Cyclic subgroups of $GL_2(F_q)$ Let $F$ be a finite field with $q = p^f$ elements for $p$ a prime.  I know that $G = GL_2(F_q)$  contains a cyclic group of order $q-1$.  It is the set of matrices of the form $\begin{pmatrix} x & 0 \\ 0 &x \end{pmatrix}$, $x \neq 0$.  
An exercise claims there is a subgroup isomorphic to $F_{q^2}^{\ast} \cong \mathbb{Z}/\mathbb{Z}(q^2-1)$.  How can I construct this?  I was thinking of something along the lines of taking the subgroup generated by $\begin{pmatrix} x & 1 \\ 0 & x^{-1} \end{pmatrix}$.
 A: As explained in Qiaochu's comment we need to locate a copy of $\Bbb{F}_{q^2}^*$ acting on
the 2-dimensional space $\Bbb{F}_q^2$. The existence is clear (probably the point of the exercise). I give a relatively generic construction as it sounds to me that you would like to see one. We do need to first pick a special element of the field. AFAICT this is unavoidable. 
Assume first that $p>2$. Pick a non-square element $\varepsilon\in\Bbb{F}_q^*$. It is then easy to show that the set of matrices
$$
C=\left\{\left(\begin{array}{cc}a&b\varepsilon\\b&a\end{array}\right)\,\bigg\vert\,a,b\in\Bbb{F}_q,\,\text{at least one of $a,b\neq0$}\right\}
$$
is a copy of $\Bbb{F}_{q^2}^*$. As 
$$
\left(\begin{array}{cc}0&\varepsilon\\1&0\end{array}\right)^2=\varepsilon I_2
$$
this is essentially just the non-zero elements of $\Bbb{F}_q[\sqrt\varepsilon]$. 
In particular when $q\equiv -1\pmod 4$ we can use $\varepsilon=-1$. In that case $C$
very much resembles the well known way of representing complex numbers by $2\times2$ real matrices. For the case $q\equiv1\pmod4$ I am not aware of such a simple generic choice. 
When $p=2$ we need to do something different as we cannot get $\Bbb{F}_{q^2}$ from $\Bbb{F}_q$ by adjoining a square root (all the elements of $\Bbb{F}_q$ are squares because the Frobenius map is an automorphism). However, we can always find an element
$\alpha\in\Bbb{F}_q$ such that the polynomial
$$
p(x)=x^2+x+\alpha
$$
is irreducible. It is known (ask, if you want more details) that this happens, if and only if
$$
tr(\alpha)=\alpha+\alpha^2+\alpha^4+\alpha^8+\cdots+\alpha^{2^{f-1}}=1,
$$
which is the case for exactly $q/2$ elements $\alpha$. Anyway, then the companion matrix of $p(x)$
$$
A=\left(\begin{array}{cc}0&1\\1&\alpha\end{array}\right)
$$
satisfies the relation
$$
A^2+A=\alpha I_2.
$$
Therefore $A$ can serve in the role of a zero of $p(x)$ and the sought after group is
$$
C=\{aI_2+bA\mid a,b\in\Bbb{F}_q, \,\text{at least one of $a,b\neq0$}\}.
$$
Here a special case is that of an odd extension degree $f$, when we can pick $\alpha=1$. In that case the roots of $p(x)$ are simply primitive cubic roots of unity, and if $\omega$
is one of those we, indeed, have $\Bbb{F}_{q^2}=\Bbb{F}_q[\omega]$.
