nLab Q/Z

Idea

The quotient group $\mathbb{Q}/\mathbb{Z}$ of the additive abelian group of rational numbers by its subgroup of integers.

Properties

Relation to cyclic groups

For each natural number $n \in \mathbb{N}$ the cyclic group $\mathbb{Z}/n\mathbb{Z}$ is a subgroup

$$\array{ \mathbb{Z}/n\mathbb{Z} &\overset{\;\;\;\;\;\;}{\hookrightarrow}& \mathbb{Q}/\mathbb{Z} \\ [k] &\mapsto& [k/n] }$$

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

$$\mathbb{Z}/m\mathbb{Z} \hookrightarrow \mathbb{Z}/n\mathbb{Z}$$

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 $\mathbb{Q}/\mathbb{Z}$.

Relation to Prüfer groups

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

$$\bigoplus_p \mathbb{Z}[1/p]/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$$

is an isomorphism.

Relation to the circle group

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

$$\array{ \mathbb{Q}/\mathbb{Z} &\overset{\;\;\;\;\;\;}{\hookrightarrow}& U(1) \\ [q] &\mapsto& \exp\big( 2\pi\mathrm{i} q\big) \,. }$$

From this point of view, $\mathbb{Q}/\mathbb{Z}$ is the torsion subgroup of U(1), whose elements are precisely the roots of unity.

Further properties

• 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,

$$A \to \prod_{f: A \to \mathbb{Q}/\mathbb{Z}} \mathbb{Q}/\mathbb{Z},$$

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

$$\prod_{p} \mathbb{Z}_p$$

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).

• For any ring $R$, its algebraic dual $\hom(R, \mathbb{Q}/\mathbb{Z})$, equipped with the left $R$-module structure defined by $r \cdot f: s \mapsto f(s r)$, is an injective cogenerator in the category of left $R$-modules. This follows easily from the corresponding property for $\mathbb{Q}/\mathbb{Z}$.

Related concepts

• cyclic group, multiplicative group of integers modulo n

• Prüfer group, root of unity

• Brown-Comenetz duality

• e-invariant