TITLE: Can we find element of order $q^2-1$ in $\text{GL}_2(\mathbb{F}_q)$?
QUESTION [8 upvotes]: How to find element of order $q^2-1$ in $\text{GL}_2(\mathbb{F}_q)$? I am hoping to find field $\mathbb F_{q^2}$ as subalgebra of $2\times 2$ matrices over field $\mathbb F_q$ where $q$ is power of prime number. I was trying with element $\pmatrix {n&1 \\ 1&0}$ but it works only for $q=2,3,4,8,16$. 
It is suggested that this is duplicate of question 
$GL_n(\mathbb F_q)$ has an element of order $q^n-1$
But I don't know how to find $\mathbb F_{q^2}$ in $M_2(\mathbb F_q)$.
Here is test in GAP for above matrix - $n$ is generator of the field multiplicative group.
gap> List([2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41],
>         k->Order([[Z(k),1],[1,0]]*Z(k)^0));
[ 3, 8, 5, 12, 16, 9, 20, 24, 28, 17, 16, 40, 22, 52, 56, 20, 64, 31, 76, 40 ]

To give my motivation - I want to prove that algebra  $M_2(\mathbb F_q)$ can be represented as ${a+bj}$ for $a,b$ belonging to $\mathbb F_{q^2}$ and multiplication given by Cayley-Dickson
$$(a+bj)(c+dj)=ac+\bar db + (da+b\bar c)j,$$
where $j$ is matrix satisfying $\bar j=-j$.

REPLY [2 votes]: The main question (the part with Cayley-Dickson) follows from the Skolem-Noether theorem.
Let's denote $K=\Bbb{F}_q, L=\Bbb{F}_{q^2}$, and for all $z\in L$ let $z^q=\overline{z}$ be the Frobenius automorphism. As the other answers have explained there exist ways of embedding $L$ into $M_2(K)$. Let's fix one such embedding $\phi:L\to M_2(K)$. Then $z\mapsto \phi(\overline{z})$ is another embedding of $L$, so by Skolem-Noether there exists an invertible matrix $u\in M_2(K)$ such that $u\phi(z)=\phi(\overline{z})u$ for all $z\in L$. Identifying $L$ with $\phi(L)$ allows us to rewrite this as
$$uz=\overline{z}u\quad\text{for all $z\in L$.}\qquad(*)$$
A priori there is no reason to think that $u^2$ would be equal to the identity. However, we can always arrange that to be the case. Namely, if $c\in L$ is arbitrary, it follows that $cz=zc$ for all $z\in L$ implying that $u'=uc$ can take the role of $u$ in equation $(*)$. Actually, the double centralizer theorem tells us that any two matrices satisfying $(*)$ are gotten from each other by multiplication by an element of $L$.
Anyway, applying $(*)$ twice tells us that
$$u^2z=u(uz)=u(\overline{z}u)=(u\overline{z})u=(zu)u=zu^2$$
for all $z\in L$. In other words $u^2$ commutes with all the matrices in $L$. Trivially $u^2$ commutes with $u$. The sum of subspaces $L+uL$ must be direct, and hence all of $M_2(K)$. We have just shown that $u^2$ is in the center of $M_2(K)$. Therefore $u^2$ is a scalar matrix, and we can think of it as an element $u^2=\alpha\in K$.
But, for all $c\in L$
$$(uc)^2=u(cu)c=u^2\overline{c}c=u^2N(c),$$
where $N:L\to K, z\mapsto z\overline{z}$ is the norm map. In the case of finite fields the norm map is surjective, so we can find an element $c_0$ such that $N(c_0)=1/\alpha$.
Therefore
$$
j=uc_0
$$
satisfies the relation $j^2=1$ as well as the relation $jz=\overline{z}j$ for all $z\in L$
as prescribed.