Can you help me and give me the proof of this statement please? And can you explain me why this statement is not true when $H$ is not a subgroup of $Z(G)$? Thank you very much
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2$\begingroup$ Sounds an awful lot like homework. Voting to close. $\endgroup$– Igor RivinCommented Jul 18, 2011 at 16:40
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2$\begingroup$ It CAN be true when H is not inside Z(G) that G is nilpotent if G/H is nilpotent but it is not GUARANTEED to be true. If H is contained in Z(G), it is guaranteed to be true that G is nilpotent if G/H is nilpotent. How to prove this may depend on the definition(s) you have used for nilpotent groups. $\endgroup$– Geoff RobinsonCommented Jul 18, 2011 at 17:33
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$\begingroup$ The question has been completely answered, even though the answer isn't (yet) marked as accepted, so it should be closed. $\endgroup$– Andreas BlassCommented Jul 19, 2011 at 16:06
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1 Answer
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Take a central series $$1 =G_0 \leq G_1 \leq \ldots \leq G_n=G/H$$ for $G/H$ and let $\overline{G}_i$ be the preimage of $G_i$ in $G$ via the natural projection $\pi \colon G \to G/H.$
Then $$1=H_0 \leq H \leq \overline{G}_1 \leq \ldots \leq \overline{G}_n=G$$ is a central series for $G$ (since $H$ is in the center of $G$), hence $G$ is nilpotent.
If $H$ is not a subgroup of $Z(G)$ the statement is clearly false: take $G=S_3$ and $H=\langle (123) \rangle$. Then $G/H \cong \mathbb{Z}/2 \mathbb{Z}$ which is clearly nilpotent, whereas $G$ is not nilpotent since its center is trivial.
answered Jul 18, 2011 at 15:40