Is $j(\tau)^{1/3}$ the hauptmodul for the congruence subgroup generated by $\tau\rightarrow\tau+3, \tau\rightarrow-1/\tau$? The 3rd root of the modular invariant $j$ is
$$ j(\tau)^{1/3}=q^{-1/3}(1+ 248q+ 4124q^2+ 34752q^3+\cdots),$$
where $q=e^{2\pi i \tau}$.
I was wondering if $j(\tau)^{1/3}$ the hauptmodul for the congruence subgroup generated by $\tau \rightarrow \tau+3, \tau \rightarrow-1/\tau$.
If this true, can we say the following assertion? 
If a function $f(\tau)$ that takes the form $f(\tau)=q^{-1/3}(1+\sum_{n=1}^{\infty} a_n q^n)$ with $a_n \geq 0$ and is invariant under $\tau \rightarrow \tau+3, \tau \rightarrow-1/\tau$, then $f(\tau)=j(\tau)^{1/3}$.
Thanks a lot!
 A: The function $x(\tau) = j(\tau)^{1/3}$ is a hauptmodul, just not for the group that you indicate. This function is also invariant under $\tau \mapsto \frac{2 \tau + 1}{\tau + 1}$ and $\tau \mapsto \frac{\tau+1}{\tau + 2}$, and this
means that $x(\tau)$ is a hauptmodul for an index $3$ subgroup of ${\rm SL}_{2}(\mathbb{Z})$, which is the normalizer of a non-split Cartan modulo 3. (This has been known for quite a while. If you want to read a proof of this, see the paper of Imin Chen titled "On Siegel's modular curve of level 5 and the Class Number One Problem" published in the Journal of Number Theory in 1999. Look in Section 4.) The subgroup you specify has index at least $18$ (and it may not even be congruence).
I think that the natural modification of your second question also has a negative answer. (Edited to fix the example.) In particular,
$$
  x(\tau) \frac{j(\tau)}{j(\tau)-1728} = q^{-1/3} (1 + 1976q + 2133020q^{2} + \cdots)
$$
is a modular function for the same group as $x(\tau)$ and also has non-negative Fourier coefficients. With some thought, one can see that $j(\tau)$ has non-negative Fourier coefficients, and so does $\frac{1}{j(\tau) - 1728} = \frac{E_{4}^{3}/E_{6}^{2} - 1}{1728}$. Here $E_{4} = 1 + 240 \sum_{n=1}^{\infty} \sigma_{3}(n) q^{n}$ and $E_{6} = 1 - 504 \sum_{n=1}^{\infty} \sigma_{5}(n) q^{n}$ are the usual Eisenstein series.
