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We already have the Milnor-Svarc Lemma, which tells us that if a group acts "nicely" on a space, then the Cayley graph of the group is quasi-isometric to the given space. This gives us a lot of hyperbolic groups, because hyperbolic manifolds are ubiquitous.

Edit: For example, the MS-lemma tells us that the surface groups are hyperbolic.

Edit:

Of course, in the large-scale geometric setting here, we consider two metric spaces equivalent if they are quasi-isometric.

There is a result of Kharlampovich and Myasnikov, in "Hyperbolic groups and free constructions" which allows us to construct a "separated HNN extension" of two hyperbolic groups to yield a hyperbolic group.

What are some other ways of constructing word-hyperbolic groups?

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    $\begingroup$ A similar question was asked on MathOverflow recently. The answer of Genevois is pretty comprehensive. Regarding the Kharlampovich-Myasnikov result, Bestvina-Feighn's combination theorem has a similar flavour (although I've never found a way to apply it, while I have applied the Kharlampovich-Myasnikov result). $\endgroup$
    – user1729
    Sep 5, 2019 at 19:51
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    $\begingroup$ In fact, the "geometric" action of $SL(2,Z)$ on the upper half-plane does not tell you that $SL(2,Z)$ is hyperbolic. But you can conclude hyperbolicity by observing that this group is commensurable with the free group of rank 2. $\endgroup$
    – Moishe Kohan
    Sep 5, 2019 at 21:13
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    $\begingroup$ The point about geometric is important because in higher dimensions you can have a super nice action on hyperbolic space, but if it isn't cocompact it might not be hyperbolic (fundamental group of hyperbolic 3-manifolds with toral cusps for example). $\endgroup$
    – user29123
    Sep 6, 2019 at 13:40
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    $\begingroup$ I don't really know what you mean to be honest $\endgroup$
    – user29123
    Sep 6, 2019 at 23:01
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    $\begingroup$ I meant that I not really sure if I knew what you mean by "growth not well behaved" or just "growth" and still don't. It might make sense but I don't know from what you have said. $\endgroup$
    – user29123
    Sep 7, 2019 at 12:29

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A similar question was asked on MathOverflow recently. The answer of Genevois is pretty comprehensive.

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