I want to prove that the Eisenstein series $$G_k(z)= \sum_{c,d}(cz+d)^{-k}, (c,d)\in \mathbb{Z}^2 \setminus \{(0,0)\}$$ satisfies the relationship $$G_k((az+b)/(cz+d))= (cz+d)^k\sum_{(c',d')} ((c'a+d'c)z+(c'b+d'd))^{-k}=(cz+d)^k G_k(z)$$. where $a,b,c,d$ are integers such that $ad-bc=1.$
i.e. as $(c'd')$ runs through $\mathbb{Z}^2 - \{(0,0)\}$, so does $(c'a+d'c,c'b+d'd)$.'
My attempt is as follows: It suffices to prove that there exists a bijection $$\mathbb{Z}^2 - \{(0,0)\} \leftrightarrow (a\mathbb{Z}+c\mathbb{Z}) \times (b\mathbb{Z}+d\mathbb{Z}) - \{(0,0)\}$$ .Now for integers $a,b,c,d$ satisfying $ad-bc=1$, we must have $gcd(a,c)=1=gcd(b,d)$. Otherwise, there would be a common divisor $p>1$ of $a,c$ or $b,d$ such that $ad-bc=p(ad/p-bc/p)>1$. So that the set $ (a\mathbb{Z}+c\mathbb{Z}) \times (b\mathbb{Z}+d\mathbb{Z}) - \{(0,0)\}= gcd(a,c)\mathbb{Z} \times gcd(b,c)\mathbb{Z}- \{(0,0)\}=\mathbb{Z}^2 - \{(0,0)\}$ and we are done.
I want to ask if my proof is correct and if there is any alternative proof, thank you so much.
Your proof seems about right. It could be made shorter:
The matrix* $M=\begin{pmatrix} a &b \\ c&d \end{pmatrix} \in \text{SL}_2 (\Bbb{Z})$ is an automorphism of (the free module) $\Bbb{Z}^2$ .
To prove this, consider the image of $S_{p,q}=\{(x,y):x \in [p,p+1), y \in [q, q+1) \}$ under $M$, namely $S'_{p,q}$. The area of $S_{p,q}, S'_{p,q}$ are the same (i.e. $1$) as $|M|=1$. As $M(\Bbb{Z}^2)\subseteq\Bbb{Z^2}$ we have that the four corners of $S'_{p,q}$ are in $\Bbb{Z}^2$, so by Pick's theorem there are no interior lattice points in $S'_{p,q}$. Thus $M$ is a surjective linear map. It is injective as $M$ is invertible, so an automorphism.
*Actually a linear map ($\Bbb{Z}^2$ is a module).
I don't know how much algebra you've done, but essentially my point is that $\exists \alpha, \beta:$ $M$ will take $(\alpha, \beta)$ to $(c,d)$, so the summation is the same.