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I'm trying to prove that $\mathbb Q/\mathbb Z$ is an infinite abelian group, the easiest part is to prove that this set is an abelian group, I'm trying so hard to prove that this group is infinite without success.

This set is defined to be equivalences classes of $\mathbb Q$, where $x\sim y$ iff $x-y\in \mathbb Z$.

I need help here.

thanks a lot

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up vote 11 down vote accepted

Hint: Prove that if $x,y\in[0,1)\cap\Bbb Q$, then $x\sim y$ if and only if $x=y$.

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Very clever, really elegant solution, I'm trying to prove it right now. Thanks a lot. – user42912 Apr 12 '13 at 0:01
Thank you very much. It's the analogue from proving that Vitali sets are uncountable. :-) – Asaf Karagila Apr 12 '13 at 0:02
That's "" simple "", @AsafKaragila : just prove the Vitali set is not measurable...:) – DonAntonio Apr 12 '13 at 0:04
@DonAntonio: Trick question! There are no Vitali sets in Solovay's model! ;-) – Asaf Karagila Apr 12 '13 at 0:06
My heart bleeds for Solovay's model, which fortunately enough I've never heard of (is that something from logic? If so pass it over to Sharon Shelakh: not for me) – DonAntonio Apr 12 '13 at 0:07

Look at $[1/n]$ in this equivalence class for $n\ge 2$ These are all distinct mod $\mathbb{Z}$ in $\mathbb{Q}$.

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A little bit more background: $\mathbb{Q}/\mathbb{Z}$ is a nontrivial divisible abelian group. The only finite divisible abelian group is the trivial group (else you cannot divide by the group order by Lagrange).

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Intuitively, you can think of the quotient of $\mathbb{Q}$ by $\mathbb{Z}$ as fractions in an interval from $0$ to $1$. What you're doing when you quotient by $\mathbb{Z}$ is you set each integer to be $0$ - it's the rationals "mod 1."

To easily argue that the group is infinite, notice the fact that $\frac{1}{s}\mathbb{Z}=\frac{1}{r}\mathbb{Z} \Leftrightarrow \frac{1}{s}-\frac{1}{t}\in\mathbb{Z}$. To verify my interpretation of $\mathbb{Q}/\mathbb{Z}$ is true, compare two arbitrary $x\mathbb{Z},y\mathbb{Z}$ with $0\leq x,y<1$.

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+1 very nice Alexander. – Babak S. Jun 19 '13 at 5:51

This sounds like a question where there many ways to solve it. The quickest I could come up with:

Count the prime numbers: $p_1, p_2, p_3,\ldots$. Now look at the equivalence classes of $\frac{1}{p_i}$ in $\mathbb{Q}/\mathbb{Z}$. For $i\neq j$ we have that $\frac{1}{p_i}-\frac{1}{p_j}=\frac{p_j-p_i}{p_ip_j}$. If this were an integer, say $n$, we would see that $p_j-p_i = np_ip_j$, or equivalently $p_j=p_i(1+np_j)$. This is somewhat impossible and so the difference is no element of $\mathbb{Z}$. This proves that all equivalence classes of $\frac{1}{p_i}$ are different. Therefore, there must be infinitely many elements in $\mathbb{Q}/\mathbb{Z}$.

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Thank you, using primes makes the solution simpler. – user42912 Apr 12 '13 at 0:25
It doesn't make it simpler, it makes it more complicated. All the $\overline{1/n}$ are pairwise distinct, by a simple distance argument, and we don't need that $n$ is prime. – Martin Brandenburg Apr 12 '13 at 15:35
The thought of using a distance argument simply hadn't occured to me, last night. I agree that it is a far simpler one. I'm nonetheless glad that one can live without needing the concept of distance here. – HSN Apr 12 '13 at 16:39

Another hint: prove that for

$$n,k\in\Bbb N\;,\;\;n\neq k\;,\;\;\;\frac{1}{n}+\Bbb Z\neq\frac{1}{k}+\Bbb Z$$

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Of course. Thanks] – DonAntonio Apr 12 '13 at 13:50

The map $\rho: \mathbb{R} \to S^1,\;t \mapsto e^{2\pi\text{i}t}$ is a surjective homomorphism of groups with kernel $\mathbb{Z}$. Hence $\rho(\mathbb{Q})$ is the group of roots of unity $\mu$ and $\mathbb{Q}/\mathbb{Z}\cong \mathbb{\mu}$. In particular, its infinite, since $\mu$ has the infinite subset $\{e^{\frac{2\pi\text{i}}{n}}\mid n \in \mathbb{Z},\;n> 0\}$.

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You could also write directly that $\overline{1/n}$ are distinct in $\mathbb{Q}/\mathbb{Z}$ without this homomorphism ;). – Martin Brandenburg Apr 12 '13 at 15:34
Of course, but I wanted to relate the group in question to a group the OP might be more familar with. – Ralph Apr 12 '13 at 19:43

By regarding to $\mathbb Z(p^{\infty})$ where in $p\in P$ the set of all prime numbers, it is not hard to see that $$\mathbb Q/\mathbb Z\cong \sum_{p\in P} \mathbb Z(p^{\infty})$$

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+1 You were busy at today! That's good for us all ;-) – amWhy Apr 14 '13 at 1:18

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