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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 $i$?

Alexander Gruber
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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. –  Apr 11 '13 at 15:20
  • You may want to check here: http://math.stackexchange.com/questions/40985/bound-for-the-rank-of-a-nilpotent-group , 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. –  Apr 12 '13 at 22:11
  • @DonAntonio. Because I can assume that $c \leq m$? –  Apr 12 '13 at 22:12
  • We have by hypothesis $G$ is a $p$-group (then is nilpotent). So, we have that the quotient $\gamma_i(G)/\gamma_{i+1}(G)$ is (finitely) generated by a function that dependent only $i$ and $m$ (because $G$ is $m$-generated). How $G$ is nilpotent, then really exists $c$ that $\gamma_{c+1}(G) = 1$. Then, we have that for all $i \leq c$ then $\gamma_i(G)$ is generated by a number of elements that depends only $i$ and $m$ (because the class of finitely generated groups is closed to a extention). And, for $i > c$ we have that $\gamma_i(G) = 1$. I asking: This argument is true? –  Apr 12 '13 at 22:17
  • @AgenorAndrade, did you ask me why can you assume $,c\le m,$ ? If you did: because of a simple inductive argument and taking into account that a group of order $,p^n,$ has , for any $,0\le k\le n,$, a normal subgroup of order $,p^k,$ (and thus we can go to quotients and etc.) – DonAntonio Apr 13 '13 at 01:22

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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 $c$.

Derek Holt
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