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Exact value of exponential growth rate depends on generating set

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0

$begingroup$

I am trying to solve an exercise from Clara Loh's Geometric Group Theory: An introduction. The problem uses the exponential growth rate of a finitely generated group $G$ with generating set $S$. The exponential growth rate is then defined as the limit
$$ rho_{G,S} = inf_{rin mathbb{N }_{>0}} (beta_{G,S}(r))^{1/r}, $$
where $beta_{G,S}$ is the growth function of $G$.

The question is then the following: Given a free group F of rank 2, freely generated by ${a,b}$ and consider the set $S:= {a,b,aba^{-1},a^2} subset F $.
Show that
$$rho_{F,{a,b}} neq rho_{F,S}. $$

I know that $rho_{F,{a,b}} = 3$, since for general free groups of rank $n$ and freely generating set $T$ we have
$$beta_{G,T} (r) = 1 + frac{n}{n-1} ( (2n-1)^r -1).$$
Moreover, I know that we can always find a subset of $S$ (in this case ${a,b}$) which will freely generating a free subgroup of $F$ of rank 2. Hence $rho_{F,S} geq rho_{F,{a,b}} = 3$.

So I am trying to prove that this inequality is strict. For this I tried to understand $beta_{F,S}(r)$ by drawing a Calay-graph up to $r=2$. For larger $r$ this gets a bit messy. But what I do observe from drawing this, is the fact that $beta_{F,S}(r) > beta_{F,{a,b}}(r)$. But taking the infimum does not guarantee strictness of this inequality.

I am somehow hoping that it is possible to show $beta_{F,S}(r)$ grows by an additional factor of $3^r$, but I am far from prooving anything like this.

Some hints on how to tackle this problem are very welcome. Thank you.

free-groups geometric-group-theory subgroup-growth

share|cite|improve this question

asked Mar 15 at 18:15

Trenzalore96Trenzalore96

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to draw a Cayley graph to, at least, compute $beta_S(3)$.
  $endgroup$
  – YCor
  Mar 15 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps I don't have to. But still, counting the number of elements in lower radius balls won't give me any insight on how the growth function exactly behaves for arbitrarely r. So this is where I'm stuck.
  $endgroup$
  – Trenzalore96
  Mar 16 at 10:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  counting them, maybe not, but listing them, possibly yes. Or drawing, in the standard Cayley graph (with respect to ${a,b}$), the $r$-ball with respect to the generating subset $S$, for $r=3$ and maybe even $r=4$. So as to have a guess about higher values of $r$.
  $endgroup$
  – YCor
  Mar 16 at 10:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright, in this fashion I'm able to see that we add in each $r$-ball, at least $4 times 7^r$ elements. Hence taking $n$ the number of generators in $S$ (i.e. $n=4$), I obtain that $$beta_{F,S}(r) > 1 + 4 sum_{i=0}^{r-1} (2n-1)^i = 1 + frac{4}{2(n-1)}((2n-1)^r-1). $$ Hence, computing the exponential growth rate of the growth function on the right gives us $2n-1 = 7$. Therefore, $rho_{F,S} geq 7 > 3 = rho_{F,{a,b}}$ and so $rho_{F,S} neq rho_{F,{a,b}}$.
  $endgroup$
  – Trenzalore96
  Mar 16 at 13:20
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I am trying to solve an exercise from Clara Loh's Geometric Group Theory: An introduction. The problem uses the exponential growth rate of a finitely generated group $G$ with generating set $S$. The exponential growth rate is then defined as the limit
$$ rho_{G,S} = inf_{rin mathbb{N }_{>0}} (beta_{G,S}(r))^{1/r}, $$
where $beta_{G,S}$ is the growth function of $G$.

The question is then the following: Given a free group F of rank 2, freely generated by ${a,b}$ and consider the set $S:= {a,b,aba^{-1},a^2} subset F $.
Show that
$$rho_{F,{a,b}} neq rho_{F,S}. $$

I know that $rho_{F,{a,b}} = 3$, since for general free groups of rank $n$ and freely generating set $T$ we have
$$beta_{G,T} (r) = 1 + frac{n}{n-1} ( (2n-1)^r -1).$$
Moreover, I know that we can always find a subset of $S$ (in this case ${a,b}$) which will freely generating a free subgroup of $F$ of rank 2. Hence $rho_{F,S} geq rho_{F,{a,b}} = 3$.

