# Kernel of a group action of rotation of a cube

Question:

Let G be the rotation group of a cube

Show that G has an action on a set of size 3.

Well, if we consider axes through each opposite faces, then this set has only 3 possible axes. The trivial rotation $R_{0}$ fixes all axes; that is, all axes are invariant under the $R_{0}$. Since the trivial rotation is the only element in G that fixes every axes elements through each opposite faces, the group action of G on a set of size 3 exists, and in particular, is a faithful group action.

Interpret this action geometrically.

Under the trivial rotation, any axes stays fixed.

What is the kernel of this action, $\mu$?

I do not understand this question.

Any help is appreciated.

• The kernel of an action is the set of elements of the group which fix all the objects on which the group acts. – Tobias Kildetoft Jul 22 '16 at 8:58
• Also, there are nontrivial rotations which preserve the axes, at least the half-turns around the axes. – arctic tern Jul 22 '16 at 9:00
• It doesn't make sense for a group element to send an axis to the identity, because the identity is a group element not a feature of the cube. A group action of $G$ on $X$ is typically formalized as a group homomorphims $G\to \mathrm{Perm}(X)$, and it makes sense to speak of kernels of group homomorphisms. – arctic tern Jul 22 '16 at 9:01
• @arctictern That was a typo. – Mathematicing Jul 22 '16 at 9:02
• There are four elements in the kernel. The identity element and the three half-turns around the three axes. I find your notation for the kernel a bit ugly, as it is more suited for the $G\to\mathrm{Perm}(X)$ definition of group action that it is for the $G\times X\to X$ definition of group action (and let's stick to left actions if possible). Also, you would need to quantify "for all $\omega$" for that definition of $\ker\mu$ to work. – arctic tern Jul 22 '16 at 11:09

1. A $90^{\circ}$ rotation around the axis between the centers of two opposite faces. These rotations have order $4$, and so any one of them "squared" has order $2$. Such a rotation fixes one of your three axes you are acting on (think of them as the $x,y,z$ axes if you imagine your cube with corners at $(\pm\frac{1}{2},\pm \frac{1}{2})$ in $\Bbb R^3$), and "swaps" the other two. So what happens with the "squares"? There are three such rotation "generators", accounting for $10$ of the $24$ cube rotations (this includes the "null rotation", or identity).
2. A $180^{\circ}$ rotation about the axis between the mid-points of two diagonally-opposite edges (this kind of rotation is the hardest for people to visualize). Convince yourself this kind of rotation fixes just one of your three axes. This accounts for $6$ more of the cube rotations.
3. A $120^{\circ}$ degree rotation about the axis between two diagonally opposite corners. Again, convince yourself that this kind of rotation fixes none of the three axes we're acting upon. There are $8$ of these ($4$ corner pairs and two directions of rotation for each corner), which thus accounts for the rest of the $24$ cube rotations.