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My problem is 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: $d_{\Gamma}(v,xv)\leq 1$ for all $v\in G=\Gamma(G,X)$ 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.

2 Answers 2

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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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    Thanks. But why is $d_\Gamma(v,xv)=|v^{-1}\cdot xv|_X=|1\cdot x|_X=d_\Gamma(1,x)=d_\Gamma(vv^{-1},xvv^{-1})$? Why does the word length in relation to X of the conjugated Element $v^{-1}xv$ equals the word length of x? So, why is it true that rightmultiplication in the Cayley-Graph is an isometry? I thougth only left multiplication is an isometry...!?2012-02-22
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    The Cayley-Graph $\Gamma(G,X)$ is the following Graph: Let G be a group with generating set X. Then the vertices are the elements of G and we got an edge between two vertices $g,g'$ iff $gx=g'$ for some $x\in X$. So the leftmultiplication with group elements of G acts in a natural way on the Cayley-Graph of a group.2012-02-22
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    we got an edge between two vertices $g,g'$ iff $gx=g'$ for some $x\in X\cup X^{-1}$.2012-02-22
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    Ah, you make a good point, I'm being very careless with the order of multiplication. You are correct that with this order only left multiplication must be an isometry, and thus my "proof" is not correct. I'll think about it some more and see if I can get anywhere. I'm now beginning to suspect that Jacob's counterexample may be correct, but I'm not certain how to prove it.2012-02-22
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    In fact I now also think I have a counter-example, so I'll update the answer to show that instead.2012-02-22
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    I checked the same Example. :-) I think it's incorrect formulated. The right formulation of the Theorem should be as follwos:2012-02-22
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    Let $G$ be a group and $\Gamma(G,X)$ the Cayley-Graph respective to the generating set X. Then the subset $X_v:=\{g\in G:d_\Gamma(v,gv)\leq 1\}\subset X$ generates $G$ f.a. $v\in G=\Gamma(G,X)$. $X_v=X$ if $v=1$.2012-02-22
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    That seems more plausible (and is also a very nice theorem). Sorry for the misinformation earlier!2012-02-22
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    This Theorem holds for an action by automorphism of a group $G$ on a connected graph $\Gamma$, with the property, that G acts transitive on the set of vertices of $\Gamma$, i.e. $Gv=\Gamma$ f.a. vertice $v$ of $\Gamma$.2012-02-22
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    This Theorem we proved in our lecture. As an example we denoted, that if $\Gamma$ is a Cayley-Graph for an generating set X of G. We get that $X=X_v$. But this is only true, if we set $v=1_G$. Because if we look at an Element of $G:=F_2:=\langle a,b \rangle$ the free group of rank 2, we can see that $X_v$ not necessary have to be equal to $\{a,b\}$, but we get $a,b\in X_v$. I checked that a few minutes before. So there was a little incorrectness in our script. Thanks for help!2012-02-22