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?



Classification of subgroups of finitely generated abelian groups
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.**](http://dx.doi.org/10.1016/j.jpaa.2005.07.002), J. Pure Appl. Algebra 205, No. 2, 412-422 (2006). [ZBL1147.16019](https://zbmath.org/?q=an:1147.16019).


and its references.


