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Let $G$ be a group with finite generating set $A$, define the distance $$d(g,h)=\mathrm{min}\{n:gh^{-1}=a_1^{\varepsilon_1}\dots a_n^{\varepsilon_n}, a_i\in A,\varepsilon_i=\pm1\}.$$ Define the ball of radius $n$, $$B_{G,A}(n)=\{g\in G:g=a_1^{\varepsilon_1}\dots a_m^{\varepsilon_m}, a_i\in A,\varepsilon_i=\pm1,m\leq n\}.$$ And define the function $\beta_{G,A}(n)=|B_{G,A}(n)|$, I want to prove that this is a linear function if and only if the group is virtually cyclic. I proved that if the group is virtually cyclic then this function is linear, I have some problems in the other direction. Looking at the example of the infinite dihedral group $D_\infty=\langle a,b|a^2=b^2=1\rangle$, I thought that maybe if we define $g$ to be the product of all the generators and $H$ the subgroup generated by $g$ then this can work, but I couldn't prove that this is true (I even don't know if this is true).

Of course we can suppose $G$ is infinite, otherwise this is trivial.

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springerlink.com/content/k63jq68u7pt76756 might be helpful. I don't know of any more elementary arguments. The one in the paper is long, and I don't feel like reproducing it :) –  user641 Mar 28 '12 at 23:29
    
I have already found that paper, but I couldn't find a free copy –  Alex M Mar 28 '12 at 23:32
    
What you propose is not true. Consider x=ab in $D_\infty$. Then $\langle a,x\vert a^2, (ax)^2\rangle=D_\infty$ is a counterexample (the group $\langle ax\rangle=\langle b\rangle$ is finite). –  tomasz May 31 '12 at 18:34
    
The approach I'm familiar with uses the ends of the Cayley graph of the group. (Pretty much) the same question has been asked here: mathoverflow.net/questions/21578/… –  tomasz May 31 '12 at 18:48

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