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 $\mathbb T \simeq \rm{SL}_2(\mathbb F_3)$?

  • $\begingroup$ I suspect that the binary tetrahedral group is conjugate to a subgroup of $\text{SL}_2(\mathbb{Z})$. If so, it's known that if $G$ is a finite subgroup of $\text{GL}_n(\mathbb{Z})$ then $G$ injects into $\text{GL}_n(\mathbb{F}_3)$, and that provides a pretty straightforward isomorphism. $\endgroup$ Apr 24, 2013 at 17:41
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    $\begingroup$ The isomorphism $\mathfrak A(4) \simeq \rm{PSL}_2(\mathbb F_3)$ is quite easy to see: the action of $\rm{PSL}_2(\mathbb F_3)$ on the 4-element projective line $\rm P^1(\mathbb F_3)$ provides an injective morphism $\rm{PSL}_2(\mathbb F_3) \to \mathfrak S(4)$ which gives easily the desired isomorphism. What is unclear to me is why the double cover gives precisely $\rm{SL}_2(\mathbb F_3)$. Maybe there are direct cohomological arguments? $\endgroup$
    – PseudoNeo
    Apr 24, 2013 at 19:35
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    $\begingroup$ @QiaochuYuan: I doubt it: finite subgroups of $\rm{GL}_2(\mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral. $\endgroup$
    – PseudoNeo
    Apr 24, 2013 at 19:38
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    $\begingroup$ Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$. $\endgroup$
    – user641
    Apr 24, 2013 at 23:24
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    $\begingroup$ There is a 2-dimensional complex representation of $Q_8$, but it cannot be written over the rationals, so $Q_8$ does not embed in ${\rm GL}_2({\mathbb Z})$. (But ${\rm SL}_2(3)$, which contains $Q_8$, does embed in ${\rm SL}_2(p)$ for all odd primes $p$.) $\endgroup$
    – Derek Holt
    Apr 25, 2013 at 8:59

3 Answers 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.)


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 $\mathbb{H}_0\simeq\mathbb{R}^3$ 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.


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