An infinite cyclic group is isomorphic to the additive group Z where Z is the set of integers. It has only two generators 1 and -1. In this article, we will study infinite cyclic groups along with its generators.

A cyclic group is a group where every element is expressed as an integral power of some element in that group. That element is called the generator of its group.

Table of Contents

## Prove that Infinite Cyclic Group is Isomorphic to Z

**Proof:**

Let G be an infinite cyclic group generated by a.

That is, G = <a>.

We will show that G ≅ (ℤ, +). Define a mapping φ: ℤ → G by

φ(n) = a^{n}.

**Step 1:**

To prove φ is a homomorphism.

Take m, n ∈ ℤ.

φ(m+n) = a^{m+n} = a^{m} a^{n} = φ(m) φ(n)

This shows that φ is a homomorphism.

**Step 2:**

To prove φ is onto.

Take x ∈ G. So x =a^{k} for some integer k. Now φ(k) = a^{k} = x, so x has a pre-image under φ. Therefore, φ is onto.

**Step 3:**

To prove φ is one-to-one.

For two integers m ≠ n, we have that a^{m} ≠ a^{n}. Otherwise, a would have finite order contradicting G = <a> is finite as |G| = order of a.

⇒ φ is one-to-one.

Combining all the above facts, by the first isomorphism theorem, we conclude that φ is an isomorphism, that is, G ≅ (ℤ, +). Therefore, an infinite cyclic group is isomorphic to the additive group (ℤ, +).

**Topics:**

## Generators of an Infinite Cyclic Group

An infinite cyclic group is isomorphic to the additive group (ℤ, +). We know that (ℤ, +) is a cyclic group with 1 and -1 as generators. This means ℤ has two generators. Hence an infinite cyclic group has only two generators.

**More Reading:** Subgroups of Cyclic Groups

## Solved Problems

Question: If G is an infinite cyclic group generated by a, then prove that a and a^{-1} are the only generators of G. |

**Solution:**

We are given that G = <a> is an infinite cyclic group. Therefore, both G and a have infinite orders.

Let b be another generator of G.

As b ∈ G and G = <a>, we have b=a^{k} for some integer k.

On the other hand, as a ∈ G and G = <b>, we have a=b^{m} for some integer m.

Combining, a = b^{m} = (a^{k})^{m}⇒ a = a ^{km}⇒ a ^{km-1} = e, the identity element in G. |

As the order of a is infinite, we must have that km-1=0 ⇒ km = 1.

So either k=1, m=1 or k=-1, m=-1.

Therefore, either b=a or b=a^{-1}.

This proves the result.

**Note:** There are only two generators of an infinite cyclic group.

**Read This:** Each subgroup of a cyclic group is cyclic: Proof

## Generators of Z

**Question:** Find the generators of (Z, +).

**Answer:**

The additive group (ℤ, +) is an infinite cyclic group. We know that (ℤ, +) is generated by 1. So by the above fact, the other generator of of ℤ is -1 only. So the group ℤ has only two generators 1 and -1.

**You Can Read:** Group of Prime Order is Cyclic: Proof

## FAQs

**Q1: How many infinite cyclic groups exist, up to isomorphism?**

Answer: Up to isomorphism, there is only one infinite cyclic group isomorphic to the additive group of integers.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.