Order of elements in quotient Q/Z So I have a question asking me to compute the order of:
$\frac 8 9 +\Bbb Z$, $\frac {14} 5 +\Bbb Z$, and $\frac {48} {28} +\Bbb Z$
in the quotient $\Bbb Q/\Bbb Z$.
I think the answer is order 9, 5, and 7 respectively but would like some clarification as to why that is the case.
Any help would be greatly appreciated as I am new to group theory.
 A: The order of a coset $a+\mathbb Z$ is the least positive integer $n$ such that $na\in \mathbb Z$. This is equivalent to saying that
$$n(a+\mathbb Z)=0+\mathbb Z=\mathbb Z$$
and $n$ is the smallest positive integer such that this holds.
A: For an element $r+\mathbb{Z}$ ($r\in\mathbb{Q}$) in $\mathbb{Q}/\mathbb{Z}$ the order of such an element would be the smallest natural number $a$ such that $ar\in \mathbb{Z}$. This is because in $\mathbb{Q}/\mathbb{Z}$ if $r\in\mathbb{Z}$ then $r+\mathbb{Z}=0+\mathbb{Z}$ which is the identity element of $\mathbb{Q}/\mathbb{Z}$.
To find this all you need do is write $r$ in lowest terms i.e. $r=\frac{p}{q}$ for coprime integers $p,q$ with $q>0$ then this $q$ is the value of $a$ you are looking for so in your example we have:


*

*$r=\frac{8}{9}$ which is already in lowest terms so the order of $\frac{8}{9}+\mathbb{Z}$ is 9.

*$r=\frac{14}{5}$, again already in lowest terms so the order of $\frac{14}{5}+\mathbb{Z}$ is 5.

*$r=\frac{48}{28}$ which in lowest terms is $\frac{12}{7}$ and so the order of $r=\frac{48}{28}+\mathbb{Z}$ is 7.


I hope this clears things up for you.
