# 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

 $\displaystyle \paren {a \circ b} \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle e$ Definition of Inverse Element $\displaystyle \leadsto \ \$ $\displaystyle a \circ \paren {b \circ \paren {a \circ b}^{-1} }$ $=$ $\displaystyle e$ Group Axiom $G1$: Associativity $\displaystyle \leadsto \ \$ $\displaystyle b \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle a^{-1}$ Group Product Identity therefore Inverses $\displaystyle \leadsto \ \$ $\displaystyle b^{-1} \circ b \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle b^{-1} \circ a^{-1}$ $\displaystyle \leadsto \ \$ $\displaystyle e \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle b^{-1} \circ a^{-1}$ Definition of Inverse Element $\displaystyle \leadsto \ \$ $\displaystyle \paren {a \circ b}^{-1}$ $=$ $\displaystyle b^{-1} \circ a^{-1}$ Definition of Identity Element

$\blacksquare$