Conditions to ensure that if every element of a finitely generated group $G$ of is of finite order, then $|G|< \infty$.

Question

In general, it's not true that every element being of finite order with $G$ finitely generated is enough to ensure that $G$ is finite.

I made some observations: the first observation is incorrect. Maybe a different argument can be used?

1. If $G$ is finitely presented, then the previous data is enough to ensure that $G$ is finite since for generators $\{a_i\}$ with order $n_i$, there should be a surjection $$\langle a_1, \dots a_n \mid a_1^{n_1}, \dots, a_n^{n_n}\rangle=H \to G$$ by taking the quotient by the normal closure of further relations. Or, since all said conditions in $H$ are true in $G$ as well, we should be able to find tietze transformations that make this work.

2. If $G$ is abelian, we should also be good by the classification theorem.

Question: Are there conditions for which we can conclude that a finitely generated group where every element is of finite order, is finite.

Answer 1

I don't expect any necessary and sufficient condition that is anything other than restating the condition (could refine to bounded torsion).

There are many classes of group which are widely studied, where the infinite examples always have infinite order elements: acylicdrically hyperbolic groups, CAT(0) groups, automatic groups, fg elementary amenable groups... etc

An interesting non-example though is that amenable groups in general do have infinite finitely generated torsion examples, in particular Grigorchuk groups are torsion, although they don't have bounded torsion. I think it is open if there are bounded torsion examples in the class of finitely generated amenable groups.