Show that $|G|=24$, where $G$ is the Rotations of the Cube in $\mathbb{R}^3$

Question

I am looking at this problem from a visual stand point, Many of the explanations that I have found online invoke the multiplication principal of counting; basically, there are $6$ faces, and $4$ possible rotations if you fix any $1$ of the faces, thus, there are $6\times 4=24$ possible rotations that fix the cube's structure.

This does not seem right. If you fix a face of the cube, then you are naturally fixing another face opposite of it. So if face $1$ is opposite face $6$, rotating face $1$ counterclockwise $90$ degrees is the same as rotating face $6$ counterclockwise $270$ degrees. Thus, this would only be $12$ of the total rotations permitted.

From there then, looking at a drawing of the cube i made, I think the next $12$ possible rotations come from the following. There are $8$ vertices. If we fix vertex $1$, then we naturally also fix vertex $8$ across from it. There are only $2$ rotations then, and with $4$ pairs of vertices, this gives another $8$ rotations different from the original $12$. However, I'm not sure where the final $4$ are coming from.

Is there an "easier" way of looking at this to visually convince me that $|G|=24$? Am I incorrect about my intuition about the multiplication principal?

Answer 1

Assume the cube as $[{-1},1]^3$. There are

• the identity,
• $3$ rotations ($\pm 90^\circ$, $180^\circ$) around each of the $3$ coordinate axes,
• $2$ rotations ($\pm 120^\circ$) around each of the $4$ space diagonals,
• $1$ rotation ($180^\circ$) around each of $6$ axes through the midpoints of opposite edges.

Answer 2

One way of looking at it is by considering the diagonal lines inside the cube. For example:

There are four diagonals within the main cube. Any symmetry of the cube will permute these diagonals. So, there are a total of $4! = 24$ total symmetries of the cube.