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# Prove that Every Cyclic Group is an Abelian Group

• Last Updated : 13 Nov, 2022

Groups, subgroups, rings, fields, integral domains, graphs, trees, cut sets, etc are one of the most important concepts in Discrete Mathematics. In this article, we are going to discuss and prove that every cyclic group is an abelian group. Before proceeding to the proof let’s first understand some basic terminologies like what is an abelian group, cyclic group, etc.

Abelian group: In a group G, if for all a, b ∈ G such that a ∗ b = b  ∗ a, that group is called the abelian group.

Non-Abelian group: In a group G, if for all a, b ∈ G such that a ∗ b  ≠  b  ∗ a, then group ( G, ∗ )is called Non – abelian group.

### Cyclic group:

A group G becomes a cyclic group if there exists an element a ∈ G, such that every element of G is in the form an, where n is any integer. The element a is called the generator of G and G is called a cyclic group generated by a and is denoted by G = <a>.

### Cyclic group example:

1. The multiplicative Group G = { 1, -1, i, -i } is cyclic generated by i and -i.

```(i)0 = 1
(i)1 = i
(i)2 = -1
(i)3 = -i```

2. The additive group Z of integers is an infinite cyclic group generated by 1 and -1.

Proof: We have to prove every cyclic group is an abelian group. Let’s take a cyclic group

Suppose G is a cyclic group that is generated by a.

Let’s take two elements x & y ∈ G.

`Suppose, x = am & y = an for some integers m, n.`

Now, the product of these two elements is

```xy = am . an
xy = am+n
xy = an+m
xy = an . am
⇒ xy = yx```

⇒ x y = y x which shows that this cyclic group is an abelian group (because it satisfies the condition of the abelian group (a ∗ b = b ∗ a) )

Therefore, G is an abelian group.

Hence it is proved that every cyclic group is an abelian group.

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