Number of Subgroups of Index m

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Homework Statement
The special linear group ##SL(2, \mathbb{Z})## over ##\mathbb{Z}## is the multiplicative group consisting of all ##2 \times 2## matrices with entries in ##\mathbb{Z}## and determinant ##1##; that is,

$$
SL(2, \mathbb{Z}) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in \mathbb{Z} \text{ and } ad - bc = 1 \right\}.
$$

Let ##G## be the quotient group ##SL(2, \mathbb{Z}) / \{\pm I\}##, where ##I## is the ##2 \times 2## identity matrix. Find with proof the number of subgroups of ##G## of index ##m## for each ##m \in \{2, 3, 4, 5, 6\}##.
Relevant Equations
Let ##G## be a group and ##H## be a subgroup of ##G##. The index of ##H## in ##G##, denoted ##[G : H]##, is the number of distinct left cosets of ##H## in ##G##. That is,
$$
[G : H] = \frac{|G|}{|H|},
$$
if ##G## is finite. For infinite groups, ##[G : H]## is the cardinality of the set of left cosets:
$$
[G : H] = |\{gH : g \in G\}|.
$$
Perhaps we can use congruence subgroups here? Or perhaps we can study SL(2,Z) using its action on the projective line over the integers modulo n? I'm pretty stumped and would appreciate any help.
 
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I searched the internet to find some information about what subgroups of ##\operatorname{SL}(2,\mathbb{Z})## look like and found https://kconrad.math.uconn.edu/blurbs/grouptheory/SL(2,Z).pdf.

The problem appears somewhat complex so I wondered which theorems and techniques your source provides to approach it. Anyway, the above paper suggests that you consider the matrices ##S,T## and ##U## and generally elements of finite order.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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