4
$\begingroup$

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. ‎

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.