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use this notation for the following $\textbf{Theorem}$ -

$\textbf{notation}-$ $G^n$ is the subgroup generated by $n$th power of elements of $G$

$\textbf{Theorem}$- In a finitely generated nilpotent group, $\cap{G^p}$, for any infinite set of primes, is finite,(follows by a paper of HIGMAN) does it help me to prove this question above. I don't think so.

Is it true, that a group generated by finite elements of finite order is finite. If it is true then to prove $G$ is finite in the problem is easy, and don not require nilpotency. But it should not be true, but what can be an example?

Shaun
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Bhaskar Vashishth
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    Tarski monsters are examples of finitely generated groups whose generators have finite order. – Teri Sep 25 '14 at 17:41
  • so we need nilpotency to say $G$ is finite. How can i approach this problem? – Bhaskar Vashishth Sep 25 '14 at 17:43
  • Use induction on the nilpotence class, and consider the central quotient in the non-abelian case. – James Sep 25 '14 at 19:24
  • I can't use induction, until I show $G$ is finite – Bhaskar Vashishth Sep 25 '14 at 19:25
  • The fact that $G$ is finite is proved in the answer by Artuto Magidin to http://math.stackexchange.com/questions/79474 Since it is finite, it is the direct product of its Sylow subgroups, and it is easy to see that the only promes dividing $|G|$ are those dividing the order of one of the generators. (And the proof does use induction!) – Derek Holt Sep 25 '14 at 19:27
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    @BhaskarVashishth The induction is on nilpotence class, which is finite by assumption. – zibadawa timmy Sep 25 '14 at 19:33
  • ohh yes. Induction on class. I am still reading Arturo's proof. let see which one will be easier. – Bhaskar Vashishth Sep 25 '14 at 19:34
  • @DerekHolt by Artuto's Proof only significant thing for me which I could find out was that in his theorem he proves that product of two elements of finite order is of finite order, which in my case tells that $G$ has all its elements of finite order. But $G$ can still be infinite? what I am missing, to see $G$ is finite?? – Bhaskar Vashishth Sep 25 '14 at 20:35
  • That's an easy induction on the class. The torsion subgroup true is finite for abelian groups and, for nonabelian groups, by induction both $G/Z(G)$ and $Z(G)$ are finite, so $G$ is finite. – Derek Holt Sep 25 '14 at 21:28

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