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

Question

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?

Answer 1

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 $\Bbb Q$.

Answer 2

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.

Answer 3

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 $$\mathbb Q/\mathbb Z\cong\sum_{p\in P}\mathbb Z(p^{\infty})$$

Answer 4

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.