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THÔNG TIN TÀI LIỆU
Nội dung
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
ˆ
ˆ
ˆ
1.1 Γ0 (N ), Γ1 (N ) and Γ(N ) . . . . . . . . . . . . . . . . . . . . . . . . 1
ˆ 0 (N ), Γ
ˆ 1 (N ) and Γ(N
ˆ ). . . . . . . . . . . . . . . . . . . 8
1.2 Cusps of Γ
1.3 Cusp Widths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2 The
2.1
2.2
2.3
2.4
2.5
Modular Group P SL2 (Z)
Γ0 (N ), Γ1 (N ) and Γ(N ) . . . . .
Indices of Subgroups of P SL2 (Z)
Cusps of Γ0 (N ), Γ1 (N ) and Γ(N )
Cusp Widths . . . . . . . . . . .
The Genus Formula . . . . . . .
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25
25
27
31
36
37
3 Genus of Γsqf
τ (m; m/d, ε, χ)
3.1 Larcher Congruence Subgroups . . . . .
3.2 Index of Γsqf
τ (m; m/d, ε, χ) in P SL2 (Z)
3.3 Number of Inequivalent Cusps . . . . .
3.4 Number of Elliptic Subgroups . . . . . .
3.5 Genus Formula of Γsqf
τ (m; m/d, ε, χ) . .
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38
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40
42
50
54
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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 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
g =1+
v2 v3 v∞
µ
−
−
−
,
12
4
3
2
where
µ = the index of G in P SL2 (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 P SL2 (Z). This treatment is
neater and more systematic than a head-on approach with P SL2 (Z). Chapter 2 also deals with the indices of subgroups of P SL2 (Z) and introduces
the genus formula.
iii
Chapter 3 introduces the notions of “Larcher congruence subgroups”
and “Γsqf
τ (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 methodology 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 conˆ 0 (N ), Γ
ˆ 1 (N ) and Γ(N
ˆ ).
gruence subgroups, Γ
1.1
ˆ 0(N ), Γ
ˆ 1(N ) and Γ(N
ˆ )
Γ
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,
SL2 (Z) =
a b
c d
|
a, b, c, d ∈ Z, ad − bc = 1 .
Definition 1.2. Let N ∈N. We define the following subgroups of SL2 (Z)
to be,
ˆ 0 (N ) =
Γ
a b
c d
| ad − bc = 1, c ≡ 0 (mod N )
ˆ 1 (N ) =
Γ
a b
c d
| ad − bc = 1, a ≡ d ≡ 1 (mod N ), c ≡ 0 (mod N )
ˆ )=
Γ(N
a b
c d
,
,
| ad − bc = 1, a ≡ d ≡ 1 (mod N ), b ≡ c ≡ 0 (mod N )
.
Definition 1.3. Let G be a subgroup of SL2 (Z). G is a congruence subˆ ) is a subgroup of
group of SL2 (Z) if there exists M ∈ N such that Γ(M
1
2
Chapter 1: Congruence Subgroups of SL2 (Z)
ˆ 0 (N ), Γ
ˆ 1 (N ), Γ(N
ˆ )
G. Otherwise, G is a non-congruence subgroup. Thus, Γ
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
c d
A=
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 = ±I. Now,
by direct calculation, the characteristic polynomial of A is,
(x) = x2 − (a + d)x + 1.
Since A = ±I, (x) is the minimal polynomial of A and thus divides xn − 1
2tπ
which has roots ei n , t = 0, 1, ..., n − 1. This means that the roots of
2lπ
2kπ
(x) = 0 can be written as ei n and ei n for some 0 ≤ k, l ≤ n − 1 and
k, l ∈ Z. Hence,
(x) = x2 − (a + d)x + 1
= (x − ei
2kπ
n
)(x − ei
i 2kπ
n
2
= x − (e
i 2lπ
n
+e
2lπ
n
)
)x + ei
2kπ
n
· ei
2lπ
n
.
By comparing coefficients, we see that
ei
ei
2kπ
n
2kπ
n
+ ei
· ei
2lπ
n
2lπ
n
= 1,
(1.1.1)
= a + d ∈ Z.
(1.1.2)
2(k+l)π
i n
We can deduce from (1.1.1) that e
= 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
2lπ
2kπ
p = 0 or 1. Substituting l = −k + pn, we get ei n = e−i n . Thus (1.1.2)
reduces to
2kπ
2kπ
2kπ
a + d = ei n + e−i 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) =
3
Chapter 1: Congruence Subgroups of SL2 (Z)
2kπ
n
n
d > 1, and we have (e±i n ) d = 1. This implies that (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 = ±I. If A is of finite order n,
then
(i) n = 3 if and only if
(x) = x2 + x + 1,
(ii) n = 4 if and only if
(x) = x2 + 1,
(iii) n = 6 if and only if
(x) = x2 − x + 1.
Proof. In the proof of Theorem 1.4, we have seen that (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 = ±I, (x) divides xn − 1. Clearly from (1.1.3), when n = 3,
(x) = x2 + x + 1. Similarly, n = 4 implies (x) = x2 + 1. Let us now
consider the case when n = 6. We notice that from (1.1.5),
(x) = x2 − x + 1, or , x2 + x + 1.
But we have earlier just shown that (x) = x2 + x + 1 implies n = 3. Thus
(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, (x), is either x2 + x + 1, x2 +
1,or x2 − x + 1.
Proof. This follows immediately from Theorem 1.4. and Corollary 1.5.
4
Chapter 1: Congruence Subgroups of SL2 (Z)
Theorem 1.7.
−1 0
0 −1
is the only element of order 2 in SL2 (Z).
Proof. Suppose g = ac db is an element of order 2 in SL2 (Z), and g = −I.
Then since g = ±I, the characteristic polynomial of g, (x), is the minimal
polynomial of g which divides x2 − 1. But (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
0
ˆ ) and
element of order 2, it suffices to check whether −1
∈ G. So, Γ(N
0
−1
ˆ 1 (N ) do not contain any element of order 2 if and only if N ≥ 3 as −1 0
Γ
0
−1
ˆ ) and Γ
ˆ 1 (N ) if and only if N ≥ 3. On the other
is not contained in Γ(N
0
ˆ 0 (N ) for all N ∈ N.
hand, −1
∈Γ
0
−1
ˆ ) has no elements of order 3, 4 or 6 if and
Theorem 1.9. Let N ∈ N. Γ(N
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
ˆ ).
A=
∈ Γ(N
cN
1 + dN
By direct calculation, the characteristic polynomial of A is, (x) = x2 −
(2 + N (a + d))x + 1. Now, by Corollary 1.6, we see that there are only three
choices for (x), that is, x2 + x + 1, x2 + 1, or x2 − x + 1. In other words,
ˆ ) will not contain any elements of order 3, 4 and 6. For
if N ≥ 4, then Γ(N
ˆ
N = 3, we have for any A ∈ Γ(3),
A=
1 + 3a
3b
, and
3c
1 + 3d
(x) = x2 − (2 + 3(a + d))x + 1.
Suppose A is of order 3, 4, or 6. Then by Corollary 1.6,
(x) = x2 + x + 1, x2 + 1, or x2 − x + 1.
Let us now look at the coefficient of x in (x). Note that 2 + 3(a + d) = 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,
A=
−2 − 3d
3b
3c
1 + 3d
and |A| = (−2 − 3d)(1 + 3d) − 9bc = 1,
which reduces to −2 − 9d − 9d2 − 9bc = 1, or −2 − 9(bc + d + d2 ) = 1, and
ˆ
we get a contradiction. For N = 2 and any A ∈ Γ(2),
A=
1 + 2a
2b
2c
1 + 2d
and
(x) = x2 − (2 + 2(a + d))x + 1.
