# Cube. Coloring. Isomorphism.

• Apr 29th 2013, 03:07 PM
vercammen
Cube. Coloring. Isomorphism.
Let $\displaystyle G$ be a group of rigid motions of cube.

a) Show that $\displaystyle G = S_4$

b) Show that the alternating subgroup $\displaystyle A_4 \le S_4$ is isomorphic to the group
of rigid motions of regular
tetrahedron.

c)Find cycle index for both $\displaystyle S_4$ and $\displaystyle A_4$.

d) Determine pattern inventory of $\displaystyle m$ coloring of a set
$\displaystyle X = \{d_1,d_2,d_3,d_4\}$, where $\displaystyle d_i$ is the $\displaystyle i$ -th diagonal of
the cube, $\displaystyle I(c_1,c_2,...,c_m)$ (representing colors $\displaystyle 1$ to $\displaystyle m$) for
both $\displaystyle S_4$ and $\displaystyle A_4$.

a) Consider how $\displaystyle G$ acts on - either the set of 6 faces, the set of 12 edges, the set of 8 vertices of a cube. I recommend using the last one. Regardless of your consideration, use the orbit-stabliser theorem to show that $\displaystyle |G|=24$.
Now note that vertices which are opposite each other remain opposite of each other under any rigid motion. Thus $\displaystyle G$ acts on the set of pairs of opposite vertices, of which there are 4. This induces a homomorphism $\displaystyle G\to S_4$, and you just need to show that the map is injective (or surjective) since the orders of $\displaystyle G$ and $\displaystyle S_4$ agree.
b) Same idea as before. This time, consider the set of the triangular faces of the tetrahedron (this has cardinality 4). Orbit-stabilizer argument shows that the order of the group is 12. Since G acts on the 4 faces, you have a homomorphism $\displaystyle G\to S_4$. Now show that the kernel of this map has order 2.