I have been trying to learn about congruence groups. Here is an example:

\begin{eqnarray*} \Gamma\big(1+2i\big) &=& \text{SL}_2\big(\mathbb{Z}[i]\big)(1+2i) \\ \\ &=& \left\{ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) : ad-bc = 1 \text{ and } \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \equiv \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \pmod{1+2i} \right\} \\ \\ &\subseteq & \text{SL}_2\big(\mathbb{Z}[i]\big) \end{eqnarray*}

While the proof is uses the theory of algebraic groups, we can prove that is finitely generated. This particular case is look elementary, these are $2 \times 2$ invertible matrices, $\mathbb{Z}[i]$ is a Eucliean domain, and we can solve $ad-bc = 1$ by finding two primes (e.g. $6+i $ or $ 2+3i$) and looking for their greatest common divisor.

Since this group is finitely generated, how can I find a generating set? What are the generators? Even computer code would be helpful.

In order to specify what I am looking for here is the result for $\text{SL}_2(\mathbb{Z})$:

$$ \text{SL}_2(\mathbb{Z}) \simeq \big\{ S,T : S^2 = 1,\; (ST)^3 = 1 \big\} \quad\text{with}\quad S = \left( \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right) \text{ and } T = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$$

At least over $\mathbb{Z}$ there an exact sequence relating the congruence groups to the special linear group over finite fields:

$$ 1 \to \Gamma(N) \to \Gamma \to \text{PSL}_2\big(\mathbb{Z}/N\mathbb{Z}\big) \to 1 $$

And so it's likely that congruence groups of $\text{SL}_2(\mathbb{Z})$ should have finite presentations. Even if I do something slightly more inefficient and use the exact sequence.

Also