subgroups of SL2(Z) 13

giSL2(Z)∞ = γgjSL2(Z)∞, ˆ ˆ Γ0(N)giSL2(Z)∞ = Γ0(N)gjSL2(Z)∞.

ˆ ˆ Conversely, suppose that i 6= j, Γ0(N)giSL2(Z)∞ = Γ0(N)gjSL2(Z)∞. Then, ˆ ˆ Γ0(N)giSL2(Z)∞∞ = Γ0(N)gjSL2(Z)∞∞, which implies ˆ ˆ Γ0(N)gi∞ = Γ0(N)gj∞.

We require the following to prove Theorem 1.27. Theorem 1.26. (Dirichlet’s Theorem) Let a and b be two integers where gcd(a, b) = 1. Then there exists inﬁnitely many primes of the form ax + b.

Theorem 1.27. (i) A complete set of the double coset representatives is as follows, ˆ aip bip Γ0(N) SL2(Z)∞, ci dip where gcd(ci, dip) = 1, ci |N, 0 ≤ dip < ci, and for each ci and p 6= q, we have dip 6≡ diq (mod gcd(N/ci, ci)).

(ii) The number of double coset representatives is equal to X φ(gcd(N/ci, ci)),

ci|N where φ(n) represents the Euler phi-function. k l Proof. Let g = ∈ SL ( ) which implies gcd(k, m) = 1, and, m n 2 Z x y k l Γˆ (N)gSL ( ) = Γˆ (N) SL ( ) , 0 2 Z ∞ 0 zN w m n 2 Z ∞ ∗ ∗ = Γˆ (N) SL ( ) , 0 kNz + mw ∗ 2 Z ∞ Chapter 1: Congruence Subgroups of SL2(Z) 14

Let gcd(kN, m) = c, N = cN0 and m = cm0. So,

kNz + mw = c(kN0z + m0w).

Since gcd(k, m) = 1 and gcd(N0, m0) = 1, gcd(kN0, m0) = 1. Thus we can find infinitely many z, w such that kN0z+m0w = 1, and gcd(N0z, w) = 1. In particular, let kN0z0 + m0w0 = 1. The general solutions for z and w are z = −m0t + z0, and w = kN0t + w0 respectively. By Dirichlet’s Theorem, there are infinitely many primes of the form kN0t + w0 because gcd(kN0, w0) = 1. So we may assume w to be a prime such that gcd(c, w) = 1. Clearly, gcd(cN0z, w) = 1 or gcd(Nz, w) = 1. Thus there exists x, y ∈ Z where x y ∈ Γˆ (N). zN w 0

We can now write, where c ≥ 1, c|N, and gcd(c, u) = 1,

s v Γˆ (N)gSL ( ) = Γˆ (N) SL ( ) 0 2 Z ∞ 0 c u 2 Z ∞ s v 1 r = Γˆ (N) SL ( ) 0 c u 0 1 2 Z ∞ ∗ ∗ = Γˆ (N) SL ( ) , 0 c cr + u 2 Z ∞ where 1 r ∈ SL ( ) . 0 1 2 Z ∞ Since r can be chosen so that it satisﬁes 0 ≤ cr + u < c, we have for any g belonging to SL2(Z), a b Γˆ (N)gSL ( ) = Γˆ (N) SL ( ) , 0 2 Z ∞ 0 c d 2 Z ∞ where c ≥ 1, c|N, gcd(c, d) = 1, and 0 ≤ d < c. If ˆ a1 b1 ˆ a2 b2 Γ0(N) SL2(Z)∞ = Γ0(N) SL2(Z)∞, c1 d1 c2 d2 Chapter 1: Congruence Subgroups of SL2(Z) 15

γ δ ±1 h then for some ∈ Γˆ (N) and ∈ SL ( ) , we have αN β 0 0 ±1 2 Z ∞

γ δ a b ±1 h a b 1 1 = 2 2 . αN β c1 d1 0 ±1 c2 d2 And we get ±αNa1 ± βc1 = c2, (1.2.2)

h(αNa1 + βc1) ± αNb1 ± βd1 = d2. (1.2.3)

Since c1, c2 divides N, (1.2.2) can be written as,

N c2 N c1 ±α a1 ± β = , or, ± α a1 ± β = 1. c1 c1 c2 c2

This implies that c1|c2 and c2|c1. As c1, c2 > 0, c1 = c2. Let c1 = c2 = c, and we also notice that ±β − 1 ≡ 0 (mod N/c). Therefore, together with (1.2.3), we obtain

d2 − d1 = d1(±β − 1) + N(hαa1 ± αb1) + hβc ≡ 0 (mod gcd(N/c, c)).

This gives us d1 ≡ d2 (mod gcd(N/c, c)). So we have the double coset representatives as stated in the theorem and it follows immediately that the X number of double cosets is equal to φ(gcd(N/ci, ci)).

ci|N

ˆ Theorem 1.28. A complete set of inequivalent cusps for Γ0(N) is given by x ip , ci where gcd(ci, xip) = 1, ci |N, 0 ≤ xip < ci, and for each ci, p 6= q, xip 6≡ xiq (mod gcd(N/ci, ci)). Proof. Evidently from Lemma 1.24. and Theorem 1.27, aip ˆ aip bip = Γ0(N) ∞ ci ci dip ˆ aip = Γ0(N) ci Chapter 1: Congruence Subgroups of SL2(Z) 16

where gcd(ci, dip) = 1, ci |N, 0 ≤ dip < ci, and for each ci, p 6= q, dip 6≡ diq (mod gcd(N/ci, ci)). Since aipdip − bipci = 1, we have

−1 −1 aip ≡ dip (mod ci), or, aip = dip + kci for some k ∈ Z. This implies that −1 aip ˆ dip + kci = Γ0(N) ci ci −1 ˆ 1 −k dip + kci 1 −k ˆ = Γ0(N) , where ∈ Γ0(N) 0 1 ci 0 1 −1 ˆ dip = Γ0(N) ci d −1 = ip . ci

−1 −1 −1 If 0 ≤ dip < ci and since dip also satisﬁes gcd(ci, dip ) = 1, then −1 −1 dip 6≡ diq (mod gcd(N/ci, ci)) for p 6= q, where ci|N, then take xip to be −1 −1 dip and we are done. Otherwise, let dip = xip + hci for some integer h and 0 ≤ xip < ci. We see that aip ˆ xip + hci = Γ0(N) ci ci ˆ 1 −h xip + hci 1 −h ˆ = Γ0(N) , where ∈ Γ0(N) 0 1 ci 0 1 ˆ xip = Γ0(N) ci x = ip . ci

−1 −1 −1 Now, as dip = xip +hci, dip 6≡ diq (mod gcd(N/ci, ci)) implies xip 6≡ xiq (mod gcd(N/ci, ci)). It remains to show that gcd(xip, ci) = 1. Suppose −1 not. Then there exists p > 1, p | ci, p | xip, and thus p | dip which is a contradiction. This completes the proof of the theorem.

ˆ X Corollary 1.29. Γ0(N) has φ(gcd(N/c, c)) inequivalent cusps. c|N Chapter 1: Congruence Subgroups of SL2(Z) 17

Proof. It is an immediate consequence of the previous theorem.

x y Lemma 1.30. Suppose that gcd(a,b)=1, and ∈ SL ( ). Then z w 2 Z gcd (ax + by, az + bw) = 1.

Proof. Since gcd (a, b) = 1, there exists u, v ∈ Z such that av − bu = 1. Hence a u ∈ SL ( ). b v 2 Z It follows that x y a u ax + by ∗ = ∈ SL ( ). z w b v az + bw ∗ 2 Z

As a consequence, gcd (ax + by, az + bw) = 1.

Theorem 1.31. Let a/b, c/d ∈ Q ∪ {∞}, with gcd(a, b) = gcd(c, d) = 1. Then [a/b] = [c/d] in Γ(ˆ N) if and only if

(i) a ≡ c, b ≡ d (mod N), or

(ii) a ≡ −c, b ≡ −d (mod N).

Proof. Suppose [a/b]=[c/d] with gcd(a, b) = gcd(c, d) = 1. Then there exists some 1 + xN yN ∈ Γ(ˆ N) zN 1 + wN such that a 1 + xN yN c (1 + xN)c + yNd = = . b zN 1 + wN d zNc + (1 + wN)d

By Lemma 1.30,

gcd((1 + xN)c + yNd, zNc + (1 + wN)d) = 1, we conclude that either Chapter 1: Congruence Subgroups of SL2(Z) 18

(i) a = (1 + xN)c + yNd, b = zNc + (1 + wN)d, or

(ii) −a = (1 + xN)c + yNd, −b = zNc + (1 + wN)d.

This implies that

(i) a ≡ c, b ≡ d (mod N) or

(ii) −a ≡ c, −b ≡ d (mod N).

Conversely, suppose that

(i) a ≡ c, b ≡ d (mod N).

Since gcd(c, d) = 1, there exists x, y ∈ Z such that cy − dx = 1. Hence c x τ = ∈ SL ( ). d y 2 Z

It is clear that c x−1 a ay − bx = . d y b −ad + bc Note that

ay − bx ≡ cy − dx = 1 (mod N), −ad + bc ≡ −cd + cd = 0 (mod N).

