I was trying to solve a problem given in Simon Donalson's book on Riemann Surfaces.
The problem asks to classify the Riemann Surfaces formed as quotients of $\mathbb{C}$.
If $\Gamma$ is the (discrete) group by which we're forming the quotient, I was able to prove that $\Gamma \subset \{z \mapsto az + b: |a| = 1\}$ (otherwise the quotient won't be Hausdorff).
Since what I need to prove is that $\mathbb{C}/\Gamma$ is isomorphic to $\mathbb{C}/\mathbb{Z}$ or $\mathbb{C}/\Lambda$ for some lattice $\Lambda$, I was trying to show that $\Gamma \subset \{z \mapsto z+b\}$
This attempt was unsuccessful. In fact, if $\Gamma_m = \{z \mapsto ze^{2i\pi \frac{k}{m}}:k \in \mathbb{Z}\}$ then I was able to equip $\mathbb{C}/\Gamma_m$ with a conformal structure, which turned out to be equivalent to $\mathbb{C}$ (via the map $z \mapsto z^m$).
Similarly, I was able to define charts on $\mathbb{C}/\Gamma$ even if some points had non-trivial stabilisers (via the technique Donaldson uses to define charts when the group action isn't free).
If only one point has a non-trivial stabiliser, the case more or less becomes the same as $\Gamma= \Gamma_m$ and so the quotient is $\mathbb{C}$.
So, the case I'm stuck is when there are multiple points (possibly infinitely many) with non-trivial stabilisers.
Thanks
Edit:
As Andrew has pointed out in the comments, it is likely that the group action was supposed to be free, in which case the conclusion is easy to see.
In case someone does have any interesting results when the group action isn't free, feel free to add an answer about it.