Question

If every element of a group is its own inverse then prove that the group is abelian.

Answer 1

Hint: If a is an element of the group then inverse of a is also a. That means, ${{a}^{-1}}=a$. For a group to be abelian, the product of two elements should follow the commutative property of multiplication. Hence we need to show that $ab=ba$, for any two elements a and b from the group. If this is true, the group is said to be abelian.

Complete step by step solution:
Let G be a group. It is given in the question that every element of a group is its own inverse. As per the properties of the group we know that for each element of a group there exist an inverse element such that:
 $a*{{a}^{-1}}=e$, where a is an element of the group and e is the identity element of the group.
In this group G, ${{a}^{-1}}=a$, for every element belongs to the group G.
A group is said to be abelian, if for any two elements a, b from the group G, it satisfies the following property:
$ab=ba$
Let us take two elements from the group G, say a, b.
Since we know that every element in group G is equal to its inverse, we can write that the product of a and b should also be equal to its inverse. So we will proceed as:
$ab={{\left( ab \right)}^{-1}}$ , since every element is its own inverse.
Therefore, we can write that:
$\Rightarrow ab={{b}^{-1}}{{a}^{-1}}$
We know that ${{b}^{-1}}=b,{{a}^{-1}}=a$. Therefore, we can write that:
$\Rightarrow ab=ba$
Since a, b are arbitrary elements of G, the above expression holds for any two elements.
Therefore, the group G is abelian. This is the required answer.

Note: Alternatively, we can prove this statement by using the fact that if every element of the group is its own inverse then the order of every element of that group is 2. That means,
$\left( ab \right)*\left( ab \right)=e$, e is the identity element of G.
  From the above expression we can show that $ab=ba$.