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I am searching for a directly proof of the fact that: If $G$ is f.g. torsion-free nilpotent group, then every (nontrivial) $x \in G$, $x \notin G^{p}$ for all, but finitely many primes $p$

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So a finintely generated, torsion free, nipotent group, it is not possible that for all primes, $x = y^p$ for some $y$. Hmmm... – cactus314 Oct 12 '12 at 16:18
If $G$ is finitely generated torsion-free nilpotent, then so is $G/Z(G)$. So you can use induction on the class of $G$. – Derek Holt Oct 13 '12 at 11:15
I think of that. But I think we need to distinguish cases on whether $x \in Z(G)$ and $x \notin Z(G)$. In the case that $x \notin Z(G)$ we can use the induction hypothesis, but I can't continue the argument in the case that $x \in Z(G)$. – Dennis Oct 13 '12 at 13:17
It can be shown that, for primes $p$ larger than the class of the group, the set of $p$-th powers forms a subgroup. (I guess I'll need to prove that or find a reference.) If you accept that, then the elements of $Z(G) \cap G^p$ must be $p$-th powers of elements in $Z(G)$ (since $G/Z(G)$ is torsion-free), and the result follows from the abelian case. – Derek Holt Oct 13 '12 at 20:14
I think that I found a reference("Lectures notes on nilpotent groups", Baumslag). Thank you very much for your time. – Dennis Oct 14 '12 at 14:31

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