# 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?

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?

-
I don't think there is much prospect of finding natural necessary and sufficient conditions for FGIP. The best you can hope for is to decide whether or not various interesting classes of groups have the property. For example free groups do. – Derek Holt Mar 26 '12 at 8:20
An only-vaguely related, but interesting all the same, fact is that if both these groups sit with finite index in the supergroup then their intersection will also be finitely generated (and of finite index in the supergroup). – user1729 Mar 26 '12 at 9:46
In addition to Derek Holt's comment, it is closed under free products! If $H$ and $K$ both have FGIP then so does $H\ast K$. – user1729 Mar 26 '12 at 9:54
Interesting results. If you guys haven't had a chance to look at it yet, I've expanded the scope of my question a bit; I'd appreciate it if you could shed any further insight on the matter. – Dan M. Katz Mar 27 '12 at 20:48

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

1. There exists a group $G$ with property $P$,
2. 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.

-
What a cool and far-reaching theorem. Thanks! – Dan M. Katz Mar 25 '14 at 3:43