does every right-angled coxeter group have a right-angled artin group as a subgroup of finite index? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:29:09Z http://mathoverflow.net/feeds/question/18558 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18558/does-every-right-angled-coxeter-group-have-a-right-angled-artin-group-as-a-subgro does every right-angled coxeter group have a right-angled artin group as a subgroup of finite index? david mccune 2010-03-18T07:55:39Z 2010-03-18T17:12:45Z <p>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.</p> http://mathoverflow.net/questions/18558/does-every-right-angled-coxeter-group-have-a-right-angled-artin-group-as-a-subgro/18596#18596 Answer by James for does every right-angled coxeter group have a right-angled artin group as a subgroup of finite index? James 2010-03-18T14:57:38Z 2010-03-18T14:57:38Z <p>You might be thinking of this paper:</p> <p><a href="http://linkinghub.elsevier.com/retrieve/pii/S0022404999001759" rel="nofollow">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.</a></p> http://mathoverflow.net/questions/18558/does-every-right-angled-coxeter-group-have-a-right-angled-artin-group-as-a-subgro/18630#18630 Answer by HW for does every right-angled coxeter group have a right-angled artin group as a subgroup of finite index? HW 2010-03-18T17:12:45Z 2010-03-18T17:12:45Z <p>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 <code>$\langle a_1,\ldots, a_5 \mid a_i^2=1, [a_i,a_{i+1}]=1\rangle$</code> where the indices are considered mod 5.</p> <p>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.</p>