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.