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How do I show that $G$ is nilpotent given that if $G$ is polycyclic and $G$ is residually finite p-group?

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closed as off-topic by Derek Holt, colormegone, user26857, zz20s, Michael Albanese Mar 12 '16 at 21:53

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Note that $H'H^p=\Phi(H)$. Thus you have $[G', G]\le \Phi(H) \le \Phi(G)$. That means $G/\Phi(G)$ is nilpotent; use this to show $G$ is nilpotent.

(Assuming $G$ is finite.)

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  • $\begingroup$ sorry, what is $\Phi(H)$? Is $\Phi(G)=G'G^p$? $\endgroup$ – BetaY Mar 12 '16 at 20:10
  • $\begingroup$ @JeremyH: the Frattini subgroup. And no, in general, that relation only holds for $p$-groups. $\endgroup$ – Steve D Mar 12 '16 at 20:11
  • $\begingroup$ why does this mean $G/\Phi(H)$ is nilpotent? I haven't learned the Frattini subgroup, so I am not too familiar with this. Hope it is alright if you could explain it a bit more? thank you! $\endgroup$ – BetaY Mar 12 '16 at 20:14
  • $\begingroup$ @JeremyH: I've shown that $[G', G]\le\Phi(G)$, so in the quotient $G/\Phi(G)$, the third entry in the commutator series vanishes. That's the definition of nilpotent. $\endgroup$ – Steve D Mar 12 '16 at 20:16
  • $\begingroup$ Right, I think I got this. how do I show that $G$ is nilpotent from there? $\endgroup$ – BetaY Mar 12 '16 at 20:20

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