# Let C be a cube and let G be its rotational symmetry group. Outline a proof that G is isomorphic to Sym(4)

Let C be a cube and let G be its rotational symmetry group. Outline a proof that G is isomorphic to Sym(4).

I am trying to use this https://www.math.lsu.edu/~verrill/teaching/math4023/spring2006/handouts/week8/cubesolution.pdf as a guide to construct the permutations for the symmetry group.

Using the given $R_1$, I can show there is a subgroup {$R_1, R_1^2$, e} but then I can't seem to construct a working rotation for $R_1'$, the second rotation of that kind. Is there an easier way to do this, I can't picture which vertices go to which.

So far I have stated that Sym(4) clearly has order 24, as does the symmetry group of a cube, but I probably need to prove this? And both groups are not abelian, which I could prove just by taking two elements in Sym(4) and showing they are not commutative and the same with two elements in the cube symmetry group.

Also, since the groups have large order (24) would I just find the subgroups and hence show they are the same group structure or would I do something else?

Finally a second question I have is: The cube C is coloured in k colours. Two colourings are equivalent if one is obtained from the other by rotational symmetry of C. Find the number of inequivalent colourings.

Thanks

• IMO you should post the second question separately (except that it is likely a duplicate, i.e. has already been answered somewhere here). Many would like to say something about one of your questions but not both. – Jyrki Lahtonen May 15 '14 at 6:54

Let $S$ be the set of pairs of diagonally opposite vertices on a cube. Note that $|S| = 4$. Let $Cube$ act on $S$. A particular element $g \in Cube$ acting on $S$ will yield some permutation on the pairs. Can you show that there exists a $g \in Cube$ that will correspond to every possible $2$-cycle in $S_4$? If so, we can conclude that $S_4 \subseteq Cube$ since, in general, $S_n$ is generated by the set of all $2$-cycles. At this point, we are done since $|S_4| = |Cube|$, so $S_4 \subseteq Cube \Longrightarrow S_4 \cong Cube$.