I'm reading the proof of a theorem that says

If $G$ is a finitely generated FC-group, then its set of torsion elements $T(G)$ is a finite normal subgroup of $G$.

I understand everything the proof says, but I don't understand why it doesn't explain why $T(G)$ is finite. Is it obvious that a finitely generated group has only finitely many torsion elements?

EDIT: Thanks to Jim Belk's answer, I know that being finitely generated isn't enough to have finitely many torsion elements for groups. Why is that true for FC-groups then? I know and can prove that that the commutator subgroup of a finitely generated FC-group is finite, but in such groups $G'\subseteq T(G)\subseteq G$, so this doesn't immidiately imply the finiteness of $T(G)$...


3 Answers 3


A finitely generated group can have infinitely many torsion elements. For example, the infinite dihedral group $\langle a,b \mid a^2=b^2=1\rangle$ has infinitely many elements of order two.

  • $\begingroup$ Thank you. I will edit the question. $\endgroup$
    – user23211
    Feb 15, 2012 at 0:17

Further to Jim Belks answer (and because it answers the title question, if not the question in the post),

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 $n\in \mathbb{N}$ there exists $g\in G$ such that $o(g)>p^n$).

The next obvious question is "do there exist finitely-generated, infinite torsion groups of bounded exponent?". This was solved pretty soon after Golod and Shafarevich (still in the 60s - I forget when).

Now, the next-again obvious question is "what kind of bound can we put on the exponent?". Tarski conjectured, and in 1982 Ol'shanskii proved, the following theorem for "sufficiently large" primes p,

There exist infinite, $2$-generated groups such that every proper, non-trivial subgroup is cyclic of order $p$.

Such groups are called Tarski monster groups, and this result is somewhat striking. I believe the question of replacing "finitely generated" with "finitely presented" is still an open problem.

All this deals with odd exponent. If your exponent is even then there were some existence results in the 90s for groups which had even exponents containing high powers of $2$ (I think by Ivanov). Note that if your exponent is $2$, then you automatically get finite (why?).

Now, a related question is,

What can be said about the number of $n$-generated finite groups of exponent $m$?

This question was solved by Efim Zelmanov in 1994, and he won a fields medal for it!

  • $\begingroup$ Very informative! Thank you. $\endgroup$
    – user23211
    Feb 15, 2012 at 15:48

You want to look up Dietzmann's lemma. Basically, $T(G)$ is finitely generated, and each generator has finitely many conjugates. So $T(G)$ is generated by a finite set of generators, closed under conjugation, and each generator has finite order. Dietzmann's lemma implies $T(G)$ is finite.


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