Intersection of conjugates of subgroups in free groups 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.
 A: 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):


*

*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. 

*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 $S^1$. 

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

*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). 
A: 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. 
