The group mentioned in the title, $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}=1\rangle$, is in between the torus fundamental group $\langle x,y|xyx^{-1}y^{-1}=1\rangle$ and the two-holed torus fundamental group $\langle x,y,z,w|xyzx^{-1}y^{-1}z^{-1}w^{-1}=1\rangle$. 


It is not the hexagonal presentation of the torus because the homology is not the same (gluing up a hexagon with that pattern gives you two vertices instead of one, giving you only two loops in the one-skeleton, so this is not the torus group).


Its Cayley graph can be realized as an infinite tiling of hexagons, each of which meet six to a vertex.


Such a tiling can be embedded in the hyperbolic plane, making the Cayley graph quasi-isometric to hyperbolic space, which means that the group is delta hyperbolic with a circle at infinity, **implying that the group is Fuchsian**, by the work of Gabai and others.


So it has a finite index surface subgroup. But this group is a subgroup of the RAAG with defining graph the diamond graph, I.e. $F\_2\times F\_2$. This can be seen by letting $a,b$ generate the first free group, $c,d$ generate the second subgroup, and letting $x=ab^{-1},y=bc^{-1}$, and $z=cd^{-1}$.


\**Edit:*\*I meant to say that $a,c$ generate the first group and $b,d$ generate the second.\


So this implies that the diamond graph contains a hyperbolic subgroup. But in all the RAAG references, they say that RAAG's contain surface subgroups if they contain 5-cycles. So why does it seem as if a four-cycle (the diamond graph) contains a surface subgroup?



Why isn't $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group?
I think Lee's and Steve's comments pretty much answer this question. Let me try to summarize, and clear up a couple of misconceptions that seem to be lurking. For convenience, I'll denote your group by $G$.


**The map $G\to F\_2\times F\_2$.**


Actually, I don't think the map $G\to F\_2\times F\_2$ that you describe is an injection. The images of $x$ and $z$ commute, so the image also satisfies the relation $[x,z]=1$. Your relator then becomes $[z^{-1}x,y]=1$. It follows from these two relations that the image is isomorphic to $F\_2\times\mathbb{Z}$.


Indeed, the theorem of Baumslag and Roseblade that I alluded to in my comments says more or less (there are some difficulties in getting the statement exactly right) that every fp subgroup of a free-cross-free group is virtually free-cross-free, so we could have guessed it would be of this form.


**The group $G$.**


If you have a presentation in which every generator (or its inverse) appears exactly twice then it is, indeed, natural to guess that it might be a surface group. And it nearly is. But there's one more thing you need to check: the link of the 1-vertex of the corresponding presentation complex.


This is called the *Whitehead graph* of the relations. It's necessarily a union of cycles (in the case when each generator appears exactly twice), but the presentation complex is a surface exactly when the Whitehead graph consists of just one cycle. Otherwise, the presentation complex is a surface with some points identified, and so the group is a free product of a surface group and a free group.


The Whitehead graph is easy to calculate; in this case, it turns out to consist of two cycles, and so we have two points identified, and $H\_1$ tells us that the surface is in fact a torus.


Therefore, $G\cong\mathbb{Z}\*\mathbb{Z}^2$, as Lee and Steve correctly said. But I hope the above gives you some idea of how one might calculate this, rather than just pluck it from thin air.


**The Convergence Group Theorem**.


By the way, here's another way to see that something must have been wrong. It follows from the Convergence Group Theorem of Tukia, Casson--Jungreis and Gabai, which you mention, that any torsion-free group which is virtually a surface group is in fact a surface group. Therefore, if $G$ really were a subgroup of $F\_2\times F\_2$ and virtually a surface group, it would have had to have been a surface group. Of course, there's only one candidate, which can be ruled out by looking at $H\_1$.


