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I am aware the Fundamental region for $PSL(2,\mathbb{Z})$ has 3 vertices, namely: $\rho = \frac{-1+i\sqrt{3}}{2}$, $\rho +1 = \frac{1+i\sqrt{3}}{2}$ and $i$, each stabilised by the cyclic subgroups generated by $z\mapsto\frac{-z-1}{2}$, $z\mapsto\frac{z-1}{2}$ and $z\mapsto\frac{-1}{z}$ respectively.

However, how can one show that:

  1. $\rho$ and $\rho +1$ are the only two elliptic elements of order 3.
  2. $i$ is the only elliptic element of order 2.

Any help would be greatly appreciated.

math189925
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  • Related. Possibly even a duplicate. – Jyrki Lahtonen Aug 18 '19 at 08:48
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    $\Im(\frac{az+b}{cz+d}) = \frac{\Im(z)}{|cz+d|^2}$ thus if $\gamma(z) = z, \Im(z) \ge 1/2$ then $cz+d \in \pm 1,\pm i$ so that $(c,d) = (0,\pm 1),\frac{az+b}{cz+d} = \pm az+b $ or $(c,d) = (\pm 1,0), z = i,\frac{az+b}{cz+d} = \pm a+1/z$ – reuns Aug 18 '19 at 14:03

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