By Lemma 1.30, gcd(ay − bx, −ad + bc) = 1. Let p, q ∈ Z be chosen such that (ay − bx) − 1 = (−ad + bc)Nq − (ay − bx)Np.

This implies that

ay − bx Nq ∈ Γ(ˆ N). −ad + bc 1 + Np

Note that ay − bx Nq 1 ay − bx = . −ad + bc 1 + Np 0 −ad + bc Chapter 1: Congruence Subgroups of SL2(Z) 19

As a consequence,

c x ay − bx Nq −1 c x−1 a c = . d y −ad + bc 1 + Np d y b d Since c x ay − bx Nq −1 c x−1 ∈ Γ(ˆ N), d y −ad + bc 1 + Np d y we conclude that hai h c i = . b d Suppose that (ii) −a ≡ c, −b ≡ d (mod N). Similar to the above, we can show that [a/b] = [c/d] . This completes the proof of the theorem.

Lemma 1.32. Let N ∈ N. Then N−1 [ 1 k Γˆ (N) = Γ(ˆ N). 1 0 1 k=0 Proof. Note that

[SL ( ): Γ(ˆ N)] [Γˆ (N): Γ(ˆ N)] = 2 Z = N. 1 ˆ [SL2(Z): Γ1(N)] ˆ ˆ The formulae for [SL2(Z): Γ1(N)] and [SL2(Z): Γ(N)] can be found in Chapter 2, Theorem 2.11. Now, let 0 ≤ x, y ≤ N − 1. Then 1 x 1 y Γ(ˆ N) = Γ(ˆ N), 0 1 0 1 if and only if 1 y−1 1 x Γ(ˆ N) = Γ(ˆ N), 0 1 0 1 if and only if 1 x − y ∈ Γ(ˆ N), 0 1 Chapter 1: Congruence Subgroups of SL2(Z) 20 if and only if x − y is a multiple of N, if and only if x = y.

Theorem 1.33. Let a/b, c/d ∈ Q ∪ {∞}, with gcd (a, b) = gcd(c, d) = 1. ˆ Then a/b and c/d are equivalent to each other in Γ1(N) if and only if (i) b − d is a multiple of N, a − c is a multiple of b modulo N, or (ii) b + d is a multiple of N, a + c is a multiple of b modulo N. Proof. Since by the previous lemma N−1 [ 1 k Γˆ (N) = Γ(ˆ N), 1 0 1 k=0 ˆ a/b and c/d are equivalent to each other in Γ1(N) if and only if there exists k ∈ Z, τ ∈ Γ(ˆ N) such that 1 k a c τ = , 0 1 b d or 1 k a −c τ = . 0 1 b −d Let a x τ = . b y This implies that a ≡ x, b ≡ y (mod N). As a consequence, c 1 k a c x + ky a + kb = τ = = ≡ (mod N), or d 0 1 b d y b

−c 1 k a c x + ky a + kb = τ = = ≡ (mod N). −d 0 1 b d y b Hence Chapter 1: Congruence Subgroups of SL2(Z) 21

(i) b − d is a multiple of N, a − c is a multiple of b modulo N, or (ii) b + d is a multiple of N, a + c is a multiple of b modulo N. Conversely, suppose that b − d is a multiple of N, a − c is a multiple of b modulo N. Then c ≡ a + kb, d ≡ b (mod N) for some k ∈ Z. By Theorem 1.31, there exists some τ ∈ Γ(ˆ N) such that c a + kb τ = . d b Therefore c 1 −k a = τ −1 . d 0 1 b ˆ Hence a/b and c/d are equivalent to each other in Γ1(N). Suppose that b + d is a multiple of N and a + c is a multiple of b modulo N. Similar to the above, one can show that a/b and c/d are equivalent to each other in ˆ Γ1(N). This completes the proof of the theorem.

Deﬁnition 1.34. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. The stabilizer of a/b in SL2(Z), SL2(Z)a/b, is deﬁned as follows, n a a o SL ( ) = g ∈ SL ( )| g = . 2 Z a/b 2 Z b b In particular,

SL2(Z)∞ = {g ∈ SL2(Z)| g∞ = ∞ } ±1 m = | m ∈ 0 ±1 Z 1 1 = ± . 0 1

Lemma 1.35. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. Then for some c, d ∈ Z, the stabilizer of a/b in SL2(Z) is given by, * + a c 1 1 a c−1 SL ( ) = ± . 2 Z a/b b d 0 1 b d Chapter 1: Congruence Subgroups of SL2(Z) 22

a c Proof. Since gcd(a, b) = 1, b d ∈ SL2(Z) for some c and d ∈ Z and by following a similar argument to Theorem 1.17, we observe that

a c ±1 m a c−1 a a = . b d 0 ±1 b d b b So * + a c 1 1 a c−1 ± ⊆ SL ( ) . b d 0 1 b d 2 Z a/b

Conversely, let γ ∈ SL2(Z)a/b. Then a a γ = , b b a c a γ ∞ = , b d b

a c−1 a c a c−1 a γ ∞ = b d b d b d b = ∞.

This implies that

a c−1 a c 1 1 γ ∈ ± , or, b d b d 0 1

a c 1 1 a c−1 γ ∈ ± , or, b d 0 1 b d * + a c 1 1 a c−1 γ ∈ ± . b d 0 1 b d Consequently, * + a c 1 1 a c−1 SL ( ) ⊆ ± , 2 Z a/b b d 0 1 b d from which the lemma readily follows. Chapter 1: Congruence Subgroups of SL2(Z) 23

1.3 Cusp Widths

Deﬁnition 1.36. Let G be a subgroup of SL2(Z) and a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. We deﬁne the G-width of a/b (also referred to as the width of the cusp a/b with respect to G) to be the smallest positive integer m such that

a c 1 m a c−1 a c ∈ G, for any ∈ SL ( ). b d 0 1 b d b d 2 Z

ˆ Remark 1.37. In accordance with Deﬁnition 1.36, we can deﬁne Γ0(N)- ˆ ˆ width, Γ1(N)-width and Γ(N)-width of a/b with gcd(a, b) = 1 in a similar manner.

Lemma 1.38. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. By considering

a c 1 m a c−1 a c , for any ∈ SL ( ), b d 0 1 b d b d 2 Z we have the following:

ˆ 2 (i) Γ0(N)-width of a/b is the smallest positive integer m such that N| b m, ˆ (ii) Γ1(N)-width of a/b is the smallest positive integer m such that N| abm and N| b2m,

(iii) Γ(ˆ N)-width of a/b is N.

Proof. By direct calculation,

a c 1 m a c−1 a c 1 m d −c = b d 0 1 b d b d 0 1 −b a a am + c d −c = b bm + d −b a 1 − abm a2m = . −b2m 1 + abm Chapter 1: Congruence Subgroups of SL2(Z) 24

ˆ ˆ Hence for the above element to be in Γ0(N) and Γ1(N) respectively, we require the conditions as listed in (i) and (ii) to be satisﬁed. For Γ(ˆ N), we need m to be the smallest positive integer such that N| abm, N| a2m, and N| b2m. This implies that

N| gcd(abm, a2m, b2m).

Clearly, gcd(abm, a2m, b2m)| m · gcd(ab, a2, b2). But gcd(ab, a2, b2) = 1 as gcd(a, b) = 1. Hence N|m. However, m is the smallest positive integer that satisﬁes the condition and thus we must have m = N for Γ(ˆ N). Chapter 2

The Modular Group PSL2(Z)

2.1 Γ0(N), Γ1(N) and Γ(N)

Deﬁnition 2.1. Let I be the identity element of SL2(Z). We deﬁne the modular group, PSL2(Z) as follows:

PSL2(Z) = SL2(Z)/{±I}.

In addition, let N ∈ N. We denote the following subgroups of PSL2(Z) by, ˆ Γ0(N) = Γ0(N)/{±I}, ˆ Γ1(N) = Γ1(N)/{±I}, Γ(N) = Γ(ˆ N)/{±I}.

Deﬁnition 2.2. Let G be a subgroup of PSL2(Z). G is a congruence subgroup of PSL2(Z) if there exists M ∈ N such that Γ(M) is a subgroup of G. Otherwise, G is a non-congruence subgroup. Thus, Γ0(N), Γ1(N), Γ(N) are all congruence subgroups of PSL2(Z).

Note that all deﬁnitions obtained from replacing SL2(Z) in the deﬁni- tions of Chapter 1 by PSL2(Z) are valid. Let us now revisit some theorems which we have proved for SL2(Z) so that we can establish similar results for PSL2(Z).

25 Chapter 2: The Modular Group PSL2(Z) 26

Remark 2.3. Since I = −I in PSL2(Z), the order of an element, A, in n PSL2(Z) is the smallest positive integer, n, such that A = ±I.

Theorem 2.4. The ﬁnite orders of elements in PSL2(Z) are 1, 2 and 3.

Proof. Recall from Theorem 1.4.that the ﬁnite orders of elements in SL2(Z) are 1, 2, 3, 4 and 6. By Remark 2.3, if the order of an element, g say, is 4 in SL2(Z), that is, g4 = I, then g2 = −I, which follows that the order of g in PSL2(Z) is 2. Similarly, an element of order 6 in SL2(Z) would be of order 3 in PSL2(Z) and the theorem thus follows.

