Question

In a group (G,*), for some element ‘a’ of G, if ${{a}^{2}}=e$, where e is the identity element, then
[a] $a={{a}^{-1}}$
[b] $a=\sqrt{e}$
[c] $a=\dfrac{1}{{{a}^{2}}}$
[d] $a=e$

Answer 1

Hint: Recall the definition of a group and the inverse of an element. Use the fact that ${{a}^{n}}$ is defined as ${{a}^{n}}={{a}^{n-1}}*a$. Hence prove that $a*a=e$. Apply operation * on both sides with
${{a}^{-1}}$ and use the associative property to find the value of $a$.

Complete step-by-step answer:
Before solving the question, we need to understand what a group is. For that, we need to understand the definitions of semigroup and monoid.
Semigroup: A semigroup S is a finite/infinite set with an operation * such that the * on S enjoys two properties
[a] Closure: S is closed under *, i.e. $\forall a\in S,b\in S,a*b\in S$
[b] Associate: * is associative, i.e. $\forall a,b,c\in S,a*\left( b*c \right)=\left( a*b \right)*c$
Monoid: A monoid is a semigroup with an identity element, i.e. there exists $e\in S$ such that $\forall a\in S,\left( a*e \right)=a$. Hence a monoid has three properties
[a] Closure
[b] Associative
[c] Existence of identity element
Group: A group is a monoid in which every element has its inverse, i.e. $\forall a\in S,\exists !{{a}^{-1}}\in S$ such that $a*{{a}^{-1}}=e$.
Now, we have
${{a}^{2}}=e$
We know that ${{a}^{n}}={{a}^{n-1}}*a$
Hence, we have
$a*a=e$
Since G is a group, there exists the inverse of a.
$\left( a*a \right)*{{a}^{-1}}=e*{{a}^{-1}}$
Since G is a group, we have * is associative.
Hence, we have
$a*\left( a*{{a}^{-1}} \right)={{a}^{-1}}$
We know that $a*{{a}^{-1}}=e$
Hence, we have
$a*e={{a}^{-1}}$
We know that $a*e=a$. Hence, we have
$a={{a}^{-1}}$
Hence, we conclude that option [a] is correct.

Note: The most common mistake done by the students in the questions of the above type is that they treat * operation same as multiplication and e as 1. Hence, they arrive that ${{a}^{2}}=e\Rightarrow a=\sqrt{e}=e$, which is incorrect. It should be noted that * is a representative of operation and not multiplication itself and normal algebra does not apply in groups.