ˆ
Similar to the case for Γ(3),
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,
A=
1 + 2a
2b
.
2c
1 + 2d
Eventually, we obtain −2(bc + d + d2 ) = 1 which is again a contradiction.
ˆ ) has no elements of order 3, 4 and 6 if N ≥ 2.
This means that Γ(N
ˆ 1 (N ) has no elements of order 3, 4 or 6 if
Theorem 1.10. Let N ∈ N. Γ
and only if N ≥ 4.
Proof. Let
A=
1 + aN
b
cN
1 + dN
ˆ 1 (N ). It is obvious that the characteristic polynomial
be an element in Γ
2
of A, (x) = x − (2 + N (a + d))x + 1. Suppose N ≥ 4. Then similar
to the proof of Theorem 1.9, if N ≥ 4, then (x) cannot be any of the
3 polynomials in Corollary 1.6, therefore A is not of order 3, 4, and 6.
ˆ 1 (N ) has no elements of order 3, 4 or 6 implies N ≥ 4 since
Conversely, Γ
3
−1
ˆ 1 (2) while 4 −7 is an element of order
is an element of order 4 in Γ
10
−3
3
−5
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.
6
Chapter 1: Congruence Subgroups of SL2 (Z)
Lemma 1.11. In SL2 (Z),
(i) All cyclic subgroups of order 3 are conjugate to
−1 −1
1
0
(ii) All cyclic subgroups of order 4 are conjugate to
0 −1
1 0
.
(iii) All cyclic subgroups of order 6 are conjugate to
1 −1
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 0 (N ) has no elements
Theorem 1.13. Let N ∈ N and p be an odd prime. Γ
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 0 (N ) admits an element of order 4. In view of Lemma
Proof. Suppose that Γ
1.11.(ii),
a b
0 −1
d −b
ˆ 0 (N ),
A=
∈Γ
c d
1 0
−c a
a b
∈ SL2 (Z) with gcd(c, d) = 1 (because ad − bc = 1).
c d
This implies that
for some
A=
b −a
d −c
d −b
−c a
=
ac + bd −a2 − b2
c2 + d2 −ac − bd
ˆ 0 (N ).
∈Γ
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
7
Chapter 1: Congruence Subgroups of SL2 (Z)
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
gcd(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,
(c·d−1 )2 ≡ −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 0 (N ) admits an element of order 6, then
Lemma 1.11.(iii), if Γ
A =
a b
c d
1 −1
1 0
d −b
−c a
=
a + b −a
c + d −c
d −b
−c a
=
ad + ac + bd −a2 − ab − b2
c2 + cd + d2 −ac − bc − bd
ˆ 0 (N ),
∈Γ
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 )
Let N = p1 e1 p2 e2 ...pi ei ...pm em . 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 if
all the prime divisors (pi > 3) of N are of the form 3k + 1. Equivalently,
8
Chapter 1: Congruence Subgroups of SL2 (Z)
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
ˆ 0(N ), Γ
ˆ 1(N ) and Γ(N
ˆ )
Cusps of Γ
Definition 1.14. Let G be a subgroup of SL2 (Z). g =
to be parabolic if trace of g, tr(g) = a + d = ±2.
a
c
b
d
∈ G is said
Remark 1.15. Let a, b, c, d ∈ Z and z ∈ C ∪ {∞}. Then, for z = ∞, we
define
az + b
a b
,
z=
c d
cz + d
and for z = ∞, we define
a
if c = 0 ,
a b
∞= 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 ∪ {∞}.
Proof. Let the set of cusps for SL2 (Z) be S. ∞ is a cusp of SL2 (Z) as
1
0
1
1
is a parabolic element in SL2 (Z), and 10 11 ∞ = ∞. So assume c = 0. For
any a/c ∈ Q and gcd(a, c) = 1, we can find infinitely many b and d such
that ad − bc = 1. As a consequence, ac db ∈ SL2 (Z). Now,
a
a b
∞= ,
c d
c
9
Chapter 1: Congruence Subgroups of SL2 (Z)
−1
a b
c d
a b
∞=
c d
a b
c d
−1
a
.
c
As a result,
−1
a
= ∞.
c
1 1
0 1
a b
c d
a b
c d
Note that
a b
c d
−1
is a non-trivial parabolic element in SL2 (Z). It can be easily checked with
the above equalities that
a b
c d
1 1
0 1
a b
c d
−1
a
a
= .
c
c
Thus we have proved that Q ∪ {∞} ⊆ S. Conversely, let x ∈ S. Then for
k
l
some A = m
∈ SL2 (Z), tr(A) = ±2, and A = ±I, we have
n
Ax = x,
kx + l
= x,
mx + n
mx2 + (n − k)x − l = 0.
(1.2.1)
Case 1: m = 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,
x=
k−n
∈ Q.
2m
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 = ∞, then we will obtain l = 0 from both of the
previous equations. But A = ±I, so x = ∞. In other words, S ⊆ Q ∪ {∞}.
This completes the proof of the theorem.
10
Chapter 1: Congruence Subgroups of SL2 (Z)
ˆ ) is Q ∪ {∞}.
Theorem 1.18. The set of cusps for Γ(N
ˆ ), and a b ∈ SL2 (Z). Since
Proof. Let S denote the set of cusps for Γ(N
c d
1 N
ˆ ), and
is a parabolic element of Γ(N
0
1
1 N
0 1
∞ = ∞,
ˆ ). Similar to the explanation of the preceding
∞ must be a cusp of Γ(N
lemma, we have
a b
c d
1 N
0 1
a b
c d
−1
a
a
= .
c
c
ˆ ) is a normal subgroup of SL2 (Z). So,
It can be easily checked that Γ(N
a b
c d
1 N
0 1
a b
c d
−1
ˆ ).
∈ Γ(N
ˆ ). So
Furthermore, it is parabolic. This implies that a/c is a cusp of Γ(N
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 1 (N ) and Γ
ˆ 0 (N ) are the same,
Corollary 1.20. The sets of cusps for Γ
which is, Q ∪ {∞}.
ˆ 1 (N ), Γ
ˆ 0 (N ) and SL2 (Z)
Proof. Let C1 , C0 , and C be the sets of cusps for Γ
respectively. Since
ˆ )⊆Γ
ˆ 1 (N ) ⊆ Γ
ˆ 0 (N ) ⊆ SL2 (Z),
Γ(N
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 = 0, and gcd(a, c)
= 1, there always exists
a b
∈ SL2 (Z)
c d
such that
a
a b
∞= .
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 0 (N ) is a subset of
Lemma 1.24. The set of inequivalent cusps of Γ
ˆ 0 (N )g1 ∞, Γ
ˆ 0 (N )g2 ∞, ..., Γ
ˆ 0 (N )gm ∞},
{Γ
m
ˆ 0 (N )gi .
Γ
where SL2 (Z) =
i=1
ˆ 0 (N ). Then
Proof. Let a/c be a cusp of Γ
a
ˆ 0 (N ) a .
=Γ
c
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
c d
such that
∈ SL2 (Z)
a
a b
∞= .
c d
c
Then,
a
ˆ 0 (N ) a b ∞.
=Γ
c d
c
However,
ˆ 0 (N ) a b
Γ
c d
ˆ 0 (N )gi , for some integer i satisfying 0 ≤ i ≤ m.
=Γ
Therefore,
a
ˆ 0 (N )gi ∞,
=Γ
c
and the lemma follows.
ˆ 0 (N ) is equal to
Theorem 1.25. The number of inequivalent cusps for Γ
ˆ
the number of double cosets of the form Γ0 (N )\SL2 (Z)/SL2 (Z)∞ , where
±1 m
SL2 (Z)∞ = {β ∈ SL2 (Z)| β∞ = ∞} =
|m ∈ Z .
