I want to prove the following statement:
PROPOSITION: Let $G$ be a closed subgroup of $SL(2,\mathbb{R})$. Suppose that G is non compact. Then the following are equal:
(i) There is no finite set $L\subset\mathbb{P}^1$ such that $M(L)=L$ for all $M\in G$.
(ii) There is no finite set $L\subset\mathbb{P}^1$ with $|L|=1$ or $2$ ($|L|$ is the cardinality of the set $L$), such that $M(L)=L$ for all $M\in G$.
Trivially (i) implies (ii).
For the converse I'm just able to prove the following result:
LEMMA: If $M\in SL(2,\mathbb{R})$ has 3 different invariant direction then $M=\pm \mathbb{I}$
Indeed if $M= \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)$ and $\bar g=\left( \begin{array}{ccc} g\\ 1\end{array} \right)\in\mathbb{P}^1$ then:
$$\left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)\bar g=\bar g \iff \frac{ag+b}{cg+d}=g \iff cg^2+(-a+d)g-b=0$$
So we have proved the lemma.
I need to prove the following lemma to conlude the proof of the proposition:
LEMMA: If there exists a finite set $L\subset\mathbb{P}^1$ with $|L|>2$ such that $M(L)=L$ for all $M\in G$ then all $M\in G$ has 3 different invariant direction.
The only think that I know is that a closed subgroup of $SL(2,\mathbb{R})$ is compact if and only if it is conjugated to a closed subgroup of $O(2,\mathbb{R})$
Any suggestion?
Suppose that $L\subset {\mathbb R}P^1$ is a finite set of cardinality $n\ge 3$ and that $G< PGL(2, {\mathbb R})$ is a subgroup preserving $L$. We thus have the restriction homomorphism $\rho: G\to Perm(L)\cong \Sigma_n$, the permutation group on $n$ elements; $\rho$ is given by restricting elements of $g$ to $L$. You already know (your 1st lemma) that the kernel of $\rho$ equals $\{1\}$; hence, $\rho: G\to \Sigma_n$ is injective, hence, $G$ has cardinality $\le n!$, hence, $G$ is compact. We conclude that a noncompact subgroup of $PGL(2, {\mathbb R})$ cannot preserve a finite subset of ${\mathbb R}P^1$ consisting of $\ge 3$ elements. Thus, (ii) $\Rightarrow$ (i) in your question. qed