In this article, we will prove that every group of order 4 is an abelian group.

**Basic Terminologies**

**Group**: A binary operation is a group if- It is closed.
- Associative.
- Existence of an identity element.
- Each element has a unique inverse.

**Abelian Group**: A group that is commutative is knows as abelian group, i.e., a*b = b*a.**Order of a Group:**Order of a group is defined as the number of elements in the group.

**Proof – Contradiction Method**

We will prove the given statement by contradiction.

Consider a group G of order 4. Assume that G is not an abelian group. This implies that there exists atleast one pair of elements a, b ∈ G such that a*b != b*a.

If a*b != b*a, then we have three cases

- a*b != e and b*a != e ( e is the identity element )

This statement implies that a and b are not inverse of each other, a^{-1}!= b. - a*b != a and b*a != a

As we know the above statement is true since a*b != b*a. This implies that b is not the identity element. - a*b != b and b*a != b

As we know the above statement is true since a*b != b*a. This implies that a is not the identity element.

From the above three cases, the set G must contain atleast 5 different elements, G = {a, b, a*b, b*a, e}. But this is a contradiction since set G only contains 4 elements. This implies that our assumption that G is not an abelian group ( or G is not commutative ) is wrong.

Therefore, we can conclude that every group G of order 4 must be an abelian group. Hence proved.

.**Proof – Direct Method**

Consider a group G of order 4. Let G = {a, b, c, e}. Let e be the identity element of G.

We need to prove that G is an abelian group. To do so, we need to show that for every pair x, y ∈ G, x*y = y*x, i.e., G is commutative.

Since e is the identity element, we can say that

- a*e = e*a = e
- b*e = e*b = e
- c*e = e*c = e

Now we just need to show that every pair x, y ∈ {a, b, c}, x*y = y*x.

We have 2 cases

**CASE 1: a, b, c are self inverse, i.e., a=a ^{-1} , b=b^{-1} and c=c^{-1}.**

Take element a and b

ab(ab)^{-1} = e

a^{-1}ab(ab)^{-1} = a^{-1}e

eb(ab)^{-1} = a^{-1}

b(ab)^{-1} = a^{-1}

b^{-1}b(ab)^{-1} = b^{-1}a^{-1}

e(ab)^{-1} = b^{-1}a^{-1}

(ab)^{-1} = b^{-1}a^{-1}

Since every element is inverse of each other, we can say that (ab)^{-1} = ab and b^{-1}a^{-1} = ba. Substituting in the above equation, we get

ab = ba

Similarly, we can prove that for every pair x, y ∈ {a, b, c}, x*y = y*x. Therefore, G is abelian group.

**CASE 2: Every element of the set is not a self-inverse**

The identity element is always self-inverse so we can ignore that.

We are now left with 3 elements, {a, b, c}. Since the inverse of every element is unique, one of {a, b, c} must be a self-inverse. Let c be self inverse and a,b be inverse of each other.

In that case, we have

- c=c
^{-1} - ab = ba = e, i.e., a
^{-1}= b.

Now we just need to show ca = ac and bc = cb to prove that G is abelian.

Suppose ca is equal to e

ca = e

c^{-1}ca = c^{-1}e

ea = c^{-1}

a = c

which is not possible since {a, b, c, e} are unique.

Suppose ca is equal to a

ca = a

caa^{-1} = aa^{-1}

c = e

which is not possible since {a, b, c, e} are unique.

Suppose ca is equal to c

ca = c

c^{-1}ca = c^{-1}c

a = e

which is not possible since {a, b, c, e} are unique.

We just showed that ca is not equal to a, c, or e. This leaves us with one option only. That is ca is equal to b. Similarly, we can show that ac is also equal to b. Therefore, ac = ca = b.

Similarly, we can prove that bc = cb = a.

We just proved that for every pair x, y ∈ {a, b, c}, x*y = y*x. Therefore, group G is abelian.