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Aug 22, 2020 at 8:41 comment added Geoff Robinson @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.
Aug 21, 2020 at 22:30 comment added LSpice 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 ….)
Aug 21, 2020 at 22:28 history edited LSpice CC BY-SA 4.0
Link to article; \DeclareMathOperator; prime p -> prime q
Aug 21, 2020 at 19:02 answer added Geoff Robinson timeline score: 2
Aug 21, 2020 at 13:50 comment added Derek Holt 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!
Aug 21, 2020 at 12:55 history edited YCor CC BY-SA 4.0
fixed question
Aug 21, 2020 at 12:51 review First posts
Aug 21, 2020 at 14:28
Aug 21, 2020 at 12:49 history asked Benedict CC BY-SA 4.0