Quotient group $\mathbb{Q}/\mathbb{Z}$

Question

How do I prove that the quotient group $\mathbb{Q}/\mathbb{Z}$ is a group? What is its unity element? How do I prove that all elements are of finite order?

Answer 1

The rationals $\mathbb{Q}$ are a group under addition and $\mathbb{Z}$ is a subgroup (normal, as $\mathbb{Q}$ is abelian). Thus there is no need to prove that $\mathbb{Q}/\mathbb{Z}$ is a group, because it is by definition of quotient group.

The identity is the coset of $0$, that is $0+\mathbb{Z}$.

Every element has finite order, because, if $a/b\in \mathbb{Q}$, then you can assume $b>0$ and you have $$ b\left(\frac{a}{b}+\mathbb{Z}\right)=a+\mathbb{Z}=0+\mathbb{Z} $$ because $a\in\mathbb{Z}$.

Answer 2

$\mathbb{Q}$ is abelian so $\mathbb{Z}$ is a normal subgroup, hence $\mathbb{Q}/\mathbb{Z}$ is a group. Its unit element is the equivalence class of $0$ modulo $\mathbb{Z}$ (all integers). Let $q\in \mathbb{Q}/\mathbb{Z}$ be the equivalence class of $\frac{a}{b}$ ($a\in\mathbb{Z}$, $b\in\mathbb{N}$), then clearly $\left(\frac{a}{b}\right)^{b}\equiv a\equiv 0$ (where $\left(\frac{a}{b}\right)^b=\frac{a}{b}+\cdots+\frac{a}{b}$ $b$ times, since addition is the group operation). So each element has finite order.

Answer 3

As uncountable pointed out, we need for the group quotient $G/H$to be a group, for $H$ to be normal in $G. \mathbb Z$ is normal in $\mathbb Q$ , because it's Abelian.

Now, by definition, given $q,q' \in \mathbb Q, q \~ q'$ iff $ q-q' \in \mathbb Z $. This means $ q + \mathbb Z =q $, so that ) the class of $\mathbb Z$ is the unit/neutral element. Thus, we get $[0,1)$ as the quotient $\mathbb Q / \mathbb Z$ , with $ \mathbb Z $ as the additive neutral element.

For any $p/q; q \neq 0$ in $\mathbb Q$, we have $ q(p/q)= p$, so every element/ class $p/q$ has finite order $q$.