Product with Inverse equals Identity iff Equality

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:

\(\ds a \circ b^{-1}\)

\(=\)

\(\ds e\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \paren {a \circ b^{-1} } \circ b\)

\(=\)

\(\ds e \circ b\)

Cancellation Laws

\(\ds \leadstoandfrom \ \ \)

\(\ds a \circ \paren {b^{-1} \circ b}\)

\(=\)

\(\ds e \circ b\)

Group Axiom $\text G 1$: Associativity

\(\ds \leadstoandfrom \ \ \)

\(\ds a\)

\(=\)

\(\ds b\)

Group Axiom $\text G 2$: Existence of Identity Element and Group Axiom $\text G 3$: Existence of Inverse Element

$\blacksquare$