Thursday, 25 December 2008

gr.group theory - Cayley graphs of finitely generated groups

The answer is no, as expected. The following proof is "joint work" with L. Scheele.
Consider G=mathbbZ, K=Dinfty and H:=mathbbZtimesmathbbZ/2mathbbZ. Then GapproxK and KapproxH, but GnotapproxH.



Indeed, the Cayley graph associated to 1,1 for G and the Cayley graph associated to s,t where Dinfty=langles,t:s2=t2=1rangle are clearly isometric.



Similarly, the Cayley graph associated to s,st,ts for K and the graph associated to (0,1),(1,0),(1,0) for H are isometric.



However, let S be some symmetric generating set for G. Then Sk, the set of vertices which have distance precisely kgeq1 from the identity, has even cardinality because Sk is invariant under the mapping xmapstox and doesn't contain 0.



Now let T be some symmetric generating set for H. Let k0 be the distance of (0,1) from the identity in the associated graph. Then Tk0, the set of vertices which have distance precisely k0 from the identity, has odd cardinality. Indeed, it is invariant under the mapping (x,y)mapsto(x,y) since T is assumed to be symmetric. But (0,1) is a fixed point.

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