Question

`"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."`

Answer 1

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