Rotational Symmetries of a Cube 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 


*

*rotation about an  axis going through this vertex and the vertex
diagonally opposite and lower down,

*rotation about $2\pi$ and

*rotation about $-2\pi$.


Is this correct? So my answer would be $24 = (8 \times 3)$ by Orbit Stabilizer. 
 A: Your reasoning isn't correct, I'm afraid, but all your conclusions are.
Yes, the orbit of a vertex has the eight vertices in it.
Yes, the stabilizer of a vertex has three rotations in it.
Two of the rotations you listed are the identity rotation, because rotating by $2\pi$ leaves the cube exactly in it's original position. If that didn't count as the identity, you would have infinitely many symmetries, one for each full turn cockwise or anticlockwise, but no, we don't consider the route, we consider the transformation from start position to end position, and since they're the same, it's the identity rotation.
Why are there three elements in the stabilizer for the vertex?  
When you rotate a cube about its longest diagonal axis, there are three faces that are permuted cyclically (in the picture, the centre of the red, yellow and blue faces are an orbit of three, and the black, white and green faces are another). This stabilizer is a cyclic group of order three, because you can do that rotation three times before getting back to where you started, as you can visualize by watching the picture.
Hence the stabilizer of a vertex under rotations of the cube consists of three elements: 1. the identity rotation (by $0$ or $2\pi$ or $-24\pi$, it's all the same symmetry), 2. rotation about the long diagonal axis by $2\pi/3$ and 3. by twice that.
A: 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.
A: 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:
$|G| = |\mathcal{O}_G(f)|*|\text{Stab}_G(f)| = 6*4 = 24$
