Let $G$ be a finite group and $N$ be a normal subgroup of $G$. Is the following true :

(1) G is nilpotent if and only if $N$ and $G/N$ are nilpotent.

(2) Let $Z(G)$ be the center of $G$. Then $G$ is nilpotent if and only if there exists a subgroup $A$ of $Z(G)$ such that $G/A$ is nilpotent.

For (1), first I think that this characterization is true for solvable group, but I am not sure if it holds for nilpotent. However, if this holds for finite nilpotent group, then the class of finite nilpotent group and finite solvable group is the same, which seems deceiving. Anyway, it is true that any subgroup and quotient of nilpotent group are nilpotent (True even $G$ is not finite). So it is left to determine if the converse statement holds. The problem is I cannot find a counter example to the statement.

For (2), I think the if $G$ is nilpotent, then the trivial subgroup $\{e\}$ perfectly serve the role of $A$. So, again, it is left to determine if the converse holds. I notice that for any subgroup $A$ of $Z(G)$, $A$ is normal in $G$. So (2) is somehow a more constraint version of (1) in the sense that $N$ in (1) is both normal and contain in the center of $G$. I guess that it might possible that this statement holds. Is my forecast true ? Any hint in proving this please.

Thank you in advanced for help.

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    $\begingroup$ I wonder why this was downvoted. $\endgroup$ – Myself Dec 13 '15 at 19:42

A counterexample for (1) is very easy to find. I mean, take the smallest non-nilpotent group you can think of. (Hint: it is the smallest non-abelian group.)

For (2), if there is $A \le Z(G)$ such that $G/A$ is nilpotent, then $$ G/Z(G) \cong \frac{G/A}{Z(G)/A} $$ is also nilpotent, as a quotient of a nilpotent group. How you continue depends on your definition of nilpotent group, but you may want to think in terms of the upper central series.

  • $\begingroup$ For (1),I guess that you recommend me to look at $G = S_3$ and $N = A_3.$. For (2), the definition of nilpotent I use is the one involving the ascending central series (preimage of center under canonical projection). I try to use the fact above for a while but I am not sure how this will confirm that for some $n$, the ascending series will reach $G$. I guess that the $n$ that make the ascending series of $\frac{G/A}{Z(G)/A}$ reach this group might confirm that the same $n$ make the ascending central series of $G$ reach $G$. Can you point out more for (2), Thank you very much. $\endgroup$ – user117375 Dec 13 '15 at 19:57
  • $\begingroup$ @user117375, in the meantime you probably got there yourself, anyway, $G/Z(G)$ nilpotent implies $G$ nilpotent, as the upper central series of $G$ starts with $Z(G)$, and then continues with the preimages in $G$ of the terms of the upper central series of $G/Z(G)$. $\endgroup$ – Andreas Caranti Dec 14 '15 at 11:35

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