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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?

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This group is the free product $\mathbb{Z} * \mathbb{Z}^2$. You can see that by identifying the two vertices of the hexagon pattern on the torus to a single vertex. – Lee Mosher Apr 28 '13 at 2:23
Also, in the "infinite tiling of hexagons" that you refer to, the link of the vertex is a disjoint union of two circles, rather than a single circle, and so this tiling is not planar. – Lee Mosher Apr 28 '13 at 2:25
Let $a=zx^{-1}$ and $b=y^{-1}z^{-1}$. Then your relator is $[a,b]=1$. Since your group is torsion-free (easy because it is a one-relator group), it contains a copy of $\mathbb{Z}\times\mathbb{Z}$, which cannot happen in hyperbolic groups. – Steve D Apr 28 '13 at 5:38
Brian, your claim that $F_2\times F_2$ contains a surface group is incorrect. A theorem of Baumslag--Roseblade says every finitely presented subgroup of of $F_2\times F_2$ is of finite index. – HJRW Apr 28 '13 at 7:28
(added to clarify)... in some free-cross-free subgroup. (Of course, the factors may be cyclic or trivial, or the subgroup may be diagonal.) – HJRW Apr 28 '13 at 8:03
up vote 8 down vote accepted

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 $H_1$.

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Just a trivial addition to the answer: One constructs a presentation complex by attaching cells to a "rose", which is a 1-vertex graph. This is where the vertex identification in the surface is coming from. I think, the confusion in the question stems from the fact that Brian forgot about this part in the construction of the presentation complex. – Misha Apr 28 '13 at 17:48
Perhaps you're right, Misha. But the same considerations apply when you're checking if an arbitrary compact 2-complex is a surface: every edge should be contained in exactly two faces, and every vertex link should be a cycle. – HJRW Apr 28 '13 at 20:53
Great answer, and thanks for the comments from everyone! My embedding of $G$ was a typo, which I've corrected above. It seems like any such group with an odd number of generators in the same pattern will be a surface with a point identified, right? – Brian Rushton Apr 29 '13 at 17:47
Brian: Odd number of generators is irrelevant; what's relevant is that in a surface equipped with cell-complex structure (with a single 2-cell), you have to identify all vertices to a single point. This gives the presentation complex, see my comment above. – Misha Apr 29 '13 at 17:59
Brian - yes, I think that's right. You should get the orientable surface of the relevant genus, with one point identified. – HJRW Apr 29 '13 at 19:40

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