Let $K$ be a simplicial graph [i.e. no loops and no double edges]. Let $G$ be a group which acts on $K$ by graph isomorphisms. Suppose that the action satisfies:

  • $K$ is connected

  • finite number of orbits of edges (and vertices)

  • each edge stabilizer is a finite subgroup of $G$

(Dis)prove that there is just a finite number of orbits of "triangles", i.e., circuits [injective path loops] of length $3$.

PS. vertices could of course have infinite degree!

Thanks to Shaun for the advice!

The problem is false if we use $4$ circuits instead of triangles: consider the graph with vertices $x, y$ (symbols) and the integers, with edges: $[x, n], [n, y]$ for every integer $n$ Let $G$ be $\mathbb{Z}$, with action $\mathbb{Z} \times K \to K$ given by $m\cdot x(y)=x(y)$ and $m\cdot n=m+n$.

This action is free on edges, with two edges orbits. But it is easy to see that there are infinite orbits of $4$ circuits!

  • 1
    $\begingroup$ Please show your working so far. It'll help you and encourage others to help you :) $\endgroup$
    – Shaun
    Nov 7, 2013 at 10:43
  • $\begingroup$ I put your question into TeX for readability - can you check I got the action right in your example? I think that's what you meant, but I'm not completely sure. $\endgroup$
    – mdp
    Nov 7, 2013 at 10:58
  • $\begingroup$ yes it is correct. Thank you! $\endgroup$
    – fritz
    Nov 7, 2013 at 11:03


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