# In a group of Möbius transformations, does discontinuity imply discreteness?

Let $G$ be a subgroup of the group of Möbius transformations $$z \mapsto \frac{az+b}{cz+d}.$$ What is the relationship between the two conditions:

(1) $G$ being discrete.

(2) $G$ acting properly discontinuously.

Partial answer: Clearly (2) implies (1), since if $G$ is non-discrete, there is a convergent sequence in $G$, say $g_n \to g\in G$. Now take the sequence $g^{-1}\circ g_n$ which converges to the identity, and the orbit of any $z\in \widehat{\mathbb{C}}$ has an accumulation point, so $G$ does not act properly discontinuously.

What can be said about the other implication ? In case the two conditions are not equivalent, can we specify additional requirements on the elements of $G$ so that (1) implies (2).

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It is a theorem in the theory of Fuchsian groups that $\Gamma$ is a Fuchsian group (discrete subgroup of the Möbius group) if and only if $\Gamma$ acts as a properly disconituous group of isometries on the upper half plane $\mathbb{H}$.
It's important to note however, that it is not true in general that for a metric space $M$, a subgroup $\Gamma$ of the isometry group $\operatorname{Isom}(M)$ is discrete if and only if it is properly discontinuous.
If $M$ is a complete Riemannian manifold, however, (1) and (2) are equivalent. –  user641 Jan 21 '13 at 13:38