# Finite simple groups with three conjugacy classes of maximal local subgroups

$$\DeclareMathOperator\PSL{PSL}$$In  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:

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

• 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! – Derek Holt Aug 21 at 13:50
• 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 ….) – LSpice Aug 21 at 22:30
• @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. – Geoff Robinson Aug 22 at 8:41
• @DerekHolt Thanks! I have solved this problem recently. I found the book by you and your co-authors on the maximal subgroups of the low-dimensional finite classical groups very helpful! – Benedict Oct 13 at 12:20

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.