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

From Groupprops
Jump to: navigation, search

This article adapts material from the main article: every nontrivial subgroup of the group of integers is cyclic on its smallest element

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)
PREVIOUS: Equivalence of definitions of cyclic group| UP: Introduction four (beginners)| NEXT: Subgroup containment relation equals divisibility relation on generators
General instructions for the tour | Pedagogical notes for the tour | Pedagogical notes for this part
WHAT YOU NEED TO DO:
  • Read the statement and understand it thoroughly.
  • Go through the proof very carefully. Make sure you understand all the steps of the proof, particularly the application of Euclidean division.

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}.

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.

This page is part of the Groupprops guided tour for beginners. Make notes of any doubts, confusions or comments you have about this page before proceeding.
PREVIOUS: Equivalence of definitions of cyclic group| UP: Introduction four (beginners)|