Let $(G, +)$ be a commutative group. The endomorphism set $\text{End}(G)$ of all group endomorphisms $f:G\to G$ is a ring, where $+$ is taken pointwise and the multiplication is the composition of endomorphisms.


Suppose $f\in \text{End}(G)$ is not bijective (i.e. not a unit in the ring) and not a zero-divisor. Can it be written as a product of irreducible elements? 


(We call $g\in\text{End}(G)$ *reducible* if there are non-units $h\_1,h\_2\in\text{End}(G)$ such that $g = h\_1\circ h\_2$; and we call $g$ *irreducible* otherwise.)



Irreducible elements in endomorphism rings
If $R$ is a commutative unital ring, then $R \cong\_{\sf Ring} {\rm End}\_{{\sf Mod}\_R}(R\_R)$, with the elements of $R$ acting on $R$ by left multiplication. Now, let $R$ be any non-atomic integral domain and note that the multiplicative monoid of ${\rm End}\_{{\sf Mod}\_R}(R\_R)$ is a divisor-closed submonoid of the multiplicative monoid of ${\rm End}\_{{\sf Grp}}(R\_R)$. Lastly, recall that a commutative unital ring is atomic (i.e., every non-unit, non-zero element is a product of some atoms) iff so is its multiplicative monoid, and that a monoid $H$ with zero is atomic only if so are all the divisor-closed submonoids of $H$ (we say that a submonoid $M$ of $H$ is divisor-closed if $x \mid\_H y$ and $y \in M$ imply $x \in M$). 


