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Let $$\Gamma=\bigg \lbrace \begin{pmatrix} a&b\\c&d\end{pmatrix}\in\Gamma(1):b\equiv c~(\text{mod }3)\text{ or } a\equiv d\equiv 0~(\text{mod }3)\bigg \rbrace.$$

Then $\Gamma$ has exactly one cusp and its hauptmodul is the modular function $g=j^{1/3}$ usually denoted by $\gamma_2$.

Now pretend that we do not know that $ g^3=j $ and let $h$ be a Hauptmodul for $\Gamma$ which is holomorphic on the upper half-plane $\mathfrak H$.

Now let's try and find a relation between $j$ and $h$ using the description of $\Gamma$.

Since $j$ is holomorphic on $\mathfrak H$, there is a polynomial $P$ such that $j=P(h)$. Because the only zero of $j$ is at $\rho=e^{2\pi i /3}$ (modulo the action of $\Gamma(1)$),the coset representatives for $\Gamma$ in $\Gamma(1)$ are $I,T,T^2$, and $h$ has a simple pole at infinity (that is, its Fourier expansion begins with $q^{-1/3}$), we can write $$j=(g-g(\rho))(g-g(T\rho))(g-g(T^2\rho)).$$ On the other hand, $ST\rho= \rho$. Therefore, as $S\in \Gamma $, we have $g(T\rho)=g(\rho)$.

However, I am at loss trying to prove that $g(\rho)=g(T^2\rho)$. What am I missing?

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    $\begingroup$ The second condition in the definition of $\Gamma$ looks wrong. $\endgroup$ May 28, 2019 at 11:47
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    $\begingroup$ @MyNinthAccount, thanks, corrected. $\endgroup$
    – Shimrod
    May 28, 2019 at 12:10

1 Answer 1

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Because $S \in \Gamma$ and $\Gamma$ is a normal subgroup of $\Gamma(1)$, we may as well take the coset representatives to be $1, ST , (ST)^2$.

Or alternately we have $\rho = ST ST \rho = S (TS T^{-1}) T^2 \rho$ and $S (TST^{-1}) \in \Gamma$ because $\Gamma$ is a normal subgroup (or by explicit calculation).

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