It is provable, using transfer theory and Thompson's J-subgroup, that if a nonabelian finite simple group $G$ has a nilpotent maximal subgroup $M$, that $M$ must be a 2-subgroup. It then follows from Sylow theory that $M$ must be a Sylow 2-subgroup of $G$. (It also follows from Sylow theory that, if this is true, any Sylow 2-subgroup of $G$ will be maximal.)
It's not unusual, to my understanding, for group theory textbooks to present $PSL_{2}(\mathbb{F}_{17})$ as an example of a nonabelian finite simple group in which the Sylow 2-subgroups are maximal. But this can be generalized: $PSL_{2}(\mathbb{F}_{p})$ has maximal Sylow 2-subgroups whenever $p$ is prime, $p \geq 17$, and $p$ is 1 away from a power of 2 (which is to say that $p$ is a Fermat prime or a Mersenne prime). (Note that the only composite prime-power that is 1 away from a power of 2 is 9. 9 does not need to be included here because $PSL_{2}(\mathbb{F}_{9}) \cong A_{6}$ does not have maximal Sylow 2-subgroups. So weakening the condition by allowing powers of primes does not help here.)
My question is: if a nonabelian finite simple group has maximal Sylow 2-subgroups, it is necessarily isomorphic to one of the $PSL_{2}$ groups described above?

  • 1
    $\begingroup$ I think the answer can be deduced from some pretty heavy duty theorems (eg M. Aschbacher's $C(G,T)$-theorem). There may be more direct proofs available, $\endgroup$ Jun 16 '17 at 8:28

Yes this follows from work of Baumann and Thompson. See the paper `On finite insoluble groups with nilpotent maximal subgroups' by John Rose https://doi.org/10.1016/0021-8693(77)90301-5

In fact, these are the only examples of simple groups with nilpotent maximal subgroups.


This site is temporarily in read only mode and not accepting new answers.

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