# (Symmetric) group acting on a graph

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)

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Well, if I wanted to write down a group $P$ which admits a map from $\pi(X)$ I'd probably do it by writing down a map out of $X$. With that group $G$ around, the obvious map is the quotient map $X \to X/G$... – Qiaochu Yuan Sep 16 '12 at 23:07
Here is an attempt: Let $f: X \rightarrow Y$ be a surjective map of graphs (vertices surject onto vertices and so do edges; basically it is a contraction of certain edges and vertices)then we can talk about the group Aut(X/Y) which are graph automorphisms of $X$ commuting with the map $f$. In your case group $P$ is the group $Aut(\tilde{X}/(X/G))$. Well here $\tilde{X}=$ universal cover of $X$ and $X/G$ the quotient graph of $X$ by $G$. – s.b Sep 16 '12 at 23:21