What are the conditions for a finitely generated group $G$ with finite ordered generators say $a_1, a_2,...,a_n$ to be finite?
Note:I know that if $G$ is abelian, then it is finite. Are there any other known results that I can use here?
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$\begingroup$ $\Bbb{Z},+$ is abelian and finitely generated (by $1$) but not finite. So it seems that every generator should have finite order, but is that sufficient? $\endgroup$ – Marc Bogaerts Sep 19 '14 at 9:16
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In general, you have to require much more than generators to have finite orders. For instance, even if you take a 2-step nilpotent group, you would have to require all the commutators of the generators to have finite order (in addition to the generators themselves).
Edit: The above is not right, as Derek Holt noted: Since the subset of finite order elements of a nilpotent group is a subgroup, if a group $G$ is nilpotent and all its generators are of finite order, then every element of $G$ has finite order, then $G$ is finite.
In order to prove this, note that all the members of the derived series of $G$ are finitely-generated. Then, we argue argue by induction on the length of the derived series of $G$: The center $Z(G)$ of $G$ is finite, since it is a finitely generated abelian and torsion. By the induction hypothesis, $G/Z(G)$ is also finite. Hence, $G$ is finite as well.
Once you leave the realm of solvable groups, things become very complicated. For instance, it is not even enough to assume that all elements of your group have finite order, for $G$ to be finite. Read about [Burnside problem] (http://en.wikipedia.org/wiki/Burnside's_problem) to find more about it.
answered Sep 19 '14 at 9:51
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$\begingroup$ I don't think that's right. A nilpotent group generated by finitely many elements of finite order is finite. This is not true for solvable groups however, as shown by the infinite dihedral group. $\endgroup$ – Derek Holt Sep 19 '14 at 11:42
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$\begingroup$ See math.stackexchange.com/questions/79474 for a proof of the claim about nilpotent groups. $\endgroup$ – Derek Holt Sep 19 '14 at 11:53
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$\begingroup$ Oh, you a right of course: I forgot that the subset of torsion element in a nilpotent group is a subgroup. $\endgroup$ – Moishe Kohan Sep 19 '14 at 13:06