0 ±1
m
ˆ 0 (N )gi . Suppose that i = j and
Γ
Proof. Let SL2 (Z) =
i=1
ˆ 0 (N )gi ∞ = Γ
ˆ 0 (N )gj ∞.
Γ
ˆ 0 (N ),
Then for some γ ∈ Γ
gi ∞ = γgj ∞, or
gj −1 γ −1 gi ∞ = ∞.
This means that gj −1 γ −1 gi is an element in SL2 (Z)∞ , and
gj −1 γ −1 gi SL2 (Z)∞ = SL2 (Z)∞ ,
13
Chapter 1: Congruence Subgroups of SL2 (Z)
gi SL2 (Z)∞ = γgj SL2 (Z)∞ ,
ˆ 0 (N )gi SL2 (Z)∞ = Γ
ˆ 0 (N )gj SL2 (Z)∞ .
Γ
ˆ 0 (N )gi SL2 (Z)∞ = Γ
ˆ 0 (N )gj SL2 (Z)∞ .
Conversely, suppose that i = j, Γ
Then,
ˆ 0 (N )gi SL2 (Z)∞ ∞ = Γ
ˆ 0 (N )gj SL2 (Z)∞ ∞,
Γ
which implies
ˆ 0 (N )gi ∞ = Γ
ˆ 0 (N )gj ∞.
Γ
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 0 (N ) aip bip SL2 (Z)∞ ,
Γ
ci dip
where gcd(ci , dip ) = 1, ci |N, 0 ≤ dip < ci , and for each ci and p = q, we
have dip ≡ diq (mod gcd(N/ci , ci )).
(ii) The number of double coset representatives is equal to
φ(gcd(N/ci , ci )),
ci |N
where φ(n) represents the Euler phi-function.
Proof. Let g =
k l
m n
∈ SL2 (Z) which implies gcd(k, m) = 1, and,
ˆ 0 (N )gSL2 (Z)∞ = Γ
ˆ 0 (N )
Γ
ˆ 0 (N )
= Γ
x y
zN w
k l
SL2 (Z)∞ ,
m n
∗
∗
SL2 (Z)∞ ,
kN z + mw ∗
14
Chapter 1: Congruence Subgroups of SL2 (Z)
Let gcd(kN, m) = c, N = cN0 and m = cm0 . So,
kN z + mw = c(kN0 z + m0 w).
Since gcd(k, m) = 1 and gcd(N0 , m0 ) = 1, gcd(kN0 , m0 ) = 1. Thus we can
find infinitely many z, w such that kN0 z+m0 w = 1, and gcd(N0 z, w) = 1. In
particular, let kN0 z0 + m0 w0 = 1. The general solutions for z and w are z =
−m0 t + z0 , and w = kN0 t + w0 respectively. By Dirichlet’s Theorem, there
are infinitely many primes of the form kN0 t + w0 because gcd(kN0 , w0 ) = 1.
So we may assume w to be a prime such that gcd(c, w) = 1. Clearly,
gcd(cN0 z, w) = 1 or gcd(N z, w) = 1. Thus there exists x, y ∈ Z where
x y
zN w
ˆ 0 (N ).
∈Γ
We can now write, where c ≥ 1, c|N, and gcd(c, u) = 1,
ˆ 0 (N )gSL2 (Z)∞ = Γ
ˆ 0 (N ) s v SL2 (Z)∞
Γ
c u
ˆ 0 (N ) s v
= Γ
c u
1 r
SL2 (Z)∞
0 1
∗
ˆ 0 (N ) ∗
= Γ
SL2 (Z)∞ ,
c cr + u
where
1 r
0 1
∈ SL2 (Z)∞ .
Since r can be chosen so that it satisfies 0 ≤ cr + u < c, we have for any g
belonging to SL2 (Z),
ˆ 0 (N )gSL2 (Z)∞ = Γ
ˆ 0 (N ) a b SL2 (Z)∞ ,
Γ
c d
where c ≥ 1, c|N, gcd(c, d) = 1, and 0 ≤ d < c.
If
ˆ 0 (N ) a1 b1 SL2 (Z)∞ = Γ
ˆ 0 (N ) a2 b2 SL2 (Z)∞ ,
Γ
c1 d1
c2 d2
15
Chapter 1: Congruence Subgroups of SL2 (Z)
then for some
γ δ
αN β
ˆ 0 (N ) and
∈Γ
γ δ
αN β
a1 b 1
c1 d1
±1 h
0 ±1
±1 h
0 ±1
=
∈ SL2 (Z)∞ , we have
a2 b 2
.
c2 d2
And we get
±αN a1 ± βc1 = c2 ,
(1.2.2)
h(αN a1 + βc1 ) ± αN b1 ± βd1 = d2 .
(1.2.3)
Since c1 , c2 divides N , (1.2.2) can be written as,
±α
c2
N
a1 ± β = ,
c1
c1
or,
±α
N
c1
a1 ± β = 1.
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
number of double cosets is equal to
φ(gcd(N/ci , ci )).
ci |N
ˆ 0 (N ) is given by
Theorem 1.28. A complete set of inequivalent cusps for Γ
xip
,
ci
where gcd(ci , xip ) = 1, ci |N, 0 ≤ xip < ci , and for each ci , p = q, xip ≡ xiq
(mod gcd(N/ci , ci )).
Proof. Evidently from Lemma 1.24. and Theorem 1.27,
aip
ci
ˆ 0 (N ) aip bip ∞
= Γ
ci dip
ˆ 0 (N ) aip
= Γ
ci
16
Chapter 1: Congruence Subgroups of SL2 (Z)
where gcd(ci , dip ) = 1, ci |N, 0 ≤ dip < ci , and for each ci , p = q, dip ≡ diq
(mod gcd(N/ci , ci )). Since aip dip − bip ci = 1, we have
aip ≡ dip −1 (mod ci ), or, aip = dip −1 + kci for some k ∈ Z.
This implies that
aip
ci
dip −1 + kci
ci
dip −1 + kci
ˆ 0 (N ) 1 −k
= Γ
0 1
ci
ˆ 0 (N )
= Γ
ˆ 0 (N )
= Γ
=
, where
1 −k
0 1
ˆ 0 (N )
∈Γ
dip −1
ci
dip −1
.
ci
If 0 ≤ dip −1 < ci and since dip −1 also satisfies gcd(ci , dip −1 ) = 1, then
dip −1 ≡ diq −1 (mod gcd(N/ci , ci )) for p = q, where ci |N , then take xip to be
dip −1 and we are done. Otherwise, let dip −1 = xip + hci for some integer h
and 0 ≤ xip < ci . We see that
aip
ci
xip + hci
ci
xip + hci
ˆ 0 (N ) 1 −h
= Γ
0 1
ci
ˆ 0 (N ) xip
= Γ
ci
xip
=
.
ci
ˆ 0 (N )
= Γ
, where
1 −h
0 1
ˆ 0 (N )
∈Γ
Now, as dip −1 = xip +hci , dip −1 ≡ diq −1 (mod gcd(N/ci , ci )) implies xip ≡ xiq
(mod gcd(N/ci , ci )). It remains to show that gcd(xip , ci ) = 1. Suppose
not. Then there exists p > 1, p | ci , p | xip , and thus p | dip −1 which is a
contradiction. This completes the proof of the theorem.
ˆ 0 (N ) has
Corollary 1.29. Γ
φ(gcd(N/c, c)) inequivalent cusps.
c|N
17
Chapter 1: Congruence Subgroups of SL2 (Z)
Proof. It is an immediate consequence of the previous theorem.
Lemma 1.30. Suppose that gcd(a,b)=1, and
x y
z w
∈ SL2 (Z). Then
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
∈ SL2 (Z).
b v
It follows that
x y
z w
a u
b v
=
ax + by ∗
az + bw ∗
∈ SL2 (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.
