Currently I am reading a paper "Infinite group actions on spheres" by Gaven Martin. I am a first year graduate students and I got lots of questions, so one of them is about the following example: (sorry in advance, if that question would be trivial)

In Example 4.2. of that paper, we let $G$ be a nonelementary Fuchsian group acting on $\mathbb{S}^2$. Then $G$ leaves the disk $D^2$ invariant, and we identify the disk to a point $x_0$ and extend the action of $G$ over this point by agreeing that every element of $G$ fixes $x_0$. This produces a group homeomorphism $H$ of $\mathbb{S}^2 / D \cong \mathbb{S}^2$ acting properly discontinuously in the complement of the point $x_0$. Thus $H$ is a convergence group and is not conjugate to any Mobius group since the stabilizer of a point in a Mobius group is virtually abelian and the stabilizer of $x_0$ is $H \cong G$.

$\textbf{Now my question:}$ As I understand, the $G$-invariant disk $D^2$ should be considered as a subset of $\mathbb{S}^2$, for example the upper hemisphere, and by definition of Fuchsian group $G$ acts properly discontinuously on $D^2$. But after the identification it to $x_0$, the complement of $x_0$ is $\mathbb{S}^2 - D^2$, and the action of $G$ on $\mathbb{S}^2 - D^2$ is not propery discontinuous because $G$ is nonelementary and $\mathbb{S}^2 - D^2$ contains some limit points. Since $H$ is produced from $G$ by $$H|_{\mathbb{S}^2 - D^2} = G|_{\mathbb{S}^2 - D^2} \quad \textrm{and} \quad Hx_0 = x_0,$$ then the action of $H$ on the complement of $x_0$ is not properly discontinuous too.

This is completely contrast to the initial example by G. Martin. Can someone show me where the mistake in my thoughts is?

Thank you in advance!

P.S. why do we need nonelementary type of group?

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    $\begingroup$ The limits points of a Fuchsian group are all on the boundary of the disk. Therefore, if we take $D^2$ to be the closed disk, then, contrary to what you say, there are no limit points in $\mathbb{S}^2 - D^2$. If $H$ is elementary, then it is virtually abelian, so there is no contradiction at the end of the proof. $\endgroup$ – Dave Witte Morris Jul 17 '16 at 22:26
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    $\begingroup$ @DaveWitteMorris thank you very much! just one more question: if we let $G$ be a nonelementary Fuchsian group acting on closed disk in complex plane. We identify the boundary circle to point $x_0$, and we get a $2$-sphere. Then the produced group $H$ fixes $x_0$ and acts on its complement properly discontinuously. and the result would be the same, am I right?? Thank you in advance! $\endgroup$ – Jane Jul 18 '16 at 0:48
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    $\begingroup$ Yes, that's another way of doing the same thing as Martin (except that you replaced the complement of $D^2$ with the interior of $D^2$). $\endgroup$ – Dave Witte Morris Jul 18 '16 at 3:49
  • $\begingroup$ @DaveWitteMorris thank you very much! $\endgroup$ – Jane Jul 18 '16 at 4:11

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