Let $q$ be a power of an odd prime, $\mathbb{F}_{q}$ be the finite field with $q$ elements, and $\mathbb{F}_{q^{n}}$ the finite field with $q^{n}$ elements. We have that $n=[\mathbb{F}_{q^{n}}:\mathbb{F}_{q}]$.
We define the following group: $$G=\{g\in\text{GL}(2,\mathbb{F}_{q^{n}})|\text{det}(g)\in\mathbb{F}_{q}^{*}\},$$
and we denote by $\overline{G}$ to the image of $G$ in $\text{PGL}(2,\mathbb{F}_{q^{n}})$.
I want to prove that $$\overline{G}=\begin{cases} \text{PGL}(2,\mathbb{F}_{q^{n}})\text{ if }n \text{ is odd}\\ \text{PSL}(2,\mathbb{F}_{q^{n}})\text{ if }n \text{ is even} \end{cases}.$$
My attempt so far has been as follows:
It is clear that $\text{PSL}(2,\mathbb{F}_{q^{n}})\subseteq \overline{G}$, and since $q$ is odd, we have that $[\text{PGL}(2,\mathbb{F}_{q^{n}}):\text{PSL}(2,\mathbb{F}_{q^{n}})]=2$, so $\overline{G}$ can only be $\text{PSL}(2,\mathbb{F}_{q^{n}})$ or $\text{PGL}(2,\mathbb{F}_{q^{n}})$.
So my idea is now to distinguish the cases by looking at the determinant of the elements of $\text{PGL}(2,\mathbb{F}_{q^{n}})$, but I do not know how to proceed or how to use the fact that the degree of the extension is either odd or even.
If someone could help me, I would appreciate it a lot!
Thanks in advance!