Given G, a fuchsian group and a finite sub set A of G. Does there exist a finite index subgroup H in G such that inter section of A with H is empty?
2 Answers
The answer is yes. The group is residually finite. This means that if $A$ is taken to be a finite set of elements of the Fuchsian group $G$ $not \quad containing\quad identity$, then there exists a finite quotient of $G$ where no element of $A$ is trivial. The kernel of this quotient map is a finite index subgroup$H$ not intersecting $A$.
On the other hand, if $A$ contains identity, then of course the statement is false.

1$\begingroup$ Residual finiteness follows from the fact that it's linear. This implication is often attributed to Mal'cev, though in essence it goes back to Selberg. For a proof see, for instance Alperin, Roger C. An elementary account of Selberg's lemma. Enseign. Math. (2) 33 (1987), no. 34, 269–273 $\endgroup$– HJRWCommented Mar 13, 2013 at 10:35

1$\begingroup$ Regarding Mal'cev vs Selberg, de la Harpe's book says that Mal'cev published it 20 years earlier (1940 vs 1960). I looked into original papers long time ago and my recollection is that de la Harpe is correct. $\endgroup$ Commented Mar 13, 2013 at 11:37

1$\begingroup$ Selberg's new result was that fg linear groups are virtually torsion free. Of course, his 1960 paper is worth reading for other things as well. $\endgroup$– MishaCommented Mar 13, 2013 at 13:53

$\begingroup$ Sorry, you're quite right, I have my dates mixed up. $\endgroup$– HJRWCommented Mar 13, 2013 at 14:01

1$\begingroup$ One more thing: Peter Scott (1978) gave a very nice and elementary geometric proof of residual finiteness for Fuchsian groups. Work of Wise and Agol implying LERF for hyperbolic 3manifold groups is also geometric, but much, much harder. $\endgroup$– MishaCommented Mar 13, 2013 at 17:25
On Yves' request, here is a proof of residual finiteness of nonfinitely generated Fuchsian groups. (I do not know a reference for this: In general, people do not like working with infinitely generated groups, so, it is possible that nobody bothered writing a proof, even though many people could. The result could be in Beardon's or Maskit's book though, I did not check.) A side remark: Residual finiteness of Fuchsian groups was a partial motivation for this question.
A Fuchsian group is a discrete group $\Phi$ of isometries of hyperbolic plane $H^2$. Then $\Phi$ contains an index 2 subgroup $\Gamma$ which preserves orientation on the hyperbolic plane and, hence, $\Gamma < PSL(2, R)$. Since residual finiteness is a commensurablity invariant, it suffices to consider the case of discrete subgroups of $PSL(2,R)$. Let $O=H^2/\Gamma$ be the quotient orbifold. Its singular set is a discrete subset $\Sigma=\{x_i: i\in I\}$ of the underlying space $X$ of $O$. I will assume that $\Gamma$ is not finitely generated, thus, $O$ is noncompact.
Let $T\subset X$ be an infinite locally finite, 1ended properly embedded tree (with geodesic edges) which contains $\Sigma$. Let $U$ be a small neighborhood (with smooth boundary) of $T$ in $X$ which admits a retraction to $T$. Every boundary component of $U$ is contractible. Let $O_U$ be the suborbifold of $O$ supported by $U$. Thus, by Seifertvan Kampen theorem for orbifolds applied to this situation, the fundamental group $\Gamma$ of $O$ splits as a free product $$ \pi_1(O_U) * \prod^*_i \pi_1(S_i) $$ where $S_i$'s are components of $X\setminus U$, and $\prod^*$ means free product. Now, since each surface $S_i$ is noncompact, by a theorem of Whitehead, it is homotopyequivalent to a graph (see the discussion here). Hence, $\Gamma \cong \pi_1(O_U) * F$, where $F$ is a free group. Lastly (again by Seifertvan Kampen theorem applied to $O_U$), the group $\pi_1(O_U)$ is a free product of finite cyclic groups $Z_{n_i}$, where $n_i$ is the order of the singular point $x_i\in \Sigma$. Thus, $\Gamma$ is the free product of countably many cyclic groups (some of which might be finite and some infinite).
The converse is also true: If $\Gamma$ is a countable product of cyclic groups $C_i$, then $\Gamma$ is isomorphic to a discrete subgroup of $PSL(2, R)$. Namley, make each $C_i$ act isometrically on $H^2$ with the fundamental domain $D_i$, so that the sets $cl(H^2 D_i)$ are pairwise disjoint. Now, take the subgroup $\Phi$ of $PSL(2, R)$ generated by the subgroups $C_i$. By Klein combination theorem (also known as the "pingpong argument") $\Phi$ is discrete (its fundamental domain is the intersection of $D_i$'s) and isomorphic to $\Gamma$.
Now, suppose that $\Gamma$ is a group which is a countable free product of residually finite groups $\Gamma_i, i\in {\mathbb N}$. Let $S\subset \Gamma$ be a finite subset not containing the identity. Then $$ S\subset \prod^*_{j\in J} \Gamma_j$$ where $J\subset I$ is a finite subset. Hence, $S$ projects bijectively to a subset $P$ in the quotient $$ \Lambda=\Gamma/\langle \langle \bigcup_{i\notin J} \Gamma_i\rangle \rangle. $$ Now, $\Lambda$ is residually finite (as a finite free product of residually finite groups). Hence, $\Lambda$ contains a finiteindex subgroup $\Lambda'$ disjoint from $P$. Taking preimage of $\Lambda'$ under the epimomorphism $\Gamma \to \Lambda$, we obtain a finite index subgroup in $\Gamma$ disjoint from $S$. This argument, of course, applies to free products of cyclic groups, thereby proving residual finiteness of nonfinitely generated Fuchsian groups.