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Consider a Coxeter group presentation $< s_1, \ldots s_n \mid (s_i s_j)^{m_{ij}}>$ with $m_{ii}=1$ for all $i$. I can prove (using http://arxiv.org/abs/1011.4255) that if for each $i$ there are at most two $j\neq i$ with $m_{ij}< \infty$, then the corresponding Cayley graph is planar. Is this fact known?

In fact I can do a bit more, namely prove that by defining two 2-cells of the cayley complex to be parallel if they have the same boundary, and deleting all but one 2-cells from each parallel class, we obtain a planar 2-complex. This might be interesting as it gives a nice action of the group on the 2-sphere.

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Let $\Gamma$ be the Coxeter group in question. Order the $s_i$ so that if $m_{ij}\neq\infty$ then $|i-j|\leq 1$. Now let $P$ be a polygon with edges $e_i$ and angles $\pi/m_{i,i+1}$ between consecutive edges. Note that some vertices are 'ideal', meaning that they have interior angle zero.

The polygon $P$ can be realized geometrically in $X$, where $X$ is one of $S^2$, $\mathbb{R}^2$ or $\mathbb{H}^2$, and $\Gamma$ is the corresponding reflection group. There is a corresponding tiling of $X$ by copies of $P$, and the Cayley graph is just the dual graph to this tiling. So one sees that the Cayley graph is indeed planar, and furthermore the action of $\Gamma$ preserves a natural metric.

This is all quite standard. For further details, I suggest looking at Peter Scott's article 'The geometries of 3-manifolds'. So, yes, it is indeed known that these Cayley graphs are planar.

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