Inverse of Group Product/Proof 3

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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

$\blacksquare$