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.

Thus, $$H = m\mathbb{Z}$$, or $$H$$ is the set of multiples of $$m$$.

An explicit isomorphism from $$\mathbb{Z}$$ to $$H$$ is given by the map sending an integer $$x$$ to the integer $$mx$$.

Textbook references

 * , Page 45, Proposition 2.4, Chapter 2, Section 2