Is the group generated by two loxodromic isometries with a fixed point in common cocompact? If you have two distinct loxodromic isometries of the hyperbolic plane $\gamma_1, \gamma_2$ such that they have a fixed point in common. For simplicity let's take the half plane model and let the fixed point in common be $\infty$. Is the group $\langle \gamma_1,\gamma_2\rangle$ cocompact? Is it even discrete? I can show that if the translation length $|\gamma_1| = |\gamma_2|$ then the group contains a parabolic element, so either it's not cocompact, or it's not proper. How horrible can this group be? I'd like to be able to prove that it's not cocompact, but I'm starting to suspect that it might be.
 A: The group $\Gamma = \langle \gamma_1,\gamma_2\rangle$ is not discrete. 
To prove this, consider the "height map" $\mathbb{H}^2 \to \mathbb{R}$ defined by $f(x,y) = \ln(y)$. The inverse images of points under this map are the horocycles $y = \text{(constant)}$ all based at $\infty$, each of which is preserved by both $\gamma_1$ and $\gamma_2$, and therefore there is an induced action of $\Gamma$ on $\mathbb{R}$, the ``height displacement''. Let $\delta_i$ denote the height displacement of $\gamma_i$. The "height displacement" subgroup $\langle\delta_1,\delta_2\rangle$ of $\mathbb{R}$ may be discrete (infinite cyclic) or indiscrete (abelian of rank 2). In either case, given $\epsilon>0$ there exist nonzero integers $m_1,m_2$ such that the absolute value of $\Delta_y = m_1\delta_1 + m_2 \delta_2$ is less than $\epsilon/2$ (in the case that the height displacement group is cyclic, one can arrange that $\Delta_y=0$).
Consider now the points $p = 0+1i$ in $\mathbb{H^2}$ and $q = \gamma_1^{m_1} \gamma_2^{m_2}(p)$, which are in the same orbit. Their vertical displacement is $\Delta_y < \epsilon/2$. Let $\Delta_x$ be the difference of their $x$-coordinates, which might still be large. As $k \to +\infty$ the points $\gamma_1^k(p)$, $\gamma_2^k(p)$ are connected by concatenation of a horizontal segment whose length $<C \Delta_x e^{-k}$ goes to zero, followed by a vertical segment of length $\Delta_y <\epsilon/2$, and so $d(\gamma_1^k(p),\gamma_2^k(q)) <\epsilon$ for large enough $k$. This shows that $\Gamma$ is not discrete.
On the other hand, $\Gamma$ is cocompact in the strong sense that there exists a compact subset $B \subset \mathbb{H}^2$ whose translates $\{\gamma B \mid \gamma \in \Gamma\}$ cover $\mathbb{H}^2$. One can choose this set $B$ to be a "horobrick" whose appearance is an $xy$-rectangle in the upper half plane, whose two vertical sides are hyperbolic geodesic segments and whose two horizontal sides are horocylic segments. Simply choose $B$ to have bottom left corner containing $p$ and top right corner containing $\gamma_2(q)$ (assuming $q$ has larger $x$-coordinate than $p$).