ˆ ) if and only if
Then [a/b] = [c/d] in Γ(N
(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
ˆ )
∈ Γ(N
zN
1 + wN
such that
a
=
b
1 + xN
yN
zN
1 + wN
c
(1 + xN )c + yN d
=
.
d
zN c + (1 + wN )d
By Lemma 1.30,
gcd((1 + xN )c + yN d, zN c + (1 + wN )d) = 1,
we conclude that either
18
Chapter 1: Congruence Subgroups of SL2 (Z)
(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
(i) a ≡ c, b ≡ d (mod N ).
Since gcd(c, d) = 1, there exists x, y ∈ Z such that cy − dx = 1. Hence
τ=
c x
d y
It is clear that
c x
d y
−1
∈ SL2 (Z).
ay − bx
a
=
.
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)N q − (ay − bx)N p.
This implies that
ay − bx
Nq
−ad + bc 1 + N p
Note that
ay − bx
Nq
−ad + bc 1 + N p
ˆ ).
∈ Γ(N
1
ay − bx
=
.
0
−ad + bc
19
Chapter 1: Congruence Subgroups of SL2 (Z)
As a consequence,
c x
d y
ay − bx
Nq
−ad + bc 1 + N p
c x
d y
ay − bx
Nq
−ad + bc 1 + N p
Since
we conclude that
−1
−1
c x
d y
c x
d y
−1
a
c
= .
b
d
−1
ˆ ),
∈ Γ(N
a
c
=
.
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
ˆ 1 (N ) =
Γ
k=0
1 k ˆ
Γ(N ).
0 1
Proof. Note that
ˆ
ˆ 1 (N ) : Γ(N
ˆ )] = [SL2 (Z) : Γ(N )] = N.
[Γ
ˆ 1 (N )]
[SL2 (Z) : Γ
ˆ 1 (N )] and [SL2 (Z) : Γ(N
ˆ )] can be found in
The formulae for [SL2 (Z) : Γ
Chapter 2, Theorem 2.11. Now, let 0 ≤ x, y ≤ N − 1. Then
1 x ˆ
Γ(N ) =
0 1
1 y ˆ
Γ(N ),
0 1
if and only if
1 y
0 1
−1
1 x ˆ
ˆ ),
Γ(N ) = Γ(N
0 1
if and only if
1 x−y
0
1
ˆ ),
∈ Γ(N
20
Chapter 1: Congruence Subgroups of SL2 (Z)
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 1 (N ) if and only if
Then a/b and c/d are equivalent to each other in Γ
(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 ˆ
Γ(N ),
0 1
ˆ 1 (N ) =
Γ
k=0
ˆ 1 (N ) if and only if there exists
a/b and c/d are equivalent to each other in Γ
ˆ ) such that
k ∈ Z, τ ∈ Γ(N
1 k
τ
0 1
a
b
or
1 k
τ
0 1
a
b
c
d
=
,
−c
−d
=
.
Let
a
b
τ
=
x
y
.
This implies that
a ≡ x, b ≡ y (mod N ).
As a consequence,
c
d
−c
−d
Hence
1 k
τ
0 1
=
=
1 k
τ
0 1
a
b
c
d
=
a
b
=
x + ky
y
=
c
d
=
x + ky
y
a + kb
b
≡
≡
a + kb
b
(mod N ), or
(mod N ).
21
Chapter 1: Congruence Subgroups of SL2 (Z)
(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 )
ˆ ) such that
for some k ∈ Z. By Theorem 1.31, there exists some τ ∈ Γ(N
τ
Therefore
a + kb
c
=
.
d
b
c
1 −k
= τ −1
0 1
d
a
.
b
ˆ 1 (N ). Suppose that
Hence a/b and c/d are equivalent to each other in Γ
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 1 (N ). This completes the proof of the theorem.
Γ
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,
SL2 (Z)a/b = g ∈ SL2 (Z)| g
a
a
=
b
b
.
In particular,
SL2 (Z)∞ = {g ∈ SL2 (Z)| g∞ = ∞ }
=
=
±1 m
|m ∈ Z
0 ±1
±
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,
SL2 (Z)a/b =
a c
±
b d
1 1
0 1
a c
b d
−1
.
22
Chapter 1: Congruence Subgroups of SL2 (Z)
Proof. Since gcd(a, b) = 1, ab dc ∈ SL2 (Z) for some c and d ∈ Z and by
following a similar argument to Theorem 1.17, we observe that
a c
b d
±1 m
0 ±1
a c
b d
1 1
0 1
So
±
−1
a c
b d
a
a
= .
b
b
−1
a c
b d
⊆ SL2 (Z)a/b .
Conversely, let γ ∈ SL2 (Z)a/b . Then
a
a
= ,
b
b
γ
a
a c
∞= ,
b d
b
γ
a c
b d
−1
−1
a c
a c
∞ =
b d
b d
= ∞.
γ
a
b
This implies that
a c
b d
γ∈
γ∈
−1
γ
a c
b d
a c
b d
∈
±
1 1
±
0 1
a c
±
b d
1 1
0 1
, or,
−1
a c
b d
1 1
0 1
, or,
a c
b d
−1
.
Consequently,
SL2 (Z)a/b ⊆
±
a c
b d
from which the lemma readily follows.
1 1
0 1
a c
b d
−1
,
23
Chapter 1: Congruence Subgroups of SL2 (Z)
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
b d
1 m
0 1
a c
b d
−1
∈ G, for any
a c
b d
∈ SL2 (Z).
ˆ 0 (N )Remark 1.37. In accordance with Definition 1.36, we can define Γ
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
b d
1 m
0 1
a c
b d
−1
, for any
a c
b d
∈ SL2 (Z),
we have the following:
ˆ 0 (N )-width of a/b is the smallest positive integer m such that N | b2 m,
(i) Γ
ˆ 1 (N )-width of a/b is the smallest positive integer m such that N | abm
(ii) Γ
and N | b2 m,
ˆ )-width of a/b is N .
(iii) Γ(N
Proof. By direct calculation,
a c
b d
1 m
0 1
a c
b d
−1
1 m
0 1
d −c
−b a
=
a c
b d
=
a am + c
b bm + d
=
1 − abm
a2 m
.
−b2 m 1 + abm
d −c
−b a
Chapter 1: Congruence Subgroups of SL2 (Z)
24
ˆ 0 (N ) and Γ
ˆ 1 (N ) respectively, we
Hence for the above element to be in Γ
ˆ ), we
require the conditions as listed in (i) and (ii) to be satisfied. For Γ(N
need m to be the smallest positive integer such that N | abm, N | a2 m, and
N | b2 m. This implies that
N | gcd(abm, a2 m, b2 m).
Clearly,
gcd(abm, a2 m, b2 m)| 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
ˆ ).
m = N for Γ(N
Chapter 2
The Modular Group P SL2(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, P SL2 (Z) as follows:
P SL2 (Z) = SL2 (Z)/{±I}.
In addition, let N ∈ N. We denote the following subgroups of P SL2 (Z) by,
ˆ 0 (N )/{±I},
Γ0 (N ) = Γ
ˆ 1 (N )/{±I},
Γ1 (N ) = Γ
ˆ )/{±I}.
Γ(N ) = Γ(N
Definition 2.2. Let G be a subgroup of P SL2 (Z). G is a congruence
subgroup 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 definitions of Chapter 1 by P SL2 (Z) are valid. Let us now revisit some theorems
which we have proved for SL2 (Z) so that we can establish similar results
for P SL2 (Z).
