I'm currently reading "Groups acting on graphs" by Dunwoody and Dicks, and I came across section 9 (p. 39), which states that if $G$ is a group which acts on a graph $X$, then there is a group $P$ which is an extension $1 \to \pi(X) \to P \to G \to 1$.
Now, I don't quite understand the proof which is given in the book, but I wanted at least to see if I could apply it in a simple case, like the example which is given with a cube.
So, if one considers $X$ to be the skeleton of a tetrahedron (with 12 points), and the symmetric group $S_4$ which acts on $X$, I then calculated that one should have the following extension:
$1 \to \ F_3 \to S_3\underset{\mathbb{Z}_2}\star (\mathbb{Z}_2\times\mathbb{Z}_2) \to S_4 \to 1$
where $F_3$ is the free group on 3 generators. Is this correct ?
(By the way, if one has at least an intuitive sketch of the proof for the extension theorem, I would be interested as well)