2 Answered question about ${\rm PSL}(2,7).$
Well, ${\rm GL}(2,q)$ has Abelian Sylow $p$-subgroups for every odd prime $p.$ The symmetric group $S_{n}$ has non-Abelian Sylow $p$-subgroups for each prime $p$ such that $p^2 \leq n.$ Hence thesymmetric group $S_{25}$ is not a subgroup of any ${\rm GL}(2,q)$ since it has non-Abelian Sylow $3$-subgroups and non-Abelian Sylow $5$-subgroups ( even the alternating group $A_{25}$ will do). To answer directly the question about ${\rm PSL}(2,7),$ if $q$ is a power of $2,$ then all odd order subgroups of ${\rm GL}(2,q)$ are Abelian, so ${\rm PSL}(2,7)$ can't be a subgroup of such a ${\rm GL}((2,q)$ since ${\rm PSL}(2,7)$ contains a non-Abelian subgroup of order $21.$ On the other hand, if $q$ is odd, then the alternating group $A_4$ is not a subgroup of ${\rm GL}(2,q)$ ( for example, any Klein 4-subgroup of such a ${\rm GL}(2,q)$ contains a cenral element of order $2,$ whereas $A_4$ has no central element of order $2$), but ${\rm PSL}(2,7)$ has a subgroup isomorphic to $A_4.$
Well, ${\rm GL}(2,q)$ has Abelian Sylow $p$-subgroups for every odd prime $p.$ The symmetric group $S_{n}$ has non-Abelian Sylow $p$-subgroups for each prime $p$ such that $p^2 \leq n.$ Hence thesymmetric group $S_{25}$ is not a subgroup of any ${\rm GL}(2,q)$ since it has non-Abelian Sylow $3$-subgroups and non-Abelian Sylow $5$-subgroups ( even the alternating group $A_{25}$ will do).