Inverse of Group Product/Proof 3

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 $\text G 1$: Associativity

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds a^{-1}\)

Group Product Identity therefore Inverses

\(\ds \leadsto \ \ \)

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

\(=\)

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

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \leadsto \ \ \)

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

\(=\)

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

Group Axiom $\text G 1$: Associativity

\(\ds \leadsto \ \ \)

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

\(=\)

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

Definition of Inverse Element

\(\ds \leadsto \ \ \)

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

\(=\)

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

Definition of Identity Element

$\blacksquare$