Inverse of Group Product/Proof 1

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

 $\displaystyle \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }$ $=$ $\displaystyle \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}$ $\quad$ Group Axiom $G \, 1$: Associativity $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}$ $\quad$ Group Axiom $G \, 1$: Associativity $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {a \circ e} \circ a^{-1}$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$ $\displaystyle$ $=$ $\displaystyle a \circ a^{-1}$ $\quad$ Group Axiom $G \, 2$: Identity $\quad$ $\displaystyle$ $=$ $\displaystyle e$ $\quad$ Group Axiom $G \, 3$: Inverses $\quad$

The result follows from Group Product Identity therefore Inverses:

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

$\blacksquare$