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Let $C$ be either class of finite cyclic groups or class of finite nilpotent groups.

$\textbf{Q:}$ Why class of finite nilpotnent groups is not closed under extension?(i.e. Given $1\to K\to G\to H\to 1$ exact sequence with $K,H\in C$, then one deduces $G\in C$.) I do not see any obvious example.

$\textbf{Q':}$ Why class of finite cyclic groups does not have the following property? If $G$ is a finite group s.t. $N_1,N_2$ are subgroups of $G$ and $G/N_i\in C$(i.e. $G/N_i$ are finite cyclic), then $G/(N_1\cap N_2)$ is finite cyclic.)

Ref. Profinite Groups Chpt 2 by Luis Ribes.

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Hint: symmetric group $S_3$ is solvable and not nilpotent, but is the extension of a cyclic group by a cyclic group. Take $A_3$ the subgroup of even permtation, $S_3/A_3=S_2$.

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