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> Is a (finitely generated) torsion-free subgroup of a right-angled Coxeter group isomorphic to a subgroup of a right-angled Artin group?
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It is well-known from the theory of special cube complexes that a subgroup $H$ of a right-angled Coxeter group $G$ contains a finite-index subgroup isomorphic to a subgroup of a right-angled Artin group (e.g. $H \cap [G,G]$), but is it true for the whole group if we assume in addition that $H$ is torsion-free?


The difficulty when applying the theory of special cube complexes comes from the fact that the induced action of $H$ on the usual CAT(0) cube complex $X$ of $G$ may invert hyperplanes (i.e. some isometries may stabilise a hyperplane and switch its halfspaces). This can be avoided by replacing $X$ with its barycentric subdivision, but then hyperplane-inversions yield self-osculations (i.e. there exist an element $h \in H$ and an oriented hyperplane $\vec{J}$ such that $\vec{J}$ and $g \vec{J}$ contain two intersecting oriented edges pointing to their common vertex and that do not span a square).


**NB:** In all the question, right-angled Artin/Coxeter groups are finitely generated. Recall that, given a (finite) simplicial graph $\Gamma$, the associated right-angled Coxeter group is defined by the presentation
$$\langle u \in V(\Gamma) \mid u^2=1 \ (u \in V(\Gamma)), \ [u,v]=1 \ (\{u,v\} \in E(\Gamma)) \rangle$$
and the associated right-angled Artin group by the presentation
$$\langle u \in V(\Gamma) \mid [u,v]=1 \ (\{u,v\} \in E(\Gamma)) \rangle$$
where $V(\Gamma)$ and $E(\Gamma)$ denote respectively the vertex- and edge-sets of $\Gamma$.



Subgroups of RAAGs vs. subgroups of RACGs
The fundamental group of the (non-orientable) closed surface of Euler characteristic -1 provides a counterexample.


On the one hand, it’s a subgroup of index 4 in the reflection group on the right-angled pentagon, so it embeds in a RACG, and of course it’s torsion-free.


On the other hand, [Crisp—Wiest](https://arxiv.org/abs/math/0303217v2) proved that it never embeds in a RAAG.


