Genus of congruence subgroups of the modular group

Then the genus, g, of H∗/Galso referred to as the genus of G is given by v2 = the number of inequivalent elliptic subgroups of order 2 of G, v3 = the number of inequivalent elliptic subg

OF THE MODULAR GROUP

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

A THESIS SUBMITTEDFOR THE DEGREE OF MASTERS OF SCIENCE

DEPARTMENT OF MATHEMATICSNATIONAL UNIVERSITY OF SINGAPORE

2003

Acknowledgements ii

1.1 Γˆ0(N ), ˆΓ1(N ) and ˆΓ(N ) 1

1.2 Cusps of ˆΓ0(N ), ˆΓ1(N ) and ˆΓ(N ) 8

1.3 Cusp Widths 23

2 The Modular Group P SL2(Z) 25 2.1 Γ0(N ), Γ1(N ) and Γ(N ) 25

2.2 Indices of Subgroups of P SL2(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

3 Genus of Γsqfτ (m; m/d, ε, χ) 38 3.1 Larcher Congruence Subgroups 38

3.2 Index of Γsqfτ (m; m/d, ε, χ) in P SL2(Z) 40

3.3 Number of Inequivalent Cusps 42

3.4 Number of Elliptic Subgroups 50

3.5 Genus Formula of Γsqfτ (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

i

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

And also a very big thank you to all the lecturers whom have taught

me, my family and friends

ii

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

Let G be a subgroup of finite index in P SL2(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

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 essentialconsiderations in this thesis and compose the content of the four chapters

of the thesis

Chapter 1 begins by establishing auxillary results about the classicalconguence 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 P SL2(Z) This treatment isneater and more systematic than a head-on approach with P SL2(Z) Chap-ter 2 also deals with the indices of subgroups of P SL2(Z) and introducesthe genus formula

iii

in Γsqfτ (m; m/d, ε, χ) would be given, and, we will determine its genus viaidentification 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

Definition 1.2 Let N ∈N We define the following subgroups of SL2(Z)

Definition 1.3 Let G be a subgroup of SL2(Z) G is a congruence group of SL2(Z) if there exists M ∈ N such that ˆΓ(M ) is a subgroup of

sub-1

G Otherwise, G is a non-congruence subgroup Thus, ˆΓ0(N ), ˆΓ1(N ), ˆΓ(N )are all congruence subgroups of SL2(Z).

Theorem 1.4 The finite orders of elements in SL2(Z) are 1, 2, 3, 4 and6

Proof Let

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− 1which has roots ei2tπn , t = 0, 1, , n − 1 This means that the roots of4(x) = 0 can be written as ei2kπn and ei2lπn for some 0 ≤ k, l ≤ n − 1 and

We can deduce from (1.1.1) that ei2(k+l)πn = 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

p = 0 or 1 Substituting l = −k + pn, we get ei2lπn = e−i2kπn Thus (1.1.2)reduces to

a + d = ei2kπn + e−i2kπn = 2 cos2kπ

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) =

d > 1, and we have (e±i2kπn )dn = 1 This implies that 4(x) divides xnd −

1 which contradicts our choice of n Consider cos(2kπ/n) = −1 Then2kπ/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 1respectively 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,

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 nowconsider 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 Thus4(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

Theorem 1.7 −1 0



is the only element of order 2 in SL2(Z)

Proof Suppose g = ac bdis 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 minimalpolynomial 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 anyelement of order 2, it suffices to check whether −1 0

0 −1



∈ G So, ˆΓ(N ) andˆ

Γ1(N ) do not contain any element of order 2 if and only if N ≥ 3 as−1 0

if N ≥ 4, then ˆΓ(N ) will not contain any elements of order 3, 4 and 6 For

N = 3, we have for any A ∈ ˆΓ(3),

, and 4(x) = x2− (2 + 3(a + d))x + 1.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= 0because a, d ∈ Z, and also, 2 + 3(a + d) = 1 implies 3(a + d) = −1 which

is impossible too So, 2 + 3(a + d) = −1 which leads to a = −d − 1.Consequently,

and |A| = (−2 − 3d)(1 + 3d) − 9bc = 1,

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 ∈ ˆΓ(2),

and 4(x) = x2− (2 + 2(a + d))x + 1

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



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

Theorem 1.10 Let N ∈ N ˆΓ1(N ) has no elements of order 3, 4 or 6 ifand only if N ≥ 4

Proof Let



be an element in ˆΓ1(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.Conversely, ˆΓ1(N ) has no elements of order 3, 4 or 6 implies N ≥ 4 since

Lemma 1.11 In SL2(Z),

(i) All cyclic subgroups of order 3 are conjugate to −1 −1



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

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

Theorem 1.13 Let N ∈ N and p be an odd prime ˆΓ0(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

Proof Suppose that ˆΓ0(N ) admits an element of order 4 In view of Lemma1.11.(ii),

