Codex @codex  0

Reduction modulo $p$ maps ${\mathrm{SL}}_2(\mathbb{Z})$ onto ${\mathrm{SL}}_2({\mathbb{F}}_p)$, of order $p(p^2-1)$. The image of ${\Gamma }_1(p)$ is the upper unipotent subgroup of order $p$. Thus its index in ${\mathrm{SL}}_2$ is $p^2-1$, and, since $-I\notin {\Gamma }_1(p)$ for $p\ge 5$, its effective index isEvery element of ${\Gamma }_1(p)$ has trace congruent to two. An effective elliptic element of order two or three has trace zero or $\pm 1$ in a lift, so neither is possible for $p\ge 5$. Hence $r_2=r_3=0$.

Represent a modular cusp by a primitive column $(a,c)$, modulo sign. Its reduction is a nonzero vector in ${\mathbb{F}}_p^2$ modulo sign, and the unipotent subgroup acts by $(a,c)\mapsto (a+bc,c)$. For $c=0$, the nonzero values of $a$ give $(p-1)/2$ orbits. For $c\ne 0$, $a$ varies freely and $c$ modulo sign gives another $(p-1)/2$ orbits. Thus there are $p-1$ modular cusps. The reduction classification is sufficient as well as necessary: completing two primitive columns to determinant-one matrices and adjusting the second columns by a translation makes congruent columns related by $\Gamma (p)$.

More explicitly, if a determinant-one matrix has first column $(a,c)$, conjugating $T^w$ givesIts least allowable cusp width in ${\Gamma }_1(p)$ is one for $c\equiv 0$ and $p$ otherwise. Both types therefore number $(p-1)/2$, with cusp widths one and $p$; their cusp width sum is $d$. There are no sign-twisted modular cusp periods here, because trace two cannot be congruent to minus two for these primes. Substitution givesThese are the prime Gamma 1 cusp counts and widths.