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Take a congruence subgroup of $\Gamma \subset PSL_{2}(Z)$ and a representation, $\rho:\Gamma \rightarrow C^{\times}$. I was wondering if it's true that some power of $\rho$ is trivial, i.e. $\rho^{N} = 1$ for $N$ large enough?

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There is a subgroup in $PSL_2(Z)$ of finite index isomorphic to $F_2$, the free group of two elements. – yeshengkui Nov 29 '12 at 22:32
/is this a congruence group? – Adam Nov 30 '12 at 0:16
You just need a torsion-free congruence-subgroup $Γ=\Gamma(N)$ in $PSL(2,Z)$, it is then automatically free. In fact, for every $N\ge 2$, the congruence subgroup $\Gamma(N)$ is torsion-free. – Misha Nov 30 '12 at 1:07

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