This is not a duplicate of Can a finitely generated group have infinitely many torsion elements?
There he asks specifically about FC-groups.
Is a finitely generated torsion group finite in general?
Using Dietzmann's lemma allows you to prove this for FC-groups:
Dietzmann's lemma says:
Let G be a torsion group, and M a normal finite subset of G.
Then $\langle M \rangle $ (the subgroup generated by M) is finite.
Now for a finitely generated group G, we take M to be the set of generators.
Then it is finite, but is it normal?
If G is an FC-group then we know every generator in M has a finite conjugacy class (by definition of an "FC group").
And taking the union of the conjugacy classes of the generators gives us a finite union of finite sets, which is finite, and this set is normal because conjugating any element will take us to another element in its conjugacy class which is also in the set.
My question is if there is a way to prove this without the FC-group requirement.
Or better yet, specify what is the required and sufficient condition for G to be finite assuming G is finitely generated and a torsio group.
Some remarks:
In the question I linked to someone specifies that the infinite dihedral group represented by $\langle a,b : a^2 = b^2 = 1\rangle$ is finitely generated but not finite, $a$ and $b$ are torsion elements, but IMO the group is not a torsion group \ so this does not disprove what I am trying to prove.
EDIT:
A correct answer was provided but I will leave the question open a bit in case someone knows a necessary and sufficient condition on G for it to be finite.
Best Answer
No. This is the General Burnside Problem -- must a finitely generated periodic (this is what you call "torsion": every element has finite order) group be finite? -- to which a negative answer was given by Golod and Shafarevich in 1964. Please see the linked article for more information.