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Let $G$ be a finite group with trivial Frattini subgroup (i.e. the intersection of all maximal subgroups of $G$ is trivial) such that $G$ has more than two maximal subgroups and at least one of its maximal subgroup isn't of prime order. Do there exist two distinct non-trivial subgroups $H_1$ and $H_2$ of $G$ such that

  • $H_1$ and $H_2$ are contained in a common maximal subgroup $M$
  • For each maximal subgroup $M'$ of $G$, either $H_1 \leq M'$ or $H_2 \leq M'$ (or both)?

If it is hard to answer the question in general, can we answer it for certain classes of finite groups (finite simple groups, symmetric groups,…)?

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  • $\begingroup$ I think that you must omit (or both) in the question, because if for all maximal subgroups this occur then clearly the frattini subgroup is non-trivial and so the answer is NO! $\endgroup$
    – majid arezoomand
    Commented Dec 8, 2012 at 20:59
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    $\begingroup$ @higwain: When you change the question so that one of the answers becomes incorrect, then you should do the change in a way that makes it clear and obvious that you changed the question. Here you have added the condition "at least one of its maximal subgroups isn't of prime order", which makes the answer that had already been posted by majid arezoomand look as if majid didn't read your question carefully enough; you have, in essence, made it so that instead of it looking like an omission on your part, it looks like an error on his part. $\endgroup$
    – Arturo Magidin
    Commented Dec 8, 2012 at 23:46

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The alternating group $A_5$ shows that this cannot hold in general. This group has five maximal subgroups of index $5$, the point stabilizers. If the stated condition holds, then at least three of these would have to contain either $H_1$ or $H_2$. But in this group, the intersection of any three point stabilizers is trivial.

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The answer in general is "No". For example consider the dihedral group $D_{10}$ of order 10. Then the frattini subgroup of $D_{10}$ is trivial and has 6 maximal subgroups that one of them is of order 5 and the others are of order 2. Now clearly, the answer is "NO".

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  • $\begingroup$ Note: The OP has changed the question to add the condition that at least one maximal subgroup not have prime order. $\endgroup$
    – Arturo Magidin
    Commented Dec 8, 2012 at 23:47
  • $\begingroup$ I apologise for the edition that held in the question after majid arezoomand's answer! $\endgroup$
    – sebastian
    Commented Dec 9, 2012 at 5:47
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It seems that there are some examples. Let $G=D_{10}\times S_2$. Then the Frattini subgroup of $G$ is trivial, and the maximal subgroups of $G$ are (this can be checked by GAP):

$M_1=\langle(6,7),(1,2,3,4,5)\rangle$,

$M_2=\langle (2,5)(3,4),(1,2,3,4,5)\rangle$,

$M_3=\langle(2,5)(3,4)(6,7),(1,2,3,4,5)\rangle$,

$M_4=\langle(6,7),(2,5)(3,4)\rangle$,

$M_5=\langle(6,7),(1,4)(2,3)\rangle$,

$M_6=\langle(6,7),(1,2)(3,5)\rangle$,

$M_7=\langle(6,7),(1,5)(2,4)\rangle$, and

$M_8=\langle(6,7),(1,3)(4,5)\rangle$.

Now consider $H_1=\langle(6,7)\rangle$ and $H_2=\langle(1,2,3,4,5)\rangle$.

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  • $\begingroup$ I think that the classification of such groups is a good problem $\endgroup$
    – majid arezoomand
    Commented Dec 9, 2012 at 10:06
  • $\begingroup$ @ majid arezoomand! many thanks for your examples, I think so! $\endgroup$
    – sebastian
    Commented Jan 22, 2013 at 17:53

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