Generators for a matrix group Let's denote $\Gamma_0(4)$ the subgroup of $\mathrm{SL}_2(\mathbb Z)$:
$$\Gamma_0(4):=\left\{\begin{pmatrix} a &b\\ 4c&d \end{pmatrix}\in \mathrm{SL}_2(\mathbb Z)\right\}.$$
We also define $A$ and $B$ in $\Gamma_0(4)$ as follow:
$$\gamma_1:=\begin{pmatrix} 1 &1\\ 0&1 \end{pmatrix},$$
$$\gamma_2:=\begin{pmatrix} 1 &0\\ 4&1 \end{pmatrix}.$$
We already know (with sophisticated arguments using group actions) that $\Gamma_0(4)/\{\pm 1\}$ is generated by $\gamma_1$ and $\gamma_2$:
$$\Gamma_0(4)/\{\pm 1\}=\langle \gamma_1,\gamma_2\rangle.$$
I would like to prove it with elementary arguments if possible.
Do you have any leads/solutions to this problem?
 A: Since I managed to find the answer, I am going to write it here in case it is useful for someone.

Let's defined the group $G=<\gamma_1,\gamma_2>$ generated by:
$$ \displaystyle \gamma_1=\begin{pmatrix} 1 & 0\\ 4 & 1 \end{pmatrix} \text{   and   } \displaystyle \gamma_2=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}.$$
First, we compute ${\gamma_1}^n$ and ${\gamma_2}^n$ by induction 
$$\forall n \in \mathbb Z\quad {\gamma_1}^n = \begin{pmatrix} 1 & 0\\ 4n & 1 \end{pmatrix} \text{ and } {\gamma_2}^n = \begin{pmatrix} 1 & n\\ 0 & 1 \end{pmatrix}.$$
Let $\gamma = \begin{pmatrix} a & b\\ 4c & d \end{pmatrix} \in \Gamma_0(4)$.
Everything we will be doing here works if we are dividing by non zero numbers. If it is not the case, there is nothing to prove. 


*

*Let $q$ and $r$ be the quotient and the rest of the euclidien division of $a$ by $4c$.
We  have $a=4cq+r$, so $r=a-4cq$.
Then we have
$$ {\gamma_2}^{-q}\gamma=\begin{pmatrix} a-4cq & b-dq \\ 4c & d \end{pmatrix} = \begin{pmatrix} r & b'\\ 4c & d \end{pmatrix}. $$

*Let $q'$ and $r'$ be the quotient and the rest of the euclidien division of $c$ by $a$.
We have $c=aq'+r'$, so $4r'=4c-4aq'$.
Then we have
$$ {\gamma_1}^{-q'}\gamma=\begin{pmatrix} a & b \\ 4c-4aq' & d-4bq' \end{pmatrix} = \begin{pmatrix} a & b \\ 4r' & d' \end{pmatrix}.$$
We now apply a slightly modified version of Euclid's algorithm (I can add the details if someone's interested) so the algorithm works on $\mathbb Z$.
We then obtain a sequence of matrixes of the form:
$$ \begin{pmatrix} a & b \\ 4c & d \end{pmatrix} \rightarrow 
\begin{pmatrix} r & \star \\ 4c & \star \end{pmatrix} \rightarrow
\begin{pmatrix} r & \star \\ 4r' & \star \end{pmatrix} \rightarrow
\ \cdots \ \rightarrow
\begin{pmatrix} \pm1 & \star \\ 0 & \star \end{pmatrix}. $$
This is beacause, $\gamma\in \text{PSL}_2(\mathbb Z)$, so $ad-4bc=1$. 
So with Bezout's theorem, $a\wedge c=1$. 
We work in the  quotient by $\{\pm 1\}$, which allow us to consider only the positive case.
Then,
$$ \exists \delta \in G,\quad \delta\gamma= \begin{pmatrix} 1 & b' \\ 0 & d' \end{pmatrix}.$$
And $\delta\gamma \in \text{PSL}_2(\mathbb Z)$ so $\det(\delta\gamma)=1$, so $d'=1$.
So,
$$ \delta\gamma= \begin{pmatrix} 1 & b' \\ 0 & 1 \end{pmatrix}.$$
Then,
$${\gamma_2}^{-b'}\delta\gamma= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
So $\gamma\in G$.
So $\Gamma_0(4) \subset G$.
Finally,
$$\Gamma_0(4) / \{\pm 1\}= G.\qquad \square$$
