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## Problem

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 \(|G|=6\times 4=24 \).

## Helpdesk

**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.