"Is there a group of order 48 in the set of" \ \ 3 xx 3 \ \ "matrices of integers ?" "If so, can you exhibit one ? If not, prove its impossibility."

1 Answer
Feb 17, 2018

Yes, for example the group generated by:

((0, 0, 1),(0, 1, 0),(-1, 0, 0)), ((-1, 0, 0),(0, 0, -1),(0,-1,0)), ((-1, 0, 0), (0, -1, 0), (0, 0, -1))

Explanation:

Consider the subgroup of GL(3, ZZ) generated by the 2 matrices:

A = ((0, 0, 1),(0, 1, 0),(-1, 0, 0))

B = ((-1, 0, 0),(0, 0, -1),(0,-1,0))

Notice that A represents a rotation of pi/2 about the y-axis, while B represents a rotation of pi about the line x = y+z = 0.

Note that det(A) = det(B) = 1, so they are elements of SL(3, ZZ).

Between them, these two geometrical operations generate a subgroup of SL(3, ZZ) of order 24, isomorphic to S_4, and the rotational symmetry group of the cube.

If we then add a third generator:

C = ((-1, 0, 0), (0, -1, 0), (0, 0, -1))

with det(C) = -1

we get a subgroup of GL(3, ZZ) of order 48, which is the full symmetry group of the cube.

Further reading

For an in depth analysis of the finite subgroups of GL(3, ZZ) you may like to read the paper at https://projecteuclid.org/download/pdf_1/euclid.nmj/1118798212