TITLE: Binary tetrahedral group and $\rm{SL}_2(\mathbb F_3)$
QUESTION [16 upvotes]: 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 $\mathbb T \simeq \rm{SL}_2(\mathbb F_3)$?

REPLY [3 votes]: Maybe the confusion arrises from the fact that $\mathbb{T}$
has no subgroup of order $12$.
Rather it has the quaternion group
$Q=\{\pm1,\pm i,\pm j, \pm k\}$
as a subgroup of order $8$.
I think a good way to grasp this intuitively is to see it in the context of the well known action of $\mathbb{H}^1$
on the pure quaternions $\mathbb{H}_0\simeq\mathbb{R}^3$
by rotation via $q(p)=qp\overline{q}$,
which gives a double cover of $\mathrm{SO}(3)$.
In particular,
$\mathbb{H}^1/\{\pm1\}\cong\mathrm{SO}(3)$.
We can use this to realize the action of $\mathbb{T}$
on a tetrahedron explicitly, as follows.
Let $T\subset\mathbb{H}_0$
be the tetrahedron with vertices
$$i+j+k,\quad i-j-k,\quad -i+j-k,\quad -i-j+k.$$
An element of $Q$
acts on this via rotation by $\pi$
about a line through the midpoints of opposite sides of $T$.
An element of $\mathbb{T}-Q$
acts on this by rotation by $2\pi/3$
about a line through a vertex and the midpoint of its opposite face.
Any pair of elements of $\mathbb{T}$
differing only by sign give the same action.
Now, by passing from $Q$
to $\mathbb{T}$
we have tripled the size of the group,
and included everything necessary to get the full tetrahedral group,
but with redundancy.
In particular, the kernel of the action of $\mathbb{T}$ on $T$ is $\{\pm1\}$.
So $\mathbb{T}$
gives a double cover of the tetrahedral group.
As you mentioned, there's a straightforward hands-on way to see that the tetrahedral group $\mathfrak{U}(4)$
is isomorphic to $\mathrm{PSL}(2,\mathbb{F}_3)$.
So if we pull back to the full group,
we've got $\mathbb{T}\cong\mathrm{SL}_2(\mathbb{F}_3)$.
This way of thinking about it is not my own invention. It's from a preprint I have of the book Quaternion Algebras by John Voight, which is expected to go into print very soon and which I highly recommend.