Following up from How do you obtain the $(2,3,7)$ triangle group as a quotient of the modular group?, I'm trying to understand the (2,3,7) triangle group as a quotient as the modular group.
So, we are taking the modular group mod $\{(sr)^7\}$, where $(sr)^7$ is $$\begin{bmatrix} 1 & 0 \\ 7 & 1 \end{bmatrix}$$ My question is given two 2 by 2 matrices, how do you determine if they are equivalent in this quotient group?
I determined that the conjugates of $(sr)^7$ are
$$\begin{bmatrix} 1 + 7bd & -7b^2 \\ 7d^2 & 1 - 7bd \end{bmatrix}$$
for any coprime $b$ and $d$. This is just the result of conjugating by $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$. There existing a matrix with determinant $1$ and a given $b$ and $d$ is a necessary and sufficient condition for $b$ and $d$ being coprime, by Bézout's identity. The group generated by those conjugates is the normal closure of $\{(sr)^7\}$, so modding the modular group by it gives us the (2,3,7)-triangle group. The issue is that I do not know a good way to describe this subgroup. It is a subgroup of $\Gamma(7)$, but that is about as far as I got. (It is also infinite index in the modular group since the triangle group is infinite.)