Finite groups with a certain Frattini subgroup Let $G$ be a finite group different from a cyclic $p-$group 
and $\Phi(G)=M_i\cap M_j$, where $M_i$ and $M_j$ are two arbitrary distinct
maximal subgroups of $G$ and $i, j \geq 1$. Is it possible to characterize such $G$? 
($\Phi(G)$ denotes the Frattini subgroup of $G$, 
which is the intersection of all maximal subgroups of $G$).
Examples: (a) Cyclic groups of order $p^nq^m$ are examples of such groups, 
where $p, q$ are distinct primes. Also quaternion group of order 8 is another example.
(b) Alternating groups $A_n; n\geq 5$, have no such property because 
as well as we know, there are different non-trivial intersections of any 
two point stabilizers in $A_n$ and on the other hand $\Phi(A_n)=1$.
Remark: According to an exercise we know that if $p$ is a 
prime divisor of $|G|$, ${\rm Syl}_p(G)$ is the set of all Sylow 
$p-$subgroups of $G$ and $P\in {\rm Syl}_p(G)$ then $P\nleq \Phi(G)$.
Combining this fact with the condition on $G$ implies that 
each Sylow subgroup of $G$ contained in exactly one 
maximal subgroup of $G$. May be this point can help.
Thank's a lot!
 A: I interpret your question as saying that any two maximal subgroups of $G$ intersect in $\Phi(G).$ I only consider the case $\Phi(G) =1,$ which is no real loss of generality, because $G/\Phi(G)$ has the same property.
If $G$ is nilpotent, then $G$ is Abelian of square free exponent as $\Phi(G) = 1$. If $G$ is a $p$-group, then $G$ is either cyclic or elementary Abelian of order $p^{2}.$ If $G$ is nilpotent but not a $p$-group, and $\Phi(G) = 1,$ then $G$ must be cyclic of order $pq$ for distinct primes $p$ and $q.$
Suppose then that $\Phi(G) = 1,$ but that $G$ is not nilpotent. Then $G$ has a maximal subgroup $H$ with $H \not \lhd G.$ Then $H = N_{G}(H),$ and $H \cap H^{g} = 1$ for all $g \in G  \backslash H.$ By Frobenius's theorem, there is $K \lhd G$ with $G = KH$ and $K \cap H = 1.$ Also ${\rm gcd}(|H|,|K|) = 1.$ Since $H$ is a maximal subgroup of $G,$ no proper non-trivial subgroup of $K$ is $H$-invariant. 
Hence $K$ is an elementary Abelian $q$-group for some prime $q.$ Now $H$ must have a unique maximal subgroup, for if $S$ and $T$ are different maximal subgroups of $H,$ then $KS$ and $KT$ are distinct maximal subgroups of $G$ with intersection $K \neq 1.$ Hence $H$ is a cyclic $p$-group for some prime $q \neq p.$ In particular, $G$ is solvable. 
(Later edit: In fact $H$ has order $p$: I thought I had an argument for this earlier, but realised it was faulty: I think this one is OK. The hypotheses imply that every element of order $p$ is contained in a unique maximal subgroup of $G$: if $|H| >p,$ then $ \Omega_{1}(H) (\neq 1,H) $ is contained in the maximal subgroup $H$, and in a maximal subgroup containing $K \Omega_{1}(H),$ and these are different).
Back to the general case (maybe $\Phi(G) \neq 1),$ we may conclude that $G$ is solvable, since $\Phi(G)$ is nilpotent. Then $|G|$ must have the form $p^{a}q^{b}$ with $p,q$ distinct primes ( but possibly $ab = 0),$ for if $G$ were divisible by $3$ different primes $p,q$ and $r,$ then a Sylow $p$-subgroup of $G$ would be contained in a Hall $q^{\prime}$-subgroup and a Hall $r^{\prime}$-subgroup, and no maximal subgroup of $G$ contains both of these.
