Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
No, if $A$ contains the identity. –  YCor Mar 13 '13 at 13:28

2 Answers 2

up vote 7 down vote accepted

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.

share|improve this answer
1  
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. 3-4, 269–273 –  HJRW Mar 13 '13 at 10:35
1  
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. –  Igor Belegradek Mar 13 '13 at 11:37
1  
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. –  Misha Mar 13 '13 at 13:53
    
Sorry, you're quite right, I have my dates mixed up. –  HJRW Mar 13 '13 at 14:01
1  
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 3-manifold groups is also geometric, but much, much harder. –  Misha Mar 13 '13 at 17:25

On Yves' request, here is a proof of residual finiteness of non-finitely 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, 1-ended 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 Seifert-van 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 homotopy-equivalent to a graph (see the discussion here). Hence, $\Gamma \cong \pi_1(O_U) * F$, where $F$ is a free group. Lastly (again by Seifert-van 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 "ping-pong 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 finite-index 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 non-finitely generated Fuchsian groups.

share|improve this answer
    
thank you. This is great! –  Venkataramana Mar 16 '13 at 11:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.