I know that if $G$ is a finitely generated group and $H$ is a subgroup of $G$, then $H$ is not necessarilly finitely generated.
However, I am asking if this property is true if $G$ is a (finitely generated abelian)-by-(finitely generated abelian) group.
My attempt is the following:
Since $G$ is (finitely generated abelian)-by-(finitely generated abelian), there are two finitely generated abelian groups $A$ and $B$ and a short exact sequence $$ 1 \rightarrow A \rightarrow G \rightarrow B \rightarrow 1,$$ and let us denote the homomorphism $G \mapsto B$ by $p$.
Let $H$ be a subgroup of $G$. Then, that short exact sequence descends to a short exact sequence $$ 1 \rightarrow A \cap H \rightarrow H \rightarrow p(H) \rightarrow 1.$$
Since $A \cap H$ is a subgroup of $A$ and $A$ is finitely generated abelian, $A\cap H$ is also finitely generated abelian. Because of the same reason, $p(H)$ is finitely generated abelian.
Therefore, $H$ is (finitely generated abelian)-by-(finitely generated abelian). In particular, $H$ is finitely generated.
Is this correct? Thanks in advance!