Cube root of the $j$-invariant

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?

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

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).