Elliptic subgroup of $PSL(2,\mathbb C)$

Question

Consider the group $PSL(2,\mathbb C)$ acting on $\mathbb{CP}^1$ via Möbius transformations. Recall that an element of $PSL(2,\mathbb C)$ different from the identity is called parabolic if it is conjugated to $z \mapsto z+1$, loxodromic if it is conjugated to $z \mapsto \lambda z$ for some $\lambda \in \mathbb C^* $ with $|\lambda| \neq 0,1$ and elliptic if it is conjugated to $z \mapsto \lambda z$ for some $\lambda \in \mathbb C^* $ with $|\lambda| =1$.

There is a natural embedding $PSU(2) \subset PSL(2,\mathbb C)$.

Question: Let $G$ be a finitelly generated subgroup of $PSL(2,\mathbb C)$, not necessarily discrete. Is it true that if $G$ contains only elliptic elements (and the identity) then $G$ is conjugated to a subgroup of $PSU(2)$?

If $G$ is discrete it can be shown that it is finite, so by averaging the standard inner product on $\mathbb C^2$ we get a $G$-invariant inner product, so $G$ is conjugated to a subgroup of $PSU(2)$. What about the case when $G$ is not discrete?

Answer 1

After reading this question I found out that the result is indeed true. It can be found in the book Complex Functions by G. Jones and D. Singerman (Theorem 2.13.1).

Answer 2

This is an addendum to the question:

There exists a 2-generated subgroup $\Gamma$ of isometries of the hyperbolic 5-space ${\mathbb H}^5$ such that every element of $\Gamma$ is elliptic but $\Gamma$ does not fix a point in ${\mathbb H}^5$.

(An isometry of a space $X$ of curvature $\le 0$ is elliptic when it fixes a point in $X$. I will use this definition for Euclidean and for hyperbolic spaces.)

Proof. The trick is to construct a group of isometries of the Euclidean 4-space with the same property. Then we use the natural inclusion $Isom({\mathbb E}^4)\subset Isom({\mathbb H}^5)$ which sends elliptic elements to elliptic elements. The key to the proof is the fact that the subgroup $SU(2)< SO(4)$ contains $F_2$, a free group on two generators, say, $A_1, A_2$. Every element $A$ of $SU(2)- \{e\}$ has exactly one fixed point in the Euclidean 4-space (the origin). From this, it follows that for every vector $b\in {\mathbb R}^4$, the affine transformation $$ x\mapsto Ax + b $$ has exactly one fixed point in ${\mathbb E}^4$ as well, hence, is elliptic.

Now, pick a nonzero vector $b\in {\mathbb R}^4$ and define a subgroup $\Gamma$ of $Isom({\mathbb E}^4)$ generated by $$ \hat{A}_1: x\mapsto A_1x+b, $$ and $A_2$. Every nontrivial element of this group will be elliptic (since its linear part is in $SU(2)$ and is $\ne I_4$ since $A_1, A_2$ generate a free subgroup). At the same time, $\Gamma$ does not fix any points in ${\mathbb E}^4$ (since $A_2$ fixes only the origin and $\hat{A}_1$ does not fix the origin as $b\ne 0$). From this, one sees that the extension of this group to ${\mathbb H}^5$ does not fix any point of ${\mathbb H}^5$ while every element $g$ fixes a point (actually, the entire geodesic connecting the point $\infty$ to the fixed point of $g$ in ${\mathbb E}^4$).

See also;

H. Bass, Groups of integral representation type, Pacific Math Journal, Volume 86, Number 1 (1980), 15-51.