📂Abstract Algebra

Uniqueness Proof of Identity Elements and Inverse Elements in groups

Theorem ¹

For a group $\left<G, \ast \right>$, the identity element $e$ that satisfies $e \ast x = x \ast e = x$ for all elements $x$ of $G$ is unique. For any element $a$ of $G$, the inverse element $a '$ that satisfies $a \ast {a’} = {a’} \ast a = e$ is unique with respect to $a$.

Explanation

Though everyone takes it for granted, the definition of a group only mentions their existence, not uniqueness. That such elements exist uniquely requires proof.

Proof

Strategy: As usual when proving uniqueness, proof by contradiction is used.

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Part 1. Identity Element

Let’s assume there exists another identity element $e'$ besides $e$. Since $e$ is an identity element, $$e \ast\ e' = e' \ast\ e = e'$$ holds. Meanwhile, $e'$, being an identity element as well, $$e' \ast\ e = e \ast\ e' = e$$ holds. Hence, $e = e'$, this contradicts the assumption $e \ne e'$.

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Part 2. Inverse Element

Similarly, let’s assume there exists a different inverse element $a’’$ other than $a '$ for $a$. Then, $a’ \ast\ a = e$ and $a’’ \ast\ a = e$, leading to $a’ \ast\ a = a’’ \ast\ a$.

Cancellation Law: For an element $a,b,c$ of a group $\left<G, \ast \right>$, $$a \ast b = a \ast c \implies b = c \\ b \ast a = c \ast a \implies b=c$$

By the cancellation law, $a’ \ast\ a = a’’ \ast\ a$ implies $a’ = a’’$.

This contradicts the assumption $a’ \ne a’’$.

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