I’m trying to prove that if every maximal subgroup of a finite group $G$ is nilpotent, then $G$ is solvable. I know already that if any two maximal subgroups of a finite group are conjugate, then it is cyclic, so it is solvable. So my question is:
If $G$ is a finite group such that every maximal subgroup is nilpotent, are any two maximal subgroups of $G$ conjugate?
I’m using the following definition of “nilpotent group”: $G$ has a lower central series terminating in the trivial subgroup after finitely many steps.
As a counterexample, let $G$ be any non-cyclic finite abelian group.
Since $G$ is abelian, every maximal subgroup of $G$ is abelian, hence nilpotent.
Since $G$ is a non-cyclic finite abelian group, $G$ has more than one maximal subgroup, no two of which are conjugate (in an abelian group, no two distinct subgroups are conjugate to each other).