I would like to know if there is a simple description of the following group. It has 2 generators whose the square of the commutator is trivial.
$$G=\langle a,b | (aba^{-1}b^{-1})^2=1\rangle$$
By advance thank you.
Edgar.



Is there a simple description of this group?
The Cayley graph is the skeleton of the order-4 octagonal tiling:


<http://en.wikipedia.org/wiki/Order-4_octagonal_tiling>


Consequently, we can construct your group $G$ as a (normal) subgroup of the symmetry group of this hyperbolic tiling, which is in turn is a subgroup of the symmetry group of the hyperbolic plane, which can be embedded in $PGL(2,\mathbb{C})$ by the Poincare disc model.


Specifically, we take $a : \mathbb{CP}^1 \rightarrow \mathbb{CP}^1$ to be the Möbius map:


$$ z \mapsto \dfrac{(p+1)z + p}{pz + (p+1)} $$


where $p = \dfrac{1}{\sqrt{2}} + \sqrt{\frac{1}{2}(1 + \sqrt{2})}$. Similarly, take $b$ to be the map obtained by conjugating $a$ by a rotation by $\frac{\pi}{2}$ (multiplication by $i$):


$$ z \mapsto \dfrac{(p+1)z - pi}{ipz + (p+1)} $$


These are two perpendicular translations of the Poincaré disc, and $G = \langle a , b \rangle$ is isomorphic to your group.


