Question

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.

Answer 1

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.

Answer 2

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.