5
$\begingroup$

$\DeclareMathOperator\PSL{PSL}$In [1] it was proved that

A finite nonsolvable group $G$ has three conjugacy classes of maximal subgroups if and only if $G/\Phi(G)$ is isomorphic to $\PSL(2,7)$ or $\PSL(2,2^q)$ for some prime $q$. This implies that, among finite simple groups, only only $\PSL(2,7)$ and $\PSL(2,2^q)$ have three conjugacy classes of maximal subgroups.

My question: I wonder if we can also find all finite simple groups with three conjugacy classes of maximal local subgroups.

A subgroup is a local subgroup if it is the normalizer of some nontrivial subgroup of prime power order. A proper local subgroup is a maximal local subgroup if it is maximal among proper local subgroups.

Maximal subgroups are not necessarily local, and maximal local subgroups are not necessarily maximal subgroups. I know that the three non-conjugate maximal subgroups of $\PSL(2,4)=A_5$ and $\PSL(2,7)$ are local respectively, but is it true that $\PSL(2,2^q)$ has three conjugacy classes of maximal local subgroups for each prime $q$? And how can I find all simple groups with such property?

Any help is appreciated!

Reference:

[1] Belonogov, V. A.: Finite groups with three classes of maximal subgroups. Math. Sb., 131, 225–239 (1986)

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Yes, the three classes of maximals subgroups of ${\rm PSL}(2,2^q)$ with $q$ prime consist of dihedral groups of orders $2(2^q \pm 1)$, and a group with structure $2^2:(2^q-1)$, and all of these are local. But there are other simple groups, such as $A_6$ and ${\rm PSL}(2,16)$, with exactly three classes of maximal subgroups that are local. I am afraid that finding them all would involve a lot of hard work on your part! $\endgroup$
    – Derek Holt
    Commented Aug 21, 2020 at 13:50
  • $\begingroup$ You twice referred to $\operatorname{PSL}(2, 2^q)$ "for some prime $p$", which I think was meant to be "… for some prime $q$"; I edited accordingly, as well as some other small changes. (Also, what does $\operatorname{PSL}(2, 2^q)$ mean? I would normally think that you are taking the quotient by the centre, but it's trivial ….) $\endgroup$
    – LSpice
    Commented Aug 21, 2020 at 22:30
  • 1
    $\begingroup$ @LSpice: ${\rm PSL}(2,2^{q})$ is just the same as ${\rm SL}(2,2^{q})$. As you say, the centre of the latter group is trivial, so this is consistent notation, but with some redundancy in this case. $\endgroup$
    – Geoff Robinson
    Commented Aug 22, 2020 at 8:41

1 Answer 1

Reset to default
2
$\begingroup$

If you want to use CFSG, I think this is doable (and may even be doable without CFSG if you use H. Bender's classification of finite groups with a strongly embedded subgroup, with some additional work).

For sporadic groups, is a matter of checking. In an alternating group $G$, there are three non-conjugate maximal local subgroups, $N_{G}(\langle (123) \rangle)$, $N_{G}( \langle (12)(34), (13)(24) \rangle )$ and $N_{G}(\langle (12345) \rangle)$, and for $n \geq 7$, it is easy to construct maximal local subgroups not conjugate to any of these.

For simple groups of Lie type of defining characteristic $p$, then for rank at last three, there are at least three conjugacy classes of maximal $p$-locals ( which are parabolics here) which are also non-conjugate maximal local subgroups. Also, (with a few exceptions), the normalizer of the maximal torus ,$T$, of the Borel is contained in a maximal local subgroup which is not conjugate to any parabolic.

Hence the real work is in dealing with simple groups of Lie type of defining characteristic $p$ and of rank at most $2$, and this should be manageable.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I have solved this problem recently. Thanks for your time and kindness! $\endgroup$
    – Benedict
    Commented Oct 13, 2020 at 12:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .