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?