The subobject poset of the integers in abelian groups

A monomorphism $f:H\to \mathbb{Z}$ in $\mathbf{A}\mathbf{b}$ is injective: the inclusion $\ker f\to H$ and the zero map $\ker f\to H$ have equal composites with $f$, so they are equal and $\ker f=0$. Its corestriction onto the image $H^{\mathrm{\prime }}=f[H]\le \mathbb{Z}$ is then an isomorphism $H\to H^{\mathrm{\prime }}$ with $f={\iota }_{H^{\mathrm{\prime }}}\circ \bar{f}$ and ${\iota }_{H^{\mathrm{\prime }}}=f\circ {\bar{f}}^{-1}$, so $f$ mutually factors with the inclusion of $H^{\mathrm{\prime }}$ and represents it. By [L1], exactly one $n\in \mathbb{N}$ has $H^{\mathrm{\prime }}=n\mathbb{Z}$, so [L2] gives precisely the displayed representatives.

The inclusion $n\mathbb{Z}\hookrightarrow \mathbb{Z}$ factors through $m\mathbb{Z}\hookrightarrow \mathbb{Z}$ exactly when $n\mathbb{Z}\subseteq m\mathbb{Z}$, which holds exactly when $m$ divides $n$. By [L2] and [L3], this is the subobject order.

At $n=0$ the subgroup is $\{0\}$ and factors through every subgroup, whereas at $n=1$ it is all of $\mathbb{Z}$ and every subgroup factors through it. These are respectively the least and greatest classes.