$\Gamma(N)$ -inequivalent cusps clarification

Question

We know that $\Gamma(N)$ has at most $[\Gamma(1):\Gamma(N)]$ inequivalent cusps given possibly by $g_i \infty$ where the $g_i$ are coset representatives of the subgroup $\Gamma(N)$. Then I don't understand why the number of inequivalent cusps is actually $[\Gamma(1):\Gamma(N)]/N$. We know that $[\Gamma(1)_{\infty}: \Gamma(N)_{\infty}]=N$ (by explicitely describing these groups) and I am not sure how to use this to conclude the argument.Here $\Gamma(1)_{\infty},\Gamma(N)_{\infty}$ are the stabilizers of $\infty$.

Answer 1

Intuitively, this should make sense: For every coset $[g_i]$, we are considering $g_i\infty$ as a potential cusp - but we do not want to count the same cusps twice. Thus, we want to ignore all the cosets $[g_i]\neq [1]$ satisfying $g_i\infty = \infty$, since these will be redundant. It turns out that only one in $N$ cosets actually moves $\infty$ - this is what $[\Gamma(1)_\infty:\Gamma(N)_\infty]=N$ means. Further, for two distinct cosets $[g]$ and $[h]$, if $g\infty = h\infty$, we see that $h^{-1}g\infty = \infty$, i.e. we only want to count one of the cosets $[g]$ and $[h]$, as $[h^{-1}g]$ acts similarly to $[1]$. Thus, it works simply dividing $[\Gamma(1):\Gamma(N)]$ by $N$.

Let me also supply a rigorous group theory argument. Take the quotient $G:=\Gamma(1)/\Gamma(N)$, and let us consider the subgroup $K:=\Gamma(1)_\infty/(\Gamma(1)_\infty\cap \Gamma(N))$. Now, it should be clear that $\Gamma(1)_\infty\cap \Gamma(N) = \Gamma(N)_\infty$, and so $\#K=N$. As you have stated, the group $G$ has a transitive group action on the cusps, and basically by definition, the stabilizer of $\infty$ is $K$. Thus, $G/K$ has a simply transitive (i.e. transitive and free) group action on the cusps, thus a bijection to the cusps, and the group has size exactly $[\Gamma(1):\Gamma(N)]/N$.