25
Chapter 2: The Modular Group P SL2 (Z)
26
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,
g 4 = I,
then
g 2 = −I,
which follows that the order of g in P SL2 (Z) is 2. Similarly, an element of
order 6 in SL2 (Z) would be of order 3 in P SL2 (Z) and the theorem thus
follows.
Definition 2.5. Let G be a subgroup of P SL2 (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 P SL2 (Z). g ∈ G is said to be
elliptic if |tr(g)| < 2.
The following result is taken from [Sh].
27
Chapter 2: The Modular Group P SL2 (Z)
Lemma 2.8. In P SL2 (Z),
(i) All cyclic subgroups of order 2 are conjugate to
0 −1
1 0
.
(ii) All cyclic subgroups of order 3 are conjugate to
0 −1
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
x2 + 1 ≡ 0(mod N ) in ZN , that is,
if 4|N ,
0
−1
v2 =
otherwise.
1
+
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
x2 + x + 1 ≡ 0(mod N ) in ZN , that is,
if 9|N ,
0
−3
v3 =
1+
otherwise.
p
p|N
2.2
Indices of Subgroups of P SL2(Z)
Let us first state the following result from [Sh].
Theorem 2.11. Let N ∈ N, N ≥ 2 and p be a prime. Then,
ˆ ) in SL2 (Z) is N 3
(i) the index of Γ(N
(1 −
p|N
ˆ 1 (N ) in SL2 (Z) is N 2
(ii) the index of Γ
1
),
p2
(1 −
p|N
1
), and
p2
28
Chapter 2: The Modular Group P SL2 (Z)
ˆ 0 (N ) in SL2 (Z) is N
(iii) the index of Γ
p|N
1
(1 + ).
p
Remark 2.12. Let p be a prime and the indices of Γ(N ), Γ1 (N ) and Γ0 (N )
in P SL2 (Z) be µ, µ1 and µ0 respectively. Since −I ∈ Γ(2) and Γ1 (2), and,
−I ∈ Γ(N ) and Γ1 (N ) for N ≥ 3, we have
3
N
1
(1 − 2 ) if N ≥ 3,
2
p
µ=
p|N
6
if N = 2.
2
N
2
µ1 =
3
(1 −
p|N
1
) if N ≥ 3,
p2
if N = 2.
Because −I ∈ Γ0 (N ) for all N ∈ N, this implies that the index of Γ0 (N ) in
ˆ 0 (N ) in SL2 (Z), that is,
P SL2 (Z) is the same as that of Γ
µ0 = N
p|N
1
(1 + ).
p
Lemma 2.13. Let N ∈ N. Then
N −1
Γ1 (N ) =
k=0
1 k
Γ(N ).
0 1
Proof. The proof of Lemma 2.13. is similar to that of Lemma 1.32.
Lemma 2.14. Let N > 2, N ∈ N. Then
Γ0 (N ) =
x,y
x 0
Γ1 (N ),
0 y
where xy ≡ 1 (mod N ), 1 ≤ x ≤ N/2 and 1 ≤ y ≤ N − 1.
29
Chapter 2: The Modular Group P SL2 (Z)
Proof. Applying the formulae in Remark 2.12, we obtain
[Γ0 (N ) : Γ1 (N )] =
=
=
[P SL2 (Z) : Γ1 (N )]
[P SL2 (Z) : Γ0 (N )]
N
2
p|N
p prime
1
(1 − )
p
φ(N )
.
2
Now, consider the following cosets of Γ(N ),
x,y
x 0
Γ1 (N ),
0 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 x1 y1 ≡ 1 (mod N ) and x2 y2 ≡ 1 (mod N ). Then
x1 0
Γ1 (N ) =
0 y1
x2 0
Γ1 (N ),
0 y2
if and only if
x2 0
0 y2
−1
x1 0
Γ1 (N ) = Γ1 (N ),
0 y1
if and only if
x1 y2
0
0
x2 y1
∈ Γ1 (N ),
if and only if
x1 y2 ≡ 1 (mod N ), and, x2 y1 ≡ 1 (mod N ),
if and only if
x1 = x2 , and, y1 = y2 .
30
Chapter 2: The Modular Group P SL2 (Z)
Lemma 2.15. Let N > 2, N ∈ N. Then
Γ0 (N ) =
x,y,z
x 0
0 y
1 z
Γ(N ),
0 1
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
Γ0 (N ) =
x,y
=
x,y
=
x,y,z
x 0
Γ1 (N )
0 y
N −1
x 0
0 y
z=0
x 0
0 y
1 z
Γ(N )
0 1
1 z
Γ(N ),
0 1
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
Γ0 (N ) = Γ1 (N ) =
k=0
1 k
Γ(N ).
0 1
Proof. By Lemma 2.15, we have
Γ0 (N ) =
x,y,z
x 0
0 y
1 z
Γ(N ),
0 1
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
Γ0 (N ) =
k=0
1 k
Γ(N ).
0 1
On the other hand, notice from Lemma 2.13. that
N −1
Γ1 (N ) =
k=0
1 k
Γ(N ).
0 1
This completes the proof of the proposition.
31
Chapter 2: The Modular Group P SL2 (Z)
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 P SL2 (Z), Γ0 (N ), Γ1 (N ) and Γ(N )
are the same, namely, Q ∪ {∞}.
Theorem 2.18. P SL2 (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
ˆ ), and
are equivalent to each other in Γ(N
ˆ ).
(ii) the number of inequivalent cusps of Γ(N ) is equal to that of Γ(N
ˆ ). Then there exists
Proof. Let x, y ∈ Q ∪ {∞}. Suppose [x] = [y] in Γ(N
ˆ ) such that gx = y. But g ∈ Γ(N ). So [x] = [y] in Γ(N ).
g ∈ Γ(N
Conversely, suppose [x] = [y] in Γ(N ). Then there exists
a b
c d
∈ Γ(N )
such that
a b
x = y.
c d
Now, we distinguish into two cases.
Case 1: a ≡ d ≡ 1 (mod N ).
Then
a b
c d
ˆ ).
and thus [x] = [y] in Γ(N
ˆ )
∈ Γ(N
32
Chapter 2: The Modular Group P SL2 (Z)
Case 2: a ≡ d ≡ −1 (mod N ).
Then
−a −b
−c −d
and
ˆ )
∈ Γ(N
−a −b
x = y.
−c −d
ˆ ). This completes the proof of the theorem.
Hence [x] = [y] in Γ(N
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 1 (N ), and
are equivalent to each other in Γ
ˆ 1 (N ).
(ii) the number of inequivalent cusps of Γ1 (N ) is equal to that of Γ
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 0 (N ), and
are equivalent to each other in Γ
ˆ 0 (N ).
(ii) the number of inequivalent cusps of Γ0 (N ) is equal to that of Γ
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 P SL2 (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 = {ˆ
x | 1 ≤ xˆ ≤ 2m, gcd(gcd (k, xˆ), 2m)) = 1}.
For each xˆ ∈ Ak , let x = xˆ + 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 = {ˆ
x | 1 ≤ xˆ ≤ m, gcd(gcd (m, xˆ), 2m)) = 1},
A2m = {ˆ
x | 1 ≤ xˆ ≤ m, gcd(gcd (2m, xˆ), 2m)) = 1}.
Define
(i) Sm = {x/m | xˆ ∈ Am }, where x = xˆ + 2my is the smallest positive
integer such that gcd (x, m) = 1,
(ii) S2m = {x/2m | xˆ ∈ A2m }, where x = xˆ + 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 = {ˆ
x | 1 ≤ xˆ ≤ 2m + 1, gcd(gcd (k, xˆ), 2m + 1)) = 1}.
For each xˆ ∈ Ak , let x = xˆ + (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 = {ˆ
x | 1 ≤ xˆ ≤ m, gcd(gcd (k, xˆ), 2m + 1)) = 1}.
