What is the order of the symmetry group of the cube?

Solution 1:

Certainly you can see that there is at least one reflection that fixes the cube, and that it does not appear in the group of rotational symmetries. Now fix one such reflection and compose it with any one of the rotational symmetries (in a fixed order, say for concreteness that you always start with the reflection). This gives $24$ symmetries, and none of them are rotational symmetries (for otherwise you could get the reflection by composing two rotational symmetries, which you cannot). So you've got $24$ new symmetries of the cube, all reversing the orientation. Moreover this is everything, since a symmetry that preserves orientation is rotational, and one $S$ that reverses orientation becomes an orientation-preserving symmetry $R$ when you precede it by you chosen reflection, and $R$ is the rotational symmetry you need to perform after your chosen reflection to get $S$. That gives $24+24=48$ symmetries in all.

A more canonical way to pair up rotational and reflectional symmetries is to compose with the central symmetry (sending points to their antipodes) which (thanks to the odd dimension) reverses orientation. This is the unique non-trivial symmetry that fixes all $4$ diagonals. Each rotational-reflectional symmetry pair then corresponds to a permutation of the the $4$ diagonals, the permutation they both induce.

Note that not all orientation-reversing symmetries are reflections (whence I used the vague "reflectional" above); there are in fact $9$ reflections (for $3$ pairs of opposite faces and $6$ pairs of opposite edges), $6$ rotary reflections with axis through a pair of faces, $8$ rotary reflections with axis a diagonal, and $1$ central symmetry. They nicely match your classification of the rotational symmetries, through the mentioned canonical pairing.

Solution 2:

I found a nice and elementary way of visualising this.

Look at linear transformations on $\mathbb{R}^3$. We want the cube to remain fixed.
The fundamental way that is done is by re-labelling the axes $(x, y, z)$. [eg: $(x,y,z)\mapsto (y,z,x)$] You can do that in $3! = 6$ ways.
Thereafter, you can also assign signs to the axes. [eg: $(x,y,z)\mapsto (y,-z,x)$]. You can assign $2$ signs in each position independently $-$ in $2^3 = 8$ ways.

This exhausts the allowed possibilities. So, the total number of transformations that keep the cube fixed is $3! \times 2^3 = 48$.

Solution 3:

The symmetry group acts on the diagonals by permutation, which again gives you the $4!=24$ you found. Again, if you add reflections, the number doubles.