# Symmetry group of a cube

Given a cube $C \subset \mathbb{R}^3$ with the vertices $(x, y, z)$, where $x=\pm 1,y=\pm 1,z=\pm 1$, and given G, the symmetry group of the cube, I've got show that G includes the subgroup M, that includes all transformations by $(u_1,u_2,u_3)\longmapsto (\pm u_1,\pm u_2,\pm u_3)$.

Furthermore, I've got to prove for $\sigma \in S_3$, that $\Phi_{\sigma}:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, $(u_1,u_2,u_3)\longmapsto (u_{\sigma(1)},u_{\sigma(2)},u_{\sigma(3)})$ is an element of G.

I really don't understand the geometric meaning of these two mappings: I'd be grateful if someone could help me.

• Just apply those transformations to some corners and see where they move. – celtschk Mar 13 '17 at 21:59

Some examples:

$(u_1, u_2, u_3) \mapsto (-u_1, u_2, u_3)$: reflection in the $u_1 = 0$ plane.

$(u_1, u_2, u_3) \mapsto (-u_1, -u_2, u_3)$: rotation through $180$ degrees about the $u_3$-axis.

$(u_1, u_2, u_3) \mapsto (-u_1, -u_2, -u_3)$: inversion.

$(u_1, u_2, u_3) \mapsto (u_2, u_1, u_3)$: reflection in the $u_1 = u_2$ plane.

$(u_1, u_2, u_3) \mapsto (u_2, u_3, u_1)$: rotation through $120$ degrees about the line $u_1 = u_2 = u_3$.

You can view the first map as drilling an axis going from the middle of an edge, through the center to the middle of the opposite edge and rotate $180°$. This interchanges to vertices on the same edge, but since the cube is rigid also swaps the remaining edges.

The second map can be viewed as drilling an axis from a vertex, through the center to the opposite vertex. Rotating $120°$ then interchanges the three edges that are concurrent in this vertex.