# action of symmetry group of cube on pairs of opposite faces.

I want to solve the following problem from Dummit & Foote's Abstract Algebra:

Explain why the action of the group of rigid motions of a cube on the set of three pairs of opposite faces is not faithful. Find the kernel of this action.

My attempt: If the action is faithful, we would have an injective group homomorphism $G \to S_A$ where $A$ is the set of three pairs of opposite faces of the cube. That means $|G| \leq |S_A|$, but $|G|=24,|S_A|=6$. Thus this action cannot be faithful.

In order to see what elements of $G$ are in the kernel, we list them all explicitly:

1. There are 3 rotations around a line through centres of two opposite faces, each of order 4.
2. There are 4 rotations around a line through opposite vertices, each of order 3.
3. There are 6 rotations around a line through midpoints of opposite edges, each of order 2

Thus we have $3(4-1)+4(3-1)+6(2-1)=23$ nonidentity rotations. If we add the identity we see that we indeed have listed all elements of $G$. It is now geometrically evident that the kernel of the action consists of the squares of rotations in part $1$, as well as the identity.

Is my solution correct? If not, please help me fix it. Thanks!

• Your solution is correct, but in stead of saying that "It is geometrically evident that...", you could give a proof of the fact that the kernel consists of the squares of rotations in part 1 and the identity, and nothing else. – Servaes Apr 17 '14 at 9:32