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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.