Line graph of Cayley graph of $\mathbb{Z}_2^3$ is $A_4$

Question

Consider the group $G=\mathbb{Z}_2^3$ with generators $S=\{e_1,e_2,e_3\}$ with $e_1=(1,0,0),e_2=(0,1,0),e_3=(0,0,1)$.

The Cayley graph $\text{Cay}(G,S)$ is the 3D hypercube graph. It's line graph $\Gamma$ is the $4$-regular Cuboctahedron graph. This graph is the Cayley graph of $A_4$ with generators $g=(1 2 3), h=(234)$ (as noted here).

I want to understand the fact the that $\Gamma$ is the Cayley graph of $A_4$ (or some other group) through the Cayley graph structure of $\text{Cay}(G,S)$. I will try to construct a group $H$ and a set $T$ of $4$ generators such that $\Gamma=\text{Cay}(S,T)$. We will probably have $H\cong A_4$.

So, as the elements of $H$ I'm taking the elements of $G$ "modulo" the generators in $S$. I use the following notation for these $12$ elements:

?00, ?01, ?10, ?11, 0?0, 0?1, 1?0, 1?1, 00?, 01?, 10?, 11?

Next, I would to define the multiplication on these $12$ elements and choose the set of $4$ generators $T\subset H$.

Who are the neighbors of 00? in $\Gamma$? They are ?00, 0?0, ?01, 0?1. So, the $4$ generators must be $00?^{-1}\cdot?00$ and $00?^{-1}\cdot0?0$ and $00?^{-1}\cdot?01$ and $00?^{-1}\cdot0?1$.

Similar relations should hold for the neighbors of all other vertices of $\Gamma$. So we have many equations describing the $4$ generators, although they don't say which generator is which.

I'm a little stuck here. I've listed some required conditions, but I don't know how to complete this to a specific group structure and $4$ generators.

What is a (relatively easy to describe) way to give those $12$ elements a group structure and choose $4$ generators such that the resulting Cayley graph is $\Gamma$ (the line graph of $\text{Cay}(G,S)$)?

I hope I made it clear that what I'm looking for is a group strucutre easy to describe in terms of the names of these 12 elements, as given.

Answer 1

Note that $\mathbb{Z}_2^3$ acts naturally on your set $H$ of 12 elements, by addition. For example, $?01+010=?11$ and $?01+100=?01$. Next, note that $S_3$ also acts on $H$, by permuting the coordinates. For example, if we apply the cyclic rotation $(123)$ to $?01$, we get $1?0$. It's not hard that, together, this gives us an action of $\mathbb{Z}_2^3\rtimes S_3\cong S_4$. Simply take $A_4$ inside there, and check that this is a regular action, so you can identify your set with $A_4$. For the connection set, you can take any two inverse pairs of elements of order $3$.