Deﬁnition 2.5. Let G be a subgroup of PSL2(Z). G is said to be torsion free if the only element (of G) of ﬁnite order is the identity element.

With the abovementioned deﬁnition and applying the same reasoning as in Theorem 2.4, we can deduce the following theorem.

Theorem 2.6. Let N ∈ N and p be a prime. Then, (i) Γ(N) is torsion free if and only if N ≥ 2.

(ii) Γ1(N) is torsion free if and only if N ≥ 4.

(iii) Γ0(N) is torsion free if and only if (a) 4|N, or ∃ p|N such that p is of the form 4k + 3, and (b) 9|N, or ∃ p|N such that p is of the form 3k + 2.

Deﬁnition 2.7. Let G be a subgroup of PSL2(Z). g ∈ G is said to be elliptic if |tr(g)| < 2.

The following result is taken from [Sh]. Chapter 2: The Modular Group PSL2(Z) 27

Lemma 2.8. In PSL2(Z), 0 −1 (i) All cyclic subgroups of order 2 are conjugate to . 1 0

0 −1 (ii) All cyclic subgroups of order 3 are conjugate to . 1 −1

The following two well known results are taken from [Sh]. Theorem 2.9. Let p be a prime. The number of inequivalent elliptic subgroups of order 2 in Γ0(N), v2, is equal to the number of solutions of 2 x + 1 ≡ 0(mod N) in ZN , that is,  0 if 4|N,  Y −1 v2 = 1 + otherwise.  p p|N

Theorem 2.10. Let p be a prime. The number of inequivalent elliptic subgroups of order 3 in Γ0(N), v3, is equal to the number of solutions of 2 x + x + 1 ≡ 0(mod N) in ZN , that is,  0 if 9|N,  Y −3 v3 = 1 + otherwise.  p p|N

2.2 Indices of Subgroups of PSL2(Z) Let us ﬁrst state the following result from [Sh].

Theorem 2.11. Let N ∈ N,N ≥ 2 and p be a prime. Then, Y 1 (i) the index of Γ(ˆ N) in SL ( ) is N 3 (1 − ), 2 Z p2 p|N Y 1 (ii) the index of Γˆ (N) in SL ( ) is N 2 (1 − ), and 1 2 Z p2 p|N Chapter 2: The Modular Group PSL2(Z) 28

Y 1 (iii) the index of Γˆ (N) in SL ( ) is N (1 + ). 0 2 Z p p|N

Remark 2.12. Let p be a prime and the indices of Γ(N), Γ1(N) and Γ0(N) in PSL2(Z) be µ, µ1 and µ0 respectively. Since −I ∈ Γ(2) and Γ1(2), and, −I 6∈ Γ(N) and Γ1(N) for N ≥ 3, we have

N 3 Y 1  (1 − ) if N ≥ 3,  2 p2 µ = p|N 6 if N = 2.

N 2 Y 1  (1 − ) if N ≥ 3,  2 p2 µ1 = p|N 3 if N = 2.

Because −I ∈ Γ0(N) for all N ∈ N, this implies that the index of Γ0(N) in ˆ PSL2(Z) is the same as that of Γ0(N) in SL2(Z), that is, Y 1 µ = N (1 + ). 0 p p|N

Lemma 2.13. Let N ∈ N. Then

N−1 [ 1 k Γ (N) = Γ(N). 1 0 1 k=0 Proof. The proof of Lemma 2.13. is similar to that of Lemma 1.32.

Lemma 2.14. Let N > 2,N ∈ N. Then [ x 0 Γ (N) = Γ (N), 0 0 y 1 x,y where xy ≡ 1 (mod N), 1 ≤ x ≤ N/2 and 1 ≤ y ≤ N − 1. Chapter 2: The Modular Group PSL2(Z) 29

Proof. Applying the formulae in Remark 2.12, we obtain

[PSL2(Z):Γ1(N)] [Γ0(N):Γ1(N)] = [PSL2(Z):Γ0(N)] N Y 1 = (1 − ) 2 p p|N p prime φ(N) = . 2 Now, consider the following cosets of Γ(N),

[ x 0 Γ (N), 0 y 1 x,y where xy ≡ 1 (mod N), 1 ≤ x ≤ N/2 and 1 ≤ y ≤ N − 1. The number of x that satisﬁes the above conditions is in fact half the number of integers that are co-prime to and not exceeding N. So, there are φ(N)/2 choices for x and thus φ(N)/2 such cosets. Now, let us show that the cosets are distinct from each other. Let 1 ≤ x1, x2 ≤ N/2 and 1 ≤ y1, y2 ≤ N − 1, where x1y1 ≡ 1 (mod N) and x2y2 ≡ 1 (mod N). Then x1 0 x2 0 Γ1(N) = Γ1(N), 0 y1 0 y2 if and only if −1 x2 0 x1 0 Γ1(N) = Γ1(N), 0 y2 0 y1 if and only if x1y2 0 ∈ Γ1(N), 0 x2y1 if and only if

x1y2 ≡ 1 (mod N), and, x2y1 ≡ 1 (mod N), if and only if x1 = x2, and, y1 = y2. Chapter 2: The Modular Group PSL2(Z) 30

Lemma 2.15. Let N > 2,N ∈ N. Then [ x 0 1 z Γ (N) = Γ(N), 0 0 y 0 1 x,y,z where 0 ≤ z ≤ N − 1, xy ≡ 1 (mod N), 1 ≤ x ≤ N/2 and 1 ≤ y ≤ N − 1. Proof. Combining the results of the above two lemmas, we get [ x 0 Γ (N) = Γ (N) 0 0 y 1 x,y N−1 [ x 0 [ 1 z = Γ(N) 0 y 0 1 x,y z=0 [ x 0 1 z = Γ(N), 0 y 0 1 x,y,z where 0 ≤ z ≤ N−1, xy ≡ 1 (mod N), 1 ≤ x ≤ N/2 and 1 ≤ y ≤ N−1.

Proposition 2.16. Let N > 2,N ∈ N. Suppose φ(N) = 2. Then N−1 [ 1 k Γ (N) = Γ (N) = Γ(N). 0 1 0 1 k=0 Proof. By Lemma 2.15, we have [ x 0 1 z Γ (N) = Γ(N), 0 0 y 0 1 x,y,z where 0 ≤ z ≤ N − 1, xy ≡ 1 (mod N), 1 ≤ x ≤ N/2 and 1 ≤ y ≤ N − 1. Since φ(N)/2 = 1, there is only one choice for x, namely, x = 1. So x = 1, y = 1 and 0 ≤ z ≤ N − 1. As a consequence, N−1 [ 1 k Γ (N) = Γ(N). 0 0 1 k=0 On the other hand, notice from Lemma 2.13. that N−1 [ 1 k Γ (N) = Γ(N). 1 0 1 k=0 This completes the proof of the proposition. Chapter 2: The Modular Group PSL2(Z) 31

2.3 Cusps of Γ0(N), Γ1(N) and Γ(N) The proofs for Theorem 2.17. and Theorem 2.18. can be procured in a similar way to that of the SL2(Z) case and we shall just state below these two theorems.

Theorem 2.17. The sets of cusps for PSL2(Z), Γ0(N), Γ1(N) and Γ(N) are the same, namely, Q ∪ {∞}.

Theorem 2.18. PSL2(Z) has only one equivalence class of cusps which is [∞].

Theorem 2.19. Let a/b, c/d ∈ Q ∪ {∞}, with gcd(a, b) = gcd(c, d) = 1. Then

(i) a/b and c/d are equivalent to each other in Γ(N) if and only if they are equivalent to each other in Γ(ˆ N), and

(ii) the number of inequivalent cusps of Γ(N) is equal to that of Γ(ˆ N).

Proof. Let x, y ∈ Q ∪ {∞}. Suppose [x] = [y] in Γ(ˆ N). Then there exists g ∈ Γ(ˆ N) such that gx = y. But g ∈ Γ(N). So [x] = [y] in Γ(N). Conversely, suppose [x] = [y] in Γ(N). Then there exists

a b ∈ Γ(N) c d such that a b x = y. c d Now, we distinguish into two cases. Case 1: a ≡ d ≡ 1 (mod N). Then a b ∈ Γ(ˆ N) c d and thus [x] = [y] in Γ(ˆ N). Chapter 2: The Modular Group PSL2(Z) 32

Case 2: a ≡ d ≡ −1 (mod N). Then −a −b ∈ Γ(ˆ N) −c −d and −a −b x = y. −c −d Hence [x] = [y] in Γ(ˆ N). This completes the proof of the theorem.

Likewise to the above theorem, we obtain the following two results.

Theorem 2.20. Let a/b, c/d ∈ Q ∪ {∞}, with gcd(a, b) = gcd(c, d) = 1. Then

(i) a/b and c/d are equivalent to each other in Γ1(N) if and only if they ˆ are equivalent to each other in Γ1(N), and ˆ (ii) the number of inequivalent cusps of Γ1(N) is equal to that of Γ1(N).

