TITLE: Example of finitely generated subgroups whose intersection is not finitely generated
QUESTION [11 upvotes]: I'm reading G.Graetzer's Lattice Theory: First Concepts and Distributive Lattices and working on its exercises.  One of them is to prove $(A, \subset)$, where $A$ is the set of finitely generated subgroups of a group $G$, is a join-semilattice, but not necessarily a lattice.  I proved the former half, but I can't think of a counterexample for the latter half, partly because I'm not familiar to group theory.  What is a simple example of a group whose finitely generated subgroups' intersection is not finitely generated?

REPLY [3 votes]: The question was also asking to show that the poset $A$ of finitely generated subgroups of a group $G$, ordered by inclusion, is not a meet-semilattice in general.
In the example from user1729 with $G=F_2\times \mathbb{Z}$, the poset $A$ contains the finitely generated subgroups $P$ and $Q$ of $G$.  $P$ and $Q$ do not have a meet (= greatest lower bound) in $A$.
Indeed, the set of lower bounds of $P$ and $Q$ in $A$ is the set of finitely generated subgroups of $R=P\cap Q$.  But $R$ is a free group on a countable set of generators, say $R=\langle z_1,z_2,...\rangle$. Every word in the generators only makes use of a finite number of generators, so is contained in $\langle z_1,...,z_m\rangle$ for some $m$.  So every finitely generated subgroup of $R$ is contained in some $\langle z_1,...,z_m\rangle$, and we can always find a strictly larger subgroup $\langle z_1,...,z_{m+1}\rangle$ in $A$.  So the set of lower bounds of $P$ and $Q$ does not have a greatest element.