Prove that if a set has an identity element, , under ∗, then is always its own inverse

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A binary operation * on a set is a mapping : × → that takes any two elements , ∈ to exactly one element (,) ∈ , which will be denoted ∗ .

We know that by definition

For a set and a binary operation ∗ on , if there is an element ∈ such that ∗ = ∗ = for all ∈ , then we call an identity element of .

Also know that any two elements of the set $S$ under the binary operation gives another element in the set $S$

since $e*x=x*e=x$ for any $x \in S$ and $e \in S$ then $e*e=e$ and therefore $e$ is its own inverse

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Your proof is correct. "$b$ is an inverse of $a$" means that $a*b=b*a=e$. So the fact that $e*e=e$ tells us that $e$ is an inverse of $e$.

If the operation is associative, then it turns out that each element will have a unique inverse. Saying that $*$ is associative means that for all $a,b,c\in S$, $a*(b*c)=(a*b)*c$. Here is a helpful exercise:

Exercise Let $*$ be an associative binary operation on a set $S$ with identity $e$. Show that if an element $a\in S$ has an inverse, then the inverse is unique. In other words, show that if $b$ and $c$ are inverses of $a$, then $b=c$.

In your proof, you noted that $e*e=e$, which tells us that $e$ is an inverse of $e$. By the above exercise, we see that if the operation is associative, then $e$ is the inverse of $e$.

Solution to the exercise: Let $b$ and $c$ be inverses of $a$. Then

$$b=b*e=b*(a*c)=(b*a)*c=e*c=c\quad\Box$$