Theorem 2.21. Let a/b, c/d ∈ Q ∪ {∞}, with gcd(a, b) = gcd(c, d) = 1. Then

(i) a/b and c/d are equivalent to each other in Γ0(N) if and only if they ˆ are equivalent to each other in Γ0(N), and ˆ (ii) the number of inequivalent cusps of Γ0(N) is equal to that of Γ0(N).

Theorem 2.22. Let a/b, c/d ∈ Q ∪ {∞}, with gcd(a, b) = gcd(c, d) = 1. Then [a/b] = [c/d] in Γ(N) if and only if

(i) a ≡ c, b ≡ d (mod N), or

(ii) a ≡ −c, b ≡ −d (mod N).

Proof. This follows directly from Theorem 1.31. and Theorem 2.19. Chapter 2: The Modular Group PSL2(Z) 33

Applying Theorem 2.22, one may write down a set of representatives of inequivalent cusps as follows: (See [L1]) (1) Cusps of Γ(2m).

For each k (1 ≤ k ≤ m − 1), deﬁne Ak to be the set

Ak = {xˆ | 1 ≤ xˆ ≤ 2m, gcd(gcd (k, xˆ), 2m)) = 1}.

For eachx ˆ ∈ Ak, let x =x ˆ + 2my be the smallest positive integer such that gcd (x, k) = 1. Deﬁne Sk to be the set

Sk = {x/k | xˆ ∈ Ak}.

Let Am = {xˆ | 1 ≤ xˆ ≤ m, gcd(gcd (m, xˆ), 2m)) = 1},

A2m = {xˆ | 1 ≤ xˆ ≤ m, gcd(gcd (2m, xˆ), 2m)) = 1}. Deﬁne

(i) Sm = {x/m | xˆ ∈ Am}, where x =x ˆ + 2my is the smallest positive integer such that gcd (x, m) = 1,

(ii) S2m = {x/2m | xˆ ∈ A2m}, where x =x ˆ + 2my is the smallest positive integer such that gcd (x, 2m) = 1.

Then S1 ∪ S2 ∪ · · · ∪ Sm−1 ∪ Sm ∪ S2m is a set of cusps of Γ(2m).

2. Cusps of Γ(2m + 1).

For each k (1 ≤ k ≤ m), deﬁne Ak to be the set

Ak = {xˆ | 1 ≤ xˆ ≤ 2m + 1, gcd(gcd (k, xˆ), 2m + 1)) = 1}.

For eachx ˆ ∈ Ak, let x =x ˆ + (2m + 1)y be the smallest positive integer such that gcd (x, k) = 1. Deﬁne Sk to be the set

Sk = {x/k | xˆ ∈ Ak}.

Let A2m+1 = {xˆ | 1 ≤ xˆ ≤ m, gcd(gcd (k, xˆ), 2m + 1)) = 1}. Deﬁne S2m+1 = {x/(2m + 1) | xˆ ∈ A2m+1}, Chapter 2: The Modular Group PSL2(Z) 34 where x =x ˆ+(2m+1)y is the smallest positive integer such that gcd (x, 2m+ 1) = 1. Then S1 ∪ S2 ∪ · · · ∪ Sm ∪ S2m+1 is a set of cusps of Γ(2m + 1).

Remark 2.23. It can be seen from Theorem 2.22 that in Γ(N), 1/N is equivalent to ∞ (note that ∞ = 1/0) while 0 is equivalent to N.

Theorem 2.24. Let a/b, c/d ∈ Q ∪ {∞}, with gcd(a, b) = gcd(c, d) = 1. Then a/b and c/d are equivalent to each other in Γ1(N) if and only if

(i) b − d is a multiple of N, a − c is a multiple of b modulo N, or

(ii) b + d is a multiple of N, a + c is a multiple of b modulo N.

Proof. This follows directly from Theorem 1.33. and Theorem 2.20.

Theorem 2.25. A complete set of inequivalent cusps for Γ0(N) is x ip , ci where gcd(ci, xip) = 1, ci |N, 0 ≤ xip < ci, and for each ci, p 6= q, xip 6≡ xiq (mod gcd(N/ci, ci)). Proof. This follows directly from Theorem 1.28. and Theorem 2.21.

X Corollary 2.26. Γ0(N) has φ(gcd(N/c, c)) inequivalent cusps. c|N Proof. It is an immediate consequence of the previous theorem.

The following result is taken from [L1].

Lemma 2.27. Let a/b be a cusp of Γ0(N) with gcd(a, b) = 1. Let gcd(b, N) = n0 and b = n0y. Choose r such that 1 ≤ r ≤ n0, r ≡ ay (mod gcd(n0, N/n0)) and gcd(r, n0) = 1. Then a/b and r/n0 are Γ0(N)-equivalent. Chapter 2: The Modular Group PSL2(Z) 35

Remark 2.28. With the above lemma, for any cusp not of the form as described in Theorem 2.25, we can always ﬁnd a cusp of such form that is equivalent to it modulo Γ0(N).

Similar to Definition 1.34, we have the following definition. Definition 2.29. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. The stabilizer of a/b in PSL2(Z), PSL2(Z)a/b, is defined as follows, n a a o PSL ( ) = g ∈ PSL ( )| g = . 2 Z a/b 2 Z b b In particular,

PSL2(Z)∞ = {g ∈ PSL2(Z)| g∞ = ∞ } ±1 m = | m ∈ 0 ±1 Z 1 1 = . 0 1

Lemma 2.30. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. Then the stabilizer of a/b in PSL2(Z) is an inﬁnite cyclic group. Proof. Likewise to Lemma 1.35, we can show that there exist c, d ∈ Z such that * + a c 1 1 a c−1 PSL ( ) = , 2 Z a/b b d 0 1 b d which is an inﬁnite cyclic group.

Proposition 2.31. Let G be a subgroup of PSL2(Z) and gcd(a, c) = 1. Then the stabilizer of a/c in G, denoted by Ga/c is generated by a c 1 m a c−1 1 − acm a2x σ = = , b d 0 1 b d −c2m 1 + acm where a, b, c, d ∈ Z, ad − bc = 1, and, m is the smallest positive integer such that σ ∈ G.

Proof. It follows directly from the previous lemma and the fact that Ga/c is a subgroup of PSL2(Z)a/c. Chapter 2: The Modular Group PSL2(Z) 36

2.4 Cusp Widths

Analogous to the SL2(Z) case, we have Deﬁnition 2.32. and Theorem 2.33. as listed below.

Deﬁnition 2.32. Let G be a subgroup of PSL2(Z) and a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. We deﬁne the G-width of a/b (also referred to as the width of the cusp a/b with respect to G) to be the smallest positive integer m such that a c 1 m a c−1 a c ∈ G, for any ∈ PSL ( ). b d 0 1 b d b d 2 Z

Theorem 2.33. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. By considering a c 1 λ a c−1 a c , for any ∈ PSL ( ), b d 0 1 b d b d 2 Z we have the following: 2 (i) Γ0(N)-width of a/b is the smallest positive integer λ such that N| b λ,

(ii) Γ1(N)-width of a/b is the smallest positive integer λ such that N| abλ and N| b2λ, and, (iii) Γ(N)-width of a/b is N.

Deﬁnition 2.34. Let G be a subgroup of PSL2(Z) and {x1, x2, ..., xh} be a set of inequivalent cusps for G. Denote the cusp widths of x1, x2, ..., xh by n(1), n(2), ..., n(h) respectively. Then the sequence (n(1), n(2), ..., n(h)) is deﬁned to be the cusp-split of G, where the order of the cusp widths in a cusp-split is arbitrary.

Theorem 2.35. (W.W. Stothers [St]) Let (n(1), n(2), ..., n(h)) be the cusp- split of G, where G is of index n in PSL2(Z). Then h X n = n(i) = n(1) + n(2) + ... + n(h). i=1

Corollary 2.36. The number of inequivalent cusps for Γ(N) is µN /N, where µN is the index of Γ(N) in PSL2(Z). Proof. It is a direct consequence of Theorem 2.33. and Theorem 2.35. Chapter 2: The Modular Group PSL2(Z) 37

2.5 The Genus Formula

Theorem 2.37. (G. Shimura [Sh]) Let G be a subgroup of ﬁnite index in ∗ PSL2(Z), and H = H ∪Q∪{∞}, where H is the upper half of the complex plane. Then the genus, g, of H∗/G is given by µ v v v g = 1 + − 2 − 3 − ∞ , (2.5.1) 12 4 3 2 where µ = the index of G in PSL2(Z), v2 = the number of inequivalent elliptic subgroups of order 2 of G, v3 = the number of inequivalent elliptic subgroups of order 3 of G, v∞ = the number of inequivalent cusps of G.

Remark 2.38. Let G be a subgroup of PSL2(Z). We shall refer to the genus of H∗/G as the genus of G and denote it by g(G).

Lemma 2.39. Let A be a subgroup of ﬁnite index in B, where A and B are subgroups of PSL2(Z). Suppose the genus of B, g(B) 6= 0. Then the genus of A, g(A) 6= 0.

