Vtyjkyuk

i have some confusion.,,,that this question has already asked here
how to find the order of an element in a quotient group

Let $mathbbQ/mathbbZ$ be the quotient group of the additive group of rational numbers.then the order of the element $frac23+mathbbZ$ in $mathbbQ/mathbbZ$.

$1) 2 $

$2) 3 $

$c) 4$

$d) 6$

My attempts : $fracmathbbQmathbbZ$ additive groups of rational numbers

$(2/3 + mathbbZ ) + (2/3 + mathbbZ) + (2/3 + mathbbZ)= 2 + mathbbZ=mathbbZ$

$(2/3 + mathbbZ ) + (2/3 + mathbbZ) + (2/3 + mathbbZ) +(2/3 + mathbbZ ) + (2/3 + mathbbZ) + (2/3 + mathbbZ)=6 + mathbbZ=mathbbZ$

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

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 $ncdot x=0$. In your case, we have $G=Bbb Q/Bbb Z$ and $x=frac23+Bbb Z$, and you've noted that $3cdot x=0$ and $6cdot x=0$. Since $3$ is smaller than $6$, and no smaller positive integer satisfies $ncdot x=0$, we have that the order of $frac23+Bbb Z$ is $3$.

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.

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. )

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 Q mod Z for any element p/q +Z the order of that element is q