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I am looking for a reference for the following

Fact 1: if $A$ and $B$ are finitely generated subgroups of infinite index in a finitely generated free group $F$ then there exists $f \in F$ such that $fAf^{-1} \cap B=\{1\}$.

I know several proofs for this, but surely this should be classical, so a reference would be desirable. Modulo a known result of W. Neumann (Neumann, Walter D. `On intersections of finitely generated subgroups of free groups.' Groups—Canberra 1989, 161–170, Lecture Notes in Math., 1456, Springer, Berlin, 1990), this fact would immediately follow from the following

Fact 2: with the assumptions of Fact 1, $F$ cannot be covered by finitely many double cosets of the form AgB, $g \in F$.

I proved a generalization of Fact 2 for hyperbolic groups some time ago, but I do not know of an explicit (classical) reference for it.

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I do not know any references, but I think Fact 1 can be strengthened: $A$ and $B^f$ generate their free product for some $f$. –  Anton Klyachko Apr 1 '13 at 14:29
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It seems like something that Karrass and Solitar might have proved in their 69 paper on finitely generated subgroups of a free group. It can be proved easily using Stallings graphs. –  Benjamin Steinberg Apr 1 '13 at 14:54
    
Yes, Stallings core graphs would give a short proof of both Fact 1 and the statement Anton Klyachko mentions above. I could not find this in the 1969 paper of Karrass-Solitar, though. –  Ashot Minasyan Apr 1 '13 at 15:04
    
If you like to avoid mentioning graphs, you may recall that any f.g. subgroup is just a free factor of a finite-index subgroup. This reduces the problem to the case where $A$ is a free factor of $F$... –  Anton Klyachko Apr 1 '13 at 15:24
    
Anton, I do not really see how $A$ being a free factor helps here... –  Ashot Minasyan Apr 1 '13 at 15:26
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2 Answers 2

Here is the proof of Fact 1 which uses only facts that were known already to Klein and Poincare (or at least to Dehn and Nielsen):

  1. Every finitely generated free group can be realized as a discrete group of isometries $F$ of the hyperbolic plane $H^2$, so that the quotient $H^2/F$ has finite area but noncompact; thus, $F$ contains parabolic elements.

  2. If $A\subset F$ is a finitely generated infinite index subgroup then the limit set of $A$ is a proper subset of the unit circle. Indeed, being finitely generated Fuchsian group, $A$ is geometrically finite, i.e., the quotient of the convex hull of its limit set by $A$ has finite area. (This is the only mildly nontrivial ingredient in the entire proof.) In our setting, this would mean that covering map $H^2/A\to H^2/F$ is between two surfaces of finite area and so has to be finite. This contradicts the assumption that $|F:A|=\infty$. Thus, the limit set $L(A)$ of $A$ has empty interior in $S^1$.

  3. Fixed points of parabolic elements of $F$ are dense in $S^1$.

  4. Now, suppose that $A, B$ are finitely generated subgroups of infinite index. By combining 2 and 3 we can find a parabolic element $g\in F$ whose fixed point $p$ is not in the union of the limit sets $L(A) \cup L(B)$ of the groups $A$ and $B$. (Similarly, one can use hyperbolic elements of $F$ since their fixed pairs are dense in $S^1\times S^1$.) Since $g^n, n\to infinity$ converges to $p$ uniformly on compacts away from $p$, there exists $n$ so that $$ g^p(L(A))\cap L(B)=\emptyset. $$ It follows that for $f=g^n$ the groups $fAf^{-1}, B$ have trivial intersection (since fixed points of infinite order elements of a Fuchsian group belong to its limit set).

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Thanks, Misha. I have thought about a similar argument that works more generally for infinite index quasiconvex subgroups in a hyperbolic group. –  Ashot Minasyan Apr 1 '13 at 22:04
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Ashot: Of course, this is what the general hyperbolic groups argument is modelled on. My point is that this staff is ancient. –  Misha Apr 1 '13 at 22:16
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I'm sorry I don't have a reference but here is a constructive proof of Fact 1 just using basic algebraic topology: Let $F$ be generated by $a_1$ and $a_2$ (the higher rank case is an easy generalization). Let $B$ be the figure eight with basepoint $b_0$. Let $X$ and $Y$ be covers corresponding to $A$ and $B$ with base points $x_0$ and $y_0$. Since $A$ and $B$ are finitely generated and of infinite-index, there exists infinite graphs that are isomorphic, with labelings, to the images of the lifts of all cyclically reduced words which start with $a_1$ (and similarly $a_1^{-1}$) in the universal cover. Call $I_X$ this infinite graph corresponding to $a_1$ (including the initial $a_1$ edge). Call $I_Y$ the infinite graph corresponding to $a_1^{-1}$ in $Y$ but do not include the edge $a_1^{-1}$ in $I_Y$ (so it does not include the initial $a_1^{-1}$ edge).

Take the graph $X$ and remove the infinite piece $I_X$ to obtain $X_0$. Take $Y$ and remove the infinite piece $I_Y$ to obtain $Y_0$. Glue $X_0$ to $Y_0$ in the obvious place. This gives a new graph $Z$ which is a cover of $B$. A path from $x_0$ to $y_0$ in $Z$ is an $f$ you want.

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