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.

  • $\begingroup$ 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. $\endgroup$ Sep 8, 2013 at 15:54
  • $\begingroup$ 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 $x \mapsto x+2$. $\endgroup$ Sep 8, 2013 at 16:03
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    $\begingroup$ 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.) $\endgroup$
    – user1729
    Sep 8, 2013 at 16:19
  • $\begingroup$ @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? $\endgroup$ Sep 8, 2013 at 22:38
  • $\begingroup$ 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. $\endgroup$
    – fiftyeight
    Sep 9, 2013 at 2:44

2 Answers 2


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.


It is a classical question whether there exist finitely generated groups where every element is torsion. This is called Burnside's Problem. It was posed c1900, and Golod and Shafarevich proved in 1964 that there existed finitely->generated infinite p-groups (for some primes p), although the groups they constructed had unbounded order (for all n∈N there exists g∈G such that o(g)>pn). (from https://math.stackexchange.com/a/109582/491561)

Please see http://www.ijma.info/index.php/ijma/article/download/5635/3299 for a detailed proof of the Burnside problem: Every Finitely Generated Torsion Group is Finite.

  • $\begingroup$ This seems to ignore the careful framing of the Question, and in any case depends largely for links to supply any useful content. While links can serve to supplement an Answer by filling background details, when one relies on them for the substance of a solution, one needs some sort of summary of what a Reader expects to find by following a link (ideally a verbatim quote of the most relevant material, with attribution). $\endgroup$
    – hardmath
    Feb 3, 2018 at 2:06
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    $\begingroup$ The question is still open and the cited publication is flawed. (It uses an implication akin to "If one animal is a cat, then every animal is a cat") $\endgroup$
    – ahulpke
    Dec 2, 2018 at 9:52
  • $\begingroup$ The cited article uses an implication akin to "If one animal is a cat, then every non-animal is not a cat" which is logically sound. $\endgroup$ Dec 3, 2018 at 15:10
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    $\begingroup$ 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." $\endgroup$
    – justadzr
    Jul 12, 2020 at 3:08
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    $\begingroup$ I have downvoted this answer as it cites flawed research. In addition, I note the "journal" IJMA is a known predatory publisher, and no work within this journal should be trusted. $\endgroup$
    – davidlowryduda
    Jul 13, 2020 at 17:26

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