# Finding a colouring of an octahedron such that the stabilizer is $S_3$, $V_4$ or $A_4$

I am a bit playing around with the octahedral symmetry, and inspired by another exercise I did on finding configurations of the cube such that the stabilizer is equal to a certain group, I tried doing the same for an octahedron. I considered two ways to 'colour' the octahedron. The first is using colouring each face either black or white (the colours do not have to be used equally often), the second is to draw on each face an arrow pointing towards one of the three adjacent vertices.

EDIT 2: My work on the colouring of the octahedron with stabilizer $S_3$ was incorrect, as pointed out by Clément Guérin. Therefore I deleted this. I also just thought about the following. If we allow all rotations that leave one middle-of-two-opposite-faces-connecting line fixed (i.e. stabilizer is $S_3$), this means that we might colour one pair of opposite faces black, and the rest of the $6$ faces white, to get that $S_3$ is the stabilizer of this colouring. To me, this seems correct, and maybe someone can verify this. If so, we found colourings with stabilizer $S_3$, $V_4$, $A_4$ and $C_4$. Similarly I conjecture that a colouring of an octahedron with all but two faces white, and the two black faces are on the bottom of the octahedron, and not adjacent, would have stabilizer $C_2$.

If this is correct, the question is completely answered for the case of a colouring. Perhaps someone also knows how to deal with arrow configurations then, and I will also think about this.

EDIT As Steven Stadnicki pointed out a 'checkerboard pattern colouring (i.e. colour one face black and each adjacent face with a the other colour then the neighbours) the stabilizer is $A_4$ indeed! I coloured the printout below.

Question: Am I right in my claims? $A_4$ was solved by Steven Stadnicki, $V_4$ and $C_4$ by Clément Guérin, and following the thoughts of Clément Guérin I have a conjecture on $C_2$ and $S_3$. The arrow configurations are still unclear to me.

If you find a configuration such that the stabilizer is isomorphic to either $A_4$, $V_4$ or $S_3$, I would find it most helpful if you draw it explicitly on a printout or an octahedron.

I hope I made my question clear, because I am really in doubt about my work and my spatial visualization ability is not that strong.

• There's an essentially unique 'checkerboard' coloring of the octahedron such that every face is adjacent along its edges only to faces of the opposite color; isn't the stabilizer of that coloring $A_4$? – Steven Stadnicki May 23 '18 at 22:00
• @StevenStadnicki: you are right! I can see directly that the stabilizer is indeed $A_4$. Thank you, I will make an edit :) – Václav Mordvinov May 24 '18 at 5:55

Up to direct isometry of the octahedron, you have $7$ colorings of the faces of the octahedron. It suffices now to compute the stabilizer of each of these colorings. Let $S_i$ be the stabilizer of the $i$-th coloring.

• Let $s$ be in $S_1$ then clearly the vertex on the bottom needs to be fixed by $s$. From this we directly see that $S_1$ needs to be isomorphic to $\mathbb{Z}/4$.

• For the second one it is the Klein group. You have one axis of $\pi$ rotation going from the middle point of the upmost leftmost edge to the opposite one, another $\pi$ axis of $\pi$ rotation going from the middle point of the upmost rightmost edge to the opposite one and an axis of $\pi$ rotation going from the front vertex to the opposite one.

• The third one is the checkerboard coloring and has a stabilizer isomorphic to $A_4$.

• For the forth one, $S_4$ needs to be trivial (the vertex on the bottom needs to be fixed)

• For the fifth one, $S_4$ needs to be trivial (the vertex on the bottom needs to be fixed)

• For the sixth one, $S_6$ needs to be trivial (the vertex on the bottom needs to be fixed)

• For the seventh one, $S_7$ needs to be trivial (the vertex on the bottom needs to be fixed)

Remark, I think that the coloring you claim has $Sym_3$ as a stabilizer is the fourth one, because both have a chain of three connected black faces and one isolated black face. From this picture it seems clear that it has no direct symmetry.

Edit: using Polya's formula you have, up to direct isometry $1$, $1$, $3$, $3$, $7$ colorings with $(n_w,n_b)=(8,0),(7,1),(6,2),(5,3),(4,4)$respectively where $n_w$ (resp. $n_b$) is the number of white (resp. blue) faces. Below are the different colorings

The respective stabilizers are $Sym_4$, $\mathbb{Z}/3$, $\mathbb{Z}/2$, the Klein group (Edit2: this is wrong, actually, it is $Sym_3$), $\mathbb{Z}/2$, trivial group, trivial group and $\mathbb{Z}/3$.

