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The quotient group $\mathbb{Q}/\mathbb{Z}$ of the additive abelian group of rational numbers by its subgroup of integers.
It is also the coequalizer of the identity function of the rational numbers and the function $x \mapsto x + 1$ on the rational numbers:
For each natural number $n \in \mathbb{N}$ the cyclic group $\mathbb{Z}/n\mathbb{Z}$ is a subgroup
In fact $\mathbb{Q}/\mathbb{Z}$ is the filtered colimit of a diagram of cyclic groups and inclusions between them. In more detail, whenever $m|n$, multiplication by $n/m$ induces an inclusion
and this gives a functor to abelian groups from the lattice of natural numbers ordered by divisibility. The colimit of this functor is the abelian group $\mathbb{Q}/\mathbb{Z}$.
For each prime $p$ the $p$-Prüfer group $\mathbb{Z}[1/p]/\mathbb{Z}$, similarly formed as the colimit of groups $\mathbb{Z}/p^k\mathbb{Z}$ and inclusions between them, embeds in $\mathbb{Q}/\mathbb{Z}$. Furthermore, it follows from the Chinese remainder theorem? that the induced map
is an isomorphism.
We have a canonical subgroup inclusion into the circle group: If the latter is identified as the canonical subgroup $U(1) \;\subset\; \mathbb{C}^\times$ in the group of units of the complex numbers, this inclusion is given by
From this point of view, $\mathbb{Q}/\mathbb{Z}$ is the torsion subgroup of U(1), whose elements are precisely the roots of unity.
The group $\mathbb{Q}/\mathbb{Z}$ is an injective object in the category Ab of abelian groups.
It is also a cogenerator in the category of abelian groups.
Proof: let $A$ be an abelian group. It suffices to check that for every $a \in A$ there exists $f: A \to \mathbb{Q}/\mathbb{Z}$ such that $f(a) \neq 0$. But if $\langle a \rangle$ is the cyclic subgroup generated by $a$, then it is easy to find a map $g: \langle a \rangle \to \mathbb{Q}/\mathbb{Z}$ such that $g(a) \neq 0$, and then we can extend $g$ to a map $f: A \to \mathbb{Q}/\mathbb{Z}$ using injectivity of $\mathbb{Q}/\mathbb{Z}$.
This means every abelian group embeds into an injective abelian group,
and into an algebraic double dual, $A \hookrightarrow \hom(\hom(A, \mathbb{Q}/\mathbb{Z}), \mathbb{Q}/\mathbb{Z})$. The algebraic double dual of $\mathbb{Z}$ is its profinite completion
where $\mathbb{Z}_p$ is the group of $p$-adic integers (this is connected with the direct sum decomposition of $\mathbb{Q}/\mathbb{Z}$ into Prüfer $p$-groups).
Last revised on December 9, 2023 at 12:14:10. See the history of this page for a list of all contributions to it.