Embedding $\mathbb{F}_{q^2}^*$ into $GL_2(\mathbb{F}_{q})$

Question

If we see $\mathbb{F}_{q^2}$ as a $2$-dimensional vector space over $\mathbb{F}_{q}$ (and pick a base) then we can identify $\operatorname{Aut}_{\mathbb{F}_{q}}(\mathbb{F}_{q^2})$ with $GL_2(\mathbb{F}_{q})$. Therefore we can embed $\mathbb{F}_{q^2}^*$ into $GL_2(\mathbb{F}_{q})$, by the natural action of $\mathbb{F}_{q^2}^*$ on $\mathbb{F}_{q^2}$.

However, I have difficulty to concretely visualise the subgroup $E$ of $GL_2(\mathbb{F}_{q})$ that should be congruent to $\mathbb{F}_{q^2}^*$. Is there a nice way to see this group $E$?

Answer 1

If $q$ is odd we can do the following. Choose a non-square element $\epsilon\in \Bbb{F}_q^*\setminus (\Bbb{F}_q^*)^2.$ With respect to a suitable basis the elements of the extension field, when represented by matrices, then have the shape $$ \Bbb{F}_{q^2}=\left\{\left( \begin{array}{rr} a&b\\ \epsilon b&a \end{array}\right)\,\bigg\vert\ a,b\in\Bbb{F}_q\right\}. $$ The set of such matrices is easily seen to be a vector space over $\Bbb{F}_q$ and closed under multiplication. Because the determinant has the form $a^2-\epsilon b^2$, our assumption implies that it vanishes only when $a=0=b$. Hence all those matrices are invertible. Therefore they form a field of cardinality $q^2$.

This is particularly simple in the case $q\equiv 3\pmod 4$, because we can then choose $\epsilon=-1$ and end up with the analogue of the familiar representation of complex numbers as $2\times2$ real matrices. The reason for this similarity is, of course, that we use the construction $$ \Bbb{F}_{q^2}=\Bbb{F}_q[\sqrt\epsilon]. $$

When $q$ is even there are similar ways, but they are not so "clean". If $a\in \Bbb{F}_q$ is such that its absolute trace is $=1$ (i.e. does not vanish), then the polynomial $p(x)=x^2+x+a$ is irreducible in $\Bbb{F}_q[x]$, and we can pick a basis $\{1,\alpha\}$, where $\alpha$ is a zero of $p(x)$.

Answer 2

It might not be a fully satisfactory description.

Note that the subgroup $E\le GL_2(\Bbb F_q)$ varies according to the choice of the basis.

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Knowing that the multiplicative group of any finite field is cyclic, choose a generator $\alpha$ of $\Bbb F_{q^2}$, and pick the following basis of $\Bbb F_{q^2}$ over $\Bbb F_q$: $$(1,\,\alpha)$$ Then the linear mapping $x\mapsto \alpha\cdot x$ becomes the matrix $M$ with columns $(\alpha,\alpha^2)$.
[Note that $\alpha=\pmatrix{0\\1}$ in our chosen basis.]

and $E$ will consist of all the powers of $M$.