Cayley graphs for all orthogonal symmetries of Platonic solids

Generator symmetries c, b, and a, applied sequentially to the octahedron, resulting in a composite operation of order 2

I've become a bit obsessed with Cayley graphs recently. I figured out a way to construct a graph for the rotational symmetries of the 5 Platonic solids, and the result was quite elegant, I thought.  And it seemed to me that extending this to the complete set of  orthogonal symmetries (which includes reflections) should be quite simple - it is just twice the number of nodes after all.  However,  it took a surprising amount of staring into the middle distance and mumbling to myself to come up with the answer. And I'm going to describe its construction in this post. 

As an example I'm using the octahedron here,  but the construction works in exactly the same way whichever solid you choose.  My generators I'm calling $a$, $b$, and $c$, where $a$ is a clockwise rotation of a face,  $b$ is a clockwise rotation around an adjoining vertex,  and $c$ is a reflection that keeps that vertex and one whole edge of the face fixed.

It follows immediately from the definitions that $a^n=1$, $b^m=1$, and $c^2=1$, where $n$ is the number of edges per face and $m$ the degree of each vertex.  I showed in the previous post that, in addition,  $(ab)^2=1$.  All that is now required is a relation that connects $c$ to $a$ and $b$. Then we can start building a Cayley graph and check it satisfies all the relations and has the required number of nodes. 

The picture above shows some operations applied to an octahedron, which is viewed looking straight onto the vertex. The composite operation $cba$ reflects the solid in a plane that bisects an edge.  Doing this twice returns it to its starting point, which implies that $(cba)^2=1$.  Now let's see if these relations lead to a representation group the same size as the symmetry group, namely $2mv$, where $v$ is the number of vertices in the Platonic solid$^\dagger$.

Partially complete Cayley graph showing transitions between reflected orientations in light grey and transitions between unreflected orientations in black. 

I constructed the (partially-complete) Cayley graph above in the following manner: First split the Platonic solid into separate $n$-gons so that in the case of the octahedron you have 8 equilateral triangles almost, but not quite, in contact. Next connect the faces with $m$-gons (squares in the case of the octahedron) where each vertex used to be.  Add clockwise arrows around the $n$-gons representing the symmetry $a$, and clockwise arrows around the $m$-gons representing the symmetry $b$. At this point you have a graph representing a group  of $mv$ elements, which is spanned by $a$ and $b$ and in which $a^n=1$, $b^m=1$, and $(ab)^2=1$. This is the same size as the rotation group and the relations used are all satisfied by the rotation group. From that we may infer it is the rotation group. 

There is one final step needed to get to the graph for the full symmetry group, including the reflections. Obviously we need double the number of nodes so we must duplicate the existing graph and shrink it so that it sits inside the original graph, Russian doll-like. However, when we do this we must also reverse all the arrows! On the outer shell we must twist each $m$-gon slightly clockwise, and on the inner shell they should be twisted slightly anticlockwise.  This brings alternate nodes in each $m$-gon pair slightly closer together, and these must be connected by edges representing the reflection symmetry,  namely $c$.

Hey presto, the final graph can be inspected to verify it is the Cayley graph for

$\langle a,b,c \vert a^n=b^m=(ab)^2=(cba)^2=1\rangle$

The graph also shows that the group corresponding to the above representation has $2mv$ nodes. As this matches the size of the full symmetry group the two groups must be the identical.

FOOTNOTES

• $\dagger$ If they do the two groups must be the same. To see why, note that the natural homomorphism from the representation $R$ to the symmetry group $G$ is onto, since the symmetries $a$, $b$, and $c$ span the symmetry group.  That implies that $G \approx R/K$ where $K$ is the kernel of the homomorphism. But $|R/K|=|R|/|K|$, and so if $|R|=|G|$ then $|K|$ must be $1$ and the homomorphism must be an isomorphism.