Finding a particular chief series for a supersolvable group Let $G$ be a finite supersolvable group.  Since it is supersolvable it has (by definition) some normal series $1=N_0 \le N_1 \le \cdots \le N_n=G$ such that each factor $N_i/N_{i+1}$ is cyclic.  Also since $G$ is finite, we know that it has a chief series.
How do we go about finding a chief series of $G$ for which the factors are cyclic?

First start with your normal series $N_0,\ldots, N_n$ that has cyclic quotients.  Since $N_i/N_{i-1}$ is a subgroup of $G/N_{i-1}$, it's solvable, so its minimal normal subgroups are elementary abelian $p$-groups.  In particular, since $N_i/N_{i-1}$ is cyclic, its minimal normal subgroups are cyclic of order $p$.
In order to have a chief series, you need each $N_i/N_{i-1}$ to be a minimal normal subgroup of $G/N_{i-1}$ (this is if and only if).  So, pick an $i$, and let $p$ be a prime such that $p^2$ divides $[N_i:N_{i-1}]$, assuming that such a $p$ exists.  Take a cyclic $M/N_{i-1}\lhd N_i/N_{i-1}$ of order $p$, and by the lattice theorem you have $N_{i-1}\lhd M \lhd N_i$ in $G$.  We have a new normal series: 
$$
\begin{array}{ccccccccccccccc}
\\
N_0 & \lhd & N_1 & \lhd & \cdots & \lhd & N_{i-1} & \lhd & M & \lhd & N_i & \lhd & \cdots & \lhd & N_n \\
\overset{||}{} & & \overset{||}{} &  &  &  & \overset{||}{} &  & \overset{||}{} &  & \overset{||}{} &  &  & & \overset{||}{} \\
M_0 & \lhd  & M_1 & \lhd & \cdots & \lhd & M_{i-1} & \lhd & M_i & \lhd & M_{i+1} & \lhd & \cdots & \lhd & M_{n+1}
\end{array}
$$
Repeat this process until no further refinements can be made.  By construction, each quotient $M_{i}/M_{i-1}$ is cyclic of prime order, and is a minimal normal subgroup of $G/M_{i-1}$, so the final refinement is a chief series of $G$.
