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Let ‎$G‎‎$ ‎be a‎ ‎finite ‎simple ‎group. ‎Is ‎it ‎possible ‎to ‎find ‎two ‎‎ distinct ‎proper non-trivial ‎subgroups ‎‎‎$H_1‎‎$ and ‎‎‎‎$‎‎H_2$ ‎of ‎‎$G‎‎$ ‎‎ such ‎that ‎‎$\langle H_1 , H_2\rangle‎\lneq G‎‎‎$ ‎and ‎for ‎each ‎maximal ‎subgroup ‎‎$M‎‎$ ‎of ‎‎$G‎‎$‎, ‎ if ‎$‎‎H_i\nleq M$, ‎then ‎‎$‎‎H_j\leq M‎$ ‎‎‎(‎‎$i , j\in \{1,2\}‎‎$‎)?‎

If‎ it is ‎dificult ‎to ‎answer ‎the ‎question in the class of finite simple groups, ‎ ‎could ‎we ‎answer ‎it ‎in ‎certain ‎classes ‎of ‎finite ‎simple ‎groups‎?‎ (such as alternating groups, minimal simple groups, ...). ‎ ‎ Many thanks. ‎

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    $\begingroup$ It doesn't seem particularly likely. Why do you think it might be true? Have you checked whether it is true in $A_5$? $\endgroup$ – Derek Holt Apr 15 '13 at 9:21
  • $\begingroup$ Dear professor Derek Holt, I guess it might be true but unfortunately until now I don't have any example or counterexample for it! $\endgroup$ – shankfei Apr 15 '13 at 10:04
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    $\begingroup$ Well I suggest you check whether it is true in $A_5$, using a computer if necessary. You could assume that $H_1$ and $H_2$ have prime order i.e. 2,3 or 5. $\endgroup$ – Derek Holt Apr 15 '13 at 11:40
  • $\begingroup$ Many yhanks Professor Holt! I verify it with your example. $\endgroup$ – shankfei Apr 15 '13 at 12:35
  • $\begingroup$ (Someone should say "non-abelian". And I think that someone just became me.) $\endgroup$ – user1729 Aug 16 '13 at 9:24
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As mentioned in the comments, this is not possible in $A_5$. Suppose $H_1$ and $H_2$ are subgroups of $A_5$ with the desired property. The point stabilizers in $A_5$ are maximal, so each of them contains either $H_1$ or $H_2$. Thus one of the $H_i$ is contained in at least three point stabilizers, which implies that $H_i$ is trivial.

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  • $\begingroup$ thanks for your answer! But how do you conlude that one of the $H_i$ is contained in at least three point stabilizers? $\endgroup$ – shankfei Aug 25 '13 at 17:59
  • $\begingroup$ @shankfei: Basically it's the pigeonhole principle. There are five point stabilizers. If both $H_i$ are contained in at most two point stabilizers, then we have at most four point stabilizers containing some $H_i$. But we assume that each point stabilizer contains some $H_i$. $\endgroup$ – Mikko Korhonen Aug 25 '13 at 18:15
  • $\begingroup$ ! excuse me! We know that the intersection of each three point stabilizers in $A_5$, contains an involution. So at the end of your answer how do you coclude that $H_i$ is trivial? $\endgroup$ – shankfei Aug 25 '13 at 20:01
  • $\begingroup$ @shankfei: Which involution? Permutations of cycle type $(ab)$ are odd. The involutions in $A_5$ are of cycle type $(ab)(cd)$. $\endgroup$ – Mikko Korhonen Aug 25 '13 at 20:25
  • $\begingroup$ @ Mikko Korhonen you are right! Many thanks for your nice answer! $\endgroup$ – shankfei Aug 26 '13 at 9:08

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