Let $\Gamma$ be a discrete group with a generating set $S$. Let $p\_c(\Gamma,S)$ be the critical probability for percolation of the Cayley graph of $\Gamma$. Is it known that if $\Gamma$ is non-amenable then there exists a generating set $S$ such that $p\_c(\Gamma,S)<\frac{1}{2}$ (or some other constant)? 


The details on the question:


Let $0<p<1$ , the Bernoulli bond percolation of the Cayley graph defined as follows:
the edge is open with probability $p$ and closed with $1-p$. The connected components of the graph are those that spanned by open edges. Let $\theta(p)$ be the probability that the origin belongs to an infinite open component. Then $p\_c(\Gamma, S)=\sup (p:\theta(p)=0)$.



The critical value of percolation on Cayley graphs.
Here's a more direct argument:


Let $S$ be a generating set and let $s\in S$. Pick some $x\_1,x\_2,\ldots , x\_k$ such that all of the $x\_i$'s and $x\_i^{-1}s$ are different from each other and are not in $S$. Do the same for all $s\in S$ in such a way that all of these elements are distinct. Let $S'$ be the resulting set.


Now, $p\_c(\Gamma,S')$ is small. Why? because we may think of percolation on $Cayley(\Gamma,S')$ as inducing a percolation on $Cayley(\Gamma,S)$, except that now the probability of some $a$ and $as$ being connected is at least $(1-(1-p^2)^k)$ since they may be connected through any pair $x\_i$ and $x\_i^{-1}s$, there are $k$ such pairs and each has a probability of $p^2$ being open. This immediately show that $p\_c$ goes to 0 with $k$.


