Classification of subgroups of finitely generated abelian groups

Question

A finitely generated abelian group $A$ is isomorphic to a direct sum of cyclic groups. I am interested in an extension of this result on couples of abelian groups $(A,B),$ where $B$ is a subgroup of $A.$ Consider the category of such couples $(A,B),$ where morphism $f:(A,B)\to (A',B')$ is a homomorphism $f:A\to A'$ such that $f(B)\subseteq B'.$ A couple $(A,B)$ is called cyclic if $A$ and $B$ are cyclic groups.

Question 1: Is it true that any couple of finitely generated abelian groups $(A,B)$ is isomorphic to a direct sum of cyclic couples?

If $A$ is free, it follows from the Smith normal form theorem. If there was a version of the Smith normal form theorem for arbitrary homomorphisms of finitely generated abelian groups, then, I believe, this result would follow.

Question 2: Is there a version of the Smith normal form theorem for arbitrary homomorphisms of finitely generated abelian groups?

Answer 1

The answer to Question 1 is no.

Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and let $B$ be the subgroup generated by $(2,1)$.

Since $B$ is cyclic of order $4$, if it were contained in a proper direct summand of $A$ then it would be contained in a cyclic subgroup of $A$ of order $8$, and so $(2,1)$ would be equal to $2a$ for some $a\in A$, which it's not.

In fact, if you consider only the case where $A$ has exponent dividing $p^n$ for some prime $p$ and natural number $n$, there are infinitely many indecomposable such couples if $n=6$, and the classification is in some sense "wild" if $n\geq7$. See, for example,

Ringel, Claus Michael; Schmidmeier, Markus, Submodule categories of wild representation type., J. Pure Appl. Algebra 205, No. 2, 412-422 (2006). ZBL1147.16019.

and its references.

Answer 2

For finite abelian groups see Sapir, M. V. Varieties with a finite number of subquasivarieties. Sibirsk. Mat. Zh. 22 (1981), no. 6, 168–187, 226.