0
$\begingroup$

I have the following question :

$G=\mathbb{R},H=\mathbb{Z}$ find the order of $(-\frac{10}{24}+H)$ and $(-\frac{10}{28}+H)$ in $K=\frac{G}{H}$

I think that I need to find $r\in \mathbb{N}$ such that $(-\frac{10}{24}+H)^r=-\frac{10}{24}^r=e$ when $e$ is the natural element yet I can't figure out what is the natural element in $K$

Any ideas/suggestions?

Any help will be appreciated.

  • 0
    What's the operation in $G/H$? Hint: It's not multiplication! – 2017-01-31
  • 0
    This is very similar to your [prior question](http://math.stackexchange.com/questions/2120768/g-frac-mathbbq14-mathbbz-h-frac-mathbbq21-mathbbz-find-o?noredirect=1#comment4362054_2120768). The same approach will work here. – 2017-01-31
  • 0
    @lulu (you have a good memory), yet I still don't understand why this method works, I really want to understand how it solve the problem instead of using a memory method with no understanding. – 2017-01-31
  • 0
    @user4894 the operation of $G$ is addition then I guess that $G/H$ is addition too? – 2017-01-31
  • 0
    Hint: The operation is clearly addition....$\mathbb Z$ isn't even a group under multiplication. So, just add your element to itself a few times until you see what you need. – 2017-01-31
  • 0
    @lulu But the order of an element is define as $a^k=e$ while $k$ is minimal meaning that $a^k=a*a*a*a...=e$ (k times) – 2017-01-31
  • 1
    @JaVaPG $a^k$ is multiplicative notation but in this case the group operation is addition. A group only has one operation. – 2017-01-31

1 Answers 1

4

Let's consider the element $-\frac{10}{24}+\mathbf Z=-\frac 5{12}+\mathbf Z$ (irreducible form of the fraction). By definition it is the smallest $n>0$ such that $$n\Bigl(-\frac5{12}+\mathbf Z\Bigr)=\frac{-5n}{12}+\mathbf Z=\mathbf Z\iff \frac{-5n}{12}\in\mathbf Z\iff12\mid 5n\iff 12\mid n,$$ since $5\wedge 12=1$. Thus the order is $12$.