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

 $\ds \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }$ $=$ $\ds \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds \paren {a \circ e} \circ a^{-1}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds a \circ a^{-1}$ Group Axiom $\text G 2$: Existence of Identity Element $\ds$ $=$ $\ds e$ Group Axiom $\text G 3$: Existence of Inverse Element

The result follows from Group Product Identity therefore Inverses:

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

$\blacksquare$