Proof. Since A is a subgroup of ﬁnite index in B, the following Riemann- Hurwitz formula holds (see [Sh]), that is, X 2g(A) = 2 + [B : A](2g(B) − 2) + (ez − 1), (2.5.2) z∈H∗/A

∗ where ez is the ramiﬁcation index of z, for z ∈ H /A. Since g(B) 6= 0, we ∗ have 2g(B) − 2 ≥ 0. Furthermore, ez ≥ 1 for all z ∈ H /A and [B : A] ≥ 1 imply that for (2.5.2), RHS > 0, and so g(A) 6= 0. Chapter 3 sqf Genus of Γτ (m; m/d, ε, χ)

The main objective of this chapter is to determine the genus formula of sqf Γτ (m; m/d, ε, χ) (which we will define formally in due course). Recall the genus formula for a subgroup of finite index in PSL2(Z) from Chapter 2, Theorem 2.37. This means that we need to find the number of inequivalent sqf sqf cusps of Γτ (m; m/d, ε, χ), the index of Γτ (m; m/d, ε, χ) in PSL2(Z), and, the number of inequivalent elliptic subgroups of order 2 and 3 respectively sqf for Γτ (m; m/d, ε, χ).

3.1 Larcher Congruence Subgroups

Deﬁnition 3.1. Let G be a subgroup of PSL2(Z). Then G is a congruence subgroup of level m if m is the least positive integer such that Γ(m) is a subgroup of G.

Deﬁnition 3.2. Let m be a positive integer and d be a positive divisor of 2 m. Then m/d = h n, for some n ∈ Z, where n is square-free. Also, let ε and χ be positive integers such that ε|h and χ|gcd(dε, m/dε2) and let τ ∈ {1, 2, ..., χ}. Then we have the following deﬁnition of Larcher congruence subgroup,  m  1 + α dβ   εχ  Γτ (m; m/d, ε, χ) = ±  m m  | γ ≡ τα (mod χ) / ± I, γ 1 + δ  χ χ 

38 sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 39 where α, β, γ and δ are integers and the elements of this subgroup are of determinant 1.

The following theorem is due to H. Larcher. (See [La].)

Theorem 3.3. Suppose Γτ (m; m/d, ε, χ) 6∈ { Γ1(4; 2, 1, 2), Γ1(8; 8, 2, 2) }. Then Γτ (m; m/d, ε, χ) is a congruence subgroup of PSL2(Z) of level m. Note that Γ1(4; 2, 1, 2) and Γ1(8; 8, 2, 2) are in fact Γ(2) and Γ0(4) respectively.

Lemma 3.4. Γτ (m; m/d, ε, χ) ⊆ Γ0(m/χ) ∩ Γ1(m/εχ) ∩ Γ(d). In particular, if m is square-free, then

Γτ (m; m/d, ε, χ) = Γ1(m) ∩ Γ(d).

Proof. Since χ|m/dε2, we have d|m/εχ and d|m/χ. Thus, it is clear that Γτ (m; m/d, ε, χ) ⊆ Γ0(m/χ) ∩ Γ1(m/εχ) ∩ Γ(d). If m is square-free, then ε = 1, which implies that gcd(dε, m/dε2) = 1. So χ = 1, and thus τ = 1. By the deﬁnition of Γτ (m; m/d, ε, χ), we can see that Γτ (m; m/d, ε, χ) is just the subgroup Γ1(m) ∩ Γ(d).

Theorem 3.5. and Lemma 3.6. are taken from [Se].

Theorem 3.5. Let G be a congruence group of level m, and let d be the least cusp width in G. Then

(i) Γτ (m; m/d, ε, χ) is a subgroup of G. (ii) The widths of any cusp (rational or ∞) are the same with respect to G and to Γτ (m; m/d, ε, χ), and we refer to Γτ (m; m/d, ε, χ) as the Larcher group corresponding to G.

Lemma 3.6. Suppose that Γτ (m; m/d, ε, χ) 6∈ { Γ1(4; 2, 1, 2), Γ1(8; 8, 2, 2) }. Then the cusp widths in Γτ (m; m/d, ε, χ) are as follows,

(i) the cusp width of ∞ is d, with d being the least cusp width and the gcd of all cusp widths,

(ii) the cusp width of 0 is m, with m being the lcm of all cusp widths, and, sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 40

(iii) the cusp width of a/b, where gcd(a, b) = 1, is

dσ m , with σ = . gcd (σ, εχ, b − τaε) gcd (db, m)

Let m, d, , χ satisfy the conditions listed in Deﬁnition 3.2. In addition, if m is square-free, we shall denote the corresponding Larcher congruence sqf sqf subgroup by Γτ (m; m/d, ε, χ). So, by Theorem 3.3, Γτ (m; m/d, ε, χ) is a Larcher congruence subgroup of level m. Furthermore, Lemma 3.4. sqf establishes that Γτ (m; m/d, ε, χ) is in fact Γ1(m) ∩ Γ(d), that is,

1 + mα dβ Γsqf (m; m/d, ε, χ) = ± I, τ mγ 1 + mδ / where α, β, γ and δ are integers and the elements of this subgroup are of determinant 1.

sqf 3.2 Index of Γτ (m; m/d, ε, χ) in PSL2(Z)

sqf Given the congruence subgroup Γτ (m; m/d, ε, χ). We are now inter- ested in calculating the index of this subgroup in PSL2(Z). Suppose m = 2. sqf Since by assumption, d|m, so d = 1 or 2. Thus Γτ (m; m/d, ε, χ) = Γ0(2) which is of index 3 by Remark 2.12. Hence we shall consider the case where m ≥ 3.

sqf Proposition 3.7. Γτ (m; m/d, ε, χ) is conjugate by d 0 0 1 to Γ0(md) ∩ Γ1(m). sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 41

1 + αm βd Proof. Let ± be an element in Γsqf (m; m/d, ε, χ). Then γm 1 + δm τ

1/d 0 1 + αm βd d 0 ± 0 1 γm 1 + δm 0 1

1 + αm β = ± γmd 1 + δm

∈ Γ0(md) ∩ Γ1(m).

Thus,

1/d 0 d 0 Γsqf (m; m/d, ε, χ) ⊆ Γ (md) ∩ Γ (m). 0 1 τ 0 1 0 1

1 + α0m β0 Conversely, let ± ∈ Γ (md) ∩ Γ (m). Then γ0md 1 + δ0m 0 1

d 0 1 + α0m β0 1/d 0 ± 0 1 γ0md 1 + δ0m 0 1

1 + α0m β0d = ± γ0m 1 + δ0m

sqf ∈ Γτ (m; m/d, ε, χ). Therefore,

d 0 1/d 0 Γ (md) ∩ Γ (m) ⊆ Γsqf (m; m/d, ε, χ), 0 1 0 1 0 1 τ from which the proposition readily follows.

Remark 3.8. Refer to Theorem 2.37. for the genus formula. Now since sqf Γτ (m; m/d, ε, χ) and Γ0(md)∩Γ1(m) are both subgroups of PSL2(Z),and, d 0 conjugation by 0 1 preserves the genus, v2, v3 and v∞, this means that the index is also preserved. Hence, it suﬃces to determine the index of Γ0(md) ∩ Γ1(m) in PSL2(Z). sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 42

Proposition 3.9. Let m ≥ 3. Then

m2d Y 1 [PSL ( ):Γ (md) ∩ Γ (m)] = (1 − ). 2 Z 0 1 2 p2 p prime p|m Proof. Since m ≥ 3, we have

[Γ0(md):Γ0(md) ∩ Γ1(m)] = φ(md)/2|G|, ˆ ˆ × where G = G{±I}/{±I} with G being the subgroup of Zmd generated by {z | z ≡ 1(mod m)} and φ(md) denotes the Euler function. Now, clearly, |G| = d, and together with Remark 2.12. and the fact that d is a divisor of m, we obtain

m2d2 Y 1 m2d Y 1 [PSL ( ):Γ (md) ∩ Γ (m)] = (1 − ) = (1 − ), 2 Z 0 1 2|G| p2 2 p2 p prime p prime p|md p|m as desired.

sqf Theorem 3.10. The index of Γτ (m; m/d, ε, χ) in PSL2(Z) is, m2d Y 1  (1 − ) if m ≥ 3,  2 p2 sqf p prime [PSL2(Z):Γτ (m; m/d, ε, χ)] =  p|m 3 if m = 2.

Proof. It is a direct consequence of Proposition 3.7, Remark 3.8. and Propo- sition 3.9.

3.3 Number of Inequivalent Cusps

sqf In this section, we will study the inequivalent cusps of Γτ (m; m/d, ε, χ). Theorem 3.11. Let m, n ∈ Z and a/b, c/d ∈ Q ∪ {∞}, with gcd(a, b) = gcd(c, d) = 1. Then a/b and c/d are equivalent to each other in Γ1(m)∩Γ(n) if and only if sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 43

(i) a − c is a multiple of nb modulo l, and, b − d is a multiple of l, where l = lcm(m, n), or

(ii) a + c is a multiple of nb modulo l, and, b + d is a multiple of l, where l = lcm(m, n).

Proof. Given that a/b is equivalent to c/d in Γ1(m) ∩ Γ(n). Then there exists 1 + αl βn γl 1 + δl in Γ1(m) ∩ Γ(n) such that 1 + αl βn a c = . γl 1 + δl b d This implies a + αla + βnb c = . γla + b + δlb d By a similar proof to Lemma 1.30, we can show that

gcd(a + αla + βnb , γla + b + δlb) = 1.

