Can we get the line graph of the $3D$ cube as a Cayley graph?

Given a graph $G=(V,E)$, the line graph of $G$ is a graph $\Gamma$ whose vertices are $E$ (the edges of $G$) and in $\Gamma$, two vertices $e_1,e_2$ are connected if, as edges in $G$, they share an endpoint.

Now let $G$ be the $3D$ cube graph. It has $8$ vertices, $12$ edges and it is $3$-regular. It is actually the Cayley graph of $\mathbb{Z}_2^3$ with generators $(1,0,0),(0,1,0),(0,0,1)$.

Let $\Gamma$ be the line graph of this $G$. Then $\Gamma$ has $12$ vertices and it is $4$-regular and highly symmetric.

Is $\Gamma$ the Cayley graph of some group $H$ (of order $12$) with a symmetric set of generator $S \subset H$ ($|S|=4$)?

(by a symmetric set I mean that $x\in S \Rightarrow x^{-1}\in S$, making the Cayley graph undirected).

• There are only 5 groups of order 12 (up to isomorphism). Have you tried any of them on for size? – Gerry Myerson Jun 12 '13 at 9:47
• @Gerry: I looked in wikipedia, but it is not just a matter of choosing the group, but also choosing the generators. So, although I didn't see how to do it, I might have chosen the generators in the wrong way. – Alex Jun 12 '13 at 10:08
• @user1729: I think I need to choose $4$ generators of of group of size $12$, because every edge is adjacent to 4 other. – Alex Jun 12 '13 at 10:20
• @Alex Sorry, I meant four, and then I realised that it should be $4/2=2$, or there are generators of order two... – user1729 Jun 12 '13 at 10:21
• OK, but maybe you can rule out some of the groups. – Gerry Myerson Jun 12 '13 at 10:35

I haven't checked this, but try $H=A_4$, with $S=\{g,h,g^{-1},h^{-1}\}$ where $g=(1,2,3)$ and $h=(2,3,4)$. So vertices of the cube are labelled alternately with $g$ and $h$, and multiplication of an element (egde) by a generator (vertex) corresponds to rotating the edge clockwise round that vertex (or anticlockwise for multiplication by the inverse).
• Do you mean $H=A_4$? – Alex Jun 12 '13 at 10:16