Define
S2m+1 = {x/(2m + 1) | xˆ ∈ A2m+1 },
Chapter 2: The Modular Group P SL2 (Z)
34
where x = xˆ+(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
xip
,
ci
where gcd(ci , xip ) = 1, ci |N, 0 ≤ xip < ci , and for each ci , p = q, xip ≡
xiq (mod gcd(N/ci , ci )).
Proof. This follows directly from Theorem 1.28. and Theorem 2.21.
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 = n0 y. 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.
35
Chapter 2: The Modular Group P SL2 (Z)
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 P SL2 (Z), P SL2 (Z)a/b , is defined as follows,
P SL2 (Z)a/b = g ∈ P SL2 (Z)| g
a
a
=
b
b
.
In particular,
P SL2 (Z)∞ = {g ∈ P SL2 (Z)| g∞ = ∞ }
=
±1 m
|m ∈ Z
0 ±1
=
1 1
0 1
.
Lemma 2.30. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. Then the stabilizer
of a/b in P SL2 (Z) is an infinite cyclic group.
Proof. Likewise to Lemma 1.35, we can show that there exist c, d ∈ Z such
that
−1
a c
1 1
a c
P SL2 (Z)a/b =
,
b d
0 1
b d
which is an infinite cyclic group.
Proposition 2.31. Let G be a subgroup of P SL2 (Z) and gcd(a, c) = 1.
Then the stabilizer of a/c in G, denoted by Ga/c is generated by
σ=
a c
b d
1 m
0 1
a c
b d
−1
=
1 − acm
a2 x
,
−c2 m 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 P SL2 (Z)a/c .
36
Chapter 2: The Modular Group P SL2 (Z)
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 P 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
b d
1 m
0 1
a c
b d
−1
∈ G, for any
a c
b d
∈ P SL2 (Z).
Theorem 2.33. Let a/b ∈ Q ∪ {∞} with gcd(a, b) = 1. By considering
a c
b d
1 λ
0 1
a c
b d
−1
, for any
a c
b d
∈ P SL2 (Z),
we have the following:
(i) Γ0 (N )-width of a/b is the smallest positive integer λ such that N | b2 λ,
(ii) Γ1 (N )-width of a/b is the smallest positive integer λ such that N | abλ
and N | b2 λ, and,
(iii) Γ(N )-width of a/b is N .
Definition 2.34. Let G be a subgroup of P SL2 (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 cuspsplit of G, where G is of index n in P SL2 (Z). Then
h
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 P SL2 (Z).
Proof. It is a direct consequence of Theorem 2.33. and Theorem 2.35.
37
Chapter 2: The Modular Group P SL2 (Z)
2.5
The Genus Formula
Theorem 2.37. (G. Shimura [Sh]) 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 is given by
g =1+
µ
v2 v3 v∞
−
−
−
,
12
4
3
2
(2.5.1)
where
µ = the index of G in P SL2 (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 P SL2 (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 P SL2 (Z). Suppose the genus of B, g(B) = 0. Then the genus
of A, g(A) = 0.
Proof. Since A is a subgroup of finite index in B, the following RiemannHurwitz formula holds (see [Sh]), that is,
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) = 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) = 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
(which we will define formally in due course). Recall the
genus formula for a subgroup of finite index in P SL2 (Z) from Chapter 2,
Theorem 2.37. This means that we need to find the number of inequivalent
sqf
cusps of Γsqf
τ (m; m/d, ε, χ), the index of Γτ (m; m/d, ε, χ) in P SL2 (Z), and,
the number of inequivalent elliptic subgroups of order 2 and 3 respectively
for Γsqf
τ (m; m/d, ε, χ).
Γsqf
τ (m; m/d, ε, χ)
3.1
Larcher Congruence Subgroups
Definition 3.1. Let G be a subgroup of P SL2 (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
m. Then m/d = h2 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
dβ
1+ α
εχ
γ ≡ τ α (mod χ)
± I,
Γτ (m; m/d, ε, χ) = ±
m
m
γ
1+ δ
χ
χ
|
38
/
Chapter 3: Genus of Γsqf
τ (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) ∈ { Γ1 (4; 2, 1, 2), Γ1 (8; 8, 2, 2) }.
Then Γτ (m; m/d, ε, χ) is a congruence subgroup of P SL2 (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) ∈ { 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,
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
40
(iii) the cusp width of a/b, where gcd(a, b) = 1, is
dσ
,
gcd (σ, εχ, b − τ aε)
with σ =
m
.
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
subgroup by Γsqf
τ (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.
establishes that Γsqf
τ (m; m/d, ε, χ) is in fact Γ1 (m) ∩ Γ(d), that is,
Γsqf
τ (m; m/d, ε, χ) =
1 + mα
dβ
mγ
1 + mδ
/ ± I,
where α, β, γ and δ are integers and the elements of this subgroup are of
determinant 1.
3.2
Index of Γsqf
τ (m; m/d, ε, χ) in P SL2 (Z)
Given the congruence subgroup Γsqf
τ (m; m/d, ε, χ). We are now interested in calculating the index of this subgroup in P SL2 (Z). Suppose m = 2.
Since by assumption, d|m, so d = 1 or 2. Thus Γsqf
τ (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.
Proposition 3.7. Γsqf
τ (m; m/d, ε, χ) is conjugate by
d 0
0 1
to Γ0 (md) ∩ Γ1 (m).
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
Proof. Let ±
41
1 + αm
βd
be an element in Γsqf
τ (m; m/d, ε, χ). Then
γm
1 + δm
±
= ±
1/d 0
0 1
1 + αm
βd
γm
1 + δm
d 0
0 1
1 + αm
β
γmd 1 + δm
∈ Γ0 (md) ∩ Γ1 (m).
Thus,
1/d 0
d 0
Γsqf
τ (m; m/d, ε, χ)
0 1
0 1
Conversely, let ±
1+αm
β
γ md 1 + δ m
±
= ±
d 0
0 1
⊆ Γ0 (md) ∩ Γ1 (m).
∈ Γ0 (md) ∩ Γ1 (m). Then
1+αm
β
γ md 1 + δ m
1/d 0
0 1
1+αm
βd
γm
1+δm
∈ Γsqf
τ (m; m/d, ε, χ).
Therefore,
d 0
1/d 0
Γ0 (md) ∩ Γ1 (m)
0 1
0 1
⊆ Γsqf
τ (m; m/d, ε, χ),
from which the proposition readily follows.
Remark 3.8. Refer to Theorem 2.37. for the genus formula. Now since
Γsqf
τ (m; m/d, ε, χ) and Γ0 (md)∩Γ1 (m) are both subgroups of P SL2 (Z),and,
conjugation by d0 10 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 P SL2 (Z).
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
42
Proposition 3.9. Let m ≥ 3. Then
[P SL2 (Z) : Γ0 (md) ∩ Γ1 (m)] =
m2 d
1
(1 − 2 ).
2 p prime
p
p|m
Proof. Since m ≥ 3, we have
[Γ0 (md) : Γ0 (md) ∩ Γ1 (m)] = φ(md)/2|G|,
ˆ
ˆ being the subgroup of Z× generated by
where G = G{±I}/{±I}
with G
md
{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
[P SL2 (Z) : Γ0 (md) ∩ Γ1 (m)] =
m2 d2
2|G|
(1 −
p prime
p|md
1
m2 d
1
)
=
(1 − 2 ),
2
p
2 p prime
p
p|m
as desired.
Theorem 3.10. The index of Γsqf
τ (m; m/d, ε, χ) in P SL2 (Z) is,
2
md
1
(1 − 2 ) if m ≥ 3,
2 p prime
p
[P SL2 (Z) : Γsqf
τ (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 Proposition 3.9.
