A finite solvable group $G$ which is not nilpotent and has at most $4$ maximal subgroups satisfies $G/\Phi(G) \cong S_{3},$ where $\Phi(G)$ is the Frattini subgroup, the intersection of all maximal subgroup of $G.$
I'm writing the answer in steps since the connection is unreliable. Suppose $G$ is solvable, not nilpotent, and has at most $4$ maximal subgroups. Suppose also that $\Phi(G) = 1,$ which is no loss of generality. Then $G$ has a maximal subgroup $M$ which is not normal. Then $M$ has at most $4$ conjugates, and there is at least one maximal subgroup of $G$ which is not conjugate to $M.$