So I am trying to prove that this inequality is strict. For this I tried to understand $beta_{F,S}(r)$ by drawing a Calay-graph up to $r=2$. For larger $r$ this gets a bit messy. But what I do observe from drawing this, is the fact that $beta_{F,S}(r) > beta_{F,{a,b}}(r)$. But taking the infimum does not guarantee strictness of this inequality.

I am somehow hoping that it is possible to show $beta_{F,S}(r)$ grows by an additional factor of $3^r$, but I am far from prooving anything like this.

Some hints on how to tackle this problem are very welcome. Thank you.

free-groups geometric-group-theory subgroup-growth

share|cite|improve this question

asked Mar 15 at 18:15

Trenzalore96Trenzalore96

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to draw a Cayley graph to, at least, compute $beta_S(3)$.
  $endgroup$
  – YCor
  Mar 15 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps I don't have to. But still, counting the number of elements in lower radius balls won't give me any insight on how the growth function exactly behaves for arbitrarely r. So this is where I'm stuck.
  $endgroup$
  – Trenzalore96
  Mar 16 at 10:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  counting them, maybe not, but listing them, possibly yes. Or drawing, in the standard Cayley graph (with respect to ${a,b}$), the $r$-ball with respect to the generating subset $S$, for $r=3$ and maybe even $r=4$. So as to have a guess about higher values of $r$.
  $endgroup$
  – YCor
  Mar 16 at 10:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright, in this fashion I'm able to see that we add in each $r$-ball, at least $4 times 7^r$ elements. Hence taking $n$ the number of generators in $S$ (i.e. $n=4$), I obtain that $$beta_{F,S}(r) > 1 + 4 sum_{i=0}^{r-1} (2n-1)^i = 1 + frac{4}{2(n-1)}((2n-1)^r-1). $$ Hence, computing the exponential growth rate of the growth function on the right gives us $2n-1 = 7$. Therefore, $rho_{F,S} geq 7 > 3 = rho_{F,{a,b}}$ and so $rho_{F,S} neq rho_{F,{a,b}}$.
  $endgroup$
  – Trenzalore96
  Mar 16 at 13:20
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to solve an exercise from Clara Loh's Geometric Group Theory: An introduction. The problem uses the exponential growth rate of a finitely generated group $G$ with generating set $S$. The exponential growth rate is then defined as the limit
$$ rho_{G,S} = inf_{rin mathbb{N }_{>0}} (beta_{G,S}(r))^{1/r}, $$
where $beta_{G,S}$ is the growth function of $G$.

The question is then the following: Given a free group F of rank 2, freely generated by ${a,b}$ and consider the set $S:= {a,b,aba^{-1},a^2} subset F $.
Show that
$$rho_{F,{a,b}} neq rho_{F,S}. $$

I know that $rho_{F,{a,b}} = 3$, since for general free groups of rank $n$ and freely generating set $T$ we have
$$beta_{G,T} (r) = 1 + frac{n}{n-1} ( (2n-1)^r -1).$$
Moreover, I know that we can always find a subset of $S$ (in this case ${a,b}$) which will freely generating a free subgroup of $F$ of rank 2. Hence $rho_{F,S} geq rho_{F,{a,b}} = 3$.

So I am trying to prove that this inequality is strict. For this I tried to understand $beta_{F,S}(r)$ by drawing a Calay-graph up to $r=2$. For larger $r$ this gets a bit messy. But what I do observe from drawing this, is the fact that $beta_{F,S}(r) > beta_{F,{a,b}}(r)$. But taking the infimum does not guarantee strictness of this inequality.

I am somehow hoping that it is possible to show $beta_{F,S}(r)$ grows by an additional factor of $3^r$, but I am far from prooving anything like this.

Some hints on how to tackle this problem are very welcome. Thank you.

free-groups geometric-group-theory subgroup-growth

share|cite|improve this question

asked Mar 15 at 18:15

Trenzalore96Trenzalore96

1

$endgroup$

share|cite|improve this question

asked Mar 15 at 18:15

Trenzalore96Trenzalore96

1

share|cite|improve this question

asked Mar 15 at 18:15

Trenzalore96Trenzalore96

1

asked Mar 15 at 18:15

Trenzalore96Trenzalore96

1

asked Mar 15 at 18:15

Trenzalore96Trenzalore96

1