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The modular discriminant has the convergent productis a weight-twelve level-one cusp form, and has order one at infinity. Applying the degeneracy map for cusp forms with $M=1,N=D=p$ gives $\boxed{\Delta (p\tau )\in S_{12}({\Gamma }_0(p))}$. The original form $\Delta$ is also a cusp form on this subgroup.

There are precisely two modular cusps for prime level, represented by infinity and zero. To see this directly, write a rational cusp as $a/c$ in lowest terms. If $p\mid c$, completing $(a,c{)}^T$ to a determinant-one matrix gives a member of ${\Gamma }_0(p)$ taking infinity there. If $p\nmid c$, the equation $Ac-aC=1$ has a solution with $p\mid C$: adjust a Bézout solution modulo $p$, using the invertibility of $c$. The matrix $\left(\begin{array}{cc}A& a\\ C& c\end{array}\right)$ then lies in ${\Gamma }_0(p)$ and takes zero there. The two classes are distinct because their denominator divisibility is preserved by the group.

At infinity the cusp width is one, so $q$ is the local parameter. The products start with $q$ and $q^p$, giving orders one and $p$. At zero use $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Since $S{T}^h{S}^{-1}=\left(\begin{array}{cc}1& 0\\ -h& 1\end{array}\right)$, the cusp width is $p$ and the parameter is $q_0=e^{2\pi i\tau /p}$. The weight-twelve inversion law of the modular discriminant givesHere determinant one makes the printed and ordinary slash operators for modular forms agree. In terms of $q_0$, the leading terms are $q_0^p$ and $p^{-12}{q}_0$. Hence the prime-level cusp orders of discriminant degeneracy forms areThe scalar $p^{-12}$ changes no order of vanishing.