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Let $p$ a prime number. Let $G$ be a $p$-group finitely generated, say, $G = \left<x_1, x_2, \ldots, x_m \right>$. Denote by $\gamma_i(G)$ the $ith$ term of the lower central series of $G$. Then, is true that $\gamma_i(G)$ is generated by a number of elements that depends only $m$ and $i$?

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In general, we have that if $G$ is a nilpotent and finitely generated group then, all subgroups of $G$ there are finitely generated. More, if $G = \left<x_1, \ldots, x_m \right>$ is finitely generated then $\gamma_i(G)$ is the normal closure of the subgroup of $G$ generated by all the commutators in the form $[b_1, \ldots, b_i]$, where $b_j \in \{x_1, \ldots, x_m\}, 1 \leq j \leq i$. But even with this information I could not still conclude my question is whether true or not. –  user59969 Apr 11 '13 at 15:20
    
You may want to check here: math.stackexchange.com/questions/40985/… , The nilpotency class $\,c\,$ seems to be key in this problem, but of course you can always argue that $\,c\le m\,$ ... –  DonAntonio Apr 11 '13 at 16:10
    
Are you assuming that $G$ is finite? –  Derek Holt Apr 11 '13 at 18:39
    
I'm not sure whom your asking, @DerekHolt, but the OP's title states "finite p-group" –  DonAntonio Apr 11 '13 at 18:59
    
@DerekHolt Yes. I supose $G$ a finite group. –  user59969 Apr 12 '13 at 22:11

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I think the answer is no. Let $G$ be a wreath product of a cyclic group of order $p$ with a cyclic group of order $p^k$. Then the derived group $\gamma_2(G)$ is elementary abelian of rank $p^k-1$. There is of course a bound as a function of $m$, $i$ and the nilpotency class $c$.

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