Reading about the symmetries of a cube, here, they talk about (in the section "Details") "3 x rotation about a four fold axis" etc. I'm not quite sure what "four fold axis" means in the context. Can someone help? Also, since a cube is face transitive, it should be possible to define a group of rotations such that starting with one face, the rotations construct all the other faces. Is there a name for and easy way to identify this group? The full Octahedral symmetries (24) are much more than the faces.
[Math] What does “n-fold axis” mean in symmetry groups
platonic-solidsrotationssolid-geometrysymmetry
Related Solutions
(1) "Take hold of diagonally opposite vertices"
Since we're only in 3d, perhaps you can get by just drawing nice 2d pictures. If you draw a picture of a cube (just the front parts that you can see, don't worry about the back), you see that each vertex is connected by an edge to exactly three other vertices. In the drawing, you should have one vertex that stands out because you can see all three of its neighbors.
Now imagine rotating the cube so that you're staring directly at that vertex, with its opposite vertex directly behind it. The picture you should come up with is a hexagon, tiled by three quadrilaterals (draw a regular hexagon, and a line to the center of the hexagon from every other vertex, three total).
This picture highlights the three rotational symmetries for that particular pair of diagonal vertices. The tiled hexagon can be rotated 0, 1/3, or 2/3 of a turn, as can the entire cube.
(2) The point here is that we're singling out specific rotations, or group elements. If you've seen the term "group action," that's exactly what we've got. So we're thinking of the group elements in terms of what they do to the cube.
By finding this rotation of order 3, we found a subgroup of order 3 generated by 1/3 of a full rotation fixing a pair of vertices. How do we know the whole group has four subgroups of order 3? Well, this rotation shuffles 6 of the vertices around, but leaves two (opposite) vertices fixed. For each pair of opposite vertices, we can find a rotation that fixes those two vertices. Thus, we have at least 4 subgroups of order 3.
Note that these subgroups can only have the identity in common; each nonidentity element of a particular subgroup of order 3 fixes a specific pair of opposite vertices. A very similar story tells us why we must have at least 3 subgroups of order 4. Instead of focusing on rotations that fix opposite vertices, we can find rotations that fix (not pointwise!) opposite faces, and reason similarly.
In time, you'll see that what you've been asked to do is a general method of examining a group, provided that group "acts" on something. At first, we have no idea how many group elements there are, or subgroups, or anything like that. However, we can find this sort of information out by thinking about the object being acted on (in this case, the cube).
If what you allow as a “geometric object” is sufficiently broad to match the kinds of groups you allow, the answer is positive. I’ll first restrict to the finite case, which from your examples seems to be the case you’re mainly interested in, and then discuss the infinite case.
For a finite group $G$, by Frucht’s theorem (linked to in a comment under the first answer you linked to), every group is isomorphic to the automorphism group of a finite undirected graph. Embed the graph $(V,E)$ in $\mathbb R^{|V|}$ by bijectively mapping the vertices to the canonical basis vectors and the edges to line segments between the vertices they are incident upon. The isometry group of the resulting geometric object is isomorphic to $G$.
The isometries of a Euclidean space are linear transformations, so specifying the images of all basis vectors under an isometry specifies the isometry. Since an automorphism of the graph specifies the images of all basis vectors, it uniquely defines an isometry; the object is invariant under this isometry; and the composition law of these isometries is the composition law of the automorphisms. Conversely, every isometry of the object corresponds to an automorphism of the graph. Hence the group of isometries is isomorphic to the group of automorphisms, which is isomorphic to $G$.
This doesn’t work in the infinite case, since there are groups of arbitrarily large cardinality (e.g. the free group over a set of arbitrarily large cardinality) and the Euclidean group only has the cardinality of the continuum. However, Frucht’s theorem was extended to infinite groups and graphs (see this section of the Wikipedia article, with references), so if we allow “geometric objects” in arbitrary powers of $\mathbb R$, we can embed an infinite graph $(E,V)$ whose automorphism group is isomorphic to $G$ in the subspace of $\mathbb R^V$ with finitely many non-zero components by again mapping the vertices to canonical basis vectors and the edges to line segments connecting them. Then a linear transformation is again uniquely determined by the images of all basis vectors (this is where we need the restriction to finitely many non-zero components), and it follows that the group of linear transformations of the resulting “geometric object” is isomorphic to $G$.
Best Answer
A four fold axis of symmetry is one for which a rotation of $\pi/2$ exists, so that rotation is of order four. The symmetries of the cube are the full octahedral group: https://en.wikipedia.org/wiki/Octahedral_symmetry
Edit. The edit to the question provides the same wikipedia link. It's all there.