Noam Elkies writes to John Baez:

Hello again,

You write:

[...]

"I'd like to wrap up with a few small comments about last Week.  There I said a bit about a 24-element group called the 'binary tetrahedral group', a 24-element group called SL(2,Z/3), and the vertices of a regular polytope in 4 dimensions called the '24-cell'.  The most important fact is that these are all the same thing! And I've learned a bit more about this thing from here:"

[...]

Here's yet another way to see this: the 24-cell is the subgroup of the unit quaternions (a.k.a. SU(2)) consisting of the elements of norm 1 in the Hurwitz quaternions - the ring of quaternions obtained from the Z-span of {1,i,j,k} by plugging up the holes at (1+i+j+k)/2 and its <1,i,j,k> translates. Call this ring A. Then this group maps injectively to A/3A, because for any g,g' in the group |g-g'| is at most 2 so g-g' is not in 3A unless g=g'. But for any odd prime p the (Z/pZ)-algebra A/pA is isomorphic with the algebra of 2*2 matrices with entries in Z/pZ, with the quaternion norm identified with the determinant. So our 24-element group injects into SL₂(Z/3Z) - which is barely large enough to accommodate it. So the injection must be an isomorphism.

Continuing a bit longer in this vein: this 24-element group then injects into SL₂(Z/pZ) for any odd prime p, but this injection is not an isomorphism once p>3. For instance, when p=5 the image has index 5 - which, however, does give us a map from SL₂(Z/5Z) to the symmetric group of order 5, using the action of SL₂(Z/5Z) by conjugation on the 5 conjugates of the 24-element group. This turns out to be one way to see the isomorphism of PSL₂(Z/5Z) with the alternating group A₅.

Likewise the octahedral and icosahedral groups S₄ and A₅ can be found in PSL₂(Z/7Z) and PSL₂(Z/11Z), which gives the permutation representations of those two groups on 7 and 11 letters respectively; and A₅ is also an index-6 subgroup of PSL₂(F₉), which yields the identification of that group with A₆.

NDE