Normalizers of intermediate congruence subgroups of the Hecke subgroups

Theorem 2.1

A matrix M is contained in 𝔑_Δ(N) only if M is represented in PGL2+(R) as

where Q||N h 2 and x, y, z, w ∈ ℤ such that det(M) = Q.

If h = 1, then we denote M by W_Q in Theorem 2.1. Such a matrix W_Q is contained in the normalizer of the group Γ₀(N) and it defines a unique automorphism on the modular curve X₀(N) which is called the Atkin-Lehner involution. However, the uniqueness doesn’t hold for general congruence groups Γ_Δ(N).

We investigate when W_Q belongs to 𝔑_Δ(N). Each γ ∈ Γ_Δ(N) is of the form

where a ∈ Δ and ā is an integer with aā ≡ 1 (mod N). For WQ = (QxyNzQw)  and  γ = (abc a ¯) ∈ ΓΔ(N) , one can easily compute that WQγWQ−1∈ΓΔ(N) if and only if the following condition holds:

From Q²xw − Nyz = Q, we have that Qxw−NQyz=1 and hence the following holds:

Note that ā is the multiplicative inverse of a modulo Q. Now we define an isomorphism t_Q : (ℤ/Nℤ)* → (ℤ/Nℤ)* by

Since (ℤ/Nℤ)* is isomorphic to the direct product Z /Q Z∗× Z / N Q Z∗ , one can show that the condition (2) holds if and only if t_Q(a) ∈ Δ. Therefore we have the following result:

Proposition 2.2

W_Q ∈ 𝔑_Δ(N) if and only if t_Q(Δ) = Δ.

If M ∈ 𝔑_Δ(N), then

Taking the trace, we see that 2 = a + d. Since d is a multiplicative inverse of a in (ℤ/Nℤ)*,

and hence

Now consider the natural homomorphism

Then ker(φ) = {1, σq + 1, 2σq + 1,..., (σ − 1)σq + 1} is the cyclic group of order σ generated by σq + 1. Thus equation (4) is equivalent to that a ∈ ker(φ).

In [11], the third author and Koo prove that 𝔑_Δ(N) is generated by the elements of Γ₀(N) and W_Q for all Q || N when N ≠ 4 and Δ = {± 1}, and its proof mainly depends on the following two conditions:

If (Δ /{± 1}) ∩ ker(φ) = {1} holds, then Eq. (3) is the same as Eq. (6). Similarly Eq. (7) is the same as the following condition:

By exactly the same arguments as those in [11], we have the following result:

Theorem 2.3

If (Δ/{±1}) ∩ ker(φ) = {1}, then 𝔑_Δ(N) is generated by the elements of Γ₀(N) and W_Q with t_Q (Δ) = Δ for each Q || N.

From Theorem 2.3, one can easily obtain the following result:

Corollary 2.4

If N is square-free, then 𝔑_Δ(N) is generated by the elements of Γ₀(N) and W_q with t_Q(Δ) = Δ for Q|N.

Proof

If N is square-free, then φ defined in (5) is an isomorphism, and hence ker(φ) is trivial.  □

Lemma 3.1

Proof

For an integer a prime to N, let [a] denote a matrix represented by γ ∈ Γ₀ (N) such that γ ≡ (a∗0∗) mod N. Consider the natural surjective homomorphism

defined by φ ([a]) = a. One can prove that the kernel of φ is equal to Γ_Δ(N), and the result follows from the first isomorphism theorem.  □

By Corollary 2.4, 𝔑_Δ(N) is the same as 𝔑₀(N) for square-free N. Then the Atkin-Lehner involutions modulo Γ₀(N) generate 𝔑_Δ(N)/Γ₀(N) which is isomorphic to (ℤ/2ℤ)^r where r is the number of prime divisors of N. Now we investigate when the exact sequence (1) splits, in which case, we have the following isomorphism:

For that we should find a group homomorphism h : 𝔑_Δ(N)/Γ₀(N) → 𝔑_Δ(N)/Γ_Δ(N) so that g ∘ h is the identity map, where g appears in (1). Note that the generators of 𝔑_Δ(N)/Γ₀(N) are the Atkin-Lehner involutions W_p for each prime divisor p of N, and so are their preimages of g in (1). Therefore the exact sequence (1) splits if and only if there exists an elementary abelian 2-subgroup of 𝔑_Δ(N)/Γ_Δ (N) generated by W_p with prime divisors p of N. Put N = p₁p₂ ... p_r with distinct primes p₁, p₂, ..., p_r. Then the exact sequence (1) splits if and only if one can find W_Pi for all i so that the following two conditions hold:

We give necessary and sufficient conditions for the splitting property of the sequence (1) in turn when r = 1, 2 and r ≥ 3.

