## Return to Answer

2 made more concise

The answer to your question is "there must be, it's just a question of doing the bookkeeping carefully". It's well-known that a subgroup of $\mathrm{PGL}(2,p)$ with order prime to $p$ is either cyclic, dihedral, tetrahedral ($A_4$), octahedral ($S_4$) or icosahedral ($A_5$). The icosahedral case only happens when $p\equiv\pm1$ (mod $5$). We now have to pull these back to $\mathrm{GL}(2,p)$, so have to count how many subgroups of $\mathrm{GL}(2,p)$ lie above a given subgroup of $\mathrm{PGL}(2,p)$ etc.

For subgroups of order divisible by $p$, the result in $\mathrm{PGL}(2,p)$ is that such a subgroup lies inside a normalizer of a Sylow $p$-subgroup or is $\mathrm{PSL}(2,p)$ or $\mathrm{PGL}(2,p)$. So in $\mathrm{GL}(2,p)$ either the group lies inside a normalizer of a Sylow $p$-subgroup or lies between $\mathrm{SL}(2,p)$ and $\mathrm{GL}(2,p)$.

1

The answer to your question is "there must be, it's just a question of doing the bookkeeping carefully". It's well-known that a subgroup of $\mathrm{PGL}(2,p)$ with order prime to $p$ is either cyclic, dihedral, tetrahedral ($A_4$), octahedral ($S_4$) or icosahedral ($A_5$). The icosahedral case only happens when $p\equiv\pm1$ (mod $5$). We now have to pull these back to $\mathrm{GL}(2,p)$, so have to count how many subgroups of $\mathrm{GL}(2,p)$ lie above a given subgroup of $\mathrm{PGL}(2,p)$ etc.

For subgroups of order divisible by $p$, the result in $\mathrm{PGL}(2,p)$ is that such a subgroup lies inside a normalizer of a Sylow $p$-subgroup or is $\mathrm{PSL}(2,p)$ or $\mathrm{PGL}(2,p)$. So in $\mathrm{GL}(2,p)$ either the group lies inside a normalizer of a Sylow $p$-subgroup or lies between $\mathrm{SL}(2,p)$ and $\mathrm{GL}(2,p)$.
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