Klein's j-invariant is automorphic w.r.t. the modular group $SL^\pm(2, \mathbb Z)$ (solution verification) I want to solve the following problem from Ahlfors' text. I think I got the solution right and I would like to verify it.

Show that the function
  $$J(\tau)=\frac{4}{27} \frac{(1-\lambda +\lambda^2)^3}{\lambda^2(1-\lambda)^2} $$
  is automorphic with respect to the full modular group.

Here $\lambda$ is the modular $\lambda$ function, defined by $$\lambda(\tau)=\frac{e_3-e_2}{e_1-e_2} $$
where $e_1=\wp(\omega_1/2),e_2=\wp(\omega_2/2),e_3=\wp((\omega_1+\omega_2)/2).$ The "modular group" is the group of all fractional linear transformations with determinant $\pm1$ with the operation of function composition.
My attempt:
I'm not sure but according to Ahlfors, page 278, it is enough to consider the transformations $$T_1(\tau)=\tau+1,T_2(\tau)=-\frac{1}{\tau} .$$
(I suppose they generate the modular group together [?])
The following identities are proven in the text:
$$\lambda(\tau+1)=\frac{\lambda(\tau)}{\lambda(\tau)-1},\lambda(-1/\tau)=1-\lambda(\tau) $$
Applying these it follows easily that $J(\tau) $ is invariant under both $T_1$ and $T_2$.
Is my solution correct? If not, please help me correct it. Thanks!
 A: No, the transformations $T_1$ and $T_2$ do not generate the modular group. For one thing, they both have determinant $1$. What is true is that the union of the congruence subgroup with $\{T_1,T_2\}$   generates the modular group. 
Let's check this: 
$$
T_1 = \begin{pmatrix} 1 & 1 \\ 0 & 1   \end{pmatrix} 
$$
$$
T_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0   \end{pmatrix} 
$$
$$
T_1T_2 = \begin{pmatrix} 1 & 1 \\ 1 & 0   \end{pmatrix} 
$$
$$
T_1T_2 = \begin{pmatrix} 0 & 1 \\ 1 & 1   \end{pmatrix} 
$$
$$
T_2T_1T_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1  \end{pmatrix} 
$$
The above, and the identity matrix, exhaust the group $GL_2(\mathbb Z/2\mathbb Z)$. (Indeed, a matrix in this group can have either one or two zeroes, and if it has two, they can't be in the same row or column.) This group is in 1-1 correspondence with the cosets of the congruence subgroup. Indeed, taking quotient of the modular group by the congruence subgroup is the same as applying the group homomorphism 
$SL_{2}^{\pm}(\mathbb Z)\to GL_{2}(\mathbb Z/2\mathbb Z)$ induced by the ring homomorphism $\mathbb Z\to \mathbb Z/2\mathbb Z$.
Note that the congruence subgroup preserves $\lambda$, hence preserves $J(\tau)$. 
A: Yes, that is a perfectly good solution!
