Show that the integers have infinite index in the additive group of rational numbers. 
Show that the integers have infinite index in the additive group of rational numbers.

Anybody in a good enough mood to tell me how this is done?
 A: Let define a group $G=(X|R)$ such that: $$X=\{x_0,x_1,\cdots,x_n,\cdots\}$$ and $$R=\{px_0,x_0-px_1,x_1-px_2,\cdots,x_{n-1}-px_n,\cdots\}$$ in which $p\in P$ is a prime. This constructed group is denoted as $\mathbb Z(p^{\infty})$. It is not hard seeing that $$\mathbb Q/\mathbb Z\cong\sum_{p\in P}\mathbb Z(p^{\infty})$$
A: From another angle: $\mathbb{Q}$ is a so-called divisible group: an (additive) abelian group $G$ is called divisible if, for every positive integer $n$ and every $g \in G$, there exists an $h \in G$ such that $nh = g$. The notion of divisibility is important in the study of abelian groups - they are the injective objects in the category of abelian groups. Now let us show that a divisible group $G$ cannot have subgroups of finite index: suppose $H$ is a subgroup of $G$ with $index[G:H]=n$. Pick a $g \in G \backslash H$. Since $G$ is divisible one can find an $h \in G$ with $nh=g$. Owing to Lagrange's Theorem, this means in $G/H$ that $\bar g = \bar {0}$, that is, $g \in H$, a contradiction.
A: I’ll point you in the right direction. The cosets of $\Bbb Z$ in $\Bbb Q$ are sets of the form $$q+\Bbb Z=\{q+n:n\in\Bbb Z\}$$ for $q\in\Bbb Q$. Showing that $[\Bbb Q:\Bbb Z]$ is infinite is just showing that there are infinitely many distinct cosets.


*

*Show that for any $p,q\in\Bbb Q$, $p+\Bbb Z=q+\Bbb Z$ if and only if $p-q\in\Bbb Z$. Conclude that if $0<|p-q|<1$, then $p+\Bbb Z\ne q+\Bbb Z$.  

*Find $\{q_n:n\in\Bbb N\}\subseteq\Bbb Q$ such that $0<|q_m-q_n|<1$ whenever $m,n\in\Bbb N$ and $m\ne n$.  

*Display an infinite family of distinct cosets of $\Bbb Z$ in $\Bbb Q$.

A: Another solution may be:


*

*Suppose by contradiction that $\mathbb{Q}/\mathbb{Z}$ is finite, ie. $\mathbb{Q}/ \mathbb{Z}= \{x_1 \mathbb{Z}, \dots, x_r \mathbb{Z}\}$. Without loss of generality, we may suppose $0<x_1 < \cdots < x_r$. 

*Then $\frac{1}{2}x_1 \in x_i \mathbb{Z}$ for some $1 \leq i \leq r$.

*Deduce that $x_1>x_i$, a contradiction.


More generally, a proper subgroup of a divisible group has infinite index. See for example the question Subgroups of finite index in divisible group.
A: Similar to Brian M. Scott's idea: all cosets of the form
$$\frac{1}{n} + \mathbb{Z} = \left\{\frac{1}{n} + i : i \in \mathbb{Z} \right\}$$
for $n \in \mathbb{N}$ are distinct, so there are infinitely many distinct cosets.
Proof: suppose for contradiction that there are distinct $n_1, n_2 \in \mathbb{N}$ (where $n_2>n_1$ without loss of generality) such that $\{\frac{1}{n_1} + i : i \in \mathbb{Z} \} = \{\frac{1}{n_2} + i : i \in \mathbb{Z} \}$. Thus $\frac{1}{n_1} \in \{\frac{1}{n_2} + i : i \in \mathbb{Z} \}$, i.e. there exists an integer $i=\frac{1}{n_1}-\frac{1}{n_2}=\frac{n_2-n_1}{n_2n_1}=\frac{p}{q}$ for $p, q \in \mathbb{N}$ where gcd$(p,q)=1$ (reduced fraction). Notice $q=1$ since $q$ is a positive common divisor of $p$ and $q$. Therefore positive $p \geq q=1$, so $n_2-n_1 \geq n_2n_1\geq n_2$, which is impossible. Thus each $n \in \mathbb{N}$ yields a distinct coset.
