Modular forms Concepts , Congruence Subgroup, Modularity definition 1) Defintion : 
A congruence subgroup of level $N$ is one that contains the principal subgroup at level $N$ which is defined as $ (a b c d) \in SL_2(Z) : a,d\equiv 1 (mod N), b,c \equiv 0 (mod N) $ (apologies $ ( a b c d)$ is a 2x2 matrix.)
The Hecke group is one such example given by $T_{0}(N) = ( *, *) ( 0, *) $ (apologies again that's the first row and second row of the matrix respecitvely).
So is my understanding of these definitions correct: that $T(N) \in T_0(N)$ since can choose $( *, *) = (1,0) $ and $(0,d) = (0,1) $ to give the required $a,d \equiv 1 (mod N), b,c \equiv 0 (mod N) $ ?or eg $(*,*)= (cN+1, cN) $, $c$ is a constant, and then both of these elements divide by $N$? (that's the top row of the matrix $\in$ $SL_2(Z)$ , apologies again. 
2) In my notes I have that the 'corresponding' condifiton for a function $f(t)$ to be modular for $T_0(N)$ differs to the $SL_2(Z)$ condition that must be holomorphic at $\infty$,( from $f(t)$can be written as an expansion as $f(t)= \sum\limits^{\infty}_{n=0} a_{n} q^n $, to :
$f$ is 'holomporphic at all cusps of $T_0 (N) $ , that is the limit 
$lim _{q \to 0} (ct+d)^{-k} f( \alpha t) $ exists for all $\alpha$ \in $SL_2(Z)$.
And that this only needs to be checked at finitely many $\alpha$, for those which map the (inequivalent) cusps to $\infty$. 
So i don't really understand why this is the condition, nor why it is sufficient to only check the $\alpha$ that map the cusps to $\infty$.
Here's what I know:


*

*In $SL_2(Z) $ all cusps are $T$-equivalent to $\infty$, since all rational numbers are, and so this is why it suffices to only check holomorphicity at  $\infty$ for $SL_2(Z)$ ? 

*Whereas for $T _0 (N) $ fewer points are $T$-equivalent  and so $T_0 $ can have other cusps at the rational numbes, that is, the cusps can no longer be mapped to $\infty$. So we check the expansion of $f$ mapped to $\infty$ from these inequivalent cusps - however I don't really under where the condition comes from. Why is the idea to map to $\infty$?  Why is the condition working with maps $\in SL_2(Z)$ , how is this ok?
3) This is proabably a stupid question but in my notes it says:
$T _0 (N) $ touches the real axis at the point $0$. In fact, such a behavior will happen for any $T_0(N)$. Any fundamental domain for$ T_0(N)$ will touch the the real axis in ﬁnitely many rational points, and we call these points (together with the inﬁnite cusp $∞$), the (equivalence classes) of cusps of $T_0(N)$. 
I don't understand how the cusp at $0$ and $\infty$ are called equivalence classes, since $0$ is mapped to $\infty$ by $S$ , however, for e.g $T_0 (p) $ $s \notin T_0(p) $ and so the points $0$ and $\infty$ are not $T$-equivalent...
I'm pretty confused.
Any clarification what so ever greatly appreciated, ta. 
 A: $\gamma \in \Gamma(N) $ if $\gamma = {\scriptstyle \begin{pmatrix} a  & b \\ c & d\end{pmatrix}} \in SL_2(\mathbb{Z})$ and $\gamma \equiv {\scriptstyle \begin{pmatrix} 1  & 0 \\ 0 & 1\end{pmatrix}} \bmod N$
Let $\Gamma$ be a congruence subgroup, that is $\Gamma(N) \subseteq \Gamma \subseteq SL_2(\mathbb{Z})$ for some $N$. This is the condition we need for ensuring that it is a normal subgroup of finite index $m=[SL_2(\mathbb{Z}):\Gamma]$.
Then $f(\tau)$ is a modular form of weight $k$ for $\Gamma$ if :


*

*$f$ is holomorphic on $\mathcal{H} = \{\tau \in \mathbb{C},Im(\tau) > 0\}$

*forall $\gamma ={\scriptstyle \begin{pmatrix} a  & b \\ c & d\end{pmatrix}}\in \Gamma$ :  $f|_k\gamma ( \tau) = (c\tau+d)^{-k} f(\frac{a\tau+b}{c\tau+d})=f(\tau)$ 
Let $g_\gamma(q) = f|_k \gamma(\frac{\log q}{2i\pi})$. Those two conditions means it is holomorphic on $0 < |q| < 1$. We'd like one more condition for ensuring it is holomorphic at $q=0$ too, so it is holomorphic on the whole unit disk and it is equal to its Taylor series $g_\gamma(q) = \sum_{n=0}^\infty a_n(\gamma) q^n$ for $|q| < 1$ and hence $f|_k\gamma ( \tau) = \sum_{n=0}^\infty a_n(\gamma) e^{2i \pi n \tau}$ for every $\tau$.

*Forall $\gamma ={\scriptstyle \begin{pmatrix} a  & b \\ c & d\end{pmatrix}}\in SL_2(\mathbb{Z})$, $f|_k \gamma (\tau)$ is holomorphic at the cusp $i\infty$, that is for $Im(\tau)$ large enough : 
$$ f|_k\gamma ( \tau) = \sum_{n=0}^\infty a_n(\gamma) e^{2i \pi n \tau}$$
Equivalently, by Riemann's theorem on removable singularities, it is  enough to ask that $\lim_{q \to 0} g_\gamma(q)$ exists.
From this there are various generalization, allowing $g_\gamma$ to be meromorphic at $0$ (example, the modular forms of weight $0$) and meromorphic on the unit disk (example $1/f(\tau)$).

If $f$ is a modular form for $SL_2(\mathbb{Z})$ and $f(\tau)$ is holomorphic at the cusp $i\infty$, then since $f|_k \gamma = f$, it is holomorphic at the cusp $i\infty$ too, and $f$ is holomorphic at every cusp. Or, in other words, there is only one cusp in $SL_2(\mathbb{Z})\setminus\mathcal{H}$ (which is obvious when looking at the fundamental domains).
If $f$ is a modular form only for the congruence subgroup $\Gamma$, then there are $m$ different $\gamma$ or cusps to check for ensuring that $\forall \gamma \in SL_2(\mathbb{Z}), f |_k \gamma$ is holomorphic at the cusp $i\infty$, and that $f$ is holomorphic at every cusp of $\Gamma\setminus\mathcal{H}$.
Note that you can refer to a cusp by $a/c$ instead of ${\scriptstyle \begin{pmatrix} a  & b \\ c & d\end{pmatrix}}$.
