$f\in M_k (\Gamma_1(N)$ and $g(z)=f(dz)$, then $g\in M_k(\Gamma_1 (dN)$ Let $N,d\in\mathbb{N}$ and $\Gamma_1(N)=\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}):a\equiv d\equiv 1\mod N,~c\equiv 0\mod N\}$. 
I need a proof (references or your ideas) of:

Let $f$ be a modular form of weight $k$ on $\Gamma_1(N)$ and $g(z):=f(dz)$. Then $g$ is a modular form of weight $k$ on $\Gamma_1(dN)$.

There are two things to show:


*

*g is holomorphic on the upper half plane

*$g\left(\dfrac{az+b}{cz+d}\right)=(cz+d)^kg(z)$ for all $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_1(dN)$

*g is holomorphic on cusps


1 is clear. Can anybody explain or name references, why 2 holds? It might follow from $f\left(\dfrac{az+b}{cz+d}\right)=(cz+d)^kf(z)$ for all $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_1(N)$, but how?
Thanks in advance.
 A: Let $\mathbb{H}$ denote the upper half plane and $M = \begin{pmatrix} a & b \\c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{R}) $. Then we define for a function $f:\mathbb{H} \to \mathbb{C}$ the slash operator as
\begin{equation}
(f|_kM)(z) := (cz+d)^{-k}f\left( \frac{az+b}{cz+d} \right).
\end{equation}
We note that $f|_k(M_1M_2) = (f|_kM_1)|_kM_2$ for all $M_1,M_2 \in \mathrm{SL}_2(\mathbb{R})$.
With this slash operator we can rewrite 
\begin{equation}
f\left( \frac{az+b}{cz+d}\right) = (cz+d)^{k}f(z)
\end{equation}
as
\begin{equation}
(f|_kM)(z) = f(z).
\end{equation}
Now let $A = \begin{pmatrix} n & 0 \\0 & 1 \end{pmatrix} $ and $\gamma = \begin{pmatrix} a & b \\c & d \end{pmatrix} \in \Gamma_1(nN)$.
Then $g(z)=f(nz)=(f|_kA)(z)$.
It is $A\gamma A^{-1} = \begin{pmatrix} a & bn \\ c/n & d \end{pmatrix} \in \Gamma_1(N)$ since $c \equiv 0 \mod nN$ and $\mathrm{det}(A\gamma A^{-1}) = 1$ and $a \equiv d \equiv 1 \mod N$.
We obtain
\begin{equation}
(f|_kA)|_k \gamma
= (f|_k(A\gamma))
= (f|_k(A\gamma A^{-1}A))
= (f|_k(A\gamma A^{-1}))|_k A)
= f|_kA
\end{equation}
for all $\gamma \in \Gamma_1(nN)$.
