Why is there a $z_0\in \mathbb H$, such that $g(z_0)=z_0\implies g=id$?

Question

Take $\mathbb H$ the Poincaré upper half plane and $G$ a Fuchsian group acting on it.

I don't see why we can pick a $z_0\in \mathbb H$ such that
$$ g(z_0)=z_0\implies g=id,\quad \forall g\in G.$$

Answer 1

You're looking for a $z_0$ such that the only element of $G$ that leaves it in place is the identity.

Consider the upper half-plane to be a model of the hyperbolic plane, and $G$ acting isometrically on it. If any $g\in G$ fixes three points that are not (hyperbolically) collinear, then it must be the identity. Thus, for every $g\ne\rm id$, the set $$ \{z \mid gz = z \} $$ is contained in a hyperbolic line (it need no be the whole line, though: it can be empty or a singleton), and therefore it has Lebesgue measure zero.

Since $G$ is countable, this means that $$ \bigcup_{g\in G\setminus\{\rm id\}} \{ z \mid gz = z \} $$ -- that is, the set of points that are fixed by any non-identity element of $G$ -- also has measure zero. In particular it cannot be the entire $\mathbb H$, so you can choose a $z_0$ outside of it.

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(Of course measure theory can be considered a bit overkill for showing that a countable collection of lines cannot cover the plane. Instead we could also look at a single line that is not among the fixpoint lines -- each fixpoint line intersects it in at most one point, so only countably many of the points on that line will be excluded).

Answer 2

A Fuchsian group is not only discrete, but also acts properly discontinously on $\mathbb H$.

So take a nontrivial compact subset $K\subset \mathbb H$ and assume that $$\forall k\in K\,\exists g\in G\setminus\{id\}\mid g(k)=k.$$ Since any nontrivial $g\in G$ has at most two fixed points, this implies that $$\#\{g\mid g(K)\cap K\neq \emptyset\}=\infty. $$ This is a contradiction, so there must be a $z_0\in K\subset \mathbb H$ such that $$g(z_0)=z_0\implies g=id.$$