weakly holomorphic modular forms with a simple pole at $\infty$ Let $l$ be a prime. Suppose that $M_0^{!}(\Gamma_0(l))$ donote the space of weakly holomorphic modular forms of weight $0$ for the congruence subgroup $\Gamma_0(l)$. Does there exist a $f\in M_0^{!}(\Gamma_0(l))$ such that $f$ just has a simple pole at $\infty$? Here I require that $f$ is holomorphic on the upper half plane and $0$.
It's known that if $l\in \{2, 3, 5, 7, 13\}$, the answer is yes and we can construct such $f$ with Dedekind eta function. What if $l=11$ or $l>13$?
 A: No such $f$ exists, except for the five specific prime levels $2,3,5,7,13$ you describe.
Any such form would give an isomorphism of Riemann surfaces between $X_0(p)$ and $\mathbf{P}^1$. So it cannot exist if $X_0(p)$ has genus $> 0$, which holds for all $p$ except these five.
EDIT. You asked about the minimal order of pole of a (nonzero) weakly holomorphic function. This is equivalent to asking for the smallest $n$ such that $\ell(n\infty) > 1$, where $\ell(D)$ is the vector space of functions whose poles are bounded by the divisor $D$. The Riemann-Roch theorem tells us that this is bounded above by $1 + g$ but it could be smaller; the exact value is not purely determined by the genus (unless $g = 0$ or $g = 1$), since it depends on whether $\infty$ is a Weierstrass point, see https://en.wikipedia.org/wiki/Weierstrass_point.
EDIT (II). According to this paper, Atkin proved in 1973 that if $\ell$ is prime, $\infty$ is never a Weierstrass point of $X_0(\ell)$. So the minimal order of pole of a non-constant modular function of level $\Gamma_0(\ell)$, holomorphic away from $\infty$, is exactly $g + 1$ where $g$ is the genus.
