Let $H$ and $K$ be two proper non-trivial subgroups of the alternating group $A_n; n\geq 5$. Then there exists a maximal subgroup $M$ of $A_n$ such that $H\not\leq M$ and $K\not\leq M$. To see this let $\Omega:=\{1,2,\cdots,n\}$ and as always ${\rm Supp}_{\Omega}(H):= \{\omega\in\Omega\mid \omega^{h}\neq \omega,$ for some $h\in H\}$. If there exists $l\in {\rm Supp}_{\Omega}(H)\cap {\rm Supp}_{\Omega}(K)$, put $M\!:={\rm Stab}_{A_n}(l)$, otherwise take $l\in {\rm Supp}_{\Omega}(H)\setminus {\rm Supp}_{\Omega}(K)$
and $t\in {\rm Supp}_{\Omega}(K)\setminus {\rm Supp}_{\Omega}(H)$ and put $M:={\rm Stab}_{A_n}\{l,t\}$. (May be other types of maximal subgroups of $A_n$, according to O'Nan-Scott Therorem, satisfy this property but I do not know).

By a GAP code I could check that this property is also true for a few finite non-abelian simple groups apart from the alternating groups, like ${\rm PSL}(2,7), {\rm PSL}(2,8), {\rm M}_{11},{\rm M}_{12}$.

Could we generalize this property for other types of finite non-abelian
simple groups?

Thank you very much!

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    $\begingroup$ Just an elementary comment. You only need to consider the case that $H$ and $K$ have prime order. $\endgroup$ – Geoff Robinson Nov 26 '14 at 22:11
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    $\begingroup$ The identical question was asked on math.stackexchange.com/questions/1004938 As I pointed out there, this property holds for all groups with a $2$-transitive permutation representation, which includes all of the examples that you have mentioned. $\endgroup$ – Derek Holt Nov 27 '14 at 1:03
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    $\begingroup$ @Derek Holt: Why does it hold for $2$-transitive actions? A $2$-set stabilizer in a $2$-transitive simple group need not be a maximal subgroup. $\endgroup$ – Peter Mueller Nov 27 '14 at 17:58

Yes, such an extensions to other simple groups exists: Let the finite group $S$ act primitively of degree $n$. We may assume that $H$ and $M$ are cyclic, generated by $x$ and $y$, respectively. We are done unless $x$ and $y$ have disjoint supports. But then one of these elements moves at most $n/2$ elements.

Guralnick and Magaard classify the primitive groups $G$ of degree $n$ which contain a nontrivial element which moves at most $n/2$ elements. The only examples where $G$ is simple have $G$ alternating, or are among the possibilities listed as cases 3(a), (b), (c), and (d) in Theorem 1. Note that $n$ is even in the cases (a), (b) and (d). We could have started with a primitive action of odd degree (by choosing an action on the cosets of a maximal subgroup of $S$ which contains a Sylow $2$-subgroup). So the only case left is 3(c). Here $S$ is the simple group $P\Omega^-_{2k}(2)$, and the only examples isomorphic to $S$ have a unique permutation degree $n=(2^k+1)(2^{k-1}-1)$. Again, apply the same trick as above: Let $p$ be a prime divisor of $n$, and $M$ be a maximal subgroup of $S$ which contains a Sylow $p$-subgroup of $S$. Then the corresponding primitive action of $S$ doesn't appear as an example in Guralnick/Magaard anymore.

Remark: Using Guralnick/Magaard, which heavily relies on the classification of the finite simple groups, probably is vast overkill for this question.

  • $\begingroup$ Dear Peter Mueller thank you so much for your nice answer! I do not know the particular case "S is simple" corresponds to What result in the paper. Could we apply the argument in my question statement about an orthogonal group over $\Bbb{F}_2$? (Is an orthogonal group over $\Bbb{F}_2$, $4-$transitive?) Please add more details if it is possible! $\endgroup$ – sebastian Nov 28 '14 at 12:14
  • $\begingroup$ I've enhanced the answer. $\endgroup$ – Peter Mueller Nov 28 '14 at 13:07

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