Theorem 4.1

Let N = 2q where q is a prime with q ≡ 5 (mod 8). Then,

In this case, 𝔑_Δ(N)/Γ_Δ(N) is isomorphic to the Dihedral group D_q−1 of order 2(q − 1).

Proof

From (10) and Euler’s criterion, we have

Let d = ord_q (2). Then q−1d should be odd. Suppose that q−1d is even, then d| q−12 which is a contradiction to (14).

Take a primitive root r ∈ (Z/Nℤ)* of q so that 2≡r q − 1 d modq , and put x₁ to be an integer satisfying x₁ = r^m(mod q) where m is the integer with q−1d + 2m = 1. Then 2x12≡r(modq),and ordq2x12=q−1. If we take W2=(2 x 1 y 1N2 z 1) for some y₁, z₁, and let w1=2x12+qy1,thenW22=w 1 in 𝔑_Δ(N)/Γ_Δ(N) by (11). Since ordNw1=ordq2x12=q−1andw1q−12≡−1(modN) , the order of W₂ in 𝔑_Δ(N)/Γ_Δ(N) is equal to q − 1. We recall that we work modulo Δ = {± 1}, so that the order q − 1 of w₁ means an order of q−12ofW22 , whence an order of q − 1 of W₂ itself.

If we take Wq=qx2y2Nqz2and letw2=qx22+2y2,thenWq2=w2 in 𝔑_Δ(N)/Γ_Δ(N) by (11). Since w₂ ≡ −1 (mod N) from (12), the order of W_q in 𝔑_Δ(N)/Γ_Δ(N) is equal to 2.

Since w₁ generate (ℤ/Nℤ)*, 𝔑_Δ(N)/Γ_Δ(N) can be generated by W₂ and W_q. From (13) we know that (W₂W_q)² = 1.

Let G = 〈a, b | a^q−1 = b² = (ab)² = 1〉. Then the map a ⟼ W₂ and b ⟼ W_q can be extended to a unique homomorphism from G to 𝔑_Δ(N) /Γ_Δ(N) because W₂ and W_q satisfy all the relations in G if we replace a and b by W₂ and W_q . Clearly, the order |G| of G is equal to 2(q − 1) which is the same as |𝔑_Δ(N)/Γ_Δ(N)|. Thus G is isomorphic to 𝔑_Δ(N)/Γ_Δ(N).  □

Theorem 4.2

Let N = pq where p and q are primes satisfying one of the following:

1. p ≡ q ≡ 3 (mod 4), in which case we choose p and q such that p q = − 1.

2. p ≡ 1 (mod 4) and q ≡ 3 (mod 4) with p q = − 1.

Then,

Proof

Let us consider the case when p ≡ q ≡ 3 (mod 4) by choosing the notation for p and q such that p q = −1. By the same reason as in the proof of Theorem 4.1, q−1 d 1 is odd where d₁ : = ord_q(p). Thus we can take a primitive root r ∈ (Z/Nℤ)* of q and an integer x₁ so that px12≡r(modq),and hence ordqpx12=q−1. Take Wp=px1y1Npz1for somey1,z1,and letw1=px12+qy1. By the Chinese Remainder Theorem, we know that

and hence ord_N(w₁) = lcm(ord_q( px12 ), ord_p(qy₁)) = lcm{q − 1, 2} = q — 1 by (12). From (12), we also know that w1 q − 1 2≡−1(modN) because q−12 is odd, and hence the order of W_p in 𝔑_Δ(N)/Γ_Δ (N) is equal to q − 1.

On the other hand, qp≡qp−12≡1(modp).Thusd2:=ordpq|p−12,andp−1d2is even. Take a primitive root s ∈ (Z/Nℤ)* of p so that q≡s q − 1 d 2 (modp) , and put x₂ to be an integer satisfying x₂ = s^m₂ (mod p) where m₂ is the integer with q−1d1+2m2=2.Thenpx22=s2(modp),andordpqx22=p−12. In fact, we cannot take x₂ so that ordpqx22=p−1.Now we takeWq=qx2y2Nqz2for somey2,z2,andletw2=qx22+py2.

From (12), we know ord_N (w₂) = 1cm( p−12 , 2); hence it is equal to p − 1 because p−12 is odd by our assumption about p. From (12) again,

and hence w2 p − 1 2 ≢ ±1 (mod N). Thus the order of w₂ modulo Δ = {±1} is equal to p − 1; hence the order of W_q in 𝔑_Δ(N)/Γ_Δ(N) is equal to 2(p − 1).