3.3
Number of Inequivalent Cusps
In this section, we will study the inequivalent cusps of Γsqf
τ (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
Chapter 3: Genus of Γsqf
τ (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
γl
1 + δl
c
a
= .
b
d
This implies
c
a + αla + βnb
= .
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
g=
1 + xl
yl
zl
1 + wl
Chapter 3: Genus of Γsqf
τ (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
=g
0 1
d
a
.
b
Now,
g
1 kn
0 1
=
1 + xl
yl
zl
1 + wl
1 kn
0 1
=
1 + xl kn + xlkn + yl
zl
zlkn + 1 + wl
∈ Γ1 (m) ∩ Γ(n),
as l is a multiple of n and
det(g) = det
1 kn
0 1
=1
implies
det g
1 kn
0 1
= 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.
Then a/b and x/y are equivalent to each other in Γsqf
τ (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.
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
45
Proof. As mentioned in the previous section, Γsqf
τ (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 P SL2 (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,
ν
a
c
= .
b
d
But ν ∈ B since A ⊆ B. So a/b and c/d are also equivalent in B.
Remark 3.14. Note that from Theorem 3.3, Γ(m) ⊆ Γsqf
τ (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
down explicitly a set of inequivalent cusps for Γsqf
τ (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
cusps of Γsqf
τ (m; m/d, ε, χ).
Proposition 3.15. Let P be a subgroup of Q, where P and Q are subgroups
of P SL2 (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.
Chapter 3: Genus of Γsqf
τ (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
y
x
y
a
= P q , for some q ∈ Q.
c
= P
Let
n
Q=
P qi .
i=1
Then
P q = P qi ,
for some 1 ≤ i ≤ n. Therefore,
a
x
= P qi .
y
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 = j and
P qi
a
a
= P qj .
c
c
Then for some γ ∈ P , we have
qi
a
a
= γqj , or,
c
c
qj−1 γ −1 qi
a
a
= ,
c
c
that is, qj−1 γ −1 qi ∈ Qa/c . Thus
qj−1 γ −1 qi Qa/c = Qa/c ,
qi Qa/c = γqj Qa/c ,
P qi Qa/c = P qj Qa/c .
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
47
Conversely, suppose that i = j and
P qi Qa/c = P qj Qa/c .
Then
a
a
= P qj Qa/c ,
c
c
a
a
P qi = P qj .
c
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 qi Qa/c
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 ]
,
[Qa/c : Pa/c ]
and so in turn yields our desired result.
|Q/P Qa/c | =
ˆ to be the subgroup of
Lemma 3.16. Let c|md and gcd(a, c) = 1. Denote G
×
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
a2 x
,
−c2 x 1 + acx
where x = md/(md, c2 ). Similarly, Γ0 (md) ∩ Γ1 (m))a/c is generated by
1 − acx
a2 x
,
−c2 x 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.
Chapter 3: Genus of Γsqf
τ (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
a
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 to be the subgroup of Z× generated by {z | z ≡ 1(mod m)} and
Denote G
md
ˆ
G = G{±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
n
md
1 − act
(md, c2 )
≡ 1 − ct
md
∈ G,
(md, c2 )
which implies that s ≤ t. On the other hand,
1 − cs
md
(md, c2 )
a
≡ 1 − acs
and so t ≤ s. Hence t = s, as desired.
md
∈ G,
(md, c2 )
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
49
Theorem 3.18. Γsqf
τ (m; m/d, ε, χ) has
c|md
φ((md/c, c))φ(md)((md/c, c), d)
2d(md/c, c)
inequivalent cusps.
Proof. Since Γsqf
τ (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 ≡ 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
c|md
φ((md/c, c))[Γ0 (md) : Γ0 (md) ∩ Γ1 (m)]
[Γ0 (md)a/c : (Γ0 (md) ∩ Γ1 (m))a/c ]
inequivalent cusps. Note that
[Γ0 (md) : Γ0 (md) ∩ Γ1 (m)] =
φ(md)
,
2|G|
ˆ
ˆ being the subgroup of Z× generated
where G = G{±I}/{±I}
with G
md
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 ],
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
50
where tc is the smallest positive integer such that 1 − ctc md/(md, c2 ) ∈ G.
In fact, one can easily see that for each c|md,
tc =
(md/c, c)
.
((md/c, c), d)
Also recall that |G| = d. Thus, we can conclude that Γ0 (md) ∩ Γ1 (m) and
therefore Γsqf
τ (m; m/d, ε, χ) has
c|md
φ((md/c, c))φ(md)((md/c, c), d)
2d(md/c, c)
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
ci N
bi
−ai
| ai ≡ ai, a2i + 1 ≡ 0, ai ≡ aj (mod N ), 1 ≤ i ≤ 2η
,
where ai , 1 ≤ i ≤ 2η are the solutions of x2 + 1 ≡ 0(mod N ), ai , bi , ci
are integers and the subgroups of the set are of determinant 1.
Proof. By Theorem 2.9, v2 is equal to the number of solutions of x2 + 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
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
51
ˆ 0 (N ) has no element of order 4 if and only if 4|N or N has a prime
that Γ
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
p
= 1,
which implies that Γ0 (N ) has
1+
p|N
−1
p
= 2η
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
of N . Since Γ0 (N ) also has 2η nonconjugating elliptic subgroups of order 2,
we may assert that for each solution a of x2 + 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
ci N −ai
| ai ≡ ai, a2i + 1 ≡ 0, ai ≡ aj (mod N ), 1 ≤ i ≤ 2η
,
where ai , 1 ≤ i ≤ 2η are the solutions of x2 + 1 ≡ 0(mod N ), ai , bi , ci ∈ Z
and the subgroups of the set are of determinant 1.
Chapter 3: Genus of Γsqf
τ (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
ci N
bi
1 − ai
| ai ≡ ai, a2i − ai + 1 ≡ 0, ai ≡ aj (mod N ), 1 ≤ i ≤ 2η
where ai , 1 ≤ i ≤ 2η are the solutions of x2 − 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 P SL2 (Z), where A is a normal
subgroup of B. Suppose that B is of finite index in P SL2 (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 P SL2 (Z), CA (g) = CB (g) =< g >, which completes the proof of
the lemma.
Theorem 3.22.
(i) Γsqf
τ (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
Γsqf
τ (m; m/d, ε, χ) admits no elliptic subgroup of order 2.
,
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
53
ˆ be the subgroup of Z× generated by {z | z ≡ 1(mod m)} and
(iii) Let G
md
ˆ
G = G{±I}/{±I}.
Suppose that all the prime divisors of md are
of the form 4k + 1. Then Γsqf
τ (m; m/d, ε, χ) has tφ(md)/2d nonconjugating elliptic subgroups of order 2, where t is the cardinality of
{x ∈ Zmd | x2 + 1 ≡ 0(mod md), x ∈ G}.
Proof. Suppose m = 2. Since by assumption d|m, d = 1 or 2. This means
that Γsqf
τ (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
Γsqf
τ (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
of md are of the form 4k + 1. Let ai be a solution of x2 + 1 ≡ 0(mod md)
and let
ai b i
∈ Γ0 (md),
ci N 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
{x ∈ Zmd | x2 + 1 ≡ 0(mod md), x ∈ G}.
Since conjugation preserves v2 , (ii) and (iii) thus holds.
Remark 3.23. Recall that Γsqf
τ (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
d ≥ 2, then Γsqf
τ (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.
(i) Γsqf
τ (3; 3/d, ε, χ) = Γ0 (3) has a unique conjugacy class of elliptic subgroup of order 3.
Chapter 3: Genus of Γsqf
τ (m; m/d, ε, χ)
54
(ii) Suppose 9|md or md has a prime divisor of the form 3k + 2. Then
Γsqf
τ (m; m/d, ε, χ) admits no elliptic subgroup of order 3.
