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According to this paper by M. D. Baker and A. W. Reid, if $\Gamma(N)$ denotes the principal congruence subgroup of $\mathrm{SL}_2(\mathbb{Z})$ of level $N\geq1$, then it should be a straightforward observation that $$\Gamma(N)\text{ is generated by parabolic elements}\iff X(N)\text{ has genus }0.$$ A parabolic element is a matrix $A\in\mathrm{SL}_2(\mathbb{Z})$ with trace $\pm2$, and $X(N)$ denotes the compactified modular curve associated to $\Gamma(N)$. I'm still learning about congruence subgroups and modular curves, so my question is:

Q. What's the explanation behind this observation?

I know that, for $N\geq3$, $X(N)$ has no elliptic points and its number of cusps $\varepsilon_\infty(N)$ is given by $$\varepsilon_\infty(N)=\frac{N^2}{2}\prod_{p\ \mid\ N}\left(1-\frac{1}{p^2}\right),$$ while its genus $g(N)$ is given by $$g(N)=1+\frac{1}{24}(N-6)N^2\prod_{p\ \mid\ N}\left(1-\frac{1}{p^2}\right).$$ Perhaps this has something to do with the fact that $\Gamma(N)^\mathrm{ab}\cong\mathbb{Z}^{2g(N)+\varepsilon_\infty(N)-1}$.

K. Makabre
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1 Answers1

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The formulas you give for $\varepsilon_\infty(N)$ and for $g(N)$ are not relevant.

Your final formula $$\Gamma(N)^\mathrm{ab}\cong\mathbb{Z}^{2g(N)+\varepsilon_\infty(N)-1} $$ is partially relevant: if $g(N) \ge 1$ then $\Gamma(N)^{\mathrm{ab}}$ is a free abelian group of rank $R = 2g(N)+\varepsilon_\infty(N)-1 > \varepsilon_\infty(N)$; but a group whose abelianization is free abelian of rank $R$ cannot be generated by fewer than $R$ elements.

For the other direction, you need the fact that $\Gamma(N)$ is a nonabelian free group of rank $R = 2g(N)+\varepsilon_\infty(N)-1$, in order to conclude that $\Gamma(N)$ is indeed generated by $R$ elements. For proving this fact, some topology helps: $\Gamma(N)$ is the fundamental group of the quotient surface which is obtained from a closed surface of genus $g(N)$ by removing $\varepsilon_\infty(N)$ points, and that makes $\Gamma(N)$ a nonabelian free group of the stated rank.

Lee Mosher
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