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A group $G$ is relatively hyperbolic relative to a collection of subgroups $\mathcal{G}$, if $G$ admits an action on a connected graph, $K$, with the following properties:

(1) $K$ is hyperbolic, and each edge of $K$ is contained in only finitely many circuits of length $n$ for any given integer, $n$,

(2) there are finitely many $G$-orbits of edges, and each edge stabiliser is finite,

(3) the elements of $\mathcal{G}$ are precisely the infinite vertex stabilisers of $K$, and

(4) every element of $\mathcal{G}$ is finitely generated.

This is taken from Bowditch. There are other equivalent definitions, and an inequivalent definition (though groups satisfying the latter are now usually called weakly relatively hyperbolic).

I have two questions about relatively hyperbolic groups. They sound elementary, but I haven't seen an explicit answer in, for example, papers by Farb or Bowditch.

  1. Is a free product of finitely many relatively hyperbolic groups itself relatively hyperbolic (relative to the collection of given parabolic subgroups)?

  2. If $G$ is hyperbolic relative to non-trivial subgroups $H_1,\ldots,H_n$, and each $H_i$ is a subgroup of $H_i'$, is $G$ hyperbolic relative to $H_1',\ldots H_n'$? (Assuming that $H_i'\cap H_j'$ is trivial for $i\neq j$.)

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  • $\begingroup$ $A * B$ is hyperbolic relative to $\{A, B\}$. $\endgroup$ – M.U. Dec 28 '15 at 19:32
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    $\begingroup$ It may good to add the definition of relative hyperbolicity you are using. Since the one used by Farb (which is now often called weak relative hyperbolicity) is essentially different (if one does not include the BCP-property) than the one of Bowditch or Osin (and others). $\endgroup$ – M.U. Dec 28 '15 at 19:43
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    $\begingroup$ I think the answer to 2 is no with either definition of relative hyperbolicity. If $G$ is hyperbolic then it is hyperbolic relative to the trivial subgroup, but it is not necessarily hyperbolic relative to all of its subgroups. Rips constructed examples of hyperbolic groups $G$ with normal subgroup $N$ such that $G/N$ has unsolvable word problem, so $G$ cannot be hyperbolic relative to $N$. $\endgroup$ – Derek Holt Dec 28 '15 at 20:02
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    $\begingroup$ That makes no difference, the answer is still no. Let $H$ be any group whatsoever and $K$ a hyperbolic group, then take $G=H \times K$, so $G$ is hyperbolic relative to $H$, and you can take $H' = H \times N$ with $N$ as the previous comment. $\endgroup$ – Derek Holt Dec 28 '15 at 22:51
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    $\begingroup$ From the discussion and comments on mathoverflow.net/questions/134091 it appears that it is unknown whether malnormality implies quasiconvexity for subgroups of hyperbolic groups. On the other hand, hyperbolic groups are hyperbolic relative to subgroup that are both malnormal and quasiconvex. So that suggests that the answer to the question in your last comment is unknown for single subgroups of hyperbolic groups. But of course there could be other types of examples, $\endgroup$ – Derek Holt Dec 29 '15 at 21:11
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The answer to question 1 is "yes", by results of Osin. Indeed one has, under some mild assumptions, an analogous statement for HNN- Extensions and free products with amalgamation. However for free products these apply trivially.

See "F. Dahmani. Combination of convergence groups" and especially "D.V. Osin, Relative Dehn functions of amalgamated products and HNN-Extensions".

Question 2 has been treated in the comments (@Derek Holt).

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    $\begingroup$ In fact, general results about combination theorem for relatively hyperbolic groups appear in Dahmani's Thesis, published in Geom Top front.math.ucdavis.edu/0203.5258, as noted by Osin arxiv.org/abs/math/0411027 $\endgroup$ – Thomas Dec 29 '15 at 12:43
  • $\begingroup$ @Thomas: you're absolutely right. However Osin's result are more general. In any case I should edit. $\endgroup$ – M.U. Dec 29 '15 at 14:34
  • $\begingroup$ Osin's Corollary 1.5 seems to be just what I need - thanks! (I can't yet upvote.) $\endgroup$ – user876789 Dec 29 '15 at 14:39
  • $\begingroup$ @user876789, exactly, just set $K = \{1\}$. $\endgroup$ – M.U. Dec 29 '15 at 14:40

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