Partition of Facets of Rubik's Cube

Theorem

Let $S$ denote the set of the facets of Rukik's cube.

Then $S$ can be partitioned as follows:

$S = \set {S_C \mid S_E \mid S_Z}$

where:

$S_C$ denotes the set of corner facets

$S_E$ denotes the set of edge facets

$S_Z$ denotes the set of center facets.

Proof

From the definition of the facets, each face is divided into $9$ facets.

A facet is either:

on the corner of a face, for example $flu$, $fru$

on the edge of a face, for example $fu$, $fr$

in the center of a face, for example $F$.

$(1):\quad$ Each facet can be either in $S_C$ or $S_E$ or $S_Z$ and can not be in more than one.

$(2):\quad$ Each facet can be either in $S_C$ or $S_E$ or $S_Z$ and there are no other possibilities.

$(3):\quad$ None of $S_C$, $S_E$ and $S_Z$ is empty.

Thus the criteria for $S = \set {S_C \mid S_E \mid S_Z}$ to be a partition are fulfilled.

$\blacksquare$

Sources

• 2008: David Joyner: Adventures in Group Theory (2nd ed.) ... (previous) ... (next): Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson: Example $1.2.3$