TITLE: Regarding subgroups of $GL_{2}(\mathbb{F}_q)$ of order $p$, where $p$ is an odd prime dividing $q+1$.
QUESTION [5 upvotes]: The problem (this is not a homework problem, though it does feel like one):
Suppose p and q are primes such that $p \vert q+1$ and p is odd. Then all subgroups of $GL_{2}(\mathbb{F}_{q})$ of order p are conjugate to one another.
What I know so far:
First of all, $\vert GL_{2}(\mathbb{F}_{q})\vert = q(q+1)(q-1)^2$. 
If $q=2$, then $p=3$ and  $\vert GL_{2}(\mathbb{F}_{q})\vert =6$. Then the only subgroup of order $p$ is the unique Sylow $3$-subgroup. Of course this subgroup is conjugate to itself.
So, now we look at the case where $q$ is an odd prime. Obviously $p \nmid q(q-1)^2$. But, $p$ may not be the largest power of $p$ dividing $q+1$. So, the subgroups of order $p$ are not necessarily Sylow $p$-subgroups. Of course, they must be conjugate to a subgroup of a Sylow $p$-subgroup.
What is the right way to complete this argument? I am hoping for an answer that pushes me in the right direction without giving it all away (at least for now).
Thanks for your time. My apologies for the very basic question.
Edit: Check the comments below Derek's hint for a solution. Please let me know if you spot any problems with my argument.

REPLY [2 votes]: The Sylow $p$-subgroups of ${\rm GL}_2(q)$ (for $p$ an odd prime dividing $q+1$) are cyclic. Using that fact, you should be able to prove using Sylow's Theorem that all subgroups of order $p$ are conjugate.
To see that the Sylow $p$-subgroups are cyclic, observe that ${\rm GL}_2(q)$ contains an element of order $q^2-1$ arising from multiplication by a multiplicative generator of the field ${\mathbb F}_{q^2}$ of order $q^2$.