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I thought that I read a paper making this claim a few months ago, but now I can't find it. If the answer is yes, is there a nice way to go from the presentation of the right-angled coxeter group to a presentation of its right-angled artin subgroup? Thanks.

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You should try arXiv, for example arXiv:0905.1282, and perhaps ask Ruth Charney (Brandeis) or look at her recent papers on artin groups. – Jim Humphreys Mar 18 '10 at 10:56
up vote 12 down vote accepted

As James points out, the paper of Davis and Januskiewicz proves the inverse. To see that the answer to your question is 'no', consider the right-angled Coxeter group whose nerve graph is a pentagon. That is, it's the group with presentation $\langle a_1,\ldots, a_5 \mid a_i^2=1, [a_i,a_{i+1}]=1\rangle$ where the indices are considered mod 5.

This group acts properly discontinuously and cocompactly on the hyperbolic plane, and it's not hard to see that it has a finite-index subgroup which is the fundamental group of a closed hyperbolic surface. Every finite-index subgroup of a right-angled Artin group is either free or contains a copy of $\mathbb{Z}^2$, but the fundamental group of a closed hyperbolic surface has no finite-index subgroups of this form.

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To summarise this answer in a slogan: there are lots of interesting RACGs that are (word)-hyperbolic, but any hyperbolic RAAG is free. – HJRW Mar 19 '10 at 16:27

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