$\mathbb Q/\mathbb Z$ is an infinite group 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
 A: Look at $[1/n]$ in this equivalence class for $n\ge 2$ These are all distinct mod $\mathbb{Z}$ in $\mathbb{Q}$.
A: 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).
A: Hint: Prove that if $x,y\in[0,1)\cap\Bbb Q$, then $x\sim y$ if and only if $x=y$.
A: 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 $\mathbb{Q}/\mathbb{Z}$.
A: Another hint: prove that for 
$$n,k\in\Bbb N\;,\;\;n\neq k\;,\;\;\;\frac{1}{n}+\Bbb Z\neq\frac{1}{k}+\Bbb Z$$
A: 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 $0\leq x,y<1$.
A: 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 $\{e^{\frac{2\pi\text{i}}{n}}\mid n \in \mathbb{Z},\;n> 0\}$. 
A: 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 $$\mathbb Q/\mathbb Z\cong \sum_{p\in P} \mathbb Z(p^{\infty})$$
