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I have the following question :

$G=\frac{\mathbb{Q}}{14\mathbb{Z}}$, $H=\frac{\mathbb{Q}}{21\mathbb{Z}}$, Find ord of $\frac{2}{5}+14\mathbb{Z}$ and $\frac{2}{5}+21\mathbb{Z}$.

I tried to use the first homomorphism theorem but still didn't manage to calculate the ord,

Any ideas?

Any help will be appreciated.

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    Well, can you produce some integer $n$ such that $n\times \frac 25 \equiv 0 \pmod {14}$? Hint: $5\times \frac 25\equiv 2$. Once you have one such $n$ then you know the order is a divisor of $n$. – 2017-01-30
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    @lulu for $n=35$ I get that $14 \equiv 0 \pmod {14}$ but I still don't understand the relation between $n*\frac{2}{5}$ and the problem we use addition ($+$) not multiplication ($*$) – 2017-01-30
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    Well, $5\times \frac 25 = \frac 25+\frac 25+\frac 25+\frac 25 +\frac 25$. – 2017-01-30
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    OH so we look for $n\in \mathbb{Z}$ such that $n(\frac{2}{5}+14\mathbb{Z}) \equiv n(\frac{2}{5}+0) \equiv n*\frac{2}{5} \equiv 0 \pmod {14}$? – 2017-01-30
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    Yes, exactly. At first, just look for one such $n$...then worry about the possibility that there might be a smaller one. – 2017-01-30
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    Thank you just one last thing that is not really clear each class in group $G$ looks like that $(...,-14,0,14,...),$(,....-13,1,15,...$,$(....,-12,2,16,...),...$ because of that you knew that the mod is $14$ right? – 2017-01-30
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    Not following you. Yes, I am writing $\frac 25$ as a shorthand for the residue class that contains it. – 2017-01-30
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    @lulu I have the impression that the OP thinks that $G$ us a finite group, because the classes he mentions are those if integer rationals. – 2017-01-30
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    @MarcBogaerts When I thought about it again I realize that it not finite. what I wrote in the comment is not true. – 2017-01-30
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    It is a nice example of an inifinite group where each element has a finite order. – 2017-01-30

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