Let, $G=\left\langle a,b\ |\ a^2,b^3,(ab)^{8}\right\rangle$.
Is the following a faithful representation?
$a\rightarrow \frac{1}{2}\begin{bmatrix}-1+\sqrt{2} & -1 & -2 \\-1 & -3+\sqrt{2} & -2+2\sqrt{2}\\-\sqrt{2} & 2-\sqrt{2} & 2-2\sqrt{2}\end{bmatrix}$
$b\rightarrow \frac{1}{2}\begin{bmatrix}-1 & -1+\sqrt{2} & -2 \\1-\sqrt{2} & -1+ 2 \sqrt{2} & -2+2\sqrt{2}\\\sqrt{2} & 2-\sqrt{2} & 2-2\sqrt{2}\end{bmatrix}$
The matrices do satisfy the above relations. And many other products I've tried don't equal the identity. I suspect this is faithful, but I don't know how to approach a proof.
I should add how I found this representation, as it looks like magic. I was thinking about how you can place $3$ triangles around a vertex to get a regular tetrahedron, $4$ for an octohedron, $5$ for icosahedron. And these live nicely in Euclidean space. With $8$ triangles around each vertex you get a tiling of a hyperbolic surface which doesn't live nicely in Euclidean space.
So let's put it there anyway.
At each vertex, instead of wrapping 8 triangles once around a vertex, if you wrapped them twice around each vertex, you would cover an octohedron. If instead you wrap them $3$ times around each vertex, you can still put the whole tiling in Euclidean $3$ space, but maybe without any triangles coinciding. It intersects itself all over the place, and is infinite, so a picture doesn't help. But, you can find matrices which represent rotation around a specific edge ($a$) or triangle ($b$).
To make things easier I required $ab$ to correspond to a rotation of $3/8$ of a turn around the z axis. Then I cleaned up the representation a bit.
Thoughts?