This means that either (i) c = a + αla + βnb, d = γla + b + δlb, or,

(ii) −c = a + αla + βnb, −d = γla + b + δlb, which give us (i) a − c is a multiple of nb modulo l, b − d is a multiple of l, or,

(ii) a + c is a multiple of nb modulo l, b + d is a multiple of l. Conversely, suppose we are given (i) or (ii). From (i), we have

c ≡ a + knb(mod l), d ≡ b(mod l) for some k ∈ Z. By Theorem 2.22, there exists g ∈ Γ(l), where 1 + xl yl g = zl 1 + wl sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 44 for some x, y, z, w ∈ Z, such that c a + knb = g . d b So, c 1 kn a = g . d 0 1 b Now,

1 kn 1 + xl yl 1 kn g = 0 1 zl 1 + wl 0 1 1 + xl kn + xlkn + yl = zl zlkn + 1 + wl

∈ Γ1(m) ∩ Γ(n), as l is a multiple of n and

1 kn det(g) = det = 1 0 1 implies 1 kn det g = 1. 0 1

Hence a/b and c/d are equivalent to each other in Γ1(m) ∩ Γ(n). Given (ii), similar to above, one can show that a/b and c/d are equivalent to each other in Γ1(m) ∩ Γ(n). This completes the proof of the theorem.

Corollary 3.12. Let a/b, x/y ∈ Q ∪ {∞}, with gcd(a, b) = gcd(x, y) = 1. sqf Then a/b and x/y are equivalent to each other in Γτ (m; m/d, ε, χ) if and only if

(i) a − x is a multiple of db modulo m, and, b − y is a multiple of m, or

(ii) a + x is a multiple of db modulo m, and, b + y is a multiple of m. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 45

sqf Proof. As mentioned in the previous section, Γτ (m; m/d, ε, χ) = Γ1(m) ∩ Γ(d). Also, by assumption, d is a divisor of m which simply implies that lcm(m, d) = m. Thus applying Theorem 3.11. gives us our desired result.

Lemma 3.13. Let A be a subgroup of B, where A and B are both subgroups of PSL2(Z). Suppose a/b and c/d, with gcd(a, b) = gcd(c, d) = 1, are inequivalent in B. Then they are also inequivalent in A.

Proof. Assume a/b and c/d are equivalent in A. Then for some ν ∈ A, a c ν = . b d But ν ∈ B since A ⊆ B. So a/b and c/d are also equivalent in B.

sqf Remark 3.14. Note that from Theorem 3.3, Γ(m) ⊆ Γτ (m; m/d, ε, χ), and Theorem 2.22. provides us the tool to write down a set of inequivalent cusps for Γ(m). Hence Corollary 3.12. and Lemma 3.13. enables us to write sqf down explicitly a set of inequivalent cusps for Γτ (m; m/d, ε, χ).

After procuring the description for the equivalence of the cusps, we proceed to ﬁnd a general formula for calculating the number of inequivalent sqf cusps of Γτ (m; m/d, ε, χ).

Proposition 3.15. Let P be a subgroup of Q, where P and Q are subgroups of PSL2(Z). Let Ω be a complete set of inequivalent cusps of Q. Then for each cusp a/c ∈ Ω, the equivalence class {q(a/c) | q ∈ Q} splits into |P \Q/Qa/c| inequivalent P cusps. Furthermore, if P is a normal subgroup of Q, then the equivalence class {q(a/c) | q ∈ Q} splits into

[Q : P ]

[Qa/c : Pa/c] inequivalent P cusps. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 46

Proof. Let x/y ∈ {q(a/c) | q ∈ Q}. Then the equivalence class of x/y in P is,

x x = P y y a = P q , for some q ∈ Q. c Let n [ Q = P qi. i=1 Then P q = P qi, for some 1 ≤ i ≤ n. Therefore,

x a = P q . y i c

This shows that for each a/c ∈ Ω, the equivalence class {q(a/c) | q ∈ Q} splits into inequivalent P cusps of the form P qi(a/c), for 1 ≤ i ≤ n. Let us now prove that the equivalence class {q(a/c) | q ∈ Q} splits into |P \Q/Qa/c| inequivalent P cusps. Suppose that i 6= j and a a P q = P q . i c j c Then for some γ ∈ P , we have a a q = γq , or, i c j c a a q−1γ−1q = , j i c c −1 −1 that is, qj γ qi ∈ Qa/c. Thus

−1 −1 qj γ qiQa/c = Qa/c,

qiQa/c = γqjQa/c,

P qiQa/c = P qjQa/c. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 47

Conversely, suppose that i 6= j and

P qiQa/c = P qjQa/c. Then a a P q Q = P q Q , i a/c c j a/c c a a P q = P q . i c j c Thus the equivalence class {q(a/c) | q ∈ Q} splits into |P \Q/Qa/c| inequivalent P cusps. Now, if P is a normal subgroup of Q, then

P \Q/Qa/c = {P qQa/c | q ∈ Q}

= {qP Qa/c | q ∈ Q}

= Q/P Qa/c. But it is clear that [Q : P ] |Q/P Qa/c| = , [Qa/c : Pa/c] and so in turn yields our desired result.

Lemma 3.16. Let c|md and gcd(a, c) = 1. Denote Gˆ to be the subgroup of × ˆ Zmd generated by {z | z ≡ 1(mod m)} and G = G{±I}/{±I}. Then

t = [Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c], where t is the smallest positive integer such that 1 − actmd/(md, c2) ∈ G.

Proof. We can deduce from Proposition 2.31 that Γ0(md)a/c is generated by 1 − acx a2x , −c2x 1 + acx 2 where x = md/(md, c ). Similarly, Γ0(md) ∩ Γ1(m))a/c is generated by 1 − acx a2x , −c2x 1 + acx where x = tmd/(md, c2), and, t is the smallest positive integer such that 1 − acx = 1 − actmd/(md, c2) ∈ G. As a consequence,

t = [Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c], where t is the smallest positive integer such that 1−actmd/(md, c2) ∈ G. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 48

Lemma 3.17. Let c|md and gcd(a, c) = 1. Then

[Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c] = [Γ0(md)1/c : (Γ0(md) ∩ Γ1(m))1/c].

Proof. Notice that for all k ∈ Z, 1 k 0 1 is an element in Γ0(md). This means that

hai a + kc = c c in Γ0(md) for all k ∈ Z. By applying Dirichlet’s Theorem, we may assume that a is a prime number such that gcd(a, md) = 1. Let

t = [Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c] and s = [Γ0(md)1/c : (Γ0(md) ∩ Γ1(m))1/c]. ˆ × Denote G to be the subgroup of Zmd generated by {z | z ≡ 1(mod m)} and G = Gˆ{±I}/{±I}. Then by the preceding lemma, t is the smallest positive integer such that 1 − actmd/(md, c2) ∈ G and s is the smallest positive integer such that 1 − csmd/(md, c2) ∈ G respectively. By our assumption of a, we can always choose n ∈ Z such that an ≡ 1(mod md). Then md n md 1 − act ≡ 1 − ct ∈ G, (md, c2) (md, c2) which implies that s ≤ t. On the other hand,

md a md 1 − cs ≡ 1 − acs ∈ G, (md, c2) (md, c2) and so t ≤ s. Hence t = s, as desired. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 49

sqf Theorem 3.18. Γτ (m; m/d, ε, χ) has X φ((md/c, c))φ(md)((md/c, c), d) 2d(md/c, c) c|md inequivalent cusps. sqf Proof. Since Γτ (m; m/d, ε, χ) and Γ0(md) ∩ Γ1(m) are conjugates to each other by Proposition 3.7. which follows that both of these subgroups have the same number of inequivalent cusps, we need only consider the case for Γ0(md) ∩ Γ1(m). Now, for each c|md, let

Ωc = {ai/c | (ai, c) = 1, ai 6≡ aj(mod (md/c, c))}

It is clear that Ωc comprises φ((md/c, c)) elements, and, [ Ω = Ωc c|md is a complete set of inequivalent cusps for Γ0(md) from Theorem 2.25. In addition, it can be easily checked that Γ0(md)∩Γ1(m) is a normal subgroup of Γ0(md). So, by the previous proposition, for each cusp a/c ∈ Ω, the equivalence class {g(a/c) | g ∈ Γ0(md)} splits into

[Γ0(md):Γ0(md) ∩ Γ1(m)]

[Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c] inequivalent Γ0(md)∩Γ1(m) cusps. Together with the fact that Ωc comprises φ((md/c, c)) elements, we can establish that Γ0(md) ∩ Γ1(m) has

X φ((md/c, c))[Γ0(md):Γ0(md) ∩ Γ1(m)] [Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c] c|md inequivalent cusps. Note that φ(md) [Γ (md):Γ (md) ∩ Γ (m)] = , 0 0 1 2|G| ˆ ˆ × where G = G{±I}/{±I} with G being the subgroup of Zmd generated by {z | z ≡ 1(mod m)}. It remains to determine [Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c]. By Lemma 3.16. and Lemma 3.17,

tc = [Γ0(md)a/c : (Γ0(md) ∩ Γ1(m))a/c], sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 50

2 where tc is the smallest positive integer such that 1 − ctcmd/(md, c ) ∈ G. In fact, one can easily see that for each c|md,

(md/c, c) t = . c ((md/c, c), d)

Also recall that |G| = d. Thus, we can conclude that Γ0(md) ∩ Γ1(m) and sqf therefore Γτ (m; m/d, ε, χ) has X φ((md/c, c))φ(md)((md/c, c), d) 2d(md/c, c) c|md inequivalent cusps.

