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Let $G$ be a limit group, and let $A,B \leq G$ be finitely generated subgroups generating $G$ (i.e. $\langle A \cup B \rangle = G$). Must $A \cap B$ be finitely generated?

Recall that a limit group is a group whose existential theory (the set of true sentences in first order theory which use only the quantifier $\exists$ and not $\forall$) is the same as that of a nonabelian free group.

Howson's theorem says that the answer is positive in case that $G$ is free.

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The answer is 'yes'. A geometric proof, showing in fact that every finitely generated subgroup of a limit group is relatively quasiconvex, was given by Dahmani.

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    $\begingroup$ I think we have the same answer $\endgroup$
    – Benjamin Steinberg
    Commented Nov 10, 2015 at 16:49
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    $\begingroup$ Looks like we do! $\endgroup$
    – HJRW
    Commented Nov 10, 2015 at 19:22
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They do, see theoerrm 4.7. http://arxiv.org/pdf/math/0203258.pdf by Dahmani.

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