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

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

$\blacksquare$