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The function field of $X_{1}(5)$ is generated by the function $t(\tau) = q\prod_{n = 1}^{\infty}(1 - q^{n})^{5\left(\frac{n}{5}\right)}$ where $q = e^{2\pi i\tau}$ and $\left(\frac{n}{5}\right)$ is the Legendre symbol mod 5. The inequivalent cusps of $\Gamma_{1}(5)$ are $0$, $1/2$, $2/5$, and $\infty$. Consider the fundamental domain of $\Gamma_{1}(5)$

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and the piece of it that lies in the first quadrant. Why does $t(\tau)$ map this piece one-to-one onto the upper half plane with $t(0) = (-11 +5\sqrt{5})/2$, $t(\infty) = 0$, $t(1/2) = (-11-5\sqrt{5})/2$, and $t(2/5) = \infty$?

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    Try this paper: Sebbar, "Modular subgroups, forms, curves and surfaces" http://mysite.science.uottawa.ca/asebbar/publi/mfcs.pdf ... 6 groups are described there, including $\Gamma_1(5)$.2017-03-22

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