Inverse of Group Product/Proof 1

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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

\(\ds \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }\) \(=\) \(\ds \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}\) 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:

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

$\blacksquare$


Sources

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