Question

Prove or disprove: Every subgroup of the integers has finite index.

Answer 1

Let $H$ be a subgroup of $\mathbb{Z}$. First of all, we study apart the case of $H=\{0\}$. In this case $H$ has an infinite index, but this is a degenerate case.

Now let $H$ be a non-trivial subgroup, then $H$ has a non-zero element $x$ with a minimal absolute value $|x|>0$. Let us prove that $H=x\mathbb{Z}$. First, $x\mathbb{Z}\subseteq H$. Second, suppose that there is $y\in H$ such that $y\neq a\cdot x$ with $a\in \mathbb{Z}$. The euclidian division of $y$ by $x$ gives us a remainder $r\in H$ (as $r=y-a\cdot x$ and both terms are in $H$), but by the properties of a euclidian division $|r|<|x| \text{ and } \; r\neq 0$ as $y\neq a\cdot x$. This condraticts the fact that $x\in H$ is of minimal absolute value. Therefore, $H=x\cdot \mathbb{Z}$ and we easily deduce that the index of $H$ is $|x|$ and so is finite.

We conclude that the index of every non-trivial subgroup of the integers is finite. For the trivial subgroup this is obviously false.