Orders of rotations in higher dimension Do we have a general method to figure out the order of the group of rotations in a higher dimension?
My first attack is on 3-d. We can figure out the order of the group of rotations of a regular solid(like tetrahedron or dodecahedron) by considering the axis of rotation.
But for higher dimensions, if we use the same method for dimension 4, we need to consider the rotations around the axis and faces. I guess we can shorten this process by taking a subgroup.
Is there a more efficient way to think about this? Thanks in advance!
 A: I assume you mean the rotational symmetry group of a polytope.
The best technique might be the orbit-stabilizer theorem. This says that given any choice of feature that the group acts on, the size of the group is the size of the stabilizer times the size of the orbit. For illustration, let's consider a cube and a triangular prism in three dimensions, with various features.
CUBE, $\,24=3\times8=2\times12=4\times6$ symmetries using any of the following "features":

*

*vertices: $8$ vertices (the orbit), and $3$ rotations fixing a given vertex (the stabilizer)

*edges:  $12$ edges (the orbit) and $2$ rotations fixing a given edge (the stabilizer)

*faces: $6$ faces (the orbit) and $4$ rotations fixing a given face (the stabilizer)

*axes: there are $3$ axes thru faces, $2\times4$ rotations fixing a given axis (the stabilizer)

*inscribed tetrahedra: there are $2$ inscribed tetrahedra, each fixed by $12$ rotations

You can decide for yourself which of the above is most practical.
TRIANGULAR PRISM, $6$ symmetries, assuming triangular faces and long rectangles.

*

*vertices: $6$ vertices (orbit) but only the identity rotation fixing it (stabilizer)

*long edges: $3$ long edges (orbit) and $2$ rotations fixing one of them (stabilizer)

*short edges: $6$ triangle edges (orbit), only identity fixing it (stabilizer)

Similar reasoning for faces.
Exercise. Show the octahedron also has $24$ rotational symmetries.
Now let's try something in four dimensions: the $24$-cell. First take the integer lattice $\mathbb{Z}^4\subseteq\mathbb{R}^4$, and notice opposite points are exactly $2$ units apart by the Pythagorean theorem, so we can take the permutations of the points $(\pm1,0,0,0)$ and the combinations of $(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2})$ (the midpoints of line segments from the origin the the hypercube antipodes away from it) to get vertices of the $24$-cell. The key is that any rotation which fixes a vector restricts to a rotation of its orthogonal complement, which has one lower dimension; this allows essentially an inductive argument.
There are $24$ vertices. The vector $(1,0,0,0)$'s orthogonal complement contains $6$ points of the $24$-cell, the octahedron with coordinate axis vertices. Therefore the $24$-cell has a total of $24^2=576$ symmetries.
