The number of holomorphic coverings (with given degree) of the punctured sphere is finite.

Question

I'm looking for a proof of the following theorem:

Fix a finite set $B=\{y_1,\ldots,y_k\}\subseteq \mathbb P^1(\mathbb C)$, then there is only a finite number of isomoprhism classes of holomorphic coverings of $\mathbb P^1(\mathbb C)$ with the following properties:

1. Their branch locus is contained in $B$
2. They have the same degree $d$

One way to proceed is by proving the correspondence between coverings and monodromy representations (see Miranda's book), but this way is a bit long and moreover it demonstrates much more than I want. I'm interested only on the cardinality of the set of all possible holomorphic coverings of given degree (up to isomorphism). Is there a 'quick way' to prove the above theorem?

Addendum: I need that result only when $k=3$, namely $B=\{y_1,y_2,y_3\}$, therefore I will accept also answers for this particular case.

Thanks in advance.

Answer 1

Your question is equivalent to: how many subgroups of index $d$ does the free group $F_n$ on $n$ elements have, up to conjugacy? This question is extremely difficult to answer precisely. However, one does have:

Theorem. Let $G$ be a finitely generated group. Then $G$ has finitely many subgroups of a given index $d$.

Proof: The left action of $G$ on cosets identifies subgroups of $G$ of index $d$ with a subset of the set of representations $G \to S_d$. Since $S_d$ is finite and $G$ is finitely generated, there are finitely many such representations.

Corollary $F_n$ has finitely many subgroups of index $d$.

However, giving estimates for these numbers is very hard. Suppose you want to count only normal subgroups of $F_n$, for instance. Then you are asking: how many finite groups of order $d$ can be generated by $n$ elements? Counting finite groups is not a light task - usually the only way of counting groups is to list them, and this problem is essentially the problem of the classification of finite groups.