How many cusps does the congruence subgroup $\Gamma(N)$ have?
Thanks;
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3$\begingroup$ I voted to close. You'd be hard pressed to find a modular forms tetbook that did not have this information. The question is hardly research-level. $\endgroup$– David LoefflerCommented Sep 18, 2012 at 7:08
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1 Answer
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This can be found in most books on modular forms, there's a lot of detailed information in Chapter 3 of Diamond-Shurman. It's $$ \frac 1 2 N^2 \prod_{p \mid N}\left(1 - \frac 1 {p^2}\right)$$ if $N \geq 3$ and $3$ if $N=2$. The factor $N^2 \prod_{p \mid N}\left(1 - \frac 1 {p^2}\right)$ arises as the number of elements of order $N$ in $(\mathbf Z/N)^2$.
answered Sep 17, 2012 at 18:31