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I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is as "metric/combinatoric" as possible.

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$.

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it).

[Edit: just ignore the following paragraph] PS: If it helps, one could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.

I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is as "metric/combinatoric" as possible.

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$.

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it).

[Edit: just ignore this PS] PS: If it helps, one could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.

[Edit: as suggested by Yves in the comments...]

$\mathbf{Possible~motivation~and~alternate~question:~}$ Given a quasi-transitive hyperbolic graph $\Gamma$, does it satisfy a "strong" isoperimetry, i.e. $|\partial F| \geq K |F|$ for some $K>0$?

Definition of quasi-transitive: there are constants $C$ and $L$ so that the set $SQI_{C,L}$ of all $(C,L)$-quasi-isometries from $\Gamma$ to itself is transitive, i.e. for any vertices $x$ and $y$ there is a $\phi \in SQI_{C,L}$ so that $\phi(x) =y$.

I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is a "metric/combinatoric" as possible.

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$.

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it).

[Edit: just ignore the following paragraph] PS: If it helps, on could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.

I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is as "metric/combinatoric" as possible.

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$.

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it).

[Edit: just ignore the following paragraph] PS: If it helps, one could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.

I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is a "metric/combinatoric" as possible.

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$.

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it).

PS: If it helps, on could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.

I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is a "metric/combinatoric" as possible.

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$.

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it).

[Edit: just ignore the following paragraph] PS: If it helps, on could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.