Let $\Gamma\subset I(\mathbb{H}^{n})$ be a finitely generated discrete group of isometries of the hyperbolic $n$-space.
Let $\Gamma_{\infty}$ be the stabilizer of $\infty$, and assume it contains only elliptic and parabolic elements.
Let $A\subset \Gamma$ be the set of elliptic elements of $\Gamma$.
My question is whether or not can the generated subgroup $B=\langle A\cap\Gamma_{\infty}\rangle$ be infinite.
My intuition says it cannot. In some manner that would imply that there are too many elliptic elements in $\Gamma$, which would contradict the Selberg lemma. But I'm not sure if it is true, and how to formalize it. I guess the Selberg lemma shows that if $B$ is infinite it must contain a parabolic element. Question is if it is a contradiction or not.
Best Answer
There's one restricted case in which this is true, namely if you assume that $n=2$ and that $\Gamma$ consists only of orientation preserving isometries. In that case, every nontrivial finite order element of $\Gamma$ is a rotation of some rational angle $2\pi k/n$ around some point of $\mathbb H^2$, and hence has no fixed points on the circle at infinite, so $A \cap \Gamma_\infty$ consists solely of the identity.
However, even when $n=2$ there are counterexamples if you allow elements of $\Gamma$ to reverse orientation. For instance, the infinite dihedral group $D_\infty = \langle a,b \mid a^2 = b^2 = ab\rangle$ acts on $\mathbb H$ by the formula $a \cdot (x+iy) = -x+iy$, $b \cdot (x+iy) = 1-x+iy$. Under this action, the point $\infty$ is fixed by every element, and all of the infinitely many conjugates of $a$ and $b$ are reflection isometries each of which has a whole line of fixed points. For example, $a$ fixes every point on the imaginary axis $x=0$, and more generally $b^k a b^{-k}$ fixes every point on the line $x=k$.
In higher dimensions there are orientation preserving counterexamples as well. The general idea is that if you use the "upper half space" model $\mathbb H^n = \mathbb R^{n-1} \times \mathbb R_+$ then the action of the group of Euclidean isometries of $\mathbb R^{n-1}$ extends to an isometric action on $\mathbb H^n$ that fixes $\infty$. Its easy to produce infinite groups of orientation preserving Euclidean isometries having infinitely many torsion elements.
For a specific example acting on $\mathbb H^3$, start with the infinite dihedral group of orientation preserving Euclidean isometries of $\mathbb R^2$ generated by the rotations $a(x,y) = (-x,-y)$ and $b(x,y) = (1-x,-y)$.