Example of finitely generated subgroups whose intersection is not finitely generated 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?
 A: 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.
A: $F_2\times \mathbb{Z}\cong \langle a, b, c; [a, c], [b, c]\rangle$ 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.
