Csdrhrt

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 $Hotimesmathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2times 2}(mathbb F_p)$ (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 $Hotimesmathbb F_p$, i.e. to $SL_2(mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $mathbb Tto SL_2(mathbb F_3)$.

(Ref.: TWF 198.)

We know that $mathbb T= <x,y,z| x^2=y^3=z^3=xyz>$. We also know that $ord(xyz)=2$, $mathbb{T}/<xyz> simeq A_4$ and $mathbb{T}$ has no subgroup of order 12.

In one of my problem sheets, we are intended to give a guided proof of the fact that $mathbb T simeq SL_2( mathbb F_3)$ by showing that the map $phi: mathbb T to SL_2(mathbb F_3)$ that sends $$x longrightarrow left( begin{array}{cc} 0 & 2 \ 1 & 0 end{array} right), y longrightarrow left( begin{array}{cc} 2 & 0 \ 1 & 2 end{array} right) text{ and } z longrightarrow left( begin{array}{cc} 0 & 1 \ 2 & 1 end{array} right)$$ extends to an isomorphism between $mathbb T$ and $SL_2( mathbb{F}_3)$. This map is clearly a homomorphism, because it is easy to see that the images of the generators respect the relations.

We can just count the elements of $SL_2(mathbb F_3)$ to see that $|SL_2(mathbb F_3)|= | mathbb{T}|=24$, so the only thing that we still have to show is either that $phi$ is injective or surjective. Anybody can think of an easy way to prove this? I think this might lead to an easy proof of the result, without even making use of the notion of semidirect products.

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 $mathbb{H}_0simeqmathbb{R}^3$
by rotation via $q(p)=qpoverline{q}$,
which gives a double cover of $mathrm{SO}(3)$.
In particular,
$mathbb{H}^1/{pm1}congmathrm{SO}(3)$.

We can use this to realize the action of $mathbb{T}$
on a tetrahedron explicitly, as follows.
Let $Tsubsetmathbb{H}_0$
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 $2pi/3$
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 $mathbb{T}congmathrm{SL}_2(mathbb{F}_3)$.

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.