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It's actually possible to analyse the nilpotent case in a similar manner, and go a litle bit further than Arturo does. If $G$ is finite nilpotent with at most $4$ maximal subgroupswe may reduce to the case $\Phi(G) = 1,$ so that $G$ is an Abelian group of squarefree exponent. If $G$ ha a non-cyclic Sylow $p$-subgroup for som prime $p,$ then there are at least $1+p$ maximal subgroups whose index is a power of $p.$ If $G$ is not a $p$-group, there is at least one maximal subgroup of $G$ containing a Sylow $p$-subgroup of $G,$ and we have then exhibited at least $2+p$ maximal subgroups of $G,$ so $p=2.$ Even when $p=2,$ it easily follows that $G$ is the direct product of the form $A \times B,$ where $B$ is a cyclic $q$-group for some odd prime $q$ and $A$ is a $2$-group which may be generated by $2$ elements. The ultimate conclusion is that a non-Abelian finite nilpotent group $G$ which has at most $4$ maximal subgroups has one of the forms: A non-Abelian $3$-group which can be generated by $2$ elements; a group of the form $A \times B,$ where $A$ is a non-Abelian $2$-group generated by $2$ elements and $B$ is a cyclic $q$-group for some odd prime $q$ ($B$ may be trivial).

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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 $G.$

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 $M.$

Now $M = N_{G}(M)$ by maximality, as $M \not \lhd G.$ We have $[G:M] < 4,$ but we can't have $[G:M]= 2$ as $M$ is not normal. Hence $[G:M] = 3$ and $G/K \cong S_{3},$ where $K$ is the intersection of all $G$-conjugates of $M.$ But then by the isomorphism theorems, there are $4$ maximal subgroups of $G$ containg $K.$ these These are the three conjugates of $M,$ together with a normal subgroup $L$ of index $2.$ But this yields $K \leq \Phi(G)$ since $G$ has at most $4$ maximal subgroups. By assumption, $\Phi(G) = 1,$ so that $K = 1$ and $G \cong S_{3}.$

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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 $M.$

Now $M = N_{G}(M)$ by maximality, as $M \not \lhd G.$ We have $[G:M] < 4,$ but we can't have $[G:M]= 2$ as $M$ is not normal. Hence $[G:M] = 3$ and $G/K \cong S_{3},$ where $K$ is the intersection of all $G$-conjugates of $M.$ But then by the isomorphism theorems, there are $4$ maximal subgroups of $G$ containg $K.$ these are the three conjugates of $M,$ together with a normal subgroup $L$ of index $2.$ But this yields $K \leq \Phi(G)$ since $G$ has at most $4$ maximal subgroups. By assumption, $\Phi(G) = 1,$ so that $K = 1$ and $G \cong S_{3}.$

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