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 Nregardless 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 thatgcd(c, d) = 1 Now, for all the odd prime divisors of N, c2+ d2 ≡ 0 (mod pi),and d−1 (mod pi) exists as d is relatively prime to pi Therefore,

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 thatboth 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 tioned above, c, d and N are relatively prime to one another, so d−1 (mod N )exists and thus

men-[d−1(3c + d)]2 ≡ −3 (mod N )Let N = p1e1p2e2 piei pmem Then,

[d−1(3c + d)]2 ≡ −3 (mod pi), for i = 1, 2, , m

So, from Lemma 1.12.(ii), we conclude that this is solvable if and only ifall the prime divisors (pi > 3) of N are of the form 3k + 1 Equivalently,

c2+ cd + d2 ≡ 0 (mod N ) is not solvable if and only if there exists a primedivisor of N which is of the form 3k + 2 For A having order 3, using Lemma1.11.(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 theoremholds

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, zsatisfies the condition gz = z

Theorem 1.17 The set of cusps for SL2(Z) is Q ∪ {∞}

Proof Let the set of cusps for SL2(Z) be S ∞ is a cusp of SL2(Z) as10 11

is a parabolic element in SL2(Z), and10 11∞ = ∞ So assume c 6= 0 Forany a/c ∈ Q and gcd(a, c) = 1, we can find infinitely many b and d suchthat ad − bc = 1 As a consequence, ac db∈ SL2(Z) Now,

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 theprevious equations But A 6= ±I, so x = ∞ In other words, S ⊆ Q ∪ {∞}.This completes the proof of the theorem

Theorem 1.18 The set of cusps for ˆΓ(N ) is Q ∪ {∞}.

Proof Let S denote the set of cusps for ˆΓ(N ), andac db∈ SL2(Z) Since

Furthermore, it is parabolic This implies that a/c is a cusp of ˆΓ(N ) So

Q ∪ {∞} ⊆ S Conversely, we can prove that S ⊆ Q ∪ {∞} by a similarargument mentioned in Theorem 1.17 Thus S = Q ∪ {∞}

Lemma 1.19 Let G1 and G2 be subgroups of SL2(Z) Denote the set ofcusps 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

Corollary 1.20 The sets of cusps for ˆΓ1(N ) and ˆΓ0(N ) are the same,which is, Q ∪ {∞}

Proof Let C1, C0, and C be the sets of cusps for ˆΓ1(N ), ˆΓ0(N ) and SL2(Z)respectively Since

ˆΓ(N ) ⊆ ˆΓ1(N ) ⊆ ˆΓ0(N ) ⊆ SL2(Z),and combining the results of Theorem 1.17, Theorem 1.18 and Lemma1.19, we have the following,

C ⊆ C1 ⊆ C0 ⊆ C

Hence we are done

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 Gand equivalent modulo G) if gx1 = x2 for some g ∈ G Moreover, we denotethe equivalence classes of x1 and x2 by [ x1] and [ x2] respectively In otherwords, 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 tosay, 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

c d



∈ SL2(Z)such that

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

“equiv-Lemma 1.24 The set of inequivalent cusps of ˆΓ0(N ) is a subset of

{ˆΓ0(N )g1∞, ˆΓ0(N )g2∞, , ˆΓ0(N )gm∞},where SL2(Z) =

i

= ˆΓ0(N )a

c.

Since a/c is a cusp of SL2(Z), by Theorem 1.22, there exists

hac

i

= ˆΓ0(N )gi∞,and the lemma follows

Theorem 1.25 The number of inequivalent cusps for ˆΓ0(N ) is equal tothe number of double cosets of the form ˆΓ0(N )\SL2(Z)/SL2(Z)∞, where

SL2(Z)∞= {β ∈ SL2(Z)| β∞ = ∞} =±10 ±1m



| m ∈ Z



giSL2(Z)∞= γgjSL2(Z)∞,ˆ

ˆ

Γ0(N )gi∞ = ˆΓ0(N )gj∞

We require the following to prove Theorem 1.27

Theorem 1.26 (Dirichlet’s Theorem) Let a and b be two integers wheregcd(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,

(ii) The number of double coset representatives is equal to

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

kN z + mw = c(kN0z + m0w)

Since gcd(k, m) = 1 and gcd(N0, m0) = 1, gcd(kN0, m0) = 1 Thus we canfind infinitely many z, w such that kN0z+m0w = 1, and gcd(N0z, w) = 1 Inparticular, let kN0z0+ m0w0 = 1 The general solutions for z and w are z =

−m0t + z0, and w = kN0t + w0 respectively By Dirichlet’s Theorem, thereare 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(N z, w) = 1 Thus there exists x, y ∈ Z where

then for some

And we get

h(αN a1+ βc1) ± αN b1 ± βd1 = d2 (1.2.3)Since c1, c2 divides N , (1.2.2) can be written as,

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 rep-resentatives as stated in the theorem and it follows immediately that thenumber of double cosets is equal to X

