Every nontrivial subgroup of the group of integers is cyclic on its smallest element

Statement

Let $H$ be a subgroup of $\mathbb{Z}$ , the group of integers under addition. Then, there are two possibilities:

$H$ is the trivial subgroup, i.e. $H = \{ 0 \}$
$H$ contains a smallest positive element, say $m$ , and $H$ is the set of multiples of $m$ . Thus, $H$ is an infinite cyclic group generated by $m$ , and is isomorphic to $\mathbb{Z}$ . We typically write $H = m\mathbb{Z}$ .

Related facts

The result actually generalizes to the additive group of any Euclidean domain, where smallest element is replaced by element of smallest norm. A difference in the general case is that we may not have a way of uniquely picking one generator for the subgroup.

Another generalization is to discrete subgroups of the reals:

Every nontrivial discrete subgroup of reals is infinite cyclic

Proof

Given: A nontrivial subgroup $H$ of $\mathbb{Z}$ , the group of integers under addition

To prove: There exists a smallest positive element $m$ in $H$ , and $H = m \mathbb{Z}$ , so $H$ is isomorphic to $\mathbb{Z}$

Proof: First, observe that if $H$ is nontrivial, then there exists a nonzero element in $H$ . This element may be either positive or negative. However, since $H$ is a subgroup, it is closed under taking additive inverses, so even if the element originally picked was negative, we have found a positive number in $H$ .

Thus, the set of positive numbers in $H$ is nonempty. Hence, there exists a smallest positive number in $H$ . Call it $m$ .

Clearly, all the integer multiples of $m$ are in $H$ . We need to prove that every element in $H$ is a multiple of $m$ .

By the Euclidean division algorithm, we can write:

$n = mq + r$

where $q,r$ are integers and $0 \le r < m$ . Since $n,q \in H$ , $r = n - mq = n - (q + q + \dots + q) \in H$ . Thus, $r$ is a nonnegative integer less than $m$ such that $r \in H$ . By the minimality of $m$ , we have $r = 0$ , so $m | n$ , as desired.