# Finding a 'colouring'/configuration of a cube such that the stabilizer is isomorphic to a given group

I'm making a few old exams to practice for a group theory exam. In every old exam I practiced, there is a question such as the following:

Suppose we have a cube and on each face, we draw an arrow starting from the middle of the face, towards one of the four vertices of that face.

Then first, I have to find the number of ways to do this up to rotation symmetry. Thus this is a kind of 'standard' exercise where you should use Burnside's counting lemma.

But then, they ask for the following: there is a twodimensional printout for a cube given: and then they ask to draw arrows on this printout such that the stabilizer$^*$ is isomorphic to $S_3$. Now I know that the rotation of a cube is $S_4$ and that we can classify these rotations as follows:

• The identity permutation
• $6$ rotations over $90$ degrees through a line through two opposite faces.
• $3$ rotations over $180$ degrees through a line through two opposite faces.
• $8$ rotations over $120$ degrees through a main diagonal of the cube.
• $6$ rotations over $180$ degrees through the middle of two opposite edges.

Now the solution should be the following

but I really don't see how I should come up with this solution. I even don't see why this solution is correct. There are also similar exercises where you have to fill in the printout such that the stabilizer is isomorphic to $V_4$ and $A_4$.

I cannot find anything silimar on the internet and this is never explained in class. Any help is much appreciated!

$^*$: of this configuration of arrows, the rotation group of the cube $S_4$ acts on the set of arrow configurations

• +1 Fun question! – Servaes May 19 '18 at 11:36

One way to see this is by visualizing the action of $S_4$ on the cube. The cube has four long diagonals, by which I mean diagonals connecting diametrically opposite vertices of the cube (see image below). The rotations of the cube correspond bijectively to the permutations of these diagonals.
Now an obvious way to realize $S_3$ as a subgroup of this $S_4$ is as the stabilizer of one of the diagonals; if one diagonal is fixed the other three can still be permuted freely by rotations fixing this diagonal. Therefore we want to draw arrows in such a way that they point out one diagonal while being symmetric with respect to the other diagonals. A straightforward way to do this is to choose a diagonal and to draw three arrows at each of its two vertices, all pointing towards their vertex.
In general, to find a configuration with stabilizer $H$, first pick a subgroup $H\subset S_4$ and then determine the orbits of the arrow positions under the action of $H$. Then within each orbit, label all position identically, and label distinct orbits distinctly. Then the stabilizer will be $H$.
• Thank you, you couldn't make it any clearer for me! Do you also know how to find a configuration such that the stabilizer is $A_4$ or $V_4$, for example? like in the other exercises I mentioned. Because I understand this case now, but I still don't get how to treat this problem in general. I was thinking here about only the even permutations of the main diagonals ($A_4$), but I don't really know what a fixed point ('configuration') would look like then. For $V_4$, I was thinking about switching the two pairs of main diagonals, but I also cannot visualize this – Václav Mordvinov May 19 '18 at 11:12
• The subgroup $A_4\subset S_4$ is generated by the $3$-cycles. These correspond to the rotations of the cube fixing a single diagonal. This divides the vertices of the cube into two orbits (draw them!). Within each orbit all vertices must be labelled identically, and symmetrically with respect to rotation. So either blank, or three arrows at each vertex, all pointing in the same direction. The two orbits must be labelled differently as otherwise the stabilizer will be $S_4$. – Servaes May 19 '18 at 11:26
• Similarly the subgroup $V_4\subset S_4$ is generated by (and consists of) the pairs of disjoint transpositions. These correspond to half-turns around the axes orthogonal to the faces of the cube. This divides the vertices of the cube into the same orbits as with $A_4$, but this time their labelling need not (and must not) be symmetric with respect to rotation. – Servaes May 19 '18 at 11:29