Conditions for the Finitely Generated Intersection Property While going through Gratzer's "General Lattice Theory", I was surprised to learn (via some exercise) that the intersection of two finitely generated subgroups is not necessarily finitely generated. Apparently, a group for which this condition holds is said to have the "Finitely-Generated Intersection Property" (FGIP). Some quick Google-ing yields some papers which have results for specific cases, but little in regard to the property in the general case.
My question is this: What can be said in the general case about the FGIP? Is there some known necessary and sufficient criteria which a group must possess for the FGIP property to hold? Or is this property too vague for consideration in the general case?
Thanks in advance!
EDIT:
I think that the following questions are also natural and related to  my original post. They may be equivalent variations of the same question, but I am not sure for my own part. I apologize if they are redundant.
(1) Given a group G and two specific finitely-generated subgroups, H and K, are there necessary and sufficient conditions as to whether the intersection of H and K is finitely-generated?
(2) Given an arbitrary group G, is it a decidable problem to determine whether it possesses FGIP?
(3) Are there any known counter-examples to (2), that is, a group for which the problem of determining whether the group possesses FGIP is undecidable?
 A: I can only answer number 2.  It is undecidable to determine if a group $G$ has FGIP, since FGIP is a Markov Property for groups.  
A property $P$ of groups (group presentations) is said to have the Markov Property if 


*

*There exists a group $G$ with property $P$,

*There exists a group $H$ which is not isomorphic to a subgroup of a group with property $P$ (i.e. $H$ cannot be embedded in a group with property $P$.)


There is a theorem by Adion and Rabin (1955/58) which states:
If $P$ has Markov property, then there exists no algorithm to decide if a presentation $G=\langle A\vert R\rangle $ has property $P$.
So, to see that FGIP is Markov, let $H$ be a group which does not have FGIP.  (An example of this is $\pi _1 (\Sigma_2 \times S^1)$, where $\Sigma_2$ is a surface.) Then suppose $H\cong J\leq K$ and $K$ has the FGIP. Now let $A,B$ be finitely generated subgroups of $H$ and $\overline{A},\overline{B}$ be the corresponding subgroups of $J$. Then $\overline{A},\overline{B}$ are also finitely generated and since they are subgroups of $K$, $\overline{A}\cap \overline{B}$ is finitely generated.  This implies that $A\cap B$ is finitely generated, which implies $H$ has FGIP, which is a contradiction. Hence FGIP is Markov and the theorem applies.
