Product with Inverse equals Identity iff Equality

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Theorem

Let $\struct {G, \circ}$ be a group whose identity element is $e$.

Then:

$\forall a, b \in G: a \circ b^{-1} = e \iff a = b$


Proof

Using various properties of groups:


\(\displaystyle a \circ b^{-1}\) \(=\) \(\displaystyle e\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {a \circ b^{-1} } \circ b\) \(=\) \(\displaystyle e \circ b\) Cancellation Laws
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle a \circ \paren {b^{-1} \circ b}\) \(=\) \(\displaystyle e \circ b\) Group Axiom $G \, 1$: Associativity
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle a\) \(=\) \(\displaystyle b\) Group Axiom $G \, 2$: Identity and Group Axiom $G \, 3$: Inverses

$\blacksquare$