Infinite subgroups of $\text{PGL}(2,\mathbb C)$ where each element has finite order

Question

I am interested in finite subgroups of $\text{PGL}(2,\mathbb C)\cong\text{Aut}\,\mathbb C(x)$.

Let $G\subset\text{PGL}(2,\mathbb C)$ be finitely generated such that each $g\in G$ has finite order. Is $G$ necessarily finite?

The "finitely generated" hypothesis is necessary here, since $$\left\{\begin{bmatrix}e^{2\pi i t} & 0\\0&1\end{bmatrix}:t\in\mathbb Q\right\}$$ is an infinite subgroup where each element has finite order.

This problem without the restriction of being a subgroup of $\text{PGL}(2,\mathbb C)$ is known as the Burnside problem, for which Golod and Shafarevich came up with a counter-example in 1964. However, my instinct is that $\text{PGL}(2,\mathbb C)$ wouldn't have such weird subgroups.

More generally, is the question true for subgroups of $\text{PGL}(n,\mathbb C)$?

Answer 1

The magic words are "Selberg's lemma" (which states that a finitely generated matrix group has a torsion free subgroup of finite index).

Answer 2

I found the following in Irving Kaplansky's book "Fields and Rings":

Theorem G. Any torsion group of matrices is locally finite.

Here, "locally finite" means every finitely generated subgroup is finite.