Proof that cube has 24 rotational symmetries

I was doing a combinatorics problem which states this definition of symmetry: for a subset $S$ of $\mathbb{R}^3$ a symmetry is "rigid motion" $f:\mathbb{R^3}\rightarrow \mathbb{R^3}$ such that any $x\in S$ has its image $f(x)$ also in S. (To provide some fodder for the problem, the author dumbs it down for me that a symmetry is some operation which leaves $S$ in the same place although the individual points may be rearranged)

From this working definition it is asked to find the number of rotational symmetries of the cube. So far I have seen two different arguments and I still dont know because drawing on paper is getting confusing. Also my solution was incorrect and I dont know where I undercounted.

Ignore my method.Drawing 4 axis perpendicular with the faces, each with 4 rotational symmetries $(0^\circ,90^\circ,180^\circ,270^\circ)$ and I get $4\times 4 =16$ symmetries. Plus the four body diagonals$=20$. This is wrong and the answer is $24$.

The book gives the short solution: "move any vertex to any other and rotate the edges leading to it in three ways". I don't get it. There are 8 vertices, and 3 ways of what?

I google and get another argument from wikipedia: "The group of orientation-preserving symmetries is S4, or the group of permutations of four objects, since there is exactly one such symmetry for each permutation of the four pairs of opposite sides of the octahedron." Indeed $4!=24$ but I cannot convince myself that aall permutations of the pair of opposite edges is all the symmetries there can be (I clearly undercounted when I got 16 and still have no insight why, so this does not help me as I dont know for sure there cant be more)

My imagination and visualization skills are very weak. Is there a good online software that allow me to visualize and play around with a labelled cube?

Solution 1:

The point of the comment in the book is this: given a cube, pick one corner and call it $A$. You can certainly rotate the cube so that $A$ either stays put or moves to any chosen other corner, of which there are 7. So you have 8 choices for where you want $A$ to be.

Once you've done this, the other corners are obviously not free to move as they wish. Let $B$ be some corner adjacent to $A$, meaning it shares an edge of the cube with $A$. If A moves to $A'$, $B$ must move to some corner adjacent to $A'$ - otherwise the edge $AB$ gets distorted. How many corners are adjacent to $A'$? That's where the 3 comes from.

Finally you'll need to convince yourself that once $A$ and $B$ have been placed, all other corners are "forced" - their final location is already fixed. See if you can do that just using the relationships of adjacency, sharing a face etc.

Solution 2:

A cube has 12 edges, so it has 24 oriented edges (each edge can be oriented in exactly two ways) It is pretty obvious, if you have a cube to play with, that the group $G$ of rotations acts simply transitively on these orientied edges. Therefore $G$ has 24 elements.

Exactly the same argument counts the number of rotational symmetries of each regular polyhedron, of course.