Let $q=p^k$ for some prime $p$, and let $GL\_n(\mathbb{F}\_q)$ be the group of invertible matrices over the finite field of $q$ elements. If $\pi$ is the set of primes not equal to $p$, does $GL\_n(\mathbb{F}\_q)$ have a Hall $\pi$-subgroup? I've looked through a lot of the literature but cannot find an explicit answer to this very natural question.



Hall subgroups of general linear group
If ${\rm GL}(n,q)$ has a Hall $p^{\prime}$-subgroup, then by Dedekind's Lemma, all its parabolic subgroups would have such a Hall subgroup. Now for $n >2,$ ${\rm GL}(n,q)$ has a parabolic subgroup $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,q) \times {\rm GL}(n-2,q).$ Now if $P$ has a Hall $p^{\prime}$-subgroup, so does $P/U,$ and so does every normal subgroup of $P/U.$ Hence whenever ${\rm GL}(2,q)$ does not have a Hall $p^{\prime}$-subgroup, the same is true for ${\rm GL}(n,q)$ for any $n \geq 3.$ This eliminates such a Hall subgroup for most (but not all) values of $q.$ A similar argument shows that whenever ${\rm GL}(3,q)$ does not have a Hall $p^{\prime}$-subgroup, the same is true for ${\rm GL}(n,q)$ for any $n \geq 4.$ From these two cases it is possible to list the rare cases where such a Hall subgroup exists (though it is necessary to check ${\rm GL}(4,2)$ as well when $q = 2$).


Later edit: To take care of the case $q =p >11,$ it was known to Galois that ${\rm PSL}(2,p)$ has no subgroup of index $p$ when $p >11,$ which implies that ${\rm GL}(2,p)$ has no Hall $p^{\prime}$-subgroup.


