Finite groups with trivial Frattini subgroup Let $G$ be a finite group with trivial Frattini subgroup (i.e. the intersection of all maximal subgroups of $G$ is trivial) such that $G$ has more than two maximal subgroups and at least one of its maximal subgroup isn't of prime order. 
Do there exist two distinct non-trivial subgroups $H_1$ and $H_2$ of $G$ such that 


*

*$H_1$ and $H_2$ are contained in a common maximal subgroup $M$ 

*For each maximal subgroup $M'$ of $G$, either $H_1 \leq M'$ or $H_2 \leq M'$ (or both)?


If it is hard to answer the question in general, can we answer it for certain classes of finite groups (finite simple groups, symmetric groups,…)?
 A: The alternating group $A_5$ shows that this cannot hold in general. This group has five maximal subgroups of index $5$, the point stabilizers. If the stated condition holds, then at least three of these would have to contain either $H_1$ or $H_2$. But in this group, the intersection of any three point stabilizers is trivial. 
A: The answer in general is "No". For example consider the dihedral group $D_{10}$ of order 10. Then the frattini subgroup of $D_{10}$ is trivial and has 6 maximal subgroups that one of them is of order 5 and the others are of order 2. Now clearly, the answer is "NO".
A: It seems that there are some examples. Let $G=D_{10}\times S_2$. Then the Frattini subgroup of $G$ is trivial, and the maximal subgroups of $G$ are (this can be checked by GAP):
$M_1=\langle(6,7),(1,2,3,4,5)\rangle$,
$M_2=\langle (2,5)(3,4),(1,2,3,4,5)\rangle$, 
$M_3=\langle(2,5)(3,4)(6,7),(1,2,3,4,5)\rangle$, 
$M_4=\langle(6,7),(2,5)(3,4)\rangle$, 
$M_5=\langle(6,7),(1,4)(2,3)\rangle$, 
$M_6=\langle(6,7),(1,2)(3,5)\rangle$,
$M_7=\langle(6,7),(1,5)(2,4)\rangle$, and 
$M_8=\langle(6,7),(1,3)(4,5)\rangle$. 
Now consider $H_1=\langle(6,7)\rangle$ and $H_2=\langle(1,2,3,4,5)\rangle$.
