Let $G$ be a finite nilpotent group. If G has a normal series $$<e>=N_0\leq N_1\leq ... \leq N_k=G$$ such that $G_i\lhd G$ and $G_{i+1}/G_i$ is cyclic for all $i$, then G is called supersolvable.
I want to prove that a finite nilpotent group $G$ has a normal series above whose factors are cyclic. How should I construct desired series?
A finite nilpotent group is the direct product of its Sylow subgroups, so it suffices to show in the case when $G$ is a $p$-group, say of order $p^n$.
To this end, we argue by induction. It is trivial when $n=1$, so let $n>1$. Then $G$ has non-trivial center (a common fact about $p$-groups), so we can choose an element $x\in Z(G)$ of order $p$. Let $N=\langle x\rangle$, and note $N$ is normal in $G$ since $N\leqslant Z(G)$. Then $G/N$ has order $p^{n-1}$, so by induction has a super-solvable series. Lifting back gives a supersolvable series for $G$, and we are done.