Can a finitely generated group have infinitely many torsion elements? 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 $T(G)$...

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.
