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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?

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 ).

**What is this topic:**Abstract Algebra**What are some of the associated concept:**Orbit-Stabilizer Theorem**Book Suggestions:**Abstract Algebra by David S. Dummit and Richard M. Foote.

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