An ordered pair $\ \mathbf X := (X\ d)\ $ is called a chain group $\ \Leftarrow:\Rightarrow\ X\ $ is an abelian group, $\ d:X\rightarrow X\ $ is an abelian group endomorphism, and $\ d\circ d= 0$.


A chain homomorphism $\ f:\mathbf X\rightarrow \mathbf X'\ $ of chain groups $\ \mathbf X := (X\ d)\ $ and $\ \mathbf X' := (X'\ d')\ $ is any abelian group homomorfizm $\ f:X\rightarrow X'\ $ such that $\ d'\circ f=f\circ d$. These are morphisms of the category of the chain groups, and chain monomorphisms (epimorphisms) are chain homomorphisms which are monomorphisms (resp. epimorphisms) as group homomorphisms.


A chain group $\ (X\ d)\ $ is chain generated by a set $\ A\subseteq X\ \Leftarrow:\Rightarrow\ $ for every chain group $\ (Y\ d')\ $ such that
$\ Y\ $ is subgroup of group $\ X,\ d' = d|Y,\ $ and
$\ A\subseteq Y,\ $ we have $Y=X$. A chain group is finitely generated if it admits a finite set of generators.


**Problem**   Provide a full classification of finitely generated chain groups.


Of course, the classification of finitely generated abelian groups is a classical result. I think that the above problem is open (is it?), and I feel that--in the view of the said classical classification, it should not be extremely hard.



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> **Remark**   A set $\ A\ $ generates a chain group $\ (X\ d)\ \Leftrightarrow\ $ the set $\ A\cup\{d(a) : a\in A\}\ $ generates the abelian group $\ X.\ $ Thus, it follows that the above definition of finitely generated chain groups is equivalent to the definition which says that $\ X\ $ should be finitely generated as an abelian group.
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Classification of finitely generated chain groups
The general classification problem seems to be a [wild](https://mathoverflow.net/questions/10481/when-is-a-classification-problem-wild/10484#10484) problem (i.e., contains the classification of pairs of square matrices over a field up to simultaneous conjugacy as a subproblem), and so is probably intractable.


Fix a prime $p$, and consider only those "chain groups" $(X,d)$ where $X=(\mathbb{Z}/p^4\mathbb{Z})^n$ is a direct product of cyclic groups of order $p^4$, and where $p^2X\leq\ker(d)$. 


Such an $(X,d)$ is determined by $n$ and the map $d':X/p^2X\to p^2X$ induced by $d$. Multiplication by $p^2$ gives an isomorphism $\alpha:X/p^2X\to p^2X$, independent of $d$. 


So $(X,d)$ is determined by $n$ and the endomorphism $d'\alpha^{-1}:p^2X\to p^2X$, which is described by some $n\times n$ matrix $A$ over $\mathbb{Z}/p^2\mathbb{Z}$. Let $X\_A$ denote the chain group corresponding to the matrix $A$.


Two chain groups $X\_A$ and $X\_B$ are isomorphic if and only if the matrices $A$ and $B$ are conjugate by an element of $\text{GL}(n,\mathbb{Z}/p^4\mathbb{Z})$, or equivalently by the image of that element in $\text{GL}(n,\mathbb{Z}/p^2\mathbb{Z})$.


I've not found a copy of the original paper, but several references claim that



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> Nagornyı, S. V. "Complex representations of the general linear group of degree three modulo a power of a prime." Zap. Naucn. Sem. Leningrad. Otdel. Mat. Inst. Steklov.(LOMI) 75 (1978): 143-150.
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contains a proof that the classification of square matrices over $\mathbb{Z}/p^2\mathbb{Z}$ is a wild problem. More precisely, that the classification of $4n\times4n$ matrices over $\mathbb{Z}/p^2\mathbb{Z}$ up to conjugacy would suffice to classify pairs of $n\times n$ matrices over $\mathbb{Z}/p\mathbb{Z}$ up to simultaneous conjugacy.