ˆ be the subgroup of Z× generated by {z | z ≡ 1(mod m)} and
(iii) Let G
md
ˆ
G = G{±I}/{±I}.
Suppose that all the prime divisors of md are
of the form 3k + 1. Then Γsqf
τ (m; m/d, ε, χ) has tφ(md)/2d nonconjugating elliptic subgroups of order 3, where t is the cardinality of
{x ∈ Zmd | x2 − 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.
3.5
Genus Formula of Γsqf
τ (m; m/d, ε, χ)
Finally, we conclude this chapter by giving the genus formula of the
congruence subgroup Γsqf
τ (m; m/d, ε, χ) which is as follows:
Theorem 3.26. Suppose that m ≥ 3. Then the genus of Γsqf
τ (m; m/d, ε, χ)
is given by
1+
1
1
m2 d
(1 − 2 ) −
24 p prime
p
2
c|md
φ((md/c, c))φ(md)((md/c, c), d) v2 v3
−
− ,
2d(md/c, c)
4
3
p|m
where v2 and v3 are given in Theorem 3.22. and 3.24 respectively.
Proof. The genus formula of Γsqf
τ (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
the genus formula of Γsqf
τ (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 be the subgroup of Z× generated by {z | z ≡ 1(mod L)} and
(iii) Let G
LN
ˆ
G = G{±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
{x ∈ ZLN | x2 + 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 be the subgroup of Z× generated by {z | z ≡ 1(mod L)} and
(iii) Let G
LN
ˆ
G = G{±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
{x ∈ ZLN | x2 − 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
1+
L2 N
24
(1 −
p prime
p|L
1
1
)−
2
p
2
c|LN
φ((LN/c, c))φ(LN )((LN/c, c), d) v2 v3
− − ,
2N (LN/c, c)
4
3
where v2 and v3 are given in Theorem 4.1. and 4.2. respectively.
57
Chapter 4: Genus of some Congruence Subgroups
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
mγ/2
1 + mδ/2
A=±
∈ Γ1 (m; 2, 1, 2),
where γ ≡ α (mod 2). Then
1 1
1 + mα/2
mβ/2
±
0 2
mγ/2
1 + mδ/2
±
1 −1/2
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),
A−1 = ±
1 + mδ/2 −mβ/2
−mγ/2 1 + mα/2
∈ Γ1 (m; 2, 1, 2)
which implies −γ ≡ δ (mod 2), that is to say, δ − γ is also even. So,
1 1
1 −1/2
A
0 2
0 1/2
Conversely, let ±
1 + mα
mγ
∈ Γ1 (m) ∩ Γ(m/4).
mβ /4
1 + mδ
be an element in Γ1 (m) ∩ Γ(m/4).
Then
±
= ±
1 −1/2
0 1/2
1 + mα
mγ
mβ /4
1 + mδ
1 1
0 2
1 + m(2α − γ )/2 4(2α + β − γ − 2δ )
mγ /2
1 + m(γ + 2δ )/2
∈ 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
Lemma 4.6. Suppose that G1 = νGν −1 , where G and G1 are subgroups of
P SL2 (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
g ∈ G such that ga/b = x/y. Since G1 = νGν −1 , there exists g1 ∈ G1 such
that g = ν −1 g1 ν. So, ν −1 g1 νa/b = x/y, which implies g1 νa/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 = p0 r0 , 2b = q0 r0 , x + y = u0 t0 , 2y = v0 t0 , 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,
Γ1 (m; 2, 1, 2) =
1 −1/2
1 1
Γ1 (m) ∩ Γ(m/4)
.
0 1/2
0 2
Furthermore, Lemma 4.6 establishes that a/b and x/y are equivalent in
Γ1 (m; 2, 1, 2) if and only if 10 12 a/b and 10 12 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
1+
m3
96
(1 −
p prime
p|m
1
)−
p2
c|m2 /4
φ((m2 /4c, c))φ(m2 /4)((m2 /4c, c), m/2) v2
v3
−
− ,
m(m2 /4c, c)
4
3
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.
60
[...]... ±I, so x = ∞ In other words, S ⊆ Q ∪ {∞} This completes the proof of the theorem 10 Chapter 1: Congruence Subgroups of SL2 (Z) ˆ ) is Q ∪ {∞} Theorem 1.18 The set of cusps for Γ(N ˆ ), and a b ∈ SL2 (Z) Since Proof Let S denote the set of cusps for Γ(N c d 1 N ˆ ), and is a parabolic element of Γ(N 0 1 1 N 0 1 ∞ = ∞, ˆ ) Similar to the explanation of the preceding ∞ must be a cusp of Γ(N lemma, we... 1 This completes the proof of the proposition 31 Chapter 2: The Modular Group P SL2 (Z) 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 P SL2 (Z), Γ0 (N ), Γ1 (N ) and Γ(N ) are the same, namely, Q ∪ {∞} Theorem 2.18 P SL2... to the number of solutions of x2 + 1 ≡ 0(mod N ) in ZN , that is, if 4|N , 0 −1 v2 = otherwise 1 + 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 x2 + x + 1 ≡ 0(mod N ) in ZN , that is, if 9|N , 0 −3 v3 = 1+ otherwise p p|N 2.2 Indices of Subgroups of P SL2(Z) Let us first state the. .. 2.2 Let G be a subgroup of P SL2 (Z) G is a congruence subgroup 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 definitions of Chapter 1 by P SL2 (Z) are valid Let us now revisit some theorems which we... not Then there exists p > 1, p | ci , p | xip , and thus p | dip −1 which is a contradiction This completes the proof of the theorem ˆ 0 (N ) has Corollary 1.29 Γ φ(gcd(N/c, c)) inequivalent cusps c|N 17 Chapter 1: Congruence Subgroups of SL2 (Z) Proof It is an immediate consequence of the previous theorem Lemma 1.30 Suppose that gcd(a,b)=1, and x y z w ∈ SL2 (Z) Then gcd (ax + by, az + bw) = 1 Proof... elliptic if |tr(g)| < 2 The following result is taken from [Sh] 27 Chapter 2: The Modular Group P SL2 (Z) Lemma 2.8 In P SL2 (Z), (i) All cyclic subgroups of order 2 are conjugate to 0 −1 1 0 (ii) All cyclic subgroups of order 3 are conjugate to 0 −1 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... if and only if all the prime divisors (pi > 3) of N are of the form 3k + 1 Equivalently, 8 Chapter 1: Congruence Subgroups of SL2 (Z) 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... Chapter 1: Congruence Subgroups of SL2 (Z) Lemma 1.11 In SL2 (Z), (i) All cyclic subgroups of order 3 are conjugate to −1 −1 1 0 (ii) All cyclic subgroups of order 4 are conjugate to 0 −1 1 0 (iii) All cyclic subgroups of order 6 are conjugate to 1 −1 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... = I, then g 2 = −I, which follows that the order of g in P SL2 (Z) is 2 Similarly, an element of order 6 in SL2 (Z) would be of order 3 in P SL2 (Z) and the theorem thus follows Definition 2.5 Let G be a subgroup of P SL2 (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... not a divisor of N Then if 2|N , all the 7 Chapter 1: Congruence Subgroups of SL2 (Z) 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 ... 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 P SL2 (Z) Suppose the genus of B, g(B) = Then the. .. inequivalent elliptic subgroups of order of G, v3 = the number of inequivalent elliptic subgroups of order of G, v∞ = the number of inequivalent cusps of G Remark 2.38 Let G be a subgroup of P SL2 (Z)... inequivalent elliptic subgroups of order of G, v3 = the number of inequivalent elliptic subgroups of order of G, v∞ = the number of inequivalent cusps of G Hence the study of v2 , v3 , v∞ , indices,