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.

  • 2
    $\begingroup$ I think that $C = A \cap B$ is self-normalizing in $G$ in your situation. Since $G = AB$ and $A,B$ are proper, we know that $A$ and $B$ are not $G$-conjugate. But $N_{G}(A \cap B)$ permutes the maximal subgroups which contain $A \cap B$, so must normalize both $A$ and $B.$ Since $A$ and $B$ are each maximal but not normal, it follows that $N_{G}(A \cap B) \leq A \cap B.$ $\endgroup$ – Geoff Robinson Jul 20 '19 at 17:10
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    $\begingroup$ @JohnJiang: Assume the existence of $g \in G$ such that $A=gBg^{-1}$. Now $g=ab$ with $a \in A$ and $b \in B$ because $G=AB$. So $A = abBb^{-1}a^{-1} = aBa^{-1}$, $A=a^{-1}Aa = B$. It follows that $A=B=G$, contradiction with $A,B$ proper subgroups. $\endgroup$ – Sebastien Palcoux Jul 20 '19 at 19:24
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    $\begingroup$ @JohnJiang : Sebastian Palcoux answered your question, but here is a different explanation : If $G = AgAg^{-1}$, then $G = Gg = AgA.$ If $g \in A$ we have $G = A$, a contradiction. If $g \not \in A$, then $A$ is disjoint from $AgA$, contrary to $G = AgA$. $\endgroup$ – Geoff Robinson Jul 21 '19 at 10:04
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    $\begingroup$ I don't know of such a classification but the original memoir by Liebeck, Praeger and Saxl often determines if $A\cap B$ is maximal in $A$ and $B$. This is necessary for your condition. $\endgroup$ – Michael Giudici Jul 21 '19 at 10:48
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    $\begingroup$ There is a column on the tables which indicates when $A\cap B$ is maximal in $A$ and $A$ is almost simple. Similarly, for $B$. There is further information hidden throughout the proof for each individual factorisation. $\endgroup$ – Michael Giudici Jul 22 '19 at 10:17

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.

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