Exercise 3.1.14 (Quotient group $\mathbb{Q}/\mathbb{Z}$)

Proof. Consider the additive quotient group $\mathbb{Q}∕\mathbb{Z}$.

(a)

A coset of $\mathbb{Z}$ in $\mathbb{Q}$ is a set $C=\alpha +\mathbb{Z}$, where $\alpha =a∕b\in \mathbb{Q}$ ( $a\in \mathbb{Z},b\in {\mathbb{N}}^{\text{∗}}$).

The Euclidean division gives $(q,r)\in {\mathbb{Z}}^2$ such that $a=\mathit{bq}+r,0\le r<b$.

Since $q$ is an integer, $q+\mathbb{Z}=\mathbb{Z}$, thus

$$C=\alpha +\mathbb{Z}=\frac{a}{b}+\mathbb{Z}=\frac{r}{b}+q+\mathbb{Z}=\frac{r}{b}+\mathbb{Z}=\beta +\mathbb{Z},\text{where }0\le \beta =\frac{r}{b}<1.$$

Therefore there is some $\beta \in \mathbb{Q},0\le \beta <1$ such that $\beta \in C=\alpha +\mathbb{Z}$, and $C=\beta +\mathbb{Z}$.

If $\gamma \in C$ satisfies $0\le \gamma <1$, then $\gamma \in \beta +\mathbb{Z}$, so there is some integer $k$ such that $\gamma -\beta =k$.

Since $-1<-\beta \le \gamma -\beta \le \gamma <1$, we obtain $-1<k<1$, where $k\in \mathbb{Z}$. Therefore $k=0$ and $\gamma =\beta$.

In conclusion, every coset of $\mathbb{Z}$ in $\mathbb{Q}$ contains exactly one representative $\beta \in \mathbb{Q}$ in the range $0\le \beta <1$.

(b)

Let $\bar{\alpha }$ denote the class of every $\alpha \in \mathbb{Q}$ in $\mathbb{Q}∕\mathbb{Z}$, so that $\bar{\alpha }=\alpha +\mathbb{Z}$, where $\alpha =a∕b\in \mathbb{Q}$ ( $a\in \mathbb{Z},b\in {\mathbb{N}}^{\text{∗}},g.c.d(a,b)=1)$).

Then $\mathit{b\alpha }=a\in \mathbb{Z}$, thus $b\bar{\alpha }=\overline{\mathit{b\alpha }}=\bar{a}=\bar{0}$, thus $\bar{\alpha }$ has a finite order $d$. Moreover, for all integers $n$,

$$\begin{array}{llllll}\hfill n\bar{\alpha }=\bar{0}& ⟺n\frac{a}{b}\in \mathbb{Z}& \hfill & & \hfill & \\ \hfill & ⟺b\mid \mathit{na}& \hfill & & \hfill & \\ \hfill & ⟺b\mid n& \hfill \text{because }g.c.d(a,b)=1.& & \hfill & & \hfill \end{array}$$

Therefore the order of $\bar{\alpha }$ is $b$, where $a∕b$ ( $b>0$) is the irreducible fraction representing the rational $\alpha$.

Let $N$ be any positive integer. Then the class of the irreducible fraction $\frac{1}{N}$ has order $N$, so there are elements of arbitrary large order.

(c)

We recall that the torsion group of an abelian group $G$ is the subgroup of $G$ of elements of finite order.

Let $T$ denote the torsion group of $ℝ∕\mathbb{Z}$.

Let $\bar{\alpha }=\alpha +\mathbb{Z}\in ℝ∕\mathbb{Z}$, where $\alpha$ is a real number.

As in part (b), if $\bar{\alpha }\in \mathbb{Q}∕\mathbb{Z}$, so that $\alpha =p∕q\in \mathbb{Q}$ (where $a\in \mathbb{Z},b\in {\mathbb{N}}^{\text{∗}}$) then $b\bar{\alpha }=\bar{0}$. Therefore $\alpha$ has finite order, so $\alpha \in T$.

Conversely, suppose that $\bar{\alpha }\in T$, so that $\alpha$ has a finite order $n>0$. Then $n\bar{\alpha }=0$, so $\mathit{n\alpha }\in \mathbb{Z}$, therefore there is some integer $k$ such that $\mathit{n\alpha }=k$, so $\alpha =k∕n\in \mathbb{Q}$, and $\bar{\alpha }\in \mathbb{Q}∕\mathbb{Z}$. This shows that $T=\mathbb{Q}∕\mathbb{Z}$.

In conclusion, $\mathbb{Q}∕\mathbb{Z}$ is the torsion group of $ℝ∕\mathbb{Z}$.

(d)

Let $$G=\{z\in ℂ\mid \exists <!--\; FUNCTION\; APPLICATION\; -->n\in {\mathbb{N}}^{\text{∗}},z^n=1\}$$

be the multiplicative group of roots of unity, in other words the torsion group of ${ℂ}^{\text{×}}$.

Consider the map

$$\phi \{\begin{array}{ccc}\hfill \mathbb{Q}∕\mathbb{Z}\hfill & \hfill \to \hfill & G\hfill \\ \hfill \bar{\alpha }\hfill & \hfill \mapsto \hfill & e^{2\mathit{\pi i\alpha }}\hfill \end{array}$$

• If $\bar{\alpha }=\alpha +\mathbb{Z}\in \mathbb{Q}∕\mathbb{Z}$, where $\alpha =p∕q$, then ${\left(e^{2\mathit{\pi ip}∕q}\right)}^q=e^{2\mathit{\pi ip}}=1$, so $e^{2\mathit{\pi i\alpha }}\in G$.

• $\phi$ is well defined: If $\bar{\alpha }=\bar{\beta }$, then $\beta -\alpha \in \mathbb{Z}$, so $\beta =\alpha +k$ for some $k\in \mathbb{Z}$. Therefore

  $$e^{2\mathit{\pi i\beta }}=e^{2\mathit{\pi i}(\alpha +k)}=e^{2\mathit{\pi i\alpha }}{e}^{2\mathit{\pi ik}}=e^{2\mathit{\pi i\alpha }}.$$

  So $e^{2\mathit{\pi i\alpha }}$ doesn’t depend of the choice of the representative $\alpha$ in the coset $\bar{\alpha }$.

  (Alternatively, we may use the future First Isomorphism Theorem to define $\phi$ from the homomorphism $\psi :\alpha \mapsto e^{2\mathit{\pi i\alpha }}$ from $\mathbb{Q}$ into $G$.)

• $\phi$ is injective: If $\bar{\alpha }\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $e^{2\mathit{\pi i\alpha }}=1$, therefore $\alpha \in \mathbb{Z}$, hence $\bar{\alpha }=\{\bar{0}\}$, so $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{\bar{0}\}$, which proves that $\phi$ is injective.
• Let $z$ be any element of $G$. Then $z^n=1$ for some integer $n\ge 1$. Solving the equation $z^n=1$ shows that $z=e^{2\mathit{\pi ik}∕n}$ for some integer $k$. Therefore $z=\phi (\overline{k∕n})$, where $\overline{k∕n}=k∕n+\mathbb{Z}\in \mathbb{Q}∕\mathbb{Z}$. This shows that $\phi$ is surjective.

This shows that $\phi$ is well defined, and is an isomorphism, so

$$G=\{z\in ℂ\mid \exists <!--\; FUNCTION\; APPLICATION\; -->n\in {\mathbb{N}}^{\text{∗}},z^n=1\}\simeq \mathbb{Q}∕\mathbb{Z}.$$

The group of roots of unity is isomorphic to $\mathbb{Q}∕\mathbb{Z}$.