Now we will show that W_p and W_q generate 𝔑_Δ(N)/Γ_Δ(N). For that it suffices to show that w₁ and w₂ generate (ℤ/Nℤ)*/{±1}. Since

from (12), we can view w₁ and w₂ as the elements (−1, r) and (s², −1) of [(ℤ/pℤ)* × (ℤ/qℤ)*] /{±(1, 1)}, respectively . Since p−12 is odd, −s² is a primitive root modulop. Thus (1,r) = (−1, 1)(− 1, r) = (−s², 1) ^p−12 (−1, r), and hence (1, r) and (−s²,1) are expressed by ±(−1, r) and ±(s², −1). Since (1, r) and (−s², 1) generate [(ℤ/pℤ)* × (ℤ/qℤ)*]/{±(1, 1)}, so do (−1, r) and (s², −1). Thus w₁ and w₂ generate (ℤ/Nℤ)*/{±1}.

From (13) we know that (W_qW_p)² = 1. For u ∈ (ℤ/Nℤ)*, by the action of the Atkin-Lehner involution W_q on (ℤ/Nℤ)* via the t_q operator which is in correspondence with conjugation by the W_q on Γ₀(N) modulo Γ₁ (N), we have the following:

Thus Wq[w1]Wq−1[w1][1,1]≡1(modq) , and clearly Wq[w1]Wq−1[w1][1,1]≡1(modp) because w₁ = −1 (mod p). Therefore, WqWp2Wq−1Wp2=1 holds.

Let G = 〈a, b | a^q−1 = b^2(p−1) = (ba)² = ba²b⁻¹a² = 1〉. Then there is a unique homomorphism from G to 𝔑_Δ(N)/Γ_Δ(N) determined by the map a ⟼ W_p, and b ⟼ W_q. From the relations in G, we know that

and hence every element of G can be expressed as aⁱ b^j with 0 ≤ i < q − 1 and 0 ≤ j < 2(p − 1). Thus the order |G| of G is less than or equal to 2(p − 1)(q − 1). Since |𝔑_Δ(N)/Γ_Δ(N)| ≤ |G| and 𝔑_Δ(N)/Γ_Δ(N) is of order 2(p − 1)(q − 1), G is isomorphic to 𝔑_Δ(N)/Γ_Δ(N).

Next consider the case when p ≡ 1 (mod 4) and q ≡ 3 (mod 4) with p q = − 1. By the quadratic reciprocity law, q p = −1 too. Under the exact same notations as in the previous case, we know that ord_q(w₁) = q − 1 and ord_p (w₂) = p − 1. The fact that ord_q(w₁) = q − 1 comes from two conditions q ≡ 3 (mod 4) and p q = − 1 which is the same as in the previous case. Since q p≡q p − 1 2 ≡−1(modp),p−1 d 2 should be odd. Thus one can take x₂ so that qx22 is a primitive root modulo p, and hence ord_p(w₂) = p − 1.

From (12), w1q−12≡−1(modN)becauseq−12is odd, butw2p−12 ≢ −1 (mod N) because p−12 is even. Thus the order of W_p and W_q in 𝔑_Δ(N)/Γ_Δ (N) are equal to q − 1 and 2(p − 1) respectively.

In this case we can view w₁ and w₂ as the elements (−1, r) and (s, −1) in [(ℤ/pℤ)* × (ℤ/qℤ)*]/{±(1, 1)}. Since p−12 is even, −s is a primitive root modulo p too. By the similar argument as in the previous case, (1, r) can be expressed by (−1, r) and (−s, 1). Since (−s, 1) and (1, r) generate [(ℤ/pℤ)* × (ℤ/qℤ)*]/{±(1, 1)}, so w₁ and w₂ generate (ℤ/Nℤ)*/{±1}. Thus the result follows.  □

Theorem 4.3

Let N = pq where p and q are primes satisfying p ≡ q ≡ 1 (mod 4) with p q = − 1, Then,

Proof

The notations are exactly the same as in the proof of Theorem 4.2. By quadratic reciprocity law, p q = q p = — 1; hence pq−12≡−1(modq)andqp−12≡−1modp,and henced−1d1andp−1d2 are odd. Thus we can take x₁ and x₂ so that px12andqx22 are primitive roots of q and p, respectively. Thus ord_N(w₁) = q − 1 and ord_N(w₂) = p − 1. Since q−12andp−12are even,w1q−12 ≢ ±1 (mod N) and w2 p − 1 2 ≢ ±1 (mod N) from (12), and hence the orders of W_p and W_q in 𝔑_Δ (N)/Γ_Δ(N) are equal to 2(q − 1) and 2(p − 1) respectively.