find the the order of the element $\frac{2}{3}+\mathbb{Z}$ in $\mathbb{Q}/\mathbb{Z}$. i have  some  confusion.,,,that  this  question has  already asked here 
how to find the order of an element in a quotient group

Let $\mathbb{Q}/\mathbb{Z}$ be the quotient group of the additive group of rational numbers.then  the order of the element $\frac{2}{3}+\mathbb{Z}$ in $\mathbb{Q}/\mathbb{Z}$.

$1) 2 $
$2) 3 $
$c) 4$
$d) 6$
My attempts : $\frac{\mathbb{Q}}{\mathbb{Z}}$ additive  groups  of  rational numbers
$(2/3  + \mathbb{Z} ) + (2/3 + \mathbb{Z})  + (2/3  + \mathbb{Z})= 2  + \mathbb{Z}=\mathbb{Z}$
$(2/3  + \mathbb{Z} ) + (2/3 + \mathbb{Z})  + (2/3  + \mathbb{Z}) +(2/3  + \mathbb{Z} ) + (2/3 + \mathbb{Z})  + (2/3  + \mathbb{Z})=6  + \mathbb{Z}=\mathbb{Z}$
now my answer  is  option  $ 2)$  and  option $4)$
But   according to  duplicate  Question  answer   is given      $3$     that  order  will be  $3$
im confusing that  why order  $6$   is not correct ???
Any hints/solutiuon will be appreciated
 A: The order of an element $x$ in a group $G$ (let's say abelian, written additively) is defined to be the smallest positive integer $n$ for which $n\cdot x=0$. In your case, we have $G=\Bbb Q/\Bbb Z$ and $x=\frac23+\Bbb Z$, and you've noted that $3\cdot x=0$ and $6\cdot x=0$. Since $3$ is smaller than $6$, and no smaller positive integer satisfies $n\cdot x=0$, we have that the order of $\frac23+\Bbb Z$ is $3$.
A: An element in a group only has one order (the question also talks about "the order"): the smallest integer $n$ such that $ng=0$ (writing the group operation additively).
With your interpretation and your example, e.g. $9$ (any multiple of $3$) would be an order but not among the options, making the options poorly chosen.
A: Remember that in general, the order of an element $g$ in a group $G$ is the smallest integer $n$ such that $g^n = e$ or ($ng = 0$ in additive notation. )
A: You have to choose the smallest one because order is the least positive integer $n$ such that $a^n =e$.
Now here $3$ is the least positive integer .it is not only $6$ but also for any multiple of $3$ you get.$a^{3n}$$=$$e$.(Where $e$ is the identity element).
In $\mathbb{Q} / \mathbb{Z}$ for any element $p/q + \mathbb{Z}$ the order of that element is $q$.Where $gcd(p,q)=1$.