3.4 Number of Elliptic Subgroups

Throughout the remaining sections, we shall denote the number of nonconjugating elliptic subgroups of order 2 and 3 by v2 and v3 respectively.

Theorem 3.19. Let N ∈ N. Then

(i) Γ0(2) has a unique conjugacy class of elliptic subgroup of order 2,

(ii) if 4|N or N has a prime divisor of the form 4k+3, then Γ0(N) admits no elliptic subgroup of order 2, and,

(iii) if all the prime divisors of N are of the form 4k + 1, then Γ0(N) has 2η nonconjugating elliptic subgroups of order 2, where η is the number of prime divisors of N. Furthermore, the following is a complete set of nonconjugating elliptic subgroups of order 2 of Γ0(N), ai bi 2 η | ai ≡ ai, ai + 1 ≡ 0, ai 6≡ aj(mod N), 1 ≤ i ≤ 2 , ciN −ai

η 2 where ai, 1 ≤ i ≤ 2 are the solutions of x + 1 ≡ 0(mod N), ai, bi, ci are integers and the subgroups of the set are of determinant 1.

2 Proof. By Theorem 2.9, v2 is equal to the number of solutions of x + 1 ≡ 0(mod N). For N = 2, x = 1 is the unique solution to x2 +1 ≡ 0(mod 2) in Z2, from which (i) readily follows. From the proof of Theorem 1.13, we see sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 51

ˆ that Γ0(N) has no element of order 4 if and only if 4|N or N has a prime divisor of the form 4k + 3. Together with Theorem 1.7. and Remark 2.3, we obtain (ii). Now, suppose all the prime divisors of N are of the form 4k + 1, then by Lemma 1.12.(i),

−1 = 1, p which implies that Γ0(N) has

Y −1 1 + = 2η p p|N nonconjugating elliptic sungroups of order 2, where η is the number of prime divisors of N. In addition,

a b cN d is an elliptic subgroup of order 2 if and only if a + d = 0. This can be deduced from Corollary 1.5, Theorem 1.7. and Remark 2.3. So, this gives us −a2 = a(−a) = ad ≡ 1(mod N), which has 2η solutions modulo N, where η is the number of prime divisors η of N. Since Γ0(N) also has 2 nonconjugating elliptic subgroups of order 2, 2 we may assert that for each solution a of x + 1 ≡ 0(mod N), Γ0(N) has an elliptic subgroup of order 2 of the form

a b , cN d where a ≡ a(mod N). Hence, the following is a complete set of nonconjugating elliptic subgroups of order 2 of Γ0(N), ai bi 2 η | ai ≡ ai, ai + 1 ≡ 0, ai 6≡ aj(mod N), 1 ≤ i ≤ 2 , ciN −ai

η 2 where ai, 1 ≤ i ≤ 2 are the solutions of x + 1 ≡ 0(mod N), ai, bi, ci ∈ Z and the subgroups of the set are of determinant 1. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 52

Similar to above, we have the following theorem.

Theorem 3.20. Let N ∈ N. Then

(i) Γ0(3) has a unique conjugacy class of elliptic subgroup of order 3,

(ii) if 9|N or N has a prime divisor of the form 3k+2, then Γ0(N) admits no elliptic subgroup of order 3, and,

(iii) if all the prime divisors of N are of the form 3k + 1, then Γ0(N) has 2η nonconjugating elliptic subgroups of order 3, where η is the number of prime divisors of N. Furthermore, the following is a complete set of nonconjugating elliptic subgroups of order 3 of Γ0(N), ai bi 2 η | ai ≡ ai, ai − ai + 1 ≡ 0, ai 6≡ aj(mod N), 1 ≤ i ≤ 2 , ciN 1 − ai

η 2 where ai, 1 ≤ i ≤ 2 are the solutions of x − x + 1 ≡ 0(mod N), ai, bi, ci ∈ Z and the subgroups of the set are of determinant ±1.

Lemma 3.21. Let A and B be subgroups of PSL2(Z), where A is a normal subgroup of B. Suppose that B is of ﬁnite index in PSL2(Z). Let g ∈ A be an elliptic element of order 2 or 3. Then ClB(g) spilts into [B : A] conjugacy classes in B.

Proof. Since A is a normal subgroup of B and g ∈ A, ClB(g) ⊂ A spilts into [B : A]/[CB(g): CA(g)] conjugacy classes in A. Since g is an elliptic element of PSL2(Z), CA(g) = CB(g) =< g >, which completes the proof of the lemma.

Theorem 3.22.

sqf (i) Γτ (2; 2/d, ε, χ) = Γ0(2) has a unique conjugacy class of elliptic subgroup of order 2.

(ii) Suppose 4|md or md has a prime divisor of the form 4k + 3. Then sqf Γτ (m; m/d, ε, χ) admits no elliptic subgroup of order 2. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 53

ˆ × (iii) Let G be the subgroup of Zmd generated by {z | z ≡ 1(mod m)} and G = Gˆ{±I}/{±I}. Suppose that all the prime divisors of md are sqf of the form 4k + 1. Then Γτ (m; m/d, ε, χ) has tφ(md)/2d nonconjugating elliptic subgroups of order 2, where t is the cardinality of 2 {x ∈ Zmd | x + 1 ≡ 0(mod md), x ∈ G}. Proof. Suppose m = 2. Since by assumption d|m, d = 1 or 2. This means sqf that Γτ (2; 2/d, ε, χ) = Γ0(2) and applying Theorem 3.19. yields (i). Let us now ﬁrst consider the case for Γ0(md) ∩ Γ1(m) which is conjugate to sqf Γτ (m; m/d, ε, χ). If 4|md or md has a prime divisor of the form 4k + 3, then by Theorem 3.19, Γ0(md) has no elliptic subgroup of order 2. Since Γ0(md)∩Γ1(m) is a subgroup of Γ0(md), Γ0(md)∩Γ1(m) also does not admit any elliptic subgroup of order 2. Now, suppose that all the prime divisors 2 of md are of the form 4k + 1. Let ai be a solution of x + 1 ≡ 0(mod md) and let ai bi ∈ Γ0(md), ciN di where ai ≡ ai(mod md). From Theorem 3.19, we know that the above subgroup is of order 2. Furthermore, one may easily see that it is a subgroup of Γ0(md) ∩ Γ1(m) if and only if ai ∈ G. By Lemma 3.21, Γ0(md) ∩ Γ1(m) has t[Γ0(md):Γ0(md) ∩ Γ1(m)] = tφ(md)/2|G| = tφ(md)/2d, where t is the cardinality of

2 {x ∈ Zmd | x + 1 ≡ 0(mod md), x ∈ G}.

Since conjugation preserves v2, (ii) and (iii) thus holds.

sqf Remark 3.23. Recall that Γτ (m; m/d, ε, χ) is the congruence subgroup Γ1(m) ∩ Γ(d). Hence, we can deduce from Theorem 2.6. that if m ≥ 4 or sqf d ≥ 2, then Γτ (m; m/d, ε, χ) is torsion free, and so t, as described in the above theorem, is fact equal to 0.

Theorem 3.24.

sqf (i) Γτ (3; 3/d, ε, χ) = Γ0(3) has a unique conjugacy class of elliptic subgroup of order 3. sqf Chapter 3: Genus of Γτ (m; m/d, ε, χ) 54

(ii) Suppose 9|md or md has a prime divisor of the form 3k + 2. Then sqf Γτ (m; m/d, ε, χ) admits no elliptic subgroup of order 3. ˆ × (iii) Let G be the subgroup of Zmd generated by {z | z ≡ 1(mod m)} and G = Gˆ{±I}/{±I}. Suppose that all the prime divisors of md are sqf of the form 3k + 1. Then Γτ (m; m/d, ε, χ) has tφ(md)/2d nonconjugating elliptic subgroups of order 3, where t is the cardinality of 2 {x ∈ Zmd | x − x + 1 ≡ 0(mod md), x ∈ G}. Proof. The proof is similar to that of Theorem 3.22.

Remark 3.25. Similar to Remark 3.23, t here equals to 0 if m ≥ 4 or d ≥ 2.

sqf 3.5 Genus Formula of Γτ (m; m/d, ε, χ) Finally, we conclude this chapter by giving the genus formula of the sqf congruence subgroup Γτ (m; m/d, ε, χ) which is as follows:

sqf Theorem 3.26. Suppose that m ≥ 3. Then the genus of Γτ (m; m/d, ε, χ) is given by

2 m d Y 1 1 X φ((md/c, c))φ(md)((md/c, c), d) v2 v3 1 + (1 − ) − − − , 24 p2 2 2d(md/c, c) 4 3 p prime c|md p|m where v2 and v3 are given in Theorem 3.22. and 3.24 respectively.

sqf Proof. The genus formula of Γτ (m; m/d, ε, χ) is obtained by applying the results in Theorem 3.10, Theorem 3.18, Theorem 3.22. and Theorem 3.24. to Theorem 2.37. Chapter 4

Genus of some Congruence Subgroups

One may observe that the approach employed in Chapter 3 to determine sqf the genus formula of Γτ (m; m/d, ε, χ) may be extended to the congruence subgroup Γ1(M) ∩ Γ(N) for any positive integer M and N. In fact, we can further adopt this approach to ﬁnd the genus formula of subgroups of a more general form Γ(M,N, G), as detailed in [L2], evincing the extensive nature of this approach. Hence, in this chapter, we shall explore and unveil some of the conditions under which the genus formula of the Larcher congruence subgroups can also be procured in a similar fashion.

