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Consider the group $(\mathbb{Q}, +)/(\mathbb{Z}, +)$, the group of rationals (under addition) modulo the subgroup of integers. So an element of this group is a coset $a + \mathbb{Z}$ where a is a rational number.

(a) Find the order of the element $3/4 + \mathbb{Z}$.

(b) Show that every element of this group has finite order.

(c) Prove that the group is infinite.

(d) Prove that every finite subgroup is cyclic.

I am having difficulty trying to write the proofs to these given that when I try to write proofs on cosets they end up being way too long and not concise at all.

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Hints, taking into account that for $\;q\in\Bbb Q\;,\;\;q+\Bbb Z=\overline 0\in\Bbb Q/\Bbb Z \iff q\in\Bbb Z\;$ :

$$\forall\,m,n\in\Bbb Z\;,\;\;n\left(\frac mn+\Bbb Z\right)=m+\Bbb Z=\ldots$$

$$\text{For any two different }\;\;p,q\in \Bbb N\;\;:\;\;\frac1p+\Bbb Z\neq\frac1q+\Bbb Z$$

With the above and some work, the last part (d) should also follow (the first three follow almost at once from the above)

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  • $\begingroup$ I am doing a similar question, can you kindly explain $\;q\in\Bbb Q\;,\;\;q+\Bbb Z=\overline 0\in\Bbb Q/\Bbb Z \iff q\in\Bbb Z\;$? Thanks! $\endgroup$ – Nighty Oct 7 '14 at 16:44
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$4(3/4+\mathbb{Z})=\mathbb{Z}$, and 4 is the least positive integer for which $4(3/4+\mathbb{Z})=\mathbb{Z}$, thus order of $3/4+\mathbb{Z}$ is 4.

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