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Is it true that an infinite group which is finitely generated must have an element of infinite order? If not, I'd like a counterexample.

[Edit] Somewhat easily proven false.

More interesting (and as it happens, more relevant to my particular problem domain), what about finitely presented groups?

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There are infinite Burnside groups. See en.wikipedia.org/wiki/Burnside%27s_problem –  user83827 Oct 17 '11 at 22:33
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See also the Grigorchuk group for a standard counterexample to many similar questions. –  t.b. Oct 17 '11 at 22:34
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More interesting to ask this question for finitely presented groups. –  user641 Oct 17 '11 at 23:21
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(Someone should wrap all these comments into an answer...) –  Mariano Suárez-Alvarez Oct 17 '11 at 23:42
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The Tarski monster groups deserve a mention - they are finitely generated groups such that every people, non-trivial subgroup is cyclic of order a fixed prime $p$. These exist for all primes greater than some number which has been calculated by I cannot recall of the top of my head. They were first proven to exist by Ol'shanskii in the early 80s. –  user1729 Oct 18 '11 at 9:50

1 Answer 1

up vote 5 down vote accepted

In full generality, this is false. As mentioned in the comments, the Burnside groups can be infinite groups, finitely generated by construction, such that every element has bounded finite order. One can even force such a group to be simple.

There are positive results for certain families of groups, however. Immediately I am reminded of solvable groups and hyperbolic groups.

Perhaps a much more interesting question is if there is a finitely presented, infinite group, with no element of infinite order. AFAIK, this question is still wide open.

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