$G$ is supersolvable if $G_{i+1}/G_i$ is cyclic for every $i\geq 0$ where $G_i$ is a normal sequence of $G=G_n$.
According to the lecture note I read, it says that in the following short exact sequence,
$1\to G_{i+1}/G_i\to G/G_i\to G/G_{i+1}\to 1$
supersolvable group implies that we have successive cyclic extensions of $\{1\}$ for all $i$. From my understanding, this means that for every mapping we have above, kernel of it is cyclic. So following this new definition, we have to have that $1$, $G_{i+1}/G_i$, $G/G_i$ should be cyclic (since short exact sequence means that the image of mapping in the former is equal to the kernel of the latter). The original definition of supersolvable demands that $G_{i+1}/G_i$ is cyclic, but not necessarily $G/G_i$. Where did I have a mistake?
Also, I have another question.
If $G$ is supersolvable, any central extension of $G$ is also super solvable. Hence nilpotent group is supersolvable group.
How do I see this? Is there any good reference to it? I would greatly appreciate any help.
Every finitely generated nilpotent group is supersolvable. It is not true that any nilpotent group is supersolvable.
In fact, all supersolvable groups are finitely generated. So take the free nilpotent group on an infinite set as a counterexample. Or any uncountable nilpotent group, which are known to exist, if we assume the generalized continuum hypothesis (CH), for instance.