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My problem ist the following: Let G be a group with generating set X. We can look the Cayley-Graph $\Gamma(G,X)$ of G. Let $x\in G$. Then it holds: $d_{\Gamma}(v,xv)\leq 1$ f.a. $v\in G=\Gamma(G,X)$ if and only if $x\in X$. Why is that true? I know, that $d_{\Gamma}(v,vx)=d_X(v,vx)=|v^{-1}vx|_X=1$, where $d_X$ is the word metric on G relative to X, iff x lies in X. I think its not very difficult, but I think I make a mistake in my thinking about the problem. Thanks for help. |
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I am wondering if it is true. Take the example of $G = \mathbb{Z}/2\mathbb{Z} \ltimes \mathbb{Z}$ where the composition law is defined by $$(x,y)\cdot(x',y') = (x + x',(-1)^xy'+y).$$ Set $S = \{(0,1),(0,-1) ,(1,0)\}$, then for example $(1,-2) = (1,0)\cdot(0,2)$, but I don't think that $d((0,2),(1,-2)) \le 1$. |
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I shall attempt to replace my incorrect "proof" with a counter-example. I should say first that the counterexample suggested by Jacob already is probably correct, but I find the graph in that example hard to visualise, so I'll give a different one. Take $G=\langle a,b\rangle$ to be the free group on generators $a$ and $b$, and let $X=\{a,b\}$. Then letting $v=ba$ and $x=b$, we find $d_\Gamma(ba,b^2a)=3$ - you can see this by looking at the graph, or by using $d_\Gamma(ba,b^2a)=d_X(ba,b^2a)=|b^2aa^{-1}b|_X=|b^3|_X=3$. However, $b\in X$. It may still be true that if $d_\Gamma(v,xv)=1$ then $x\in X$, but I'm not sure. |
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