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Let $G$=$P$$\rtimes$$\langle$$x$$\rangle$ a finite group where $P$ denotes a normal $p$-group, $\langle$$x$$\rangle$ is a cyclic $q$-group (p,q distinct primes) |$\langle$$x$$\rangle$|=$q^b$ but $x^q$$\in$$Z(G)$. Why can't I conclude that $G$ is a minimal non nilpotent group? The condition $x^q$$\in$$Z(G)$ guarantees that every proprer subgroup is direct product of its Sylow's subgroups. Isn't it? Where is my mistake? Thank you very much for every kind of suggestion!

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  • $\begingroup$ "Minimal nilpotent group"...in what family of groups?? For one, $\;P\le G\;$ is nilpotent, too... $\endgroup$ – DonAntonio Dec 2 '16 at 15:54
  • $\begingroup$ @DonAntonio Excuse me for the mistake: $G$ is not nilpotent but every proper non trivial subgroup is nilpotent. $\endgroup$ – Sibilla Dec 2 '16 at 15:57
  • $\begingroup$ The could be a non-nilpotent subgroup of the form $Q \rtimes \langle x \rangle$ for some $Q < P$. $\endgroup$ – Derek Holt Dec 2 '16 at 18:14
  • $\begingroup$ @DerekHolt That's right! Sorry for the stupid question. Thank you very much for the help! $\endgroup$ – Sibilla Dec 2 '16 at 18:16

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