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Not only these subgroups are finitely presented, they are all finite free products of cyclic groups; most of them (for sufficiently large $n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). For instance, you can easily check that $\Gamma(n)$ is torsion-free for all $n\ge 2$ by looking at traces for $n\ge 3$ (since $1\ne 2$ mod $n\ge 2$ and $2\ne 0$ mod $n\ge 3$) and by looking at matrix coefficients for $n=2$. Rank is easily computable if you know index of the congruence subgroup in the modular group. The magic formula is multiplicativity of the Euler characteristic: For the modular group $\Gamma$, $\chi=-1+\frac{1}{2} + \frac{1}{3}=-\frac{1}{6}$. If $\Gamma'\subset \Gamma$ is a subgroup of index $i$ then $\chi(\Gamma')=i \chi(\Gamma)$. If $\Gamma$ is free of rank $r$ then $\chi(\Gamma)= 1-r$. For instance, to find index $i$ for $\Gamma(n)$, compute the order of the quotient group $SL(2, Z_n)/\pm I$. There is a closed formula for the order of this group (in terms of prime factors of $n$) which will tell you what the index is:

If $n$ is the product of powers of primes $\prod_i p_i^{k_i}$ then $$|PSL(2,Z_n)|= \frac{n^3}{2} ~~~\prod_i (1- p_i^{2})p_i^{-2}).$$

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Not only these subgroups are finitely presented, they are all finite free products of cyclic groups; most of them (for sufficiently large $n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). For instance, you can easily check that $\Gamma(n)$ is torsion-free for all $n\ge 2$ by looking at traces for $n\ge 3$ (since $1\ne 2$ mod $n\ge 2$ and $2\ne 0$ mod $n\ge 3$) and by looking at matrix coefficients for $n=2$. Rank is easily computable if you know index of the congruence subgroup in the modular group. The magic formula is multiplicativity of the Euler characteristic: For the modular group $\Gamma$, $\chi=-1+\frac{1}{2} + \frac{1}{3}=-\frac{1}{6}$. If $\Gamma'\subset \Gamma$ is a subgroup of index $i$ then $\chi(\Gamma')=i \chi(\Gamma)$. If $\Gamma$ is free of rank $r$ then $\chi(\Gamma)= 1-r$. For instance, to find index $i$ for $\Gamma(n)$, compute the order of the quotient group $PSL(2, Z_n)$SL(2, Z_n)/\pm I$. There is a closed formula for the order of this group (in terms of prime factors of$n$) which will tell you what the index is: If$n$is the product of powers of primes$\prod_i p_i^{k_i}$then $$|PSL(2,Z_n)|= \frac{n^3}{2} ~~~\prod_i (1- p_i^{2}).$$ 4 deleted 17 characters in body Not only these subgroups are finitely presented, they are all finite free products of cyclic groups; most of them (for sufficiently large$n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). Rank is easily computable if you know index of the congruence subgroup in the modular group. The magic formula is multiplicativity of the Euler characteristic: For the modular group$\Gamma$,$\chi=-1+\frac{1}{2} + \frac{1}{3}=-\frac{1}{6}$. If$\Gamma'\subset \Gamma$is a subgroup of index$i$then$\chi(\Gamma')=i \chi(\Gamma)$. If$\Gamma$is free of rank$r$then$\chi(\Gamma)= 1-r$. For instance, to find index$i$for$\Gamma(n)$, compute the order of the quotient group$PSL(2, Z_n)$. There is a (somewhat ugly) closed formula for the order of this group (in terms of prime factors of$n$) which will tell you what the index is: If$n$is the product of powers of primes$\prod_i p_i^{k_i}\$ then $$|PSL(2,Z_n)|= \frac{1}{2} n^3 \prod_i frac{n^3}{2} ~~~\prod_i (1- p_i^{2}).$$

3 products ==> free products
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