Is $\operatorname{Hom}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z})$ isomorphic to any "known" group? I suppose what I mean is, is it isomorphic to a group that isn't a Hom group? If such an isomorphism is known, is there a reference which sketches it out?
I know $\mathbb{Q}/\mathbb{Z}=\bigoplus_p\mathbb{Z}_{p^\infty}$, but using the various properties to pull direct sums out doesn't seem fruitful.
My motivation is that I know $\operatorname{Ext}^0_\mathbb{Z}(A,B)$ is naturally isomorphic to $\operatorname{Hom}_\mathbb{Z}(A,B)$, and pairs like $\operatorname{Hom}(\mathbb{Z},\mathbb{Z})$, $\operatorname{Hom}(\mathbb{Q},\mathbb{Q})$, etc., come up frequently in algebraic topology/basic examples in homological algebra. Since $\mathbb{Q}/\mathbb{Z}$ shows up commonly in topology, I'm curious about that case as well.
Best Answer
$\mathbb{Q}/\mathbb{Z}$ can be described as the filtered colimit of its $n$-torsion subgroups, each isomorphic to $\mathbb{Z}/n\mathbb{Z}$. Hence $\text{Hom}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})$ is the cofiltered limit of the groups $\text{Hom}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{Z}/n\mathbb{Z}$ (a simple form of Pontryagin duality), and a little inspection to check that the morphisms in the corresponding diagram are what you expect them to be shows that you get the profinite integers $\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$ (a priori as a group but in fact as a ring).