Hyperbolic boundaries of infinitely generated groups I was wondering the following: in every book and paper that I looked into the definition of word hyperbolic groups (in the sense of Gromov) contains the condition that the group is finitely generated. I do not know too much about this topic but as far as I know it is quite common to speak about hyperbolic graphs and their compactifications even for graphs which are not locally finite or countable. That means that for groups which are not finitely generated the notion of hyperbolicity and the definition of the corresponding boundary/compactification would still make sense, right? Can somebody explain why one still imposes the condition?
 A: It does not follow that for groups which are not finitely generated the notion of hyperbolicity makes sense.
What follows, instead, is that for any pair $(G,S)$ such that $G$ is a group and $S$ is a generating set, the notion of hyperbolicity is well-defined: the Cayley graph $\Gamma(G,S)$ is certainly defined (and it is connected), and hyperbolicity of $\Gamma(G,S)$ certainly makes sense.
The trouble is that without any control on the cardinality of $S$, it is possible to have a group $G$ and two generating sets $S_1,S_2$ such that $\Gamma(G,S_1)$ is hyperbolic whereas $\Gamma(G,S_2)$ is not.
In fact, here's an example: $G = \mathbb Z^2$ with $S_1 = \{(\pm 1,0),(0,\pm 1)\}$ and with $S_2 = \mathbb Z^2$. The Cayley graph $\Gamma(G,S_1)$ is not hyperbolic whereas $\Gamma(G,S_2)$ is bounded and is therefore hyperbolic. That's an example where $G$ is in fact finitely generated, but one can easily cook up examples where $G$ is not finitely generated, e.g. $G = \mathbb Z^\infty$ with the "standard" generating set $S_1 = \{\pm e_i \mid i \ge 1\}$ is not hyperbolic, but with the generating set $S_2=G$ it is hyperbolic.
What happens in the finitely generated case is that if $S_1,S_2$ are two finite generating sets for the same group $G$, then $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are quasi-isometric to each other, and hyperbolicity is a quasi-isometry invariant, so $\Gamma(G,S_1)$ is hyperbolic if and only if $\Gamma(G,S_2)$ is hypebolic. Thus one says that a finitely generated group $G$ is hyperbolic if and only if some (hence any) finite generating set $S$ results in a hyperbolic Cayley graph $(G,S)$.
The summary statement is that hyperbolicity of a finitely generated group is well-defined independent of the choice of finite generating set, but take away those bold faced words and the statement becomes false.

ADDED: In answer to a followup question in the comments, every Gromov hyperbolic space (i.e. every geodesic metric space satisfying the thin triangle property) has a Gromov boundary.
And, given a group $G$ and two generating sets $S_1$, $S_2$, if $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are both hyperbolic, then both of them do have a Gromov boundary. And if $S_1,S_2$ are both finite then one can use the quasi-isometry between $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ to deduce that their Gromov boundaries are homeomorphic; so in the finitely generated case you are free to speak about THE (homeomorphism class of the) Gromov boundary.
However, without the assumption that $S_1,S_2$ are both finite, it need not be true that the Gromov of $\Gamma(G,S_1)$ is homeomorphic to the Gromov boundary of $\Gamma(G,S_2)$.
For a counterexample, take your favorite infinite, finitely generated, hyperbolic group $G$. With any finite generating set $S_1$, the Gromov boundary is nonempty. With the infinite generating set $S_2=G$, the Gromov boundary is empty.
Anyway, it follows that to speak of THE Gromov boundary of $G$ is nonsensical in the infinitely generated case.
A: If you read Gromov's original book, "hyperbolic group" means a group endowed with a left-invariant metric, that is hyperbolic. Quite equivalently, it's a group endowed with an isometric action on a hyperbolic space. The boundary can be defined in this broad context. For instance, if $\mathbf{Z}$ acts on the hyperbolic plane by a parabolic isometry, its boundary is a singleton.
He then especially studied the case of finitely generated groups with a word metric (in which case the boundary is independent of such a choice).
A more general realm is that of locally compact groups: if the word metric with respect to some compact generating subset is hyperbolic, then the boundary doesn't depend on the particular choice. This includes a rich supply of examples that do not resemble the finitely generated ones, e.g., connected Lie group with left-invariant negatively curved metric and millefeuille spaces (below the boundary of a millefeuille space, borrowed from here; it purports to represent the one-point compactification of $\mathbf{Q}_p\times\mathbf{R}$).

There's even broader settings where there's a canonical such metric, in the context of Polish groups (see Rosendal's work), but I'm not aware it's ever been studied in the angle of hyperbolicity.
