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

Question

I figured out that there are 24 rotational symmetries, shown below. Rotation fixing: faces = 9 diagonals = 8 edges = 6 identity = 1 total = 24

Now, I don't know why the symmetry group of the 3-cube has 48 elements; I know it has to do something with reflection but am unable to picture this. Also, my teacher attached the following hint that I have hard time understanding: "Consider the action of the symmetry group on the set of four diagonals." How is this relevant?

Answer 1

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.

Answer 2

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.

Answer 3

Here's an alternative to what I see posted so far. I think. Choose one special corner $A$. Note that $A$ has three adjacent vertices $B_1,B_2,B_3$.

Any permutation of the cube maps $A$ to one of the corners. So there are $8$ possibilities for where to send $A$ to.

Once $A$ is repositioned to $\varphi(A)$, you need to decide how $B_1,B_2,B_3$ will map to the three neighbors of $\varphi(A)$. There are $3!=6$ ways to do this and they are all valid. So now there are $8\cdot6=48$ possibilities.

At this point you have no flexibility for where the remaining four vertices get mapped to. Each one is the antipode for one of $A,B_1,B_2,B_3$ and so must land at the corresponding antipode to $\varphi(A),\varphi(B_1),\varphi(B_2),\varphi(B_3)$. So there are $48$ symmetries total and that's it.

Now when you think back, of the $6$ ways to relocate $B_1,B_2,B_3$, half are orientation preserving and half are not. Thus you get the $24$ rotations and the other $24$ symmetries that cannot be obtained through 3D spatial rotation alone. (Some of those are like what you might see if you held the cube up to a mirror.)

Also, each one of $A,B_1,B_2,B_3$ starts on a different diagonal. And the same is true of $\varphi(A),\varphi(B_1),\varphi(B_2),\varphi(B_3)$. So another way to look back on this is that we permuted the four corresponding diagonals $d,d_1,d_2,d_3$, coupled with a choice for which end of $\varphi(d_1)$ on which $\varphi(A)$ will land. Once all that is decided, the exact location of $\varphi(B_1),\varphi(B_2),\varphi(B_3)$ is determined. For example, $\varphi(B_1)$ is on the diagonal $\varphi(d_1)$, but only one of its two endpoints is adjacent to $\varphi(A)$. That comes to $4!\cdot2=48$ as well.