Let $ G $ be a group and $ N \lhd G $. Why is $G $ supersoluble if $ N $ is cyclic and $ G/N $ is supersoluble?

Question

Let $ G $ be a group and $ N \lhd G $. If $ N $ and $ G/N $ are soluble, then $ G $ is a soluble group. But it is not true for supersoluble groups. Why is $G $ supersoluble if $ N $ is cyclic and $ G/N $ is supersoluble?

Answer 1

Since $G/N$ is supersolvable, there is a normal series $$ \{N\} = H_0 \triangleleft H_1 \triangleleft \cdots \triangleleft H_{s-1} \triangleleft H_s = G/N $$ with all quotients $H_{i+1}/H_i$ cyclic.

By the correspondence theorem the normal subgroups of $G/N$ correspond to normal subgroups of $G$ containing $N$, so each $H_i$ is uniquely determined by some normal subgroup $N_i\supseteq N$ as $H_i = N_i/N$. By the third isomorphism theorem the quotients are $$ H_{i+1}/H_i = (N_{i+1}/N)/(N_i/N) \cong N_{i+1}/N_i. $$ So the successive quotients of the $N_i$ are again cyclic. Can you take it from here?