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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!

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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 $\mathbb{Z}^2$.

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  • $\begingroup$ 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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