• Thank you very much for your answer! I really appreciate your help. You considered all the case of colouring with two colours that are used equally often. I will leave the question open for a bit longer because maybe someone also knows of an answer for the case the colours are not equally often, or for an arrow configuration. – Václav Mordvinov May 24 '18 at 15:46
• By the way, with what programn did you make these coloured octahedrons? They look really good. – Václav Mordvinov May 24 '18 at 15:49
• @VáclavMordvinov, I made them with tikz (the integrated drawing tool of LaTeX). – Clément Guérin May 24 '18 at 20:00
• Thanks for your edit! (And sorry if the question was not clear in the first case)I figured the case for arrow configurations out myself, and I will try to post an answer for this myself this afternoon, and afterwards I will accept your answer and award you the bounty – Václav Mordvinov May 25 '18 at 8:35
• Isn't the stabilizer of your fourth octahedron (of the added ones in your edit) isomorphic to $S_3$? Any rotation which fixes the diagonal of the two coloured opposite faces fixes this colouring. – Václav Mordvinov May 25 '18 at 9:22

The 'main' case, for colouring the octahedron with two colours is thoroughly covered by Clément Guérins answer, and this yields colourings with stabilizers $C_1$, $C_2$, $C_3$, $C_4$, $V_4$, $S_3$, $A_4$ and $S_4$. So for all possible subgroups of $S_4$, we only haven't seen $D_4$ here (yet?).

In this answer, for the sake of completeness, I will also post my solution for arrow configurations. Since I am the asker of this question, I am not $100\%$ sure if this solution is correct. If you think there is a mistake, please let me know.

Note that an 'arrow configuration' is a way to colour the octahedron such that we to draw on each face exactly one arrow pointing towards one of the three adjacent vertices.

So for all the possible subgroups of $S_4$: $C_1$, $C_2$, $C_3$, $C_4$, $V_4$, $S_3$, $D_4$, $A_4$ and $S_4$, I first show that there cannot be an arrow configuration with a stabilizer that has an order divisible by $3$. Suppose to the contrary that we find such a configuration. By Cauchy's theorem, now there is a rotation of order $3$ that maps the coloured octahedron onto itself. Since $\gcd(3,8)=1$, with $8$ the number of faces, there are (at least two) faces that are mapped onto themselves. Such a face is an equilateral triangle with an arrow (without loss of generality) towards the top of the triangle. After rotation over $120^{\circ}$ this is mapped onto itself; but now the arrow points either left or right, a contradiction. So the order of the stabilizer is not divisible by $3$. So for any configuration, the stabilizer should be isomorphic to one of the following groups: $C_1$, $C_2$, $C_4$, $V_4$ or $D_4$.

A configuration such that the stabilizer is trivial is really easy to find. I will give examples of configurations such that the stabilizer is $C_2$, $C_4$, and $D_4$. The black arrows are on faces you can see from the given perspective; the grey ones on faces you cannot. These configurations with their stabilizers are shown below.

To give some explanation: because of the above reasoning the stabilizer of any of the colourings below must be isomorphic to one of the groups $C_1$, $C_2$, $C_4$, $V_4$ or $D_4$. The first is only fixed by a rotation over $180^{\circ}$ through the line through the top and bottom vertices. The second is clearly fixed by the (the generated group of) a rotation of order $4$ through the line through the top and bottom vertices, but not by any other rotation. The third is clearly fixed by any rotation with order $4$ through the line through the top and bottom vertices, but also by a rotation over $180^{\circ}$ through the line through the left and right vertices. Say this is configuration $C$. Since $|\mathrm{Stab}(C)|>4$, $\mathrm{Stab}(C)\neq C_1,C_2,C_4,V_4$, thus $\mathrm{Stab}(C)=D_4$.

There does not seem to be a configuration so that the stabilizer is $V_4$. I checked the case for a configuration that is only fixed by $180^{\circ}$ opposite vertices rotation, so for $V_4=\{\mathrm{Id},(12)(34),(13)(24),(14)(23)\}$. I'm not sure about the case where we would set $V_4=\{\mathrm{Id},(12)(34),(12),(34)\}$.