Example of finitely generated subgroups whose intersection is not finitely generated

Question

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?

Answer 1

$F_2\times \mathbb{Z}\cong \langle a, b, c; [a, c], [b, c]\rangle$ works.

Let $P:=\langle a, bc\rangle$ and $Q:=\langle a, b\rangle$. Then $P\cap Q=\langle b^iab^{-i}:i\in\mathbb{Z}\rangle$, which is a free group as subgroups of free groups are free. It is therefore clearly infinitely generated. Write $R:=\langle b^iab^{-i}:i\in\mathbb{Z}\rangle$.

To see that $P\cap Q$ is this, note that $b^iab^{-i}=b^ic^iac^{-i}b^{-i}=(bc)^ia(bc)^{-i}$ so these generators generate a subgroup of the intersection ($R\leq P\cap Q$), while taking a word, $W(a, bc)$ say, then $W(a, bc)=W(a, b)c^x$ where $x$ is the exponent sum of $b$ in $W(a, b)$. Thus, if $W(a, b)\in P\cap Q$ we must have that $W$ has exponent sum zero in $b$, and so $W(a, b)\in R$. Thus, $P\cap Q\leq R$. Therefore, $P\cap Q=R$ as required.

This proof is from a paper of D. I. Moldavanskii, entitled "Intersection of finitely generated subgroups", 1968.