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,

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

aip≡ dip−1(mod ci), or, aip= dip−1+ kci for some k ∈ Z

This implies that

If 0 ≤ dip−1 < ci and since dip−1 also satisfies gcd(ci, dip−1) = 1, then

dip−1 6≡ diq−1 (mod gcd(N/ci, ci)) for p 6= q, where ci|N , then take xip to be

dip−1 and we are done Otherwise, let dip−1 = xip+ hci for some integer hand 0 ≤ xip < ci We see that

Now, as dip−1 = xip+hci, dip−1 6≡ diq−1(mod gcd(N/ci, ci)) implies xip6≡ xiq(mod gcd(N/ci, ci)) It remains to show that gcd(xip, ci) = 1 Supposenot Then there exists p > 1, p | ci, p | xip, and thus p | dip−1 which is acontradiction This completes the proof of the theorem

Corollary 1.29 ˆΓ0(N ) has X

c|N

φ(gcd(N/c, c)) inequivalent cusps

Proof It is an immediate consequence of the previous theorem.

Lemma 1.30 Suppose that gcd(a,b)=1, and x y



∈ SL2(Z) Thengcd (ax + by, az + bw) = 1

Proof Since gcd (a, b) = 1, there exists u, v ∈ Z such that av − bu = 1.Hence

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 ˆΓ(N ) if and only if

(i) a = (1 + xN )c + yN d, b = zN c + (1 + wN )d, or

(ii) −a = (1 + xN )c + yN d, −b = zN c + (1 + wN )d

This implies that

(i) a ≡ c, b ≡ d (mod N ) or

(ii) −a ≡ c, −b ≡ d (mod N )

Conversely, suppose 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)N q − (ay − bx)N p

This implies that

=hcd

i.Suppose that

(ii) −a ≡ c, −b ≡ d (mod N )

Similar to the above, we can show that [a/b] = [c/d] This completes theproof of the theorem

Lemma 1.32 Let N ∈ N Then

Proof Note that

0 1

ˆΓ(N ),

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

Proof Since by the previous lemma

a/b and c/d are equivalent to each other in ˆΓ1(N ) if and only if there exists



=

cd

,or

1 k

0 1



τ ab

τ ab



y

.This implies that

ab



=

cd



=



x + kyy



≡



a + kbb

(mod N ), or

ab



=

cd



=



x + kyy



≡



a + kbb

(mod N )

Hence

(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 bmodulo N Then

c ≡ a + kb, d ≡ b (mod N )for some k ∈ Z By Theorem 1.31, there exists some τ ∈ ˆΓ(N ) such that

b + d is a multiple of N and a + c is a multiple of b modulo N Similar tothe above, one can show that a/b and c/d are equivalent to each other inˆ

Γ1(N ) This completes the proof of the theorem

Definition 1.34 Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1 The stabilizer ofa/b in SL2(Z), SL2(Z)a/b, is defined as follows,

SL2(Z)a/b=ng ∈ SL2(Z)| ga

b



= ab

o

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,

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

from which the lemma readily follows

1.3 Cusp Widths

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

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

we have the following:

(i) ˆΓ0(N )-width of a/b is the smallest positive integer m such that N | b2m,

(ii) ˆΓ1(N )-width of a/b is the smallest positive integer m such that N | abmand N | b2m,

(iii) ˆΓ(N )-width of a/b is N

Proof By direct calculation,

Hence for the above element to be in ˆΓ0(N ) and ˆΓ1(N ) respectively, werequire the conditions as listed in (i) and (ii) to be satisfied For ˆΓ(N ), weneed m to be the smallest positive integer such that N | abm, N | a2m, and

N | b2m This implies that

The Modular Group P SL 2 (Z)

Definition 2.2 Let G be a subgroup of P SL2(Z) G is a congruencesubgroup of P SL2(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 P SL2(Z)

Note that all definitions obtained from replacing SL2(Z) in the tions of Chapter 1 by P SL2(Z) are valid Let us now revisit some theoremswhich we have proved for SL2(Z) so that we can establish similar resultsfor P SL2(Z)

defini-25

Remark 2.3 Since I = −I in P SL2(Z), the order of an element, A, in

P SL2(Z) is the smallest positive integer, n, such that An= ±I

Theorem 2.4 The finite orders of elements in P SL2(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 P SL2(Z) is 2 Similarly, an element oforder 6 in SL2(Z) would be of order 3 in P SL2(Z) and the theorem thusfollows

Definition 2.5 Let G be a subgroup of P SL2(Z) G is said to be torsionfree 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 P SL2(Z) g ∈ G is said to beelliptic if |tr(g)| < 2

The following result is taken from [Sh]

The following two well known results are taken from [Sh].

Theorem 2.9 Let p be a prime The number of inequivalent elliptic groups of order 2 in Γ0(N ), v2, is equal to the number of solutions of

sub-x2+ 1 ≡ 0(mod N ) in ZN, that is,

Let us first state the following result from [Sh]

Theorem 2.11 Let N ∈ N, N ≥ 2 and p be a prime Then,

(i) the index of ˆΓ(N ) in SL2(Z) is N3Y