Question

Is center of a fundamental group of finite volume-hyperbolic n-orbifold trivial? Is there a good reference that the proof is wriiten?

Answer 1

Any lattice in a hyperbolic space acts as a convergence group on the sphere at infinity. Thus it suffices to prove the following:

Lemma If $G$ has a minimal convergence group action on a set $S$ of cardinality $>2$ (e.g. a sphere), then any normal abelian subgroup $A$ of $G$ is finite.

Proof. If $A$ is infinite, then $A$ fixes one or two points of $S$, and this is the whole fixed point set of $A$. Now any group that contains $A$ as a normal subgroup must stabilize the fixed point set of $A$, but $G$ has a dense orbit in $S$, a contradiction.

Basic facts about convergence groups that were used above can be found e.g. in "Convergence groups and configuration spaces" by B.H. Bowditch available here.

Answer 2

Let $G$ be the fundamental group of your orbifold, and recall that $G$ is a lattice in the group of isometries of hyperbolic space. An element in the center of $G$ has to fix every fixed point of any loxodromic and any parabolic element of $G$. The fixed points of such elements are dense in the boundary at infinity of hyperbolic space (this is a consequence of the fact that the orbifold has finite volume, I think that a proof of this fact can be found in Ratcliffe's book, for example).

This implies that any element in the center of $G$ acts as the identity on the boundary of hyperbolic space, so the center of $G$ is trivial.

Answer 3

The center of your group $G$ has to be finite (since it is finitely generated, abelian, and has no element of infinite order -- if it did, the group would fail to be word-hyperbolic, since that element together with some other element of $G$ of infinite order would generate a $\mathbb{Z} \oplus \mathbb{Z}$). But now, Lemma 3.3 of http://arxiv.org/pdf/1106.4595 states that no discrete group of motions of an Hadamard manifold ($\mathbb{H}^n$ certainly qualifies) has a finite normal subgroup.

EDIT If the orbifold has cusps, and the center has an element of infinite order $\gamma,$ then all the elements of $G$ fix the fixed point (which has to be on the sphere at infinity) of $\gamma$ (since they all commute with $\gamma$) which contradicts the finite volume assumption. In fact, if $\gamma$ has finite order, it has a fixed point in $\mathbb{H}^n,$ that fixed point has to be fixed by every element of $G,$ and so $G$ is a subgroup of $SO(n)$ (and the quotient is, again, not finite volume) -- the argument before the edit is unnecessary, but is left there for historical reason.