4.1 Genus Formula of Γ1(M) ∩ Γ(N) As mentioned in the beginning of this chapter, by following closely to the approach used in Chapter 3, we are able to determine the genus formula of congruence subgroup of the form Γ1(M) ∩ Γ(N) for any positive integers M and N, which we will state below.

Theorem 4.1. Let M,N ∈ N and L = lcm(M,N).

(i) Suppose LN = 2. Then Γ1(M)∩Γ(N) = Γ0(2) has a unique conjugacy class of elliptic subgroup of order 2.

(ii) Suppose 4|LN or LN has a prime divisor of the form 4k + 3. Then Γ1(M) ∩ Γ(N) admits no elliptic subgroup of order 2.

55 Chapter 4: Genus of some Congruence Subgroups 56

ˆ × (iii) Let G be the subgroup of ZLN generated by {z | z ≡ 1(mod L)} and G = Gˆ{±I}/{±I}. Suppose that all the prime divisors of LN are of the form 4k + 1. Then Γ1(M) ∩ Γ(N) has tφ(LN)/2N nonconjugating elliptic subgroups of order 2, where t is the cardinality of 2 {x ∈ ZLN | x + 1 ≡ 0(mod LN), x ∈ G}.

Theorem 4.2. Let M,N ∈ N and L = lcm(M,N).

(i) Suppose LN = 3. Then Γ1(M)∩Γ(N) = Γ0(3) has a unique conjugacy class of elliptic subgroup of order 3.

(ii) Suppose 9|LN or LN has a prime divisor of the form 3k + 2. Then Γ1(M) ∩ Γ(N) admits no elliptic subgroup of order 3. ˆ × (iii) Let G be the subgroup of ZLN generated by {z | z ≡ 1(mod L)} and G = Gˆ{±I}/{±I}. Suppose that all the prime divisors of LN are of the form 3k + 1. Then Γ1(M) ∩ Γ(N) has tφ(LN)/2N nonconjugating elliptic subgroups of order 3, where t is the cardinality of 2 {x ∈ ZLN | x − x + 1 ≡ 0(mod LN), x ∈ G}.

Remark 4.3. Similar to Remark 3.23, if M ≥ 4 or N ≥ 2, then t = 0 in both Theorem 4.1. and 4.2.

Theorem 4.4. Let M,N ∈ N and L = lcm(M,N). Suppose that LN ≥ 3. Then the genus of Γ1(M) ∩ Γ(N) is given by

2 L N Y 1 1 X φ((LN/c, c))φ(LN)((LN/c, c), d) v2 v3 1 + (1 − ) − − − , 24 p2 2 2N(LN/c, c) 4 3 p prime c|LN p|L where v2 and v3 are given in Theorem 4.1. and 4.2. respectively. Chapter 4: Genus of some Congruence Subgroups 57

4.2 Genus Formula of Γ1(m; 2, 1, 2) Lemma 4.5. Let m = 2d, ε = 1, χ = 2, where m, d, ε and χ are positive integers that satisfy the conditions stated in Deﬁnition 3.2. Then the corresponding Larcher congruence subgroup Γ1(m; 2, 1, 2) is conjugate by 1 −1/2 0 1/2 to Γ1(m) ∩ Γ(m/4). Proof. Let 1 + mα/2 mβ/2 A = ± ∈ Γ (m; 2, 1, 2), mγ/2 1 + mδ/2 1 where γ ≡ α (mod 2). Then 1 1 1 + mα/2 mβ/2 1 −1/2 ± ± 0 2 mγ/2 1 + mδ/2 0 1/2

1 + m(α + γ)/2 m(−α + β − γ + δ)/4 = ± mγ 1 + m(δ − γ)/2,

Note that α+γ is even as γ ≡ α (mod 2). Moreover, since A ∈ Γ1(m; 2, 1, 2), 1 + mδ/2 −mβ/2 A−1 = ± ∈ Γ (m; 2, 1, 2) −mγ/2 1 + mα/2 1 which implies −γ ≡ δ (mod 2), that is to say, δ − γ is also even. So, 1 1 1 −1/2 A ∈ Γ (m) ∩ Γ(m/4). 0 2 0 1/2 1 1 + mα0 mβ0/4 Conversely, let ± be an element in Γ (m) ∩ Γ(m/4). mγ0 1 + mδ0 1 Then 1 −1/2 1 + mα0 mβ0/4 1 1 ± 0 1/2 mγ0 1 + mδ0 0 2

1 + m(2α0 − γ0)/2 4(2α0 + β0 − γ0 − 2δ0) = ± mγ0/2 1 + m(γ0 + 2δ0)/2

0 0 0 ∈ Γ1(m; 2, 1, 2), since 2α − γ ≡ γ (mod 2). This completes the proof of the lemma. Chapter 4: Genus of some Congruence Subgroups 58

−1 Lemma 4.6. Suppose that G1 = νGν , where G and G1 are subgroups of PSL2(Z). Let a/b, x/y ∈ Q ∪ {∞}, with gcd(a, b) = gcd(x, y) = 1. Then a/b and x/y are equivalent in G if and only if νa/b and νx/y are equivalent in G1. Proof. Suppose that a/b and x/y are equivalent in G. Then there exists −1 g ∈ G such that ga/b = x/y. Since G1 = νGν , there exists g1 ∈ G1 such −1 −1 that g = ν g1ν. So, ν g1νa/b = x/y, which implies g1νa/b = νx/y, that is, νa/b and νx/y are equivalent in G1. Similarly, we can prove the converse and hence the lemma follows.

Theorem 4.7. Let a/b, x/y ∈ Q ∪ {∞}, with gcd(a, b) = gcd(x, y) = 1. Also, let a + b = p0r0, 2b = q0r0, x + y = u0t0, 2y = v0t0, where r0 = gcd(a + b, 2b) and t0 = gcd(x + y, 2y). Then a/b and x/y are equivalent in Γ1(m; 2, 1, 2) if and only if

(i) p0 − u0 is a multiple of mq0/4 modulo m, q0 − v0 is a multiple of m, or,

(ii) p0 + u0 is a multiple of mq0/4 modulo m, q0 + v0 is a multiple of m,

Proof. By Lemma 4.5,

1 −1/2 1 1 Γ (m; 2, 1, 2) = Γ (m) ∩ Γ(m/4) . 1 0 1/2 1 0 2

Furthermore, Lemma 4.6 establishes that a/b and x/y are equivalent in 1 1 1 1 Γ1(m; 2, 1, 2) if and only if 0 2 a/b and 0 2 x/y are equivalent in Γ1(m)∩ Γ(m/4), that is, if and only if, p0/q0 and u0/v0 are equivalent in Γ1(m) ∩ Γ(m/4). Now, apply Theorem 3.11. to the two cusps, p0/q0 and u0/v0, which will give us the required result.

Let m = 2d, ε = 1, χ = 2, where m, d, ε and χ are positive integers that satisfy the conditions stated in Deﬁnition 3.2. By Lemma 4.5, the corresponding Larcher congruence subgroup Γ1(m; 2, 1, 2) is conjugate to Γ1(m) ∩ Γ(m/4). This means that both of these subgroups have the same genus formula, which is as follows. Chapter 4: Genus of some Congruence Subgroups 59

Theorem 4.8. The genus formula of Γ1(m; 2, 1, 2) is given by

3 2 2 2 m Y 1 X φ((m /4c, c))φ(m /4)((m /4c, c), m/2) v2 v3 1 + (1 − ) − − − , 96 p2 m(m2/4c, c) 4 3 p prime c|m2/4 p|m where v2 and v3 are obtained by subsituting M = m and N = m/4 in Theorem 4.1. and 4.2. respectively. Bibliography

[K] P. G. Kluit, Doctoral Disseration, Antwerp, 1979.

[L1] M. L. Lang, Congruence Subgroups Associated to the Monster, preprint.

[L2] M. L. Lang, Genus Formula of Γ(M,N, G)\H , preprint. [La] H. Larcher, The Cusp Amplitudes of the Congruence Subgroups of the Classical Modular Groups, II. Illinois J. Math. 28 (1984), 312-338.

[Se] A. Sebbar, Classiﬁcation of Torsion-Free Genus Zero Congruence Groups, Proc. of AMS, 129(2001), 2517-2527.

[Sh] G. Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Iwanami Shoten, Tokyo, and Princeton University Press, Princeton, NJ, 1971.

[St] W. W. Stothers, Level and Index in the Modular Group, Proc. Roy. Soc. Edinburgh, 99A(1984), 115-126.