Show that in the factor group $\Bbb Q / \Bbb Z$, there is an element for every $n \in \Bbb N_+$ such that the order of that element is $n$

Question

Task:

Show that in the factor group $\Bbb Q / \Bbb Z$, there is an element for every $n \in \Bbb N_+$ such that the order of that element is $n$.

Solution:

We take a look at the residue class $[$$1 \over n$$]$.

It is

$n$ $[$$1 \over n$$]$ = $[$$n \over n$$]$ = [$1$] = [$0$]

since $1$ belongs to $\Bbb Z$. On the other hand, for

$1 \le k \le n - 1$,

we receive

$k$ $[$$1 \over n$$]$ = $[$$k \over n$$]$,

which is a rational number, but not an integer. So it is

$k$ $[$$1 \over n$$]$ $\neq 0$,

and therefore, $n$ is the order of $[$$1 \over n$$]$.

There are some things that I don't understand since I have my issues with residue classes in general.

1. What kind of elements does $\Bbb Q / \Bbb Z$ contain?

2. Why is [$1$] = [$0$]?

3. How does the proof function in general? How did he conclude that $n$ is indeed the order of $[$$1 \over n$$]$? I don't see the point he is making.

Answer 1

1. $\mathbb{Q} / \mathbb{Z}$ contains equivalence classes of rationals under the binary relation "the difference between the two numbers is integer". Since every rational can be expressed as an integer plus a number in $[0,1)$, we may view the elements of $\mathbb{Q}/\mathbb{Z}$ as being simply elements of $[0,1) \cap \mathbb{Q}$, with addition performed modulo $1$: that is, subtracting $1$ if necessary to bring the answers back into $[0,1)$. If we do view them like this, we'll write $[x]$ instead of $x$ to remind us that we're talking about classes of rational numbers with this strange addition operator, rather than about the rationals themselves.
2. $[1] = [0]$ because $1$ and $0$ are related by the relation "the difference between the two numbers is integer".
3. The order of $[\frac{1}{n}]$ is the least positive number of times we need to add $\frac{1}{n}$ to itself to get an integer: that is, the least positive number $m$ such that $m \times \frac{1}{n}$ is integer. If $m < n$ then $m \times \frac{1}{n}$ is strictly between $0$ and $1$, so the order can't be less than $n$; and if $m = n$ then $m \times \frac{1}{n}$ is precisely the integer $1$. Hence the order is $n$ and not less than $n$.

Answer 2

I like to think of the elements of $\mathbb{Q}/\mathbb{Z}$ as the numbers in $[0,1) \cap \mathbb{Q}$. However this just picks some representatives. That is, as a set the elements are the residue classes of the quotient map. To each $q\in \mathbb{Q}$ we associate the set $\{q+k \mid k \in \mathbb{Z}\}$. Now several elements in $\mathbb{Q}$ get associated with the same set, for example 0 and 1, as $\{0+k \mid k \in \mathbb{Z}\} =\{1+(k-1) \mid k \in \mathbb{Z}\} = \{1+(k-1) \mid (k-1) \in \mathbb{Z}\}= \{1+k \mid k \in \mathbb{Z}\}$.

But in $[0,1)$ each number gets a unique set of residue classes. I hope this answers questions 1. and 2.

For 3. we want a number such that $n$ time the number is mapped to the same residue class a $[0]$. As we have already seen, $[0]=[1]$, and thus $\frac{1}{n}$ does the trick.

Answer 3

1. The elements of $\mathbb{Q}/\mathbb{Z}$ are cosets of the form $a+\mathbb{Z}=\left\{a+y\mid y\in \mathbb{Z}\right\}\subset\mathbb{Q}$, $a\in\mathbb{Q}$. We denote the element $a+\mathbb{Z}$ by $[a]$. If you take the equivalence relation $\sim$ on $\mathbb{Q}$ given by $a\sim b$ if $a-b\in\mathbb{Z}$, then $[a]$ is actually an equivalence class $[a]=\left\{x\in \mathbb{Q}\mid x-a\in\mathbb{Z}\right\}$ that contains all the elements $x$ such that $x\sim a$. The $[0]$ class is just the set $\mathbb{Z}$.
2. Two elements $[a]=a+\mathbb{Z}$ and $[b]=b+\mathbb{Z}$ are equal iff $a\sim b$, i.e. iff $a-b\in\mathbb{Z}$. For example $[\frac{1}{2}]=[\frac{3}{2}]$ because $\frac{1}{2}-\frac{3}{2}=-1\in\mathbb{Z}$. Here $1-0\in\mathbb{Z}$, so $[0]=[1]$. For all $a\in\mathbb{Z}$, $[a]=[0]$ and conversely if $[a]=0$ for some $a\in\mathbb{Q}$, then $a\in\mathbb{Z}$.
3. Recall the definition of the order of an element $x$ of a group $G$: $n\in\mathbb{N}_+$ is the order of $x$ if $x^n$ is the identity element of $G$ and $x^k$ is not the identity element for any $k$ smaller than $n$. Here, we are in abelian situation and we write the operation additively, so $x^n$ is $n\cdot x$. First, he shows that $n\cdot[\frac{1}{n}]=[1]=[0]$. He then shows that $k\cdot[\frac{1}{n}]\neq [0]$, because $\frac{k}{n}\not\in \mathbb{Z}$.

Answer 4

It may be easier to deal with homomorphisms with a given kernel than with factor groups.

The map $t \mapsto e^{2\pi i t}$ is a group homomorphism $(\mathbb R,+) \to (\mathbb C, \times)$ whose kernel is $\mathbb Z$.

Under this map, $\mathbb Q$ is sent onto the set $U$ of roots of unity and so $\mathbb Q/\mathbb Z \cong U$.

It is clear that there are elements of order $n$ in $U$ for every $n$.