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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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2 Answers 2

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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1  
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...!? –  Peter Feb 22 '12 at 14:34
    
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. –  Peter Feb 22 '12 at 14:37
    
we got an edge between two vertices $g,g'$ iff $gx=g'$ for some $x\in X\cup X^{-1}$. –  Peter Feb 22 '12 at 14:55
    
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. –  Matt Pressland Feb 22 '12 at 16:21
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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! –  Peter Feb 22 '12 at 17:03

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