I am trying to show that given a finitely generated group $G$ and generators $S$ then $G$'s action on a graph is the full symmetry group of the graph . I try to show it for the Cayley graph of $G$ and $S$, $\cal{G}$ , given by $g.s=gs$.

I may have choose a wrong action. (or a wrong graph, it may be a full symmetry group on a other graph) But i am stuck, Given a graph automorphism $\phi:\cal{G}\to\cal{G}$ I try to show that if $\phi(e)=g$ then $\phi (h)=g.h$. I tried with induction on the word length but having trouble with the step. I can see why in the step we get neighbourhood relations of order 2, But not equality. I think Maybe I need to change the graph a little bit so the "neighbourhood permutation" won't be aposible symmetry. maybe by adding some vertexes or something.

Help will be highly appreciated!


What you are trying to show is false.

For a counterexample, take the Cayley graph of the standard presentation of $\mathbb{Z}^2$ given by $$\mathbb{Z}^2 = \langle a,b \bigm| [a,b]=1\rangle $$ The Cayley graph is just the union of horizontal lines in $\mathbb{R}^2$ with integer $y$-coordinate and vertical lines with integer $x$ coordinate. The rotational and reflective symmetries of this graph are not in the action of $\mathbb{Z}^2$.

  • $\begingroup$ Well part of my question is if I can make modifications to the graph to brack those symmetrys? $\endgroup$ – sha Dec 3 '16 at 10:09
  • $\begingroup$ That's rather vague. One modification is simply this: for the edge from $g$ to $gs$, add an arrow pointing from $g$ to $gs$ and write the letter "$s$" next to the arrow. But I have no idea if those modifications are what you have in mind. Probably you can simulate those modifications by attaching a complicated little subgraph to the middle of the edge which looks like an arrow and the letter "$s$". $\endgroup$ – Lee Mosher Dec 3 '16 at 14:05

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