Proving the basic properties of the Hecke Operator I am want to ask about the proof of some of the basic properties of Hecke operators.  Reading Serre, I was really confused as I am not really familiar with things like homothety operators and lattices.  So, I am going to ask from a website which does not use these kind of vocab.  The site I will be referring to is http://wstein.org/books/modform/modform/level_one.html
I have several questions.  First is the proof of Proposition 2.28.  I am not sure how the uses of $\gamma, \delta, \sigma$ show the weakly modular property but going back to definition, I want to see if I understand what I want to prove.  A weakly modular function f, along with being meromorphic, has to satisfy the property 
$f(z)= (cz+d)^{-k}f(gz)$ where g is the matrix in $SL_{2}Z$ wich components a,b,c,d.  In this case, our function is $T_{n}(f)$ so we want to show 
$T_{n}(f)(z)=(cz+d)^{-k}T_{n}(f)(gz)$   (*)
Since $T_{n}(f)(gz)=\sum_{\gamma \in X_{n}}det(\gamma)^{k-1}(cz+d)^{-k}f(\gamma (gz))$, I guess the issue is to make the $f(\gamma (gz))=Cf(\gamma (z))$ to get something similar to the left hand side of (*)?
My second question is on the proof of the 1st part of Proposition 2.29.  I believe I get the beginning. Since $Z^{2}/L$ is an abelian group of order mn, and m and n are coprime, by Fundamental Theorem of Abelian Groups, this decomposes to a subgroup F of order n, and E of order m.  And by Fourth Isomorphism Theorem, F corresponds to some L' where L' has index n.  So we solve L'A=L.  So I get that we can get any element in $X_{nm}$ as some product of an element in $X_{n}$ with element and element in $X_{m}$
But from the next part of the proof, where does the $SL_{2}Z$ invariance come into play in the proof above and how does $X_{n}*X_{m}=X_{mn}$ imply that $T_{m}T_{n}=T_{mn}$
 A: $T_l f$ is modular for $\Gamma_1(n)$ if $f$ is modular for $\Gamma_1(n)$


*

*Look at $$A_l = \{ \frac{\alpha}{\det(\alpha)}, \ \alpha \in \mathbb{Z}^{2 \times 2},\ \ \det(\alpha) \ |\ l\ \}$$ 
Imagine we can find a finite set $\{\alpha_j\}_{j=1}^m$ of elements of $A_l$ such that we have the disjoint union $$A_l = \bigcup_{j=1}^m \Gamma_1(n)\alpha_j $$

*Let $f$ being modular $$\forall \gamma \in \Gamma_1(n),\qquad f[\gamma] = f$$
where the $[.]$ operator is $f[\gamma](\tau) = (c\tau+d)^{-k} f(\gamma \tau)$ so that $f[\gamma \gamma'] = f[\gamma][\gamma']$ whenever $\det(\gamma)=\det(\gamma') = 1$

*Define the operator $$T_l f = \sum_{j=1}^m f[\alpha_j]$$

*For any $j$ and $\gamma \in \Gamma_1(n)$,  there is $j'$ and $\gamma' \in \Gamma_1(n)$ such that
$$f[\alpha_j \gamma] = f[\gamma' \alpha_{j'}] = f[\gamma'] [\alpha_{j'}]=f[\alpha_{j'}]$$
(proof : $\det(\alpha_j \gamma) = \det(\alpha_j)$ so $\alpha_j \gamma \in A_l$, thus it is in one of the cosets $\Gamma_1(n)\alpha_{j'} $ ie. $\alpha_j \gamma=\gamma' \alpha_{j'} $)

*Assume that $j \ne j_2$ and $j' = j_2'$ then the $A_l$ subsets $\Gamma_1(n)\alpha_j\gamma$ wouldn't be disjoint, contradicting that the $\Gamma_1(n)\alpha_j$ are disjoint. 

*Hence $$Tf[\gamma] = \sum_{j=1}^m f[\alpha_j \gamma] = \sum_{j'=1}^m f[\alpha_{j'}] = Tf$$

Generalization
Replace $A_l$ by $\Gamma_1(n) {\scriptstyle\begin{pmatrix} 1/\sqrt{p} & 0 \\ 0 & \sqrt{p} \end{pmatrix}} \Gamma_1(n)$. Everything works the same way when $p$ is prime. The associated operator $T_p$ is the Hecke operator.

About $$T_l T_{l'} f = T_{ll'} f, \qquad gcd(l,l')=1$$ 
It seems quite straightforward with the above defnition,
We can also look at the Fourier expansion, obtained from an explicit description of the $\alpha_j$
