Let $M$ be a complex manifold on which the group $G$ acts properly and holomorphic. Is $M/G$ then a complex manifold?
My attempt:
We need charts for $M/G$. For any $m \in M$ we take $U\subset M$ small enough such that the restriction of the quotient map $q$ is a homeomorphism. (Is this possible because of proper discontinuity?) We have a chart $\phi_U: U \to \mathbb C^n$. As $q|_U$ is a homeomorphism, this gives us a chart $\phi_U \circ (q|_U)^{-1}: q(U)\to \mathbb C^n$.
We need to check that the functions $[\phi_V \circ (q|_V)^{-1}] \circ [\phi_U \circ (q|_U)^{-1}]^{-1}: \mathbb C^n \to \mathbb C^n$ are holomorphic. But this is the case if $(q|_U)^{-1} \circ (q|_U)$ is holomorphic. But this is simply the translation by a group element which is holomorphic as we assumed the group action to be holomorphic.
Is this right and is this enough?
Edit: Actually I want $G$ to also be discrete and act freely.
You’re right but the idea is more general.
When your action is properly discontinuous you get that $\pi: M\to M/G$ is a covering map that is open so you can define the local charts in your natural way and you get that $M/G$ is a local Euclidean complex space;
In general $M/G$ is not an Hausdorff space;
$M/G$ will be second countable because $\pi$ is surjective and open and $X$ is second countable;
It is possible built an olomorphic structure on $M/G$? I think that it is a really difficult question because in general your composition
$q|_V^{-1}\circ q|_U$ is not simply a moltiplication with respect a fixed $g\in G$