# Show that $\Bbb Q/\Bbb Z$ is an infinite group?

How do I show that $\Bbb Q/\Bbb Z$ is an infinite group? I've been thinking that all elements in $\Bbb Q/\Bbb Z$ can be written as $\left[\frac ab\right]$, where a $\in \Bbb Z$ and b $\in \Bbb N \setminus\{0\}$, is this right? Then I've been thinking, that since there exists elements of arbitrary large order(I've already proven this), then it's a infinite group? I'm just not sure whether this is right, or enough proof. I also have to proof that $\Bbb Q/\Bbb Z$ is not cyclic. Again, I think it has something to do with the order of the group being arbitrary large, but I'm not sure.

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When you write, "I have to prove", this suggests your question is part of a homework assignment. Nothing wrong with that, but if that's the case, may I encourage you to add the "homework" tag? –  Gerry Myerson Nov 19 '12 at 2:08
If $\mathbb{Q}/\mathbb{Z}$ is finite of order $n$, then the order of every element divides $n$. You've already shown that there exist elements of arbitrarily large order, so... –  Qiaochu Yuan Nov 19 '12 at 2:08
I'm sorry, I'm very new on this site, I wasn't aware that there is a "homework" tag. I will use this from now on. –  Cathrine Nov 19 '12 at 2:10

If you had already shown that $\mathbb {Q} /\mathbb {Z}$ has elements of arbitrarily large order then from this already follows that the group is infinite. One way to reason would be that in a finite group the orders of elements are bounded by the size of the group (by Largange's Theorem), so if the orders are not bounded then the group is not finite.

Showing that $\mathbb {Q} /\mathbb {Z}$ is not cyclic has little to do with it's size but a lot to do with orders of elements. Since you already showed that $\mathbb {Q} /\mathbb {Z}$ is infinite were it to be cyclic it would be isomorphic to $\mathbb {Z}$. However, in $\mathbb {Z}$ all elements are of infinite order, except for $0$. But (as you have shown) in $\mathbb {Q} /\mathbb {Z}$ there are many elements of various finite orders. Since order of elements is preserved under an isomorphism it is impossible for $\mathbb {Q} /\mathbb {Z}$ to be isomorphic to $\mathbb {Z}$, and so the former is not cyclic.

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Let $\pi\colon \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}$ be the canonical homomorphism. Let $r, s \in [0, 1) \cap \mathbb{Q}$. Then $|r - s| < 1$. Hence, if $r \neq s$, $\pi(r) \neq \pi(s)$. Hence $\pi|[0, 1) \cap \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}$ is injective. Hence $\mathbb{Q}/\mathbb{Z}$ is an infinite group.

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Nice and clean solution! –  John Smith Oct 14 at 19:45

I think we can have the following approaches:

1. $\mathbb Q$ is a divisible group so every quotient of it is the same. It means that $\frac{\mathbb Q}{\mathbb Z}$ is a divisible group. And a finite group cannot be divisible.

2. $\frac{\mathbb Q}{\mathbb Z}$ is a torsion group so we can write it as: $$\frac{\mathbb Q}{\mathbb Z}=\sum_{p\in P}\left(\frac{\mathbb Q}{\mathbb Z}\right)_p$$ where in $P$ is the set of all primes. But $\left(\frac{\mathbb Q}{\mathbb Z}\right)_p\cong\mathbb Z(p^{\infty}),\; \; \forall p\in P$.

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I'd change the wording of "...so every quotient of it is the same", since I think you meant "...so every quotient of it is also divisible". –  DonAntonio Nov 19 '12 at 2:37
By the way, \left(...\right) is much better than \big(...\big), as it automatically fixes the parentheses size. –  DonAntonio Nov 19 '12 at 2:39
@DonAntonio: Thanks Don for the edit and time. I don't know about the parentheses. :-) –  Babak S. Nov 19 '12 at 2:43
+1 ${}{}{}{}{}{}{}$ –  amWhy Apr 6 '13 at 0:16