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Use the Orbit Stabilizer Theorem to deduce the number of elements in the rotational symmetry group of the cube.

I can write $\operatorname{Stab}_G(v) = \left\{g \in G \mid g \cdot v = v\right\}$ and $\operatorname{Orb}_G(v) = \left\{g \cdot v \mid g \in G\right\}$

The orbit has size 8. Is it enough to say that it is 8 simply because there exists a symmetry such that a specific vertex can somehow get mapped to any of the others.

For the stabilizer, I considered a vertex on the top face of the cube. So I can see three rotations that would fix this vertex in place. Those are

  1. rotation about an axis going through this vertex and the vertex diagonally opposite and lower down,
  2. rotation about $2\pi$ and
  3. rotation about $-2\pi$.

Is this correct? So my answer would be $24 = (8 \times 3)$ by Orbit Stabilizer.

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Well, the answer 24 certainly seems to be correct. – Joe Z. Feb 13 '13 at 19:28
I know the answer is correct, but is my reasoning correct? – CAF Feb 13 '13 at 19:29
I've never used orbit stabilizers, so unfortunately I don't know. – Joe Z. Feb 13 '13 at 19:30
Am I correct in saying that $2\pi$ and $-2\pi$ are two of the three rotations that fix a vertex? – CAF Feb 13 '13 at 19:59
Also, how can I be sure there does not exist other rotations that keep a vertex in place? – CAF Feb 13 '13 at 20:13
up vote 1 down vote accepted

As a check, it might be helpful to note the rotational symmetry group of the cube (which we will call $G$) also acts on the set of 6 faces of the cube.

As anyone who has played with dice is aware, it is possible for any of the six faces to be "up", that is, there exists a rotational symmetry of the cube that maps a given face to any other (in a quaint display of harmony between math and ordinary English, we call the exhibiton of such a symmetry "rolling the die"). If one imagines one's cube with its center at $(0,0,0)$, with sides of length $2$, and one considers the face bounded by $(1,-1,1),(1,1,1),(1,1,-1)$ and $(1,-1,-1)$, it should be clear that the stabilizer of this face consists of the $4$ rotations (including the identity) about the $x$-axis in the $yz$-plane (which in fact is the rotational symmetry group of the SQUARE with vertices $(0,-1,1),(0,1,1),(0,1,-1),(0,-1,-1)$ in the $yz$-plane). The orbit-stabilizer theorem then tells us that:

$|G| = |\mathcal{O}_G(f)|*|\text{Stab}_G(f)| = 6*4 = 24$

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Yes, your argument is correct.

The simplest approach is to let the group act on the set of oriented edges. An oriented edge is an edge of the cube with a selected orientation; there are 24 of them, as each edge gives two oriented edges.

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I've never heard of oriented edges before. Thanks for confirming my answer, but I have a question: Am I right in saying that a rotation by $2\pi$ and $-2\pi$ are different rotations? (in a mathematical sense) – CAF Feb 13 '13 at 19:32
No, you measured the angles wrong. – Mariano Suárez-Alvarez Feb 13 '13 at 19:38
How? Rotate it 4 times means 4 times 90 degrees = 360 degrees? – CAF Feb 13 '13 at 19:43
I have no idea what that argument is supposed to prove, really ;-) For one, a rotation of angle $2\pi$ is the identity map, just as one of angle $-2\pi$, so that is surely not the angle of a non-identity symmetry of the cube. There are three rotations which fix a vertex: they rotate the cube around the diagonal going through that vertex an angle which is or $0$, or $2\pi/3$ or $-2\pi/3$. – Mariano Suárez-Alvarez Feb 13 '13 at 20:20
Independently of anything, you should stop everything you are doing and convince yourself that a rotation of angle $2\pi$ is the identity. – Mariano Suárez-Alvarez Feb 13 '13 at 20:43

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