When does a group act effectively and holomorphically on some Riemann surface?

Question

Given a Riemann surface $X$, we have some fairly standard methods for identifying which groups $G$ admit an effective and holomorphic action $G \times X \to X$. For instance, some fairly elementary arguments show that the only finite groups acting effectively and holomorphically on $\mathbb{P}^1$ are finite cyclic groups, $D_4, A_4, S_4$, and $A_5$. (See e.g. Miranda, Ch. III p. 80 for these arguments.) Another result is Hurwitz' theorem, which says that given a Riemann surface $X$, the only such finite groups must have order $\leq 84(g(X) - 1)$.

Now let's instead fix a group $G$. (Maybe you want $G$ to be a Lie group, maybe finite.) What results tell us about the (non)existence of a Riemann surface $X$ on which $G$ acts effectively and holomorphically? I'd be surprised to hear that given $G$, we can always construct such an $X$.

Unlike the problem of finding groups given $X$, I have not heard anything about this kind of problem while learning about Riemann surfaces.

Answer 1

In fact:

Theorem: Any finite group $G$ is the automorphism group of a compact Riemann surface, and more generally a smooth projective algebraic curve over any algebraically closed field.

The Riemann surface case is proved by L. Greenberg, Maximal groups and signatures, Ann. Math, Studies 1973, and second by M. Madan and M. Rosen, The automorphism group of a function field. Proc. Amer. Math. Soc(1992)

Answer 2

Donu's answer is correct but amounts to killing a fly with a gun shot: Greenberg proves a harder result than the one needed for the problem.

Theorem. Let $G$ be a countable group. Then there exists a Riemann surface $X$ such that $G$ is isomorphic to a subgroup of $Aut(X)$ acting properly discontinuously and freely on $X$.

Proof. Since $G$ is countable, it has (at most) countably many generators. Let $F_\infty$ denote the free group of countably infinite rank. Then there exists an epimorphism $\phi: F_\infty\to G$. There exists a Riemann surface $Y$ with the fundamental group $F_\infty$, namely, the complex plane ${\mathbb C}$ with a discrete closed countably infinite subset removed (say, the subset of integers). We obtain an epimorphism $\phi: \pi_1(Y)\to G$. Let $N$ denote the kernel of $\phi$ and $p: X\to Y$ the (connected) covering space corresponding to the subgroup $N< \pi_1(Y)$. The group $G$ acts as the Galois group of the covering $p$. Lifting the complex structure from $Y$ to $X$ via $p$, makes $X$ a Riemann surface and the group $G$ acts effectively and biholomorphically on $X$. qed

Incidentally, Greenberg's proof starts similarly, except he would use a compact Riemann surface $Y$, hence, $X$ will be also compact since $G$ is finite. The hard part of his proof is to observe (by a dimension count for suitable Teichmuller spaces) that if $Y$ is chosen generically, then $X$ has no conformal automorphisms other than those coming from $G$.

Lastly, if one allows, say, connected, nontrivial Lie groups $G$ in this problem then only few groups $G$ appear as subgroups of conformal automorphism groups: $S^1$, $S^1\times S^1$, ${\mathbb R}$, ${\mathbb C}, {\mathbb C}^*$, $PSL(2, {\mathbb R}), PSL(2, {\mathbb C})$, $SO(3)$, and few more solvable groups: The group of affine automorphisms of the real line, the group of complex affine automorphisms of the complex plane ${\mathbb C}$, and its subgroup ${\mathbb C}\rtimes S^1$.