Maximal factorization of finite simple groups and no extra intermediate The book The maximal factorizations of the finite simple groups and their automorphism groups (by Martin W. Liebeck, Cheryl E. Praeger and Jan Saxl) provides a classification of all the triples $(G,A,B)$ such that:  


*

*$G$ is a finite simple group,  

*$A$ and $B$ are maximal subgroups of $G$,   

*$AB=G$.  


Question: What is the classification with the following additional assumption?  


*

*No extra intermediate: if $(A \cap B) < H < G$ then $H \in \{A,B\}$.   


Remark: by GAP computation, below is the classification for $|G|<2\cdot 10^6$:    


*

*$(A_6, \ A_5, \ A_5)$,  

*$(A_8, \ A_7, \ 2^3:A_1(7))$,  

*$(M_{12}, \ M_{11}, \ M_{11})$,   

*$(C_2(2^2), \ A_1(2^4):2, \ A_1(2^4):2)$,  

*$(C_3(2), \ A_8:2, \ ^2A_2(3^2):2)$.   



This post was inspired by an exchange with Pablo Spiga.
 A: I'm not sure how much it contributes to answering the question, but it is true that if $G$ is a non-Abelian finite simple group which admits the factorization 
$G = AB$ with $A,B$ maximal subgroups with the additional property that $A \cap B$ is properly contained in no proper subgroup other than $A$ or $B$, then one of the following occurs:
i) $|A \cap B|$ has at least two distinct prime divisors.
ii) $A \cap B$ is  a Sylow $2$-subgroup of $G.$
iii) $A \cap B$ is  a Sylow $3$-subgroup of $G.$
Proof: Suppose that i) does not hold. Then $A \cap B$ is a (non-trivial) $p$-group for some prime $p.$ 
(As I remarked in comments) $A \cap B$ is self-normalizing in $G$ because $N_{G}(A \cap B)$ permutes the subgroups containing $A \cap B$ by conjugation, but $A$ and $B$ are not conjugate (since $G = AB$). Hence $N_{G}(A \cap B) \leq N_{G}(A) \cap N_{G}(B) = A \cap B$ since $G$ is simple and $A,B$ are maximal.
Let $P$ be a Sylow $p$-subgroup of $G$ containing $A \cap B.$ If $A \cap B < P,$ then $N_{P}(A \cap B) > A \cap B$, a contradiction. Hence $A \cap B = P$ is a Sylow $p$-subgroup of $G$ for some prime $p$.
If $p \geq 5,$ then a theorem of G. Glauberman implies that $G \neq O^{p}(G)$ since $N_{G}(P) = P,$ contrary to the simplicity of $G$.  Hence $p \leq 3,$ and either ii) or iii) holds.
