Let $G$ be a simple group which has a maximal subgroup of the form $p:q$ (semidirect product of ${\Bbb Z}_{p}$ and ${\Bbb Z}_{q}$) where $p$ and $q$ are some primes and $q\vert(p-1)$. Also suppose that every Sylow $p$-subgroup of $G$ is ${\Bbb Z}_{p}$ and every Sylow $p$-subgroup is contained properly in exactly one maximal subgroup of $G$.

Can we get that $G\cong L_{2}(p)$.


1 Answer 1


The Mathieu group $M_{23}$ is a counterexample with $p=23$ and $q=11$.

But there are many (almost certainly infinitely many) counterexamples that are semilinear maximal subgroups of classical groups.

As John Shareshian pointed out, for $p=2^k+1$ a Fermat prime, there is an example with $G={\rm PSL}(2,2^k)$ and $q=2$. There are other examples with ${\rm PSL}(2,r)$, such as $r=25$ with $p=13$.

Then for $G={\rm PSL}(3,r)$ we get examples with $q=3$, including $r=3$, $p=13$; $r=5$, $p=31$; $r=7$, $q=19$; etc.

For $G={\rm PSL}(5,2)$ we have an example with $q=31$, $q=5$.

  • $\begingroup$ Maybe I am being very dense, but if $2^k+1$ is a Fermat prime, isn't $L_2(2^k)$ a counterexample to the question as stated? The Coxeter torus is a Sylow subgroup of prime order and is properly contained only in its dihedral normalizer. $\endgroup$ Jul 16, 2017 at 14:18
  • $\begingroup$ More generally, what about a Coxeter torus of prime order in a linear group of prime degree? $\endgroup$ Jul 16, 2017 at 14:21

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