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

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.

• Your group H on (1) is not finite. – Mariano Suárez-Álvarez Sep 26 '17 at 18:12
• @MarianoSuárez-Álvarez a critical error, I am too used to commutative things, sorry. – Andres Mejia Sep 26 '17 at 18:20
• It is a long-standing open problem whether a finitely presented group in which all elements have finite order is necessarily finite. See mathoverflow.net/questions/78410 – Derek Holt Sep 26 '17 at 18:47
• @DerekHolt oh my. That is truly surprising. – Andres Mejia Sep 26 '17 at 18:49
• I think it is basically hopeless to come up with a necessary and sufficient condition that is anything other than basically rephrasing the property – Paul Plummer Sep 27 '17 at 1:37