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Consider a cube that exactly fills a certain cubical box. As in Examples 8.7 and 8.10, the ways in which the cube can be placed into the box corresponds to a certain group of permutations of the vertices of the cube. This is the group of group of rigid motions (or rotations) of the cube. (It should not be confused with th e group of symmetries of the figure, which will be discussed in the exercises of Section 12.) How many elements does this group have? Argue geometrically that this group has at least three different subgroups of order 4 and at least four different subgroups of order 3.

Fraleigh Solution The group has $24$ elements.

The first subgroup of order $4.$ For any one of the $6$ faces can be on top, and for each such face on top, the cube can be rotated in $4$ different positions leaving that face on top. The $4$ such rotations, leaving the top face on top and the bottom face on the bottom, form a cyclic subgroup of order $4$.

The second rotation group of order $4$ is formed by the rotations leaving the front and back faces in those positions.

The third rotation groups of order $4$ is formed by the rotations leaving the side faces in those positions.

One exhibits a subgroup of order $3$ by $\color{red}{\text{taking hold of a pair of diagonally opposite vertices and rotating through the three possible positions}}$, corresponding to the three edges emanating from each vertex.
There are $4$ such diagonally opposite pairs of vertices, giving the desired $4$ groups of order $3$.

Question 1. I feel this is easier than my other post. But I can't see how 'taking hold a pair of diagonally opposite vertices' causes 'three possible positions'? I tried the animation at that other post but no luck.

Question 2. How do you decide on the classifications of the rotations of a shape? These two solutions just say what they are. They never revealed how they prefigured this group has at least three different subgroups of order 4 and at least four different subgroups of order 3. I don't mean just playing around with the shape. I tried that and it didn't help me here.

Question 3. Does this solution break down the type of rotations differently than the other post? That solution talks about " a line joining the centers of opposite faces" and "a line joining diagonally opposite vertices ". This one doesn't.

This is from John B. Fraleigh page 86 exercise 8.45 A First Course in Abstract Algebra

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(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).

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I just want to add to the excellent answer (and to the question) that there are some groups of order 2 that deserve mention as well: if you look at an edge of the cube so that your line-of-sight through the edge passes through the cube's center, then there's another parallel edge, which you might say was "diagonally across from" your edge, also lined up with the center. Let's hold the cube so that the chosen edge sits horizontally in front of your face, as does its diagonal opposite. Let's call the left end of your edge "A" and the right end "B". You can now rotate the cube 180 degrees about the line of sight so that "A" and "B" are swapped. Clearly doing this twice gives the identity. But it's also clear that this element of order 2 isn't in any of the order-3 subgroups (2 doesn't divide 3), nor any of the order-4 subgroups, since in those subgroups, each non-identity element moves every edge to a new position.

Anyhow, what I'm saying is that there are the vertex-fixing subgroups, the face-fixing subgroups, and, as I've just described, the edge-fixing subgroups.

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