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.)

reducibleif there are non-units $h_1,h_2$ such that $g = h_1 h_2$, andirreducibleotherwise) would imply that every unit of a Dedekind-finite ring $R$ is irreducible (see mathoverflow.net/questions/261982), and so is, in particular, the identity of $R$ (which is not so good a thing). Usually, we take an irreducible element in a unital ring $R$ to be an atom of its multiplicative monoid, and an atom in a monoid $H$ is anon-unit(...) $\endgroup$ – Salvo Tringali Mar 8 '17 at 10:46