3
$\begingroup$

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.

| cite | improve this question | |
$\endgroup$
  • 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ @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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.