The Details:
A normal subgroup $N$ of a group $G$, written $N\unlhd G$, is a subgroup $N\le G$ such that for all $g\in G$, we have $$gN=Ng.$$
A chief series of a group $G$ is a finite set of normal subgroups $N_i\subseteq G$ such that
$$1=N_0\unlhd N_1\unlhd \dots\unlhd N_n=G$$
and each chief factor $N_{i+1}/N_i$ is a minimal normal subgroup of $G/N_i$.
A simple group $G$ is a nontrivial group whose only normal subgroups are the trivial group and $G$ itself.
A supersolvable group $G$ is a group with a normal series
$$1=N_0\unlhd N_1\unlhd N_2\unlhd \dots\unlhd N_n=G$$
such that each factor $N_{i+1}/N_i$ is cyclic and $N_i\unlhd G$.
The Question:
According to this answer, if a chief factor of a group $G$ is not simple, then $G$ is not supersolvable. Is this true? If not, then what is needed to make $G$ not supersolvable?
Thoughts:
Here is the contrapositive:
If $G$ is supersolvable, then each chief factor of $G$ is simple.
That might be easier to prove.
I'm not sure it's true. (I couldn't find the theorem anywhere for a start.) The reason I think this is that cyclic groups need not be simple. The smallest counterexample is
$$H\unlhd \Bbb Z_4$$
for $H\cong \Bbb Z_2$.
It could be that the extra stipulation that "the chief factor $N_{i+1}/N_i$ in question is not cyclic" that, together with it being not simple, is needed to conclude that no normal series exists with cyclic chief factors (so, in other words, the group is not supersolvable); not just the given one. This makes sense to me intuitively but I'm lost as to how to prove it.
Please help 🙂
Best Answer
If the supersolvable group is finite, then you can refine its normal series with cyclic factors to a chief series, which will also have cyclic factors, so the chief factors are simple.
In an infinite supersolvable group, at least one of the cyclic factors must be infinite, but then I don't think the group has a chief series, so I guess the claim is still true.