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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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    $\begingroup$ You should try arXiv, for example arXiv:0905.1282, and perhaps ask Ruth Charney (Brandeis) or look at her recent papers on artin groups. $\endgroup$
    – Jim Humphreys
    Commented Mar 18, 2010 at 10:56

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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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    $\begingroup$ To summarise this answer in a slogan: there are lots of interesting RACGs that are (word)-hyperbolic, but any hyperbolic RAAG is free. $\endgroup$
    – HJRW
    Commented Mar 19, 2010 at 16:27
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You might be thinking of this paper:

Michael W. Davis and Tadeusz Januszkiewicz, Right-angled Artin groups are commensurable with right-angled Coxeter groups, J. Pure Applied Algebra, 153, No. 3 (2000), 229-235.

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