Let $G$ a f.g. torsion-free nilpotent group which has the property $Z_{i} \cap G^{p}= Z_{i}^{p}$, where p a prime, $Z_{i}$ is the i-th term of upper central series of $G$ and the $A^{p} = $. I would like to prove that each upper central quotient of $G/G^{p}$ is the corresponding upper central quotient of $G$ reduced modulo p. I think that it means $Z_{i+1}(G/G^{p})/ Z_{i}(G/G^{p}) \simeq (Z_{i+1}/Z_i)/(Z_{i+1}/Z_{i})^{p}$. Can anyone help me with the induction?
each upper central quotient of $G/G^{p}$ is the corresponding upper central quotient of $G$ reduced modulo p.
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group-theory
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0It is a lemma(5.6) in the paper of D.T.Wise and T.Hsu, "Ascending HNN extensions of polycyclic group are residually finite". – 2012-09-21
1 Answers
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What you wrote is not true in the group $\langle x,y,z \mid [x,z]=[y,z]=1, [x,y]=z^p \rangle$, in which $Z(G/G^p) = G/G^p$ (which is elementary abelian of order $p^3$), but $Z(G)/Z(G)^p$ has order $p$, since $Z(G) = \langle z \rangle$.
I think you are misinterprreting what Wise and Hsu mean, which admittedly is not very clearly expressed. I think they mean something like $Z_{i+1}G^p/Z_iG^p \cong Z_{i+1}/Z_{i+1}^pZ_i$.