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
Best Answer
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 $3$.