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.
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5$\begingroup$ For future reference: "title" refers to the title of the question, not your professional title. $\endgroup$– user9072Commented May 6, 2015 at 14:37
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5$\begingroup$ @JonMarkPerry It isn't abelian. In fact it's hyperbolic. $\endgroup$– Derek HoltCommented May 6, 2015 at 15:11
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3$\begingroup$ For example, the non-Abelian group $A_{4}$ is a (proper) homomorphic image of the group $G$ ( $A_{4}$ satisfies these relations with $a = (123), b = (124)$). $\endgroup$– Geoff RobinsonCommented May 6, 2015 at 16:01
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2$\begingroup$ @JonMarkPerry No, we can't assume that $G$ is not abelian. It has to be proved that $G$ is not abelian. Geoff Robinson's comment provides an elementary proof of this. I probably unwittingly confused you by stating this without proof - in fact the assertions I made in my comment were the result of a quick computer calculation. $\endgroup$– Derek HoltCommented May 6, 2015 at 16:36
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2$\begingroup$ Your group is the fundamental group of an orbifold $O$, with genus one and a single cone point of order 2. See Peter Scott's article `The geometries of 3-manifolds' for a comprehensive discussion of 2-dimensional orbifolds. The fact that the (rational) Euler characteristic $\chi(O)=-1/2$ is negative implies that the universal cover of $O$ is the hyperbolic plane, and your group is a cocompact Fuchsian group. $\endgroup$– HJRWCommented May 7, 2015 at 12:05
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1 Answer
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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.
answered May 6, 2015 at 15:39