# Inverse of Group Product/Proof 3

## Theorem

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

Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.

Then:

$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$

## Proof

 $\ds \paren {a \circ b} \circ \paren {a \circ b}^{-1}$ $=$ $\ds e$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds a \circ \paren {b \circ \paren {a \circ b}^{-1} }$ $=$ $\ds e$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds b \circ \paren {a \circ b}^{-1}$ $=$ $\ds a^{-1}$ Group Product Identity therefore Inverses $\ds \leadsto \ \$ $\ds b^{-1} \circ \paren {b \circ \paren {a \circ b}^{-1} }$ $=$ $\ds b^{-1} \circ a^{-1}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds \paren{b^{-1} \circ b} \circ \paren {a \circ b}^{-1}$ $=$ $\ds b^{-1} \circ a^{-1}$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds e \circ \paren {a \circ b}^{-1}$ $=$ $\ds b^{-1} \circ a^{-1}$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds \paren {a \circ b}^{-1}$ $=$ $\ds b^{-1} \circ a^{-1}$ Definition of Identity Element

$\blacksquare$