If $\gamma\in\Gamma(N)$, then $(x,y)\gamma\equiv(x,y)\pmod N$, so part c gives invariance. It remains to check the cusps. For any $\sigma\in SL_2(\mathbb Z)$,
$$
G_k^{(x,y)}|_k\sigma=G_k^{(x,y)\sigma}.
$$
As $\operatorname{Im}\tau\to\infty$, the terms with $c=0$ give a finite constant and the locally uniform estimate for the terms with $c\ne0$ gives boundedness. A periodic holomorphic function bounded at infinity has a Fourier expansion with no negative powers. Thus every slash transform is holomorphic at infinity, which is holomorphy at every cusp. Hence $G_k^{(x,y)}$ is the <congruence-class Eisenstein series> in $M_k(\Gamma(N))$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 137 / 1 / d / Solution

{title2=$G_k^{(x,y)}$}

For $k>2$ and $(x,y)\in(\mathbb Z/N\mathbb Z)^2$,
$$
G_k^{(x,y)}(\tau)=
\sum_{(c,d)\ne(0,0)\atop(c,d)\equiv(x,y)\bmod N}(c\tau+d)^{-k}
$$
is a modular form of weight $k$ for the principal congruence subgroup $\Gamma(N)$. Right multiplication of $(x,y)$ by $\gamma\in SL_2(\mathbb Z)$ describes its slash transformation.


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 Incoming links: Congruence-class Eisenstein series