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Question:

Let $$C$$ denote the cube $$[-1,1]^3\subset \mathbb{R}^3$$. How many rotations are there in $$\mathbb{R}^3$$ which take $$C$$ to itself?

Discussion:

Let us label the six faces of the cube by $$F_1,F_2,…,F_6$$.

Let $$G$$ be the set consisting of all the rotations of $$\mathbb{R}^3$$ which take $$C$$ to itself. This set forms a group under the binary operation composition. In fact, $$G$$ is a subgroup of the group of isometries of $$\mathbb{R}^3$$. The identity is the identity transformation (rotation by 0 degree).

This group $$G$$ acts on the set {$$F_1,…F_6$$}. The action is the most obvious one. A face $$F_i$$ goes to face $$F_j$$ under any rotation that preserves $$C$$. That is, every rotation is a permutation of the faces $$F_1,…,F_6$$. Said in other words: $$\sigma . F_i = F_j$$ is the action where the face $$F_i$$ goes to $$F_j$$ under the rotation $$\sigma$$.

We recall the Orbit-Stabilizer theorem:

$$|G|=|orbit(x)||stab(x)|$$. Where $$G$$ acts on the set $$X$$ and $$x\in X$$.

Where $$orbit(x)=\{ g.x | g\in G\}$$ and $$stab(x)=\{g\in G | g.x=x \}$$ is the stabilizer of $$x$$. Here “.” denotes the action.

In this context, we have our group $$G$$ as described above. And $$X=\{F_1,…,F_6\}$$.

In order to apply the orbit-stabilizer theorem, we look at $$orbit(F_1)$$ and $$stab(F_1)$$.

We can always rotate the cube in order that the face $$F_1$$ goes to the face $$F_2,..,F_6$$ and also the zero-rotation takes $$F_1$$ to $$F_1$$. Therefore $$orbit(F_1)=\{F_1,…,F_6\}$$.

What is the stabilizer of $$F_1$$? Well, the only rotations which fix the face $$F_1$$ are the $$0,\pi/2,2\pi/2,3\pi/2$$ rotations in the plane of $$F_1$$ (i.e, having axis perpendicular to $$F_1$$. Hence $$|stab(F_1)|=4$$.

By orbit-stabilizer theorem we now have $$|G|=6\times 4=24$$.