Binary tetrahedral group and $\rm{SL}_2(\mathbb F_3)$ The binary tetrahedral group $\mathbb T$ is an interesting 24-element group. For instance it can be expressed as the subgroup
$$ \mathbb T = \left\{ \pm 1, \pm i, \pm j, \pm k, \dfrac{\pm 1 \pm i \pm j \pm k}2 \right\} \subseteq \mathbb{H}^\times$$
of the multiplicative group of the quaternions (which also gives you a fascinating regular 24-vertex 4-polyhedron).
Alternatively, the binary tetrahedral group can be thought of as the inverse image of $\mathfrak A(4)$ through the 2:1 morphism $\rm{SU}_2 \to \rm{SO}_3$ ($\mathfrak A(4) \subseteq \rm{SO}_3$ being the group of direct isometries of the regular tetrahedron).
The property which puzzles me is the following: $\mathbb T$ is isomorphic to the matrix group $\rm{SL}_2(\mathbb F_3)$.
I'm able to give a proof of this result, but not an enlightening one. Essentially, a 24-element group with a normal subgroup isomorphic to the 8-element quaternion group $\rm Q_8$ is either isomorphic to $\rm Q_8 \times \rm C_3$ or to the unique nontrivial semidirect product $\rm Q_8 \rtimes \rm C_3$; it's not that hard to prove that both $\mathbb T$ and $\rm{SL}_2(\mathbb F_3)$ satisfy this property and belong to the second case. The only interesting thing about this proof is the description of that normal $\rm Q_8$: on the one hand $\{\pm 1, \pm i, \pm j, \pm k\} \subset \mathbb T$ is the inverse image of the quite exceptional Vierergruppe $V_4 = \{ \rm{id}, (12)(34), (13)(24), (14)(23)\} \lhd \mathfrak A(4)\subseteq \rm{SO}_3$; on the other hand, the elements of $\mathrm{SL}_2(\mathbb F_3)$ which are diagonalisable over $\mathbb F_9$ — or, equivalently, $\pm I_2$ and the matrices whose trace is zero — form a normal subgroup isomorphic to $\rm Q_8$ (and it's quite astonishing they do form a subgroup!) whose elements are
$$\begin{array}{l} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},\\ \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix}. \end{array}$$
As I said earlier, this proof isn't very satisfying. Hence my question:

Is there an enlightening proof of the isomorphism $\mathbb T \simeq \rm{SL}_2(\mathbb F_3)$?


Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficients are integers or all coefficients are half-integers). The ring $H\otimes\mathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2\times 2}(\mathbb F_p)$ (with determinant as the norm).
Now the binary tetrahedral group is the group of norm $1$ elements in $H$ and it maps injectively (for $p>2$) to the group of norm $1$ elements in $H\otimes\mathbb F_p$, i.e. to  $SL_2(\mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $\mathbb T\to SL_2(\mathbb F_3)$.
(Ref.: TWF 198.)
