# Is a finitely generated torsion group finite in general?

This is not a duplicate of Can a finitely generated group have infinitely many torsion elements?

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.

• Some MathJax advice: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. Sep 8, 2013 at 15:54
• By the way, whether the infinite dihedral group is a torsion group is torsion is not a matter of opinion. You are correct that it is not a torsion group: e.g. the element $ab$ has infinite order. You can see this by viewing it as the infinite dihedral group, i.e., as a group of isometries of the real line. More precisely, you can take $a$ to be reflection through $1$ and $b$ to be reflection through $0$. Then $ab$ is the translation $x \mapsto x+2$. Sep 8, 2013 at 16:03
• The third answer to your linked question answers this!...(As for conditions, you might want to look up "free Burnside" groups. Every such group is a quotient of a free Burnside group (so long as the torsion is bounded, which I don't think was the case in the Golod-Shafarevich group). Classify the free ones then you can talk about their quotients.) Sep 8, 2013 at 16:19
• @fiftyeight: A necessary and sufficient for a finitely generated periodic group to be finite is that it be finite. Obviously you are looking for a different sort of criterion: could you say something about what? Sep 8, 2013 at 22:38
• For example, using Dietzmann's lemma we can see if the group is finitely generated by a normal set of generators then it is finite. Is there something equivalent to having a normal finite set of generators? It suffices for this set to be normal in itself (The set need not be a subgroup so it is not obvious that it is normal in itself). I'm searching for a criterion which is as weak as possible on the group. Sep 9, 2013 at 2:44

• Lemma 2 in the cited "paper" is absolutely false. The claim "$\exists x\notin H, x$ is of infinite order" is not the contraposition of the claim "$\forall x\in H$, $x$ is of finite order." Also, Theorem 4 is definitely not equivalent to the statement "some non-finitely presented torsion group is infinite." This paper is basically saying "If one non-animal is not a cat, then every animal is a cat." Jul 12, 2020 at 3:08